From: W. Trevor King Date: Mon, 8 Jun 2009 18:12:57 +0000 (-0400) Subject: Began versioning. X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=65d32ad227affcf75d7094e74a36414357f9e49a;p=course.git Began versioning. This course-website framework developed over the first half of 2009 while TAing for Phys201 (Modern physics for engineers) at Drexel University. After a few false starts at versioning, I'm starting this new repository because I think I've figured out a stable scheme. When you start a new course, clone this repository to create a working copy. Seperate your commits on the clone into course-specific commits (e.g. Makefile changes when adding homework 2, atom.xml updates, etc.) and general commits (corrections to README files, additional problem source in /latex/problems/, etc.). Then, cherry pick the course-specific commits back into this repo with git remote add phys201 /home/bob/phys201 git fetch phys201 git cherry-pick 1d8fb1fe41dfc1b1eb38c7b5d574577c4b341c58 git remote rm phys201 git remote prune phys201 The benefit of cloning an independent repo over just starting up a new branch is that most people don't care about the particular per-course details, but lots of people may want the framework, and not want to worry about the disk space needed for all the per-course cruft. From a more philosophical perspective, this repo will track the history of 'what you want for building a course website', while the per-course repos track the history of a particular course's website (and student grades, TA emails, etc.). --- 65d32ad227affcf75d7094e74a36414357f9e49a diff --git a/Makefile b/Makefile new file mode 100644 index 0000000..69702dd --- /dev/null +++ b/Makefile @@ -0,0 +1,28 @@ +INSTALL_HOST = einstein +INSTALL_USER = wking +INSTALL_DIR = public_html/courses/phys201_s09 +SOURCE_DIR := $(INSTALL_DIR)/source # := to avoid shifting with $INSTALL_DIR + +FRAMEWORK_SUBDIR = html +CONTENT_SUBDIRS = announcements latex +SUBDIRS = $(FRAMEWORK_SUBDIR) $(CONTENT_SUBDIRS) + +export INSTALL_HOST +export INSTALL_USER +export INSTALL_DIR +export SOURCE_DIR + +install : + @for i in $(SUBDIRS); do \ + echo "make install in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install); done + +clean : + @for i in $(SUBDIRS); do \ + echo "make clean in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) clean); done + +echo : + @for i in $(SUBDIRS); do \ + echo "make echo in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) echo); done diff --git a/README b/README new file mode 100644 index 0000000..7a2efac --- /dev/null +++ b/README @@ -0,0 +1,27 @@ +This is an attempt at an organized, open source course website. The +idea is that a course website consists of a static HTML framework, and +a bunch of content that is gradually filled in as the semester/quarter +progresses. I've put the HTML framework in the HTML directory, along +with some of the write-once-per-course content (e.g. Prof & TA info). +See html/README for more information on the layout of the HTML. + +The rest of the directories contain the code for compiling material +that is deployed as the course progresses. The announcements +directory contains the atom feed for the course, and possibly a list +of email addresses of people who would like to (or should) be notified +when new announcements are posted. The latex directory contains LaTeX +source for the course documents for which it is available, and the pdf +directory contains PDFs for which no other source is available +(e.g. scans, or PDFs sent in by Profs or TAs who neglected to include +their source code). + +Installation is though a recursive Makefile framework. You'll need to +set the INSTALL_* variables once at the beginning of the course to +make sure all the files go to the right place, and I'd strongly +recommend setting up a ssh-keyed login from your work machine to your +hosted web account (see http://www.physics.drexel.edu/~wking/code/#SSH ). + +Not posted on the website but also important to the course are the +students' grades, which I keep in the grades directory. See the +README files in any of the subdirectories for more details on that +particular portion. diff --git a/announcements/Makefile b/announcements/Makefile new file mode 100644 index 0000000..63361ac --- /dev/null +++ b/announcements/Makefile @@ -0,0 +1,9 @@ +INSTALL_FILES = atom.xml +INSTALL_DIR := $(INSTALL_DIR)/xml + +clean : + rm -f install + +install : $(INSTALL_FILES) + scp -p $^ $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR) + @date > $@ diff --git a/announcements/README b/announcements/README new file mode 100644 index 0000000..f3a7f7e --- /dev/null +++ b/announcements/README @@ -0,0 +1,26 @@ +Construct the atom feed using + atomgen -o atom.xml +See + atomgen --help +for more information. + +For example: + atomgen -o atom.xml new --title 'Physics 201' --author 'W. Trevor King' \ + http://www.physics.drexel.edu/~wking/phys201 + echo "Changes to the Phys201 website will be noted in this feed." | \ + atomgen -o atom.xml add -i atom.xml 'Feed purpose' \ + http://www.physics.drexel.edu/~wking/phys201 + + +You can send automatic emails to your students when you publish new +announcements in the atom feed. The best way I have found to date +consists of monitoring the posted atom.xml file with rss2email + http://rss2email.infogami.com/ +I setup r2e to monitor the feed and email me when there's a change. +r2e runs every 15 minutes in a cron job + 15 * * * * /usr/bin/r2e run +Then I set up a procmail rule to forward the mail off to the list + :0 + * ^From: "Physics 201, W. Trevor King" $ + ! `grep -v '^#' $HOME/course/announcements/addresses.txt` + # ^--- Make sure those are backticks, BTW (ASCII 96) ---^ diff --git a/announcements/addresses.txt.examples b/announcements/addresses.txt.examples new file mode 100644 index 0000000..5647f35 --- /dev/null +++ b/announcements/addresses.txt.examples @@ -0,0 +1,6 @@ +# Administrators +John JJ Smith +Johnny Appleseed +# Students +Jack +Jill diff --git a/announcements/script/course-details.sh b/announcements/script/course-details.sh new file mode 100644 index 0000000..cfc58f9 --- /dev/null +++ b/announcements/script/course-details.sh @@ -0,0 +1,9 @@ +#!/bin/sh +# +# Set-up some course-specific details for announcement/script/*.sh +# Not intended to be used directly + +# No trailing slash on the course website, please. +COURSE_WEBSITE='http://www.physics.drexel.edu/~wking/course' +COURSE_TITLE='Physics 201' +ATOM_AUTHOR='W. Trevor King' diff --git a/announcements/script/initial-post.sh b/announcements/script/initial-post.sh new file mode 100755 index 0000000..b92afd8 --- /dev/null +++ b/announcements/script/initial-post.sh @@ -0,0 +1,19 @@ +#!/bin/sh +# +# Create the initial posting for a course. Use carefully, this is +# just meant to save some typing, and it might not fit into your +# layout without a bit of tweaking. +# +# From the announcements directory +# usage: script/initial-post.sh + +. script/course-details.sh + +echo "atomgen -o atom.xml new --title \"$COURSE_TITLE\" --author \"$ATOM_AUTHOR\" \"$COURSE_WEBSITE\"" +atomgen -o atom.xml new --title "$COURSE_TITLE" --author "$ATOM_AUTHOR" "$COURSE_WEBSITE" + +echo "echo \"Changes to the $COURSE_TITLE website will be noted in this feed.\" | \\ + atomgen -o atom.xml add -i atom.xml 'Feed purpose' \\ + \"$COURSE_WEBSITE\"" +echo "Changes to the $COURSE_TITLE website will be noted in this feed." | \ + atomgen -o atom.xml add -i atom.xml 'Feed purpose' "$COURSE_WEBSITE" diff --git a/announcements/script/post.sh b/announcements/script/post.sh new file mode 100755 index 0000000..2047000 --- /dev/null +++ b/announcements/script/post.sh @@ -0,0 +1,57 @@ +#!/bin/bash +# +# Create a post pointing out a change in the website. Use carefully, +# this is just meant to save some typing, and it might not fit into +# your layout without a bit of tweaking. +# +# From the announcements directory +# usage: script/post.sh [] +# where is one of hwk, lec, rec, lab, exam; +# is an integer; and is an optional explaination. +# +# For example, +# script/post.sh rec 2 solutions +# points out the posting of Recitation 2 solutions, and targets the URL +# $COURSE_WEBSITE/recitations.shtml#s2 + +. script/course-details.sh + +if [ "$#" -lt 2 ]; then + echo "usage: script/post.sh []" + exit 1 +fi + +DIR="$1" +INDEX="$2" +NOTE="$3" + +if [ -n "$NOTE" ] && [ "${NOTE:0:1}" != " " ]; then + NOTE=" $NOTE" # ensure leading space +fi + +if [ "$DIR" == "hwk" ]; then + NAME="Homework" + PAGE="homeworks.shtml" +elif [ "$DIR" == "lec" ]; then + NAME="Lecture" + PAGE="lectures.shtml" +elif [ "$DIR" == "rec" ]; then + NAME="Recitation" + PAGE="recitations.shtml" +elif [ "$DIR" == "lab" ]; then + NAME="Lab" + PAGE="labs.shtml" +elif [ "$DIR" == "exam" ]; then + NAME="Exam" + PAGE="exams.shtml" +else + echo "Unrecognized dir '$DIR'" + exit 1 +fi + +echo "echo \"$NAME $INDEX$NOTE posted.\" | \\ + atomgen -o atom.xml add -i atom.xml \"$NAME $INDEX$NOTE posted.\" \\ + \"$COURSE_WEBSITE/$PAGE#s$INDEX\"" +echo "$NAME $INDEX$NOTE posted." | \ + atomgen -o atom.xml add -i atom.xml "$NAME $INDEX$NOTE posted." \ + "$COURSE_WEBSITE/$PAGE#s$INDEX" diff --git a/html/.htaccess b/html/.htaccess new file mode 100644 index 0000000..f5ff39e --- /dev/null +++ b/html/.htaccess @@ -0,0 +1 @@ +AddType application/x-httpd-php .shtml diff --git a/html/Makefile b/html/Makefile new file mode 100644 index 0000000..53fcf32 --- /dev/null +++ b/html/Makefile @@ -0,0 +1,45 @@ +HTML_FILES = $(shell echo *.shtml) +EMPTY_DIRS = doc source +DEEP_EMPYT_DIRS = doc/exam doc/hwk doc/lab doc/lec doc/rec +HTML_DIRS = shared php xml $(EMPTY_DIRS) $(DEEP_EMPTY_DIRS) +SOURCE_FILES = $(HTML_FILES) $(HTML_DIRS) README .htaccess +OTHER_FILES = Makefile +DIST_FILES = $(SOURCE_FILES) $(OTHER_FILES) +DIST_FILE = website_framework.tar.gz +DIST_DIR = html + +clean : + rm -rf $(DIST_FILE) $(DIST_DIR) install* + +$(DIST_FILE) : $(DIST_FILES) $(EMPTY_DIRS) + mkdir $(DIST_DIR) + cp -rp $^ $(DIST_DIR) + tar -cozf $@ $(DIST_DIR) + rm -rf $(DIST_DIR) + +install : install-html install-source + +# Create a new directory for the installation +install-dir : + ssh $(INSTALL_USER)@$(INSTALL_HOST) mkdir $(INSTALL_DIR) + @date > $@ + +# Avoid the install-dir step, but allow installation to continue +install-override : + @date > install-dir + +# The transform removes the preceeding DIST_DIR (e.g. `html/') +install-html : $(DIST_FILE) install-dir + cat $< | ssh $(INSTALL_USER)@$(INSTALL_HOST) \ + tar --transform="s,$(DIST_DIR),.," -xzvC $(INSTALL_DIR) + ssh $(INSTALL_USER)@$(INSTALL_HOST) \ + cd $(INSTALL_DIR) '&&' rm -rf $(OTHER_FILES) $(DIST_DIR) + @date > $@ + +install-source : $(DIST_FILE) install-html + scp -p $(DIST_FILE) $(INSTALL_USER)@$(INSTALL_HOST):$(SOURCE_DIR) + @date > $@ + +# Create empty directories if neccessary (Git doesn't track dirs) +$(EMPTY_DIRS) : + mkdir $@ diff --git a/html/README b/html/README new file mode 100644 index 0000000..b01d1a9 --- /dev/null +++ b/html/README @@ -0,0 +1,76 @@ +To manage this website: + +**** Static information **** + +This information should only need to be setup at the beginning of the +course. + +** Introductory Blurbs ** + +You should change the introductory patter in all the *.shtml files +as necessary to suit your course. + +** Contact Information ** + +Add appropriate entries to the profs.xml, TAs.xml, and webmaster.xml +files in the xml directory. + +The prof and TA files are formatted into XHTML by people.php, and the +output is included in contact.shtml. The webmaster xml file is +formatted by webmaster.php. Because the formatting occurs at +run-time, the served page will always reflect the current xml data. + +** Course Information ** + +Add appropriate entries to the course.xml file in the xml directory. +The course xml file is formatted by quarter.php, and maybe a few more +in the future. + + +**** Dynamic information **** + +** Assignments / Section documents ** + +This material gets added and updated as the course progresses. +Basically, it consists of a lab, rectitation, exam, etc. files +(e.g. problems, solutions, procedures, etc.). The contents of the +directories in doc (e.g. labs) are scanned by section_docs.php to +create a list of all the files in the directory (e.g. 'doc/lab/' +beginning with a certain prefix (e.g. lab[0-9], starts with 'lab'). +So simply dumping the appropriate files into the appropriate directory +should get them displayed on the website. + +If you want to add a comment to a section document section (e.g. to +point out the due date for hwk3, or the date/time/room of exam1), just +add the a relevant XHTML snippet to the *_comment file. For example: + +

The Interference lab starts on Wed., April 08, 2009.

+

+Wednesday, April 8, Thursday, April 9 and Friday, April 10, 2009: +EVEN number sections 060, 062, 064, and 66. +

+

+Wednesday, April 15, Thursday, April 16, and Friday, April 17, 2009: +ODD number sections 063, 065, 071, and 073. +

+ + +If you have something that you want to put up later (e.g. solutions) +you can chmod it 640 (so that it's not world-readable), and the php +parser will ignore the file. After you have collected the +homework/exam/etc., you chan chmod 644 the file (so that it is world +readable) and it will show up on the website. + +** Announcements / Atom feed ** + +Course announcements should be formatted in an atom feed + http://en.wikipedia.org/wiki/Atom_(standard) +Atom feeds may be generated with a number of tools, but a simple +command-line generator for linux is + http://www.physics.drexel.edu/~wking/code#atomgen +Once you've generated the atom.xml file, copy it into the xml +directory so atom.php can find it, and the announcements page will be +filled in automatically. Another benefit of this approach is that a +feed monitor such as rss2email can be used to monitor the feed and +send classwide emails out whenever a new announcement is posted +(http://rss2email.infogami.com/). diff --git a/html/announcements.shtml b/html/announcements.shtml new file mode 100644 index 0000000..0175860 --- /dev/null +++ b/html/announcements.shtml @@ -0,0 +1,9 @@ + + + + + diff --git a/html/homeworks.shtml b/html/homeworks.shtml new file mode 100644 index 0000000..20ef959 --- /dev/null +++ b/html/homeworks.shtml @@ -0,0 +1,10 @@ + + +

Homeworks

+ + + + diff --git a/html/inactive/contact.shtml b/html/inactive/contact.shtml new file mode 100644 index 0000000..58ce033 --- /dev/null +++ b/html/inactive/contact.shtml @@ -0,0 +1,21 @@ + + +
+ + + +

+For any questions about the webpage, contact +. +

+ + diff --git a/html/inactive/exams.shtml b/html/inactive/exams.shtml new file mode 100644 index 0000000..d5927ee --- /dev/null +++ b/html/inactive/exams.shtml @@ -0,0 +1,10 @@ + + +

Exams

+ + + + diff --git a/html/inactive/labs.shtml b/html/inactive/labs.shtml new file mode 100644 index 0000000..3abfa6a --- /dev/null +++ b/html/inactive/labs.shtml @@ -0,0 +1,30 @@ + + +

Labs

+ +
    +
  • All students should be present in Disque-820A - Phys201 + Laboratory - a few minutes before the start of their scheduled lab. +
    • +
  • Please note that pre-lab work must be completed before you go to + the lab to perform the experiment. +
    • +
  • All students in a group must actively participate in the lab work. +
    • +
  • + All pages in each report must be completed and submitted to your lab + instructor before you leave the laboratory. To get a feeling for + our expectations, consider this + sample lab report. +
    • +
  • + No grade will be given for incomplete pre-lab work or lab reports. +
    • +
+ + + + diff --git a/html/inactive/lectures.shtml b/html/inactive/lectures.shtml new file mode 100644 index 0000000..9afd513 --- /dev/null +++ b/html/inactive/lectures.shtml @@ -0,0 +1,10 @@ + + +

Lectures

+ + + + diff --git a/html/index.shtml b/html/index.shtml new file mode 100644 index 0000000..5a31893 --- /dev/null +++ b/html/index.shtml @@ -0,0 +1,27 @@ + + +

Physics 201 - Fundamentals of Physics III

+ + +

+This is my personal Phys 201 course page, for things specific to my +two recitations. I'll probably just be posting homework solutions, +but I'll email you and post an announcement if that changes. +

+ +

Resources

+ + + +

Source code

+

+For those who are interested, the source code used to generate the +content of this page is available here. +

+ + diff --git a/html/php/atom.php b/html/php/atom.php new file mode 100644 index 0000000..5990683 --- /dev/null +++ b/html/php/atom.php @@ -0,0 +1,43 @@ +\n$content\n

\n"; +} + +function printAtom($feed_title, $feed, $ignored_title){ + echo "

$feed_title

\n\n"; + + /* convert entries to an array (from some sort of iterable) */ + $entries = array(); + foreach($feed->entry as $entry) + $entries[] = $entry; + + /* print the entries */ + foreach(array_reverse($entries) as $entry) { + $title=$entry->title; + $link=$entry->link['href']; + $tpub_string=$entry->published; + $tpub=strtotime($tpub_string); + $time=strftime("%r %A, %B %d, %Y", $tpub); + if ($title == $ignored_title) + continue; + echo "
$title $time
\n"; + printContent($entry->content->asXML()); + } +} + +?> diff --git a/html/php/people.php b/html/php/people.php new file mode 100644 index 0000000..e3e97d3 --- /dev/null +++ b/html/php/people.php @@ -0,0 +1,62 @@ +\n"; +} +function printPeopleEnd($use_pictures){ + if ($use_pictures == false) + echo "\n"; +} +function printPeople($title, $people, $use_pictures){ + if ($use_pictures == true) + echo "

$title

\n"; + else + echo " $title\n"; + foreach($people->person as $person){ + $name=$person->name; + if (strlen($person->url) > 0){ + $href=" href=\"".$person->url."\""; + } else { + $href=""; + } + $office=$person->office; + $email=$person->email; + $hours=$person->hours; + $picture=$person->picture; + $extension=$person->extension; + if ($use_pictures == true) { + echo "
\n"; + echo " \"$name\"\n"; + echo "

\n"; + echo " $name
\n"; + echo " Email: $email
\n"; + echo " Office: $office
\n"; + echo " Hours: $hours
\n"; + echo " Extension: $extension
\n"; + echo "

\n"; + echo "
\n"; + } else { + echo " $name$office$email\n"; + if (strlen($hours) > 0){ + echo " Office hours: $hours\n"; + } + } + } +} +?> diff --git a/html/php/quarter.php b/html/php/quarter.php new file mode 100644 index 0000000..59d38f2 --- /dev/null +++ b/html/php/quarter.php @@ -0,0 +1,4 @@ + diff --git a/html/php/section_docs.php b/html/php/section_docs.php new file mode 100644 index 0000000..ab20df1 --- /dev/null +++ b/html/php/section_docs.php @@ -0,0 +1,44 @@ +=0; $i-=1) { + $front = "$directory/$startswith".$i."_"; + $frontlen = strlen($front); + $files = glob($front."*"); + $comment_file = $front.'comment'; // (X)HTML fragment commenting on doc. + $num_readable = 0; + foreach ($files as $file) { + if (is_readable($file) && $file != $comment_file) { + $num_readable += 1; + } + } + if ($num_readable > 0 or is_readable($comment_file)) { + echo "

$title $i

\n"; + } // 's' b/c id attributes must begin with a letter, not a digit. + if (is_readable($comment_file)) { + readfile($comment_file); + } + if ($num_readable > 0) { + echo "
    \n"; + foreach ($files as $file) { + if (!is_readable($file) or $file == $comment_file) { + continue; + } + $len = strlen($file); + $url = $file; + $name = substr($file, $frontlen, $len-$frontlen); // remove $front + + echo "
  • $name$mode
    • \n"; + } + echo "
\n"; + } + } +} +?> diff --git a/html/php/webmaster.php b/html/php/webmaster.php new file mode 100644 index 0000000..fba3f4e --- /dev/null +++ b/html/php/webmaster.php @@ -0,0 +1,15 @@ +name; +$email=$s->email; +if (strlen($email) > 0) + echo "$name ($email)"; +else + echo "$name"; +?> diff --git a/html/recitations.shtml b/html/recitations.shtml new file mode 100644 index 0000000..cb1a671 --- /dev/null +++ b/html/recitations.shtml @@ -0,0 +1,10 @@ + + +

Recitations

+ + + + diff --git a/html/shared/favicon.ico b/html/shared/favicon.ico new file mode 100644 index 0000000..028b0f5 Binary files /dev/null and b/html/shared/favicon.ico differ diff --git a/html/shared/feed-icon-14x14.png b/html/shared/feed-icon-14x14.png new file mode 100644 index 0000000..b3c949d Binary files /dev/null and b/html/shared/feed-icon-14x14.png differ diff --git a/html/shared/footer.shtml b/html/shared/footer.shtml new file mode 100644 index 0000000..3d5e456 --- /dev/null +++ b/html/shared/footer.shtml @@ -0,0 +1,25 @@ +
+ + + + + + + + diff --git a/html/shared/header.shtml b/html/shared/header.shtml new file mode 100644 index 0000000..a525627 --- /dev/null +++ b/html/shared/header.shtml @@ -0,0 +1,63 @@ + + + + + + + + + + + + + + + + +Phys201 + + + +

+ +

+ + +
diff --git a/html/shared/style.css b/html/shared/style.css new file mode 100644 index 0000000..0a6a3fb --- /dev/null +++ b/html/shared/style.css @@ -0,0 +1,120 @@ +/* General styling */ +body { + background: #FEC; + color: #000; + font: 100% Veranda,Tahoma,sans-serif; + border: 0; + margin: 0; + padding: 0; +} + +/* Arranging content */ +#header { + border: 0; + margin: 0; + padding: 0 1em 0 1em; +} + +#footer { + border: 0; + margin: 0; + padding: 0 1em 0 1em; +} + +#content { + background: #FEC; + color: #000; + border: 0; + margin: 0; + padding: 0 1em 0 1em; +} + +div.figure_auto { + float: right; + margin: 0.5em; + padding: 0.5em; + text-align: center; +} + +div.figure_frac { + float: right; + width: 35%; + margin: 0 0.5em 0 0.5em; + padding: 0 0.5em 0 0.5em; + text-align: center; +} + +div.figure_frac_odd { + float: left; + width: 35%; + margin: 0 0.5em 0 0.5em; + padding: 0 0.5em 0 0.5em; + text-align: center; +} + +div.note_frac { + float: right; + width: 40%; + margin: 0.5em; + padding: 0.5em; +} + +div.figure_big { + margin: 0.5em auto 0.5em auto; + padding: 0.5em; + text-align: center; +} + +img.scaled { + width: 100%; +} + +/* Details */ +a:link {color: #00F;} +a:visited {color: #F00;} +a:hover {color: #44F;} +a:active {color: #FFF;} + +h1, h2, h3, h4, h5, h6 { + /* Removing the top margin and replacing it with padding + keeps the appropriate container in the background. */ + margin: 0; + padding: 1em 0 0 0; + font-variant: small-caps; +} + +h5, h6 { + margin: 0; + margin-top: 1em; + padding: 0; +} + +td { + padding: 0 0.5em 0 0.5em; +} + +p { + margin: 1em 0 0 0; + padding: 0; +} + +ul, ol { + padding: 0 0 0 2em; + margin: 0 0 0 0; +} + +#header h3 { /* remove the pre padding on the #header headings */ + padding: 0; +} + +.code, pre { + font-family: monospace; + font-style: normal; + color: #333; +} + +.feed-small { + height: 14px; + padding-left: 15px; + background: url('shared/feed-icon-14x14.png') no-repeat 0% 50%; +} diff --git a/html/xml/TAs.xml b/html/xml/TAs.xml new file mode 100644 index 0000000..d906647 --- /dev/null +++ b/html/xml/TAs.xml @@ -0,0 +1,75 @@ + + + + Farnaza Neville Batliwalla + fnb23 at drexel dot edu + /students/courses/current/physics-201/faculty/farnaza.jpg + + + Benjamin Coy + btc24 at drexel dot edu + Disque 916 + 1546 + /directory/graduate/small/coy.benjamin.jpg + + + Edward Damon + ead54 at drexel dot edu + Disque 705 + 2732 + /directory/graduate/small/damon.edward.jpg + + + Nandita Ganesh + ng97 at drexel dot edu + /students/courses/current/physics-201/faculty/nandita.jpg + + + Travis Hoppe + travis.aaron.hoppe at drexel dot edu + Disque 819A + 2057 + http://www.physics.drexel.edu/~thoppe + /directory/graduate/small/hoppe.travis.jpg + + + Trevor King + wtk25 at drexel dot edu + Disque 927 + 1818 + http://www.physics.drexel.edu/~wking + /directory/graduate/small/king.trevor.jpg + + + Anitha Manohar + Disque 912 + am627 at drexel dot edu + /students/courses/current/physics-201/faculty/anitha.jpg + + + Greeshma Manomohan + gm322 at drexel dot edu + /students/courses/current/physics-201/faculty/greeshma.jpg + + + Ryan Michaluk + rmm622 at drexel dot edu + Disque 916 + 1546 + /directory/graduate/small/michaluk.ryan.jpg + + + Pavithra Ramakrishnan + Disque 915 + 2739 + pr323 at drexel.edu + /students/courses/current/physics-201/faculty/pavithra.jpg + + + John Schreck + jss74@drexel.edu + Disque 705 + 2732 + /directory/graduate/small/schreck.john.jpg + + diff --git a/html/xml/admin_mailing_list.py b/html/xml/admin_mailing_list.py new file mode 100755 index 0000000..17134d1 --- /dev/null +++ b/html/xml/admin_mailing_list.py @@ -0,0 +1,27 @@ +#!/usr/bin/python +# +# Generate a mailing list from my people XML format e.g. for TAs.xml. +# +# usage: ./admin_mailing_list.py profs.xml TAs.xml + +import xml.etree.ElementTree as ET + +def mailing_list(filename): + tree = ET.parse(filename) + root = tree.getroot() + addresses = [] + for person in root.findall("person"): + name = person.findtext("name") + email = person.findtext("email") + email = email.replace(" at ", "@") + email = email.replace(" dot ", ".") + addresses.append('"%s" <%s>' % (name, email)) + return addresses + +if __name__ == "__main__": + import sys + + addresses = [] + for filename in sys.argv[1:]: + addresses.extend(mailing_list(filename)) + print '\t' + ', \\\n\t'.join(addresses) diff --git a/html/xml/department_xml_to_people.py b/html/xml/department_xml_to_people.py new file mode 100755 index 0000000..88b6f90 --- /dev/null +++ b/html/xml/department_xml_to_people.py @@ -0,0 +1,60 @@ +#!/usr/bin/python +# +# Convert e.g. graduate.xml from the department directory into my +# people XML format e.g. for TAs.xml. +# +# usage: ./department_xml_to_people.py graduate.xml | xml_pp > TAs.xml +# +# xml_pp is included in the Ubuntu xml-twig-tools package. + +import elementtree.ElementTree as ET + +def convert_file(infile, outfile): + root = ET.Element("people") + in_tree = ET.parse(infile) + in_root = in_tree.getroot() + for in_person in in_root.findall("person"): + firstname = in_person.findtext("firstname") + lastname = in_person.findtext("lastname") + name_text = "%s %s" % (firstname, lastname) + email_text = in_person.findtext("email") + email_text = email_text.replace(" [at] ", " at ") + email_text = email_text.replace(".edu ", " dot edu") + office_text = in_person.findtext("office") + hours_text = None + phone_text = in_person.findtext("phone") + extension_text = phone_text.replace("(215) 895 - ","") + url_text = in_person.findtext("webpage") + picture_text = "/directory/graduate/small/%s.%s.jpg" \ + % (lastname.lower(), firstname.lower()) + + person = ET.SubElement(root, "person") + name = ET.SubElement(person, "name") + name.text = name_text + if email_text != None: + email = ET.SubElement(person, "email") + email.text = email_text + if office_text != None: + office = ET.SubElement(person, "office") + office.text = office_text + if hours_text != None: + hours = ET.SubElement(person, "hours") + hours.text = hours_text + if extension_text != None: + extension = ET.SubElement(person, "extension") + extension.text = extension_text + if url_text != None: + url = ET.SubElement(person, "url") + url.text = url_text + if picture_text != None: + picture = ET.SubElement(person, "picture") + picture.text = picture_text + tree = ET.ElementTree(root) + tree.write(outfile) + +if __name__ == "__main__": + import sys + + #infile = sys.argv[1] + #outfile = sys.argv[2] + convert_file(sys.stdin, sys.stdout) diff --git a/html/xml/profs.xml b/html/xml/profs.xml new file mode 100644 index 0000000..4a41a7d --- /dev/null +++ b/html/xml/profs.xml @@ -0,0 +1,12 @@ + + + + T. S. Venkataraman + Disque 912 + Open Door Policy Every Term + venkat at drexel dot edu + 2721 + /directory/faculty/homepage.html?name=Venkat + /students/courses/current/physics-201/faculty/venkatMATE.bmp + + diff --git a/html/xml/webmaster.xml b/html/xml/webmaster.xml new file mode 100644 index 0000000..ba06d43 --- /dev/null +++ b/html/xml/webmaster.xml @@ -0,0 +1,5 @@ + + + Trevor King + + diff --git a/latex/Makefile b/latex/Makefile new file mode 100644 index 0000000..19c783b --- /dev/null +++ b/latex/Makefile @@ -0,0 +1,11 @@ +SUBDIRS = hwk rec + +install : + @for i in $(SUBDIRS); do \ + echo "make install in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install); done + +clean : + @for i in $(SUBDIRS); do \ + echo "make clean in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) clean); done diff --git a/latex/README b/latex/README new file mode 100644 index 0000000..6638893 --- /dev/null +++ b/latex/README @@ -0,0 +1,14 @@ +problems Contains LaTeX source for a bunch of problems +hwk/rec Contain particular homework and recitation subdirs, e.g. hwk1 + +The homework subdirs link back to the particular problems in the +problems directory so the problems can live in a central location and +be reused in future quarters. + +There are also non-problem directories such as `syllabus' for LaTeX +source for other course documents. With these added to Makefile's +`SUBDIRS', the source they contain is compiled and installed in their +`INSTALL_DIR'. Take a look at syllabus/Makefile for the particulars, +which are not too complicated. The directory `old-source' contains +the LaTeX source from previous courses, since so much of the layout +is standardized and much of the text is boilerplate. diff --git a/latex/hwk/Makefile b/latex/hwk/Makefile new file mode 100644 index 0000000..ff632c7 --- /dev/null +++ b/latex/hwk/Makefile @@ -0,0 +1,21 @@ +# give numbers for assigned homeworks +HWK_NUMS = +# give numbers for homeworks whose solutions should be posted +# (don't install source until the solutions should be published) +SOLN_NUMS = + +INSTALL_DIR := $(INSTALL_DIR)/doc/hwk +export INSTALL_DIR + +install : + @for i in $(HWK_NUMS:%=hwk%); do \ + echo "make install-probs in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install-probs); done + @for i in $(SOLN_NUMS:%=hwk%); do \ + echo "make install-solns in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install-solns); done + +clean : + @for i in $(HWK_NUMS:%=hwk%); do \ + echo "make clean in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) clean); done diff --git a/latex/hwk/hwk1/Makefile b/latex/hwk/hwk1/Makefile new file mode 100644 index 0000000..375c9ad --- /dev/null +++ b/latex/hwk/hwk1/Makefile @@ -0,0 +1,41 @@ +THIS_DIR = $(shell basename $(PWD)) +HOMEWORK_NUMBER = $(THIS_DIR:hwk%=%) +SOURCE_FILES = all_problems.tex probs.tex sols.tex problem[0-9].tex +OTHER_FILES = Makefile +DIST_FILES = $(SOURCE_FILES) $(OTHER_FILES) +DIST_FILE = $(THIS_DIR)_source.tar.gz +DIST_DIR = hwk + +all : sols.pdf probs.pdf + +view : all + xpdf probs.pdf & + xpdf sols.pdf & + +%.pdf : %.tex $(SOURCE_FILES) + pdflatex $(patsubst %.tex,%,$<) + asy $(patsubst %.tex,%,$<) + pdflatex $(patsubst %.tex,%,$<) + +semi-clean : + rm -f rm -f *.log *.aux *.out *.thm *.toc *.pre *_[0-9]_.tex *.asy + +clean : semi-clean + rm -f *.pdf $(DIST_FILE) $(DIST_DIR) install* + +$(DIST_FILE) : $(DIST_FILES) + mkdir $(DIST_DIR) + cp -Lrp $^ $(DIST_DIR) + tar -chozf $@ $(DIST_DIR) + rm -rf $(DIST_DIR) + +install : install-probs install-solns + +install-probs : probs.pdf + scp -p $< $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR)/hwk$(HOMEWORK_NUMBER)_problems.pdf + @date > $@ + +install-solns : sols.pdf $(DIST_FILE) + scp -p $< $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR)/hwk$(HOMEWORK_NUMBER)_solutions.pdf + scp -p $(DIST_FILE) $(INSTALL_USER)@$(INSTALL_HOST):$(SOURCE_DIR) + @date > $@ diff --git a/latex/hwk/hwk1/all_problems.tex b/latex/hwk/hwk1/all_problems.tex new file mode 100644 index 0000000..b6d517e --- /dev/null +++ b/latex/hwk/hwk1/all_problems.tex @@ -0,0 +1,19 @@ +\usepackage[author={W. Trevor King}, + coursetitle={Physics 201}, + classtitle={Homework 1}, + subheading={Chapter 12}]{problempack} +\usepackage[inline]{asymptote} +\usepackage{graphicx} +\usepackage{wtk_cmmds} +\usepackage{hyperref} +\usepackage{url} + +\begin{document} + +\maketitle + +\input{problem1} +\input{problem2} +\input{problem3} + +\end{document} diff --git a/latex/hwk/hwk1/problem1.tex b/latex/hwk/hwk1/problem1.tex new file mode 120000 index 0000000..3999af0 --- /dev/null +++ b/latex/hwk/hwk1/problem1.tex @@ -0,0 +1 @@ +../../problems/problem12.V1.tex \ No newline at end of file diff --git a/latex/hwk/hwk1/problem2.tex b/latex/hwk/hwk1/problem2.tex new file mode 120000 index 0000000..33a1893 --- /dev/null +++ b/latex/hwk/hwk1/problem2.tex @@ -0,0 +1 @@ +../../problems/problem12.V2.tex \ No newline at end of file diff --git a/latex/hwk/hwk1/problem3.tex b/latex/hwk/hwk1/problem3.tex new file mode 120000 index 0000000..18d2916 --- /dev/null +++ b/latex/hwk/hwk1/problem3.tex @@ -0,0 +1 @@ +../../problems/problem12.V3.tex \ No newline at end of file diff --git a/latex/hwk/hwk1/probs.tex b/latex/hwk/hwk1/probs.tex new file mode 100644 index 0000000..fa21ac1 --- /dev/null +++ b/latex/hwk/hwk1/probs.tex @@ -0,0 +1,5 @@ +\documentclass[letterpaper, 10pt]{article} + +\PassOptionsToPackage{nosolutions}{problempack} + +\input{all_problems} diff --git a/latex/hwk/hwk1/sols.tex b/latex/hwk/hwk1/sols.tex new file mode 100644 index 0000000..3ba5c0a --- /dev/null +++ b/latex/hwk/hwk1/sols.tex @@ -0,0 +1,5 @@ +\documentclass[letterpaper, 10pt]{article} + +\PassOptionsToPackage{solutions}{problempack} + +\input{all_problems} diff --git a/latex/old-source/s09-phys201-syllabus.tex b/latex/old-source/s09-phys201-syllabus.tex new file mode 100644 index 0000000..3ecf347 --- /dev/null +++ b/latex/old-source/s09-phys201-syllabus.tex @@ -0,0 +1,141 @@ +\documentclass[10pt]{article} +\usepackage[colorlinks]{hyperref} +\usepackage{url} + +\topmargin -0.5in +\headheight 0.5in +\headsep 0.0in +\textheight 9.5in % leave a bit of extra room for page numbers +\oddsidemargin -0.5in +\textwidth 7.5in +\setlength{\parindent}{0pt} + +\usepackage{fancyhdr} +\pagestyle{fancy} +\fancyhf{} % delete the current section for header and footer +\lfoot{Dr.~Venkat --- Disque--912 --- x2721} +\rfoot{Venkat at drexel dot edu} +\renewcommand{\headrulewidth}{0pt} % remove the head-rule + +\newcommand{\Tstrut}{\rule{0pt}{2.6ex}} % strut for table spacing +\newcommand{\toprule}{} +\newcommand{\colrule}{\hline\Tstrut} +\newcommand{\botrule}{} + +\begin{document} + +\begin{center} +{\Large PHYS--201: FUNDAMENTALS OF PHYSICS III} \\ +{\large Spring Term 2009 --- Sophmore Class} \\ +{\it Principles of Physics} by Raymond A. Serway and John W. Jewett Jr. +Fourth Edition + +\bigskip +\bigskip +SYLLABUS (as of \today% +\footnote{Assigned problems/course material schedule subject to + changes determined by the progress of the course.}) +\end{center} + +\begin{tabular}{llll} +\toprule +Week & +Material & %\footnotemark & +Reading & +Recitation Problems \\ +\colrule +1 & +Oscillatory Motion and Waves & +Ch.~12 (1-8) & +Ch.~12: 2, 5, 12, 15, 18. \\ +30-Mar-09 & +Superposition of waves & +Ch.~13 (1-7) & +No homework due this week. \\ +\colrule +2 & +Interference and Standing waves & +Ch.~14 (1-5) & +Ch.~12: 20, 40. Ch.~13: 5, 7, 10, 11, 40. \\ +06-Apr-09 & +Maxwell's Equations and EM waves & +Ch.~24 (1-6) & +Homework 1: Due Wed.~April 08 in Lecture. \\ +\colrule +3 & +Interference and Diffraction & +Ch.~24 (1-10) & +Ch.~14: 5, 6, 8, 9, 20. Ch.~24: 7, 8, 9, 22, 25. \\ +13-Apr-09 & +Lasers and Holography & + & +Homework 2: Due Wed.~April 15 in Lecture. \\ +\colrule +4 & +Quantum Physics, Planck's Theory & +Ch.~28 (1-2) & +Ch.~27: 1, 3, 4, 9, 15, 22, 23, 32. \\ +20-Apr-09 & +Philoelectric effect & + & +Exam 1 : 8 AM--8:50 AM Tuesday, April 21, 2009. \\ +\colrule +5 & +Bohr Model of Hydrogen & +Ch.~11 (5-6) & +Ch.~28: 4, 6, 9, 10, 13, 56, 57. \\ +27-Apr-09 & + & + & +Homework 3: Due Wed.~April 29 in Lecture. \\ +\colrule +6 & +Special Relativity & +Ch.~9 (1-5) & +Ch.~11: 34, 35, 36, 37, 38, 41. \\ +04-May-09 & + & + & +Exam 2: 8 AM--8:50 AM Tuesday, May 05, 2009. \\ +\colrule +7 & +Special Relativity--Mass Energy & +Ch.~9 (6-8) & +Ch.~9: 18, 19, 22, + Handout Problem. \\ +11-May-09 & + & + & +Homework 4: Due Wed.~May 13 in Lecture. \\ +\colrule +8 & +X-Rays - Compton Effect & +Ch.~28 (3-8) & +Ch.~9: 23, 30, 31, 35, 50, + Handout Problem. \\ +18-May-09 & +\multicolumn{2}{l}{Wave-Particle Duality, Quantum particle} & +Exam 3: 8 AM-8:50 AM Tuesday, May 19, 2009. \\ +\colrule +9 & +Heisenberg's Uncertainty Principle & +Ch.~28 (7-8) & +Ch.~28: 14, 15, 16, + Handout Problem. \\ +25-May-09 & +Schrodinger Equation and applications & +Ch.~28 (11-12) & +Homework 5: Due Wed.~May 27 in Lecture. \\ +\colrule +10 & +Atomic Physics and Hydrogen atom & +Ch.~29 (1-5) & +Ch.~28: 18, 21, 23, 25, 32, 34. \\ +01-Jun-09 & +Nuclear Physics-Binding Fusion, Fission & +Ch.~30 (1-5) & \\ +\botrule +\end{tabular} +%\footnotetext{Assigned problems/course material schedule subject +% to changes determined by the progress of the course.} +% Normal \footnote{}s can't escape tabulars. This fix from +% http://www.tex.ac.uk/cgi-bin/texfaq2html?label=footintab + +\end{document} diff --git a/latex/problems/README b/latex/problems/README new file mode 100644 index 0000000..e179ce2 --- /dev/null +++ b/latex/problems/README @@ -0,0 +1,4 @@ +A collection of LaTeX source (using my problempack.sty and wtk_cmmds.sty) +for intro-physics problems that I've covered over the years. + +References are to Serway & Jewett, 4th Edition unless otherwise specified. diff --git a/latex/problems/equation27.07.tex b/latex/problems/equation27.07.tex new file mode 100644 index 0000000..e0e40e7 --- /dev/null +++ b/latex/problems/equation27.07.tex @@ -0,0 +1,69 @@ +\begin{problem} +\emph{BONUS PROBLEM}. +Derive the Equation 27.8, which gives the average intensity on a +screen far from a single slit relative to the maximum intensity +$I_\text{max}$ at $\theta=0$. +\begin{equation} + I_\text{avg} = I_\text{max} \cos^2\p({\frac{\pi d \sin(\theta)}{\lambda}}) +\end{equation} + +\emph{HINT}. Remember from Chapter 24 that for plane waves +\begin{equation} + I = \frac{1}{2\mu_0 c} E_\text{max}^2 +\end{equation} +where the electric field is perpendicular to the direction of +propogation. Assume the screen is far enough away that the waves +emanating from the slits can be treated as plane waves. +\end{problem} % combines the phase difference from Equation 27.7 with vector +% addition for the Electric field amplitudes. + +\begin{solution} +From the path-length argument we've used in Problems 2 and 3, we know +the phase difference between the light from each slit will be +\begin{equation} + \Delta \phi = \frac{2\pi}{\lambda} d \sin(\theta) +\end{equation} + +Drawing a phasor diagram for the Electric field, we have +\begin{center} +\begin{asy} +import Mechanics; +import ElectroMag; + +real u = 1cm; +transform t=scale(u); + +real E = 1; +pair P = (0,0); + +Vector Et = EField(t*P, u*E, 0, L="$E_t$"); +Vector Eb = EField(t*P, u*E, 140, L="$E_b$"); +Angle a = Angle(Et.pTip(), Et.center, Eb.pTip(), Et.mag/2, L="$\Delta \phi$"); + +Et.draw(); +Eb.draw(); +a.draw(); +dot(P); +\end{asy} +\end{center} +\begin{equation} + \Delta \phi = \frac{2\pi}{\lambda} d \sin(\theta) +\end{equation} +The maximum electric field is thus given by +\begin{equation} + E_\text{max} = 2 E_0 \cos\p({\frac{\Delta\phi}{2}}) +\end{equation} +where $E_0 = E_t = E_b$. +The intensity is then given by +\begin{align} + I &= \frac{1}{2\mu_0 c} E_\text{max}^2 \\ + &= \frac{1}{2\mu_0 c} 4 E_0^2 \cos^2\p({\frac{\Delta\phi}{2}}) \\ + &= I_\text{max} \cos^2\p({\frac{\Delta\phi}{2}}) \;, +\end{align} +where we made the substitution $I_\text{max} = I(\Delta\phi=0)$. +Plugging in for $\Delta\phi$, +\begin{equation} + I = I_\text{max} \cos^2\p({\frac{\pi d \sin(\theta)}{\lambda}}) \;, +\end{equation} +which is what we set out to show. +\end{solution} diff --git a/latex/problems/example13.06.T.tex b/latex/problems/example13.06.T.tex new file mode 100644 index 0000000..56b684f --- /dev/null +++ b/latex/problems/example13.06.T.tex @@ -0,0 +1,25 @@ +\begin{problem} +Jack and Jill are broadcasting kazoo music from their treehouse to a +picnic below using a tin can telephone +(\url{http://en.wikipedia.org/wiki/Tin_can_telephone}). Their +transmitting string weighs $m = 140\U{g}$, is $L = 30\U{m}$ long, and +is stretched to a tension of $T = 45\U{N}$. At what amplitude must +Jack and Jill vibrate their end of the string to drive the far can at +$P = 1\U{W}$ while playing the musical note A at $f = 440\U{Hz}$? +\end{problem} % Example 13.6 backwards + +\begin{solution} +From Equation 13.23 we know that energy transfer in a sinusoidally +oscillating string follows +\begin{equation} + P = \frac{1}{2} \mu \omega^2 A^2 v +\end{equation} +The mass density of the string is $\mu = m/L \approx 4.667\U{g/m}$, +the speed of propagation is $v = \sqrt{T/\mu} \approx 98.20\U{m/s}$, +and the angular velocity of vibration is $\omega = 2\pi f \approx +2765\U{rad/s}$. Solving the power formula for the amplitude we have +\begin{equation} + A = \frac{1}{\omega}\sqrt{\frac{2P}{\mu v}} \approx \ans{0.76\U{mm}} +\end{equation} + +\end{solution} diff --git a/latex/problems/figure27.03.T.tex b/latex/problems/figure27.03.T.tex new file mode 100644 index 0000000..5fab2a2 --- /dev/null +++ b/latex/problems/figure27.03.T.tex @@ -0,0 +1,120 @@ +\begin{problem} +A plane wave of monochromatic light moving to the right through a +strange material with index of refraction $n_a$ hits a barrier with +two slits seperated by $d$. Some light passes through each slit, and +after moving a distance $r_a$ to right, leaves the strange material +and enters air with a refractive index $n_b$. The interface bends the +light rays according to Snell's law (bending shown on the diagram, so +you don't need Snell's law to solve the problem). A distance $r_b$ to +the right of the interface is a screen, on which an interference +pattern appears. The first minimum of the interference pattern +appears at point $P$, directly to the right of the top slit. Find the +frequency of light. + +\emph{HINT}. I've drawn the paths taken to $P$ from each slit. The +vertical distance between the two rays at the strange-material/air +interface is $d_i$. You should use a geometric argument to determine +the phase difference between the two paths. +\begin{center} +\begin{asy} +real ux=2cm; +real uy=100cm; +transform t=xscale(ux)*yscale(uy); + +pen cLight=red; +pen cA=blue+white; +pen cB=white; + +real nA=2.5; +real nB=1; + +real xAB = 1; +pair P=(3,0); +pair T=(-1,0); +pair B=(-1,-0.007); + +real xMin = T.x - 1; +real xMax = P.x + 0.5; +real yMin = B.y - (T.y-B.y); +real yMax = T.y + (T.y-B.y); + +path pA = (xMin,yMin)--(xAB,yMin)--(xAB,yMax)--(xMin,yMax)--cycle; +path pB = (xMax,yMin)--(xAB,yMin)--(xAB,yMax)--(xMax,yMax)--cycle; +path pScreen = (T.x, yMax)--(T.x, yMin); +path pScreenB = (P.x, yMax)--(P.x, yMin); +path pIncident = (xMin, yMin)--(T.x, yMin)--(T.x, yMax)--(xMin, yMax)--cycle; +path pLightT = T--P; + +/* For small angles, Snell's law is + * nA thetaA = nB thetaB + * comparing the Y displacement, + * yA = rA thetaA = rA/rB rB thetaB nB/nA = (rA nB)/(rB nA) yB + * yA/yB = (rA nB)/(rB nA) + * for a total Y displacement yT = yA + yB, we have + * yA/(yT-yA) = (rA nB)/(rB nA) + * yT/yA - 1 = (rB nA)/(rA nB) + * yA = yT ((rB nA)/(rA nB) + 1)^-1 = (yT rA nB)/(rB nA + rA nB) + */ +real yT = T.y-B.y; +real rA = xAB - T.x; +real rB = P.x - xAB; +real yA = yT * rA * nB / (rB * nA + rA * nB); +real yBP = B.y + yA; +path pLightB = B--(xAB, yBP)--P; + +fill(t*pA, cA); +fill(t*pB, cB); + +fill(t*pIncident, cLight); +draw(t*pLightT, cLight); +draw(t*pLightB, cLight); + +draw(t*pScreenB); +draw(t*pScreen); +dot(t*T, cA); // make slits in the screen +dot(t*B, cA); + +dot(t*P); +label("$P$", t*P, E); +label(format("$d=%f\U{mm}$", yT*1000), t*(T+B)/2, W); +label(format("$r_a=%f\U{m}$", rA), t*(T+(xAB,T.y))/2, N); +label(format("$r_b=%f\U{m}$", rB), t*((xAB,T.y)+P)/2, N); +label(format("$d_i=%f\U{mm}$", (yT-yA)*1000), t*(xAB, (T.y+yBP)/2), W); +label(format("$n_a=%f$", nA), t*(xAB, B.y), SW); +label(format("$n_b=%f$", nB), t*(xAB, B.y), SE); +\end{asy} +\end{center} +\end{problem} % based on the double-slit interference derivation +% (Figure 27.3), with the added tweak of an index-of-defraction +% altered wavelength (Equation 25.3 or 27.9). + +\begin{solution} +The path length in mediums $a$ and $b$ for the top $t$ and bottom $b$ +slits are given by +\begin{align} + L_{ta} &= 2\U{m} & L_{tb} &= 2\U{m} \\ + L_{ba} &= \sqrt{2^2+(0.007-0.005)^2}\U{m} = (2 + 1.00\E{-6})\U{m} + & L_{bb} &= \sqrt{2^2+0.005^2}\U{m} = (2 + 6.25\E{-6})\U{m} +\end{align} + +The wavelength depends on the index of refraction according to +$\lambda = \frac{c}{nf}$, and the phase along a path is given by $\phi += \frac{2\pi}{\lambda} L$, so +\begin{equation} + \frac{\Delta \phi}{2\pi} = \frac{\phi_b - \phi_t}{2\pi} + = \frac{L_{ba}}{\lambda_a} + \frac{L_{bb}}{\lambda_b} + -\frac{L_{ta}}{\lambda_a} + \frac{L_{tb}}{\lambda_b} + = \frac{f}{c} \p[{n_a (L_{ba}-L_{ta}) + n_b(L_{bb}-L_{tb})}] + \;. +\end{equation} +At frequency for which this is the first minimum, +\begin{equation} + \Delta \phi = \pi \;, +\end{equation} +so +\begin{equation} + f = \frac{c}{2 \p[{n_a(L_{ba}-L_{ta}) + n_b(L_{bb}-L_{tb})}]} + = \frac{3.00\E{8}\U{m/s}}{2 \p({2.5 \cdot 1.00\E{-6}\U{m} + 6.25\E{-6}\U{m}})} + = \ans{17.1\U{THz}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem01.60.tex b/latex/problems/problem01.60.tex new file mode 100644 index 0000000..119e815 --- /dev/null +++ b/latex/problems/problem01.60.tex @@ -0,0 +1,44 @@ +\begin{problem*}{1.60} +The consumption of natural gas by a company satisfies the empirical +equation +\begin{equation} + V = 1.50 t + 0.00800 t^2, +\end{equation} +where $V$ is the volume in millions of cubic feet and $t$ is the time in +months. Express this equation in units of cubic feet and seconds. +Assign proper units to the coefficients. Assume that a month is $30.0$ +days. +\end{problem*} % Probem 1.60 + +\begin{solution} +Adding units to the equation coefficients: +\begin{equation} + V = 1.50 \p[{\frac{\text{million ft}^3}{\text{month}}}] t + + 8.00\E{-3} \cdot\p[{\frac{\text{million ft}^3}{\text{month}^2}}] t^2 +\end{equation} + +We prepare some conversions: +\begin{align} +1 &= \p[{\frac{10^6 \text{ ft}^3}{\text{million ft}^3}}] \\ +1 &= \p[{\frac{1 \text{ month}}{30 \text{ days}} + \cdot \frac{1 \text{ day}}{24 \text{ hours}} + \cdot \frac{1 \text{ hour}}{60 \text{ minutes}} + \cdot \frac{1 \text{ minute}}{60 \text{ s}}}] \\ + &= \p[{\frac{ 1 \text{ month}}{ 2.592 \cdot 10^6 \text{ s}}}] +\end{align} + +So converting the units in our equation to ft$^3$ and s: +\begin{align} + V &= 1.50 \p[{\frac{\text{million ft}^3}{\text{month}} + \cdot \frac{10^6\U{ft}^3}{\text{million ft}^3} + \cdot \frac{1\U{month}}{2.592\E{6}\U{s}}}] + t + + 8.00\E{-3} \p[{\frac{\text{million ft}^3}{\text{month}^2} + \cdot \frac{10^6\U{ft}^3}{\text{million ft}^3} + \cdot \p{(\frac{1\U{month}}{2.592\E{6}\U{s}}})^2 }] + t^2 \\ + V &= 0.579 \p[{\frac{\text{ft}^3}{\text{s}}}] t + + 1.19\E{-9} \cdot\p[{\frac{\text{ft}^3}{\text{s}^2}}] t^2 +\end{align} + +\end{solution} diff --git a/latex/problems/problem01.62.tex b/latex/problems/problem01.62.tex new file mode 100644 index 0000000..447af74 --- /dev/null +++ b/latex/problems/problem01.62.tex @@ -0,0 +1,37 @@ +\begin{problem*}{1.62} +In physics, it is important to use mathematical approximations. +Demonstrate that for small angles ($< 20\dg$) +\begin{equation} + \tan \alpha \approx \sin \alpha \approx \alpha = \pi \alpha ' / 180\dg +\end{equation} +where $\alpha$ is in radians and $\alpha '$ is in degrees. +Use a calculator to find the largest angle for which $\tan \alpha$ may +be approximated by $\alpha$ with an error less than $10.0$\%. +\end{problem62} % Problem 1.62 + +\begin{solution} +To kill both birds with one stone, + a table to show the approximations hold + and show the \% error of the approximation:\\ +\begin{tabular}{|r|r|r|r|r|} +\hline +$\alpha '$&$\alpha$ [rad]&$\sin \alpha$&$\tan \alpha$&\% error\\ +\hline + $0\dg$&0.000&0.000&0.000&$\emptyset$\\ + $5\dg$&0.087&0.087&0.087&$-0.25$\%\\ +$10\dg$&0.175&0.174&0.176&$-1.02$\%\\ +$15\dg$&0.262&0.259&0.268&$-2.30$\%\\ +$20\dg$&0.349&0.342&0.354&$-4.09$\%\\ +$31\dg$&0.541&0.515&0.601&$-9.95$\%\\ +$32\dg$&0.599&0.530&0.625&$-10.62$\%\\ +\hline +\end{tabular}\\ +where the \% error is given by +\begin{equation} + \text{\% error} = \frac{\text{approx.} - \text{actual}}{\text{actual}} + = \frac{\alpha - \tan \alpha}{\tan \alpha}. +\end{equation} + +So $31\dg$ is the largest whole-degree angle with $< 10$\% error. +\end{solution} + diff --git a/latex/problems/problem02.10.tex b/latex/problems/problem02.10.tex new file mode 100644 index 0000000..410eedd --- /dev/null +++ b/latex/problems/problem02.10.tex @@ -0,0 +1,20 @@ +\begin{problem*}{2.10} +A $50.0\U{g}$ super-ball traveling at $25.0\U{m/s}$ bounces off a +brick wall and rebounds at $22.0\U{m/s}$. A high-speed camera records +this event. If the ball is in contact with the wall for $3.50\U{ms}$, +what is the magnitude of the average acceleration of the ball during +this time interval. (Note: $1\U{ms} = 10^{-3}\U{s}$.) +\end{problem*} % Problem 2.10 + +\begin{solution} +Pick a coordinate system (e.g. rebound direction is positive). Then +$v_0 = -25.0\U{m/s}$ and $v_1 = 22.0\U{m/s}$. +\begin{equation} + a \equiv \frac{\Delta v}{\Delta t} + = \frac{v_1 - v_0}{\Delta t} + = \frac{[22.0 - (-25.0)][\mbox{m/s}]} + {3.50\U{ms} \cdot \frac{1\mbox{s}}{10^3\U{ms}}} + = \frac{47.0\mbox{m/s}}{3.5\E{-3}\U{s}} + = 13400\U{m/s}^2 +\end{equation} +\end{solution} diff --git a/latex/problems/problem02.16.tex b/latex/problems/problem02.16.tex new file mode 100644 index 0000000..d47b2d6 --- /dev/null +++ b/latex/problems/problem02.16.tex @@ -0,0 +1,16 @@ +\begin{problem*}{2.16} +Draw motion diagrams... +\end{problem*} % problem 2.16 + +\begin{solution} +Look at Figure 2.11 on page 50 of the text. +For question parts (e) and (f), draw figures (b) and (c) respectively +but with the $x$-axis reversed. + +Note the different spacing in the figure because the `strobe' is going +off at a constant frequency (same time between pictures). +If you didn't vary the spacing when the velocity changed, you'd need to +point out somewhere that your time intervals were not constant. + +I also accepted plots of velocity or position vs time. +\end{solution} diff --git a/latex/problems/problem02.40.tex b/latex/problems/problem02.40.tex new file mode 100644 index 0000000..2847e6f --- /dev/null +++ b/latex/problems/problem02.40.tex @@ -0,0 +1,72 @@ +\begin{problem*}{2.40} +A woman is reported to have fallen $144\U{ft}$ from the 17th floor of +a building, landing on a metal ventilator box that she crushed to a +depth of $18.0\U{in}$. She suffered only minor injuries. Ignoring +air resistance, calculate +\Part{a} the speed of the woman just before she collided with the box and +\Part{b} her average acceleration while in contact with the box. +\Part{c} Modeling her acceleration as constant, calculate the time interval +it took to crush the box. +\end{problem*} % problem 2.40 + +\begin{solution} +\Part{a} +First deal with the portion from the top (point $P_0$) to the point of +collision with the box (point $P_1$). +Pick a coordinate system pointing down, with $x_0 = 0\U{m}$. +Converting the distance into meters: +\begin{equation} + x_1 = 144\U{ft} \cdot \frac{1\U{m}}{3.28\U{ft}} + = 43.9\U{m} +\end{equation} +So\\ +\begin{tabular}{r || r | r |} + & $P_0$ & $P_1$ \\ + \hline + \hline + $a$ & \multicolumn{2}{|c|}{$9.8\U{m/s}^2$} \\ + \hline + $v$ & $0\U{m/s}$ & ? \\ + \hline + $x$ & $0\U{m}$ & $43.9\U{m}$ \\ + \hline + $t$ & $0\U{s}$ & ? \\ + \hline +\end{tabular} + +We want $v_1$ and we don't know $t_1$ so we use +\begin{align} + v_1^2 &= v_0^2 + 2 a \Delta x_{01} \\ + v_1 &= \sqrt{2 a (x_1 - x_0)} + = \sqrt{2 \cdot 9.8\U{m/s} \cdot 43.9\U{m}} + = 29.3\U{m/s} +\end{align} + +Some people wanted to leave $\Delta x$ in ft, and this works if you also +use $a$ in ft/s$^2$. You run into trouble if you use ft for one and m +for the other... + +\Part{b} +Converting the change in $x$ over the box into m we have +\begin{equation} + \Delta x_{12} = 18\mbox{in} + \cdot \frac{1\U{ft}}{12\U{in}} + \cdot \frac{1\U{m}}{3.28\U{m}} + = 0.457\U{m} +\end{equation} +Calling the point just after she crushed the box $P_2$, we have +\begin{align} + v_2^2 &= v_1^2 + 2 a_{12} \Delta x_{12} \\ + a_{12} &= \frac{-v_1^2}{2 \Delta x_{12}} + = \frac{-(29.3\U{m/s})^2}{2 \cdot 0.457\U{m}} + = -941\U{m/s}^2 +\end{align} + +\Part{c} +\begin{align} + v_2 &= a_{12} t_{12} + v_1 \\ + t_{12} &= -v_1/a_{12} + = \frac{-29.3\U{m/s}}{-941\U{m/s}^2} + = 31.2\U{ms} +\end{align} +\end{solution} diff --git a/latex/problems/problem02.49.tex b/latex/problems/problem02.49.tex new file mode 100644 index 0000000..1ce9702 --- /dev/null +++ b/latex/problems/problem02.49.tex @@ -0,0 +1,76 @@ +\begin{problem*}{2.49} +Setting a world record in a $100\U{m}$ race, Maggie and Judy cross the +finish line in a dead heat, both taking $10.2\U{s}$. Accelerating +uniformly, Maggie took $2.00\U{s}$ and Judy $3.00\U{s}$ to attain +maximum speed, which they maintained for the rest of the race. \\ +\Part{a} What was the acceleration of each sprinter? \\ +\Part{b} What were their respective maximum speeds? \\ +\Part{c} Which sprinter was ahead at the $6.00\U{s}$ mark and by how much? +\end{problem*} % problem 2.49 + +\begin{solution} +\Part{a}\Part{b} +Consider Maggie first. +Let $P_0$ be the Maggie leaving the starting line, +$P_1$ be Maggie finishing her acceleration phase, +and $P_2$ be Maggie finishing the race. + +\begin{tabular}{r || r | r | r | r |} + & $P_0$ & \multicolumn{2}{|c|}{$P_1$} & $P_2$ \\ + \hline + \hline + $a$ & \multicolumn{2}{|c|}{?} & \multicolumn{2}{|c|}{$0\U{m/s}^2$} \\ + \hline + $v$ & $0\U{m/s}$ & \multicolumn{3}{|c|}{?} \\ + \hline + $x$ & $0\U{m}$ & \multicolumn{2}{|c|}{?} & $100\U{m}$ \\ + \hline + $t$ & $0\U{s}$ & \multicolumn{2}{|c|}{$2.00\U{s}$} & + $10.2\U{s}$ \\ + \hline +\end{tabular}\\ +Using the 2nd equation from Table 2.2 on page 53 on the first leg: +\begin{equation} + x_1 = 0.5 (v_0 + v_1) t_1 + x_0 = 0.5 v_1 t_1 +\end{equation} +And again on the second leg: +\begin{align} + x_2 &= 0.5 (v_2 + v_1) (t_2 - t_1) + x_1 + = 0.5(v_1 + v_1) (t_2 - t_1) + 0.5 v_1 t_1 + = v_1 (t_2 - t_1 + 0.5 t_1) + = v_1 (t_2 - 0.5 t_1) \\ + v_1 &= \frac{x_2}{t_2 - 0.5 t_1} + = \frac{100\U{m}}{10.2\U{s} - 0.5 \cdot 2.00\U{s}} + = 10.9\U{m/s} +\end{align} +Which is the answer for Maggie in \Part{b}. So +\begin{align} + v_1 &= a_12 t_1 + v_0 \\ + a_12 &= v_1 / t_1 + = \frac{10.9\U{m/s}}{2.00\U{s}} + = 5.43\U{m/s}^2 +\end{align} +Which answers Maggie in \Part{a}. + +Applying the formulas to Judy, +\begin{align} + v_1 &= \frac{x_2}{t_2 - 0.5 t_1} + = \frac{100\U{m}}{10.2\U{s} - 0.5 \cdot 3.00\U{s}} + = 11.5\U{m/s} \\ + a_12 &= v_1 / t_1 + = \frac{11.5\U{m/s}}{3.00\U{s}} + = 3.83\U{m/s}^2 +\end{align} + +\Part{c} +\begin{align} + x_M(t) &= v_{M1} (t - t_{M1}) + x_{M1} \\ + x_J(t) &= v_{J1} (t - t_{J1}) + x_{J1} \\ + \Delta x &= x_M(6\U{s}) - x_J(6\U{s}) + = 10.9\U{m/s} \cdot 4.00\U{s} + + 0.5 \cdot 10.9\U{m/s} \cdot 2.00\U{s} + -11.5\U{m/s} \cdot 3.00\U{s} + - 0.5 \cdot 11.5\U{m/s} \cdot 3.00\U{s} + = 2.62\U{m} +\end{align} +\end{solution} diff --git a/latex/problems/problem03.09.tex b/latex/problems/problem03.09.tex new file mode 100644 index 0000000..5f4d66f --- /dev/null +++ b/latex/problems/problem03.09.tex @@ -0,0 +1,63 @@ +\begin{problem*}{3.9} +The mountain lion can jump to a height of $h = 12.0\U{ft}$ when +leaving the ground at an angle of $\theta = 45.0\dg$. With what +speed, in SI units, does it leave the ground to make the leap? +\end{problem*} % problem 3.9 + +\begin{solution} +First, we'll convert the height into SI units: +\begin{equation} + h = 12.0\U{ft} \p[{ \frac{1\U{m}}{3.28\U{ft}} }] + = 3.659\U{m} +\end{equation} + +Next, arrange the information we know in a table, calling the launch +point $P_0$ and the peak point $P_1$.\\ +\begin{tabular}{r||r|r|} + Point & $P_0$ & $P_1$ \\ + \hline + \hline + $a_x$ & \multicolumn{2}{|c|}{$0\U{m}$} \\ + \hline + $a_y$ & \multicolumn{2}{|c|}{$-9.8\U{m}$} \\ + \hline + $v_x$ & \multicolumn{2}{|c|}{?} \\ + \hline + $v_y$ & ? & $0\U{m/s}$ \\ + \hline + $x$ & $0\U{m}$ & ? \\ + \hline + $y$ & $0\U{m}$ & $3.66\U{m}$ \\ + \hline + $t$ & $0\U{s}$ & ? \\ + \hline +\end{tabular}\\ +Where we know $v_1 = 0\U{m/s}$ because $P_1$ is at the apex of the jump + and $x_0$, $y_0$, and $t_0$ through our choice of coordinate frame. + +We want to pick an equation to tell us something about the initial velocity + (because they told us $\theta$, either $v_{x0}$ or $v_{y0}$ will suffice.). +Looking at our 4 constant acceleration equations (text p. 53), +\begin{align} + v_{xf} &= v_{xi} + a_x t \\ + x_f &= x_i + \frac{1}{2}(v_{xf} + v_{xi}) t \\ + x_f &= x_i + v_{xi} t + \frac{1}{2} a_x t^2 \label{eqn.t_sqr}\\ + v_{xf}^2 &= v_{xi}^2 + 2 a_x (x_f - x_i) \label{eqn.v_sqr} +\end{align} +We see that eqn.~\ref{eqn.v_sqr} applied to the $y$ direction is +perfect, because it has no information in it that we don't already +know except $v_{y0}$. +\begin{align} + v_{y1}^2 &= (0\U{m/s})^2 = v_{y0}^2 + 2 a_y (y_1 - y_0) \\ + v_{y0}^2 &= -2 a_y y_1 \\ + v_{y0} &= \sqrt{ -2 a_y y_1 } +\end{align} +And we use the angle to solve for the magnitude of the inital velocity: +\begin{align} + v_{y0} &= v_0 \sin \theta \\ + v_0 &= \frac{v_{y0}}{\sin \theta} + = \frac{\sqrt{ -2 a_y y_1 }}{\sin \theta} + = \frac{\sqrt{-2 \cdot (-9.8\U{m/s}^2) \cdot 3.66\U{m}}}{\sin 45^o} + = \ans{12.0\U{m/s}} +\end{align} +\end{solution} diff --git a/latex/problems/problem03.19.tex b/latex/problems/problem03.19.tex new file mode 100644 index 0000000..297dffa --- /dev/null +++ b/latex/problems/problem03.19.tex @@ -0,0 +1,78 @@ +\begin{problem*}{3.19} +A soccer player kicks a rock horizontally off a $h = 40.0\U{m}$ high +cliff into a pool of water. If the player hears the sound of the +splash $\Delta t = 3.00\U{s}$ later, what was the initial speed given +to the rock? Assume that the speed of sound in air is $v_s = +343\U{m/s}$. +\end{problem*} % problem 3.19 + +\begin{solution} +Let us call the point when the player kicks the ball $P_0$, the point +where the ball lands $P_1$, and the point where the sound hits the +players ear $P_2$. + +Arranging the information we know about the ball in a table,\\ +\begin{tabular}{r||r|r|} + Point & $P_0$ & $P_1$ \\ + \hline + \hline + $a_x$ & \multicolumn{2}{|c|}{$0\U{m}$} \\ + \hline + $a_y$ & \multicolumn{2}{|c|}{$-9.8\U{m}$} \\ + \hline + $v_x$ & \multicolumn{2}{|c|}{?} \\ + \hline + $v_y$ & $0\U{m/s}$ & ? \\ + \hline + $x$ & $0\U{m}$ & ? \\ + \hline + $y$ & $40.0\U{m}$ & $0\U{m}$ \\ + \hline + $t$ & $0\U{s}$ & ? \\ + \hline +\end{tabular}\\ +Where we know $v_{y0} = 0\U{m/s}$ because the ball is kicked horizontally + and $x_0$, $y_0$, and $t_0$ through our choice of coordinate frame. + +We want to pick an equation to tell us something about the time, + (because they told us $t_2$.). +Looking at our 4 constant acceleration equations +We see that eqn. \ref{eqn.t_sqr} applied to the $y$ direction is perfect, + because it has no information in it that we don't already know except $t_1$. +\begin{align} + y_1 &= 0\U{m} = y_0 + v_{y0} t_1 + \frac{1}{2} a_y t_1^2 \\ + \frac{1}{2} a_y t_1^2 &= -y_0 \\ + t_1^2 &= \frac{ -2 y_0}{a_y} \\ + t_1 &= \sqrt{ \frac{ -2 y_0}{a_y} } + = \sqrt{ \frac{ -2 \cdot 40.0\U{m}}{-9.8\U{m/s}^2} } + = 2.86\U{s} +\end{align} +So using eqn. \ref{eqn.t_sqr} applied to the $x$ direction, we have +\begin{equation} + x_1 = x_0 + v_{x0} t_1 + \frac{1}{2} a_x t_1^2 + = v_{x0} t_1 +\end{equation} + +Now that we know $x_1$ and $y_1$, we can find the distance $\Delta x_s$ +that the sound takes returning to $P_2$. Because it moves in a straight line +(more or less), we have +\begin{equation} + \Delta x_s = \sqrt{ (x_1-x_0)^2 + (y_1-y_0)^2 } + = \sqrt{ v_{x0}^2 t_1^2 + y_0^2 } +\end{equation} +So the time $(t_2 - t_1)$ taken for the sound to return is +\begin{equation} + \Delta x_s = v_s (t_2 - t_1) \\ +\end{equation} +And they give us $t_2$ in the problem, so we just solve this for $v_{x0}$. +\begin{align} + \Delta x_s &= \sqrt{ v_{x0}^2 t_1^2 + y_0^2 } + = v_s (t_2 - t_1) \\ + v_{x0}^2 t_1^2 + y_0^2 &= \p[{ v_s (t_2 - t_1) }]^2 \\ + v_{x0}^2 &= \p[{ \p[{ v_s (t_2 - t_1) }]^2 - y_0^2 }] / t_1^2 \\ + v_{x0} &= \sqrt{ \p[{v_s (t_2 - t_1)}]^2 - y_0^2} / t_1 + = \sqrt{ \p[{343\U{m/s} (3.00\U{s} - 2.86\U{s}}]^2 + - (40.0\U{m})^2} / 2.86\U{s} + = \ans{9.91\U{m/s}} +\end{align} +\end{solution} diff --git a/latex/problems/problem03.24.tex b/latex/problems/problem03.24.tex new file mode 100644 index 0000000..ce7b983 --- /dev/null +++ b/latex/problems/problem03.24.tex @@ -0,0 +1,52 @@ +\begin{problem*}{3.24} +Suppose a copper sleeve of inner radius $r_1=2.10\U{cm}$ and outer +radius $r_2=2.20\U{cm}$ is to be cast. To eliminate bubbles and give +high structural integrity, the centripetal acceleration of each bit of +metal should be at least $a_{c\U{min}}=100g$. What rate of rotation +is required? State the answer in revolutions per minute (rpm). +\end{problem*} % problem 3.24 + +\begin{solution} +For circular motion +\begin{align} + a_c &= v^2/r \\ + v &= \sqrt{a_c r} +\end{align} +And if the circular motion is uniform, the time $T$ taken to complete +one revolution is given by +\begin{align} + \Delta x &= v T \\ + T &= \Delta x / v = \frac{2 \pi r}{v} +\end{align} +So the frequency $f = 1/T$ of rotation is +\begin{equation} + f = 1/T = \frac{v}{2 \pi r} + = \frac{\sqrt{a_c r}}{2 \pi r} + = \frac{1}{2 \pi} \sqrt{ \frac{a_c}{r} } +\end{equation} + +Because for a given frequency $a_c \propto r$, + we must use the inner radius $r_1$ to compute the $f$ required + to create an acceleration $a_c \geq 100g$. +\begin{equation} + f = \frac{1}{2 \pi} \sqrt{ \frac{100 * 9.8\U{m/s}^2}{0.0210\U{m}} } + \cdot \frac{60\U{s}}{1\U{min}} + = \ans{2060\U{rpm}} +\end{equation} + +To reassure ourselves that this frequency creates enough centripetal +acceleration $a_{c2}$ at $r_2$, we can compute $a_{c2}$. +\begin{align} + f &= \frac{1}{2 \pi} \sqrt{ \frac{a_c}{r} } \\ + 2 \pi f &= \sqrt{ \frac{a_c}{r} } +\end{align} +The left hand side is constant, so applying this equation to both $r_1$ +and $r_2$ we get, +\begin{align} + \sqrt{ \frac{a_{c\U{min}}}{r_1} } &= \sqrt{ \frac{a_{c2}}{r_2} } \\ + a_{c2} &= a_{c\U{min}} \frac{r_2}{r_1} + = a_{c\U{min}} \frac{2.20\U{cm}}{2.10\U{cm}} + = 1.05 \cdot a_{c\U{min}} +\end{align} +So the acceleration is indeed $\geq 100g$ throughout the cylinder. +\end{solution} diff --git a/latex/problems/problem03.43.tex b/latex/problems/problem03.43.tex new file mode 100644 index 0000000..28d7ac2 --- /dev/null +++ b/latex/problems/problem03.43.tex @@ -0,0 +1,55 @@ +\begin{problem*}{3.43} +A ball on the end of a string is whirled around in a horizontal cirlce +of radius $r=0.300\U{m}$. The plane of the circle is $h=1.20\U{m}$ +above the ground. The string breaks, and the ball lands $L=2.00\U{m}$ +(horizontally) away from the point on the ground directly beneath the +ball's location when the string breaks. Find the radial acceleration +of the ball during its circular motion. +\end{problem*} % problem 3.43 + +\begin{solution} +The airborne ball portion of the problem is a 2-D projectile motion +problem. Calling the point where the string breaks $P_0$ and the +point where the ball lands $P_1$, we have \\ +\begin{tabular}{r||r|r|} + Point & $P_0$ & $P_1$ \\ + \hline + \hline + $a_x$ & \multicolumn{2}{|c|}{$0\U{m}$} \\ + \hline + $a_y$ & \multicolumn{2}{|c|}{$-9.8\U{m}$} \\ + \hline + $v_x$ & \multicolumn{2}{|c|}{?} \\ + \hline + $v_y$ & $0\U{m/s}$ & ? \\ + \hline + $x$ & $0\U{m}$ & $2.00\U{m}$ \\ + \hline + $y$ & $1.20\U{m}$ & $0\U{m}$ \\ + \hline + $t$ & $0\U{s}$ & ? \\ + \hline +\end{tabular}\\ +We use eqn. \ref{eqn.t_sqr} applied to the $y$ direction to find $t_1$ +\begin{align} + y_1 &= 0\U{m} = y_0 + v_{y0} t_1 + \frac{1}{2} a_y t_1^2 \\ + \frac{1}{2} a_y t_1^2 &= -y_0 \\ + t_1 &= \sqrt{ \frac{-2 y_0}{a_y} } +\end{align} +As we saw in problem 19. + +Now applying eqn. \ref{eqn.t_sqr} to the $x$ direction to find $v_{x}$ +\begin{align} + x_1 &= x_0 + v_{x} t_1 + \frac{1}{2} a_x t_1^2 \\ + v{x} &= x_1/t_1 +\end{align} + +And plugging these into our circular motion equation +\begin{equation} + a_c = v^2 / r = \frac{(x_1/t_1)^2}{r} + = \frac{x_1^2 a_y}{-2 y_0 r} + = \frac{(2.00\U{m})^2 \cdot (-9.8\U{m/s}^2)} + {-2 \cdot 1.20\U{m} \cdot 0.300\U{m}} + = \ans{54.4\U{m/s}^2} +\end{equation} +\end{solution} diff --git a/latex/problems/problem04.08.tex b/latex/problems/problem04.08.tex new file mode 100644 index 0000000..f98960b --- /dev/null +++ b/latex/problems/problem04.08.tex @@ -0,0 +1,67 @@ +\begin{problem*}{4.8} +Three forces, given by + $\vect{F}_1 = (-2.00\vect{i} + 2.00\vect{j})\U{N}$, + $\vect{F}_2 = ( 5.00\vect{i} - 3.00\vect{j})\U{N}$, and + $\vect{F}_3 = -45.0\vect{i}\U{N}$, + act on an object to give it an acceleration of magnitude + $a = 3.75\U{m/s}^2$ + \Part{a} What is the direction of the acceleration? + \Part{b} What is the mass of the object? + \Part{c} If the object is initially at rest, what is its speed $v$ + after $t = 10.0\U{s}$? + \Part{d} What are the velocity components of the object after + $t = 10.0\U{s}$? +\end{problem*} % problem 4.8 + +\begin{solution} +\Part{a} +Summing the forces we have +\begin{align} + \sum F_x &= F_1x + F_2x + F_3x = (-2.00 + 5.00 - 45.0)\U{N} = -42.0\U{N} \\ + \sum F_y &= F_1y + F_2y + F_3y = (+2.00 - 3.00 + 0)\U{N} = -1.00\U{N} +\end{align} +We know from Newtons second law that +\begin{equation} + \sum \vect{F} = m \vect{a} +\end{equation} +So the acceleration $\vect{a}$ will be in the same direction as the +force $\vect{F}$. +The direction $\theta$ of the force is given by +\begin{equation} + \theta = \arctan \left( \frac{F_y}{F_x} \right) + = \arctan \left( \frac{-1}{-42} \right) + = (1.36 + 180)^o = \ans{181.36^o} +\end{equation} +Measured counter-clockwise from the $\vect{x}$ axis + (where we have added $180^o$ because $F_x < 0$ so we have a backside + $\arctan$). + +\Part{b} +From Newton's second law +\begin{align} + \sum \vect{F} &= m \vect{a} \\ + \left|\sum \vect{F}\right| &= m \left|\vect{a}\right| \\ + m &= \frac{\left|\sum \vect{F}\right|}{a} + = \frac{ \sqrt{(-41.0)^2 + (-1.00)^2}\U{N}}{3.75\U{m/s}^2} + = \ans{11.2\U{kg}} +\end{align} + +\Part{c} +This section is constant acceleration review. +\begin{equation} + v = a t + v_0 + = 3.75\U{m/s}^2 \cdot 10.0\U{s} = \ans{37.5\U{m/s}} +\end{equation} + +\Part{d} +Using our velocity $v = |\vect{v}|$ from \Part{c} and our angle + $\theta$ from \Part{a} (we know that $\vect{v}$ is in the same +direction as $\vect{a}$ and $\vect{F}$) we have +\begin{align} + v_x &= v \cos \theta = 37.5\U{m/s} \cdot \cos 181.36^o + = -37.49\U{m/s} \\ + v_y &= v \sin \theta = 37.5\U{m/s} \cdot \sin 181.36^o + = -0.893\U{m/s} \\ + \vect{v} &= \ans{(-37.49\vect{i} - 0.893\vect{j})\U{m/s}} +\end{align} +\end{solution} diff --git a/latex/problems/problem04.22.tex b/latex/problems/problem04.22.tex new file mode 100644 index 0000000..18b480d --- /dev/null +++ b/latex/problems/problem04.22.tex @@ -0,0 +1,82 @@ +\begin{problem*}{4.22} +The systems shown in Fig. P4.22 are in equilibrium. +If the spring scales are calibrated in newtons, what do they read? +(Ignore the masses of the pulleys and strings, and assume that the +incline is frictionless.) +\end{problem*} % problem 4.22 + +\begin{solution} +Remember that what a spring scale does is measure the tension pulling +on {\it one} of it's sides when in equilibrium. To see this, imagine +a spring scale in it's normal use, hanging from the ceiling with a +mass $m$ suspended from it. $m$ is in eqilibrium, so the tension +$T_1$ in the string connecting $m$ to the scale must be $T_1=mg$. The +(massless) scale is also in equilibrium, so the tension $T_2$ in the +string connecting the scale to the ceiling must be $T_2=T_1=mg$. The +scale has $T_1=mg$ pulling down and $T_2=mg$ pulling up, and gives a +reading of $mg$, the weight of the suspended mass. + +The text tries to remind you of this somewhat tricky concept on page +108 in quick quiz 4.7. + +\Part{a} +Starting from the left, the ball of mass $m=5.00\U{kg}$ has two forces +acting on it: gravity $F_g=mg$ and tension $T_1$. Summing forces in +the upwards direction we have +\begin{align} + \sum F &= T_1 - F_g = T_1 - mg \label{eqn.sum_forces}\\ + &= ma = 0 \label{eqn.newtons_2nd} \\ + T_1 &= mg \label{eqn.T1} +\end{align} +Where \ref{eqn.sum_forces} comes from our free-body diagram of the +forces on the ball, \ref{eqn.newtons_2nd} come from Newton's second law +and the fact that the particle is in equilibrium, and \ref{eqn.T1} comes +from combining \ref{eqn.sum_forces} and \ref{eqn.newtons_2nd}. + +Moving on to the scale, we see that the scale has two forces on it: + tension from the left ball $T_1$ and tension from the right ball $T_2$. +Summing the forces in the rightwards direction we have +\begin{align} + \sum F &= T_2 - T_1 \\ + &= m_s a = 0 \\ + T_2 &= T_1 = mg \\ +\end{align} +Following exactly the same reasoning we applied to the left ball. + +The scale has $mg$ pulling on both sides, so it will read +$mg=5.00\U{kg} \cdot 9.8\U{m/s}^2=\ans{49.0\U{N}}$ (see the note above). +(Because of the slight sneaky-ness, I didn't take off if you gave an +answer of $2mg$.) + +\Part{b} +Following the same reasoning we applied to the left ball (eqns +\ref{eqn.sum_forces} to \ref{eqn.T1}) in \Part{a}, we have +$T_1=T_2=mg$ for both hanging masses. + +So the pulley has three forces on it: the tensions of the cord +connecting the two masses $T_1$ and $T_2$, and the tension cord +connecting it to the scale $T_3$. Summing the forces in the upwards +direction we have +\begin{align} + \sum F &= T_3 - T_2 - T_1 \\ + &= m_p a = 0 \\ + T_3 &= T_2 + T_1 = 2mg +\end{align} +And the scale is in equilibrium, so as in \Part{a} it has $T_3$ pulling +on both sides, and it will read + $2mg = 2 \cdot 5.00\U{kg} \cdot 9.8\U{m/s}^2 = \ans{98.0\U{N}}$. + +\Part{c} +The block has 3 forces acting on it: gravity $F_g = mg$, tension $T$, +and a normal force $F_N$. Summing forces in the tension direction we +have +\begin{align} + \sum F &= T - F_g \cdot \sin \theta= T_1 - mg \\ + &= ma = 0 \\ + T_1 &= mg \cdot \sin \theta +\end{align} +And the scale is in equilibrium, so it has $T_1$ pulling on both sides, +and it will read + $mg \cdot \sin \theta = 5.00\U{kg} \cdot 9.8\U{m/s}^2 \cdot \sin 30^o + = \ans{24.5\U{N}}$. +\end{solution} diff --git a/latex/problems/problem04.24.tex b/latex/problems/problem04.24.tex new file mode 100644 index 0000000..9337bce --- /dev/null +++ b/latex/problems/problem04.24.tex @@ -0,0 +1,70 @@ +\begin{problem*}{4.24} +Fig. P4.24 shows loads hanging from the ceiling of an elecator that is +moving at a constant velocity. Find the tension in each of the three +strands of cord supporting each load. +\end{problem*} % problem 4.24 + +\begin{solution} +First, we need to understand the effect of the elevator. +It is moving at a constant {\it velocity} so we know that +the acceleration $\vect{a}$ of all the elements must be $0$. +So the elevator's constant motion has no effect on the tensions. + +\Part{a} +Let $m = 5.00\U{kg}$ be the mass of the ball, + $\theta_1 = 40.0^o$ be the angle between $\vect{T}_1$ and the horizontal, + and $theta_2 = 50.0^o$ be the angle between $\vect{T}_2$ and the horizontal. +Following identical reasoning to Problem 22 \Part{a}, we know +that the tension + $T_3 = mg = 5.00\U{kg} \cdot 9.8\U{m/s}^2 = \ans{49\U{N}}$. + +Now looking at the knot where the three cords come together. +There are three forces acting on the knot: $T_1$, $T_2$, and $T_3$. +Letting the upwards direction be $+\vect{x}$ + and the rightwards direction be $+\vect{y}$ we can break our tensions +into components +\begin{align} + T_{1x} &= -T_1 \cos \theta_1 \\ + T_{1y} &= T_1 \sin \theta_1 \\ + T_{2x} &= T_2 \cos \theta_2 \\ + T_{2y} &= T_2 \sin \theta_2 \\ + T_{3x} &= 0\U{N} \\ + T_{3y} &= -mg +\end{align} +Now summing the forces on the knot we have +\begin{align} + \sum F_x &= T_{3x} + T_{2x} + T_{1x} + = 0 + T_2 \cos \theta_2 - T_1 \cos \theta_1 \\ + &= m_k a_{kx} = 0 \\ + T_2 &= T_1 \frac{\cos \theta_1}{\cos \theta_2} \\ + \sum F_y &= T_{3y} + T_{2y} + T_{1y} + = -mg + T_2 \sin \theta_2 + T_1 \sin \theta_1 \\ + T_2 &= \frac{mg - T_1 \sin \theta_1}{\sin \theta_2} + = T_1 \frac{\cos \theta_1}{\cos \theta_2} \\ + \frac{mg}{\cos \theta_1} - T_1 \frac{sin \theta_1}{\cos \theta_1} + &= T_1 \frac{\sin \theta_2}{\cos \theta_2} \\ + T_1 (\tan \theta_1 + \tan \theta_2) &= \frac{mg}{\cos \theta_1} \\ + T_1 &= \frac{mg}{\cos \theta_1 (\tan \theta_1 + \tan\theta_2)} + = \frac{49\U{N}}{\cos 40^o (\tan 40^o + \tan 50^o)} + = \ans{31.5\U{N}} \\ + T_2 &= T_1 \frac{\cos \theta_1}{\cos \theta_2} + = \frac{mg}{\cos \theta_2 (\tan \theta_1 + \tan\theta_2)} + = \frac{49\U{N}}{\cos 50^o (\tan 40^o + \tan 50^o)} + = \ans{37.5\U{N}} +\end{align} + +\Part{b} +The only changes from \Part{a} are + $m = 10\U{kg}$, $\theta_1 = 60.0^o$, and $\theta_2 = 0^o$. +Plugging the new values into our symbolic equation from \Part{a}: +\begin{align} + T_3 &= mg = 10.0\U{kg} \cdot 9.8\U{m/s}^2 = \ans{98\U{N}} \\ + T_1 &= \frac{mg}{\cos \theta_1 (\tan \theta_1 + \tan\theta_2)} + = \frac{98\U{N}}{\cos 60^o (\tan 60^o + \tan 0^o)} + = \ans{113\U{N}} \\ + T_2 &= T_1 \frac{\cos \theta_1}{\cos \theta_2} + = \frac{mg}{\cos \theta_2 (\tan \theta_1 + \tan\theta_2)} + = \frac{98\U{N}}{\cos 0^o (\tan 60^o + \tan 0^o)} + = \ans{56.6\U{N}} +\end{align} +\end{solution} diff --git a/latex/problems/problem04.51.tex b/latex/problems/problem04.51.tex new file mode 100644 index 0000000..bca87d3 --- /dev/null +++ b/latex/problems/problem04.51.tex @@ -0,0 +1,62 @@ +\begin{problem*}{4.51} +If you jump from a desktop and land stiff-legged on a concrete floor, +you run a significant rist that you will break a leg. To see how that +happens, consider the average force stopping your body when you drop +from rest from a height of $h = 1.00\U{m}$ and stop in a much shorter +distance $d$. Your leg is likely to break at the point where the +cross-sectional area of the tibia is smallest. This point is just +above the anke, where the cross sectional area of one bone is about $A += 1.60\U{cm}^2$. A bone will fracture when the compressive stress on +it exceeds about $\sigma_b = 1.60\E{8}\U{N/m}^2$. If you land on both +legs, the maximum force $F_{max}$ that your ankles can safely exert on +the rest of your body is then about +\begin{equation} + F_{max} = 2 F_b = 2 \sigma_b A = 5.12\E{4}\U{N} +\end{equation} +Calculate the minimum stopping distance $d$ that will not result in a + broken leg if your mass is $m = 60.0\U{kg}$. +\end{problem*} % problem 4.51 + +\begin{solution} +The problem breaks down into two constant-acceleration problems. +Call the top of the desk dropping-off-point $P_0$, the point of maximum +velocity when you are just starting to contact the floor $P_1$, and the +point where your shoe soles have compressed a distance $d$ and brought +you back to rest $P_2$. + +First consider the constant acceleration portion from $P_0$ to $P_1$. +Your final velocity $v_1$ is given by +\begin{equation} + v_1^2 = v_0^2 + 2 a_{01} \Delta y_{01} + = 2 a_{01} \Delta y_{01} +\end{equation} +Now applying the same equation to + the second constant acceleration portion from $P_1$ to $P_2$. +\begin{align} + v_2^2 &= v_1^2 + 2 a_{12} \Delta y_{12} = 0 \\ + v_1^2 &= -2 a_{12} \Delta y_{12} = 2 a_{01} \Delta y_{01} \\ + d &= \Delta y_{12} = -\frac{a_{01}}{a_{12}} \Delta y_{01} +\end{align} +The acceleration $a_{12}$ is given by Newton's second law +(picking down as the $+\vect{x}$ direction) +\begin{align} + \sum F_x &= m a_{12x} \\ + a_{12x} &= (\sum F_x)/m + = \frac{mg - F_{max}}{m} + = g - \frac{F_{max}}{m} +\end{align} +(I forgot to include $mg$ in the sum of the forces when I was doing +the problem, so I didn't take off points if you forgot it as well.) +So +\begin{align} + d &= -\frac{a_{01}}{a_{12}} \Delta y_{01} + = -\frac{g}{g - \frac{F_{max}}{m}} h + = -\frac{1}{1 - \frac{F_{max}}{mg}} h \\ + &= -\frac{1}{1 - \frac{5.12\E{4}}{60 \cdot 9.8}} \cdot 1.00\U{m} + = \ans{1.16\U{cm}} +\end{align} + +(Ignoring gravity in your sum of forces, you would have gotten + $d = \frac{mg}{F_{max}} h = 1.15\U{cm}$. +The correction is very small because $F_{max} \gg mg$.) +\end{solution} diff --git a/latex/problems/problem05.16.tex b/latex/problems/problem05.16.tex new file mode 100644 index 0000000..739ce43 --- /dev/null +++ b/latex/problems/problem05.16.tex @@ -0,0 +1,24 @@ +\begin{problem*}{5.16} +In the Bohr model of the hydrogen atom, the speed of the electron is +approximately $v = 2.20\E{6} \U{m/s}$. Find \Part{a} the force acting +on the electron as it revolves in a circular orbit of radius $r = +0.530\E{-10} \U{m}$ and \Part{b} the centripetal acceleration of the +electron. +\end{problem*} % problem 5.16 + +\begin{solution} +Doing \Part{b} first, +\begin{equation} + a_c = v^2 / r \\ + = \frac{(2.20\E{6}\U{m/s})^2}{0.530\E{-10}\U{m}} \\ + = \ans{9.13\E{22}\U{m/s}^2} +\end{equation} + +And going back to \Part{a}, (where the mass of an electron $m_e = +9.109\E{-31} \U{kg}$ came from the inside front cover of the text.) +\begin{equation} + F_c = m_e a_c + = 9.109\E{-31}\U{kg} \cdot 9.13\E{22}\U{m/s}^2 + = \ans{8.32\E{-8}\U{N}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem05.18.tex b/latex/problems/problem05.18.tex new file mode 100644 index 0000000..b1e399c --- /dev/null +++ b/latex/problems/problem05.18.tex @@ -0,0 +1,33 @@ +\begin{problem*}{5.18} +Whenever two {\em Apollo} astronauts were on the surface of the Moon, +a third astronaut orbited the Moon. Assume the orbit to be circular +and $r_1=100\U{km}$ above the surface of the Moon. At this altitude, +the free-fall acceleration is $g=1.52\U{m/s}^2$. The radius of the +Moon is $r_0=1.70\E{6}\U{m}$. Determine + \Part{a} the astronaut's orbital speed $v$ and + \Part{b} the period of the orbit. +\end{problem*} % problem 5.18 + +\begin{solution} +\Part{a} +Using the basic formula for circular motion +\begin{align} + a_c &= \frac{v^2}{r} \\ + v &= \sqrt{a_c r} + =\sqrt{g (r_1 + r_0)} + =\sqrt{1.52\U{m/s}^2 \cdot (1.00\E{5} + 1.70\E{6})\U{m}} + =\ans{1.65\U{km/s}} +\end{align} + +\Part{b} +The astronaut travels the circumference at a constant speed so +\begin{align} + \Delta_x &= v \Delta_t \\ + T &= \frac{2 \pi r}{v} + =\frac{2 \pi r}{\sqrt{a_c r}} + =2 \pi \sqrt{\frac{r}{a_c}} + =2 \pi \sqrt{\frac{(1.00\E{5} + 1.70\E{6})\U{m}}{1.52\U{m/s}^2}} + =6.84\U{ks} + =\ans{1.90\U{hrs}} +\end{align} +\end{solution} diff --git a/latex/problems/problem05.23.tex b/latex/problems/problem05.23.tex new file mode 100644 index 0000000..6adbf76 --- /dev/null +++ b/latex/problems/problem05.23.tex @@ -0,0 +1,18 @@ +\begin{problem*}{5.23} +A pail of water is rotated in a vertical circle of radius +$r=1.00\U{m}$. What is the minimum speed of the pail, upside down at +the top of the circle, if no water is to spill out? +\end{problem*} % problem 5.23 + +At the critical low speed, all of the centerward acceleration comes +from gravity (no tension/normal force, like Chapter 5, Problem 47 from +recitation). So +\begin{align} + a_c &= \frac{v^2}{r} \\ + v &= \sqrt{a_c r} + = \sqrt{g r} + = \sqrt{9.8\U{m/s}^2 \cdot 1.00\U{m}} + = \ans{3.13U{m/s}} +\end{align} +Just like in Problem 18. +\end{solution} diff --git a/latex/problems/problem05.24.tex b/latex/problems/problem05.24.tex new file mode 100644 index 0000000..b162cee --- /dev/null +++ b/latex/problems/problem05.24.tex @@ -0,0 +1,55 @@ +\begin{problem*}{5.24} +A roller coaster has vertical loops shaped like tear drops +(Fig.~P5.24). The cars ride on the inside of the loop at the top, and +the speeds are high enough to ensure that the cars remain on the +track. The biggest loop is $h = 40.0\U{m}$ high, with a maximum speed +$v_b = 31.0\U{m/s}$ at the bottom. Suppose the speed at the top is +$v_t = 13.0\U{m/s}$ and the corresponding centripetal acceleration is +$a_{ct} = 2g$. \Part{a} What is the radius $r_t$ of the arc of the +teardrop at the top? \Part{b} If the total mass of a car plus the +riders is $M$, what force $F_N$ does the rail exert on the car at the +top? \Part{c} Suppose the roller coaster had a circular loop of +radius $r = 20\U{m}$. If the cars have the same speed $v_t$ at the +top, what is the centripetal acceleration $a_{cc}$ at the top? +Comment on the normal force at the top in this situation. +\end{problem*} % problem 5.24 + +\begin{solution} +\Part{a} +\begin{align} + a_{ct} &= v_t^2 / r_t \\ + r_t &= v_t^2 / a_{ct} + = (13.0\U{m/s})^2 / (2 \cdot 9.8\U{m/s}^2) + = \ans{8.62\U{m}} +\end{align} + +\Part{b} +The central force $F_t = 2 M g$. +This force is a combination of the force of gravity $F_g = Mg$ + and the normal force $F_N$ from the rail: +\begin{align} + F_t &= \sum F_{central} + = F_g + F_N \\ + F_N &= F_t - F_g + = 2Mg - Mg + = Mg + = \ans{9.8\U{m/s}^2 \cdot M} +\end{align} + +\Part{c} +\begin{equation} + a_{cc} = v_t^2 / r + = (13.0\U{m/s})^2 / 20\U{m} + = \ans{8.45\U{m/s}^2} +\end{equation} +So the the new normal force $F_{Nc}$ is going to be: +\begin{equation} + F_{Nc} = F_{cc} - F_g + = (8.45 - 9.8)\U{m/s}^2 \cdot M + = \ans{-1.35\U{m/s}^2 \cdot M} +\end{equation} +Where the - sign indicates the normal force is the track pulling the +car \emph{away} from the center. The teardrop shape allows the loop to +be $40\U{m}$ high while always keeping the track's normal force in the +center-ward direction. +\end{solution} diff --git a/latex/problems/problem05.32.tex b/latex/problems/problem05.32.tex new file mode 100644 index 0000000..0ff3048 --- /dev/null +++ b/latex/problems/problem05.32.tex @@ -0,0 +1,31 @@ +\begin{problem*}{5.32} +Find the order of magnitude of the gravitational force that you exert +on another person $r = 2\U{m}$ away. In your solution, state the +quantities you measure or estimate and their values. +\end{problem*} % problem 5.32 + +\begin{solution} +We'll be using Newton's law for gravitation (text p. 144): +\begin{equation} + F_g = G \frac{mM}{r^2} +\end{equation} +with $G = 6.673\E{-11}\U{Nm$^2$/kg$^2$}$. + +We need to estimate $m$ and $M$. +Both bodies are people, so I'll use my weight for both: +\begin{equation} + m \approx M + \approx 165\U{lbs} \cdot \left[ \frac{1\U{kg}}{ \sim 2 \U{lbs}} \right] + \approx 82.5\U{kg} +\end{equation} +So +\begin{equation} + F_g \approx G \frac{m^2}{r^2} + = G (m/r)^2 + \approx 6.673\E{-11}\U{Nm$^2$/kg$^2$} \left(82.5\U{kg}/2\U{m}\right)^2 + = 1.14\E{-7}\U{N} + \approx \ans{1\E{-7}\U{N}} +\end{equation} +Where I reduced the answer to one sig. fig. because of my rough mass +approximation. +\end{solution} diff --git a/latex/problems/problem05.34.tex b/latex/problems/problem05.34.tex new file mode 100644 index 0000000..b3e5a00 --- /dev/null +++ b/latex/problems/problem05.34.tex @@ -0,0 +1,24 @@ +\begin{problem*}{5.34} +In a thundercloud, there may be electric charges of $q_t = +40.0\U{C}$ +near the top of the cloud and $q_b = -40.0\U{C}$ near the bottom of +the cloud. These charges are separated by $r = 2.00\U{km}$. What is +the electric force on the top charge? +\end{problem*} % problem 5.34 + +\begin{solution} +We'll be using Coulomb's law for the electro-magnetic force (text p. 144): +\begin{equation} + F_e = k_e \frac{q_1 q_2}{r^2} +\end{equation} +with $k_e = 8.99\E{9}\U{Nm$^2$/C$^2$}$. + +So +\begin{equation} + F_e = 8.99\E{9}\U{Nm$^2$/C$^2$} + \frac{-40.0\U{C} \cdot 40.0\U{C}}{(2000\U{m})^2} + = -8.99\E{9}\U{Nm$^2$/C$^2$} + \left( \frac{40\U{C}}{2000\U{m}} \right)^2 + = \ans{-3.596\E{6}\U{N}} +\end{equation} +where the $-$ sign indicates an attractive force. +\begin{solution} diff --git a/latex/problems/problem05.45.tex b/latex/problems/problem05.45.tex new file mode 100644 index 0000000..624a054 --- /dev/null +++ b/latex/problems/problem05.45.tex @@ -0,0 +1,77 @@ +\begin{problem*}{5.45} +A car rounds a banked curve as in Fig. 5.13. The radius of curvature +of the road is $R$, the banking angle is $/theta$, and the coefficient +of static friction is $\mu_s$. +\Part{a} Determine the range of speeds the car can have without +slipping up or down the road. +\Part{b} Find the minimum value of $\mu_s$ such that the minimum +speed is zero. +\Part{c} What is the range of speeds possible if $R = 100\U{m}$, +$\theta = 10.0\dg$, and $\mu_s = 0.100$ (slippery conditions). +\end{problem*} % problem 5.45 + +\begin{solution} +Looking at Fig. 5.13 (text page 137) and adding friction, we see that +the forces on the car are friction $\vect{F}_f$, gravity $\vect{F}_g$, +and a normal force $\vect{F}_N$. Let the vertical direction be +\jhat\ and the centerward direction to be \ihat, and the direction +centerward-down parallel to the surface of the road by \khat. Let us +assume at first that $\vect{F}_N$ is in the $-\khat$ direction and at +its maximum possible value of $F_f = \mu_s F_N$. +\begin{align} + \sum F_\jhat &= F_N\cos\theta - mg + F_f\sin\theta= 0 \\ + F_N (\cos\theta + \mu_s\sin\theta) &= mg \\ + F_N &= \frac{mg}{\cos\theta + \mu_s\sin\theta} \\ + \sum F_\ihat &= F_N\sin\theta - F_f\cos\theta + = F_N (\sin\theta - \mu_s\cos\theta) \\ + &= \frac{mg}{\cos\theta + \mu_s\sin\theta} (\sin\theta - \mu_s\cos\theta) \\ + &= mg \frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta} + = m\frac{v^2}{R} \label{eqn.45.Fi}\\ + v &= \sqrt{ Rg\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta} } \label{eqn.45.v} +\end{align} + +\Part{a} +The work above shows that the minimum speed a car can have while going +around the turn is given by eqn \ref{eqn.45.v}, because that is the +case when friction is maximized in the $-\khat$ direction. The +maximum speed that the car can have can be found by simply reversing +the sign of the frictional force above (so that $\vect{F}_f$ points in +the $+\khat$ direction), which we achieve by replacing any $\mu_s$s in +eqn \ref{eqn.45.v} with $(-\mu_s)$. For any speeds between these +$F_f$ will be less than its maximum value of $\mu_s F_N$, and the car +will still not slip. So +\begin{equation} + \ans{ \sqrt{ Rg\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta} } + \le v \le + \sqrt{ Rg\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta} } } +\end{equation} + +\Part{b} +If the speed is 0, then $\vect{F}_f$ will be in the $-\khat$ direction +(opposing the $+\khat$ portion of $\vect{F}_g$). Summing the forces +in the \khat direction we have +\begin{align} + \sum F_\khat &= F_g\sin\theta - F_N + = mg(\sin\theta - \mu_s\cos\theta) + = 0 \\ + \mu_s &= \ans{\tan\theta} +\end{align} +Or we could go use eqn \ref{eqn.45.Fi}, our sum of forces in the +\ihat\ direction. +\begin{equation} + \sum F_\ihat = mg \frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta} + = m\frac{v^2}{R} = 0 +\end{equation} +And set the numerator to $0$, which gives the same formula for $\mu_s$. + +\Part{c} +Plugging into our ans for \Part{a} we have +\begin{align} + \sqrt{ 100\U{m} \cdot 9.8\U{m/s}^2 \frac{\tan 10.0\dg - 0.100} + {1 + 0.100 \cdot \tan 10.0\dg} } + &\le v \le + \sqrt{ 100\U{m} \cdot 9.8\U{m/s}^2 \frac{\tan 10.0\dg + 0.100} + {1 - 0.100 \cdot \tan 10.0\dg} } \\ + \ans{ 8.57\U{m/s\ } } & \ans{ \le v \le 16.6\U{m/s} } +\end{align} +\end{solution} diff --git a/latex/problems/problem05.47.tex b/latex/problems/problem05.47.tex new file mode 100644 index 0000000..c8cc0d7 --- /dev/null +++ b/latex/problems/problem05.47.tex @@ -0,0 +1,41 @@ +\begin{problem*}{5.47} +In a home laundry dryer, a cylindrical tub containing wet clothes is +rotated steadily about a horizontal axis as shown in Fig. P5.47. The +clothes are made to tumble so that they will dry uniformly. The rate +of rotation of the smooth-walled tub is chosen so that a small piece +of cloth will lose contact with the tub when the cloth is at an angle +of the $\theta = 68.0^o$ above the horizontal. If the radius of the +tub is $r = 0.330\U{m}$, what rate of revolution is needed? +\end{problem*} % problem 5.47 + +\begin{solution} +Looking at the figure, we see that there are fins sticking out of the +drum wall, so that clothes do not slip along the surface. Because of +this, we can ignore forces in the tangential direction. Focusing on +the center-ward direction, we see that the angle between the force of +gravity $F_g = mg$ and the center-ward direction is +\begin{equation} + \theta' = 90.0^o - \theta = 90.0^o - 68.0^o = 22.0^o +\end{equation} +The sum of forces in the center-ward direction is then +\begin{equation} + F_c = F_g \cos \theta' = mg \cos \theta' +\end{equation} + +In order for this center-ward force to provide separation from the +drum, this force must be the center-ward force needed for uniform +circular motion +\begin{align} + F_c &= m v^2/r = mg \cos \theta' \\ + v^2/r &= g \cos \theta' \\ + v &= \sqrt{r g \cos \theta'} +\end{align} +and the frequency of rotation $f$ is given by +\begin{align} + f = \frac{ v }{2 \pi r} + = \frac{ \sqrt{r g \cos \theta'} }{2 \pi r} + = \frac{1}{2 \pi}\sqrt{ \frac{g \cos \theta'}{r} } + = \frac{1}{2 \pi}\sqrt{ \frac{9.8\U{m/s}^2 \cos 22.0^o}{0.330\U{m}} } + = \ans{0.835\U{Hz}} +\end{align} +\end{solution} diff --git a/latex/problems/problem05.50.tex b/latex/problems/problem05.50.tex new file mode 100644 index 0000000..b5678da --- /dev/null +++ b/latex/problems/problem05.50.tex @@ -0,0 +1,38 @@ +\begin{problem*}{5.50} +An air puck of mass $m_1$ is tied to a string and allowed to revolve +in a circle of radius $R$ on a frictionless horizontal table. The +other end of the string passes through a hole in the center of the +table, and a counterweight of mass $m_2$ is tied to it (Fig. P5.50). +The suspended object remains in equilibrium while the puck on the +tabletop revolves. What are + \Part{a} the tension in the string, + \Part{b} the radial force acting on the puck, and + \Part{c} the speed of the puck? +\end{problem*} % problem 5.50 + +\Part{a} +Constructing a free body diagram for $m_2$, we see that the only +forces on it are the tension \vect{T} and gravity $\vect{F}_{g2}$. +Summing the forces in the downward direction we have +\begin{align} + \sum F &= F_{g2} - T = m_2 g - T \\ + &= m_2 a = 0 \\ + T &= \ans{m_2 g} +\end{align} +Where the first line is summing the forces, the second is Newton's +second law, and the third is combining the previous two and solving +for tension. + +\Part{b} +The only radial force acting on the puck is tension so +$F_c=T=\ans{m_2 g}$. + +\Part{c} +We find the speed of the puck using the circular motion formula +\begin{align} + a_c &= \frac{v^2}{r} \\ + v &= \sqrt{a_c r} + = \sqrt{\frac{F_c r}{m_1}} + = \ans{\sqrt{\frac{m_2}{m_1} g r}} +\end{align} +\end{solution} diff --git a/latex/problems/problem05.52.tex b/latex/problems/problem05.52.tex new file mode 100644 index 0000000..0cb8f8b --- /dev/null +++ b/latex/problems/problem05.52.tex @@ -0,0 +1,107 @@ +\begin{problem*}{5.52} +An amusement park ride consists of a rotating circular platform +$d=8.00\U{m}$ in diameter from which $m=10\U{kg}$ seats are suspended +at the end of $l=2.50\U{m}$ massless chains (Fig. P5.52). When the +system rotates, the chains make an angle of $\theta=28.0\dg$ with the +vertical. +\Part{a} What is the speed of each seat? +\Part{b} Draw a free-body diagram of a $m_c=40.0\U{kg}$ child riding +in a seat, and find the tension in the chain. +\end{problem*} % problem 5.52 + +\begin{solution} +\Part{a} +We will eventually use $v=\sqrt{a_c r}$ as we have in all the other +problems in this homework assignment to find $v$. + +First, we need to find the radius $r$ of the path that the seat takes +around the ride. +\begin{equation} + r=\frac{d}{2} + l \sin \theta + =(4.00 + 2.50\sin 28.0\dg)\U{m} + =5.1736\ldots\U{m} +\end{equation} + +Now we need to find the centerward acceleration $a_c$. Drawing a free +body diagram of our seat, we see that the only forces acting upon it +are the tension \vect{T} and gravity $\vect{F}_g$. We know that the +seat does not rise or fall in the vertical (\vect{y}) direction, so +summing the forces we have +\begin{align} + \sum F_y &= T \cos \theta - mg=m a_y=0 \\ + T &= \frac{mg}{\cos\theta} \label{eqn.T}\\ + \sum F_c &= T \sin \theta + =mg\tan\theta + =m a_c \\ + a_c &= g\tan\theta + =9.8\U{m/s}^2 \cdot \tan 28.0\dg + =5.2108\ldots\U{m/s}^2 \label{eqn.ac} +\end{align} +So +\begin{equation} + v=\sqrt{a_c r} + =\sqrt{g \tan \theta \cdot (\frac{d}{2} + l \sin \theta)} + =\sqrt{ 5.2108\ldots\U{m/s}^2 \cdot 5.1736\ldots\U{m}} + =5.19\U{m/s} +\end{equation} + +\Part{b} +Our free body diagram with a child in the seat will be the same as our +diagram from \Part{a} but with a new mass $m'=m + m_c=50\U{kg}$. + +Before we find the tension in the chain, we should check to see if the +chain angle changes. The angular velocity $\omega=v/r$ does not +change when people get into the seats (because they are of negligible +mass compared to the platform), so we can relate our new velocities +$v'$ and $r'$ using the same $\omega$ that we had in \Part{a}. +\begin{equation} + \omega=\frac{v'}{r'} + =\frac{v}{r} + =\frac{\sqrt{ g r \tan \theta}}{r} + = \sqrt{\frac{g\tan\theta}{r}} + =1.00\U{rad/s} +\end{equation} +Not that the numerical value is important, just that it is a constant. +We can plug $v'=\omega r'$ into our centerward acceleration equation +\begin{equation} + a_c'=\frac{v'^2}{r'} + =\frac{\omega^2 r'^2}{r'} + =r' \omega^2 +\end{equation} +And applying this to eqn. \ref{eqn.ac}(which hasn't changed except for +the need to substitute primed variables) +\begin{align} + a_c' &= g \tan\theta'=r' \omega^2 \\ + g \tan\theta' &= \left( \frac{d}{2} + l \sin\theta' \right) \omega^2 +\end{align} +The only unknown in this equation is $\theta'$, but the equation is +analytically unsolvable. We know $\theta'=28.0\dg$ is one solution, +because there are no masses in this equation, so is must also hold for +case \Part{a}. Then we have to decide if there will be any other +solutions. We know intuitively that any solutions will have $0\dg < +\theta' < 90\dg$. Considering the $\sin$ and $\tan$ functions on that +interval, we see that $\sin\theta'$ is concave down and continuous +over the entire interval, and that $\tan\theta;$ is concave up and +continuous over the entire interval. Therefore, the left hand side of +this equation only equals the right hand side at a single value of +$\theta'$ so our $28.0\dg$ solution is unique. If this doesn't make +sense to you, you can graph the right and left hand sides to check. + +Having proved that $\theta'=\theta$ we can move on to solve for the +tension. Using eqn \ref{eqn.T}. +\begin{equation} + T'=\frac{m' g}{\cos \theta'} + =\frac{50\U{kg} \cdot 9.8\U{m/s}^2}{\cos 28.0\dg} + =\ans{555\U{N}} +\end{equation} + +As far as grading is concerned I will accept anything where you did +any of the following: +\begin{itemize} +\item assumed $\theta$ didn't change (skipping the whole $\theta'=\theta$ step) +\item assumed $\omega$ didn't change, and you went on to show + $\theta'=\theta$ is a valid solution (skipping the uniqueness step). +\item assumed $\omega$ didn't change, and proved that $\theta'=\theta$ + is valid and unique. +\end{itemize} +\end{solution} diff --git a/latex/problems/problem06.09.tex b/latex/problems/problem06.09.tex new file mode 100644 index 0000000..6ba0307 --- /dev/null +++ b/latex/problems/problem06.09.tex @@ -0,0 +1,39 @@ +\begin{problem*}{6.9} +Using the definition of the scalar product, find the angles between + \Part{a} $\vect{A} = 3\ihat - 2\jhat$ and $\vect{B} = 4\ihat - 4\jhat$, + \Part{b} $\vect{A} = -2\ihat + 4\jhat$ and $\vect{B} = 3\ihat - 4\jhat + 2\khat$, and + \Part{c} $\vect{A} = \ihat - 2\jhat + 2\khat$ and $\vect{B} = 3\jhat + 4\khat$. +\end{problem*} % problem 6.9 + +\begin{solution} +From the definition of the scalar (or dot) product on pages 160 and +161, we see +\begin{align} + \vect{A} \cdot \vect{B} &= AB\cos\theta = A_x B_x + A_y B_y + A_z B_z \\ + \theta &= \arccos \left( \frac{A_x B_x + A_y B_y + A_z B_z}{AB} \right) +\end{align} + +\Part{a} +\begin{align} + A &= \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \\ + B &= \sqrt{4^2 + 4^2} = 4 \sqrt{2} \\ + \sum A_i B_i &= 3 \cdot 4 + (-2) \cdot (-4) = 12 + 8 = 20 \\ + \theta &= \arccos \left( \frac{20}{4\sqrt{26}} \right) = \ans{11.3\dg} +\end{align} + +\Part{b} +\begin{align} + A &= \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \\ + B &= \sqrt{3^2 + 4^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29} \\ + \sum A_i B_i &= (-2) \cdot 3 + 4 \cdot (-4) = -6 + -16 = -22 \\ + \theta &= \arccos \left( \frac{-22}{\sqrt{580}} \right) = \ans{156\dg} +\end{align} + +\Part{c} +\begin{align} + A &= \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \\ + B &= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \\ + \sum A_i B_i &= (-2) \cdot 3 + 2 \cdot 4 = -6 + 8 = 2 \\ + \theta &= \arccos \left( \frac{2}{15} \right) = \ans{82.3\dg} +\end{align} +\end{solution} diff --git a/latex/problems/problem06.24.tex b/latex/problems/problem06.24.tex new file mode 100644 index 0000000..98d0241 --- /dev/null +++ b/latex/problems/problem06.24.tex @@ -0,0 +1,31 @@ +\begin{problem*}{6.24} +An $m = 4.00\U{kg}$ particle is subject to a total force that varies +with position as shown in Fig. P6.11. The particle starts from rest +at $x = 0$. What is the speed at + \Part{a} $x = 5.00\U{m}$, + \Part{b} $x = 10.0\U{m}$, and + \Part{c} $x = 15.0\U{m}$? +\end{problem*} % problem 6.24 + +\begin{solution} +In each of these cases we'll be conserving energy. The energy put in +by the force all goes into the particle's kinetic energy. So +conserving energy we have +\begin{align} + W_{0,5} &= \int_{0\U{m}}^{5\U{m}} \vect{F} \cdot \vect{dx} + = \frac{1}{2} m v^2 \\ + v_5 &= \sqrt{ \frac{2 W_{0,5} }{m} } + = \sqrt{ \frac{2 \cdot 1.5\U{N} \cdot 5\U{m}}{4.00\U{kg}} } + = \ans{1.94\U{m/s}} +\end{align} +Doing the same for the other works +\begin{align} + W_{0,10} &= W_{0,5} + 3\U{N} \cdot 5\U{m} = 22.5\U{J} \\ + W_{0,15} &= W_{0,10} + 1.5\U{N} \cdot 5\U{m} = 30\U{J} +\end{align} +So the other velocities are +\begin{align} + v_{10} &= \sqrt{ \frac{2 W_{0,10} }{m} } = \ans{3.35\U{m/s}} \\ + v_{15} &= \sqrt{ \frac{2 W_{0,15} }{m} } = \ans{3.87\U{m/s}} +\end{align} +\end{solution} diff --git a/latex/problems/problem06.29.tex b/latex/problems/problem06.29.tex new file mode 100644 index 0000000..f234fd8 --- /dev/null +++ b/latex/problems/problem06.29.tex @@ -0,0 +1,51 @@ +\begin{problem*}{6.29} +An $m = 40.0\U{kg}$ box initially at rest is pushed $x = 5.00\U{m}$ +along a rough, horizontal floor with a constant applied horizontal +force of $F = 130\U{N}$. The coefficient of friction between box and +floor is $\mu = 0.300$. Find + \Part{a} the work done by the applied force, + \Part{b} the increase in internal energy in the box-floor sysem as a result of the friction, + \Part{c} the work done by the normal force, + \Part{d} the work done by the gravitational force, + \Part{e} the change in kinetic energy of the box, and + \Part{f} the final speed of the box. +\end{problem*} + +\begin{solution} +\Part{a} +\begin{equation} + W_F = \vect{F} \cdot \vect{x} = 130\U{N} \cdot 5.00\U{m} = \ans{650\U{J}} +\end{equation} + +\Part{b} +\begin{equation} + U_f = - \vect{F}_f \cdot \vect{x} = \mu m g x + = 0.300 \cdot 40.0\U{kg} \cdot 9.8\U{m/s}^2 \cdot 5.00\U{m} + = \ans{588\U{J}} +\end{equation} + +\Part{c} +\begin{equation} + W_N = \vect{F}_N \cdot \vect{x} = \ans{0\U{J}} +\end{equation} + +\Part{d} +\begin{equation} + W_g = \vect{F}_g \cdot \vect{x} = \ans{0\U{J}} +\end{equation} + +\Part{e} +Conserving energy +\begin{align} + W_F &= K_f + U_f \\ + \Delta K &= W_F - U_f = (650 - 588)\U{J} = \ans{62\U{J}} +\end{align} + +\Part{f} +\begin{align} + \frac{1}{2} m v^2 &= \Delta K \\ + v &= \sqrt{\frac{2 \Delta K}{m}} + = \sqrt{\frac{2 \cdot 62\U{J}}{40.0\U{kg}}} + = \ans{1.76\U{m/s}} +\end{align} +\end{solution} diff --git a/latex/problems/problem06.30.tex b/latex/problems/problem06.30.tex new file mode 100644 index 0000000..b25885e --- /dev/null +++ b/latex/problems/problem06.30.tex @@ -0,0 +1,70 @@ +\begin{problem*}{6.30} +An $m = 2.00\U{kg}$ block is attached to a spring of force constant $k += 500\U{N/m}$ as shown in Active Figure 6.8 on page 164. The block is +pulled $A = 5.00\U{cm}$ to the right of equilibrium and released from +rest. Find the speed the block has as it passes through equilibrium +if +\Part{a} the horizontal surface is frictionless and +\Part{b} the coefficient of friction between block and surface is +$\mu = 0.350$. +\end{problem*} % problem 6.30 + +\begin{solution} +For both cases we will use conservation of energy. Call the point +where the block is released $P_0$ and the point where the block passes +through equilibrium $P_1$. At $P_0$, the block has spring potential +energy $U_{s0} = 1/2\cdot k A^2$ and no kinetic or gravitational +potential energy. At $P_1$, the block has kinetic energy $K_1 = +1/2\cdot m v^2$ and no potential energy. + +\Part{a} +Without friction, the energy at $P_1$ is the same as that at $P_0$ +because there is no energy lost to friction. +So +\begin{align} + P_0 = P_1 + \frac{1}{2} k A^2 &= \frac{1}{2} m v^2 \\ + v &= A \sqrt{\frac{k}{m}} + = 5\U{cm} \sqrt{\frac{500\U{kg/s}^2}{2\U{kg}}} + = \ans{79.1\U{cm/s}} +\end{align} + +\Part{b} +With friction, part of the initial energy $P_0$ bleeds out into internal +heat energy. +The work done by friction is given by +\begin{equation} + W_f = \vect{F} \cdot \vect{\Delta x} +\end{equation} +Because the block is sliding the whole way in, the frictional force is +always maxed out at the constant +\begin{equation} + F_f = \mu F_N = \mu mg +\end{equation} +In the direction opposite to the motion. +So friction from the table does +\begin{equation} + W_f = -F_f A = -\mu mgA +\end{equation} +Where the negative sign denotes the frictional force sucking energy +from the block. + +Knowing the frictional work, the velocity at the equilibrium position +is given by +\begin{align} + E_0 + W_f &= U_{s0} + W_f = E_1 = K_1 \\ + \frac{1}{2} k A^2 - \mu mgA &= \frac{1}{2} m v^2\\ + m v^2 &= k A^2 - 2 \mu mgA \\ + v &= \sqrt{ \frac{k}{m} A^2 - 2 \mu g A} \\ + &= \sqrt{ \frac{500\U{kg/s}^2}{2\U{kg}} (0.05\U{m})^2 + - 2 \cdot 0.35 \cdot 9.8\U{m/s}^2 \cdot 0.05\U{m}} \\ + &= \ans{0.531\U{m/s}} +\end{align} + +What I was doing for \Part{b} in class on Wednesday was more +complicated because I had misread the question. I thought it was +asking us to find the \emph{maximum} speed, when it just asks for the +speed at equilibrium. Figuring out when the maximum speed occurs +requires more knowledge of differential equations than you guys are +responsible for. +\end{solution} diff --git a/latex/problems/problem06.43.tex b/latex/problems/problem06.43.tex new file mode 100644 index 0000000..67dbe68 --- /dev/null +++ b/latex/problems/problem06.43.tex @@ -0,0 +1,18 @@ +\begin{problem*}{6.43} +While running, a person transforms about $0.600\U{J}$ of chemical +energy to mechanical energy per step per kilogram of body mass. If a +$m = 60.0\U{kg}$ runner transforms energy at a rate of $P = 70.0\U{W}$ +during a race, how fast is the person running? Assume that a running +step is $s = 1.50\U{m}$ long. +\end{problem*} % problem 6.43 + +\begin{solution} +This is simply a units conversion problem +\begin{equation} + \frac{70.0\U{J/s}}{\mbox{runner}} + \cdot \frac{\mbox{step kg}}{0.600\U{J}} + \cdot \frac{\mbox{runner}}{60.0\U{kg}} + \cdot \frac{1.50\U{m}}{\mbox{step}} + = \ans{2.94\U{m/s}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem06.57.tex b/latex/problems/problem06.57.tex new file mode 100644 index 0000000..777c9d4 --- /dev/null +++ b/latex/problems/problem06.57.tex @@ -0,0 +1,50 @@ +\begin{problem*}{6.57} +In diatomic molecules, the consituent atoms exert attractive forces on +each other at large distances, and repulsive forces at short +distances. For many molecules, the Lennard-Jones law is a good +approximation to the magnitude of these forces: +\begin{equation} + F = F_0 \left[2\left(\frac{\sigma}{r}\right)^{13} - + \left(\frac{\sigma}{r}\right)^7 \right] +\end{equation} +Where $r$ is the center-to-center distance between the atoms in the +molecule, $\sigma$ is a length parameter, and $F_0$ is the force when +$r = \sigma$. For an oxygen molecule, $F_0 = 9.60\E{-11}\U{N}$ and +$\sigma = 3.50\E{-10}\U{m}$. Determine the work done by this force as +the atoms are pulled apart from $r_0 = 4.00\E{-10}\U{m}$ to $r_1 = +9.00\E{-10}\U{m}$. +\end{problem*} % problem 6.57 + +\begin{solution} +The work done by the force is given by +\begin{align} + W &= \int_{r_0}^{r_1} \vect{F} \cdot d\vect{r} \\ + &= \int_{r_0}^{r_1} F \cdot dr \\ + &= \int_{r_0}^{r_1} \left\{ F_0 \left[ 2 \left(\frac{\sigma}{r}\right)^{13} - + \left(\frac{\sigma}{r}\right)^7 \right] \right\} \cdot dr \\ + &= 2 F_0 \int_{r_0}^{r_1} \left(\frac{\sigma}{r}\right)^{13} dr + - F_0 \int_{r_0}^{r_1} \left(\frac{\sigma}{r}\right)^7 dr \\ +\end{align} +Then we note that +\begin{align} + \int \left(\frac{a}{x}\right)^n dx + &= a^n \int x^{-n} dx + = a^n \frac{x^{-n+1}}{-n+1} \\ + \int_{r_0}^{r_1} \left(\frac{a}{x}\right)^n dx + &= \frac{a^n}{1-n} \left(r_1^{1-n} - r_0^{1-n}\right) +\end{align} +And plug this into our equation for $W$ +\begin{align} + W &= 2 F_0 \frac{\sigma^{13}}{-12} \left(r_1^{-12} - r_0^{-12}\right) + - F_0 \frac{\sigma^7}{-6} \left(r_1^{-6} - r_0 ^{-6}\right) \\ + &= \frac{-F_0 \sigma}{6} \left[ \left(\frac{\sigma}{r_1}\right)^{12} + - \left(\frac{\sigma}{r_1}\right)^6 \right] + + \frac{F_0 \sigma}{6} \left[ \left(\frac{\sigma}{r_0}\right)^{12} + - \left(\frac{\sigma}{r_0}\right)^6 \right] \\ + &= \frac{-F_0 \sigma}{6} \left[ \sigma^{12} \left( r_1^{-12} - r_0^{-12} \right) - \sigma^6 \left( r_1^{-6} - r_0^{-6} \right) \right] \\ + &= \frac{ -9.50\E{-11}\U{N} \cdot 3.50\E{-10}\U{m}}{6} + \left\{ (3.50\U{\AA})^{12} \left[ (9.00\U{\AA})^{-12} - (4.00\U{\AA})^{-12} \right] + - (3.50\U{\AA})^{6} \left[ (9.00\U{\AA})^{-6} - (4.00\U{\AA})^{-6} \right] \right\} \\ + &= \ans{1.35\E{-25}\U{J}} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.02.tex b/latex/problems/problem07.02.tex new file mode 100644 index 0000000..3105d77 --- /dev/null +++ b/latex/problems/problem07.02.tex @@ -0,0 +1,38 @@ +\begin{problem*}{7.2} +A $F_g = 400\U{N}$ child is in a swing attached to $r = 2.00\U{m}$ +ropes. Find the gravitional potential energy $U_g$ of the child-Earth +system relative to the child's lowest position when + \Part{a} the ropes are horizontal, + \Part{b} the ropes make a $\theta = 30.0\dg$ angle with the vertical, and + \Part{c} the child is at the bottom of the cirvular arc. +\end{problem*} % problem 7.2 + +\begin{solution} +$U_g$ is given by +\begin{equation} + U_g = mgh = F_g h +\end{equation} +Therefor, we need to determine the vertical distance between the +child's location for a given part of the question and the child's +lowest position. + +\Part{a} +The child is one radius above the lowest position, so +\begin{equation} + U_{gA} = F_g r + = 400\U{N} \cdot 2.00\U{m} + = \ans{800\U{J}} +\end{equation} + +\Part{b} +The child has height $h_b = (1-\cos\theta)r$, so +\begin{equation} + U_{gB} = F_g (1-\cos\theta) r + = U_{gA} \cdot (1-\cos\theta) + = 800\U{J} \cdot (1 - \cos 30.0\dg) + = \ans{107\U{J}} +\end{equation} + +\Part{c} +The child has a height of 0, so $U_{gC} = \ans{0\U{J}}$. +\end{solution} diff --git a/latex/problems/problem07.04.tex b/latex/problems/problem07.04.tex new file mode 100644 index 0000000..28732a9 --- /dev/null +++ b/latex/problems/problem07.04.tex @@ -0,0 +1,48 @@ +\begin{problem*}{7.4} +At 11:00AM on September 7, 2001, more than one million British school +children jumped up and down for one minute. The curriculum focus of +the ``Giant Jump'' was on earthquakes, but it was integrated with many +other topics, such as exercise, geography, cooperation, testing +hypothesis, ans setting world records. Children built their own +seismographs that registered local effects. +\Part{a} Find the mechanical energy released in the experiment. +Assume that $N_c=1,050,000$ children of an average mass $m=36.0\U{kg}$ +jump $N_j=12$ times each, raising their centers of mass by +$h=25.0\U{cm}$ each time and briefly resting between one jump and the +next. The free gall acceleration in Britain is $g=9.81\U{m/s}^2$. +\Part{b} Most of the energy is converted very rapidly into internal +energy within the bodies of the children and the floors of the school +buildings. Of the energy that propagates into the ground, most +produces high frequency ``microtremor'' vibrations that are rapidly +damped and cannot travel far. Assume that $p=0.01\U{\%}$ of the +energy is carried away by a long-range seismic wave. The magnitude of +an earthquake on the Richter scale is given by +\begin{equation} + M=\frac{\log E - 4.8}{1.5} +\end{equation} +Where E is the seismic wave energy in joules. +According to this model, what is the magnitude of the demonstration quake? +It did not register above background noise overseas or on the seismograph of the Wolverton Seismic Vault, Hampshire. +\end{problem*} % problem 7.4 + +\begin{solution} +\Part{a} +From ``briefly resting between each jump'' we are to conclude that +each collision is perfectly inelastic (that all the mechanical energy +the student had during the jump was lost into internal energies). The +energy lost by one student preforming a single jump is just $U_g=mgh$, +so the energy lost during the entire experiment is +\begin{equation} + E_T=N_c N_j mgh + =1.05\E{6} \cdot 12 \cdot 36.0\U{kg} \cdot 9.81\U{m/s}^2 \cdot 0.250\U{m} + =\ans{1.11\E{9}\U{J}} +\end{equation} + +\Part{b} +The energy in long-range seismic waves is given by $E=pE_t/100$ so the +magnitude of the ``quake'' is +\begin{equation} + M=\frac{\log( 0.01 \cdot 1.11\E{9} ) - 4.8}{1.5} + =\ans{0.164} +\end{equation} +\end{solution} diff --git a/latex/problems/problem07.10.tex b/latex/problems/problem07.10.tex new file mode 100644 index 0000000..8061acd --- /dev/null +++ b/latex/problems/problem07.10.tex @@ -0,0 +1,55 @@ +\begin{problem*}{7.10} +A particle of mass $m = 5.00\U{kg}$ is released from point $A$ and +slides on the frictionless track shown in Figure P7.10. Determine + \Part{a} the particle's speed at points $B$ and $C$ and + \Part{b} the net work done by the gravitaional force as the particle + moves from $A$ to $C$. +\end{problem*} % problem 7.10 + +\begin{solution} +Reading heights from the figure, \\ +\begin{tabular}{|c|c|c|} + Point & Height & Energy\\ + \hline + \hline + A & $5.00\U{m}$ & $U_{gA}$ \\ + \hline + B & $3.20\U{m}$ & $U_{gB} + K_B$ \\ + \hline + C & $2.00\U{m}$ & $U_{gC} + K_C$ \\ + \hline +\end{tabular} \\ +Where the energies are simply the sum of the particle's kinetic and +gravitational potential energies. The particle has no kinetic energy +at $A$ because is is released from rest. + +The track is frictionless so there are no non-conservative forces +acting on the particle. Therefore the particle's energy is conserved. + +\Part{a} +Conserving energy, we have +\begin{align} + E_A = U_{gA} &= E_B = U_{gB} + K_B \\ + K_B &= U_{gA} - U_{gB} \\ + \frac{1}{2} m v_B^2 &= mgh_A - mgh_B = -mg\Delta h_{AB} \\ + v_B &= \sqrt{ -2g\Delta_{hAB} } + = \sqrt{ -2 \cdot 9.8\U{m/s}^2 \cdot (3.20 - 5.00)\U{m} } + = \ans{ 5.94\U{m/s}} +\end{align} +And applying the same symbolic formula to point $C$, we have +\begin{equation} + v_C = \sqrt{ -2g\Delta_{hAC} } + = \sqrt{ -2 \cdot 9.8\U{m/s}^2 \cdot (2.00 - 5.00)\U{m} } + = \ans{ 7.67\U{m/s}} +\end{equation} + +\Part{b} +The net work done by gravity is simply the change in gravitational +potential energy with the sign reversed. So +\begin{align} + W_g &= -(U_{gC} - U_{gA}) = U_{gA} - U_{gC} = K_C \\ + &= -mg\Delta_{hAC} + = -5.00\U{kg} \cdot 9.8\U{m/s}^2 \cdot (2.00 - 5.00)\U{m} + = \ans{147\U{J}} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.16.tex b/latex/problems/problem07.16.tex new file mode 100644 index 0000000..8970252 --- /dev/null +++ b/latex/problems/problem07.16.tex @@ -0,0 +1,20 @@ +\begin{problem*}{7.16} +An object of mass $m$ starts from rest and slides a distance $d$ down +a frictionless incline of angle $\theta$. While sliding, it contacts +an unstressed spring of negligable mass as shown in Figure P7.16. The +object slides an additional distance $x$ as it is brought momentarily +to rest by the compression of the spring (of force constant $k$). +Find the initial seperation $d$ between the object and the spring. +\end{problem*} % problem 7.16 + +\begin{solution} +This is just the symbolic form of Chapter 6 Problem 27 from last +week's recitation. + +There are no non-conservative forces, so +\begin{align} + E_i = U_{gi} = mgh &= E_f = U_{sf} = \frac{1}{2} k x^2 \\ + h = (x+d) \cdot \sin\theta &= \frac{k x^2}{mg} \\ + d &= \ans{\frac{k x^2}{mg\sin\theta} - x} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.22.tex b/latex/problems/problem07.22.tex new file mode 100644 index 0000000..6e0d377 --- /dev/null +++ b/latex/problems/problem07.22.tex @@ -0,0 +1,60 @@ +\begin{problem*}{7.22} +In a needle biopsy, a narrow strip of tissue is extracted from a +patient using a hollow needle. Rather than being pushed by hand, to +ensure a clean cut the needle can be fired into the patient's body by +a spring. Assume that the needle has mass $m = 5.60\U{g}$, the light +spring has force constant $k = 375\U{N/m}$, and the spring is +originally compressed $d_0 = 8.10\U{cm}$ to project the needle +horizontally without friction. After the needle leaves the spring, +the tip of the needle moves through $d_1 = 2.40\U{cm}$ of sking and +soft tissue, which exerts a force $F_1 = 7.60\U{N}$. Next, the needle +cuts $d_2 = 3.50\U{cm}$ into an organ, which exerts on it a backward +force of $F_2 = 9.20\U{N}$. Find + \Part{a} the maximum speed of the needle and + \Part{b} the speed at which a flange on the back end of the needle + runs into a stop that is set to limit the penetration to $p=5.90\U{cm}$. +\end{problem*} % problem 7.22 + +\begin{solution} +Let us label the various points as follows. \\ +\begin{tabular}{|r|l|} + \hline + Needle at rest, spring maximally compressed & $A$ \\ + \hline + Spring extended, needle just about to enter body & $B$ \\ + \hline + Needle at boundary between soft tissue and organ & $C$ \\ + \hline + Needle at maximal penetration into organ & $D$ \\ + \hline +\end{tabular} \\ + +\Part{a} +The needle will have its maximum speed at point $B$. The kinetic +energy $K_B$ at $B$ will be equal to the spring potential energy +$U_{sA}$ at point $A$, because the launcher is frictionless, and there +are no other relavent potentials. Therefore, +\begin{align} + K_B &= \frac{1}{2} m v_B^2 = U_{sA} = \frac{1}{2} k d_0^2 \\ + v_B &= d_0 \sqrt{\frac{k}{m}} + = 0.0810\U{m} \sqrt{\frac{375\U{N/m}}{5.60\E{-3}\U{kg}}} + = \ans{21.0\U{m/s}} +\end{align} + +\Part{b} +The kinetic energy remaining in the needle just before the flange +strikes the stop is given by +\begin{equation} + K_D = U_{sA} + W_1 + W_2 +\end{equation} +Where $W_1$ and $W_2$ are the work done by the soft tissue and organ +resistance respectively. In each case $W = -F\Delta_x$ because the +forces are constant in the opposite direction to the motion of the +needle. So +\begin{align} + \frac{1}{2} m v_D^2 &= \frac{1}{2} k d_0^2 - F_1 d_1 - F_2 d_2 \\ + v_D &= \sqrt{\frac{k d_0^2 - 2(F_1 d_1 + F_2 d_2)}{m}} \\ + &= \sqrt{\frac{375\U{N/m} \cdot (0.0810\U{m})^2 - 2\cdot(7.60\U{N} \cdot 0.0240\U{m} + 9.20\U{N} \cdot 0.0350\U{m})}{5.60\E{-3}\U{kg}}} \\ + &= \ans{16.1\U{m/s}} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.28.tex b/latex/problems/problem07.28.tex new file mode 100644 index 0000000..b452502 --- /dev/null +++ b/latex/problems/problem07.28.tex @@ -0,0 +1,43 @@ +\begin{problem*}{7.28} +An $m_1 = 50.0\U{kg}$ block and an $m_2 = 100\U{kg}$ block connected by a string as shown in Figure P7.28. +The pulley is frictionless and of negligible mass. +The coefficient of kinetic friction between $m_1$ and the incline is $\mu = 0.250$. +The incline is at an angle of $\theta = 37.0\dg$ from the horizontal. +Determine the change in the kinetic energy of $m_1$ as it moces from point $A$ to point $B$, a distance of $d = 20.0\U{m}$. +\end{problem*} % problem 7.28 + +\begin{solution} +Again, we use conservation of energy. Defining our gravitational +potential energy to be zero at $A$ we have +\begin{equation} + E_A + W_f = K_{1A} + K_{2A} + W_f = E_B = K_{1B} + K_{2B} + U_{g1B} + U_{g2B} +\end{equation} +The blocks are tied together, so they must have the same velocity +(since the string remains taught). So the change in velocity $v$ is +given by +\begin{align} + \frac{1}{2}(m_1 + m_2) v_A^2 - F_f d &= \frac{1}{2}(m_1 + m_2) v_B^2 + m_1 g d \sin \theta - m_2 g d \\ + \Delta(v^2) = v_B^2 - v_A^2 &= \frac{2}{m_1+m_2} \cdot \left[ gd \cdot (m_2 - m_1 \sin\theta) - F_f d \right] \\ +\end{align} +So the change in kinetic enery of $m_1$ is given by +\begin{equation} + \Delta(K_1) = \frac{d m_1}{m_1+m_2} \cdot \left[ g (m_2 - m_1 \sin\theta) - F_f \right] +\end{equation} + +We still need to find the force of fiction, which we do by +constructing a free body diagram of $m_1$. We see that the forces on +$m_1$ are friction $\vect{F}_f$, tension \vect{T}, normal +$\vect{F}_N$, and gravitational $\vect{F}_{g1}$. Summing the forces +in the direction perpendicular to the incline (\vect{y}), we have +\begin{align} + \sum F_y &= F_N - F_{g1} \cos \theta = 0 \\ + F_N &= m_1 g \cos\theta +\end{align} +The block is always sliding so $F_f = \mu F_N = \mu m_1 g \cos\theta$. +Plugging this into our equation for $\Delta(K_1)$ we have +\begin{align} + \Delta(K_1) &= \frac{d g m_1}{m_1+m_2} \cdot \left( m_2 - m_1 \sin\theta - \mu m_1 \cos \theta \right) \\ + &= \frac{20.0\U{m} \cdot 9.8\U{m/s}^2 \cdot 50.0\U{kg}}{150\U{kg}} \cdot \left[ 100\U{kg} - 50\U{kg}( \sin 27.0\dg - 0.250 \cos 27.0\dg ) \right] \\ + &= \ans{5.78\U{kJ}} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.47.tex b/latex/problems/problem07.47.tex new file mode 100644 index 0000000..0246cba --- /dev/null +++ b/latex/problems/problem07.47.tex @@ -0,0 +1,39 @@ +\begin{problem*}{7.47} +The system shown in Fig.~P7.47 consists of a light inextensible cord, +light frictionless pulleys, and blocks of equal mass. It is initially +held at rest so that the blocks are at the same height above the +ground. The blocks are then released. Find the speed of block $A$ at +the moment when the vertical separation of the blocks is $h$. +\end{problem*} % problem 7.47 + +\begin{solution} +Let $m$ be the mass of one block, and \ihat be the vertical direction, +with $x=0$ for both blocks at their initial position. After some +consideration, we decide that block $A$ will fall and block $B$ will +rise (if you are not convinced, find the mass that $B$ must have in +order for the system to remain stationary). In order to get a +quantitative relationship between the motion of the two blocks, +imagine that $B$ moves up a distance $x$ in some time $\Delta t$. +Then block $B$ will have an average velocity of $v_B = x/\Delta t$, +and block $A$ will have gone a distance $-2x$ with an average velocity +of $v_A = -2x/\Delta t$. So $x_A = -2x_B$ and $v_A = -2v_B$. + +When they are a distance $h$ apart +\begin{align} + h &= x_B - x_A = x_B + 2 x_B = 3 x_B \\ + x_B &= \frac{h}{3} \\ + x_A &= \frac{-2h}{3} +\end{align} +So conserving energy +\begin{align} + E_i = K_i + U_{gi} = 0 + &= E_f = K_f + U_{gf} + = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 + + mg\frac{h}{3} + mg\frac{-2h}{3} \\ + g\frac{2h}{3} &= v_A^2 + v_B^2 + = v_A^2 + \left(\frac{-v_A}{2}\right)^2 + = v_A^2 \left( 1 + \frac{1}{4} \right) + = \frac{5}{4} v_A^2 \\ + v_A &= \ans{\sqrt{\frac{8gh}{15}}} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.50.tex b/latex/problems/problem07.50.tex new file mode 100644 index 0000000..d1e116a --- /dev/null +++ b/latex/problems/problem07.50.tex @@ -0,0 +1,69 @@ +\begin{problem*}{7.50} +A child's pogo stick (Fig~P7.50) stores energy in a spring with a +force constant of $k = 2.50\E{4}\U{N/m}$. At position $A$ ($x_A = +-0.100\U{m}$), the spring compression is a maximum and the child is +momentarily at rest. At position $B$ ($x_B = 0$), the spring is +relaxed and the child is moving upward. At position $C$, the child is +again momentarily at rest at the top of the jump. The combined mass +of child and pogo stick is $m = 25.0\U{kg}$. +\Part{a} Calculate the total energy of the child-stick-Earth system, +taking both gravitational and elastic potential energy as zero for +$x=0$. +\Part{b} Determine $x_C$. +\Part{c} Calculate the speed of the child at $B$. +\Part{d} Determine the value of $x$ for which the kinetic energy of +the system is a maximum. +\Part{e} Calculate the child's maximum upward speed. +\end{problem*} % problem 7.50 + +\begin{solution} +\Part{a} +We know the most about point $A$, so we'll calculate the total energy there. +\begin{equation} + E_A = U_{sA} + U_{gA} = \frac{1}{2} k x_A^2 + mgx_A + = \frac{1}{2} \cdot 2.50\E{4}\U{N/m} \cdot (-0.100\U{m})^2 + + 25.0\U{kg} \cdot 9.8\U{m/s}^2 \cdot (-0.100\U{m}) + = \ans{100.5\U{J}} +\end{equation} + +\Part{b} +Conserving energy between $A$ and $C$ +\begin{align} + E_A &= E_C = U_{gC} = mgx_C \\ + x_C &= \frac{E_A}{mg} = \frac{100.5\U{J}}{25.0\U{kg}\cdot9.8\U{m/s}^2} + = \ans{0.410\U{m}} +\end{align} + +\Part{c} +Conserving energy between $A$ and $B$ +\begin{align} + E_A &= E_B = K_B = \frac{1}{2}mv_B^2 \\ + v_B &= \sqrt{\frac{2 E_A}{m}} = \sqrt{\frac{2 \cdot 100.5\U{J}}{25.0\U{kg}}} + = \ans{2.84\U{m/s}} +\end{align} + +\Part{d} +The kinetic energy is maximized when the speed is maximized which +occurs at the point where the accelerating spring force balances the +decelerating gravitational force. Before this point, the spring force +exceeded the gravitational force and the child was speeding up. +Afterward, the gravitation force exceeded the spring force and the +child was slowing down. +\begin{align} + F_s = -kx &= -F_g = mg \\ + x &= \frac{mg}{k} = \frac{25.0\U{kg}\cdot9.8\U{m/s}^2}{2.50\E{4}\U{N/m}} + = -0.00980\U{m} = \ans{-9.80\U{mm}} +\end{align} + +\Part{e} +Conserving energy between $A$ and the point of maximum velocity $D$ +\begin{align} + E_A &= E_D = U_{sD} + U_{gD} + K_D \\ + K_D = \frac{1}{2}mv_D^2 + &= E_A - U_{sD} - U_{gD} + = E_A - \frac{1}{2} k x_D^2 - mgx_D \\ + v_D &= \sqrt{ \frac{2E_A - k x_D^2 - 2mgx_D}{m} } \\ + &= \sqrt{ \frac{2\cdot100.5\U{J} - 2.50\E{4}\U{N/m}\cdot(-9.80\E{-3}\U{m})^2 - 2\cdot 25.0\U{kg}\cdot9.8\U{m/s}^2\cdot(-9.80\E{-3}\U{m})}{25.0\U{kg}} } \\ + &= \ans{2.85\U{m/s}} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.54.tex b/latex/problems/problem07.54.tex new file mode 100644 index 0000000..93ae628 --- /dev/null +++ b/latex/problems/problem07.54.tex @@ -0,0 +1,46 @@ +\begin{problem*}{7.54} +An $m = 1.00\U{kg}$ object slides to the right on a surface having a +coefficient of kinetic friction of $\mu = 0.250$ (Fig.~P7.54). The +object has a speed of $v_i = 3.00\U{m/s}$ when t makes contact with a +light spring that has a force constant of $k = 50.0\U{N/m}$ (point +$A$). The object comes to rest after the spring has been compressed a +distance $d$ (point $B$). The object is then forced toward the left +by the spring and continues to move in that direction beyond the +spring's unstretched position. The object finally comes to rest a +distance $D$ to the left of the unstretched spring (point $D$). Find + \Part{a} the distance of compression $d$, + \Part{b} the speed $v$ at the unstretched position when the object + is moving to the left (point $C$), and + \Part{c} the distance $D$ where the object comes to rest. +\end{problem*} % problem 7.54 + +\begin{solution} +\Part{a} +Conserving energy between points $A$ and $B$ +\begin{align} + E_A + W_{AB} &= \frac{1}{2} m v_i^2 - \mu m g d = E_B = \frac{1}{2} k d^2 \\ + 0 &= \frac{k}{m} d^2 + 2 \mu g d - v_i^2 \\ + d &= \frac{-2 \mu g \pm \sqrt{(2 \mu g)^2 - 4(k/m)(-v_i^2)}}{2k/m} \\ + &= \frac{-2 \cdot 0.250 \cdot 9.8\U{m/s}^2 \pm \sqrt{(2 \cdot 0.250 \cdot 9.8\U{m/s}^2)^2 + 4(50.0\U{N/m}/1.00\U{kg})(3.00\U{m/s})^2}}{2\cdot 50.0\U{N/m}/1.00\U{kg}} \\ + & = (-0.0490 \pm 0.427)\U{m} + = \ans{0.378\U{m}} +\end{align} + +\Part{b} +Conserving energy between $B$ and $C$ +\begin{align} + E_B + W_{BC} &= \frac{1}{2} k d^2 - \mu m g d = E_B = \frac{1}{2} m v^2 \\ + v &= \sqrt{\frac{k}{m} d^2 - 2 \mu g d} + = \sqrt{\frac{50.0\U{N/m}}{1.00\U{kg}} (0.378\ldots\U{m})^2 - 2 \cdot 0.250 \cdot 9.8\U{m/s}^2 \cdot 0.378\ldots\U{m}} + = \ans{2.67\U{m/s}} +\end{align} + +\Part{c} +Conserving energy between $C$ and $D$ +\begin{align} + E_C + W_{CD} &= \frac{1}{2} m v^2 - \mu m g D = E_D = 0\U{J} \\ + D &= \frac{v^2}{2 \mu g} + = \frac{(2.67\ldots\U{m/s})^2}{2 \cdot 0.250 \cdot 9.8\U{m/s}^2} + = \ans{1.46\U{m}} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.55.tex b/latex/problems/problem07.55.tex new file mode 100644 index 0000000..aed0876 --- /dev/null +++ b/latex/problems/problem07.55.tex @@ -0,0 +1,47 @@ +\begin{problem*}{7.55} +A block of mass $m = 0.500\U{kg}$ is pushed against a horizontal +spring of negligable mass until the spring is compressed a distance +$x$ (Fig. P7.55) (point $A$). The force constant of the spring is $k += 450\U{N/m}$. When it is released, the block travels along a +frictionless, horizontal surface to point $B$, the bottom of a +vertical circular track of radius $R = 1.00\U{m}$, and continues to +move up the track. The speed of the block at the bottom of the track +is $v_B = 12.0\U{m/s}$, and the block experiences an average friction +force of $F_f = 7.00\U{N}$ while sliding up the track. +\Part{a} What is $x$? +\Part{b} What speed $v_T$ do you predict for the block at the top of +the track (point $T$)? +\Part{c} Does the block actually reach the top of the track, or does +it fall off before reaching the top? +\end{problem*} % problem 7.55 + +\begin{solution} +\Part{a} +Conserving energy between $A$ and $B$ +\begin{align} + E_A &= \frac{1}{2} k x^2 = E_B = \frac{1}{2} m v^2 \\ + x &= v \sqrt{\frac{m}{k}} + = 12.0\U{m/s} \sqrt{\frac{0.500\U{kg}}{450\U{N/m}}} + = \ans{0.400\U{m}} +\end{align} + +\Part{b} +Conserving energy between $B$ and $T$ +\begin{align} + E_B + W_f &= \frac{1}{2} m v_B^2 - \pi R F_f = E_T = \frac{1}{2} m v_T^2 + m g 2 R \\ + v_T &= \sqrt{ v_B^2 - 2R (\pi F_f/m + 2g) } + = \sqrt{ (12.0\U{m/s})^2 - 2\cdot 1.00\U{m}(\pi 7.00\U{N}/0.500\U{kg} + 2 \cdot 9.8\U{m/s}^2) } + = \ans{4.10\U{m/s}} +\end{align} + +\Part{c} +The centerward acceleration of the block if it passes through $T$ is +\begin{equation} + a_c = \frac{v_T^2}{r} + \sim 16\U{m/s}^2 + > g +\end{equation} +So the block reaches the top and is still attached to the ramp, +because it is still pushing out with some normal force against the +track. +\end{solution} diff --git a/latex/problems/problem07.61.tex b/latex/problems/problem07.61.tex new file mode 100644 index 0000000..6067b34 --- /dev/null +++ b/latex/problems/problem07.61.tex @@ -0,0 +1,41 @@ +\begin{problem*}{7.61} +A pendulum, comprising a light string of length $L$ and a small +sphere, swings in a vertical plane. The string hits a peg located a +distance $d$ below the point of suspension (Fig.~P7.61). +\Part{a} Show that if the sphere is released from a height below that +of the peg (point $A$), it will return to this height after the string +strikes the peg (point $B$). +\Part{b} Show that if the pendulum is released from the horizontal +position ($\theta = 90\dg$) and is to swing in a complete circle +centered on the peg, the minimum value of $d$ must be $3L/5$. +\end{problem*} % problem 7.61 + +\begin{solution} +\Part{a} +Conserving energy between points $A$ and $B$ (the sphere is at rest at +both points). +\begin{align} + E_A &= mgh_A = E_B = mgh_B \\ + h_A &= h_B +\end{align} + +\Part{b} +The radius of the smaller circle is $r = L-d$. The critical point is +when sphere is in the vertical position of it's circle around the peg. +The higher the peg is, the slower it's velocity will be at this point +and the less the tension will be. At the smallest possible $d$ for a +complete circle, there will be no tension at this point and we can +find the velocity with +\begin{align} + a_c &= g = \frac{v^2}{r} \\ + v^2 &= g(L-d) +\end{align} +Then conserving energy between this point $D$ and the release point +$C$ (letting $h=0$ be the level of the peg) +\begin{align} + E_C &= mgd = E_D = \frac{1}{2} m v^2 + m g (L-d) = \frac{1}{2} mg (L-d) + mg (L-d) \\ + d &= \frac{3}{2}(L-d) \\ + \frac{5}{2}d &= \frac{3}{2}L \\ + d &= \frac{3L}{5} +\end{align} +\end{solution} diff --git a/latex/problems/problem07.62.tex b/latex/problems/problem07.62.tex new file mode 100644 index 0000000..1a9bece --- /dev/null +++ b/latex/problems/problem07.62.tex @@ -0,0 +1,47 @@ +\begin{problem*}{7.62} +A roller coaster car is released from rest at the top of the first +rise and then moves freely with negligible friction. The roller +coaster shown in Fig.~P7.62 has a circular loop of radius $R$ in the +vertical plane. +\Part{a} First, suppose the car barely makes it around the loop; at +the top of the loop the riders are upside down and feel weightless. +Find the required height of the release point above the bottom of the +loop, in terms of $R$. +\Part{b} Now assume that the release point is at or above the minimum +required height. Show that the normal force on the car at the bottom +of the loop exceeds the normal force at the top by six times the +weight of the car. The normal force on each rider follows the same +rule. Such a large normal force is dangerous and very uncomfortable +for the riders. Roller coasters are therefore not build with circular +loops in vertical planes. Figure P5.24 and the photograph on page 134 +show two actual designs. +\end{problem*} % problem 7.62 + +\begin{solution} +\Part{a} +Because the riders ``feel weightless'' at the top of the loop (point +$T$), we will assume that they are in free fall with a centerwards +acceleration of $g = v_T^2/R$. Conserving energy between $T$ and the +release point $A$ +\begin{align} + E_A = mgh &= E_T = \frac{1}{2} m v_T^2 + mg(2R) \\ + h &= \frac{1}{2g}v_T^2 + 2R = \frac{Rg}{2g} + 2R = \ans{2.5 R} +\end{align} + +\Part{b} +If the release comes from a higher point, there will be some normal +force at the top $N_T$ and at the bottom $N_B$. Summing forces at +both points +\begin{align} + \sum F_{cT} &= mg + N_T = m \frac{v_T^2}{R} \\ + v_T^2 &= gR + R\frac{N_T}{m} \\ + \sum F_{cB} &= -mg + N_B = m \frac{v_B^2}{R} \\ + v_B^2 &= -gR + R\frac{N_B}{m} +\end{align} +And conserving energy between the top and bottom +\begin{align} + E_B = \frac{1}{2}mv_B^2 &= E_T = \frac{1}{2}mv_T^2 + mg(2R) \\ + v_B^2 &= v_T^2 + 4gR = -gR + R\frac{N_B}{m} = gR + R\frac{N_T}{m} + 4gR \\ + N_B &= N_T + 6mg +\end{align} +\end{solution} diff --git a/latex/problems/problem08.05.tex b/latex/problems/problem08.05.tex new file mode 100644 index 0000000..87e6ebb --- /dev/null +++ b/latex/problems/problem08.05.tex @@ -0,0 +1,27 @@ +\begin{problem*}{8.5} +Two blocks with masses $M$ and $3M$ are plaved on a horizontal, +frictionless surface. A light spring is attached to one of them, and +the blocks are pushed to gether with the spring between them +(Fig. P8.5). A cord initially holding the blocks together is burned; +after this, the block of mass $3M$ moves to the right with a speed of +$v_B = 2.00\U{m/s}$. +\Part{a} What is the speed $v_S$ of the block of mass $M$? +\Part{b} Find the original elastic potential energy $U_s$ in the +spring, taking $M = 0.350\U{kg}$ +\end{problem*} % problem 8.5 + +\begin{solution} +\Part{a} +Conserving momentum +\begin{align} + P_i = 0 &= P_f = P_B - P_S = 3Mv_B - Mv_S \\ + v_S &= 3v_B = 3 \cdot 2.00\U{m/s} = \ans{6.00\U{m/s}} +\end{align} + +\Part{b} +Conserving energy +\begin{align} + E_i = U_s &= E_f = K_B + K_S = \frac{1}{2}3Mv_B^2 + \frac{1}{2}Mv_S^2 \\ + U_s &= \frac{1}{2}3Mv_B^2 + \frac{1}{2}M(3v_B)^2 = \frac{1}{2}Mv_B^2 (3 + 9) = 6Mv_B^2 = 5 \cdot 0.350\U{kg} \cdot (2.00\U{m/s})^2 = \ans{8.40\U{J}} +\end{align} +\end{solution} diff --git a/latex/problems/problem08.06.tex b/latex/problems/problem08.06.tex new file mode 100644 index 0000000..03f947b --- /dev/null +++ b/latex/problems/problem08.06.tex @@ -0,0 +1,18 @@ +\begin{problem*}{8.6} +A friend claims that as long as he has his seat belt on, he can hold +on to an $m = 12.0\U{kg}$ child in a $v_i = 60.0\U{mph}$ head-on +collision with a brick wall in which the car passenger compartment +comes to a stop in $\Delta t = 0.050\U{s}$. Show that the violent +force during the collision will tear the child from his arms. +\end{problem*} + +\begin{solution} +The force needed to hold on to the child is given by +\begin{equation} + F = \frac{\Delta p}{\Delta t} = - m \frac{v_i}{\Delta t} + = -12.0\U{kg}\frac{60\U{mph}}{0.050\U{s}} + \cdot \frac{1609\U{m}}{1\U{mi}} \cdot \frac{1\U{hr}}{3600\U{s}} + = -6436\U{N} +\end{equation} +Which is much larger than what the friend is capable of applying. +\end{solution} diff --git a/latex/problems/problem08.17.tex b/latex/problems/problem08.17.tex new file mode 100644 index 0000000..111f3fa --- /dev/null +++ b/latex/problems/problem08.17.tex @@ -0,0 +1,29 @@ +\begin{problem*}{8.17} +Suppose a truch and car with initial speeds of $v_i = 8.00\U{m/s}$ +collide in a perfectly inelastic head on collision. Each driver has a +mass of $m = 80.0\U{kg}$. Including the drivers, the total vehicle +masses are $m_c = 800\U{kg}$ for the car and $m_t = 4000\U{kg}$ for +the truck. If the collision time is $\Delta t = 0.120\U{s}$, what +force does the seat belt exert on each driver. +\end{problem*} % problem 8.17 + +\begin{solution} +To find the final velocity $v_f$ of the crumpled mass, we conserve momentum. +\begin{align} + P_i = m_t v_i - m_c v_i = (m_t - m_c) v_i &= P_f = (m_t + m_c) v_f \\ + v_f &= v_i \frac{m_t - m_c}{m_t + m_c} + = 8.00\U{m/s} \frac{4000\U{kg} - 800\U{kg}}{4800\U{kg}} + = 5.333\ldots\U{m/s} +\end{align} +In the same direction the truck was initially going. The average +force $F$ on each driver is then given by +\begin{align} + F &= m a = m \frac{\Delta v}{\Delta t} = m \frac{v_f - v_i}{\Delta t} \\ + F_t &= m \frac{v_f - v_i}{\Delta t} + = 80.0\U{kg} \frac{5.333\U{m/s} - 8.00\U{m/s}}{0.120\U{s}} + = \ans{-1780\U{N}} \\ + F_c &= m \frac{v_f + v_i}{\Delta t} + = 80.0\U{kg} \frac{5.333\U{m/s} + 8.00\U{m/s}}{0.120\U{s}} + = \ans{8890\U{N}} +\end{align} +\end{solution} diff --git a/latex/problems/problem08.18.tex b/latex/problems/problem08.18.tex new file mode 100644 index 0000000..9f5306c --- /dev/null +++ b/latex/problems/problem08.18.tex @@ -0,0 +1,31 @@ +\begin{problem*}{8.18} +As show in Fig.~P8.18, a bullet of mass $m$ and speed $v$ passes +completely through a pendulum bob of mass $M$. The bullet emerges +with a speed $v_f = v/2$. The pendulum bob is suspended by a stiff +rod of length $l$ and a negligable mass. What is the minimum value of +$v$ such that the pendulum bob will barely swing through a complete +vertical circle? +\end{problem*} % problem 8.18 + +\begin{solution} +Let us break the problem up into two steps: the collision where we'll +conserve momentum, and the pendulum swinging upside down where we'll +conserve energy. Call the point before the collision $A$, the point +just after the collision before the bob has started to swing $B$, and +the point where the pendulum is completely inverted $C$. We just need +to give the bob enough energy that it has no speed at $C$ (just barely +coasting through), so conserving energy back to $B$ +\begin{align} + E_C = Mg(2l) &= E_B = \frac{1}{2}Mv_B^2 \\ + v_B^2 &= 4gl \\ + v_B &= 2\sqrt{gl} +\end{align} +Where we've left out the kinetic energy of the bullet since it doesn't +change from $B$ to $C$ + +Now conserving momentum back to $A$ +\begin{align} + P_B = Mv_B + m(v/2) &= P_A = mv \\ + v &= \frac{2Mv_B}{m} = \ans{ \frac{4M}{m}\sqrt{gl} } +\end{align} +\end{solution} diff --git a/latex/problems/problem08.24.tex b/latex/problems/problem08.24.tex new file mode 100644 index 0000000..38752f0 --- /dev/null +++ b/latex/problems/problem08.24.tex @@ -0,0 +1,33 @@ +\begin{problem*}{8.24} +An $m_1 = 90\U{kg}$ fullback running east (\ihat) with a speed of $v_1 += 5.00\U{m/s}$ is tackled by an $m_2 = 95\U{kg}$ opponent running +north (\jhat) with a speed of $v_2 = 3.00\U{m/s}$. Noting that the +collision is perfectly inelastic, + + \Part{a} calculate the speed $v_f$ and direction $\theta$ of the + players just after the tackle and + \Part{b} determine the mechanical energy lost as a result of the + collision. Account for the missing energy. +\end{problem*} % problem 8.24 + +\begin{solution} +\Part{a} +Conserving momentum in the \ihat\ and \jhat\ directions +\begin{align} + P_{ix} = m_1 v_1 &= P_{fx} = (m_1 + m_2) v_{fx} \\ + v_{fx} &= v_1 \frac{m_1}{m_1 + m_2} = 2.43\U{m/s} \\ + P_{iy} = m_2 v_2 &= P_{fy} = (m_1 + m_2) v_{fy} \\ + v_{fy} &= v_2 \frac{m_2}{m_1 + m_2} = 1.54\U{m/s} \\ + v_f &= \sqrt{v_{fx}^2 + v_{fy}^2} = \ans{2.88\U{m/s}} \\ + \theta &= \arctan\left(\frac{v_{fy}}{v_{fx}}\right) = \ans{32.3\dg} +\end{align} + +\Part{b} +\begin{align} + \Delta K &= K_f - K_i = \left( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \right) - \frac{1}{2} (m_1 + m_2) v_f^2 \\ + &= \frac{1}{2} \left[ 90.0\U{kg} (5.00\U{m.s})^2 + 95.0\U{kg} (3.00\U{kg})^2 - (90.0\U{kg} + 95.0\U{kg}) (2.88\U{m/s})^2 \right]\\ + &= \ans{-786\U{J}} +\end{align} +All of which has been lost as mechanical energy, and is now thermal +energy (warmer football players), noise (a loud crunch), etc. +\end{solution} diff --git a/latex/problems/problem08.25.tex b/latex/problems/problem08.25.tex new file mode 100644 index 0000000..8f8ee87 --- /dev/null +++ b/latex/problems/problem08.25.tex @@ -0,0 +1,47 @@ +\begin{problem*}{8.25} +Two shuffleboard disks of equal mass, one orange and the other yellow, +are involved in an elastic, glancing collision. The yellow disk is +initially at rest and is struck by the orange disk moving with a speed +$v_i$. After the collision, the orange disk moves along a direction +that makes an angle $\theta$ with its initial direction of motion. +The velocities of the two disks are perpendicular after the collision. +Determine the final speed of each disk. +\end{problem*} + +\begin{solution} +Let the final speed of the orange disk be $v_o$, the final speed of +the yellow disk be $v_y$, and $m$ be the mass of one disk. Calling +the initial direction of the orange disk \ihat, and the direction +perpendicular to that \jhat\ (such that the final direction of +$\vect{v}_o$ has positive components in both directions), we see +\begin{align} + v_{o\ihat} &= v_o \cos\theta \\ + v_{o\jhat} &= v_o \sin\theta +\end{align} +For the orange puck, and that since the motion of the yellow is +perpendicular the the orange, the angle between the final motion of +the yellow and the $-\jhat$ direction is also $\theta$, so +\begin{align} + v_{y\ihat} &= v_y \sin\theta \\ + v_{y\jhat} &= -v_y \cos\theta +\end{align} + +Conserving momentum in both directions we have +\begin{align} + P_{i\jhat} = 0 &= P_{f\jhat} = m v_{y\jhat} + m v_{o\jhat} = m v_o \sin\theta - m v_y\cos\theta \\ + v_y &= v_o \frac{\sin\theta}{\cos\theta}\\ + P_{i\ihat} = m v_i &= P_{f\ihat} = m v_{y\ihat} + m v_{o\ihat} = m v_o \cos\theta + m v_y \sin\theta \\ + v_i \cos\theta &= v_o \cos^2 \theta + \left(v_o \frac{\sin\theta}{\cos\theta}\right)\sin\theta\cos\theta + = v_o \cos^2 \theta + v_o \sin^2 \theta + = v_o +\end{align} +Because +\begin{equation} + \sin^2 \theta + \cos^2 \theta = 1 +\end{equation} +So +\begin{align} + v_o &= \ans{v_i \cos\theta} \\ + v_y &= v_o \frac{\sin\theta}{\cos\theta} = \ans{v_i \sin\theta} +\end{align} +\end{solution} diff --git a/latex/problems/problem08.26.tex b/latex/problems/problem08.26.tex new file mode 100644 index 0000000..ca35a84 --- /dev/null +++ b/latex/problems/problem08.26.tex @@ -0,0 +1,25 @@ +\begin{problem*}{8.26} +Two automobiles of equal mass approach an intersection. One vehicle +is traveling with a velocity $v_1 = 13.0\U{m/s}$ towards the east +(\ihat), and the other is traveling north (\jhat) with a speed $v_2$. +Neither driver sees the other. The vehicles collide in the +intersection and stick together, leaving parallel skid marks at an +angle of $\theta = 55.0\dg$ north of east. The speed limit for both +roads is 35\U{mph}, and the driver of the northward-moving vehicle +claims that he was within the speed limit when the collision occurred. +Is he telling the truth? +\end{problem*} % problem 8.26 + +\begin{solution} +Let $m$ be the mass of one car and $\vect{v}_f$ be the final velocity +of the wreck. Conserving momentum in both directions +\begin{align} + P_{i\ihat} = m v_1 &= P_{f\ihat} = (m+m) v_f \cos\theta \\ + P_{i\jhat} = m v_2 &= P_{f\jhat} = (m+m) v_f \sin\theta \\ + \tan\theta &= \frac{\sin\theta}{\cos\theta} = \frac{m v_2}{m v_1} \\ + v_2 &= v_1 \tan\theta + = 13.0\U{m/s} \tan 55.0\dg \cdot \frac{1\U{mi}}{1609\U{m}} \cdot \frac{3600\U{s}}{1\U{h}} + = \ans{41.6\U{mph}} +\end{align} +So he was speeding. +\end{solution} diff --git a/latex/problems/problem08.28.tex b/latex/problems/problem08.28.tex new file mode 100644 index 0000000..393ec39 --- /dev/null +++ b/latex/problems/problem08.28.tex @@ -0,0 +1,47 @@ +\begin{problem*}{8.28} +A proton, moving with a velocity of $v_i\ihat$, collides elastically +with another proton that is initially at rest. Assuming that the two +protons have equal speeds after the collision, find + \Part{a} the speed $v_f$ of each proton after the collision in terms + of $v_i$ and + \Part{b} the directions of the velocity vectors after the collision. +\end{problem*} % problem 8.28 + +\begin{solution} +\Part{a} +Looking at the front inside cover of the text we see that the the mass +of a proton is given by $m_p = 1.672\E{-27}\U{kg}$. Conserving energy +(because the collision is elastic) we have +\begin{align} + K_i = \frac{1}{2} m_p v_i^2 + &= K_f = \frac{1}{2} m_p v_f^2 + \frac{1}{2} m_p v_f^2 \\ + v_f &= \sqrt{\frac{v_i^2}{2}} = \ans{\frac{v_i}{\sqrt{2}}} +\end{align} + +\Part{b} +Let $\vect{v}_{f1}$ be the final velocity for the incident proton, and +$\vect{v}_{f2}$ be the final velocity for the proton initially at +rest. Conserving momentum in the \jhat\ direction +\begin{align} + P_{iy} = 0 &= P_{fy} = m_p v_{f1y} + m_p v_{f2y} \\ + v_{f1y} &= -v_{f2y} +\end{align} +So the protons have equal magnitude speeds in the \jhat\ direction. +Because the speed of the particles are equal, the magnitude of their +speeds in the \ihat\ direction should also be equal $|v_{f1x}| = +|v_{f2x}|$. Conserving momentum in the \ihat\ direction. +\begin{align} + P_{ix} = m_p v_i &= P_{fx} = m_p v_{f1x} + m_p v_{f2x} = 2 m_p v_{fx} \\ + v_{fx} &= \frac{v_i}{2} +\end{align} +Using the Pythagorean theorem to solve for the magnitude of $v_{fy}$ +\begin{align} + v_f^2 = \frac{v_i^2}{2} &= v_{fx}^2 + v_{fy}^2 = \frac{v_i^2}{4} + v_{fy}^2 \\ + v_{fy} &= v_i \sqrt{\frac{1}{2} - \frac{1}{4}} = \frac{v_i}{2} = v_{fx} +\end{align} +So because the \ihat\ and \jhat\ components of $\vect{v}_f$ are the +same, both protons are deflected away at an angle of $\theta = +\ans{45\dg}$ from the \ihat\ direction, with opposite +\jhat\ components (so the angle between $\vect{v}_{f1}$ and +$\vect{v}_{f2}$ is $90\dg$). +\end{solution} diff --git a/latex/problems/problem08.43.tex b/latex/problems/problem08.43.tex new file mode 100644 index 0000000..0aed036 --- /dev/null +++ b/latex/problems/problem08.43.tex @@ -0,0 +1,31 @@ +\begin{problem*}{8.43} +A rocket for use in deep space is to be capable of boosting a total +load (payload plus rocket frame and engine) of $M_f = 3.00\U{metric + tons}$ to a speed of $v_f = 10.0\U{km/s}$. +\Part{a} It has an engine and fuel designed to produce an exhaust +speed of $v_{ea} = 2.000\U{km/s}$. How much fule plus oxidizer is +required? +\Part{b} If a different fuel and engine design could give an exhaust +speed of $v_{eb} = 5.000\U{km/s}$, what amount of fuel and oxidizer +would be required for the same task? +\end{problem*} + +\begin{solution} +\Part{a} +Starting with equation 8.43 from page 248, and letting $M_e = M_i - +M_f$ be the mass of the fuel and oxidizer +\begin{align} + v_f - v_i &= v_e \ln \left(\frac{M_i}{M_f}\right) \\ + M_i &= M_f \exp^{\frac{v_f - v_i}{v_e}} \\ + M_e &= M_f \left(\exp^{\frac{v_f - v_i}{v_e}} - 1 \right) \label{43.M_e} \\ + &= 3.00\U{metric tons} \left(\exp^{\frac{10}{2}} - 1\right) + = \ans{ 442\U{metric tons}} +\end{align} + +\Part{b} +Using eqn. \ref{43.M_e} with our new exhaust velocity, +\begin{align} + M_e &= 3.00\U{metric tons} \left(\exp^{\frac{10}{5}} - 1\right) + = \ans{ 19.2\U{metric tons}} +\end{align} +\end{solution} diff --git a/latex/problems/problem08.45.tex b/latex/problems/problem08.45.tex new file mode 100644 index 0000000..f6e5512 --- /dev/null +++ b/latex/problems/problem08.45.tex @@ -0,0 +1,26 @@ +\begin{problem*}{8.45} +An orbiting spacecraft is described not as a ``zero-g'' but rather as +a ``microgravity'' environment for its occupants and for onboard +experiments. Astronouts experience slight lurches due to the motions +of the equipment and other astronauts and as a result of venting of +materials from the craft. Assume that an $M_i = 3500\U{kg}$ +spacecraft underoes an acceleration of $a = 2.50\U{$\mu$g} = +2.45\E{-5}\U{m/s}^2$ due to a leak from one of its hydraulic control +systems. The fluid is know to escape with a speed of $70.0\U{m/s}$ +into the vacuum of space. How much fluid will be lost in $\Delta t = +1.00\U{h}$ if the leak is not stopped. +\end{problem*} % problem 8.45 + +\begin{solution} +If the acceleration of the spaceship remains constant, and the rate of +fluid escape remains constant, the mass of escaping fluid must be much +less than the mass of the spaceship. Conserving momentum according to +the conservation of momentum equation 8.42 in the text, +\begin{align} + M dv &= -v_e dM \\ + dM &= -M \frac{dv}{v_e} + = -M \frac{a\Delta t}{v_e} + = -3500\U{kg} \frac{2.45\E{-5}\U{m/s}^2 \cdot 3600\U{s}}{70.0\U{m/s}} + = \ans{-4.41\U{kg}} +\end{align} +\end{solution} diff --git a/latex/problems/problem08.48.tex b/latex/problems/problem08.48.tex new file mode 100644 index 0000000..13929be --- /dev/null +++ b/latex/problems/problem08.48.tex @@ -0,0 +1,31 @@ +\begin{problem*}{8.48} +A bullet of mass $m$ is fired horizontally into a block of mass $M$ +initially at rest at the edge of a frictionless table of height $h$ +(Fig. P8.48). The bullet remains in the block, and after impact the +block lands a distance $d$ from the bottom of the table. Determine +the intial speed of the bullet. +\end{problem*} % problem 87.48 + +\begin{solution} +Breaking the problem up into two parts (like problem 18), call the +point before the collision $A$, the point just after the collision $B$ +and the point when the block-bullet hits the floor $C$. + +From $B$ to $C$ is a standard projectile motion problem, which we'll +solve for the horizontal velocity $v_B$ of the block-bullet at point +$B$. Because $v_B$ is purely horizontal (the \ihat direction), we'll +use the vertical (\jhat) direction to find the time it took the ball +to fall. +\begin{align} + y_f = -h &= \frac{1}{2}a t^2 + v_{y0}t + y_0 = \frac{-g}{2}t^2 \\ + t &= \sqrt{\frac{2h}{g}} \\ + x_f = d &= v_B t + x_0 = v_B t \\ + v_B &= \frac{d}{t} = d\sqrt{\frac{g}{2h}} \\ +\end{align} + +Now conserving momentum back to $A$ +\begin{align} + P_B = (m+M) v_B &= P_A = m v \\ + v &= \frac{m+M}{m} v_B = \ans{\frac{m+M}{m}d\sqrt{\frac{g}{2h}}} +\end{align} +\end{solution} diff --git a/latex/problems/problem08.51.tex b/latex/problems/problem08.51.tex new file mode 100644 index 0000000..ca8910b --- /dev/null +++ b/latex/problems/problem08.51.tex @@ -0,0 +1,35 @@ +\begin{problem*}{8.51} +A small block of mass $m_1 = 0.500\U{kg}$ is released from rest at the +top of a curve-shaped, frictionless wedge of mass $m_2 = 3.00\U{kg}$, +which sits on a frictionless, horizontal surface as sown in +Fig. P8.51a. When the block leaves the wedge, its velocity is +measured to be $v_1 = 4.00\U{m/s}$ to the right (\ihat) as shown in +Fig. P8.51b. +\Part{a} What is the velocity $\vect{v}_2$ of the wedge after the block reaches the horizontal surface? +\Part{b} What is the height $h$ of the wedge? +\end{problem*} % problem 8.51 + +\begin{solution} +\Part{a} +Conserving momentum +\begin{align} + P_i = 0 &= P_f = m_1 v_1 + m_2 v_2 \\ + v_2 &= -v_1 \frac{m_1}{m_2} = -4.00\U{m/s} \frac{0.500\U{kg}}{3.00\U{kg}} + = \ans{-0.667\U{m/s}} +\end{align} +Where the $-$ sign denotes motion in the $-\ihat$ direction (to the +left in Fig.~P8.51b). + +\Part{b} +Let $y = 0$ be the level of the table for the purpose of calculating +gravitational potential energy. Conserving energy (since none is lost +to friction or other internal energies) +\begin{align} + E_i = m_1 g h &= E_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \\ + m_1 g h &= \frac{1}{2}\left(m_1 v_1^2 + m_2 \frac{m_1^2 v_2^2}{m_2^2}\right) \\ + &= \frac{m_1 v_1^2}{2}\left(1 + \frac{m_1}{m_2}\right) \\ + h &= \frac{v_1^2}{2g}\left(1 + \frac{m_1}{m_2}\right) + = \frac{(0.667\U{m/s})^2}{2\cdot9.80\U{m/s}^2}\left(1 + \frac{0.500\U{kg}}{3.00\U{kg}}\right) + = \ans{0.952\U{m}} +\end{align} +\end{solution} diff --git a/latex/problems/problem09.18.tex b/latex/problems/problem09.18.tex new file mode 100644 index 0000000..5752de5 --- /dev/null +++ b/latex/problems/problem09.18.tex @@ -0,0 +1,33 @@ +\begin{problem*}{9.18} +An observer in reference frame $S$ measures two events as +simultaneous. Event $A$ occurs at the point $(50.0\U{m},0,0)$ at the +instant 9:00:00 Universal time on January 15, 2005. Event $B$ occurs +at the point $(150\U{m},0,0)$ at the same moment. A second observer, +moving past with a velocity of $0.800c\ihat$, also observes the two +events. In her reference frame $S'$, which event occured first and +what time interval elapsed between the events? +\end{problem*} % problem 9.18 + +\begin{solution} +The $(x,t)$ coordinates in the $S$ frame are $(50.0\U{m},t_0)$ and +$(150\U{m},t_0)$. In the $S'$ frame, the coordinates are given by +the Lorentz transformations +\begin{align} + \gamma &= \frac{1}{\sqrt{1-v^2/c^2}} + = \frac{1}{\sqrt{1-0.800^2}} + = 1.67 \\ + t' &= \gamma \p({t - v x/c^2}) \\ + x' &= \gamma \p({x - v t}) \\ + y' &= y \\ + z' &= z +\end{align} +We see from the formula for $t'$ that the event with the largest $x$ +value will have the earliest time. Therefore, \ans{event $B$ happens +first in $S'$}. The time interval is +\begin{align} + \Delta t = t_A' - t_B' + = \gamma \p({\Delta t - v \Delta x/c^2}) + = \gamma \p({0 - 0.800 \cdot (-100\U{m}) / 3.00\E{8}\U{m/s}}) + = \ans{444\U{ns}} +\end{align} +\end{solution} diff --git a/latex/problems/problem09.19.tex b/latex/problems/problem09.19.tex new file mode 100644 index 0000000..dcb75d8 --- /dev/null +++ b/latex/problems/problem09.19.tex @@ -0,0 +1,41 @@ +\begin{problem*}{9.19} +A red light flashes at position $x_R=3.00\U{m}$ and time +$t_R=1.00\E{-9}\U{s}$, and a blue light flashes at $x_B=5.00\U{m}$ and +$t_B=9.00\E{-9}\U{s}$, all measured in the $S$ reference frame. +Reference frame $S'$ has its origin at the same point as $S$ at +$t=t'=0$; frame $S'$ moves uniformly to the right. Both flashes occur +at the same place in $S'$. \Part{a} Find the relative speed between +$S$ and $S'$. \Part{b} Find the location of the two flashes in frame +$S'$. \Part{c} At what time does the red flash occur in the $S'$ +frame? +\end{problem*} % problem 9.19 + +\begin{solution} +\Part{a} +From the Lorentz transformations +\begin{align} + \Delta x' &= \gamma \p({\Delta x - v \Delta t}) = 0 \\ + v &= \frac{\Delta x}{\Delta t} = \frac{5.00-3.00}{9-1}\U{m/ns} + = \ans{0.833c} +\end{align} + +\Part{b} +\begin{align} + \gamma &= \frac{1}{\sqrt{1-v^2/c^2}} + = \frac{1}{\sqrt{1-0.833^2}} + = 1.81 \\ + x' &= \gamma \p({x - v t}) \\ + x_R' &= 1.81 \p({3.00\U{m/s} - 0.833\cdot3.00\E{8}\U{m/s}\cdot1.00\E{-9}\U{s}}) + = \ans{4.97\U{m}} \\ + x_B' &= 1.81 \p({5.00\U{m/s} - 0.833\cdot3.00\E{8}\U{m/s}\cdot9.00\E{-9}\U{s}}) + = \ans{4.97\U{m}} +\end{align} + +\Part{c} +From the Lorentz transformations +\begin{align} + t' &= \gamma \p({t - v x/c^2}) \\ + t_R' &= 1.81 \p({1.00\E{-9}\U{s} - 0.833\frac{3.00\U{m}}{3.00\E{8}\U{m/s}}}) + = \ans{-13.3\U{ns}} +\end{align} +\end{solution} diff --git a/latex/problems/problem09.22.tex b/latex/problems/problem09.22.tex new file mode 100644 index 0000000..ffd5d25 --- /dev/null +++ b/latex/problems/problem09.22.tex @@ -0,0 +1,46 @@ +\begin{problem*}{9.22} +A spacecraft is launched from the surface of the Earth with a velocity +of $0.600\U{c}$ at an angle of $50.0\dg$ above the horizontal positive +$x$ axis. Another spacecraft is moving past with a velocity of +$0.700c$ in the negative $x$ direction. Determine the magnitude and +direction of the velocity of the first spacecraft as measured by the +pilot of the second spacecraft. +\end{problem*} % problem 9.22 + +\begin{solution} +The Lorentz transformations +\begin{align} + t' &= \gamma \p({t - v x/c^2}) \\ + x' &= \gamma \p({x - v t}) \\ + y' &= y +\end{align} +yield the velocity transformations +\begin{align} + u_x' &= \pderiv{t'}{x'} + = \frac{\gamma\p({\partial x - v \partial t})}{\gamma\p({\partial t - v \partial x/c^2})} + = \frac{\pderiv{x}{t} - v}{1 - v \pderiv{t}{x}/c^2} + = \frac{u_x - v}{1 - v u_x/c^2} \\ + u_y' &= \pderiv{t'}{y'} + = \frac{\partial y}{\gamma\p({\partial t - v \partial x/c^2})} + = \frac{\pderiv{t}{y}}{\gamma\p({1 - v \pderiv{t}{x}/c^2})} + = \frac{u_y}{\gamma\p({1 - v u_x/c^2})} \\ +\end{align} + +We can use these velocity transformations on our spacecraft's velocity. +\begin{align} + \gamma &= \frac{1}{\sqrt{1-v^2/c^2}} + = \frac{1}{\sqrt{1-(-0.700)^2}} = 1.40 \\ + u_x &= 0.600c\cos(50.0\dg) = 0.386c \\ + u_y &= 0.600c\sin(50.0\dg) = 0.460c \\ + u_x' &= \frac{u_x - v}{1 - vu_x/c^2} + = \frac{0.321c + 0.700c}{1 + 0.700\cdot0.321} + = 0.855c \\ + u_y' &= \frac{u_y}{\gamma(1 - vu_x/c^2)} + = \frac{0.383c}{1.40\cdot(1 + 0.700\cdot0.321)} + = 0.258c \\ + u' &= \sqrt{u_x'^2 + u_y'^2} + = \ans{0.893c} \\ + \theta' &= \arctan(u_y'/u_x') + = \ans{16.8\dg} +\end{align} +\end{solution} diff --git a/latex/problems/problem09.23.tex b/latex/problems/problem09.23.tex new file mode 100644 index 0000000..fe03489 --- /dev/null +++ b/latex/problems/problem09.23.tex @@ -0,0 +1,59 @@ +\begin{problem*}{9.23} +Calculate the momentum of an electron moving with a speed of \Part{a} +$0.0100c$, \Part{b} $0.500c$, \Part{c} $0.900c$. +\end{problem*} % problem 9.23 + +\begin{solution} +The momentum (using relativistic mass) is given by +\begin{equation} + p = mv = \gamma m_0 v +\end{equation} + +%\begin{python} +%import latex +%import sys +%tempmod = file('nine_twentythree.py', 'w') +%tempmod.write(""" +%def gamma(v_over_c): +% return 1.0/(1-v_over_c**2)**.5 +%def p(v_over_c): +% return gamma(v_over_c)*511e3*v_over_c # in eV/c +%""") +%tempmod.close() +%\end{python} + +\Part{a} +%\begin{equation} +% \gamma = +%\begin{python} +%from nine_twentythree import *; print gamma(0.01) +%\end{python} \\ +% p = +%\begin{python} +%from nine_twentythree import *; print p(0.01) +%\end{python} +%\end{equation} + +\begin{align} + \gamma &= \ans{1.0000500} \\ +%\begin{python} +%from nine_twentythree import *; print gamma(0.01) +%\end{python} \\ + p &= \ans{5.11\U{keV/c}} +%\begin{python} +%from nine_twentythree import *; print p(0.01) +%\end{python} +\end{align} + +\Part{b} +\begin{align} + \gamma &= \ans{1.155} \\ + p &= \ans{295\U{keV/c}} +\end{align} + +\Part{c} +\begin{align} + \gamma &= \ans{2.294} \\ + p &= \ans{1.06\U{MeV/c}} +\end{align} +\end{solution} diff --git a/latex/problems/problem09.30.tex b/latex/problems/problem09.30.tex new file mode 100644 index 0000000..2ac53bb --- /dev/null +++ b/latex/problems/problem09.30.tex @@ -0,0 +1,41 @@ +\begin{problem*}{9.30} +Show that, for any object moving at less than one-tenth the speed of +light, the relativistic kinetic energy agrees with the result of the +classical equation $K=\frac{1}{2}mv^2$ to within less than $1\%$. +Therefore, for most purposes the classical equation is good enough to +describe these objects, whose motion we call \emph{nonrelativistic}. +\end{problem*} % problem 9.30 + +\begin{solution} +The kinetic energy is the energy that is due to the objects motion. +In other words, the increase in the total energy over the rest mass +(in the absence of potential energies etc.). In math +\begin{equation} + K = E - m_0 c^2 = mc^2 - m_0c^2 = (\gamma - 1)m_0 c^2 + = \p({\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - 1}) m_0 c^2 + = \p[{\p({1-\frac{v^2}{c^2}})^{-0.5} - 1}] m_0 c^2 +\end{equation} +Using the Taylor series expansion around $x=0$, +\begin{equation} + (1+x)^b \approx 1 + bx + \mathcal{O}(x^2) +\end{equation} +we see that for small $x=-v^2/c^2$, the kinetic energy looks like +\begin{equation} + K = \p[{\p({1+0.5\frac{v^2}{c^2} + \mathcal{O}(v^4/c^4)}) - 1}] m_0 c^2 + = 0.5 m_0 v^2 + \mathcal{O}(m_0 v^4/c^2) + \approx \frac{1}{2} m_0 v^2 +\end{equation} +with the relative error on the order of $v^2/c^2 < 0.01 = 1\%$ for +$v<0.1c$, which is what we set out to show. + +Alternatively, we can compare the exact kinetic energy with the +nonrelativistic form for $v=0.1c$. From the above analysis, we see +that the relative error will be less for slower $v$. +\begin{align} + \gamma &= \frac{1}{\sqrt{1-0.1^2}} = 1.00503782 \\ + K_\text{rel} &= (\gamma-1) m_0 c^2 = 0.00503782 \gamma m_0 c^2 \\ + K_\text{non} &= \frac{1}{2} m_0 v^2 = 0.5 \cdot m_0 \cdot 0.01 c^2 = 0.005 m_0 c^2 \\ + \frac{K_\text{non}}{K_\text{rel}} &= \frac{0.005}{0.00503782} = 0.992494 +\end{align} +for a $7.51\%$ relative underestimate. +\end{solution} diff --git a/latex/problems/problem09.31.tex b/latex/problems/problem09.31.tex new file mode 100644 index 0000000..6b8ed22 --- /dev/null +++ b/latex/problems/problem09.31.tex @@ -0,0 +1,24 @@ +\begin{problem*}{9.31} +An electron has a kinetic energy five times greater than its rest +energy. Find \Part{a} its total energy and \Part{b} its speed. +\end{problem*} % problem 9.31 + +\begin{solution} +The total energy is +\Part{a} +\begin{equation} + E = K + m_0 c^2 = (5+1)m_0 c^2 + = 6\cdot 511\U{keV} + = \ans{3.07\U{MeV}} +\end{equation} + +\Part{b} +\begin{align} + E &= m c^2 = \gamma m_0 c^2 = 6 m_0 c^2 \\ + \gamma &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = 6 \\ + \frac{1}{6^2} &= 1-\frac{v^2}{c^2} \\ + \frac{v}{c} &= \sqrt{1-\frac{1}{6^2}} + = 0.986 \\ + v &= \ans{0.986c} +\end{align} +\end{solution} diff --git a/latex/problems/problem09.35.tex b/latex/problems/problem09.35.tex new file mode 100644 index 0000000..255725f --- /dev/null +++ b/latex/problems/problem09.35.tex @@ -0,0 +1,49 @@ +\begin{problem*}{9.35} +The rest energy of an electron is $0.511\U{MeV}$. The rest energy of +a proton is $938\U{MeV}$. Assume that both particles have kinetic +energies of $2.00\U{MeV}$. Find the speed of \Part{a} the electron +and \Part{b} the proton. \Part{c} By how much does the speed of the +electron exceed that of the proton? \Part{d} Repeat the calculations +assuming that both particles have kinetic energies of $2,000\U{MeV}$. +\end{problem*} % problem 9.35 + +\begin{solution} +First we'll work out the solution symbolically, since we'll need it +twice. The total energy yields $\gamma$, which in turn yields $v$. +\begin{align} + E &= K + m_0 c^2 = \gamma m_0 c^2 \\ + \gamma &= \frac{K}{m_0 c^2} + 1 \\ + \gamma &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\ + \frac{1}{\gamma^2} &= 1-\frac{v^2}{c^2} \\ + \frac{v}{c} &= \sqrt{1-\frac{1}{\gamma^2}} \\ + v &= \sqrt{1-\frac{1}{\gamma^2}} \cdot c +\end{align} + +Applying these formula to our various situations, we get + +\Part{a} +\begin{align} + \gamma_a &= \frac{2.00\U{MeV}}{0.511\U{MeV}} + 1 = 4.91 \\ + v_a &= \ans{0.979c} +\end{align} + +\Part{b} +\begin{align} + \gamma_b &= \frac{2.00\U{MeV}}{938\U{MeV}} + 1 = 1.00213 \\ + v_b &= \ans{0.0652c} +\end{align} + +\Part{c} +\begin{equation} + v_a-v_b = \ans{0.914c} +\end{equation} + +\Part{d} +\begin{align} + \gamma_e &= \frac{2.00\U{GeV}}{0.511\U{MeV}} + 1 = 3.91e3 \\ + v_e &= \ans{(1-3.26\E{-8})c} \\ + \gamma_p &= \frac{2.00\U{GeV}}{938\U{MeV}} + 1 = 3.16 \\ + v_p &= \ans{0.948c} \\ + v_e-v_p &= \ans{0.0523c} +\end{align} +\end{solution} diff --git a/latex/problems/problem09.50.tex b/latex/problems/problem09.50.tex new file mode 100644 index 0000000..84ecb15 --- /dev/null +++ b/latex/problems/problem09.50.tex @@ -0,0 +1,68 @@ +\begin{problem*}{9.50} +Ted and Mary are playing a game of catch in frame $S'$, which is +moving at $0.600\U{c}$ with respect to frame $S$, while Jim, at rest +in frame $S$, watches the action (Fig.~P9.50). Ted throws the ball to +Mary at $0.800c$ (according to Ted), and their seperation (measured in +$S'$) is $1.80\E{12}\U{m}$. \Part{a} According to Mary, how fast is +the ball moving? \Part{b} According to Mary, how long does the ball +take to reach her? \Part{c} According to Jim, how far apart are Ted +and Mary, and how fast is the ball moving? \Part{d} According to Jim, +how long does it take the ball to reach Mary? +\end{problem*} % problem 9.50 + +\begin{solution} +\Part{a} +Mary is in the same frame as Ted, so she also feels the ball is moving +at $\ans{-0.800c}$. + +\Part{b} +Mary thinks the ball takes +\begin{equation} + \Delta t' = \frac{\Delta x'}{v'} + = \frac{1.80\E{12}\U{m}}{0.800\cdot3.00\E{8}\U{m/s}} + = \ans{7500\U{s}} = \ans{2.08\U{hours}} +\end{equation} + +\Part{c} +Jim sees the $S'$ frame as length contracted, so he sees the proper +length $\Delta x'$ between Mary and Ted (proper length is length +measured in the frame where the two ends are stationary, here the $S'$ +frame) as +\begin{align} + \gamma &= \frac{1}{\sqrt{1-v^2/c^2}} + = \frac{1}{\sqrt{1-0.6^2}} = 1.25 \\ + \Delta x &= \frac{\Delta x'}{\gamma} = \frac{1.8\E{12}\U{m}}{1.25} + = \ans{1.44\E{12}\U{m}} \;. +\end{align} +You can use the Lorentz velocity transformations to find the ball +speed in Jim's frame. +\begin{equation} + u_x = \frac{u_x'+v}{1+\frac{vu_x'}{c^2}} + = \frac{-0.8c+.6c}{1-0.8\cdot0.6} + = \ans{-0.385c} +\end{equation} + + +\Part{d} +You can do this several ways. One way is to use the total distance +between Mary and Ted in the $S$ frame +\begin{align} + (v - u_x)\Delta t &= \Delta x \\ + \Delta t &= \frac{\Delta x}{v - u_x} + = \frac{1.44\E{12}\U{m}}{0.600c + 0.385c} + = \ans{4870\U{s}} = \ans{1.35\U{hours}} +\end{align} + +Another way is to convert from the $S'$ frame to the ball frame to get +proper time +\begin{equation} + \Delta t_0 = \frac{\Delta t'}{\gamma_b'} + = 2.08\U{hours}\cdot\sqrt{1-0.8^2} + = 1.25\U{hours} +\end{equation} +Then you can time dilate from the ball frame into the $S$ frame +\begin{equation} + \Delta t = \gamma_b \Delta t_0 = \frac{1.25\U{hours}}{\sqrt{1-0.385^2}} + = \ans{1.35\U{hours}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem09.V12.tex b/latex/problems/problem09.V12.tex new file mode 100644 index 0000000..e2c2c3d --- /dev/null +++ b/latex/problems/problem09.V12.tex @@ -0,0 +1,39 @@ +\begin{problem*}{9.V12} +The $S'$ frame moves along the $x$ axis of the $S$ frame with a +constant velocity $v$. Observers in $S$ see an event $A$ occuring at +$x_1=0$ at time $t_1=0$ and another event $B$ at $x_2=1.5\U{km}$ at +time $t_2=2.7\U{$\mu$s}$. In the $S'$ frame, the two events are +simultaneous. \Part{a} Find $v$. \Part{b} For which $v$ between the +two frames will the event $B$ precede the event $A$ for observers +in the $S'$ frame? +\end{problem*} % Prof. Venkat lecture problem #12 + +\begin{solution} +\Part{a} +From the Lorentz transformation, +\begin{align} + t_1' &= \gamma\p({t_1-vx_1/c^2}) \\ + t_2' &= \gamma\p({t_2-vx_2/c^2}) \\ + \Delta t' &= (t_2'-t_1') + = \gamma\p({t_2-t_1 - \frac{v}{c^2}(x_2-x_1)}) + = \gamma\p({\Delta t - \frac{v}{c^2}\Delta x}) + = 0 \\ + 0 &= \Delta t - \frac{v}{c^2}\Delta x \\ + v &= \frac{c^2\Delta t}{\Delta x} + = \frac{3.00\E{8}\U{m/s}\cdot2.7\U{$\mu$s}}{1.5\U{km}}c + = \ans{0.54c} +\end{align} + +\Part{b} +From \Part{a} we have +\begin{equation} + \Delta t' = \gamma\p({\Delta t - \frac{v}{c^2}\Delta x}) \;, +\end{equation} +$\Delta t>0$, so $\Delta t'<0$ (i.e. $B$ precedes $A$) only if +\begin{align} + \frac{v\Delta x}{c^2} &> \Delta t \\ + v &> c^2 \frac{\Delta t}{\Delta x} = 0.54c \;. +\end{align} +So $B$ precedes $A$ for $\ans{0.54= 2: + filename = sys.argv[1] + + +figure(facecolor='w', figsize=[4,3]) +border = 0.15 +ax = axes([border,border,1-2*border,1-2*border]) +ax.set_xticks(range(-10,11,1)) +ax.set_yticks(range(-10,11,1)) +ax.grid() + +ax.set_title("Limo paradox, garage frame") +ax.set_xlabel("Distance (light seconds)") +ax.set_ylabel("Time (seconds)") + +# add light beams +ax.plot([-10,10],[-10,10], 'b-') +ax.plot([10,-10],[-10,10], 'b-') + +# switch to a moving observer +v = 0.661 +width = 1 +gamma = 1.0/sqrt(1-v**2) +t= arange(-5,10,1) + +# plot his spaceship +spaceship = Polygon([(v*t[0]-width,t[0]), (v*t[0], t[0]), + (v*t[-1],t[-1]), (v*t[-1]-width, t[-1])], + alpha=0.2, facecolor='k') +ax.add_patch(spaceship) + +# plot the position track of a moving observer +ax.plot(v*t,t, 'r-') + +# plot the position track of the garage +ax.plot(0*t,t, 'g-') +ax.plot(-width+0*t,t, 'g-') + +# plot his xprime axis at his 1 second intervals +tp = gamma*t +def plot_xp(tp, v, xmin=-10, xmax=10): + x = arange(xmin, xmax, 0.01) + xp_y = x*v + for t in tp: + origin_x = v*t + origin_y = t + ax.plot(x+origin_x, xp_y+origin_y, 'r:') +plot_xp(tp, v) + +# plot his tprime grids at his 1 light-second intervals +xp = tp +def plot_tp_grid(xp, v, tmin=-10, tmax=10): + t = arange(tmin, tmax, 0.01) + tp_x = t*v + for x in xp: + origin_x = x + origin_y = v*x + ax.plot(tp_x+origin_x, t+origin_y, 'r:') +plot_tp_grid(xp, v) + +proper_width = gamma*width +x_back_limo_frame = -proper_width +t_back_limo_frame = 0 +x_back_limo_frame_p = gamma*(x_back_limo_frame+v*t_back_limo_frame) +t_back_limo_frame_p = gamma*(t_back_limo_frame+v*x_back_limo_frame) +x_gar_limo_frame = -width/gamma +t_gar_limo_frame = 0 +x_gar_limo_frame_p = gamma*(x_gar_limo_frame+v*t_gar_limo_frame) +t_gar_limo_frame_p = gamma*(t_gar_limo_frame+v*x_gar_limo_frame) +points = [(0,0), (-width,0), (x_back_limo_frame_p, t_back_limo_frame_p), + (x_gar_limo_frame_p, t_gar_limo_frame_p)] +labels = ["A", "B", "C", "D"] + +ax.plot([p[0] for p in points], [p[1] for p in points], 'ko') +for p,L in zip(points, labels): + ax.text(p[0], p[1], ' '+L, + horizontalalignment='left', + verticalalignment='center') + +ax.set_ybound(-2,0.5) +ax.set_xbound(-2,0.5) +ax.set_aspect('equal') # Make aspect ratio equal to 1 + +if filename == None: + show() +else: + savefig(filename, dpi=200) diff --git a/latex/problems/problem09.limo.T.tex b/latex/problems/problem09.limo.T.tex new file mode 100644 index 0000000..cef7a3d --- /dev/null +++ b/latex/problems/problem09.limo.T.tex @@ -0,0 +1,59 @@ +\begin{problem} +\Part{a} How fast would you have to be driving a $20\U{ft}$ long limo +to fit into $15\U{ft}$ deep garage? \Part{b} How deep would the +garage appear to the driver of the limo? Is the limo ever really +entirely in the garage? Explain any apparent paradoxes. + +Both of the lengths given are proper lengths. +\end{problem} % Based on undergrad memories + +\begin{solution} +\Part{a} +The proper length of the limo $L_0$ needs to be contracted to the +length of the garage (equations compressed for space, read right to +left, then top to bottom). +\begin{align} + L &= \frac{L_0}{\gamma} & + \frac{L}{L_0} &= \frac{1}{\gamma} = \sqrt{1-\p({\frac{v}{c}})^2} \\ + \p({\frac{L}{L_0}})^2 &= 1 - \p({\frac{v}{c}})^2 & + \p({\frac{v}{c}})^2 &= 1 - \p({\frac{L}{L_0}})^2 \\ + \frac{v}{c} &= \sqrt{1 - \p({\frac{L}{L_0}})^2} = 0.661 & + v &= \ans{0.661 c} = \ans{198\E{6}\U{m/s}} \;. +\end{align} + +\Part{b} +The garage $L_{g0}$ is length contracted in the limo frame +\begin{align} + L_g &= \frac{L_{g0}}{\gamma} = L_{g0}\sqrt{1-\p({\frac{v}{c}})^2} + = 15\U{ft}\sqrt{1-.661^2} = \ans{11.2\U{ft}} = \ans{3.43\U{m}} \;. +\end{align} + +Wait, how can the limo fit into a garage that appears even shorter +than its proper length of $15\U{ft}$? This is a +relative-simultanaeity effect like the muon clock running slower than +an earth clock while the earth clock runs slower than the muon clock. +In the garage frame, the limo-nose-passes-crashes-into-back-wall event +$A$ and the limo-tail-passes-door event $B$ occur at the same time. +In the limo frame, event $A$ happens some time before event $B$. + +Drawing a space-time diagram in the garage frame may help clarify the +different events. The limo is the grey smear. The red dotted lines +represent a $1\U{ls}$ time and space grid for the limo driver. The +blue lines show the speed of light. The green lines show the garage. +I've rescaled the garage and limo to make them $1\U{ls}$ and +$1.33\U{ls}$ long respectively so that the axes have simple labels. +The limo driver thinks that at the same time as event $A$, the tail of +the limo is back at event $C$, while the door of the garage is up at +event $D$. Note that the garage is less than $1\U{ls}$ deep in the +limo frame. +\begin{center} +\includegraphics[height=2in]{limo} +\end{center} +``Entirely in the garage'' is something of a trick question, since it +means ``all of the limo is in the garage at same time'' and ``at the +same time'' depends on your reference frame. In the garage frame, the +limo is entirely in the garage for a single instant. In the limo +frame, the limo is never entirely in the garage. An absolute, +true-no-matter-what answer to the question is not possible in the +relativistic world-view. +\end{solution} diff --git a/latex/problems/problem10.13.tex b/latex/problems/problem10.13.tex new file mode 100644 index 0000000..15c7f91 --- /dev/null +++ b/latex/problems/problem10.13.tex @@ -0,0 +1,36 @@ +\begin{problem*}{10.13} +A car traveling on a flat (unbanked) circular track accelerates +uniformly from rest with a tangential acceleration of $a_t = +1.70\U{m/s}^2$. The car makes it one forth of the way around the +circle before it skids off the track. Determine the coefficient of +static friction between the car and track from these data. +\end{problem*} % problem 10.13 + +\begin{solution} +In order to find the coefficient of static friction, we must examine +the force of friction at the slipping point where $F_f = \mu_s F_N = +\mu_s m g$. We don't know the mass of the car, but hopefully it will +cancel out somewhere along the way. The only force on the car that is +not completely in the vertical direction is friction, so let us +consider the sums of forces in the tangential and centerward +directions. First the tangential direction +\begin{equation} + \sum F_t = F_{ft} = m a_t +\end{equation} +And then in the centerward direction +\begin{equation} + \sum F_c = F_{fc} = m a_c = m \frac{v_t^2}{r} +\end{equation} +Going back to our constant acceleration equations we see that +\begin{equation} + v_t^2 = v_{ti}^2 + 2 a_t \Delta x = 2 a_t \frac{\pi r}{2} +\end{equation} +So going backwards and plugging in +\begin{align} + F_{fc} &= m \frac{2 a_t \pi r}{2 r} = \pi m a_t \\ + F_f &= \sqrt{F_{ft}^2 + F_{fc}^2} = m a_t \sqrt{1 + \pi^2} \\ + \mu_s &= \frac{F_f}{m g} = \frac{a_t}{g} \sqrt{1 + \pi^2} + = \frac{1.70\U{m/s}^2}{9.80\U{m/s}^2} \sqrt{1 + \pi^2} + = \ans{0.572} +\end{align} +\end{solution} diff --git a/latex/problems/problem10.21.tex b/latex/problems/problem10.21.tex new file mode 100644 index 0000000..82e637b --- /dev/null +++ b/latex/problems/problem10.21.tex @@ -0,0 +1,25 @@ +\begin{problem*}{10.21} +Find the net torque $\tau$ on the wheel in Fig. P10.21 about the axle +through $O$, taking $a = 10.0\U{cm}$ and $b = 25.0\U{cm}$. +\end{problem*} % problem 10.21 + +\begin{solution} +Torque is defined as $\vect{\tau} = \vect{r} \times \vect{F}$, so we +have (defining the counter clockwise direction to be positive) +\begin{equation} + \sum \tau = - b \cdot 10.0\U{N} + a \cdot 12.0\U{N} - b \cdot 9.00\U{N} +\end{equation} +Where the $-$ sign on the first and third terms denote torques in the +$-$ direction. There are no $\sin$ terms, because all three forces +are in the tangential direction. The $12.0\U{N}$ force is slightly +suspicious, since they tell you it makes an angle of $30\dg$ with the +horizontal, but if you look closely, you'll see that it isn't actually +applied to the top of the circle, and it {\it is} tangential to it's +application radius. + +Plugging in for $a$ and $b$ we have +\begin{equation} + \sum \tau = -0.250\U{m} \cdot 10.0\U{N} + 0.100\U{m} \cdot 12.0\U{N} - 0.250\U{m} \cdot 9.00\U{N} + = \ans{-3.55\U{J}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem10.44.tex b/latex/problems/problem10.44.tex new file mode 100644 index 0000000..809d9ab --- /dev/null +++ b/latex/problems/problem10.44.tex @@ -0,0 +1,56 @@ +\begin{problem*}{10.44} +A space station is constructed in the shape of a hollow ring of mass +$m = 5.00\E{4}\U{kg}$. Members of the crew walk on a deck formed by +the inner surface of the outer cylindrical wall of the ring, with a +radius of $r = 100\U{m}$. At rest when constructed, the ring is set +rotating about its axis so that the people inside experience an +effective free-fall acceleration equal to $g$. The rotation is +achieved by firing two small rockets attached tangentially at opposite +points on the outside of the ring. +\Part{a} What angular momentum does the space station acquire? +\Part{b} How long must the rockets be fired if each exerts a thrust +of $F = 125\U{N}$? +\Part{c} Prove that the total torque on the ring, multiplied by the +the time interval found in \Part{b}, is equal to the change in angular +momentum found in \Part{a}. This equality represents the +\emph{angular impulse-angular momentum theorem}. +\end{problem*} % problem 10.44 + +\begin{solution} +\Part{a} +The certerward acceleration of people on the wall of the space station +is given by +\begin{align} + g = a_c &= \frac{v^2}{r} = r \omega^2 \\ + \omega &= \sqrt{\frac{g}{r}} +\end{align} +Where we used $v = r\omega$ to replace the linear velocity $v$. The +moment of inertia of a ring is given by $I = mr^2$ from table 10.2 on +page 300. The angular momentum is then given by +\begin{equation} + L = I \omega = m r^2 \sqrt{\frac{g}{r}} = \ans{m r \sqrt{gr}} = \ans{1.57\E{8}\U{Js}} +\end{equation} + +\Part{b} +The torque on the station is given by +\begin{align} + \sum \tau &= 2 \cdot r \cdot F = I \alpha = m r^2 \alpha\\ + \alpha &= \frac{2 F}{m r} +\end{align} +Going back to our constant acceleration equations, we see that +\begin{align} + \omega &= \alpha t + \omega_0 = \alpha t \\ + t &= \frac{\omega}{\alpha} = \sqrt{\frac{g}{r}} \cdot \frac{m r}{2 F} = \sqrt{g r} \frac{m}{2F} + = \sqrt{ 9.80\U{m/s}^2 \cdot 100\U{m}} \frac{5\E{4}\U{kg}}{2 \cdot 125\U{N}} + = \ans{ 6.26\U{ks} = 1.74\U{hr} } +\end{align} + +\Part{c} +\begin{align} + \tau t &= I \alpha t = I \omega = L \\ + 2 r F t &= 2 \cdot 100\U{m} \cdot 125\U{N} \cdot 6.26\E{3}\U{s} + = 1.57\E{8}\U{Js} = L +\end{align} +So they are equal both symbolically and numerically which means I +probably didn't make any algebra mistakes (we can hope). +\end{solution} diff --git a/latex/problems/problem10.72.tex b/latex/problems/problem10.72.tex new file mode 100644 index 0000000..eac9d60 --- /dev/null +++ b/latex/problems/problem10.72.tex @@ -0,0 +1,44 @@ +\begin{problem*}{10.72} +A wad of sticky clay with mass $m$ and velocity $\vect{v}_i$ is fired +at a solid cylinder of mass $M$ and radius $R$ (Fig. P10.72). The +cylinder is initially at rest and is mounted on a fixed horizontal +axle that runs through its center of mass. The line of motion of the +projectile is perpendixular to the axle and at a distance $d < R$ from +the center. +\Part{a} Find the angular speed of the system just after the clay +strikes and sticks to the surface of the cylinder. +\Part{b} Is the mechanical energy of the clay-cylinder system +conserved in this process? Explain your answer. +\end{problem*} % problem 10.72 + +\begin{solution} +\Part{a} +Conserving angular momentum (letting the \ihat\ direction be into the page) +\begin{align} + \vect{L}_i &= \vect{R} \times \vect{p} + = R p \sin\theta \ihat + = R m v_i \frac{d}{R} \ihat = m v_i d \ihat \\ + &= \vect{L}_f = I \omega \ihat = \left(\frac{1}{2} M R^2 + m R^2\right) \omega \ihat = \left(\frac{M}{2} + m\right)R^2 \omega \ihat \\ + \omega &= \ans{v_i \frac{m d}{I_{tot}}} +\end{align} +Where $I_{tot} \equiv \left( M/2 + m \right) R^2$. + +\Part{b} +The change in energy is +\begin{align} + \Delta K &= K_f - K_i = \frac{1}{2} I_{tot} \omega^2 - \frac{1}{2} m v_i^2 \\ + &= \frac{1}{2} \left(\frac{m^2 v_i^2 d^2}{I_tot} - mv_i^2\right) + = \frac{1}{2}m v_i^2 \left(\frac{m d^2}{I_tot} - 1\right) \\ + &= K_i \left(\frac{m d^2}{\left(M/2 + m\right)R^2} - 1\right) \\ +\end{align} +So for maximum final energy $d=R$ and the 2nd term on the right hand +side reduces to +\begin{equation} + \frac{m}{M/2 + m} - 1 = \frac{m - (M/2 + m)}{M/2 + m} + = \frac{-M}{M+2m} < 0 +\end{equation} +So the final energy is less than the initial energy unless $m = 0$, in +which case the cylinder just sits still for eternity. The lost energy +goes to the same types of internal energy that we had in Problem 24 in +Chapter 8: warmer clay, thwacking sound, etc. +\end{solution} diff --git a/latex/problems/problem11.34.tex b/latex/problems/problem11.34.tex new file mode 100644 index 0000000..7b8cf83 --- /dev/null +++ b/latex/problems/problem11.34.tex @@ -0,0 +1,39 @@ +\begin{problem*}{11.34} +Within the Rosette Nebula shown in the photograph opening this +chapter, a hydrogen atom emits light as it undergoes a transition from +the $n=3$ state to the $n=2$ state. Calculate \Part{a} the +energy, \Part{b} the wavelength, and \Part{c} the frequency of the +radiation. +\end{problem*} % problem 11.34 + +\begin{solution} +\Part{a} +From the Rydberg equation (Equation 11.28) +\begin{align} + \frac{1}{\lambda} &= R_H \p({\frac{1}{n_i^2}-\frac{1}{n_f^2}}) \\ + E &= \frac{hc}{\lambda} = hcR_H \p({\frac{1}{n_i^2}-\frac{1}{n_f^2}}) + = hcR_H \p({\frac{1}{n_i^2}-\frac{1}{n_f^2}}) + = 13.6\U{eV} \cdot \p({\frac{1}{2^2} - \frac{1}{3^2}}) + = \ans{1.89\U{eV}} \;, +\end{align} +where $R_H=1.097\E{7}\U{m$^{-1}$}$. + +\Part{b} +Inverting the Rydberg equation +\begin{equation} + \lambda = \p[{R_H \p({\frac{1}{n_i^2}-\frac{1}{n_f^2}})}]^{-1} + = \ans{656\U{nm}} \;. +\end{equation} +Alternatively, you could use the energy from \Part{a} +\begin{equation} + \lambda = \frac{hc}{E} = \frac{1240\U{eV$\cdot$nm}}{1.89\U{eV}} + = \ans{656\U{nm}} \;. +\end{equation} + +\Part{c} +Using the wavelength from \Part{b} +\begin{equation} + f = \frac{c}{\lambda} = \frac{3.00\E{8}\U{m/s}}{656\U{nm}} + = \ans{457\U{THz}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem11.35.tex b/latex/problems/problem11.35.tex new file mode 100644 index 0000000..a85399e --- /dev/null +++ b/latex/problems/problem11.35.tex @@ -0,0 +1,34 @@ +\begin{problem*}{11.35} +\Part{a} +What value of $n_i$ is associated with the $94.96\U{nm}$ spectral line +in the Lyman series of hydrogen? +\Part{b} +Could this wavelength be associated with the Paschen series or the +Balmer series? +\end{problem*} % problem 11.35 + +\begin{solution} +\Part{a} +The Lyman series transitions all end in the ground state $n_f=1$ +Using the generalized Rydberg equation (Equation 11.28) +\begin{align} + \frac{1}{\lambda} &= R_\text{H} \p({\frac{1}{n_f^2} - \frac{1}{n_i^2}}) \\ + -\frac{1}{n_i^2} &= \frac{1}{\lambda R_\text{H}} - \frac{1}{n_f^2} \\ + n_i &= \p({\frac{1}{n_f^2} - \frac{1}{\lambda R_\text{H}}})^{-1/2} + = \p({1 - \frac{1}{94.96\E{-9}\U{m} \cdot 1.10\E{7}\U{m$^{-1}$}}})^{-1/2} + = \ans{5} \;. +\end{align} + +\Part{b} +This wavelength cannot have come from the Balmer ($n_f=2$) or Paschen +($n_f=3$) series because the shortest wavelength for any series is given +in the limit that $n_i \rightarrow \infty$ +\begin{align} + \frac{1}{\lambda_\text{min}} &= \frac{R_\text{H}}{n_f^2} \\ + \lambda_\text{min} &= \frac{n_f^2}{R_\text{H}} \;. +\end{align} +For the Balmer series $\lambda_\text{min} = 365\U{nm}$, and for the +Paschen series $\lambda_\text{min} = 820\U{nm}$. Both of these +series-minimum wavelengths are larger than the wavelength of our +spectral line. +\end{solution} diff --git a/latex/problems/problem11.36.tex b/latex/problems/problem11.36.tex new file mode 100644 index 0000000..9cf283e --- /dev/null +++ b/latex/problems/problem11.36.tex @@ -0,0 +1,35 @@ +\begin{problem*}{11.36} +For a hydrogen atom in its ground state, use the Bohr model to +compute \Part{a} the orbital speed of the electron, \Part{b} the +kinetic energy of the electron, and \Part{c} the electric potential +energy of the atom. +\end{problem*} % problem 11.36 + +\begin{solution} +\Part{a} +The Bohr radius is given by +\begin{align} + r_n &= \frac{n^2\hbar^2}{mke^2} = n^2 r_1 \\ + r_1 &= 0.529\E{-10}\U{m} +\end{align} +where $\hbar=h/(2\pi)=1.055\E{-34}\U{J$\cdot$s}$ and +$k=9.00\E{9}\U{Nm$^2$/C$^2$}$. So the velocity is +\begin{align} + v_n &= \frac{n\hbar}{mr_n} \\ + v_1 &= \frac{\hbar}{mr_1} = \ans{2.19\U{Mm/s}} \;. +\end{align} +\begin{equation} +\end{equation} + +\Part{b} +The kinetic energy is then +\begin{equation} + K = \frac{1}{2} m v^2 = \ans{13.6\U{eV}} \;. +\end{equation} + +\Part{c} +The electric potential energy is +\begin{equation} + U = -k\frac{e^2}{r} = -27.2\U{eV} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem11.37.tex b/latex/problems/problem11.37.tex new file mode 100644 index 0000000..5b0f6ce --- /dev/null +++ b/latex/problems/problem11.37.tex @@ -0,0 +1,45 @@ +\begin{problem*}{11.37} +Four possible transitions for a hydrogen atom are as follows: +\begin{center} +\begin{tabular}{l l c l l} + (i) & $n_i=2$; $n_f=5$ && (ii) & $n_i=5$; $n_f=3$ \\ + (iii) & $n_i=7$; $n_f=4$ && (iv) & $n_i=4$; $n_f=7$ +\end{tabular} +\end{center} +\Part{a} In which transition is light of the shortest wavelength emitted? +\Part{b} In which transition does the atom gain the most energy? +\Part{c} In which transition(s) does the atom lose energy? +\end{problem*} % problem 11.37 + +\begin{solution} +\Part{a} +Looking at the sized of the changes in $n$, we have +\begin{center} +\begin{tabular}{l l c l l} + (i) & $\Delta n=3$ && (ii) & $\Delta n=-2$ \\ + (iii) & $\Delta n=-3$ && (iv) & $\Delta n=3$ +\end{tabular} +\end{center} +We know we want to release energy, so we need $\Delta n < 0$ (relaxing +from a more excited state). That leaves (ii) and (iii). We have to +crunch some numbers for this, since each case has a point in its +favor: (iii) jumps more levels, but (ii) is closer in, where the +levels are further apart. +\begin{align} + E_{ii} &= A \p({\frac{1}{5^2}-\frac{1}{3^2}}) = -71\E{-3}\cdot A \\ + E_{iii} &= A \p({\frac{1}{7^2}-\frac{1}{4^2}}) = -42\E{-3}\cdot A \;, +\end{align} +so \ans{(ii)} releases the most energy. Note that +$A=hcR_H=13.6\U{eV}$, but it's actual value doesn't matter because it +is a constant value for hydrogen. + +\Part{b} +Now we need to compare (i) and (iv), but this is easier, since they +both increase by three levels, but (i) starts closer in (where the +levels are further apart). Therefore, \ans{(i)} will gain the most +energy. + +\Part{c} +The atom loses energy in \ans{(ii)} and \ans{(iii)} as mentioned +in \Part{a}. +\end{solution} diff --git a/latex/problems/problem11.38.tex b/latex/problems/problem11.38.tex new file mode 100644 index 0000000..e016c76 --- /dev/null +++ b/latex/problems/problem11.38.tex @@ -0,0 +1,18 @@ +\begin{problem*}{11.38} +How much energy is required to ionize hydrogen \Part{a} when it is in +the ground state and \Part{b} when it is in the state for which $n=3$? +\end{problem*} % problem 11.38 + +\begin{solution} +Ionization can be considered to be the state where $n_f=\infty$. +Using the Rydberg equation and $E=hc/\lambda$ we have +\begin{center} +\begin{tabular}{r r r} + & \Part{a} & \Part{b} \\ +$n_i$ & $1$ & $3$ \\ +$E_\text{absorbed}=hcR_H\p({\frac{1}{n_i^2}-\frac{1}{n_f^2}})$ + & $\ans{13.6\U{eV}}$ + & $\ans{1.51\U{eV}}$ +\end{tabular} +\end{center} +\end{solution} diff --git a/latex/problems/problem11.41.tex b/latex/problems/problem11.41.tex new file mode 100644 index 0000000..0b65edf --- /dev/null +++ b/latex/problems/problem11.41.tex @@ -0,0 +1,29 @@ +\begin{problem*}{11.41} +Two hydrogen atoms collide head-on and end up with zero kinetic +energy. Each atom then emits light with a wavelength of $121.6\U{nm}$ +($n=2$ to $n=1$ transition). At what speed where the atoms moving +before the collision? +\end{problem*} % problem 11.41 + +\begin{solution} +Ending up with zero kinetic energy after the collision means the atoms +must have zero velocity and momentum after the collision as well. +Conserving momentum, we see that having zero momentum after the +collision, means their momenta before the collision must have been +equal and opposite. Because both atoms are hydrogens, their masses +are the same, so their velocities must have also been equal and +opposite. Thus, their initial kinetic energies must have been equal. + +Conserving energy +\begin{align} + E_i &= 2 \cdot \frac{1}{2} m v_i^2 = m v_i^2 + = E_f = 2\cdot\frac{hc}{\lambda} \\ + v_i &= \sqrt{\frac{2hc}{\lambda m}} + = \sqrt{\frac{2\cdot6.63\E{-34}\U{J$\cdot$s}\cdot3.00\E{8}\U{m/s}} + {121.6\E{-9}\U{m}\cdot1.67\E{-27}\U{kg}}} + = \ans{44.3\U{km/s}} \;. % ((2*6.63e-34*3e8)/(121.6e-9*1.67e-27))**.5 +\end{align} +Note that the mass of a hydrogen atom is the sum of the masses of a +proton and electron, but $m_e\ll m_p$, which is why $m_H\approx +m_p=1.67\E{-27}\U{kg}$. +\end{solution} diff --git a/latex/problems/problem11.V1.tex b/latex/problems/problem11.V1.tex new file mode 100644 index 0000000..4443434 --- /dev/null +++ b/latex/problems/problem11.V1.tex @@ -0,0 +1,191 @@ +\begin{problem} +A hypothetical particle (a ``drexelon'') is captured by a proton to +form a ``drexelonic atom''. A drexelon is identical to an electron +except for its rest mass, which is $m_d=117.3\U{MeV/$c^2$}$. Note +that the rest mass of an electron is $m_e=0.511\U{keV/$c^2$}$. A +proton is approximately eight times more massive than a drexelon. It +is possible to use the equations of the Bohr model of Hydrogen to +discuss the drexelonic atom. Since the mass of the nucleus is of the +same order as that of the drexelon, you should use the reduced mass +$m_d'=m_dM_p/(m_d+m_P)$ for the rotational mass of the drexelon. The +mass of a proton is $M_p=938.28\U{MeV/$c^2$}$. + +For the first Bohr-orbit of the drexelonic atom, determine +\Part{a} the radius of the orbit, +\Part{b} the speed (in m/s) of the drexelon. +\Part{c} the kinetic energy (in eV) of the drexelon, and +\Part{d} the electrostatic potential energy (in eV) of the drexelon. + +\Part{e} What is the total energy (in eV) of the drexelon in the +first exited state of the drexelonic atom? +\Part{f} What is the smallest wavelength (in nm) of radiation in the +Balmer series of the drexelonic atom? +\Part{g} What is the wavelength (in nm) of the least energetic photon +in the Balmer series of the drexelonic atom? + + +\Part{h} Determine the largest wavelength (in nm) of the photons +emitted in the Lyman series of the drexelonic atom. +For this emitted photon, determine +\Part{i} the linear momentum (in eV/c) and +\Part{j} the angular momentum (in units of $\hbar=h/2\pi$, assuming +the angular momentum of the system is conserved). + +The linear momentum of the photon is $E/c$. If we assume the +conservation of linear momentum, find for the atom emitting the photon +\Part{k} the velocity (in multiples of $c$) and +\Part{l} the recoil kinetic energy (in eV). + +\Part{m} What is the ratio of recoil kinetic energy to the energy of +the emitted photon? Do we have to correct the energy of the emitted +photon to account for this recoil energy? What percentage energy +correction is needed? +\end{problem} % Prof. Venkat homework problem 4.1 + +\begin{solution} +\Part{a} +First we find the reduced mass of the drexelonic atom in kg. +\begin{equation} + m_d' = \frac{m_d M_p}{m_d+M_p} + = \frac{117.3\cdot938.28}{117.3+938.28}\U{MeV/$c^2$} + = 104.3\U{MeV/$c^2$} + = 104.3\E{6}\U{eV}\cdot\frac{1.60\E{-19}\U{J/eV}}{(3.00\E{8}\U{m/s})^2} + = 1.86\E{-28}\U{kg} +\end{equation} +The radius of the first orbital is then +\begin{align} + r_n &= \frac{n^2\hbar^2}{m_d'Zke^2} \\ + r_1 &= \frac{\hbar^2}{m_d'ke^2} + = \frac{\hbar^2}{m_d'ke^2} + = \frac{(1.05\E{-34}\U{J$\cdot$s})^2}{1.86\E{-28}\U{kg}\cdot8.99e9\U{Nm$^2$/$C^2$}\cdot(1.60\E{P-19}\U{C})^2} + = \ans{2.59\E{-13}\U{m}} +\end{align} + +\Part{b} +From the Bohr-model assumptions, angular momentum comes in chunks of $\hbar$ +\begin{align} + m_d'v_n'r_n &= n\hbar \\ + v_1' &= \frac{\hbar}{m_d'r_1} + = \frac{1.05\E{-34}\U{J$\cdot$s}}{1.86\E{-28}\U{kg}\cdot2.59\E{-13}\U{m}} + = 2.19\U{Mm/s} +\end{align} +$v_1'$ is the speed of the reduced mass particle, not quite that of +the drexelon. We can get the speed of the drexelon with +\begin{equation} + v_1 = v_1'\frac{r_d}{r} + = v_1'\frac{M_p}{m_d+M_p} + = 2.19\U{Mm/s}\cdot\frac{8}{1+8} + = \ans{1.94\U{Mm/s}} +\end{equation} + +\Part{c} +As in \Part{b}, we're only interested in the \emph{drexelon's} kinetic +energy, so we use the drexelon's full mass. Using the reduced mass +$m_d'$ and its associated velocity $v_1'$ would give the kinetic +energy of the whole atom (drexelon + proton). +\begin{align} + m_d &= 117.3\E{6}\U{eV}\cdot\frac{1.60\E{-19}\U{J/eV}}{(3.00\E{8}\U{m/s})^2} + = 2.09\E{-28}\U{kg} \\ + K &= \frac{1}{2} m_d v_1^2 + = \frac{1}{2}\cdot2.09\E{-28}\U{kg}\cdot(1.94\U{Mm/s})^2 + = 3.95\E{-16}\U{J}\cdot\frac{1}{1.60\E{-19}\U{J/eV}} + = \ans{2.47\U{keV}} +\end{align} + + +\Part{d} +Using the equation for the potential of a point charge $V=kq/r$, we +have +\begin{equation} + U_1= -eV_1 = \frac{-kZe^2}{r_1} + = \frac{-8.99\E{9}\U{Nm$^2$/C$^2$}\cdot(1.60\E{-19}\U{C})^2}{2.59\E{-13}\U{m}} + = -8.90\E{-16}\U{J}\cdot\frac{1}{1.60\E{-19}\U{J/eV}} + = \ans{-5.55\U{keV}} +\end{equation} + +\Part{e} +From the Bohr-model +\begin{align} + E_n &= \frac{-m_d'(kZe^2)^2}{2n^2\hbar^2} \\ + E_2 &= \frac{-m_d'(ke^2)^2}{8\hbar^2} + = \frac{-1.86\E{-28}\U{kg}[8.99\E{9}\U{Nm$^2$/C$^2$}\cdot(1.60\E{-19}\U{C})^2]^2}{8(1.05\E{-34}\U{J$\cdot$s})^2} + = -1.11\E{-16}\U{J}\cdot\frac{1}{1.60\E{-19}\U{J/eV}} + = \ans{-693\U{eV}} +\end{align} + +\Part{f} +For the Balmer series, $n_f=2$. The smallest wavelength comes from +the largest energy transition, so $n_i=\infty$. The wavelength is then +\begin{align} + E &= E_\infty-E_2 = -E_2 = 693\U{eV} \\ + \lambda &= \frac{hc}{E} + = \frac{1240\U{eV$\cdot$nm}}{693\U{eV}} = \ans{1.79\U{nm}} +\end{align} + +\Part{g} +The least energetic photon comes from the smallest energy transition, +so $n_i=n_f+1=3$. +\begin{align} + E &= E_3-E_2 + = \frac{-m_d'(ke^2)^2}{2\hbar^2}\p({\frac{1}{3^2}-\frac{1}{2^2}}) + = 385\U{eV} \\ + \lambda &= \frac{hc}{E} + = \frac{1240\U{eV$\cdot$nm}}{385\U{eV}} = \ans{3.22\U{nm}} +\end{align} + +\Part{h} +For the Lyman series, $n_f=1$. The largest wavelength photon comes +from the smallest energy transition, so $n_i=n_f+1=2$. +\begin{align} + E &= E_2-E_1 + = \frac{-m_d'(ke^2)^2}{2\hbar^2}\p({\frac{1}{2^2}-\frac{1}{1^2}}) + = 2.08\U{keV} \\ + \lambda &= \frac{hc}{E} + = \frac{1240\U{eV$\cdot$nm}}{2.08\U{keV}} = \ans{0.596\U{nm}} +\end{align} + +\Part{i} +The linear momentum of the photon is given by +\begin{equation} + p = \frac{E}{c} = \ans{2.08\U{keV/$c$}} +\end{equation} + +\Part{j} +Conserving angular momentum, +\begin{align} + L_i &= 2\hbar = L_f = \hbar + L_\text{photon} \\ + L_\text{photon} &= \ans{\hbar} +\end{align} +Remember that the Bohr model \emph{assumes} that the angular momentum +of the atom comes in chunks of $\hbar$ ($L_n=m'v_n'r_n'=n\hbar$). + +\Part{k} +If the photon leaves with momentum $p_\text{photon} = 2.08\U{keV/$c$}$ +(from \Part{h}), the atom must have the same momentum in the opposite +direction, so +\begin{align} + p &= mv \\ + v &= \frac{p}{m} + = \frac{2.08\U{keV/$c$}}{(938.28+117.3)\U{MeV/$c^2$}} + = \ans{1.97\E{-6}c} = 591\U{m/s} +\end{align} +Note that we are dealing with the momentum of the \emph{atom}, so we +use the full mass of the atom, not the reduced mass which we had used +to understand the atom's internal orbits. + +\Part{l} +The recoil kinetic energy is +\begin{equation} + K = \frac{1}{2}mv^2 + = \frac{1}{2}(938.28+117.3)\U{MeV/$c^2$}\cdot(1.97\E{-6}c)^2 + = \ans{2.05\U{meV}} +\end{equation} + +\Part{m} +The recoil kinetic energy is small enough to ignore +\begin{equation} + \frac{K}{E_\text{photon}} = \frac{2.05\U{meV}}{2.08\U{keV}} + = \ans{9.86\E{-7}} +\end{equation} +This is a $\ans{9.86\E{-5}\%}$ correction. +\end{solution} diff --git a/latex/problems/problem11.V2.tex b/latex/problems/problem11.V2.tex new file mode 100644 index 0000000..34a505a --- /dev/null +++ b/latex/problems/problem11.V2.tex @@ -0,0 +1,37 @@ +\begin{problem} +In a Lithium atom (atomic number $Z=3$) there are two electrons in the +first orbit, and, due to Pauli's exclusion principle (Serway Ch.~29, +Sec.~5), the third electron is in the second orbit $n=2$. The +interaction of the inner electrons with the one in the second orbit +can be approximated by writing the energy of the outer electron +$E_n'=-Z'^2E_1/n^2$, where $E_1=13.6\U{eV}$, $n=2$, and $Z'$ is the +effective atomic number. Note that $Z'$ is less than three because of +the screening effect of the two electrons in the first orbit. +\Part{a} If we need $5.39\U{eV}$ of energy to remove the outer +electron from its binding with the nucleus, determine $Z'$. +\Part{b} What would be the longest wavelength (in nm) of the photon +that can be absorbed by the electron in the $n=2$ state? +\end{problem} % Prof. Venkat homework problem 4.2 + +\begin{solution} +\Part{a} +\begin{align} + E' &= E_\infty'-E_2' = -E_2' = \frac{-Z'^2 E_1}{2^2} \\ %\frac{m_e(kZ'e^2)^2}{2\cdot2^2\hbar^2} \\ + Z' &= 2\sqrt{\frac{E'}{E_1}} %\frac{2\hbar}{ke^2}\cdot\sqrt{\frac{2E}{m_e}} + = 2\sqrt{\frac{5.39\U{eV}}{13.6\U{eV}}} + = \ans{1.26} +\end{align} +$Z'<3$ which is good, since there are only three protons available. +$Z'>1$ which is also good, because the screening of the inner two +electrons shouldn't be perfect. + +\Part{b} +The longest wavelength photon comes from the smallest energy +transition, so $n_i=n_f+1=3$. +\begin{align} + E' &= -(E_2'-E_3') = Z'^2E_1 \p({\frac{1}{2^2}-\frac{1}{3^2}}) + = 1.26^2\cdot13.6\U{eV} \p({\frac{1}{4}-\frac{1}{9}}) + = 2.99\U{eV} \\ + \lambda &= \frac{hc}{E'} = \ans{414\U{nm}} +\end{align} +\end{solution} diff --git a/latex/problems/problem12.01.tex b/latex/problems/problem12.01.tex new file mode 100644 index 0000000..d21b665 --- /dev/null +++ b/latex/problems/problem12.01.tex @@ -0,0 +1,10 @@ +\begin{problem*}{12.1} +A ball dropped from a height of $4.00\U{m}$ makes an elastic collision +with the ground. Assuming that no mechanical energy is lost due to +air resistance, \Part{a} show that the ensuing motion is periodic +and \Part{b} determine the period of the motion. \Part{c} Is the +motion simple harmonic? Explain. +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/problem12.02.tex b/latex/problems/problem12.02.tex new file mode 100644 index 0000000..ceb9942 --- /dev/null +++ b/latex/problems/problem12.02.tex @@ -0,0 +1,40 @@ +\begin{problem*}{12.2} +In an engine, a piston oscillates with simple harmonic motion so that +its position varies according to the expression +\begin{equation} + x = (5.00\U{cm}) \cos(2t + \pi/6) +\end{equation} +where $x$ is in centimeters and $t$ is in seconds. At $t = 0$, +find \Part{a} the position of the pistion, \Part{b} its velocity, +and \Part{c} its acceleration. \Part{d} Find the period and amplitude +of the motion. +\end{problem*} + +\begin{solution} +\Part{a} +Simply plugging in $t=0$ into the given position equation we have +\begin{equation} + x(t=0) = 5.00\U{cm} \cdot \cos(\pi/6) = \ans{4.33\U{cm}} +\end{equation} + +\Part{b} +Then we take the derivative to get the velocity +\begin{align} + v &= \pderiv{t}{x} = -2\U{s$^{-1}$} \cdot 5.00\U{cm} \cdot \sin(2t+\pi/6) \\ + v(t=0) &= -10\U{cm/s} \cdot \sin(\pi/6) = \ans{-5.00\U{cm/s}} +\end{align} + +\Part{c} +Taking the derivative again yields the acceleration +\begin{align} + a &= \pderiv{v}{x} = -2\U{s$^{-1}$}\cdot 10.00\U{cm/s}\cdot \cos(2t+\pi/6) \\ + a(t=0) &= -20\U{cm/s$^2$} \cdot \cos(\pi/6) = \ans{-17.3\U{cm/s$^2$}} +\end{align} + +\Part{d} +The period is the time it takes for a complete revolution, i.e.~for a +phase change of $2\pi$, so $\ans{T = \pi\U{s}}$. The amplitude is the +maximum distance from equilibrium. Looking at the original equation +for $x(t)$, we realize that $|\cos(\theta)| \le 1$ for all $\theta$, +so the biggest $x$ will ever get is $\ans{A = 5.00\U{cm}}$. +\end{solution} diff --git a/latex/problems/problem12.05.tex b/latex/problems/problem12.05.tex new file mode 100644 index 0000000..5edcce3 --- /dev/null +++ b/latex/problems/problem12.05.tex @@ -0,0 +1,72 @@ +\begin{problem*}{12.5} +A particle moving along the $x$ axis in simple harmonic motion starts +from its equilibrium position, the origin, at $t=0$ and moves to the +right. The amplitude of its motion is $2.00\U{cm}$ and the frequency +is $1.50\U{Hz}$. \Part{a} show that the position of the particle is +given by +\begin{equation} + x = (2.00\U{cm}) \sin(3.00\pi t) +\end{equation} +Determine \Part{b} the maximum speed and the earliest time ($t > 0$) +at which the particle has this speed, \Part{c} the maximum +acceleration and the earliest time ($t > 0$) at which the particle has +this acceleration, and \Part{d} the total distance traveled between +$t=0$ and $t = 1.00\U{s}$. +\end{problem*} + +\begin{solution} +\Part{a} +Because $x(t=0) = 0$, we can express the motion +\begin{equation} + x(t) = A \sin(\omega t) \;. +\end{equation} +(this is Equation 12.6 with $\phi = -\pi/2$, because +$\cos(\theta-\pi/2) = \sin(\theta)$.) + +To find $A$, note that $\sin(\theta)$ increases as $\theta$ increases +from $0$, so the particle's initially rightward motion requires $A > +0$. $|\sin(\theta)| \le 1$ so $|x| \le A$, and the amplitude is gives +as $2.00\U{cm}$ so $A = 2.00\U{cm}$. + +To find $\omega$, simply compute +\begin{equation} + \omega = 2\pi f = 2\pi\cdot1.50\U{Hz} = 3\pi\U{rad/s} \;. +\end{equation} + +Plugging our $A$ and $\omega$ into our $x(t)$ yields the equation of +motion we set out to find. + +\Part{b} +To find the maximum speed, we could either take the derivative of +$x(t)$ (like we did in 12.2), or realize that the derivative will have +another factor of $\omega$ in it's amplitude and jump to the answer +$v_\text{max}=A\omega=\ans{6\pi\U{cm/s}}$. + +The maximum speed occurs when the position is zero. Our particle +starts at $x=0$, so it has maximum speed at $t = 0, T/2, T, 3T/2, +\ldots$. We're asked for the first occurence for $t>0$, so +$t=T/2=1/2f=\ans{0.333\U{s}}$ + +\Part{c} +To find the maximum acceleration, we could either take the derivative +of $v(t)$ (like we did in 12.2), or realize that the derivative will +have another factor of $\omega$ in it's amplitude compared to the +velocity and jump to the answer +$a_\text{max}=A\omega^2=\ans{18\pi^2\U{cm/s$^2$}}$. + +The maximum acceleration occurs at the minimum position, because +$F=ma=-kx$. Our particle starts at $x=0$, so it at minimum extension +at $t = 3T/4, 7T/4, \ldots$. We're asked for the first +occurence for $t>0$, so $t=3T/4=3/4f=\ans{0.500\U{s}}$ + +Note that in \Part{b} we were looking for the maximum scalar +\emph{speed}, so the direction didn't matter, but in \Part{c} we were +looking for the maximum vector \emph{acceleration}, so the direction +did matter. + +\Part{d} +The period of our particle is $T = 1/f = 2/3 \U{s}$. $t = 1.00\U{s} = +1.5T$. That means it travels $0 \rightarrow A \rightarrow -A +\rightarrow A \rightarrow 0$, for a grand total of +$d=A+2A+2A+A=6A=12.0\U{cm}$. +\end{solution} diff --git a/latex/problems/problem12.06.T.tex b/latex/problems/problem12.06.T.tex new file mode 100644 index 0000000..e57f6ec --- /dev/null +++ b/latex/problems/problem12.06.T.tex @@ -0,0 +1,85 @@ +\begin{problem} +A thin, rigid rod $L = 8.4\U{m}$ long pivots freely about one end. +The rod is initially deflected $\theta_i = 6.4\dg$ from the vertical +with an angular velocity of $\dt\theta_i = 2.7\dg\text{/s}$. +\Part{a} Determine the time dependence $\theta(t)$. +\Part{b} By what angle is the rod deflected at $t=8.9\U{s}$? + +Hint: you might want to review torque and moments of inertia in +Chapter 10. +\end{problem} % combines the second part of P12.6 with Examp 12.6 and Tbl 10.2. + +\begin{solution} +\begin{center} +\begin{asy} +import Mechanics; +real u = 1cm; + +real a=6.4; // degrees +real force=2u; // magnitude +Pendulum p = makePendulum(angleDeg=a, length=2u, stringL="$L$"); +p.mass.radius = 0; +Vector fg = Force(p.mass.center/2, dir=-90, mag=force, L="$F_g$"); +Vector fgtan = Force(p.mass.center/2, dir=a-180, mag=force*sin(a), L=Label("$F_{\tan}$")); +Vector v = Velocity(p.mass.center, dir=a, mag=0.5u, L="$v$"); + +fg.draw(); +fgtan.draw(labelOffset=(-2mm,1mm)); +v.draw(); +p.draw(drawVertical=true); +\end{asy} +\end{center} +The only force on the rod is from gravity, with $mg$ pulling the rods +center of mass downward. Only the portion of this force that is +perpendicular to the rod itself (tangential to the circle the rod +sweeps out) affects its rotation. The torque on the rod is thus +\begin{equation} + \tau = -F_{\tan} \cdot \frac{L}{2} + = -F_g \sin(\theta) \cdot \frac{L}{2} + = -\frac{mgL}{2}\sin(\theta) + \approx -\frac{mgL}{2}\theta\;, +\end{equation} +where we used the small angle approximation $\sin(\theta) \approx +\theta$ for the last step. + +The equation of motion is then +\begin{equation} + \tau = I\ddt\theta = \frac{1}{3}mL^2\ddt\theta\;, +\end{equation} +because the moment of inertia of a rod rotating about it's end is $I = +\frac{1}{3}mL^2$ (Table 10.2). + +Combining the two expressions of $\tau$ we have +\begin{align} + -\frac{mgL}{2}\theta &= \frac{1}{3}mL^2\ddt\theta \\ + -\frac{3g}{2L}\theta &= \ddt\theta\;. +\end{align} +Comparing this formula to Equation 12.5 for a general simple harmonic oscillator +\begin{equation} + \ddt x = -\omega^2 x\;, +\end{equation} +we see by matching that +\begin{equation} + \omega = \sqrt{\frac{3g}{2L}} \approx \ans{1.323\U{rad/s}}\;. +\end{equation} +We can plug this $\omega$ into Equation 12.6 +\begin{equation} + \theta(t) = A \cos(\omega t + \phi)\;, +\end{equation} +where $A$ and $\psi$ are determined by the initial conditions (see Example 12.3) +\begin{align} + \theta_i &= 6.4\dg\text{/s} \cdot \frac{\pi}{180\dg} \approx 111.7\U{mrad} \\ + \dt\theta_i &= 2.7\dg\text{/s} \cdot \frac{\pi}{180\dg} \approx 47.12\U{mrad/s} \\ + \phi &= \arctan\p({\frac{-\dt\theta_i}{\omega\theta_i}}) \approx -308.7\U{mrad} \approx -17.69\dg \\ + A &= \sqrt{\theta_i^2 + \p({\frac{\dt\theta_i}{\omega}})^2} \approx 117.2\U{mrad} \approx 6.718\dg\\ + \theta(t) &\approx \ans{0.1172\cos\p({1.323t-0.3087})}\;. +\end{align} + +\Part{b} +Plugging in $t=8.9\U{s}$ yields +\begin{equation} + \theta(t=8.9\U{s}) = 0.1172\cos\p({1.323 \cdot 8.9 - 0.3087}) + \approx \ans{53\U{mrad}} + \approx \ans{3.0\dg}\;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem12.07.tex b/latex/problems/problem12.07.tex new file mode 100644 index 0000000..2e0db53 --- /dev/null +++ b/latex/problems/problem12.07.tex @@ -0,0 +1,17 @@ +\begin{problem*}{12.7} +The initial position, velocity, and acceleration of an objectmoving in +simple harmonic motion are $x_i$, $v_i$, and $a_i$; the angular +frequency of oscillation is $\omega$. \Part{a} Show that the position +and velocity of the object for all time can be written as +\begin{align} + x(t) &= x_i \cos \omega t + \p({\frac{v_i}{\omega}}) \sin \omega t \;, \\ + v(t) &= -x_i\omega \sin \omega t + v_i \cos \omega t \;. +\end{align} +\Part{b} Using $A$ to represent the amplitude of the motion, show that +\begin{equation} + v^2 - ax = v_i^2 - a_i x_i = \omega^2 A^2 \;. +\end{equation} +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/problem12.12.tex b/latex/problems/problem12.12.tex new file mode 100644 index 0000000..aba65c9 --- /dev/null +++ b/latex/problems/problem12.12.tex @@ -0,0 +1,41 @@ +\begin{problem*}{12.12} +A $1.00\U{kg}$ glider attached to a spring with a force constant of +$25.0\U{N/m}$ oscillates on a horizontal, frictionless air track. At +$t=0$, the glider is released from rest at $x=-3.00\U{cm}$ (that is, +the spring is compressed by $3.00\U{cm}$). Find \Part{a} the period +of its motion, \Part{b} the maximum values of its speed and +acceleration, and \Part{c} the position, velocity, and acceleration as +functions of time. +\end{problem*} + +\begin{solution} +\Part{a} +\begin{equation} + \omega = \sqrt{\frac{k}{m}} = \ans{5.00\U{N/m}} \;, +\end{equation} +and +\begin{equation} + T = \frac{1}{f} = \frac{2\pi}{\omega} = \ans{1.26\U{s}} \;. +\end{equation} + +\Part{b} +The maximum speed of the object is (see 12.5) +\begin{equation} + v_\text{max} = \omega A = \ans{15.0\U{cm/s}} \;, +\end{equation} +where $A = 3.00\U{cm}$. + +The maximum acceleration of the object is (see 12.5) +\begin{equation} + a_\text{max} = \omega^2 A = \ans{75.0\U{cm/s$^2$}} \;. +\end{equation} + +\Part{c} +The position starts at the minimum value of $x$, so a $-\cos$ based +expression for $x$ is in order +\begin{align} + x &= -A \cos(\omega t) = -A \cos(\omega t) \\ + v &= \omega A\sin(\omega t) = v_\text{max} \sin(\omega t) \\ + a &= \omega^2 A\cos(\omega t) = a_\text{max} \cos(\omega t) \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem12.15.tex b/latex/problems/problem12.15.tex new file mode 100644 index 0000000..9a2a171 --- /dev/null +++ b/latex/problems/problem12.15.tex @@ -0,0 +1,35 @@ +\begin{problem*}{12.15} +A block of unknown mass is attached to a spring with a spring constant +of $6.50\U{N/m}$ and undergoes simple harmonic motion with an +amplitude of $10.0\U{cm}$. When the block is halfway between its +equilibrium position and the end point, its speed is measured to be +$30.0\U{cm/s}$. Calculate \Part{a} the mass of the block, \Part{b} +the period of the motion, and \Part{c} the maximum acceleration of the +block. +\end{problem*} + +\begin{solution} +\Part{a} +Conserving energy +\begin{align} + E &= \frac{1}{2} k A^2 = 32.5\U{mJ} \\ + E &= \frac{1}{2} k \p({\frac{A}{2}})^2 + \frac{1}{2} m v^2 + = \frac{E}{4} + \frac{1}{2} m v^2 \\ + m &= \frac{2}{v^2} \cdot \frac{3}{4}E = \frac{3E}{2v^2} + = \ans{542\U{g}} \;. +\end{align} + +\Part{b} +\begin{equation} + T = \frac{1}{f} = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}} + = \ans{1.81\U{s}} \;. +\end{equation} + +\Part{c} +This is just Hooke's law +\begin{align} + F &= ma_\text{max} = kA \\ + a &= \frac{k}{m}A = \ans{1.20\U{m/s$^2$}} \;. +\end{align} + +\end{solution} diff --git a/latex/problems/problem12.18.tex b/latex/problems/problem12.18.tex new file mode 100644 index 0000000..1e6b4ce --- /dev/null +++ b/latex/problems/problem12.18.tex @@ -0,0 +1,98 @@ +\begin{problem*}{12.18} +A $2.00\U{kg}$ object is attached to a spring and placed on a +horizontal, smooth surface. A horizontal force of $20.0\U{N}$ is +required to hold the object at rest when it is pulled $0.200\U{m}$ +from its equilibrium position (the origin of the $x$ axis). The +object is now released from rest with an initial position of +$x_i=0.200\U{m}$, and it subsequently undergoes simple harmonic +oscillations. Find \Part{a} the force constant of the +spring, \Part{b} the frequency of oscillations, and \Part{c} the +maximum speed of the object. Where does this maximum speed +occur? \Part{d} Find the maximum acceleration of the object. Where +does it occur? \Part{e} Find the total energy of the oscillating +system. Find \Part{f} the speed and \Part{g} the acceleration of the +object when its position is equal to one third of the maximum value. +\end{problem*} + +\begin{solution} +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1cm; +real a = u/2; + +Surface surf = Surface(pFrom=(-.7u,-a/2), pTo=(2u,-a/2)); +Block b = Block(center=(0,0), side=a); +Spring spring = Spring(pFrom=(b.center+(a/2,0)), pTo=(2u,0)); +Vector Fspring = Force(b.center, mag=2a, dir=0, L="$F_s$"); +Vector Fexternal = Force(b.center, mag=2a, dir=180, L="$F_e$"); + +Fspring.draw(); +Fexternal.draw(); +surf.draw(); +spring.draw(); +b.draw(); +\end{asy} +\end{center} +\Part{a} +The external force $F_e$ must exactly balance the spring force $F_s$ +to hold the object at rest on a frictionless surface, so +\begin{align} + |F_s| &= k|x| = |F_e| \\ + k &= \p|{\frac{F_e}{x}}| = \ans{100\U{N/m}} \;. +\end{align} + +\Part{b} +The frequency of oscillation is then +\begin{equation} + f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{m}} + = \ans{1.13\U{Hz}} \;, +\end{equation} +and +\begin{equation} + \omega = 2 \pi f = 7.07\U{rad/s} \;. +\end{equation} + +\Part{c} +The maximum speed of the object is (see 12.5) +\begin{equation} + v_\text{max} = \omega A = 7.07\U{rad/s} \cdot 0.200\U{m} + = \ans{1.41\U{m/s}} \;, +\end{equation} +which occurs at $x=0$, when all the oscillation energy is kinetic. + +\Part{d} +The maximum acceleration of the object is (see 12.5) +\begin{equation} + a_\text{max} = \omega^2 A = (7.07\U{rad/s})^2 \cdot 0.200\U{m} + = \ans{10\U{m/s$^2$}} \;, +\end{equation} +which occurs at $x=-A=-0.200\U{m}$. If you are only interested in the +peaks in the \emph{magnitude} of the acceleration, they occur for +$x=\pm A$. + +\Part{e} +Lets find the energy in the initial situation, right after the object +was released. It's at rest, so its kinetic energy is $0$ and all the +enegy is spring-potential energy +\begin{equation} + E = \frac{1}{2} k A^2 = \ans{2.00\U{J}} +\end{equation} + +\Part{f} +Conserving energy +\begin{align} + E &= \frac{1}{2} k \p({\frac{A}{3}})^2 + \frac{1}{2} m v^2 + = \frac{E}{9} + \frac{1}{2} m v^2 \\ + v &= \pm\sqrt{\frac{2}{m} \cdot \frac{8}{9}E} = \ans{\pm1.33\U{m/s}} \;. +\end{align} + +\Part{g} +This is very similar to Problem 12.15 \Part{c}. +\begin{align} + F &= ma = kx \\ + a &= \frac{k}{m}x = \pm \frac{kA}{3m} = \ans{\pm3.33\U{m/s$^2$}} \;. +\end{align} + +\end{solution} diff --git a/latex/problems/problem12.20.tex b/latex/problems/problem12.20.tex new file mode 100644 index 0000000..80b3192 --- /dev/null +++ b/latex/problems/problem12.20.tex @@ -0,0 +1,163 @@ +\begin{problem*}{12.20} +A $65.0\U{kg}$ bungee jumper steps off a bridge with a light bungee +cord tied to her and the bridge (Figure P12.20). The unstretched +length of the cord is $11.0\U{m}$. She reaches the bottom of her +motion $36.0\U{m}$ below the bridge before bouncing back. Her motion +can be seperated into an $11.0\U{m}$ free-fall and a $25.0\U{m}$ +section of simple harmonic oscillation. \Part{a} For what time +interval is she in free-fall? \Part{b} Use the principle of +consevation of energy to find the spring constant of the bungee +cord. \Part{c} What is the location of the equilibrium point where +the spring force balances the gravitational force acting on the +jumper? Note that this point is takes as the origin in our +mathematical description of simple harmonic oscillation. \Part{d} +What is the angular frequency of the oscillation? \Part{e} What time +interval is required for the cord to stretch by $25.0\U{m}$? \Part{f} +What is the total time interval for the entire $36.0\U{m}$ drop? +\end{problem*} + +\begin{solution} +It always helps me to draw a picture of what's going on: +\begin{center} +\begin{asy} +import graph; +size(5cm,3cm, IgnoreAspect); + +real g = 9.8; // m/s^2, gravitational acceleration. +real m = 65; // kg, mass of jumper. +real L = 11; // m, length of the bungee cord. +real dH = 36; // m, peak-to-peak amplitude. +real tF = sqrt(2*L/g); //s, time to fall from peak to bungee-tension. y=-gt^2/2 +// E = mg*dH = 1/2 k Lmax^2 -> k = 2mg*dH/Lmax^2 +real dLmax = dH-L;// m, maximum stretch on bungee cord. +real k = 2*m*g*dH/dLmax**2; // N/m, bungee cord spring constant. +real dLeq = m*g / k; //m, stretch while still accelerating down. +real A = dLmax - dLeq; //m, amplitude of SHO. +real w = sqrt(k/m); //rad/s, angular speed of SHO. +real thetaEngage = asin(dLeq/A); //rad, phase angle with horizontal at dL=0. +real tS = (2*thetaEngage+pi)/w; //s, bungee-tension time per cycle; + +real y(real t) { // y=0 is the transition region, when dL=0; + // valid for t <= 3*tF + tS = second bungee phase + if (t <= 0) // standing on bridge + return L; + if (t <= tF) // free-fall + return L-g*t**2/2; + t -= tF; + if (t <= tS) // bungee oscillation + return -dLeq-A*sin(w*t-thetaEngage); + t -= tS; + return -g*t**2/2 + (g*tF)*t; // second parabola +} + +real ySHO(real t) { // if there was no free-fall section. + return -dLeq-A*sin(w*(t-tF)-thetaEngage); +} + +real tmin = -tF; +real tmax = 2.5*tF+tS; +//pair[] z={(tmin,0),(tmax,0)}; +//draw(graph(z), dashed); + +draw(graph(ySHO, tmin, tmax), blue); +draw(graph(y, tmin, tmax), red); +yequals(0, xmin=tmin, xmax=tmax, Dotted); +yequals(-dLeq, xmin=tmin, xmax=tmax, Dotted); +xequals(0, Dotted); +xequals(tF, Dotted); +xequals(tF+tS, Dotted); +\end{asy} +\end{center} +The dotted horizontal lines are, top to bottom, the position where the +bungee cord begins resisting the fall and the equilibrium position of +the simple harmonic oscillation. The dotted vertical lines are, left +to right, the times of the person jumping off the bridge, the bungee +cord beginning to resist the fall, and the bungee cord ceasing to +resist the fall as the jumper is launched back up into the air. The +red curve tracks the position of the jumper as a function of time, and +the blue curve extrapolates the simple harmonic oscillation to draw +attention to the difference between the simple harmonic oscillation +and free-fall portions of the actual trajectory. + +\Part{a} +The free-fall phase follows the parabolic $y = -\frac{1}{2}gt^2$ +behavior you all know and love from your Freshman Mechanics class. +This fall continues until the jumper drops the $L=11\U{m}$ needed to +take the slack out of the bungee cord. The time is thus +\begin{align} + -L &= -\frac{1}{2}gt_f^2 & t_f &= \sqrt{\frac{2L}{g}} = \ans{1.50\U{s}} \;. +\end{align} + +\Part{b} +We don't know anything about velocity yet, so lets conserve energy +between the points where the velocity is zero. Defining $h_b=0$ to be +the height of the lowest point in the trajectory, the energy at that +point is all spring potential energy. Of course, the energy at the +jumping-off point ($h_t=36\U{m}$)is all gravitational potential +energy. Setting these equal, we have +\begin{align} + E &= mgh_t = \frac{1}{2} k \Delta L^2 = \frac{1}{2} k \p({h_t-L})^2 \\ + k &= \frac{2mgh_t}{\p({h_t-L})^2} = \ans{73.4\U{N/m}} +\end{align} +where $\Delta L = h_t - L = 25\U{m}$ is the maximum stretch in the +bungee cord. Of course, in the real world energy is lost to heating +the bungee cord, shaking the jumper, etc., which is good, since +otherwise the jumper would bump into the bridge in the second +free-fall phase on the right side of the drawing above. + +\Part{c} +Here we just use Hooke's law and balance the vertical forces on the +jumper. +\begin{align} + mg &= k\Delta L_{eq} & \Delta L_{eq} &= \frac{mg}{k} = 8.68\U{m} \;, +\end{align} +where $\Delta L_{eq}$ is the distance from bungee-engage to equilibrium. +The equilibrium point is thus +\begin{align} + L + \Delta L_{eq} = \ans{19.7\U{m}} +\end{align} +below the bridge. + +\Part{d} +\begin{equation} + \omega = \sqrt{\frac{k}{m}} = \ans{1.06\U{rad/s}} +\end{equation} + +\Part{e} +Let's take a look at the reference circle for the harmonic oscillation +phase +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1.2cm; +real A = 16.32; +real dLeq = 8.68; +real theta = asin(dLeq/A); + +draw(scale(u)*unitcircle, dotted); +pair engage = (-u*cos(theta), u*sin(theta)); + +Angle a = Angle(engage, (0,0), (-1,0), radius=6mm, "$\theta$"); +a.draw(); +draw((0,0)--engage, red); +draw(engage--realmult(engage, (1,0)), blue); +label("$\Delta L_{eq}$", (engage.x, engage.y/2), W); +draw((0,0)--(-u,0), dashed); +draw((0,0)--(0,-u), blue); +label("$A$", (0,-u/2), E); +\end{asy} +\end{center} +so +\begin{align} + \theta &= \arcsin\p({\frac{\Delta L_{eq}}{A}}) = 0.561\U{rad} \\ + t_s &= \frac{\theta+\frac{\pi}{2}}{\omega} = \ans{2.00\U{s}} \;. +\end{align} + +\Part{f} +Combining the free-fall time from \Part{a} with the harmonic time +from \Part{e} we have +\begin{equation} + t_{tb} = t_f + t_s = \ans{3.50\U{s}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem12.31.tex b/latex/problems/problem12.31.tex new file mode 100644 index 0000000..590b2f0 --- /dev/null +++ b/latex/problems/problem12.31.tex @@ -0,0 +1,8 @@ +\begin{problem*}{12.31} +A pendulum with a length of $1.00\U{m}$ is released from an initial +angle of $15.0\dg$. After $1000\U{s}$, its amplitdue has been reduced +by friction to $5.50\dg$. What is the value of $b/2m$? +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/problem12.33.tex b/latex/problems/problem12.33.tex new file mode 100644 index 0000000..697f389 --- /dev/null +++ b/latex/problems/problem12.33.tex @@ -0,0 +1,10 @@ +\begin{problem*}{12.33} +A $2.00\U{kg}$ object attached to a spring moves without friction and +is driven by an external force $F=(3.00\U{N})\sin(2\pi t)$. Assuming +that the force constant of the spring is $20.0\U{N/m}$, +determine \Part{a} the period and \Part{b} the amplitude of the +motion. +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/problem12.38.tex b/latex/problems/problem12.38.tex new file mode 100644 index 0000000..7bd8914 --- /dev/null +++ b/latex/problems/problem12.38.tex @@ -0,0 +1,14 @@ +\begin{problem*}{12.38} +Four people, each with a mass of $72.4\U{kg}$, are in a car with a +mass of $1130\U{kg}$. An eathquake strikes. The vertical +oscillations of the ground surface make the car bounce up and down on +its suspension springs, but the driver manaages to pull off the road +and stop. When the frequency of the shaking is $1.80\U{Hz}$, the car +exhibits a maximum amplitude of vibration. The earthquake ends and +the four people leave the car as fast as they can. By what distance +does the car's undamaged suspension lift the car's body as the people +get out? +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/problem12.42.tex b/latex/problems/problem12.42.tex new file mode 100644 index 0000000..87e673d --- /dev/null +++ b/latex/problems/problem12.42.tex @@ -0,0 +1,100 @@ +\begin{problem*}{12.42} +\Part{a} A hanging spring stretches by $35.0\U{cm}$ when an object of +mass $450\U{g}$ is hung on it at rest. In this situation, we define +its position as $x=0$. The object is pulled down an additional +$18.0\U{cm}$ and released from rest to oscillate without friction. +What is its position $x$ at a time $84.4\U{s}$ later? \Part{b} A +hanging spring stretches by $35.5\U{cm}$ when an object of mass +$440\U{g}$ is hung on it at rest. We define this new position as +$x=0$. This object is also pulled down an additional $18.0\U{cm}$ and +released from rest to oscillate without friction. Find its position +$84.4\U{s}$ later. \Part{c} Why are the answers to parts \Part{a} +and \Part{b} different by such a large percentage when the data are so +similar? Does this circumstance reveal a fundamental difficulty in +calculating the future? \Part{d} Find the distance traveled by the +vibrating object in \Part{a}. \Part{e} Find the distance traveled by +the object in \Part{b}. +\end{problem*} + +\begin{solution} +There's a lot of repetition in this problem, so let's do everything +symbolically first. A stretch of $\Delta x$ due to hanging a mass $m$ +implies a spring constant of $k = F/\Delta x = mg/\Delta x$ and an +angular frequency of $\omega = \sqrt{k/m} = \sqrt{g/\Delta{x}}$. +Notice that $\omega$ does not depend on the hanging mass. At this +point we know everything about how the position changes as a function +of time, namely +\begin{equation} + x(t) = -A\cos(\omega t) = -A\cos(\sqrt{g/\Delta{x}} t)\;, +\end{equation} +where the $-\cos$ part came from letting up be the positive $x$ +direction and noticing that the mass starts at an extreme low point in +its oscillation. Finally, we note that the total distance traveled as +a function of time is going to be +\begin{equation} + d = 2A \cdot N_{T/2} + (A+x(t_\text{frac})) +\end{equation} +where $N_{T/2} = floor(2t/T)$ is the number of completed half-periods +and $t_\text{frac} = t - N_{T/2}\cdot T/2$ is the fractional +half-period left over. The extra $A$ in the right-hand term ensures +that the right-hand term is zero when $t_\text{frac}=0$. Now we just +have to plug in the two cases\ldots + +\Part{a} +\begin{align} + \Delta x &= 0.350\U{m} & A &= 0.180\U{m} & t &= 84.4\U{s} \\ + \omega &= 5.29\U{rad/s} & x(t) &= \ans{-15.8\U{cm}} +\end{align} + +\Part{b} +\begin{align} + \Delta x &= 0.355\U{m} & A &= 0.180\U{m} & t &= 84.4\U{s} \\ + \omega &= 5.25\U{rad/s} & x(t) &= \ans{15.9\U{cm}} +\end{align} + +\Part{c} +The answers to \Part{a} and \Part{b} are so different because the +angular velocities are slightly different. The phase starts out the +same but grows slightly faster in \Part{a}. The difference seems so +large because the phase loops around at $2\pi$, so a small difference +in the ``total phase'' $\omega t$ can produce a large difference in +the ``effective phase'' $\omega t \mod 2\pi$. + +This sensisivity to the initial conditions (here to $\Delta x$) is the +same sort of effect that gives rise to chaotic behavior +(see \url{http://en.wikipedia.org/wiki/Chaos_theory#Chaotic_dynamics}). +The simple harmonic oscillator is not chaotic though, because $\Delta +x$ is not a dimension that changes with time. This means that +predicting the future for a simple harmonic oscillator is not +\emph{that} difficult. You may have to keep adjusting your model +parameters as you get more data, but with time your model will get +better and better. + +Compare this system to one that \emph{is} chaotic, e.g.~the +damped-forced pendulum. In this situations, the sensitivity to +initial conditions also involves the changing parameters, like the +position of the pendulum. Not only does your model have to be +perfect, but your understanding of where the pendulum is now must also +be perfect. This is a much more difficult problem, since the position +of the pendulum is changing with time, so you can't just wait and +aquire a arbitarily precise time average. This is why it is so hard +to make long term predictions for chaotic systems. + +\Part{d} +\begin{align} + T &= \frac{2\pi}{\omega} = 1.19\U{s} & N_{T/2} &= 142 + & t_\text{frac} &= 93.9\U{ms} \\ + d &= 51.1\U{m} + 2.17\U{cm} = \ans{51.1\U{m}} +\end{align} + +\Part{e} +\begin{align} + T &= \frac{2\pi}{\omega} = 1.20\U{s} & N_{T/2} &= 141 + & t_\text{frac} &= 91.7\U{ms} \\ + d &= 50.8\U{m} + 2.05\U{cm} = \ans{50.8\U{m}} +\end{align} +Notice that the distance $d$ doesn't loop back on itself like the +position $x$, so the small relative difference in ``total phase'' +leads to a small relative difference in distance between \Part{d} +and \Part{e}. +\end{solution} diff --git a/latex/problems/problem12.47.T.tex b/latex/problems/problem12.47.T.tex new file mode 100644 index 0000000..3b4d809 --- /dev/null +++ b/latex/problems/problem12.47.T.tex @@ -0,0 +1,98 @@ +\begin{problem} +\emph{BONUS PROBLEM}. Find the resonant frequency in Hz of the sprung +pendulum for small $\theta$ on both Earth and the Moon. The mass of +the bob is $m = 2.3\U{kg}$, the length of the light rod is $r = +3.0\U{m}$, and the spring constant is $k = 1.4\U{N/m}$. The system is +at equilibrium when the pendulum rod is vertical. + +Hints: drawing a free body diagram may help determine the restoring +forces. You will need to use the small angle approximation. +\end{problem} % Developed from Ch. 12, Prob. 47. + +\begin{nosolution} +\begin{center} +\begin{asy} +import Mechanics; +real u = 1cm; + +Pendulum p = makePendulum(angleDeg=25, length=2u, angleL="$\theta$"); +Spring s = Spring(pFrom=p.mass.center, pTo=p.mass.center+2u, + unstretchedLength=2u); + +s.draw(); +p.draw(drawVertical=true); +\end{asy} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{asy} +import Mechanics; +real u = 1cm; + +Pendulum p = makePendulum(angleDeg=25, length=2u, + angleL="$\theta$", stringL="$r$"); +Spring s = Spring(pFrom=p.mass.center, pTo=p.mass.center+2u, + unstretchedLength=2u, L="$k$"); +Vector fs = Force(p.mass.center, dir=180, mag=5mm, L="$F_s$"); +Vector fg = Force(p.mass.center, dir=-90, mag=7mm, L="$F_g$"); + +s.draw(); +fs.draw(); +fg.draw(); +p.draw(drawVertical=true); + +label("$m$", p.mass.center); +\end{asy} +\end{center} +The spring is stretched or compressed by $x \approx r\sin(\theta)$ +where the approximation is exact in the limit of small angles. The +total force is the sum of the spring force $F_s$ and the gravitation +force $F_s$ acting on the bob. The portion of this total force that +is tangent to the bob's path is +\begin{align} + \sum F_{\tan} &= F_s\cos(\theta) - F_g\sin(\theta) + = -kx \cos(\theta) - mg\sin(\theta) + \approx -kr \sin(\theta) \cos(\theta) - mg\sin(\theta) + = -\p[{kr \cos(\theta) + mg}]\sin(\theta) \\ + &\approx -\p[{kr + mg}] \cdot \theta\;, +\end{align} +where we have used the small angle approximation again in the last +step. We also know from Newton's laws that +\begin{equation} + \sum F_{\tan} = ma_{\tan} = m\nderiv{2}{t}{x_{\tan}} + = mr\nderiv{2}{t}{\theta}\;. +\end{equation} + +Combining these two formulas for $\sum F$ we have +\begin{align} + mr\nderiv{2}{t}{\theta} &= -{kr + mg} \cdot \theta \\ + \nderiv{2}{t}{\theta} &= -\frac{kr + mg}{mr} \cdot \theta + = -\p({\frac{k}{m} + \frac{g}{r}}) \cdot \theta \;. +\end{align} +Looking at the last form, we see that it looks a lot like the +equation of motion for simple harmonic motion +\begin{equation} + \nderiv{2}{t}{\theta} = -\omega^2 \theta\;, +\end{equation} +and we see that the equations are equal when +\begin{equation} + \frac{k}{m} + \frac{g}{r} = \omega^2\;. +\end{equation} +Plug in to the frequency formula for a simple harmonic oscillator, we +have +\begin{equation} + f = \frac{\omega}{2\pi} + = \ans{\frac{1}{2\pi}\sqrt{\frac{k}{m}+\frac{g}{r}}}\;. +\end{equation} + +On Earth $g = 9.8\U{m/s$^2$}$, and on the Moon $g = 1.6\U{m/s$^2$}$, so +\begin{align} + f_\text{Earth} &= \frac{1}{2\pi}\sqrt{\frac{1.4}{2.3}+\frac{9.8}{3.0}} + = \ans{0.31\U{Hz}} \\ + f_\text{Moon} &= \frac{1}{2\pi}\sqrt{\frac{1.4}{2.3}+\frac{1.6}{3.0}} + = \ans{0.17\U{Hz}}\;. +\end{align} + +\end{solution} diff --git a/latex/problems/problem12.T.bombardier.jpg b/latex/problems/problem12.T.bombardier.jpg new file mode 100644 index 0000000..778c1c2 Binary files /dev/null and b/latex/problems/problem12.T.bombardier.jpg differ diff --git a/latex/problems/problem12.T.tex b/latex/problems/problem12.T.tex new file mode 100644 index 0000000..e06084d --- /dev/null +++ b/latex/problems/problem12.T.tex @@ -0,0 +1,45 @@ +\newcommand{\D}{\Delta} + +\begin{problem} +Hydraulic shock absorbers typically consist of a piston in an oil +filled reservoir (see +\url{http://en.wikipedia.org/wiki/Shock_absorber}). Orifices in the +piston allow oil to flow from one side of the piston to the other, so +piston movement stirs the oil. The stirring transforms the piston's +mechanical and kinetic energy into heat, damping any piston +oscillation. + +You are asked to design a shock absorber for a motor-unicycle +(suspended by a single shock and spring). With a rider the +motor-unicyle weighs $m = 140\U{kg}$ and is sprung with a $k = +2.0\U{kN/m}$ spring. + +\Part{a} +Determine the undamped resonant frequency in Hz. +\Part{b} +Determine the damping coefficient for critical damping. + +Hint: treat the shock absorber as a damped simple harmonic oscillator. +\begin{center} +\includegraphics[width=1.25in]{bombardier} % http://www.unicycling.org/btdt/unique.html +\end{center} +\end{problem} % application of Eqns 12.9 and def of crit damping in Sec12.6. + +\begin{solution} +\Part{a} +The frequency of underdamped vibration is given by +\begin{equation} + \omega = \sqrt{\frac{k}{m} - \p({\frac{b}{2m}})^2} +\end{equation} +Without damping ($b = 0$), +\begin{equation} + f = \frac{1}{2\pi}\omega = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \ans{0.602\U{Hz}} +\end{equation} + +\Part{b} +Critical damping occurs when $\omega = 0$ or +\begin{align} + \frac{b_c}{2m} &= \sqrt{\frac{k}{m}} \\ + b_c &= 2\sqrt{km} = \ans{1.1\U{kNs/m}} +\end{align} +\end{solution} diff --git a/latex/problems/problem12.V1.tex b/latex/problems/problem12.V1.tex new file mode 100644 index 0000000..057ba87 --- /dev/null +++ b/latex/problems/problem12.V1.tex @@ -0,0 +1,126 @@ +\begin{problem} +A mass $m$ is attached to the free end of a light vertical spring +(unstretched length $l$) of spring constant $k$ and suspended from a +ceiling. The spring stretches by $\Delta l$ under the load and comes +to equilibrium at a height $h$ above the ground level (y = 0). The +mass is pushed up vertically by $A$ from its equilibrium position and +released from rest. The mass-spring executes vertical +oscillations. Assuming that gravitational potential energy of the mass +$m$ is zero at the ground level, show that the total energy of the +spring-mass system is $\frac{1}{2}k(\Delta l^2 + A^2) + mgh$. + +Hint: Draw four clear sketches of the vertical free spring, +spring-mass in equilibrium and the two extreme positions of +oscillation of the mass. Below each of the above sketches (except free +spring), write equations for kinetic energy, elastic potential energy, +gravitational potential energy, and total energy using symbols $k$, +$\Delta l$, $m$, $A$, $g$, $h$, and $v_m$. +\end{problem} + +\begin{solution} +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1cm; +real L = 1.5u; +real dL = 0.7L; +real A = 0.5L; +real h = 2L; + +draw((0,0)--(4L,0), green); +label("ground", (0,0), W); + +draw((0,h+dL+L)--(4L,h+dL+L)); +label("fixed end", (0,h+dL+L), W); + +draw((0,h)--(4L,h), gray(0.6)+dashed); +label("equilibrium", (0,h), W); + +draw((0,h+dL)--(4L,h+dL), gray(0.6)+dashed); +label("unstretched", (0,h+dL), W); + +draw((0,h+A)--(4L,h+A), gray(0.6)+dashed); +label("maximum", (0,h+A), W); + +draw((0,h-A)--(4L,h-A), gray(0.6)+dashed); +label("minimum", (0,h-A), W); + +Mass m; +Spring s; + +real x = 0; +s = Spring(pFrom=(x,h+dL+L), pTo=(x,h+dL)); +s.draw(); + +real x = 2L; +m = Mass((x,h), radius=1.5mm); +s = Spring(pFrom=(x,h+dL+L), pTo=m.center); +s.draw(); +m.draw(); + +real x = 3L; +m.center = (x,h+A); +s = Spring(pFrom=(x,h+dL+L), pTo=m.center); +s.draw(); +m.draw(); + +real x = 4L; +m.center = (x,h-A); +s = Spring(pFrom=(x,h+dL+L), pTo=m.center); +s.draw(); +m.draw(); + +real dx = 0.5u; +Distance Dh = Distance(pFrom=(0,0), pTo=(0,h), "$h$"); +Dh.draw(rotateLabel=false); +Distance DdL = Distance(pFrom=(dx,h), pTo=(dx,h+dL), "$\Delta l$ "); +// inline asymptote crowds the label. this space intentional -^ +DdL.draw(rotateLabel=false); +Distance DL = Distance(pFrom=(2dx,h+dL), pTo=(2dx,h+dL+L), "$L$"); +DL.draw(rotateLabel=false); +Distance Aup = Distance(pFrom=(3dx,h), pTo=(3dx,h+A), "$A$"); +Aup.draw(rotateLabel=false); +Distance Adn = Distance(pFrom=(4dx,h-A), pTo=(4dx,h), "$A$"); +Adn.draw(rotateLabel=false); +\end{asy} +\end{center} + +\begin{tabular}{l l l l} + & + Equilibrium & + Top & + Bottom \\ +Kinetic energy & + $\frac{1}{2}mv_m^2$ & + $0$ & + $0$ \\ +Elastic energy & + $\frac{1}{2}k\Delta l^2$ & + $\frac{1}{2}k(\Delta l-A)^2$ & + $\frac{1}{2}k(\Delta l+A)^2$ \\ +Gravitational energy & + $mgh$ & + $mg(h+A)$ & + $mg(h-A)$ \\ +Total energy & + $\frac{1}{2}mv_m^2+mgh+\frac{1}{2}k\Delta l^2$ & + $mg(h+A)+\frac{1}{2}k(\Delta l-A)^2$ & + $mg(h-A)+\frac{1}{2}k(\Delta l+A)^2$ +\end{tabular} + +These energies are all equivalent to the suggested formula, because +\begin{align} + k\Delta l &= mg \\ + v_m &= \omega A = \sqrt{\frac{k}{m}} A \\ + E_e &= \frac{1}{2}mv_m^2 +mgh+\frac{1}{2}k\Delta l^2 + = mgh + \frac{1}{2}m\frac{k}{m}A^2 +\frac{1}{2}k\Delta l^2 + = mgh + \frac{1}{2}k(\Delta l^2 + A^2) \\ + E_t &= mg(h+A)+\frac{1}{2}k(\Delta l-A)^2 + = mgh + k\Delta l A + \p({\frac{1}{2}k\Delta l^2 - k\Delta l A + \frac{1}{2}kA^2}) + = mgh + \frac{1}{2}k(\Delta l^2 + A^2) \\ + E_b &= mg(h-A)+\frac{1}{2}k(\Delta l+A)^2 + = mgh - k\Delta l A + \p({\frac{1}{2}k\Delta l^2 + k\Delta l A + \frac{1}{2}kA^2}) + = mgh + \frac{1}{2}k(\Delta l^2 + A^2) +\end{align} +\end{solution} diff --git a/latex/problems/problem12.V2.tex b/latex/problems/problem12.V2.tex new file mode 100644 index 0000000..b65b070 --- /dev/null +++ b/latex/problems/problem12.V2.tex @@ -0,0 +1,182 @@ +\begin{problem} +For the problem above, % problem12.V1.tex +given: $m = 0.8\U{kg}$, $k = 128\U{N/m}$, $A = 0.12\U{m}$, $h = 0.25\U{m}$, and acceleration due to +gravity $g = 10\U{m/s$^2$}$. \\ +\Part{a} What is the extension $\Delta l$ in meters? \\ +\Part{b} What is the angular frequency $\omega$ of oscillation of the system? \\ +\Part{c} What are the time period and frequency of oscillation of the system? \\ + +Find the kinetic, elastic, gravitational, and total energies of the system \\ +\Part{d} for the equilibrium position, \\ +\Part{e} when the mass at top extreme position, \\ +\Part{f} when the mass is at the lowest vertical position, \\ + +\Part{g} Now, at $t=0$, imagine that, instead of releasing the mass +from rest at a height $A$ above equilibrium, you ``flick'' it with a +speed of $+0.8\U{m/s}$ vertically up when the mass is $0.10\U{m}$ +below equilibrium position. Under these circumstances, determine the +following quantities: +\begin{packed_enum} +\item the angular frequency $\omega$ \\ +and, using both kinematic and energy methods, +\item the amplitude and +\item the initial phase of oscillation. +\end{packed_enum} +\Part{h} Write an equation for the position $y$ of the mass as a +function of time $t$. +\end{problem} + +\begin{solution} +\Part{a} +\begin{align} + k\Delta l &= mg & \Delta l &= \frac{mg}{k} = \ans{6.25\U{cm}} +\end{align} + +\Part{b} +\begin{equation} + \omega = \sqrt{\frac{k}{m}} = \ans{12.6\U{rad/s}} +\end{equation} + +\Part{c} +\begin{align} + f &= \frac{\omega}{2\pi} = \ans{2.01\U{Hz}} & + T &= \frac{1}{f} = \ans{0.497\U{s}} +\end{align} + +\Part{d}\Part{e}\Part{f} + +\begin{tabular}{l l l l} + & + Equilibrium & + Top & + Bottom \\ +Kinetic energy & + $\frac{1}{2}mv_m^2 = \frac{1}{2}kA^2 = \ans{0.922\U{J}}$ & + $0$ & + $0$ \\ +Elastic energy & + $\frac{1}{2}k\Delta l^2 = \ans{0.250\U{J}}$ & + $\frac{1}{2}k(\Delta l-A)^2 = \ans{0.212\U{J}}$ & + $\frac{1}{2}k(\Delta l+A)^2 = \ans{2.13\U{J}}$ \\ +Gravitational energy & + $mgh = \ans{2.00\U{J}}$ & + $mg(h+A) = \ans{2.96\U{J}}$ & + $mg(h-A) = \ans{1.04\U{J}}$ \\ +Total energy & + $mgh+\frac{1}{2}k(\Delta l^2+A^2) = \ans{3.17\U{J}}$ & + $mgh+\frac{1}{2}k(\Delta l^2+A^2) = \ans{3.17\U{J}}$ & + $mgh+\frac{1}{2}k(\Delta l^2+A^2) = \ans{3.17\U{J}}$ +\end{tabular} + +\Part{g} +$\omega$ only depends on $k$ and $m$, so it is unaffected by the new +initial conditions. + +{\bf Energy approach}: conserve energy between the intial state and +the top position +\begin{align} + E_i &= mg(h-x) + \frac{1}{2}mv^2 + \frac{1}{2}k(\Delta l + x)^2 + = E_t = mgh+\frac{1}{2}k(\Delta l^2+A^2) \\ + \frac{1}{2}kA^2 &= -mgx + \frac{1}{2}mv^2 + \frac{1}{2}kx^2 + kx\Delta l + = -mgx + \frac{1}{2}mv^2 + \frac{1}{2}kx^2 + kx\frac{mg}{k} + = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \\ + A &= \sqrt{\frac{mv^2 + kx^2}{k}} + = \sqrt{\frac{mv^2}{k} + x^2 } + = \ans{11.8\U{cm}} +\end{align} +Of course if you realize from the start that the gravitational force +just changes the equilibrium position of the spring and has no effect +on the relative motion, you can save yourself the algebra we used to +eliminate the gravitational energy in the first two lines above. + +Once you have $A$, you can switch over to kinematics to solve for +$\phi$. +\begin{align} + x(t=0) &= -10\U{cm} = A\cos(\omega t + \phi) + = 11.8\U{cm}\cdot\cos(\phi) \\ + \phi &= \arccos\p({\frac{-10\U{cm}}{11.8\U{cm}}}) + = 2.58, 3.71\U{rad} +\end{align} +Note that there are two times during a full cycle when $x(t)$ is +$10\U{cm}$ below the equilibrium position, one going up and one going +down. +\begin{center} +\begin{asy} +import graph; + +size(4cm,3cm,IgnoreAspect); + +draw(graph(cos, 0, 2pi)); +xaxis("$\theta$", 0, 2pi); +yaxis("$\cos(\theta)$", -1, 1); +yequals(-10/11.8,red+Dotted); +dot(Label("2.58", align=SW), Scale((2.58,cos(2.58)))); +dot(Label("3.71", align=SE), Scale((3.71,cos(3.71)))); +\end{asy} +\end{center} +Since the problem specifies a positive initial velocity, +$\phi=\ans{3.71\U{rad}}$ is the correct solution. + +{\bf Kinetic approach}: define the time dependence of $x$ and $v$ +\begin{align} + x(t) &= A\cos(\omega t + \phi) \\ + v(t) &= -A\omega\sin(\omega t + \phi) \;. +\end{align} +So the initial coordinates are +\begin{align} + x(0) &= A\cos(\phi) \\ + v(0) &= -A\omega\sin(\phi) \;. +\end{align} +so +\begin{align} + x(t)^2 + \p({\frac{v(t)}{\omega}})^2 + &= A^2\cos^2(\omega t + \phi) + A^2\sin^2(\omega t +\phi) + = A^2 [\cos^2(\omega t + \phi) + \sin^2(\omega t + \phi)] + = A^2 \\ + A &= \sqrt{x(t)^2 + \p({\frac{v(t)}{\omega}})^2} + = \ans{11.8\U{cm}} +\end{align} +Finally, use a bit of trig to find $\phi$. +\begin{center} +\begin{asy} +import Mechanics; + +real u = 5cm; +real w = 12.6; //rad/s +real x = -0.1; //m +real v = 2; //m/s + +real A = sqrt(x**2 + (v/w)**2); +pair p = (x*u, -v/w*u); +real r = A*u; + +Vector xaxis = Vector((-1.4r,0), mag=2.8r, dir=0, gray(0.6), "$x$"); +xaxis.draw(); +Vector yaxis = Vector((0,1.4r), mag=2.8r, dir=-90, gray(0.6), "$v/\omega$"); +yaxis.draw(); +draw(scale(r)*unitcircle); + +Angle a = Angle(p, (0,0), (1,0), radius=-4mm, "$\phi$"); +a.draw(); +draw((0,0)--p, red); +label("$A$", p/2, W); +\end{asy} +\end{center} +Note that the $y$ axis is scaled and flipped to accound for the +$-\omega$ joining $A$ out front. +\begin{align} + \phi = \arctan\p({\frac{-v}{\omega x}}) + \pi = \ans{3.71\U{rad}} +\end{align} +The $+\pi$ is needed to deal with the ``backside'' arc-tangent +problem, since the arc-tangent only has a range of $180\dg$. + +\Part{h} +We already wrote this for \Part{g}, but here it is with everything +plugged in. We also including the equilibrium offset $h$ from our +$y=0$ ground level, which we had ignored before, and rename +$x\rightarrow y$. +\begin{equation} + y(t) = A\cos(\omega t + \phi) + h + = 11.8\U{cm}\cdot\cos(12.6\U{rad/s}\cdot t + 3.71\U{rad}) + 25.0\U{cm} +\end{equation} +\end{solution} diff --git a/latex/problems/problem12.V3.tex b/latex/problems/problem12.V3.tex new file mode 100644 index 0000000..dfd74c9 --- /dev/null +++ b/latex/problems/problem12.V3.tex @@ -0,0 +1,72 @@ +\begin{problem} +A force of $64\U{N}$ stretches a certain spring by $16\U{cm}$. A block +of mass $4\U{kg}$ is hung from the spring and the spring-mass +combination executes vertical simple harmonic oscillations when at +$t=0$, the mass is pulled down by $12\U{cm}$ below its equilibrium +position and released from rest. Assume gravitational potential energy +is zero at the highest position ($y=0$) of oscillation of the mass +$m$. \\ +\Part{a} Write down an equation for position of the mass assuming the +clock reads $t=0$ at the instant the mass is released (i.e~at the +lowest point of vertical oscillations). \\ +\Part{b} Find the shortest time needed for the mass to go through a +point $8\U{cm}$ above the equilibrium position from the instant it is +released. \\ +\Part{c} Find the position and the magnitude and direction of its +velocity of motion at $t=0.10\pi\U{s}$. \\ +\Part{d} What is the total energy of the spring mass combination at +any instant of oscillation? +\end{problem} + +\begin{solution} +\Part{a} +From the initial condition, we know that the equation must look like +\begin{equation} + y(t) = -A[\cos(\omega t) - 1] \;, +\end{equation} +where the $-1$ shifts the $\cos$ down so $y(t)_\text{max}=0$. +$A=12\U{cm}$ is given in the problem, but we need to find $\omega$ +\begin{align} + k &= \frac{F}{\Delta l} = 400\U{N/m} \\ + \omega &= \sqrt{\frac{k}{m}} = \sqrt{\frac{F}{m\Delta l}} + = 10\U{rad/s} \\ + y(t) &= \ans{-12\U{cm}\cdot[\cos(10\U{rad/s}\cdot t)-1]} +\end{align} + +\Part{b} +Let $\Delta y=8\U{cm}$. Then +\begin{align} + y(t)+A &= -A\cos(\omega t) = \Delta y \\ + \omega t &= \arccos\p({\frac{-\Delta y}{A}}) \\ + t &= \frac{\arccos\p({\frac{-\Delta y}{A}})}{\omega} + = \ans{0.230\U{s}} +\end{align} + +\Part{c} +We get the position by plugging in +\begin{equation} + y(t=0.10\pi\U{s}) = -12\U{cm}\cdot[\cos(10\U{rad/s}\cdot 0.10\pi\U{s}) + 1] + = -12\U{cm}\cdot[\cos(\pi\U{rad}) + 1] + = -12\U{cm}\cdot[-1 + 1] + = \ans{0} +\end{equation} +This is the high point of the oscillation so $v=\ans{0}$ + +\Part{d} +Energy is conserved, so we can pick our favorite point. How about +when all the energy is stored in the spring. We'll need to know +the total amount the spring stretched though. At equilibrium +\begin{align} + mg &= k\Delta L_\text{eq} \\ + \Delta L_\text{eq} &= \frac{mg}{k} = 9.8\U{cm} \;, +\end{align} +so +\begin{equation} + E = \frac{1}{2} k (\Delta L_\text{eq} - A)^2 + = \frac{1}{2}\cdot 400\U{N/m}\cdot(0.098\U{m} - 0.120\U{m})^2 + = \ans{96.8\U{mJ}} +\end{equation} + +Note: I used $g = 9.8\U{m/s$^2$}$ throughout, while Prof.~Venkat used +$10\U{m/s$^2$}$, which is why some of our answers are slightly different. +\end{solution} diff --git a/latex/problems/problem13.05.tex b/latex/problems/problem13.05.tex new file mode 100644 index 0000000..5221c02 --- /dev/null +++ b/latex/problems/problem13.05.tex @@ -0,0 +1,50 @@ +\begin{problem*}{13.5} +The wave function for a traveling wave on a taut string is (in SI units) +\begin{equation} + y(x,t) = (0.350\U{m}) \sin(10\pi t - 3\pi x + \pi/4) \;. +\end{equation} +\Part{a} What are the speed and direction of travel of the +wave? \Part{b} What is the vertical position of an element of the +string at $t = 0$, $x = 0.100\U{m}$? \Part{c} What are the wavelength +and frequency of the wave? \Part{d} What is the maximum transverse +speed of an element of the string? +\end{problem*} + +\begin{solution} +\Part{a} +When $t=0$ and $x=0$, the phase of $y$ is $\pi/4$. If we look some +small time $\Delta t$ later, we'll need some small positive +displacement $\Delta x = 10\Delta t/3$ to find that same phase. So +the direction is \ans{to the right (positive $x$)}, and the speed is +\begin{equation} + v = \frac{\Delta x}{\Delta t} = \frac{10}{3} = \ans{3.33\U{m/s}} +\end{equation} + +\Part{b} +Plugging in +\begin{equation} + y(x=0.100\U{m}, t=0) = 0.350\U{m}\cdot\sin(-0.3\pi+\pi/4) = \ans{-5.48\U{cm}} +\end{equation} + +\Part{c} +To change the phase of the wave by $2\pi$, we need either +\begin{equation} + \lambda = \Delta x_{2\pi} = \frac{2}{3} = \ans{0.667\U{m}} +\end{equation} +or +\begin{align} + T &= \Delta t_{2\pi} = \frac{2}{10} = 0.200\U{s} \\ + f &= \frac{1}{T} = \ans{5\U{Hz}} +\end{align} + +\Part{d} +We can find the speed at any time and position by differentiating +$y(x,t)$ with respect to time. After last week's recitation, we have +lots of practice with such derivatives, so we see right off that the +only effect on the amplitude will be to chain-rule out a copy of the +time coefficient $\omega = 10\pi\U{rad/s}$, so +\begin{equation} + \p({\pderiv{t}{y}})_\text{max} = A\omega = 0.350\U{m} \cdot 10\pi\U{1/s} + = \ans{11.0\U{m/s}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem13.07.tex b/latex/problems/problem13.07.tex new file mode 100644 index 0000000..7a467c1 --- /dev/null +++ b/latex/problems/problem13.07.tex @@ -0,0 +1,47 @@ +\begin{problem*}{13.7} +The string shown in Active Figure 13.8 is driven at a frequency of +$5.00\U{Hz}$. The amplitude of the motion is $12.0\U{cm}$ and the +wave speed is $20.0\U{m/s}$. Furthermore, the wave is such that $y=0$ +at $x=0$ and $t=0$. Determine \Part{a} the angular frequency +and \Part{b} wave number for this wave. \Part{c} Write an expression +for the wave function. Calculate \Part{d} the maximum transverse +speed and \Part{e} the maximum transverse acceleration of an element +of the string. +\end{problem*} + +\begin{solution} +\Part{a} +This is just a unit conversion. +\begin{align} + \frac{rad}{s}&=\frac{2\pi\U{rad}}{\text{cycle}}\cdot\frac{\text{cycles}}{s} \\ + \omega &= 2\pi f = \ans{31.4\U{rad/s}} +\end{align} + +\Part{b} +Another units conversion +\begin{align} + \frac{rad}{m} &= \frac{rad}{s} \cdot \frac{s}{m} \\ + k &= \frac{\omega}{v} = \ans{1.57\U{rad/m}} +\end{align} + +\Part{c} +Now we get to plug in those values for $k$ and $\omega$. +\begin{equation} + y(x,t) = A \sin(kx-\omega t) + = 12.0\U{cm}\cdot\sin(1.57\U{rad/m}\cdot x - 31.4\U{rad/s}\cdot t) +\end{equation} + +\Part{d} +Differentiating with respect to time pulls out a chain-rule $\omega$. +\begin{equation} + \p({\pderiv{t}{y}})_\text{max} = A\omega = 12.0\U{cm}\cdot 31.4\U{1/s} + = \ans{3.77\U{m/s}} +\end{equation} + +\Part{e} +Differentiating again with respect to time pulls out another $\omega$. +\begin{equation} + \p({\npderiv{2}{t}{y}})_\text{max} = A\omega^2 + = 12.0\U{cm}\cdot (31.4\U{1/s})^2 = \ans{118\U{m/s$^2$}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem13.10.tex b/latex/problems/problem13.10.tex new file mode 100644 index 0000000..57771d5 --- /dev/null +++ b/latex/problems/problem13.10.tex @@ -0,0 +1,28 @@ +\begin{problem*}{13.10} +A transverse wave on a string is described by the wave function +\begin{equation} + y = (0.120\U{m}) \sin\p({\frac{\pi}{8}x + 4\pi t}) \;. +\end{equation} +\Part{a} Determine the transverse speed and acceleration of an +element of the string at $t = 0.200\U{s}$ for the point on the string +located at $x = 1.60\U{m}$. \Part{b} What are the wavelength, period, +and speed of propagation of this wave? +\end{problem*} + +\begin{solution} +\Part{a} +Notice that the phase $\phi = \pi x/8 + 4\pi t = 0.2\pi + 0.8\pi=\pi$. +Since the transverse displacement is a $\sin$ curve, that means that +the transverse velocity will be most negative and the acceleration +will be zero. +\begin{align} + \pderiv{t}{y}(x=1.60\U{m}, t=0.200\U{s}) &= A\omega = \ans{-1.51\U{m/s}} \\ + \npderiv{2}{t}{y}(x=1.60\U{m}, t=0.200\U{s}) &= \ans{0\U{m/s$^2$}} \;. +\end{align} + +\Part{b} +To increase the phase by $2\pi$, we need a $\Delta x_{2\pi} = +\ans{\lambda = 16.0\U{m}}$. Similarly, we would need a $\Delta +t_{2\pi} = \ans{T = 0.500\U{s}}$. The wave speed then is $v = \Delta +x / \Delta t = \lambda / T = \ans{32.0\U{m/s}}$. +\end{solution} diff --git a/latex/problems/problem13.11.tex b/latex/problems/problem13.11.tex new file mode 100644 index 0000000..9b41186 --- /dev/null +++ b/latex/problems/problem13.11.tex @@ -0,0 +1,88 @@ +\begin{problem*}{13.11} +A transverse sinusoidal wave on a string has a period $T=25.0\U{ms}$ +and travels in the negative $x$ direction with a speed of +$30.0\U{m/s}$. At $t=0$, an element of the string at $x=0$ has a +tranverse position of $2.00\U{cm}$ and is traveling downward with a +speed of $2.00\U{m/s}$. \Part{a} What is the amplitude of the +wave? \Part{b} What is the initial phase angle? \Part{c} What is +the maximum transverse speed of an element of the string? \Part{d} +Write the wave function for the wave. +\end{problem*} + +\begin{solution} +I think the easiest way to do this is backwards. +\Part{d} +The transverse position and velocity must look something like +\begin{align} + y &= A\cos[k(x+vt)+\phi] = A\cos(\omega t+kx+\phi) \\ + \pderiv{t}{y} &= -\omega A \sin(\omega t+kx+\phi) \;, +\end{align} +where the $x+vt$ comes from the requirement for motion in the $-x$ +direction, and $\pderiv{t}{y}$ comes from differentiating. In the +next part, we consider the situation at $t=x=0$. + +\Part{b} +\begin{align} + y(0,0) &= A\cos(\phi) \\ + \pderiv{t}{y}(0,0) &= -\omega A \sin(\phi) \\ + \frac{\pderiv{t}{y}(0,0)}{y(0,0)} &= -\omega\frac{\sin(\phi)}{\cos(\phi)} + = -\omega\tan(\phi) \\ + \phi &= \arctan\p({\frac{-\pderiv{t}{y}(0,0)}{\omega y(0,0)}}) \;. +\end{align} +From the period, we can find the angular velocity and wavenumber +\begin{align} + \omega &= \frac{2\pi}{T} = 251\U{rad/s} & + k &= \frac{\omega}{v} = 8.38\U{m/s} +\end{align} +Plugging in for $\phi$ yields +\begin{equation} + \phi = \arctan\p({\frac{-(-2\U{m/s})}{251\U{1/s} \cdot 0.0200\U{m}}}) + = \arctan(0.398) = \ans{0.379\U{rad}} \;. +\end{equation} +Note that the tangent only has a range of $[-\pi/2, pi/2]$, so you +need to check and convince yourself that you aren't on the +``backside'' of the circle. We want a positive position and a +negative velocity, which happens when both our $\sin$ and $\cos$ terms +are positive (see functions above), so we \emph{do} want the +first-quadrant answer for $\phi$ given above. +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1.2cm; +real theta = 21.7; //degrees, example angle + +draw(scale(u)*unitcircle, dotted); +draw((-1.1u,0)--(1.1u,0)); +draw((0,-1.1u)--(0,1.1u)); +pair p = u*dir(theta); + +Angle a = Angle((1,0), (0,0), p, radius=7mm, L=" $\theta$"); +// space before text in the label because inline-asymptote doesn't know +// how big the label will be when it sets the positions. See +// http://asymptote.sourceforge.net/doc/LaTeX-usage.html +// So we add some space by hand. +a.draw(); +draw((0,0)--p, red); + +label("$+$", u*dir(45), NE); +label("$-$", u*dir(135), NW); +label("$+$", u*dir(-135), SW); +label("$-$", u*dir(-45), SE); +label(rotate(90)*Label("Backside 180\dg"), 1.1u*dir(-180), W); +\end{asy} +\end{center} + +\Part{a} +Now we can find the amplitude using trig +\begin{align} + y(0,0) &= A\cos(\phi) \\ + A &= \frac{y(0,0)}{\cos(\phi)} = \ans{2.15\U{cm}} \;. +\end{align} + +\Part{c} +And we can find the maximum transverse speed with +\begin{equation} + \p({\pderiv{t}{y}})_\text{max} = A\omega = \ans{5.41\U{m/s}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem13.12.T.tex b/latex/problems/problem13.12.T.tex new file mode 100644 index 0000000..83ac32d --- /dev/null +++ b/latex/problems/problem13.12.T.tex @@ -0,0 +1,42 @@ +\begin{problem} +Show that the all functions of the form $y(x,t) = f(x \pm vt)$ for any +function $f(z)$ satisfy the linear wave equation (Equation 13.19) +\begin{equation} + \npderiv{2}{x}{y} = \frac{1}{v^2}\npderiv{2}{t}{y}\;. +\end{equation} +\end{problem} % generalized form of Problem 13.12 + +\begin{solution} +Taking the partial derivatives with respect to space +\begin{align} + \pderiv{x}{y} &= \pderiv{z}{f} \cdot \pderiv{x}{z} + = \pderiv{z}{f} \cdot \pderiv{x}{}\p({x \pm vt}) + = \pderiv{z}{f} \\ + \npderiv{2}{x}{y} &= \pderiv{x}{}\p({\pderiv{z}{f}}) + = \npderiv{2}{z}{f} \cdot \pderiv{x}{z} + = \npderiv{2}{z}{f} \cdot \pderiv{x}{}\p({x \pm vt}) + = \npderiv{2}{z}{f}\;, +\end{align} +where we have used the chain rule +\begin{equation} + \pderiv{x}{}\p({f(z(x))}) = \pderiv{z}{f}\cdot\pderiv{x}{z} +\end{equation} +with $z(x) = x \pm vt$. + +Taking the partial derivitives with respect to time +\begin{align} + \pderiv{t}{y} &= \pderiv{z}{f} \cdot \pderiv{t}{z} + = \pderiv{z}{f} \cdot \pderiv{t}{}\p({x \pm vt}) + = \pm v \pderiv{z}{f} \\ + \npderiv{2}{t}{y} &= \pderiv{t}{}\p({\pm v \pderiv{z}{f}}) + = \pm v \npderiv{2}{z}{f} \cdot \pderiv{t}{z} + = \pm v \npderiv{2}{z}{f} \cdot \pderiv{t}{}\p({x \pm vt}) + = (\pm v)^2 \npderiv{2}{z}{f} + = v^2 \npderiv{2}{z}{f}\;. +\end{align} +So +\begin{equation} + \npderiv{2}{x}{y} = \npderiv{2}{z}{f} = \frac{1}{v^2}\npderiv{2}{t}{y} +\end{equation} +which is what we set out to show. +\end{solution} diff --git a/latex/problems/problem13.16.tex b/latex/problems/problem13.16.tex new file mode 100644 index 0000000..0df674e --- /dev/null +++ b/latex/problems/problem13.16.tex @@ -0,0 +1,44 @@ +\begin{problem*}{13.16} +A light string with a mass per unit length of $8.00\U{g/m}$ has its +ends tied to two walls seperated by a distance equal to three-fourths +the length of the string (Figure P13.16). An object of mass +$m$ is suspended from the center of the string, putting a tension in +the string. \Part{a} Find an expression for the transverse wave speed +in the string as a function of the mass of the hanging +object. \Part{b} What should be the mass of the object suspended from +the string so as to produce a wave speed of $60.0\U{m/s}$. +\begin{center} +\begin{asy} +import Mechanics; + +real u=1cm; +real L=3u; // length of string +real D=3L/4; // wall separation +real dx = 12pt; // length of bonus string for hanging mass & wall above string +real r = 12pt; // radius of hanging mass + +Mass m = Mass(radius=r, L="m"); +pair junction = m.center + (0, r+dx); +real theta = asin(D/L) * 180 / pi; +pair wall_join_r = junction + L/2*dir(90-theta); +pair surf_ur = wall_join_r + (0, dx); +pair surf_lr = (surf_ur.x, m.center.y - r - dx); +Surface s_r = Surface(pFrom=surf_lr, pTo=surf_ur); +pair wall_join_l = xscale(-1)*wall_join_r; +pair surf_ul = xscale(-1)*surf_ur; +pair surf_ll = xscale(-1)*surf_lr; +Surface s_l = Surface(pFrom=surf_ul, pTo=surf_ll); +Distance d = Distance(pFrom=wall_join_l, pTo=wall_join_r, L="$3L/4$"); + +draw(m.center -- junction); +draw(wall_join_l -- junction -- wall_join_r); +s_r.draw(); +s_l.draw(); +d.draw(); +m.draw(); +\end{asy} +\end{center} +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/problem13.18.tex b/latex/problems/problem13.18.tex new file mode 100644 index 0000000..96c302a --- /dev/null +++ b/latex/problems/problem13.18.tex @@ -0,0 +1,24 @@ +\begin{problem*}{13.18} +A series of pulses, each of amplitude $0.150\U{m}$, are sent down a +string that is attached to a post at one end. The pulses are +reflected at the pulse and travel back along the string without loss +of amplitude. When two waves are present on the same string, the net +displacement of a particular element of the string is the sum of the +displacements of the individual waves at that point. What is the net +displacement of an element at a point on the string where two pulses +are crossing \Part{a} if the string is rigidly attached to the post +and \Part{b} if the end at which reflection occurs is free to slide up +and down. +\end{problem*} + +\begin{solution} +\Part{a} +With a rigid connection, the reflected pulses will be inverted. +Therefor, overlapping pulses will cancel out (destructive +interference) and the total amplitude will be \ans{$0\U{m}$}. + +\Part{b} +With a sliding connection, the reflected pulses will not be inverted. +Therefor, overlapping pulses will add together (constructive +interference) and the total amplitude will be \ans{$0.300\U{m}$}. +\end{solution} diff --git a/latex/problems/problem13.27.tex b/latex/problems/problem13.27.tex new file mode 100644 index 0000000..6f74fec --- /dev/null +++ b/latex/problems/problem13.27.tex @@ -0,0 +1,42 @@ +\begin{problem*}{13.27} +An ultrasonic tape measure uses frequencies above $20\U{MHz}$ to +determine the dimensions of structures such as buildings. It does so +by emitting a pulse of ultrasound into the air and then measuring the +time interval for an echo to return from a reflecting surface whose +distance away is to be measured. The distance is displayed as a +digital readout. For a tape measure that emits a pulse of ultrasound +with a frequency of $22\U{MHz}$, \Part{a} what is the distance to an +object from which the echo pulse returns after $24.0\U{ms}$ when the +air temperature is $26\dg$? \Part{b} What should be the duration of +the emitted pulse if it is to include ten cycles of the ultrasonic +wave? \Part{c} What is the spatial length of such a pulse? +\end{problem*} + +\begin{solution} +The speed of sound in dry air is approximately +\begin{equation} + v = 331\U{m/s} + 0.6\U{m/s$\cdot$\dg C}\;, +\end{equation} +so at $26\dg\text{C}$, $v = 347\U{m/s}$. + +\Part{a} +The elapsed time is the time taken for the pulse to travel from the +tape measure to the object and back, traveling the distance between +the tape measure and the object twice. +\begin{align} + 2d &= vt & d &= \frac{vt}{2} = \ans{4.16\U{m}} +\end{align} + +\Part{b} +A pulse containing ten cycles is ten periods long +\begin{equation} + \Delta t = 10T = \frac{10}{f} = \frac{10}{22\E{6}\U{s}} = \ans{0.455\U{$\mu$s}} +\end{equation} + +\Part{c} +\begin{equation} + L = v\Delta t = 347\U{m/s}\cdot0.455\U{$\mu$s} = \ans{0.158\U{mm}} +\end{equation} + + +\end{solution} diff --git a/latex/problems/problem13.40.tex b/latex/problems/problem13.40.tex new file mode 100644 index 0000000..da74b43 --- /dev/null +++ b/latex/problems/problem13.40.tex @@ -0,0 +1,61 @@ +\begin{problem*}{13.40} +Two points $A$ and $B$ on the surface of the Earth are at the same +longitude and $60.0\dg$ apart in latitude. Suppose an earthquake at +point $A$ creates a $P$ wave that reaches point $B$ by traveling +straight through the body of the Earth at a constant speed of +$7.80\U{km/s}$. The earthquake also radiates a Rayleigh wave, which +travels along the surface of the Earth at $4.50\U{km/s}$. \Part{a} +Which of these two seismic waves arrives at $B$ first. \Part{b} What +is the time difference between the arrivals of the two waves at $B$? +Take the radius of the Earth to be $6370\U{km}$. +\end{problem*} + +\begin{solution} +\Part{a} +The $P$ wave reaches $B$ first because it travels faster along a more +direct route. + +\Part{b} +To find the distances traveled by each wave, take a crossectional view of the Earth. +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1.9cm; +real theta = 60; //degrees, between the source and point B + +draw(scale(u)*unitcircle, dotted); +pair source = u*dir(theta/2); +pair target = u*dir(-theta/2); + +Angle a = Angle(source, (0,0), (source+target)/2, radius=1cm, L=" $\theta/2$"); +// space before text in the label because inline-asymptote doesn't know +// how big the label will be when it sets the positions. See +// http://asymptote.sourceforge.net/doc/LaTeX-usage.html +// So we add some space by hand. +a.draw(); +draw(target--(0,0)--source); +draw((0,0)--((source+target)/2)); +draw(source--target, red); +draw(arc((0,0), u, theta/2, -theta/2), blue); + +label("$r$", source/2, NW); +label("$A$", source, E); +label("$B$", target, E); +\end{asy} +\end{center} +Where the red line is the $P$ wave path and the blue line is the +Rayleigh wave path. From this picture it is clear that the path +lengths are +\begin{align} + d_P &= 2r\sin(\theta/2) & d_R = r\theta +\end{align} +So the travel times are +\begin{align} + t_P &= \frac{d_P}{v_P} =\frac{2\cdot6370\U{km}\cdot\sin(\pi/6)}{7.80\U{km/s}} + = 817\U{s} = 13.6\U{min} \\ + t_R &= \frac{d_R}{v_R} = \frac{6370\U{km}\cdot\pi/3}{4.50\U{km/s}} + = 1480\U{s} = 24.7\U{min} \\ + \Delta t &= t_R - t_P = \ans{665\U{s}} = \ans{11.1\U{min}} +\end{align} +\end{solution} diff --git a/latex/problems/problem14.05.tex b/latex/problems/problem14.05.tex new file mode 100644 index 0000000..c2f57f4 --- /dev/null +++ b/latex/problems/problem14.05.tex @@ -0,0 +1,61 @@ +\begin{problem*}{14.5} +Two traveling sinusoidal waves are described by the wave functions +\begin{equation} + y_1 = (5.00\U{m})\cdot\sin[\pi(4.00x-1200t)] +\end{equation} +and +\begin{equation} + y_2 = (5.00\U{m})\cdot\sin[\pi(4.00x-1200t-0.250)] +\end{equation} +where $x$, $y_1$, and $y_2$ are in meters and $t$ is in seconds. +\Part{a} What is the amplitude of the resultant wave? +\Part{b} What is the frequency of the resultant wave? +\end{problem*} % problem 14.5 + +\begin{solution} +\Part{a} +\begin{equation} + y = y_1 + y_2 + = 5.00\U{m}\cdot\p\{{\sin[\pi(4.00x-1200t)]+\sin[\pi(4.00x-1200t-0.250)}\} + = 5.00\U{m}\cdot\p[{\sin(\theta)+\sin(\theta-\pi/4)}] \;, +\end{equation} +where $\theta\equiv\pi(4.00x-1200t)$. So $y_2$ trails $y_1$ by +$\pi/4=45\dg$. In terms of the reference circle, that looks like +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1.2cm; +pair p = (u,0); +pair q = rotate(-45)*p; + + +draw(scale(u)*unitcircle, dotted); + +Angle a = Angle(p, (0,0), q, radius=6mm, "$\pi/4$"); +a.draw(); +draw(p--(p+q), blue+dashed); +draw(q--(p+q), red+dashed); +draw((0,0)--p, red); +draw((0,0)--q, blue); +draw((0,0)--(p+q), green); +dot(p); +dot(q); +dot(p+q); +\end{asy} +\end{center} +The amplitude of $y$ is then given by vector addition +\begin{equation} + A = 2\cdot A\cos(\phi/2) + = 2\cdot 5.00\U{m}\cdot\cos(\pi/8) = \ans{9.24\U{m}} \;, +\end{equation} +where $\phi=\pi/4$ is the phase difference between $y_1$ and $y_2$. + +\Part{b} +Both $y_1$ and $y_2$ rotate around the reference circle with a frequency of +\begin{equation} + f = \frac{\omega}{2\pi} = \frac{1200\pi\U{rad/s}}{2\pi\U{rad/cycle}} + = 600\U{Hz} \;, +\end{equation} +so $y$ must also rotate around the reference circle at $\ans{600\U{Hz}}$. +\end{solution} diff --git a/latex/problems/problem14.06.tex b/latex/problems/problem14.06.tex new file mode 100644 index 0000000..b1e6a66 --- /dev/null +++ b/latex/problems/problem14.06.tex @@ -0,0 +1,27 @@ +\begin{problem*}{14.6} +Two identical sinusoidal waves with wavelengths of $3.00\U{m}$ travel +in the same direction at a speed of $2.00\U{m/s}$. The second wave +originates from the same point as the first, but at a later time. The +amplitude of the resultant wave is the same as that of each of the two +initial waves. Determine the minimum possible time interval between +the starting moments of the two waves. +\end{problem*} % problem 14.6 + +\begin{solution} +This is Problem 14.6 backwards. From the amplidude of the resultant +wave, we can find the phase difference between the two constituent +waves. +\begin{align} + A &= 2\cdot A\cdot \cos(\phi) \\ + \phi &= \arccos\p({\frac{1}{2}}) = \frac{\pi}{3} \\ + \Delta \theta &= 2\phi = \frac{2\pi}{3}\U{rad} \;. +\end{align} +The angular speed of the waves is given by +\begin{equation} + \omega = 2\pi f = 2\pi \frac{v}{\lambda} = \frac{4\pi}{3}\U{rad/s} +\end{equation} +So the time interval is +\begin{equation} + \Delta t = \frac{\Delta \theta}{\omega} = \ans{0.500\U{s}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem14.08.tex b/latex/problems/problem14.08.tex new file mode 100644 index 0000000..9529e5d --- /dev/null +++ b/latex/problems/problem14.08.tex @@ -0,0 +1,74 @@ +\begin{problem*}{14.8} +Two loudspeakers are placed on a wall $2.00\U{m}$ apart. A listener +stands $3.00\U{m}$ from the wall directly in front of one of the +speakers. A single oscillator is driving the speakers at a frequency +of $300\U{Hz}$. \Part{a} What is the phase difference between the two +waves when they reach the observer? \Part{b} What is the frequency +closest to $300\U{Hz}$ to which the oscillator may be adjusted so that +the observer hears minimal sound? +\end{problem*} % problem 14.8 + +\begin{solution} +\begin{center} +\begin{asy} +import Mechanics; + +real u=1cm; +pair SpeakerA = (0,0); +pair SpeakerB = (0,2u); +pair Listener = (3u,0); + +// Extra label text for spacing with inline asymptote +Distance dAB = Distance(SpeakerA, SpeakerB, Label("$L=2\U{m}$", "L=2 m")); +Distance dAL = Distance(Listener, SpeakerA, Label("$d_a=3\U{m}$", "da=3 m")); +Distance dBL = Distance(SpeakerB,Listener, + Label("$\qquad\qquad\qquad\qquad\qquad d_b=\sqrt{L^2+d_a^2}=3.606\U{m}$", + "$d_b=\sqrt{L^2+d_a^2}=3.606mm$")); + +dot(SpeakerA); +label("$S_a$", SpeakerA, W); +dot(SpeakerB); +label("$S_b$", SpeakerB, W); +dot(Listener); +label("Listener", Listener, E); +dAB.draw(); +dAL.draw(rotateLabel=false); +dBL.draw(rotateLabel=false); +\end{asy} +\end{center} +\Part{a} +From Table 13.1 we find that the speed of sound in air at $20\dg C$ is +$v=343\U{m/s}$. The wavelength of this sound is +\begin{equation} + \lambda = \frac{v}{f} = 1.143\U{m} +\end{equation} +The phase change from speaker $S_a$ is therefore +\begin{equation} + \theta_a = k d_a = \frac{2\pi d_a}{\lambda} = 16.49\U{rad} \;, +\end{equation} +and from speaker $S_b$ is +\begin{equation} + \theta_b = k d_b = \frac{2\pi d_b}{\lambda} = 19.81\U{rad} \;. +\end{equation} +The phase difference is +\begin{equation} + \Delta\theta = \theta_b - \theta_a = \ans{3.33\U{rad}} +\end{equation} + +\Part{b} +For minimal sound, we want the phase difference to be exactly $\pi$ +(or some odd multiple of $\pi$). We see that it's already close to +$\pi$ with our initial frequency of $300\U{Hz}$, only a bit high. +Decreasing the freqency a bit will reduce the rate of dephasing +between the two waves, reducing $\Delta\theta$, so we're looking for a +frequency slightly less than $300\U{Hz}$. +\begin{align} + \Delta\theta &= \theta_b-\theta_a + = \frac{2\pi (d_b-d_a)}{\lambda} + = \frac{2\pi f (d_b-d_a)}{v} \\ + f &= \frac{v\Delta\theta}{2\pi (d_b-d_a)} + = \frac{v \pi}{2\pi (d_b-d_a)} + = \frac{v}{2(d_b-d_a)} + = \frac{343\U{m/s}}{2\cdot 0.606\U{m}} = \ans{283\U{Hz}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem14.09.tex b/latex/problems/problem14.09.tex new file mode 100644 index 0000000..d0a3102 --- /dev/null +++ b/latex/problems/problem14.09.tex @@ -0,0 +1,41 @@ +\begin{problem*}{14.9} +Two sinusoidal waves in a string are defined by the functions +\begin{equation} + y_1 = (2.00\U{cm})\cdot\sin(20.0x-32.0t) +\end{equation} +and +\begin{equation} + y_2 = (2.00\U{cm})\cdot\sin(25.0x-40.0t) +\end{equation} +where $y$ and $x$ are in centimeters and $t$ is in seconds. +\Part{a} What is the phase difference between there two waves at the +point $x=5.00\U{cm}$ at $t=2.00\U{s}$? +\Part{b} What is the positive $x$ value closest to the origin for +which the two phases differ by $\pm\pi$ at $t=2.00\U{s}$? (This +location is where the two waves add to zero.) +\end{problem*} % problem 14.9 + +\begin{solution} +\Part{a} +This is just plugging in +\begin{align} + \theta_1 &= 20.0 \cdot 5.00 - 32.0 \cdot 2.00 = 36\U{rad} \\ + \theta_2 &= 25.0 \cdot 5.00 - 40.0 \cdot 2.00 = 45\U{rad} \\ + \Delta\theta &= \theta_2 - \theta_1 = \ans{9\U{rad}} + = 516\dg = \ans{156\dg} +\end{align} + +\Part{b} +We're supposed to find some $x$ for a given $t$ such that +\begin{align} + \Delta\theta &= \theta_2 - \theta_1 + = (25.0\cdot x - 80.0) - (20.0\cdot x - 64.0) + = 5.0\cdot x - 16.0 = n\pi \\ + x &= \frac{n\pi + 16.0}{5.0} +\end{align} +for some odd $n$. $16/\pi=5.09$, so $n=-5$ will give the smallest +positive $x$ for which this is true, and +\begin{equation} + x = \frac{-5\pi + 16.0}{5.0} = 0.0584\U{cm} = \ans{584\U{$\mu$m}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem14.11.T.tex b/latex/problems/problem14.11.T.tex new file mode 100644 index 0000000..0bc2d26 --- /dev/null +++ b/latex/problems/problem14.11.T.tex @@ -0,0 +1,253 @@ +\begin{problem} +\emph{BONUS PROBLEM}. Two identical speakers $d=10.0\U{m}$ apart are +driven by the same oscillator with a frequency of +$f=21.5\U{Hz}$. \Part{a} Explain why a reciever at point $A$ records +a minimum in sound intensity from the two speakers. \Part{b} If the +reciever is moved in the plane of the speakers, what path should it +take so that the intensity remains at a minimum? That is, determine +the relationship between $x$ and $y$ (the coordinates of the reciever) +that causes the receiver to record a minimum in sound intensity. Take +the speed of sound to be $v=344\U{m/s}$. + +\begin{center} +\begin{asy} +import Mechanics; +real u = 0.3cm; + +real d = 10; // m +real f = 21.5; // Hz +real v = 344; // m/s +real L = v/f; // m, wavelength + +Vector x = Vector(center=(-.7d*u,0), mag=1.4d*u, dir=0, L="$x$"); +Vector y = Vector(center=(0,0), mag=0.6d*u, dir=90, L="$y$"); + +x.draw(); +y.draw(); +dot((-d/2*u,0)); +dot((d/2*u,0)); +label("Speaker 1", (-d/2*u, 0), S); +label("Speaker 2", (d/2*u, 0), S); + +dot((L/4*u,0)); +label("$A$", (L/4*u,0), N); +\end{asy} +\end{center} +\end{problem} % based on chapter 14, problem 11 + +\begin{nosolution} +\end{nosolution} + +\begin{solution} +\Part{a} +To get a feeling for why a minimum exists, consider the intensity of +output sound when the amplitude of the wave generated by a speaker is +zero at the generating speaker (the wavelength is given by $\lambda = +v/f = 16.0\U{m}$). +\begin{center} +\begin{asy} +// Draw wave interference between the two speakers +import Mechanics; +real u = 1.2cm; + +real d = 10; // m +real f = 21.5; // Hz +real v = 344; // m/s +real L = v/f; // m, wavelength +write("wavelength ", L); + +Vector x = Vector(center=(-.7d*u,0), mag=1.4d*u, dir=0, L="$x$"); + +x.draw(); +dot((-d/2*u,0)); +dot((d/2*u,0)); + +int n=200; +real amp=u/3; +real xstart=-d/2; +real dx=d/n; +path pLeft = (xstart*u, 0); // Left speaker output wave +for (int i=1; i<=n; ++i) { + pLeft = pLeft..(xstart*u+i*dx*u, amp*sin(2pi*dx*i/L)); +} +draw(pLeft, p=red); +draw(xscale(-1)*pLeft, p=blue); // Right speaker output wave +path pSum = (xstart*u, 0+amp*sin(2pi*d/L)); // Total +for (int i=1; i<=n; ++i) { + pSum = pSum..(xstart*u+i*dx*u, amp*(sin(2pi*dx*i/L)+sin(2pi*(d-dx*i)/L))); +} +draw(pSum, p=green); +\end{asy} +\end{center} +Note the existence of two nodes in the sum (green). This snapshot +occurs just before the standing wave reaches it's maximum amplitude, +because a small time later the wave from the left speaker (red) will +have moved to the right, the wave from the left speaker (blue) will +have moved to the left, and the peaks will both arive in the middle, +forming +\begin{center} +\begin{asy} +// Draw wave interference between the two speakers +import Mechanics; +real u = 1.2cm; + +real d = 10; // m +real f = 21.5; // Hz +real v = 344; // m/s +real L = v/f; // m, wavelength +write("wavelength ", L); +write("phase at far speaker ", 2pi*d/L); +write("x_n at +/- ", L/4); + +Vector x = Vector(center=(-.7d*u,0), mag=1.4d*u, dir=0, L="$x$"); + +x.draw(); +dot((-d/2*u,0)); +dot((d/2*u,0)); + +int n=200; +real amp=u/3; +real xstart=-d/2; +real dx=d/n; +real phi_orig_center = d/2/L*2pi; +real phi_intended_center = pi/2; +real phi= phi_intended_center-phi_orig_center; +path pLeft = (xstart*u, amp*sin(phi)); // Left speaker output wave +for (int i=1; i<=n; ++i) { + pLeft = pLeft..(xstart*u+i*dx*u, amp*sin(2pi*dx*i/L+phi)); +} +draw(pLeft, p=red); +draw(xscale(-1)*pLeft, p=blue+dashed); // Right speaker output wave +path pSum = (xstart*u, amp*(sin(phi)+sin(2pi*d/L+phi))); // Total +for (int i=1; i<=n; ++i) { + pSum = pSum..(xstart*u+i*dx*u, amp*(sin(2pi*dx*i/L+phi)+sin(2pi*(d-dx*i)/L+phi))); +} +draw(pSum, p=green); +//dot((L/4*u,0)); // Check my answer. +\end{asy} +\end{center} + +The easiest wat to think about this is to focus on the phase +difference between the two waves as a function of position. The two +speakers are broadcasting in phase, but by the time the second wave +reaches the first speaker it is $2\pi \cdot d/\lambda = 3.93\U{rad}$ +out of phase. The nodes come when the sound from the two speakers are +out of phase by $\pi\U{rad} = 180\dg$. Letting $x=\pm d/2$ be the +position of the two speakers, we have +\begin{align} + \phi_L &= 2\pi\frac{x+d/2}{\lambda} \mod 2\pi \\ + \phi_R &= -2\pi\frac{x+d/2}{\lambda} \mod 2\pi \\ + \Delta \phi &= \phi_L - \phi_R = \frac{4\pi x}{\lambda} \mod 2\pi \;, +\end{align} +so the sound is in phase at $x=0$ (as you'd expect), and we get our +peak amplitude there. The nodes occur at +\begin{align} + \pi &= \Delta \phi (x_n) = \frac{4\pi x_n}{\lambda} \mod 2\pi \\ + x_n &= \pm\frac{\lambda}{4} = \ans{\pm 4.00\U{m}} +\end{align} + +\Part{b} +Extending this reasoning into the plane of the speakers, we know +beforehand that nodes will lie along hyperbola, because hyperbola are +``the locus of points where the difference of the distances to the two +foci is a constant''. If $\lambda/4$ is the distance from the origin to +the right hand node on the line between the two foci (which we found +in \Part{a}), and $d/2$ is the distance from the origin to either +focus, then the appropriate hyperbola is +\begin{equation} + \ans{1 = \frac{x^2}{\lambda^2/16} - \frac{y^2}{d^2/4-\lambda^2/16} + = \frac{x^2}{16\U{m$^2$}} - \frac{y^2}{25\U{m$^2$}-16\U{m$^2$}} + = \frac{x^2}{16\U{m$^2$}} - \frac{y^2}{9\U{m$^2$}} } \;. +\end{equation} +\begin{center} +\begin{asy} +import Mechanics; +real u = 0.3cm; + +real d = 10; // m +real f = 21.5; // Hz +real v = 344; // m/s +real L = v/f; // m, wavelength +write("wavelength ", L); + +Vector x = Vector(center=(-.7d*u,0), mag=1.4d*u, dir=0, L="$x$"); +Vector y = Vector(center=(0,0), mag=0.6d*u, dir=90, L="$y$"); + +x.draw(); +y.draw(); +dot((-d/2*u,0)); +dot((d/2*u,0)); +label("Speaker 1", (-d/2*u, 0), S); +label("Speaker 2", (d/2*u, 0), S); + +real x, y; +real x_scale = L**2/16; +real y_scale = d**2/4 - L**2/16; +write("x scaler ", x_scale); +write("y scaler ", y_scale); + +real xstart = L/4; +int n = 100; +real dx = (0.7d-xstart)/n; +path hyp = (xstart,0)*u; +for (int i=1; i<=n; ++i) { + x = xstart + i*dx; + y = sqrt(y_scale*(x**2/x_scale-1)); + hyp = hyp..(x, y)*u; +} +draw(hyp, red); +\end{asy} +\end{center} + +If you don't remember that much about hyperbolas, you can grind +through the algebra and get the same result. +\begin{align} + \phi_L &= 2\pi\frac{\sqrt{\p({x+d/2})^2+y^2}}{\lambda} \mod 2\pi \\ + \phi_R &= 2\pi\frac{\sqrt{\p({x-d/2})^2+y^2}}{\lambda} \mod 2\pi \\ + \Delta \phi &= \phi_L - \phi_R + = \frac{2\pi}{\lambda} + \p[{\sqrt{\p({x+d/2})^2+y^2}-\sqrt{\p({x-d/2})^2+y^2}}] \mod 2\pi \;. +\end{align} +If we decide to follow only the node on the right (where $\Delta\phi = +\pi$), we can dispense with the $\text{mod } 2\pi$ yielding +\begin{align} + \pi &= \Delta \phi + = \frac{2\pi}{\lambda} + \p[{\sqrt{\p({x+d/2})^2+y^2} + -\sqrt{\p({x-d/2})^2+y^2}}] \\ + \frac{\lambda}{2} + &= \sqrt{\p({x+d/2})^2+y^2}-\sqrt{\p({x-d/2})^2+y^2} \;. +\end{align} +Focusing on the right hand side of the equation, we see +\begin{equation} + \sqrt{\p({x \pm d/2})^2+y^2} + = \sqrt{x^2 \pm dx + d^2/4 +y^2} + = \sqrt{a \pm b} \;, +\end{equation} +with +\begin{align} + a &\equiv x^2 + \frac{d^2}{4} + y^2 & b &\equiv dx \;. +\end{align} +Going back to our main equation in terms of $a$ and $b$ +\begin{align} + \frac{\lambda}{2} &= \sqrt{a+b} - \sqrt{a-b} \\ + \frac{\lambda^2}{4} &= \p({\sqrt{a+b} - \sqrt{a-b}})^2 \\ + &= \sqrt{(a+b)^2} + \sqrt{(a-b)^2} - 2\sqrt{(a+b)(a-b)} \\ + &= (a+b) + (a-b) - 2\sqrt{a^2 - b^2} \\ + &= 2a - 2\sqrt{a^2 - b^2} \\ + \frac{\lambda^2}{4} - 2a &= -2\sqrt{a^2 - b^2} \\ + \frac{\lambda^4}{16} - \lambda^2 a + 4a^2 &= 4(a^2 - b^2) \\ + \frac{\lambda^4}{16} - \lambda^2 a + 4b^2 &= 0 \;. +\end{align} +Now that we've gotten rid of the square roots, we go back and plug in +for $a$ and $b$. +\begin{align} + \frac{\lambda^4}{16} - \lambda^2 x^2 - \frac{\lambda^2 d^2}{4} - \lambda^2 y^2 + 4d^2 x^2 &= 0 \\ + \frac{\lambda^2}{16}\p({\lambda^2 - 4d^2}) - \lambda^2 y^2 + (4d^2-\lambda^2) x^2 &= 0 \\ + (4d^2-\lambda^2) x^2 - \lambda^2 y^2 &= \frac{\lambda^2}{16}\p({4d^2 - \lambda^2}) \\ + x^2 - \frac{\lambda^2 y^2}{4d^2 - \lambda^2} &= \frac{\lambda^2}{16} \\ + \frac{16 x^2}{\lambda^2} - \frac{16 y^2}{4d^2 - \lambda^2} &= 1 \\ + \frac{x^2}{\lambda^2/16} - \frac{y^2}{d^2/4 - \lambda^2/16} &= 1 \;, +\end{align} +which is the same equation our knowledge of the hyperbola gave us earlier. +\end{solution} diff --git a/latex/problems/problem14.19.T.tex b/latex/problems/problem14.19.T.tex new file mode 100644 index 0000000..f186bc4 --- /dev/null +++ b/latex/problems/problem14.19.T.tex @@ -0,0 +1,37 @@ +\begin{problem} +A string with a mass of $5\U{g}$ and a length of $1\U{m}$ has one end +attached to a wall. $70\U{cm}$ from the wall, the string passes over +a pully and hangs, supporting a $1\U{kg}$ mass. \Part{a} What is the +fundamental frequency of vibration? \Part{b} What is the frequency of +second harmonic? +\end{problem} % based on P14.19 + +\begin{solution} +\Part{a} +The linear density of the string is +\begin{equation} + \mu = \frac{m}{L} = \frac{5\U{g}}{1\U{m}} = 5\U{g/m} \;. +\end{equation} +The tension of the string is +\begin{equation} + T = Mg = 1\U{kg} \cdot 9.8\U{m/s$^2$} = 9.8\U{N} \;. +\end{equation} +The speed of wave-propagation in the string is +\begin{equation} + v = \sqrt{\frac{T}{\mu}} = 44.3\U{m/s} \;. +\end{equation} + +For the fundamental frequency, the distance between the fixed ends is half a wavelength, so a moving wave crosses it in half a period. +\begin{align} + \Delta x &= v \Delta t = v \cdot \frac{1}{2f_1} \\ + f_1 &= \frac{v}{2\Delta x} = \frac{44.3\U{m/s}}{2\cdot 0.70\U{m}} + = \ans{31.6\U{Hz}} +\end{align} + +\Part{b} +For the second harmonic, the distance between the fixed ends is a full wavelength, so a moving wave crosses it in a full period. +\begin{align} + \Delta x &= v \cdot \frac{1}{f_2} \\ + f_2 &= 2 f_1 = \ans{63.2\U{Hz}} +\end{align} +\end{solution} diff --git a/latex/problems/problem14.20.tex b/latex/problems/problem14.20.tex new file mode 100644 index 0000000..a7281ff --- /dev/null +++ b/latex/problems/problem14.20.tex @@ -0,0 +1,105 @@ +\begin{problem*}{14.20} +In the arrangement shown in Figure P14.20, an object can be hung from +a string (with a linear mass density $\mu=2.00\U{g/m}$) that passes +over a light pulley. The string is connected to a vibrator (of +constant frequency $f$), and the length of the string between point +$P$ and the pulley is $L=2.00\U{m}$. When the mass $m$ of the object +is either $16.0\U{kg}$ or $25.0\U{kg}$, standing waves are observed, +but no standing waves are observed with any mass between these +values. \Part{a} What is the frequency of the vibrator? (Note: The +greater the tension in the string, the smaller the number of nodes in +the standing wave.) \Part{b} What is the largest object mass for +which standing waves could be observed? +\begin{center} +\begin{asy} +import Mechanics; + +real u=1cm; +real L=3u; // length between P and pulley +real A=.15u; // amplitude of vibrations +real N=3; // number of full vibrations +real w=1.7u; // width of box + +real n = 100; +real dx = L/n; +real dtheta = N*2*pi/n; +int i; +path p; +for (i=0; i<=n; ++i) { + p = p..(i*dx, A*sin(i*dtheta)); +} + +Block vib = Block((-w/2,0), width=w, height=w/3, L="vibrator"); +Block obj = Block((L+(1+cos(pi/4))*A, -L/3), width=vib.height, L="$m$"); +Distance dL = Distance((0,vib.height/2), (L,vib.height/2), L="$L$"); + +draw(p, blue); +draw(yscale(-1)*p, blue+dotted); +draw((L+(1+cos(pi/4))*A, -A*sin(pi/4))--obj.center, blue); +dL.draw(); +filldraw(shift(L+A*cos(pi/4),0-A*sin(pi/4))*scale(A)*unitcircle); +vib.draw(); +dot((0,0)); +label("$P$", (0,vib.height/2), dir(90)); +label("$\mu$", (L/2,-vib.height/2), dir(-90)); +obj.draw(); +\end{asy} +\end{center} +\end{problem*} % problem 14.20 + +\begin{solution} +\Part{a} +For a patricular hanging mass $m$, the tension in the string balances +the gravitational force on the mass, so +\begin{equation} + T = mg \;, +\end{equation} +so the speed of sound in the string is +\begin{equation} + v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{mg}{\mu}} \;. +\end{equation} + +Standing waves on strings occur when a wave completes some number of +full cycles in a round trip. In mathematical terms +\begin{align} + n\cdot 2\pi &= k\cdot 2L = \frac{2\pi}{\lambda} \cdot 2L \\ + n &= \frac{2L}{\lambda} \\ + \lambda &= \frac{2L}{n} +\end{align} +for integer $n$ (the number of the particular vibrational mode). With the +generator operating at a fixed frequency $f$, the wavelength is also +related to the wave speed by +\begin{equation} + \lambda = \frac{v}{f} \;, +\end{equation} +so +\begin{align} + \frac{2L}{n} &= \lambda = \frac{v}{f} = \frac{1}{f}\cdot\sqrt{\frac{mg}{\mu}} \\ + 2Lf\sqrt{\frac{\mu}{mg}} &= n \;, +\end{align} + +To put it all together we notice that the two masses, $m_1=16.0\U{kg}$ +and $m_2=25.0\U{kg}$ are said to produce consecutive modes. From the +last equation, we see that increasing mass $m$ decreases the mode +number $n$, so the heavier mass must be one mode lower than the +lighter, or $n_1=n_2+1$. From here on out, it's all algebra to find +$f$. +\begin{align} + n_2 &= 2Lf\sqrt{\frac{\mu}{m_2g}} \\ + n_1 &= 2Lf\sqrt{\frac{\mu}{m_1g}} = n_2+1 = 2Lf\sqrt{\frac{\mu}{m_2g}} + 1\\ + 1 &= 2Lf\sqrt{\frac{\mu}{g}}\p({\frac{1}{\sqrt{m_1}}-\frac{1}{\sqrt{m_2}}}) \\ + f &= \p[{2L\sqrt{\frac{\mu}{g}}\p({\frac{1}{\sqrt{m_1}}-\frac{1}{\sqrt{m_2}}})}]^{-1} + = \ans{350\U{Hz}} +\end{align} + +\Part{b} +As we saw in \Part{a}, increasing the mass decreased the vibrational +mode number. The largest mass that can sustain standing waves is the +one for which the vibration is in the first mode, so +\begin{align} + 1 &= n = 2Lf\sqrt{\frac{\mu}{m_\text{max} g}} \\ + \sqrt{\frac{m_\text{max} g}{\mu}} &= 2Lf \\ + \frac{m_\text{max} g}{\mu} &= (2Lf)^2 \\ + m_\text{max} &= \frac{\mu(2Lf)^2}{g} = \ans{400\U{kg}} +\end{align} +\end{solution} diff --git a/latex/problems/problem14.22.tex b/latex/problems/problem14.22.tex new file mode 100644 index 0000000..6c9e623 --- /dev/null +++ b/latex/problems/problem14.22.tex @@ -0,0 +1,14 @@ +\begin{problem*}{14.22} +The top string of a guitar has a fundamental frequency of $330\U{Hz}$ +when it is allowed to vibrate as a whole, along all its $64.0\U{cm}$ +length from the neck to the bridge. A fret is provided for limiting +vibration to just the lower two thirds of the string. \Part{a} If the +string is pressed down at this fret and plucked, what is he new +fundamental frequency? \Part{b} The guitarist can play a ``natural +harmonic'' by gently touching the string at the location of this fret +and plucking the string at about one sixth of the way along its length +from the bridge. What frequency will be herd then? +\end{problem*} % problem 14.22 + +\begin{solution} +\end{solution} diff --git a/latex/problems/problem14.51.T.tex b/latex/problems/problem14.51.T.tex new file mode 100644 index 0000000..b45938f --- /dev/null +++ b/latex/problems/problem14.51.T.tex @@ -0,0 +1,69 @@ +\begin{problem} +A student uses an audio oscillator of adjustable frequency to measure +the depth of the well. The student hears two successive resonances at +$122.0\U{Hz}$ and $127.1\U{Hz}$. How deep is the well? +\end{problem} % based on P14.51 + +\begin{solution} +We can approximate the well as a cylinder closed at the bottom and +open at the top. As a result, the well resonates at (Equation 14.11) +\begin{equation} + f_n = \frac{nv}{4L} \;. \label{eqn.pipe_freq} +\end{equation} +It's $5\text{\dg C}$ outside my window at the moment, so the speed of +sound is around (Equation 13.27) +\begin{equation} + v = 311\U{m/s} + (0.6\U{m/s$\cdot$\dg C})T = 314\U{m/s} \;. +\end{equation} + +Let the resonance at $f_L = 122\U{Hz}$ be the $n^\text{th}$ harmonic. +Then the $f_H = 127.1\U{Hz}$ harmonic is the $(n+2)^\text{th}$ +harmonic (as the next successive resonance in a pipe closed at one +end). Plugging these into Equation~\ref{eqn.pipe_freq} we have +\begin{align} + \frac{f_H}{f_L} &= \frac{\frac{(n+2)v}{4L}}{\frac{nv}{4L}} + = \frac{n+2}{n} = 1 + \frac{2}{n} \\ + \frac{2}{n} &= \frac{f_H}{f_L} - 1 \\ + n &= \frac{2}{\frac{f_H}{f_L} - 1} = \frac{2f_L}{f_H-f_L} + = 48 \;. +\end{align} +Hmm, 48 is not an odd harmonic, and all harmonics of pipes open at one +end and closed at the other are odd. This means that \emph{your TA + made a mistake} writing the homework problem, at which point you +should start emailing questions and complaints. + +\Part{corrected problem} +If you want two numbers that work, you can use $126\U{Hz}$ and $138\U{Hz}$, +in which case +\begin{equation} + n = \frac{2f_L}{f_H-f_L} = 21 \;. +\end{equation} + +Now that we know the $f_L$ is the $21^\text{st}$ harmonic, we can plug into +Equation~\ref{eqn.pipe_freq} and find the well depth +\begin{align} + f_L &= \frac{nv}{4L} \\ + L &= \frac{nv}{4 f_L} = \frac{21 \cdot 314\U{m/s}}{4 \cdot 126\U{Hz}} + = \ans{13.1\U{m}} +\end{align} + +\Part{open well} +On the other hand, perhaps the reason for the even result for $n$ is +that the well is not a well after all, but a pipe passing down into a +vertical cystern (\url{http://en.wikipedia.org/wiki/Cistern}) where +there is a layer of air before the level of the water. In that case, +the pipe is open to air at both ends and even harmonics are allowed, +so +\begin{align} + f_n &= \frac{nv}{2L} \\ + \frac{f_H}{f_L} &= \frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}} + = \frac{n+1}{n} = 1 + \frac{1}{n} \\ + \frac{1}{n} &= \frac{f_H}{f_L} - 1 \\ + n &= \frac{1}{\frac{f_H}{f_L} - 1} = \frac{f_L}{f_H-f_L} + = 24 \\ + L &= \frac{nv}{2 f_L} = \frac{24 \cdot 314\U{m/s}}{2 \cdot 122\U{Hz}} + = \ans{30.9\U{m}} +\end{align} + +Apologies for the mistake. +\end{solution} diff --git a/latex/problems/problem19.03.tex b/latex/problems/problem19.03.tex new file mode 100644 index 0000000..7188acb --- /dev/null +++ b/latex/problems/problem19.03.tex @@ -0,0 +1,29 @@ +\begin{problem*}{19.3} +Nobel laureate Richard Feynman once said that if two persons stood at +arm's length from each other and each person had $p=1$\% more +electrons than protons, the force of repulsion between them would be +enough to lift a ``weight'' equal to that of the entire Earth. Carry +out an order of magnitude calculation to substantiate this assertion. +\end{problem*} + +\begin{solution} +Let $m = 70\U{kg}$ be the mass of one person, and $q_e$ be the charge +of one electron. Assume that there are approximately equal numbers of +protons, electrons, and neutrons in a person. Electrons have much +less mass than protons or neutrons, so we ignore their mass +contribution. Protons and neutrons have very similar masses, so $N = +(m/2)/m_p$ is the number of protons, and $N_q = N \cdot p$ is the +number of extra electrons in each person. Assume they are seperated +by $r = 1\U{m}$. The force of repulsion $F$ is given by +\begin{equation} + F = k_e \frac{q^2}{r^2} + = k_e \left(\frac{m p q_e}{2 m_p r}\right)^2 + = 9.0\E{9}\U{N$\cdot$m$^2$/C$^2$} + \left(\frac{70\U{kg} \cdot 0.01 \cdot 1.6\E{-19}\U{C}} + {2 \cdot 1.7\E{-27}\U{kg} \cdot 1\U{m}}\right)^2 + \approx 1\E{10} \left( \frac{0.3\E{-19}}{1\E{-27}} \right)^2 \U{N} + = \ans{1\E{25}\U{N}} +\end{equation} +And a ``weight'' the mass of the earth would be $F_g = Mg \approx +6\E{24}\U{kg}\cdot 9.8\U{m/s$^2$} \approx 6\E{25}\U{N} \sim F$. +\end{solution} diff --git a/latex/problems/problem19.04.tex b/latex/problems/problem19.04.tex new file mode 100644 index 0000000..95736d3 --- /dev/null +++ b/latex/problems/problem19.04.tex @@ -0,0 +1,45 @@ +\begin{problem*}{19.4} +Two protons in an atomic nucleus are typically seperated by a distance +of $r = 2.00\E{-15}\U{m}$. The electric repulsion force $F$ between +the protons is huge, but the attractive nuclear force is even stronger +and keeps the nucleus from bursting apart. What is the magnitude of +$F$? +\end{problem*} % problem 19.4 + +\empaddtoprelude{ + pair A, B; + A := origin; + B := (3cm, 0); + def drawB = + draw_pcharge(A, 6pt); + draw_pcharge(B, 6pt); + label.bot("r", draw_length(A, B, 10pt)); + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[2ns] +\begin{emp}(0cm, 0cm) + drawB; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[2] +\begin{emp}(0cm, 0cm) + label.top("F", draw_force(A, B, 20pt)); + drawB; +\end{emp} +\end{empfile} +\end{center} + +\begin{equation} + F = k_e \frac{q^2}{r^2} + = 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} \left(\frac{1.60\E{-19}\U{C}}{2.00\E{-15}\U{m}}\right)^2 + = \ans{57.7\U{N}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem19.07.tex b/latex/problems/problem19.07.tex new file mode 100644 index 0000000..fdfcf4d --- /dev/null +++ b/latex/problems/problem19.07.tex @@ -0,0 +1,61 @@ +\begin{problem*}{19.7} +Two identical conducting small spheres are placed with their centers +$r = 0.300\U{m}$ apart. One is given a charge of $q_1 = 12.0\U{nC}$ +and the other a charge of $q_2 = -18.0\U{nC}$. +\Part{a} Find the electric force exerted by one sphere on the other. +\Part{b} Next, the spheres are connected by a conducting wire. +Find the electric force between the two after they have come to equilibrium. +\end{problem*} % problem 19.7 + +\empaddtoprelude{ + pair A, B; + A := origin; + B := (3cm, 0); +} + +\begin{solution} +\Part{a} +\begin{center} +\begin{empfile}[1a] +\begin{emp}(0cm, 0cm) + label.top("F", draw_force(A, B, -30pt)); + draw_pcharge(A, 5pt); + label.llft(btex $q_1$ etex, A+6pt*dir(-135)); + draw_ncharge(B, 6pt); + label.lrt(btex $q_2$ etex, B+6pt*dir(-45)); + label.bot("r", draw_length(A, B, 10pt)); +\end{emp} +\end{empfile} +\end{center} +\begin{equation} + F = k_e \frac{q_1 q_2}{r^2} + = 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} \frac{12.0\E{-9}\U{C}\cdot(-18.0)\E{-9}\U{C}}{(0.300\U{m})^2} + = \ans{-2.16\E{-5}\U{N}} +\end{equation} +And the force is towards the other sphere for each sphere because +opposites attract. + +\Part{b} +\begin{center} +\begin{empfile}[1b] +\begin{emp}(0cm, 0cm) + label.top("F", draw_force(A, B, 10pt)); + draw A--B withcolor (.7,.7,.7) withpen pencircle scaled 1pt; + draw_ncharge(A, 3pt); + label.llft(btex $Q/2$ etex, A+6pt*dir(-135)); + draw_ncharge(B, 3pt); + label.lrt(btex $Q/2$ etex, B+6pt*dir(-45)); + label.bot("r", draw_length(A, B, 10pt)); +\end{emp} +\end{empfile} +\end{center} +The total charge on the both spheres is $Q = q_1 + q_2 = -6.0\U{nC}$. +The spheres are identical, so at equilibrium, there will be $Q/2 = +-3.0\U{nC}$ on each sphere. The repulsive (since now they have the +same charge sign) force between them is given by +\begin{equation} + F = k_e \frac{(Q/2)^2}{r^2} + = 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} \left(\frac{-3.0\E{-9}\U{C}}{0.300\U{m}}\right)^2 + = \ans{8.99\E{-7}\U{N}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem19.09.tex b/latex/problems/problem19.09.tex new file mode 100644 index 0000000..0b230ce --- /dev/null +++ b/latex/problems/problem19.09.tex @@ -0,0 +1,66 @@ +\begin{problem*}{19.9} +In the Bohr theory of the hydrogen atom, an electron moves in a +circular orbit about a proton, where the radius of the orbit is $r = +0.529\E{-10}\U{m}$. +\Part{a} Find the magnitude of the electric force each exerts on the other. +\Part{b} If this force causes the centripetal acceleration of the +electron, what is the speed of the electron? +\end{problem*} % problem 19.9 + +\empaddtoprelude{ + pair A, B, C; + numeric r; + r := 1cm; + A := origin; + B := (r, 0); + C := (-r, 0); + draw A--B; + def drawCbot = + draw A--C; + label.bot("r", (A+C)/2); + enddef; + def drawCtop = + draw_pcharge(A, 6pt); + label.rt("v", draw_velocity(B-(0,1), B, .7*r)); + draw fullcircle scaled (2*r) shifted A; + draw_ncharge(B, 3pt); + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[3ns] +\begin{emp}(0cm,0cm) + drawCbot; + drawCtop; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[3] +\begin{emp}(0cm,0cm) + drawCbot; + label.top("F", draw_force(A, B, -r/2)); + drawCtop; +\end{emp} +\end{empfile} +\end{center} + +\Part{a} +\begin{equation} + F = k_e \frac{q^2}{r^2} + = 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} \left(\frac{1.60\E{-19}\U{C}}{0.529\E{-10}\U{m}}\right)^2 + = \ans{8.22\E{-8}\U{N}} +\end{equation} + +\Part{b} +Using $F_c = m a_c = m v^2/r$ +\begin{equation} + v = \sqrt{\frac{F r}{m}} + = \sqrt{\frac{8.24\E{-8}\U{N} \cdot 0.529\E{-10}\U{m}}{9.11\E{-31}\U{kg}}} + = \ans{2.19\E{6}\U{m/s}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem19.11.tex b/latex/problems/problem19.11.tex new file mode 100644 index 0000000..74c9518 --- /dev/null +++ b/latex/problems/problem19.11.tex @@ -0,0 +1,80 @@ +\begin{problem*}{19.11} +In Figure P19.11, determine the point (other than infinity) at which +the electric field is zero. +$q_1 = -2.50\U{$\mu$C}$ and $q_2 = 6.00\U{$\mu$C}$. +\end{problem*} % problem 19.11 + +\empaddtoprelude{ + pair A, B; + A := origin; + B := (2cm, 0); + C := (-3.6cm, 0); + numeric orig_labeloffset; + orig_labeloffset := labeloffset; + def drawD = + label.bot("x", draw_arrow((-2cm,0), (-1cm,0), 4cm, 0pt, black)); + label.bot("0", draw_ltic(A, -90, 0, 10pt, 0pt, black)); + label.bot("1m", draw_ltic(B, -90, 0, 10pt, 0pt, black)); + draw_ncharge(A, 4pt); + draw_pcharge(B, 6pt); + labeloffset := 8pt; + label.top(btex $q_1$ etex, A); + label.top(btex $q_2$ etex, B); + labeloffset := orig_labeloffset; + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[4ns] +\begin{emp}(0cm,0cm) + drawD; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[4] +\begin{emp}(0cm,0cm) + drawD; + draw_tic(A, -180, 0, 4.5cm, 0pt, black); % extend x axis + label.top(btex $r_1$ etex, draw_length(A, C, 20pt)); + label.bot(btex $r_2$ etex, draw_length(C, B, 20pt)); + label.bot(btex $E_1$ etex, draw_Efield(A, C, -10pt)); + label.bot(btex $E_2$ etex, draw_Efield(B, C, 10pt)); + dotlabel("",C); +\end{emp} +\end{empfile} +\end{center} + +First, we need a coordinate system. Let $q_1$ be the origin +($x_1=0$), and $q_2$ be at $x_2 = 1.00\U{m}$. + +The electric field of a finite number of point charge is given by +(p. 612, 19.6) +\begin{equation} + \vect{E} = k_e \sum_i \frac{q_i}{r_i^2}\rhat_i +\end{equation} +For any point off the $x$ axis, there would be some force moving the +charge in the vertical $y$ direction, so we only need to look at +positions on the $x$ axis. + +A positive test charge placed between the two charges would be pulled +to the left by $q_1$ and pushed to the left by $q_2$. A positive test +charge placed to the right of $q_2$ would be pushed to the right by +$q_2$ more strongly (because $q_2 > q_1$ and $r_2 < r_1$) than it +would be pulled to the left by $q_1$. So the only place to look for +equilibrium is to the left of $q_1$, ($x < 0$, where $r_2=r_1+x_2$). +\begin{align} + \vect{E} &= k_e \left(-\frac{q_1}{r_1^2} - \frac{q_2}{(r_1 + x_2)^2}\right) + = 0 \\ + \frac{q_1}{r_1^2} &= -\frac{q_2}{(r_1 + x_2)^2} \\ + \frac{r_1+x_2}{r_1} = 1 + \frac{x_2}{r_1} &= \pm\sqrt{\frac{-q_2}{q_1}} \\ + r_1 &= \frac{x_2}{\sqrt{\pm\frac{-q_2}{q_1}} - 1} = 1.82\U{m}, -0.392\U{m} +\end{align} +But $r_1 = -0.392\U{m}$ is between the two charges (where our +assumption about the electric fields opposing each other doesn't +hold), so $\vect{E} = 0$ only at a $r_1 = 1.82\U{m}$ ($x=-1.82\U{m}$). +\end{solution} diff --git a/latex/problems/problem19.13.tex b/latex/problems/problem19.13.tex new file mode 100644 index 0000000..b722de8 --- /dev/null +++ b/latex/problems/problem19.13.tex @@ -0,0 +1,73 @@ +\begin{problem*}{19.13} +Three point charges are arranged as shown in Figure P19.13.\\ +%\begin{center} +% \begin{tabular}{|l|r|r|r|} +% Name & Charge (nC) & x (m) & y (m) \\ +% \hline +% $q_1$ & $5.00$ & $0$ & $0$ \\ +% $q_2$ & $6.00$ & $0.300$ & $0$ \\ +% $q_3$ & $-3.00$ & $0$ & $-0.100$ \\ +% \hline +% \end{tabular} +%\end{center} +\Part{a} Find the vector electric field \vect{E} that $q_2$ and $q_3$ +together create at the origin. +\Part{b} Find the vector force \vect{F} on $q_1$. +\end{problem*} % problem 19.13 + +\empaddtoprelude{ + numeric a; + pair A, B, C; + a := 3cm; + A := origin; % q1 + B := (a,0); % q2 + C := (0,-a/3); % q3 + def drawB = + label.top(btex 0.300\mbox{ m} etex, draw_length(B, A, 8pt)); + label.lft(btex 0.100\mbox{ m} etex, draw_length(A, C, 8pt)); + labeloffset := 6pt; + draw_pcharge(A, 4pt); + label.rt(btex $q_1 = 5\mbox{ nC}$ etex, A); + draw_pcharge(B, 4.2pt); + label.rt(btex $q_2 = 6\mbox{ nC}$ etex, B); + draw_ncharge(C, 3pt); + label.rt(btex $q_3 = -3\mbox{ nC}$ etex, C); + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[2p] +\begin{emp}(0cm,0cm) + drawB; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[2] +\begin{emp}(0cm,0cm) + label.lft(btex $E_{21}$ etex, draw_Efield(B, A, a/8)); + label.rt(btex $E_{31}$ etex, draw_Efield(C, A, -a/5)); + draw_ijhats(-(a, a/3), 0, a/6); + drawB; +\end{emp} +\end{empfile} +\end{center} +\Part{a} +\begin{equation} +\begin{equation} + \vect{E} = k_e \sum_i \frac{q_i}{r_i^2}\rhat_i + = k_e \left[\frac{q_2}{x_2^2}(-\ihat) + \frac{q_3}{y_3^2}\jhat\right] + = 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} \left(\frac{-6.00\ihat}{0.300^2} - \frac{3.00\jhat}{0.100^2}\right)\E{-9}{C/m^2} + = \ans{\left( -0.599\ihat - 2.70\jhat \right)\U{kN/C}} +\end{equation} + +\Part{b} +\begin{equation} + \vect{F} = q_1 \vect{E} + = \ans{\left( -3.00\ihat -13.5\jhat\right)\U{$\mu$N}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem19.15.tex b/latex/problems/problem19.15.tex new file mode 100644 index 0000000..e927f4a --- /dev/null +++ b/latex/problems/problem19.15.tex @@ -0,0 +1,78 @@ +\begin{problem*}{19.15} +Four point charges are at the corners of a square of side $a$ as shown +in Figure P19.15, with $q_1=2q$, $q_2=3q$, $q_3=4q$, and $q_4=q$. +\Part{a} Determine the magnitude and direction of the electric field +at the location of charge $q_4$. +\Part{b} What is the resultant force on $q_4$? +\end{problem*} % problem 19.15 + +\empaddtoprelude{ + numeric a; + pair A, B, C, D; + a := 1cm; + A := (0, a); % q1, labeled CCW from upper left + B := origin; % q2 + C := (a, 0); % q3 + D := (a, a); % q4 + def drawE = + label.lft("a", draw_length(A, B, 8pt)); + label.bot("a", draw_length(B, C, 8pt)); + labeloffset := 5pt; + draw_pcharge(A, 3pt); + label.bot(btex $q_1$ etex, A); + draw_pcharge(B, 3pt); + label.top(btex $q_2$ etex, B); + draw_pcharge(C, 3pt); + label.top(btex $q_3$ etex, C); + draw_pcharge(D, 3pt); + label.bot(btex $q_4$ etex, D); + enddef; +} +\begin{nosolution} +\begin{center} +\begin{empfile}[5ns] +\begin{emp}(0cm,0cm) + drawE; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[5] +\begin{emp}(0cm,0cm) + label.rt(btex $E_{14}$ etex, draw_Efield(A, D, 0.5cm)); + label.urt(btex $E_{24}$ etex, draw_Efield(B, D, 0.53cm)); + label.lft(btex $E_{34}$ etex, draw_Efield(C, D, 1cm)); + draw_ijhats((-1.5*a, a/3), 0, a/3); + drawE; +\end{emp} +\end{empfile} +\end{center} +Let \ihat\ point to the right and \jhat\ point up. +\begin{equation} + \vect{E} = k_e \sum_i \frac{q_i}{r_i^2}\rhat_i + = k_e \left(\frac{2q}{a^2}\ihat + + \frac{3q}{(\sqrt{2}a)^2}\frac{\ihat + \jhat}{\sqrt{2}} + + \frac{4q}{a^2}\jhat \right) + = k_e \frac{q}{a^2} \left( 2\ihat + + \frac{3}{2\sqrt{2}}(\ihat+\jhat) + + 4\jhat \right) +\end{equation} +So the magnitude of \vect{E} is given by +\begin{equation} + E = k_e \frac{q}{a^2} \sqrt{ \left(2+\frac{3}{2\sqrt{2}}\right)^2 + \left(4+\frac{3}{2\sqrt{2}}\right)^2 } = \ans{5.91 k_e \frac{q}{a^2}} +\end{equation} +And the direction $\theta$ (measured counter clockwise from \ihat) of +\vect{E} is given by +\begin{equation} + \theta = \arctan\left(\frac{4+\frac{3}{2\sqrt{2}}}{2+\frac{3}{2\sqrt{2}}}\right) + = \ans{58.8\dg} +\end{equation} + +\Part{b} +$\vect{F} = q \vect{E}$ so the direction of \vect{F} is the same as +the direction of \vect{E}. The magnitude of \vect{F} is given by $F = +5.91 k_e q^2 / a^2$ +\end{solution} diff --git a/latex/problems/problem19.16.tex b/latex/problems/problem19.16.tex new file mode 100644 index 0000000..9e8e9d3 --- /dev/null +++ b/latex/problems/problem19.16.tex @@ -0,0 +1,68 @@ +\begin{problem*}{19.16} +Consider the electric dipole shown in Figure P19.16. Show that the +electric field at a distant point on the $+x$ axis is $E_x \approx +4k_eqa/x^3$. +\end{problem*} % problem 19.16 + +\empaddtoprelude{ + pair A, B; + numeric a; + a := 1cm; + A := (-a,0); + B := (a, 0); + C := (6a, 0); + def drawC = + drawarrow (A-(a,0))--(C+(a,0)) withpen pencircle scaled 0pt; + draw_ncharge(A, 6pt); + draw_pcharge(B, 6pt); + label.top("0", draw_ltic(origin, 90, 0, 3pt, 0pt, black)); + dotlabel.bot("x", C); + label.bot("a", draw_length(A, origin, 10pt)); + label.bot("a", draw_length(origin, B, 10pt)); + labeloffset := 8pt; + label.top("-q", A); + label.top("q", B); + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[3p] +\begin{emp}(0cm, 0cm) + drawC; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[3] +\begin{emp}(0cm, 0cm) + label.top(btex $E_{q}$ etex, draw_Efield(B, C, 18pt)); + label.top(btex $E_{-q}$ etex, draw_Efield(A, C, -16pt)); + drawC; +\end{emp} +\end{empfile} +\end{center} +Let us assume the point in question has a positive $x$ value (just +reverse the sign if $x < 0$). +\begin{equation} + \vect{E} = k_e \sum_i \frac{q_i}{r_i^2}\rhat_i + = k_e \left[\frac{q}{(x-a)^2}\ihat + \frac{-q}{(x+a)^2}\ihat\right] +\end{equation} +For $|x| \gg |c|$, +\begin{equation} + (x+c)^n = x^n \left(1+\frac{c}{x}\right)^n + = x^n \left[1 + n\frac{c}{x} + \frac{n(n-1)}{2}\cdot\left(\frac{c}{x}\right)^2 + \ldots\right] + \approx x^n (1 + n\frac{c}{x}) \;, +\end{equation} +because $(c/x)^2$ is very, very small. (We are Taylor expanding +$(x+c)^n$ as a function of $c/x$, and keeping only the first two +terms.) In our case, $n = -2$ and $c = \mp a$ +\begin{equation} + \vect{E} = k_e \left[\frac{q}{x^2}\left(1-2\frac{-a}{x}\right) + \frac{-q}{x^2}\left(1-2\frac{a}{x}\right)\right]\ihat + = k_e \frac{q}{x^2}\left(1+2\frac{a}{x} - 1+2\frac{a}{x}\right)\ihat + = \ans{\frac{4 k_e q a}{x^3}\ihat} +\end{equation} +\end{solution} diff --git a/latex/problems/problem19.19.tex b/latex/problems/problem19.19.tex new file mode 100644 index 0000000..dbccac1 --- /dev/null +++ b/latex/problems/problem19.19.tex @@ -0,0 +1,49 @@ +\begin{problem*}{19.19} +A uniformly charged ring of radius $r = 10.0\U{cm}$ has a total charge +of $q = 75.0\U{$\mu$C}$. Find the electric field on the axis of the +ring at + \Part{a} $x_a = 1.00\U{cm}$, + \Part{b} $x_b = 5.00\U{cm}$, + \Part{c} $x_c = 30.0\U{cm}$, and + \Part{d} $x_d = 100\U{cm}$ from the center of the ring. +\end{problem*} % problem 19.19 + +\begin{solution} +\begin{center} +\begin{empfile}[6] +\begin{emp}(0cm,0cm) + pair A, B, C; + numeric a; + a := 0.75cm; + A := (0,a); + B := (0,-a); + C := (1cm,0); + draw_ijhats((-1cm,a/3), 0, a/3); + draw_ring(origin, a, 0, 3cm, 1cm, red, "q", "x"); + label.bot("0", draw_ltic(origin, -90, 0, 3pt, 0pt, black)); + label.top("A", A); + label.bot("B", B); + draw A--C; label.urt(btex $d_A$ etex, (A+C)/2); + draw B--C; label.lrt(btex $d_B$ etex, (B+C)/2); + label.lrt("E", draw_Efield(origin, C, 18pt)); + label.top(btex $E_B$ etex, draw_Efield(B, C, 15pt)); + label.bot(btex $E_A$ etex, draw_Efield(A, C, 15pt)); +\end{emp} +\end{empfile} +\end{center} + +From Example 19.5 (p.~616) we see the electric field along the axis +(\ihat) of a uniformly charged ring is given by +\begin{equation} + E = \frac{k_e x q}{(x^2 + r^2)^{3/2}} \ihat +\end{equation} + +So applying this to our 4 distances (rembering to convert the +distances to meters), we have +\begin{align} + E_a &= \ans{6.64\E{6}\U{N/C}\;\ihat} \\ + E_b &= \ans{24.1\E{6}\U{N/C}\;\ihat} \\ + E_c &= \ans{6.40\E{6}\U{N/C}\;\ihat} \\ + E_d &= \ans{0.664\E{6}\U{N/C}\;\ihat} +\end{align} +\end{solution} diff --git a/latex/problems/problem19.31.tex b/latex/problems/problem19.31.tex new file mode 100644 index 0000000..61bcee9 --- /dev/null +++ b/latex/problems/problem19.31.tex @@ -0,0 +1,16 @@ +\begin{problem*}{19.31} +A $d = 40.0\U{cm}$ diameter loop is rotated in a uniform electric +field until the position of maximum electric flux is found. +The flux in this position is measured to be + $\Phi_E = 5.20\E{5}\U{N$\cdot$m$^2$/C}$. +What is the magnitude of the electric field? +\end{problem*} % problem 19.31 + +\begin{solution} +\begin{align} + \Phi_E &= EA \\ + E &= \frac{\Phi_E}{A} = \frac{\Phi_E}{\pi (d/2)^2} + = \frac{5.20\E{5}\U{N$\cdot$m$^2$/C}}{\pi \cdot (0.200\U{m})^2} + = \ans{4.14\E{6}\U{N/C}} +\end{align} +\end{solution} diff --git a/latex/problems/problem19.35.tex b/latex/problems/problem19.35.tex new file mode 100644 index 0000000..dec939b --- /dev/null +++ b/latex/problems/problem19.35.tex @@ -0,0 +1,60 @@ +\begin{problem*}{19.35} +A solid sphere of radius $R = 40.0\U{cm}$ has a total charge of $q = +26.0\U{$\mu$C}$ uniformly distributed throughout its volume. +Calculate the magnitude $E$ of the electric field + \Part{a} $r_a = 0\U{cm}$, + \Part{b} $r_b = 10.0\U{cm}$, + \Part{c} $r_c = 40.0\U{cm}$, and + \Part{d} $r_d = 60.0\U{cm}$ + from the center of the sphere. +\end{problem1} + +\begin{solution} +The charge distribution is symmetric under rotations and reflections +about the center of the sphere , so the electric field must also be +symmetric under rotations and reflections about the center of the +sphere. So the electric field can only be a function of the radius +$\vect{E}(r)$ (if it was a f'n of the angle, it wouldn't be symmetric +under rotations), and it must be only in the radial direction +$\vect{E}(r) = E(r)\rhat$\ (if it had non-radial components, it +wouldn't be symmetric under reflections). + +Because we have these insights from symmetry, we can use Gauss's Law +to solve for $E(r)$ +\begin{align} + \oint \vect{E}(\vect{r}) \cdot d\vect{A} &= \frac{q_{in}}{\epsilon_0} \\ + E(r) \oint \rhat \cdot d\vect{A} &= \frac{q_{in}}{\epsilon_0} +\end{align} +because $r$ is a constant over our surface of integration, $E(r)$ must +also be constant, so we pull it out of the integral. We also note +that \rhat\ is going to be perpendicular to our surface at every point +on it, so +\begin{align} + E(r) \oint dA = E(r)A &= \frac{q_{in}}{\epsilon_0} \label{eqn.symm_gauss} \\ + E(r) 4 \pi r^2 &= \frac{q_{in}}{\epsilon_0} \\ + E(r) &= \frac{q_{in}}{4 \pi \epsilon_0 r^2} \label{eqn.sphere_gauss} +\end{align} +(If this is confusing, you can look at the first bit of the Gauss's +law section 19.9 page 624 in the book for their derivation, and +Example 19.9 on page 627 for their take on this problem.) + +For $r \le R$ (points $A$, $B$, and $C$) we have +\begin{equation} + q_{in} = q \frac{ 4/3 \cdot \pi r^3 }{ 4/3 \cdot \pi R^3 } + = q \left(\frac{r}{R}\right)^3 +\end{equation} +so +\begin{align} + E_\le(r) &= \frac{q r^3 / R^3}{4 \pi \epsilon_0 r^2} + = \frac{q r}{4 \pi \epsilon_0 R^3} \\ + E_a &= \ans{0} \qquad \text{because $r = 0$} \\ + E_b &= \frac{26.0\E{-6}\U{C} \cdot 0.100\U{m}}{4 \pi \cdot 8.854\E{-12}\U{C$^2$/N$\cdot$m$^2$} \cdot (0.400\U{m})^3} + = \ans{3.65\E{5}\U{N/C}} +\end{align} +And for $r \ge R$ (points $C$ and $D$) we have $q_{in} = q$, so +\begin{align} + E_\ge(r) &= \frac{q}{4 \pi \epsilon_0 r^2} \label{eqn.sphere_gauss_out} \\ + E_c &= \ans{1.46\E{6}\U{N/C}} \\ + E_d &= \ans{6.49\E{5}\U{N/C}} +\end{align} +\end{solution} diff --git a/latex/problems/problem19.36.tex b/latex/problems/problem19.36.tex new file mode 100644 index 0000000..f3c37a5 --- /dev/null +++ b/latex/problems/problem19.36.tex @@ -0,0 +1,17 @@ +\begin{problem*}{19.36} +An $m = 10.0\U{g}$ piece of Styrofoam carries a net charge of $q = +-0.700\U{$\mu$C}$ and floats above the center of a large horizontal +sheet of plastic that has a uniform charge density $\sigma$ on it's +surface. Find $\sigma$. +\end{problem*} % problem 19.36 + +\begin{solution} +Because the Styrofoam is floating in equilibrium, the sum of forces in +the vertical direction must be zero. So +\begin{align} + F_g = mg &= F_E = q E = q \frac{\sigma}{2 \epsilon_0} \\ + \sigma &= \frac{2\epsilon_0 mg}{q} + = \frac{2 \cdot 8.54\E{-12}\U{C$^2$/N$\cdot$m$^2$} \cdot 0.0100\U{kg} \cdot 9.80\U{m/s$^2$}}{-0.700\E{-6}\U{C}} + = \ans{2.39\E{-6}\U{C/m$^2$}} +\end{align} +\end{solution} diff --git a/latex/problems/problem19.38.tex b/latex/problems/problem19.38.tex new file mode 100644 index 0000000..1047a4a --- /dev/null +++ b/latex/problems/problem19.38.tex @@ -0,0 +1,25 @@ +\begin{problem*}{19.38} +Consider a thin spherical shell of radius $R = 14.0\U{cm}$ with a +total charge of $q = 32.0\U{$\mu$C}$ distributed uniformly on its +surface. Find the electric field + \Part{a} $r = 10.0\U{cm}$ and + \Part{b} $r = 20.0\U{cm}$ + from the center of the charge distribution. +\end{problem*} % problem 19.38 + +\begin{solution} +Again, the problem is symmetric under rotations and reflections about +the center, so following the same reasoning as in Problem 35 we can +use Equation \ref{eqn.sphere_gauss}. + +\Part{a} +Inside the shell there is no charge ($q_{in} = 0$), so $E_a = \ans{0}$. + +\Part{b} +Outside the shell we can use Equation \ref{eqn.sphere_gauss_out} +\begin{equation} + E_b = \frac{q}{4 \pi \epsilon_0 r_b^2} + = \frac{32\E{-6}\U{C}}{4 \pi \cdot 8.853\E{-12}\U{C$^2$/N$\cdot$m$^2$} \cdot (0.200\U{m})^2} \\ + = \ans{7.19\E{6}\U{N/C}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem19.40.tex b/latex/problems/problem19.40.tex new file mode 100644 index 0000000..ca047a3 --- /dev/null +++ b/latex/problems/problem19.40.tex @@ -0,0 +1,38 @@ +\begin{problem}{19.40} +An insulating solid sphere of radius $a$ has a uniform volume charge +density $\rho$ and carries a total positive charge $Q$. A spherical +gaussian surface of radius $r$, which shares a common center with the +insulating sphere, is inflated starting from $r=0$. +\Part{a} Find an expression for the electric flux $\Phi_E$ passing through the +surface of the gaussian sphere as a function of $r$ for $r < a$. +\Part{b} Find an expression for the electric flux $\Phi_E$ for $r > a$. +\Part{c} Plot $\Phi_E$(r). +\end{problem*} % problem 19.40 + +\begin{solution} +\Part{a} +\begin{equation} + \Phi_E = \frac{q_{in}}{\epsilon_0} + = \frac{Q \cdot 4/3 \cdot \pi r^3}{\epsilon_0 \cdot 4/3 \cdot \pi R^3} + = \ans{\frac{Q r^3}{\epsilon_0 R^3}} +\end{equation} + +\Part{b} +\begin{equation} + \Phi_E = \frac{q_{in}}{\epsilon_0} + = \ans{\frac{Q}{\epsilon_0}} +\end{equation} + +\Part{c} +\empaddtoprelude{input graph} +\begin{center} +\begin{empfile}[3] +\begin{empgraph}(5cm, 3cm) + % scaled so x=r/R, y=Phi/(Q/e0) + glabel.bot(btex $r/R$ etex, OUT); + glabel.lft(btex $\displaystyle \frac{\Phi_E \epsilon_0}{Q}$ etex, OUT); + gdraw (0,0){right}..(.2,.008)..(.5, .125)..(.8,.512)..(1,1)--(2,1) withcolor green; +\end{empgraph} +\end{empfile} +\end{center} +\end{solution} diff --git a/latex/problems/problem19.55.tex b/latex/problems/problem19.55.tex new file mode 100644 index 0000000..c03f0c0 --- /dev/null +++ b/latex/problems/problem19.55.tex @@ -0,0 +1,111 @@ +\begin{problem*}{19.55} +Four identical point charges ($q = +10.0\U{$\mu$C}$) are located on +the corners of a rectangle as shown in Figure P19.55. The dimensions +of the rectangle are $L = 60.0\U{cm}$ and $W = 15.0\U{cm}$. Calculate +the magnitude and direction of the resultant electric force exerted on +the charge at the lower left corner by the other three charges. +\end{problem*} % problem 19.55 + +\empaddtoprelude{ + numeric a; + pair A, B, C, D; + a := 1cm; + A := (0, a); % q1, labeled CCW from upper left + B := origin; % q2 + C := (4a, 0); % q3 + D := (4a, a); % q4 + def drawE = + label.rt("W", draw_length(C, D, 8pt)); + label.bot("L", draw_length(B, C, 8pt)); + labeloffset := 5pt; + draw_pcharge(A, 3pt); + label.bot(btex $q_1$ etex, A); + draw_pcharge(B, 3pt); + label.top(btex $q_2$ etex, B); + draw_pcharge(C, 3pt); + label.top(btex $q_3$ etex, C); + draw_pcharge(D, 3pt); + label.bot(btex $q_4$ etex, D); + enddef; +} +\begin{nosolution} +\begin{center} +\begin{empfile}[3p] +\begin{emp}(0cm,0cm) + drawE; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[3] +\begin{emp}(0cm,0cm) + label.rt(btex $F_{12}$ etex, draw_force(A, B, 2cm)); + label.lft(btex $F_{32}$ etex, draw_force(C, B, .5cm)); + label.bot(btex $F_{42}$ etex, draw_force(D, B, .48cm)); + draw A--(D+(a/2,0)) dashed evenly; + draw A--B dashed evenly; + draw B--D dashed evenly; + label.top(btex $\theta$ etex, draw_lout_angle(D+(1,0),D,B, a/3)); + draw_right_angle(B, A, D, a/3); + draw_ijhats((5.5*a, a/3), 0, a/3); + drawE; +\end{emp} +\end{empfile} +\begin{center} +\begin{empfile}[3] +\begin{emp}(0cm,0cm) + label.rt(btex $F_{12}$ etex, draw_force(A, B, 2cm)); + label.lft(btex $F_{32}$ etex, draw_force(C, B, .5cm)); + label.bot(btex $F_{42}$ etex, draw_force(D, B, .48cm)); + draw A--(D+(a/2,0)) dashed evenly; + draw A--B dashed evenly; + draw B--D dashed evenly; + label.top(btex $\theta$ etex, draw_lout_angle(D+(1,0),D,B, a/3)); + draw_right_angle(B, A, D, a/3); + draw_ijhats((5.5*a, a/3), 0, a/3); + drawE; +\end{emp} +\end{empfile} +The unit vector \rhat\ diagonally across from the upper right is given by +\begin{align} + \rhat &= \cos\theta \ihat + \sin\theta \jhat \\ + \theta &= \arctan{W/L} + 180\dg = 194\dg \\ + \cos\theta &= -0.970 \\ + \sin\theta &= -0.243 +\end{align} +So the electric field in the lower left corner is given by +\begin{align} + \vect{E} &= k_e \sum_i \frac{q_i}{r_i^2}\rhat_i + = k_e \left(\frac{q}{L^2}(-\ihat) + + \frac{q}{(L^2 + W^2)}(\cos\theta\ihat + \sin\theta\jhat) + + \frac{q}{W^2}(-\jhat) \right) \\ + &= -k_e q \left[ + \left(\frac{1}{L^2} - \frac{\cos\theta}{L^2+W^2}\right)\ihat + + \left(\frac{1}{W^2} - \frac{\sin\theta}{L^2+W^2}\right)\jhat + \right] +\end{align} + +So the magnitude of \vect{E} is given by +\begin{equation} + E = k_e q \sqrt{ \left(L^{-2} - \frac{\cos\theta}{L^2+W^2}\right)^2 + + \left(W^{-2} - \frac{\sin\theta}{L^2+W^2}\right)^2 } + = 4.08\E{6}\U{N/C} +\end{equation} +(Remembering to convert $L$ and $W$ to meters.) And the direction +$\phi$ (measured counter clockwise from \ihat) of \vect{E} is given by +\begin{equation} + \phi = \arctan\left(\frac{-W^{-2}+\frac{\sin\theta}{L^2+W^2}}{-L^{-2}+\frac{\cos\theta}{L^2+W^2}}\right) + 180\dg + = \ans{263\dg} +\end{equation} +Where the $+180\dg$ is because the tangent has a period of $180\dg$, +and the angle we want is in the backside $180\dg$. + +$\vect{F} = q \vect{E}$ so the direction of \vect{F} is the same as +the direction of \vect{E}. The magnitude of \vect{F} is given by +\begin{equation} + F = 10.0\E{-6}\U{C} \cdot 4.08\E{6}\U{N/C} = \ans{40.8\U{N}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem19.57.tex b/latex/problems/problem19.57.tex new file mode 100644 index 0000000..3cec744 --- /dev/null +++ b/latex/problems/problem19.57.tex @@ -0,0 +1,69 @@ +\begin{problem*}{19.57} +Two identical metallic blocks resting on a frictionless horizontal +surface are connected by a light metallic spring having the spring +constant $k = 100\U{N/m}$ and an unstretched length of $L_0 = +0.300\U{m}$ as shown in Figure P19.57a. A total charge of $Q$ is +slowly placed on the system, causing the spring to stretch to an +equilibrium length of $L_1 = 0.400\U{m}$ as shown in Figure P19.57b. +Determine the value of $Q$, assuming that all charge resides in the +blocks and modeling the blocks as point charges. +\end{problem*} % problem 19.57 + +\empaddtoprelude{ + pair A, B, Ac, Bc; + vardef draw_sprung_blocks(expr center, r, labelgraphic, lengthlabelgraphic) = + numeric h; + h := 6pt; + A := center+(-r,+h); % spring attachment points + B := center+(r, +h); + draw_spring(A, B, h/2, 20); + draw_spring(A, B, h/2, 20); + label.top(lengthlabelgraphic, draw_length(B,A,h+5pt)); + Ac := A - (h,0); % block centers + Bc := B + (h,0); + draw_block(Ac,2h,2h,(.7,.7,.7)); + draw_block(Bc,2h,2h,(.7,.7,.7)); + draw_table(center, 3.7cm, 1.5 h); + label.bot(labelgraphic, center-(0,2h)); + enddef; + def drawD = + draw_sprung_blocks(origin, 0.9cm, btex (a) etex, btex $L_0 = 0.300\mbox{ m}$ etex); + draw_sprung_blocks((4cm, 0), 1.2cm, btex (b) etex, btex $L_1 = 0.400\mbox{ m}$ etex) + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[4p] +\begin{emp}(0, 0) + drawD; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[4] +\begin{emp}(0, 0) + drawD; + label.top(btex $F_s$ etex, draw_force(A,B,-16pt)); + label.rt(btex $F_E$ etex, draw_force(A,Bc,16pt)); +\end{emp} +\end{empfile} +\end{center} +Looking at the right hand block (it doesn't matter which one you +pick), we see that the only relevant forces are the attractive spring +force, and the repulsive electrostatic force. Because the blocks are +at equilibrium, these forces must cancel, so +\begin{align} + F_s &= k(L_1 - L_0) = F_E = k_e \frac{(Q/2)^2}{L_1^2} + = k_e \left(\frac{Q}{2L_1}\right)^2 \\ + Q &= 2L_1 \sqrt{k(L_1 - L_0)/k_e} \\ + &= 2 \cdot 0.400\U{m} + \cdot \sqrt{100\U{N/m} + \cdot 0.100\U{m} + / 8.99\E{9}\U{N$\cdot$m$^2$/C$^2$} } + = \ans{2.67\E{-5}\U{C}} +\end{align} +\end{solution} diff --git a/latex/problems/problem19.59.tex b/latex/problems/problem19.59.tex new file mode 100644 index 0000000..8a439dd --- /dev/null +++ b/latex/problems/problem19.59.tex @@ -0,0 +1,113 @@ +\begin{problem5}{19.59} +Two small spheres of mass $m$ are suspended from strings of length $l$ +that are connected at a common point. One sphere has charge $Q$, and +the other has charge $2Q$. The strings make angles $\theta_1$ and +$\theta_2$ with the vertical. +\Part{a} How are $\theta_1$ and $\theta_2$ related? +\Part{b} Assume that $\theta_1$ and $\theta_2$ are small. + Show that the distance $r$ between the spheres is given by + \begin{equation} + r \approx \left( \frac{4 k_e Q^2 l}{mg} \right)^{1/3} + \end{equation} +\end{problem*} % problem 19.59 + +\empaddtoprelude{ + pair A, B, C, D, mFe, mFg; + numeric L, theta, mFe, mFg; + L := 3cm; theta := 15; + mFe := 10pt; mFg := 20pt; % in the big picture (b) + A := L*dir(-90-theta); B := (0cm,0cm); C := L*dir(-90+theta); D := L*dir(-90); +} + +\begin{solution} +\begin{center} +\begin{empfile}[5a] +\begin{emp}(0, 0) + pair Ae, At, pFe, pFg; + %Ap := A / cosd theta; % extend R as deep as the vertical + Ae := A + unitvector(A) * 4mFg / cosd theta; % extend R as deep as Fg + At := A + (unitvector(A) rotated -90) * 4mFe / cosd theta; % extend as far out as Fe + pFe := A-(4mFe,0); % tip of electric force + pFg := A-(0,4mFg); % top of gravitational force + dotlabel.bot(" ", B); + draw A -- B; % broken + draw B -- D dashed evenly; % drop vertical from the pivot + label.bot(btex $\theta_1$ etex, draw_langle(A, B, D, .6L)); + % draw forces + label.bot(btex $F_E$ etex, draw_force(C,A,4mFe)); + label.bot(btex $F_g$ etex, draw_force(A+dir(90),A,4mFg)); + % draw axes + drawarrow A -- Ae; % extend string radius + drawarrow A -- At; % draw tangential axis head + draw A -- (At rotatedabout(A, 180)); % and tangential axis tail + label.top(btex $x_t$ etex, At); + label.lft(btex $x_r$ etex, Ae); + % label force angles + label.lft(btex $\theta_1$ etex, draw_langle(At, A, pFe, 3mFe)); + label.bot(btex $\theta_1$ etex, draw_langle(Ae, A, pFg, .6L)); + draw_right_angle(At, A, Ae, .7mFe); + draw_right_angle(pFe, A, pFg, 1mFe); + % draw charge + draw_pcharge(A, 4pt); +\end{emp} +\end{empfile} +\hspace{1cm} +\begin{empfile}[5b] +\begin{emp}(0, 0) + dotlabel.bot(" ", B); + draw A -- B -- C; + draw B -- D dashed evenly; + label.bot(btex $\theta_1$ etex, draw_langle(A, B, D, .6L)); + label.bot(btex $\theta_2$ etex, draw_langle(D, B, C, .6L)); + label.lft(btex $F_E$ etex, draw_force(C,A,mFe)); + label.bot(btex $F_g$ etex, draw_force(A+dir(90),A,mFg)); + draw_pcharge(A, 4pt); + draw_pcharge(C, 6pt); + labeloffset := 8pt; + label.rt(btex $m$ etex, A); + label.rt(btex $m$ etex, C); + label.ulft(btex $Q$ etex, A); + label.bot(btex $2Q$ etex, C); +\end{emp} +\end{empfile} +\end{center} +\Part{a} +Assuming that the charges are not rotating about each other, the +forces on each charge must cancel. The forces on each sphere are +gravity $F_g = mg$, electrostatic $F_E = k_e 2Q^2/r^2$, and tension +$T$. The tension will automatically handle canceling forces in the +radial direction, so we need only consider the tangential direction. +Let us assume that $F_E$ is purely in the horizontal direction +(see \Part{Note}). Summing the tangential forces on the first sphere +\begin{align} + 0 &= F_E \cos\theta_1 - F_g \sin\theta_1 \\ + \tan\theta_1 &= \frac{F_E}{F_g} +\end{align} +And on the second sphere $\tan\theta_2 = \frac{F_E}{F_g}$ + so $\theta_1 = \theta_2 = \theta$. + +\Part{b} +\begin{align} + r &= 2 l \sin\theta + \approx 2 l \tan\theta + = 2 l \frac{F_E}{F_g} + = 2 l \frac{k_e 2 Q^2 / r^2}{mg} \\ + r &\approx \left( \frac{4 l k_e Q^2}{mg} \right)^{1/3} \;, +\end{align} +where we used the small angle approximation +$\sin\theta\approx\tan\theta$ for small $\theta$. + +\Part{Note} Why $\vect{F}_E$ is horizontal. + +Let $q$ be the charge on the first mass and $Q$ be the charge on the +second. The force of $1$ on $2$ is given by $F_{12} = k_e +qQ\rhat_{12}/r^2$. This is identical to the force of $1$ on $2$ that +we would get if we had put $Q$ on $1$ and $q$ on $2$ (let us say ``the +electric force does not care about which mass has which charge''). +The only difference between the two masses is the charge, and the only +effect of that difference (the electrostatic force) does not care +about the difference, so the final situation must be symmetric +($\theta_1 = \theta_2$ [no calculation required :p] and $\vect{r}$ is +horizontal). Because $\vect{F_E} \propto \rhat_{12}$ it must also be +horizontal. +\end{solution} diff --git a/latex/problems/problem19.62.tex b/latex/problems/problem19.62.tex new file mode 100644 index 0000000..6c6c57c --- /dev/null +++ b/latex/problems/problem19.62.tex @@ -0,0 +1,105 @@ +\begin{problem*}{19.62} +Two infinite, nonconducting sheets of charge are parallel to each +other as shown in Figure P19.62. The sheet on the left has a uniform +surface charge density $\sigma$, and the one on the right has a +uniform charge density $-\sigma$. Calculate the electric field at +points + \Part{a} to the left of, + \Part{b} in between, and + \Part{c} to the right of +the two sheets. +\end{problem*} % problem 19.62 + +\empaddtoprelude{ + numeric r, rf, mf, dy, height, depth, thick, theta; + r := 1cm; + height := 2cm; + depth := 1cm; + thick := 3pt; + theta := 45; + def draw_plates = + draw_plate((-r,0), height, depth, thick, theta, .5(white+red), (.7white+.3red), black); + draw_plate(( r,0), height, depth, thick, theta, .5(white+blue), (.7white+.3blue), black); + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[6] +\begin{emp}(0, 0) + draw_plates; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\hspace{\stretch{1}} +\begin{empfile}[6a] +\begin{emp}(0, 0) + numeric rE, mE; + rE := 2r/3; mE := 2r/3; + draw_ihat((2r/3,r), 0, 2r/3); + draw_gauss_half_cyl((-thick/2,0),rE-thick/2,height/6,0.5,180); + draw_gauss_half_cyl((-thick/2,0), 0,height/6,0.5,180); + draw_plate(origin, height, depth, thick, theta, .5(white+red), (.7white+.3red), black); + draw_gauss_half_cyl(( thick/2,0), 0,height/6,0.5,180); + draw_gauss_half_cyl(( thick/2,0),rE-thick/2,height/6,0.5,0); + label.top(btex $E_+$ etex, draw_arrow(origin,(-rE, 0), mE, 2pt, red)); + label.top(btex $E_+$ etex, draw_arrow(origin,( rE, 0), mE, 2pt, red)); +\end{emp} +\end{empfile} +\hspace{\stretch{1}} +\begin{empfile}[6b] +\begin{emp}(0, 0) + numeric rE, mE, dy; + rE := 2r/3; mE := 2r/3; dy := 3pt; + draw_plates; + draw_ihat((-r/3,r), 0, 2r/3); + label.top(btex $E_+$ etex, draw_arrow((-r, dy),(-r-rE, dy), mE, 2pt, red)); + label.top(btex $E_+$ etex, draw_arrow((-r, dy),(-r+rE, dy), mE, 2pt, red)); + label.top(btex $E_+$ etex, draw_arrow((-r, dy),( r+rE, dy), mE, 2pt, red)); + label.bot(btex $E_-$ etex, draw_arrow(( r,-dy),(-r-rE-mE,-dy),-mE, 2pt, blue)); + label.bot(btex $E_-$ etex, draw_arrow(( r,-dy),(-r+rE, -dy),-mE, 2pt, blue)); + label.bot(btex $E_-$ etex, draw_arrow(( r,-dy),( r+rE+mE,-dy),-mE, 2pt, blue)); +\end{emp} +\end{empfile} +\hspace{\stretch{1}} + +Let \ihat\ be the direction to the right perpendicular to the sheets. +Because the problem has is symmetric to translations in the plane of +the sheets and reflections through planes perpendicular to the sheets, +the electric field must be of the form $\vect{E}(\vect{r}) = +E(x)\ihat$. + +Using Gauss's law to find the electric field due to a single plate, we +imagine a cylinder that extends through the plate a length $L$ out +either side. $\vect{E} = E\ihat$, so no flux passes through the side +walls of the cylinder. The single sheet is symmetric to reflection in +it's plane, so (defining $x=0$ to be the $x$ value of the plane) +$\vect{E}(x) = -\vect{E}(x)$ (positive charges are repelled from both +sides). So, letting the area of a single end cap be $A$, the charge +enclosed by the cylinder is $\sigma A$ and the flux through the +end-caps of the cylinder is given by +\begin{align} + \Phi_E = 2EA &= \frac{q_{in}}{\epsilon_0} = \frac{\sigma A}{\epsilon_0} \\ + E &= \frac{\sigma}{2\epsilon_0} \label{eqn.plane_E} +\end{align} +A constant! (See Example 19.12 on page 629 for the book's version) + +\Part{a} +Using Equation \ref{eqn.plane_E} and superposition, we see +\begin{equation} + E_L = \frac{\sigma}{2\epsilon_0} + \frac{-\sigma}{2\epsilon_0} = \ans{0} +\end{equation} + +\Part{b} +\begin{equation} + \vect{E}_B = \frac{\sigma \ihat}{2\epsilon_0} + + \frac{-\sigma \cdot (-\ihat)}{2\epsilon_0} + = \frac{\sigma}{\epsilon_0}\ihat +\end{equation} + +\Part{c} +Identically to \Part{a}, $E_R = 0$. +\end{solution} diff --git a/latex/problems/problem20.01.tex b/latex/problems/problem20.01.tex new file mode 100644 index 0000000..525ce30 --- /dev/null +++ b/latex/problems/problem20.01.tex @@ -0,0 +1,24 @@ +\begin{problem*}{20.1} +\Part{a} Calculate the speed of a proton that is accelerated from +rest through a potential difference of $\Delta V = 120\U{V}$. +\Part{b} Calculate the speed of an electron that is accelerated +through the same potential difference. +\end{problem*} % problem 20.1 + +\begin{solution} +\Part{a} +Conserving energy +\begin{gather} + E_0 = \frac{1}{2}m_p v^2 = E_1 = e\Delta V \\ + v = \sqrt{\frac{2 e \Delta V}{m_p}} + = \sqrt{\frac{2 \cdot 1.60\E{-19}\U{C} \cdot 120\U{V}}{1.67\E{-27}\U{kg}}} + = \ans{152\U{km/s}} +\end{gather} + +\Part{b} Replacing $m_p$ with $m_e$ +\begin{equation} +v = \sqrt{\frac{2 e \Delta V}{m_e}} + = \sqrt{\frac{2 \cdot 1.60\E{-19}\U{C} \cdot 120\U{V}}{9.11\E{-31}\U{kg}}} + = \ans{6.49\U{Mm/s}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.03.tex b/latex/problems/problem20.03.tex new file mode 100644 index 0000000..17e02dc --- /dev/null +++ b/latex/problems/problem20.03.tex @@ -0,0 +1,31 @@ +\begin{problem*}{20.3} +A uniform electric field of magnitude $E = 250\U{V/m}$ is directed in +the positive $x$ direction (\ihat). A $q = +12.0\U{$\mu$C}$ charge +moves from the origin to the point $(x,y) = (20.0\U{cm}, 50.0\U{cm})$. +\Part{a} What is the change in the potential energy $\Delta U$ of the +charge-field system? +\Part{b} Through what potential difference $\Delta V$ does the charge move? +\end{problem*} % problem 20.3 + +\begin{solution} +\Part{a} +From the text Equation 20.1 (page 643) we see +\begin{equation} + \Delta U = -q \int_A^B \vect{E} \cdot d\vect{s} + = -q \int_A^B E\ihat \cdot d\vect{s} + = -q E \int_{x_1}^{x_2} dx + = -q E \Delta x +\end{equation} +Which is the same process the book used to get to their Equation 20.9. +Plugging in our numbers +\begin{equation} + \Delta U = -12.0\E{-6}\U{C} \cdot 250\U{V/m} \cdot 0.200\U{m} + = \ans{-6.00\E{-4}\U{J}} +\end{equation} + +\Part{b} +The change in electric potential is given by +\begin{equation} + \Delta V = \frac{\Delta U}{q} = \ans{-50.0\U{V}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.08.tex b/latex/problems/problem20.08.tex new file mode 100644 index 0000000..d100b73 --- /dev/null +++ b/latex/problems/problem20.08.tex @@ -0,0 +1,94 @@ +\begin{problem*}{20.8} +Given two $q_0 = 2.00\U{$\mu$C}$ charges as shown in Figure P20.8 and +a positive test charge of $q = 1.28\E{-18}\U{C}$ at the origin, + \Part{a} what is the net force exerted by the two $q_0$ charges on + the test charge $q$? + \Part{b} What is the electric field at the origin due to the two + $q_0$ charges? + \Part{c} What is the electrical potential at the origin due to the two + $q_0$ charges?\\ +%\begin{center} +% \begin{tabular}{|l|r|r|} +% Name & Charge & x(m) \\ +% \hline +% $q_{0A}$ & $q_0$ & $-a$ \\ +% $q_{0B}$ & $q_0$ & $a$ \\ +% $q$ & $q$ & $0$ \\ +% \hline +% \end{tabular} +%\end{center} +%Where $a = 0.800\U{m}$. +\end{problem*} % problem 20.8 + +\empaddtoprelude{ + pair A, B, C; + numeric a; + a := 2cm; + A := (-a,0); + B := (a, 0); + C := (0, 0); + def drawA = + drawarrow (A-(a/2,0))--(B+(a,0)) withpen pencircle scaled 0pt; + label.bot(btex x etex, B+(a,0)); + draw_ihat(B+(a/4,8pt), 0, a/2); + draw_pcharge(A, 5pt); + draw_pcharge(B, 5pt); + draw_pcharge(C, 3pt); + labeloffset := 8pt; + label.bot(btex $x = -0.800\mbox{ m}$ etex, A); + label.bot(btex $x = 0.800\mbox{ m}$ etex, B); + label.bot(btex $x = 0$ etex, C); + label.top(btex $q_0$ etex, A); + label.top(btex $q_0$ etex, B); + label.top(btex q etex, C); + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[2p] +\begin{emp}(0cm, 0cm) + drawA; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[2] +\begin{emp}(0cm, 0cm) + label.top(btex $E_{x_+}$ etex, draw_Efield(B, C, 18pt)); + label.top(btex $E_{x_-}$ etex, draw_Efield(A, C, 18pt)); + drawA; +\end{emp} +\end{empfile} +\end{center} + +\Part{a} +Letting $a = 0.800\U{m}$ and summing the forces from Coulomb's law +\begin{equation} + \vect{F} = \vect{F}_A + \vect{F}_B + = k_e \left[ \frac{q_0 q}{a^2}\ihat + + \frac{q_0 q}{a^2}(-\ihat)\right] + = \ans{0} +\end{equation} +Which makes sense because the situation is symmetric. + +\Part{b} +Summing the electric fields +\begin{equation} + \vect{E}(0) = \vect{E}_A + \vect{E}_B + = k_e \left[ \frac{q_0}{a^2}\ihat + \frac{q_0}{a^2}(-\ihat)\right] + = \frac{\vect{F}}{q} = \ans{0} +\end{equation} + +\Part{c} +Summing the potentials +\begin{equation} + V(0) = V_A + V_B = k_e \frac{q_0}{a} + k_e \frac{q_0}{a} + = 2 k_e \frac{q_0}{a} + = 2 \cdot 8.99\E{9}{Vm/C} \frac{2.00\E{-6}\U{C}}{0.800\U{m}} + = \ans{45.0U{kV}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.11.tex b/latex/problems/problem20.11.tex new file mode 100644 index 0000000..646d8cf --- /dev/null +++ b/latex/problems/problem20.11.tex @@ -0,0 +1,14 @@ +\begin{problem*}{20.11} +The three charges in Figure P20.11 are at the vertices of an isosceles +triangle. Calculate the electric potential at the midpoint of the +base, taking $q = 7.00\U{$\mu$C}$. +\end{problem*} % problem 20.11 + +\begin{solution} +\begin{equation} + V = k_e \left( \frac{-q}{1.00\U{cm}} + \frac{-q}{1.00\U{cm}} + + \frac{q}{\sqrt{4.00^2-1.00^2}\U{cm}}\right) + \cdot \frac{100\U{cm}}{1\U{m}} + = \ans{-11.0\U{MV}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.19.tex b/latex/problems/problem20.19.tex new file mode 100644 index 0000000..f37b9b8 --- /dev/null +++ b/latex/problems/problem20.19.tex @@ -0,0 +1,53 @@ +\begin{problem*}{20.19} +A light, unstressed spring has a length $d$. Two identical particles, +each with charge $q$, are connected to the opposite ends of the +spring. The particles are held stationary a distance $d$ apart and +are then released at the same time. The spring has a bit of internal +kinetic friction, so the oscillation is damped. The particles +eventually stop vibrating when the distance between them is $3d$. +Find the increase in internal energy $\Delta E_i$ that appears in the +spring during the oscillations. Assume that the system of the spring +and the two charges is isolated. +\end{problem*} % problem 20.19 + +\begin{solution} +From last quarter, we remember that spring potential energy is given +by $U_s = 1/2\cdot k x^2$. To plug in for $k$, we balance the forces +at equilibrium +\begin{align} + F_e = k_e \left(\frac{q}{3d}\right)^2 &= F_s = k \cdot 2d \\ + k &= k_e \frac{q^2}{9 \cdot 2 \cdot d^3} +\end{align} + +From this chapter (Equation 20.13), we see that the electrical +potential energy of two charges is given by +\begin{equation} + U_e = k_e \frac{q_1 q_2}{r_{12}} +\end{equation} + +So the total potential energy of the system in it's final state is +given by the sum of the electric $U_e$ and spring $U_s$ potentials. +\begin{equation} + U = U_e + U_s = \frac{1}{2}k(r-d)^2 + k_e \frac{q^2}{r} +\end{equation} + +The total energy at a point in time is the sum of potential and +internal energies +\begin{equation} + E_t = U_t + E_{it} \;. +\end{equation} +Since we were only interested in the change in internal energy, we can +set the initial internal energy to $0$, and call the final internal +energy $E_i$. + +Conserving energy $E_0 = E_1$ (because the system is isolated) +\begin{align} + E_0 = U_0 &= E_1 = U_1 + E_i \\ + E_i &= U_0 - U_1 + = k_e \frac{q^2}{d} - \frac{1}{2}k(2d)^2 - k_e \frac{q^2}{3d} + = k_e \frac{q^2}{d} \left( 1 - \frac{1}{3}\right) - 2kd^2 \\ + &= k_e \frac{2q^2}{3d} - 2 \left(k_e \frac{q^2}{9 \cdot 2 \cdot d^3}\right)d^2 + = k_e \frac{2q^2}{3d} - k_e \frac{q^2}{9d} + = \ans{\frac{5 k_e q^2}{9d}} \;, +\end{align} +\end{solution} diff --git a/latex/problems/problem20.20.tex b/latex/problems/problem20.20.tex new file mode 100644 index 0000000..e7db0fd --- /dev/null +++ b/latex/problems/problem20.20.tex @@ -0,0 +1,36 @@ +\begin{problem*}{20.20} +In 1911, Ernest Rutherford and his assistants Hans Geiger and Ernest +Mardsen conducted an experiment in which they scattered alpha +particles from thin sheets of gold. An alpha particle, having a +charge of $q_\alpha = +2e$ and a mass of $m = 6.64\E{-27}\U{kg}$ is a +product of certain radioactive decays. The results of the experiment +lead Rutherford to the idea that most of the mass of an atom is in a +very small nucleus, whith electrons in orbit around it, in his +planetary model of the atom. Assume that an alpha particle, initially +very far from a gold nucleus, is fired with a velocity $v = +2.00\E{7}\U{m/s}$ directly toward the nucleus (charge $Q = +79e$). +How close does the alpha particle get to the nucleus before turning +around? Asume that the gold nucleus remains stationary. +\end{problem*} % problem 20.20 + +\begin{solution} +Let $r$ be the distance between the alpha particle and the gold +nucleus. Conserving energy between the initial point at $r=\infty$ +where the energy is all kinetic +\begin{equation} + E_0 = \frac{1}{2}m v_0^2 +\end{equation} +And the point of closest approach where the energy is all electric potential +\begin{equation} + E_1 = k_e \frac{(2e)(79e)}{r} +\end{equation} +We have +\begin{align} + E_0 = \frac{1}{2}mv^2 &= E_1 = k_e \frac{158 e^2}{r} \\ + r &= \frac{2 \cdot 158 \cdot k_e e^2}{m v^2} + = \frac{316 \cdot 8.99\E{9}\U{N m$^2$/C$^2$} \cdot (1.60\E{-19}\U{C})^2}{6.64\E{-27}\U{kg} \cdot (2.00\E{7}\U{m/s})^2} + = \ans{2.74\E{-14}\U{m}} +\end{align} +Which is significantly less than the $r_e \sim 10^{-10}\U{m}$ radius +of the gold atom. +\end{solution} diff --git a/latex/problems/problem20.21.tex b/latex/problems/problem20.21.tex new file mode 100644 index 0000000..43d4ad8 --- /dev/null +++ b/latex/problems/problem20.21.tex @@ -0,0 +1,25 @@ +\begin{problem*}{20.21} +The potential in a region between $x=0$ and $x=6.00\U{m}$ is $V = +a+bx$, where $a = 10.0\U{V}$ and $b = -7.00\U{V/m}$. Determine + \Part{a} the potential at $x = 0$, $3.00\U{m}$, and $6.00\U{m}$; and + + \Part{b} the magnitude and direction of the electric field at $x = +0$, $3.00\U{m}$, and $6.00\U{m}$. +\end{problem5} % problem 20.21 + +\begin{solution} +\Part{a} +Simply plugging into their $V(x)$ formula +\begin{align} + V(0\U{m}) &= \ans{10.0\U{V}} \\ + V(3.00\U{m}) &= 10.0\U{V} - 21.0\U{V} = \ans{-11\U{V}} \\ + V(6.00\U{m}) &= 10.0\U{V} - 42.0\U{V} = \ans{-32\U{V}} +\end{align} + +\Part{b} +Using $E_x = -dV/dx$ we have +\begin{equation} + E = - \frac{d}{dx}(a+bx) = -b = \ans{7.00\U{V/m}} +\end{equation} +At any point for $0 \le x \le 6.00\U{m}$. +\end{solution} diff --git a/latex/problems/problem20.22.tex b/latex/problems/problem20.22.tex new file mode 100644 index 0000000..16b1323 --- /dev/null +++ b/latex/problems/problem20.22.tex @@ -0,0 +1,25 @@ +\begin{problem*}{20.22} +The electric potential insize a charged spherical conductor of radius +$R$ is given by $V_i = k_e Q / R$, and the outside potential is given +by $V_o = k_e Q/r$. Using $E_r = -dV/dx$, determine the electric +field + \Part{a} inside and + \Part{b} outside + this charge distribution. +\end{problem*} % problem 20.22 + +\begin{solution} +\Part{a} +\begin{equation} + E_i = - \frac{d}{dx}\left(\frac{k_e Q}{R}\right) = 0 +\end{equation} +Because $V_i$ is constant with respect to $r$. + +\Part{b} +\begin{equation} + E_o = - \frac{d}{dx}\left(\frac{k_e Q}{r}\right) + = -k_e Q \frac{d}{dx}\left(\frac{1}{r}\right) + = -k_e Q \frac{-1}{r^2} + = \ans{\frac{k_e Q}{r^2}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.24.tex b/latex/problems/problem20.24.tex new file mode 100644 index 0000000..267a967 --- /dev/null +++ b/latex/problems/problem20.24.tex @@ -0,0 +1,33 @@ +\begin{problem*}{20.24} +Consider a ring of radius $R$ with the total charge $Q$ spread +uniformly over its perimeter. What is the potential difference +between the point at the center of the ring and a point on its axis a +distance $d=2R$ from the center? +\end{problem*} % problem 20.24 + +\begin{solution} +From the first week's recitation (P19.19), we have the electric field +along the axis due to the ring as +\begin{equation} + \vect{E} = \frac{k_e x Q}{(x^2 + R^2)^{3/2}}\ihat +\end{equation} +So the potential drop from $0$ to $d$ is given by +\begin{equation} + \Delta V = -\int_0^d E_x dx + = - k_e Q \int_0^d \frac{x \cdot dx}{(x^2 + R^2)^{3/2}} +\end{equation} +Substituting $u = x^2 + R^2$ so $du = 2x dx$ we have +\begin{equation} + \Delta V = - k_e Q \int \frac{1/2 \cdot du}{u^{3/2}} + = - \frac{1}{2} k_e Q \frac{-2}{\sqrt{u}} + = \frac{k_e Q}{\sqrt{u}} +\end{equation} +And plugging back in in terms of $x$ +\begin{equation} + \Delta V = \left.\frac{k_e Q}{\sqrt{x^2 + R^2}}\right|_0^d + = \frac{k_e Q}{\sqrt{d^2 + R^2}} - \frac{k_e Q}{R} + = k_e Q \left(\frac{1}{\sqrt{(4+1)R^2}} - \frac{1}{R}\right) + = \frac{k_e Q}{R} \left(\frac{1}{\sqrt{5}}-1\right) + = \ans{-0.533 \frac{k_e Q}{R}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.27.tex b/latex/problems/problem20.27.tex new file mode 100644 index 0000000..21a7865 --- /dev/null +++ b/latex/problems/problem20.27.tex @@ -0,0 +1,29 @@ +\begin{problem*}{20.27} +A uniformly charged insulating rod of length $L = 14.0\U{cm}$ is bent +to form a semicircle. The rod has a total charge of $Q = +-7.50\U{$\mu$C}$. Find the electric potential at the center of the +semicircle $0$. +\end{problem*} % problem 20.27 + +\begin{solution} +As in problem 20.11, we'll sum over all the charge bits, but in this +case our bits are infinitesimal, so our sum is technically an +integral. Defining the charge density $\lambda = Q/L$ we have +\begin{equation} + V = \int_0^L k_e \frac{\lambda dL}{r} = k_e \frac{\lambda}{r}\int_0^L dL + = k_e \frac{Q}{r} +\end{equation} +The same as for a point charge $Q$! This is because electric +potential is a scalar, and all the charges are the same distance from +$O$. It doesn't matter if they are all gathered together at one +point, or smeared out in a semicircle, spherical shell, or whatever, +as long as they are all the same distance $r$ from $O$. + +We still need to find $r$, but we know that the arc length of a +semicircle is $\pi r$, so $r = L/\pi$, and +\begin{equation} + V = k_e \frac{\pi Q}{L} + = 8.99\E{9}\U{N m$^2$/C$^2$} \frac{\pi \cdot (-7.50\E{-6}\U{C})}{0.140\U{m}} + = \ans{-1.51\U{MV}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.40.tex b/latex/problems/problem20.40.tex new file mode 100644 index 0000000..3f88ece --- /dev/null +++ b/latex/problems/problem20.40.tex @@ -0,0 +1,45 @@ +\begin{problem*}{20.40} +Two capacitors, $C_1 = 5.00\U{$\mu$F}$ and $C_2 = 12.0\U{$\mu$F}$, are +connected in series, and the resulting combination is connected to a +$\Delta V = 9.00\U{V}$ battery. Find + \Part{a} the equivalent capacitance of the combination, + \Part{b} the potential difference across each capacitor, and + \Part{c} the charge on each capacitor. +\end{problem*} % problem 20.40 + +\begin{solution} +\Part{a} +The wire connecting the inner plates of $C_1$ and $C_2$ contains no +net charge, so we know that any charge on the inner plate of $C_1$ +must have come from the inner plate of $C_2$. Because these charges +are equal and opposite, the total charge $Q$ on each capacitor +seperately is the same for both ($Q_1 = Q_2$). So using the +definition of capacitance for both cases we have +\begin{align} + \Delta V_1 &= Q / C_1 \label{eqn.V1} \\ + \Delta V_2 &= Q / C_2 \label{eqn.V2} \\ + \Delta V &= \Delta V_1 + \Delta V_2 + = Q \left( \frac{1}{C_1} + \frac{1}{C_2} \right) + = \frac{Q}{C_{eq}} \label{eqn.VQC} +\end{align} +So +\begin{equation} + C_{eq} = \left(\frac{1}{C_1}+\frac{1}{C_2}\right)^{-1} + = \left(\frac{1}{5.00\E{-6}\U{F}}+\frac{1}{12.0\E{-6}\U{F}}\right)^{-1} + = \ans{3.53\U{$\mu$F}} +\end{equation} + +\Part{c} +Plugging back into equation \ref{eqn.VQC} we have +\begin{equation} + Q = \Delta V \cdot C_{eq} + = 3.53\U{$\mu$F}\cdot 9.00\U{V} = \ans{31.8\U{$\mu$C}} +\end{equation} + +\Part{b} +And plugging into equations \ref{eqn.V1} and \ref{eqn.V2} we have +\begin{align} + \Delta V_1 &= \frac{31.8\E{-6}\U{C}}{5.00\E{-6}\U{F}} = \ans{6.35\U{V}} \\ + \Delta V_2 &= \frac{31.8\E{-6}\U{C}}{12.0\E{-6}\U{F}} = \ans{2.65\U{V}} \\ +\end{align} +\end{solution} diff --git a/latex/problems/problem20.41.tex b/latex/problems/problem20.41.tex new file mode 100644 index 0000000..e4fb2f4 --- /dev/null +++ b/latex/problems/problem20.41.tex @@ -0,0 +1,49 @@ +\begin{problem*}{20.41} +Four capacitors are connected as shown in Figure P20.41. +\Part{a} Find the equivalent capacitance between points $a$ and $b$. +\Part{b} Calculate the charge on each capacitor, taking + $\Delta V_{ab} = 15.0\U{V}$ +\end{problem*} % problem 20.41 + +\begin{solution} +\Part{a} +First consider the top two capacitors, $C_1 = 15.0\U{$\mu$F}$ and + $C_2 = 3.00\U{$\mu$F}$. +They are in series, so the effective capacitance of the top line is + given by +\begin{equation} + C_t = \left(\frac{1}{C_1} + \frac{1}{C_2}\right)^{-1} + = 2.50\U{$\mu$F} +\end{equation} + +We can find the effective capacitance of the box, because $C_t$ is +in parallel with $C_3 = 6.00\U{$\mu$F}$. +\begin{equation} + C_b = C_t + C_3 = 8.50\U{$\mu$C} +\end{equation} + +We can find the total equivalent capacitance, because $C_b$ is in +series with $C_4 = 20.0\U{$\mu$F}$. +\begin{equation} + C_{eq} = \left(\frac{1}{C_b} + \frac{1}{C_4}\right)^{-1} + = \ans{ 5.96\U{$\mu$F}} +\end{equation} + +\Part{b} +Working backwards to find the charges, using $Q = CV$, we have +\begin{equation} + Q_4 = Q_b = C_{eq} V_{ab} = \ans{89.5\U{$\mu$C}} +\end{equation} +So the voltage across the box is +\begin{equation} + V_b = \frac{Q_b}{C_b} = 10.5\U{V} +\end{equation} +So +\begin{equation} + Q_3 = C_3 V_b = \ans{63.2\U{$\mu$C}} +\end{equation} +and +\begin{equation} + Q_1 = Q_2 = C_t V_b = \ans{26.3\U{$\mu$C}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.43.tex b/latex/problems/problem20.43.tex new file mode 100644 index 0000000..7923aa6 --- /dev/null +++ b/latex/problems/problem20.43.tex @@ -0,0 +1,70 @@ +\begin{problem*}{20.43} +Consider the circuit shown in Figure P20.43, where $C_1 = +6.00\U{$\mu$F}$, $C_2 = 3.00\U{$\mu$F}$, and $\Delta V = 20.0\U{V}$. +Capacitor $C_1$ is first charged with $Q_1$ by the closing of switch +$S_1$. Switch $S_1$ is then opened, and the charged capacitor is +connected to the uncharged capacitor by the closing of $S_2$. +Calculate $Q_1$ and the final charge on each capacitor ($Q_1'$ and +$Q_2'$). +\begin{center} +\begin{empfile}[2p] +\begin{emp}(0cm, 0cm) + input makecirc; % circuit drawing functions + initlatex(""); + numeric a, cAy, cBy; + a = 3cm; + % add elements + battery.B(origin, 90, "V", ""); + centreof.c(B.B.n, B.B.p, cap); + capacitor.A(c.c+(a,0), normal, phi.c, "C_1", ""); + capacitor.B(c.c+(2a,0), normal, phi.c, "C_2", ""); + centreof.d(B.B.n, (xpart C.A.l, ypart B.B.n), swt); + switch.S(c.d, NO, phi.d, "S_1", ""); + centreof.e((xpart C.A.l, ypart B.B.n), (xpart C.B.l, ypart B.B.n), swt); + switch.s(c.e, NO, phi.e, "S_2", ""); + % add wiring along the bottom + wire(B.B.n, st.S.l, nsq); + wire(st.S.r, C.A.l, rlsq); + wire(st.S.r, st.s.l, nsq); + wire(st.s.r, C.B.l, rlsq); + % add wiring along the top + wire(B.B.p, C.A.r, rlsq); + wire(B.B.p, C.B.r, rlsq); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 20.43 + +\begin{solution} +The first situation with $S_1$ closed and $S_2$ open is just a +standard capacitor charging problem. Using the definition of +capacitance +\begin{equation} + Q_1 = C_1 \Delta V = 6.00\E{-6}\U{F} \cdot 20.0\U{V} = \ans{120\U{$\mu$C}} +\end{equation} + +After disconnecting the battery and connecting the two capacitors, we +have a net charge of $Q_1$ in the upper wire that we can distribute as +we desire between $C_1$ and $C_2$. Because charge is conserved, we +know +\begin{equation} + Q_1 = Q_1' + Q_2' \label{eqn.q12_one} +\end{equation} +We also know that at equilibrium the voltage across each capacitor +must be equal (because if there was a voltage difference beween the +upper plates of the two capacitors, it would push current through the +upper wire until the voltage difference dissapeared, etc.). So +\begin{equation} + \Delta V_1' = \frac{Q_1'}{C_1} = \Delta V_2' = \frac{Q_2'}{C_2} \label{eqn.q12_two} +\end{equation} +Now we have two equations relating our two unknowns $Q_1'$ and $Q_2'$. +Solving equation \ref{eqn.q12_two} for $Q_2'$ and plugging into +equation \ref{eqn.q12_one} we get +\begin{align} + Q_2' &= \frac{C_2}{C_1} Q_1' \\ + Q_1 &= \left(1 + \frac{C_2}{C_1}\right)Q_1' \\ + Q_1' &= \frac{Q_2}{1 + C_2/C_1} = \frac{120\U{$\mu$C}}{1.5} + = \ans{80\U{$\mu$C}} \\ + Q_2' &= 0.5 \cdot 80\U{$\mu$C} = \ans{40\U{$\mu$C}} +\end{align} +\end{solution} diff --git a/latex/problems/problem20.47.tex b/latex/problems/problem20.47.tex new file mode 100644 index 0000000..58eaee0 --- /dev/null +++ b/latex/problems/problem20.47.tex @@ -0,0 +1,18 @@ +\begin{problem*}{20.47} +\Part{a} A $C = 3.00\U{$\mu$F}$ capacitor is connected to a $\Delta +V_a = 12.0\U{V}$ battery. How much energy $U_a$ is stored in the +capacitor? +\Part{b} If the capacitor had been connected to a $\Delta V_b = +6.00\U{V}$ battery, how much energy would have been stored? +\end{problem*} % problem 20.47 + +\begin{solution} +Simply plugging into the formula for energy stored in a capacitor we have +\begin{align} + U_a &= \frac{1}{2} C (\Delta V)^2 + = \frac{1}{2} (3.00\E{-6}\U{F}) \cdot (12.0\U{V})^2 + = \ans{ 216 \U{$\mu$J}} \\ + U_b &= \frac{1}{2} (3.00\E{-6}\U{F}) \cdot (6.0\U{V})^2 + = \ans{ 54 \U{$\mu$J}} +\end{align} +\end{solution} diff --git a/latex/problems/problem20.49.tex b/latex/problems/problem20.49.tex new file mode 100644 index 0000000..da7437f --- /dev/null +++ b/latex/problems/problem20.49.tex @@ -0,0 +1,33 @@ +\begin{problem*}{20.49} +Two capacitors, $C_1 = 25.0\U{$\mu$F}$ and $C_2 = 5.00\U{$\mu$F}$, are +connected in parallel and charged with a $\Delta V = 100\U{V}$ power +supply. +\Part{a} Draw a circuit diagram and calculate the total energy stored in the +two capacitors. +\Part{b} What potential difference would be required across the same +two capacitors connected in series so that the combination stores the +same energy as in \Part{a}? +Draw a circuit diagram for this circuit. +\end{problem*} % problem 20.49 + +\begin{solution} +The diagrams are given in the back of the book. + +\Part{a} +The equivalent capacitance is $C_{eq} = C_1 + C_2 = 30.0\U{$\mu$F}$, +so the stored energy is +\begin{equation} + U = \frac{1}{2} C_{eq} (\Delta V)^2 = \ans{0.150\U{J}} +\end{equation} + +\Part{b} +The equivalent capacitance is now +\begin{equation} + C_{eq} = \left(\frac{1}{C_1}+\frac{1}{C_2}\right)^{-1} = 4.17\U{$\mu$F} +\end{equation} +So the necessary voltage is given by +\begin{align} + U &= \frac{1}{2} C_{eq} (\Delta V)^2 \\ + \Delta V &= \sqrt{\frac{2 U}{C_{eq}}} = \ans{ 268\U{V}} +\end{align} +\end{solution} diff --git a/latex/problems/problem20.51.tex b/latex/problems/problem20.51.tex new file mode 100644 index 0000000..439666b --- /dev/null +++ b/latex/problems/problem20.51.tex @@ -0,0 +1,16 @@ +\begin{problem*}{20.51} +Show that the force between two plates of a parallel-plate capacitor +each have an attractive force given by +\begin{equation} + F = \frac{Q^2}{2\epsilon_0 A} +\end{equation} +\end{problem*} % problem 20.51 + +\begin{solution} +The electric field generated by the plate $A$ is given by $E_A = Q/2 +\epsilon_0 A$ (which we derived for P19.62, along with $\sigma = +Q/A$). So the force on plate $B$ due to plate $A$ is given by +\begin{equation} + F = QE_A = \frac{Q^2}{2 \epsilon_0 A} +\end{equation} +\end{solution} diff --git a/latex/problems/problem20.54.tex b/latex/problems/problem20.54.tex new file mode 100644 index 0000000..6a27ce4 --- /dev/null +++ b/latex/problems/problem20.54.tex @@ -0,0 +1,46 @@ +\begin{problem*}{20.54} +\Part{a} How much charge $Q_c$ can be placed on a capacitor with air +between the plates before it breaks down if the area of each plate is +$A=5.00\U{cm}^2$? +\Part{b} Find the maximum charge assuming polystyrene is used between +the plates instead of air. +\end{problem*} % problem 20.54 + +\begin{solution} +From Chapter 19, the voltage difference due to a constant electric +field \vect{E} over a displacement \vect{d} is given by $\Delta V = +\vect{E} \cdot \vect{d}$. So for two plates a distance $d$ apart, the +breakdown voltage is given by +\begin{equation} + V_c = E_c d \label{eqn.Vc_Ec} +\end{equation} +where $E_c$ is the dielectric strength of the material. + +The capacitance of a parallel-plate capacitor is given by +\begin{equation} + C = \frac{\kappa \epsilon_0 A}{d} \label{eqn.pp_cap} +\end{equation} + +Combining these two formula with the definition of capacitance we have +\begin{equation} + E_c d = V = \frac{Q}{C} = \frac{Q d}{\kappa \epsilon_0 A} \\ + Q = \kappa E_c \epsilon_0 A +\end{equation} + +Looking up the values for air and polystyrene in Table 20.1 on page +699 of the text we see: +\begin{center} + \begin{tabular}{l r r} + Name & Dielectric constant $\kappa$ & Dielectric strength $E_c$ \\ + \hline + \Tstrut Air & $1.00059$ & $3\E{6}\U{V/m}$ \\ + Polystyrene & $2.56$ & $24\E{6}\U{V/m}$ \\ + \end{tabular} +\end{center} + +So plugging into our formula for the charge +\begin{align} + Q_a &= 1.00 \cdot (3\E{6}\U{V/m}) \cdot (8.85\E{-12}\U{C$^2$/N m$^2$}) \cdot 5\E{-4}\U{m$^2$} = \ans{1.33\E{-8}\U{C}} \\ + Q_b &= 2.56 \cdot (24\E{6}\U{V/m}) \cdot (8.85\E{-12}\U{C$^2$/N m$^2$}) \cdot 5\E{-4}\U{m$^2$} = \ans{2.72\E{-7}\U{C}} +\end{align} +\end{solution} diff --git a/latex/problems/problem20.69.tex b/latex/problems/problem20.69.tex new file mode 100644 index 0000000..24599d7 --- /dev/null +++ b/latex/problems/problem20.69.tex @@ -0,0 +1,29 @@ +\begin{problem*}{20.69} +The $x$ axis is the symmetry axis of a stationary, uniformly charged +ring of radius $R$ and charge $Q$ (Fig.~P20.69). A particle with +charge $Q$ and mass $M$ is located at the center of the ring. When it +is displaced slightly, the point charge accelerates along the $x$ axis +to infinity. Show that the ultimate speed of the point charge is +\begin{equation} + v = \left(\frac{2 k_e Q^2}{MR}\right)^{1/2} +\end{equation} +\end{problem*} % problem 20.69 + +\begin{solution} +Conserving energy, the inital energy is entirely electric, +\begin{equation} + E_i = U_e = k_e \frac{Q^2}{R} +\end{equation} +because all the ring charge is a distance $R$ from the particle. + +The final energy is entirely kinetic +\begin{equation} + E_f = K = \frac{1}{2} M v^2 +\end{equation} + +So +\begin{align} + k_e \frac{Q^2}{R} = E_i &= E_f = \frac{1}{2} M v^2 \\ + v &= \ans{\sqrt{\frac{2 k_e Q^2}{M R}}} +\end{align} +\end{solution} diff --git a/latex/problems/problem20.73.tex b/latex/problems/problem20.73.tex new file mode 100644 index 0000000..31a5b64 --- /dev/null +++ b/latex/problems/problem20.73.tex @@ -0,0 +1,31 @@ +\begin{problem*}{20.73} +A parallel-plate capacitor is constructed using a dielectric material +whose dielectric constant is $\kappa = 3.00$ and whose dielectric +strength is $E_c = 2.00\E{8}\U{V/m}$. The desired capacitance is $C = +0.250\U{$\mu$F}$, and the capacitor must withstand a maximum potential +difference of $V_c = 4000\U{V}$. Find the minimum area $A$ of the +capacitor plates. +\end{problem*} + +\begin{solution} +Using equation \ref{eqn.Vc_Ec}, we have +\begin{equation} + d \ge \frac{V_c}{E_c} +\end{equation} +Where equality represents a breakdown at $V_c$ and larger $d$ give us +more protection with larger breakdown voltages. + +From equation \ref{eqn.pp_cap} we have +\begin{equation} + A = \frac{d C}{\kappa \epsilon_0} +\end{equation} +From which we can see that the smaller $d$ is, the smaller $A$ can be, +and we pick $d = V_c/E_c$, the smallest possible value we can. + +Then the smallest area is given by +\begin{equation} + A = \frac{V_c C}{E_c \kappa \epsilon_0} + = \frac{(4000\U{V})\cdot(0.25\E{-6}\U{F})}{(2.00\E{8}\U{V/m})\cdot3.00\cdot(8.85\E{-12}\U{C$^2$/Nm$^2$})} + = \ans{0.188\U{m$^2$}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem21.01.tex b/latex/problems/problem21.01.tex new file mode 100644 index 0000000..b92f1a1 --- /dev/null +++ b/latex/problems/problem21.01.tex @@ -0,0 +1,19 @@ +\begin{problem*}{21.1} +In a particular cathode-ray tube, the measured beam current is + $I = 30.0\U{$\mu$A}$. +How many electrons strike the tube screen every $\Delta t = 40.0\U{s}$ +\end{problem*} % problem 21.1 + +\begin{solution} +Current is defined as \emph{charge passing through a given surface per + unit time} or in SI units: +\begin{equation} + 1\U{A} = \frac{1\U{C}}{1\U{s}} \;. +\end{equation} +So +\begin{align} + \Delta Q &= I \Delta t = 1.20\U{mC} \\ + N_e &= \frac{\Delta Q}{e} = \ans{7.50\E{15}} \;, +\end{align} +where $e = 1.60\E{-19}\U{C}$ is the charge on one electron. +\end{solution} diff --git a/latex/problems/problem21.04.tex b/latex/problems/problem21.04.tex new file mode 100644 index 0000000..298b647 --- /dev/null +++ b/latex/problems/problem21.04.tex @@ -0,0 +1,21 @@ +\begin{problem*}{21.4} +The quantity of charge $q$ (in coulombs) that has passed through a +surface of area $A = 2.00\U{cm$^2$}$ varies with time according to the +equation $q = 4t^3 + 5t + 6$, where $t$ is in seconds. +\Part{a} What is the instantaneous current across the surface at + $t_a = 1.00\U{s}$? +\Part{b} What is the value of the current density? +\end{problem*} % problem 21.4 + +\begin{solution} +\Part{a} +\begin{align} + I(t) = \frac{dQ}{dt} &= 12 t^2 + 5 \\ + I(t_a) &= \ans{17.0\U{A}} +\end{align} + +\Part{b} +\begin{equation} + j(t_a) = I(t_a)/A = \ans{8.50\U{A/cm$^2$}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem21.14.tex b/latex/problems/problem21.14.tex new file mode 100644 index 0000000..49cd088 --- /dev/null +++ b/latex/problems/problem21.14.tex @@ -0,0 +1,18 @@ +\begin{problem*}{21.14} +A toaster is rated at $P = 600\U{W}$ when connected to a $V = +120\U{V}$ source. What current $I$ does the toaster carry, and what +is its resistance $R$? +\end{problem*} % problem 21.14 + +\begin{solution} +(Assuming the voltage is DC). +The power through a resistor is given by $P = IV$ so +\begin{equation} + I = \frac{P}{V} = \frac{600\U{W}}{120\U{V}} = \ans{5\U{A}} +\end{equation} + +The voltage across a resistor is given by $V = IR$ so +\begin{equation} + R = \frac{V}{I} = \frac{120\U{V}}{5\U{A}} = \ans{24\U{$\Omega$}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem21.17.tex b/latex/problems/problem21.17.tex new file mode 100644 index 0000000..c3fb052 --- /dev/null +++ b/latex/problems/problem21.17.tex @@ -0,0 +1,23 @@ +\begin{problem*}{21.17} +Suppose a voltage surge produces $V_s = 140\U{V}$ for a moment. By +what percentage $p$ does the power output of a $V = 120\U{V}$, $P = +100\U{W}$ lightbulb increase? Assume the resistance does not change. +\end{problem*} % problem 21.17 + +\begin{solution} +The voltage across a resistor is +\begin{equation} + V = IR +\end{equation} +So power absorbed by a resistor is +\begin{equation} + P = IV = \frac{V^2}{R} +\end{equation} + +And the fractional change in power $f$ is given by +\begin{equation} + f = \frac{P_s}{P} = \frac{V_s^2 / R}{V^2 / R} + = \left(\frac{V_s}{V}\right)^2 = 1.361 +\end{equation} +So $p = \ans{36.1\U{\%}}$. +\end{solution} diff --git a/latex/problems/problem21.25.inversion.tex b/latex/problems/problem21.25.inversion.tex new file mode 100644 index 0000000..0ab2aea --- /dev/null +++ b/latex/problems/problem21.25.inversion.tex @@ -0,0 +1,63 @@ +\begin{align} + \begin{pmatrix} + -1 & 1 & 1 & \big| & 1 & 0 & 0 \\ + R_1 & R_2 & 0 & \big| & 0 & 1 & 0 \\ + R_1 & 0 & R_3 & \big| & 0 & 0 & 1 + \end{pmatrix} + &\rightarrow + \begin{pmatrix} + 1 & -1 & -1 & \big| & -1 & 0 & 0 \\ + 1 & R_2/R_1 & 0 & \big| & 0 & 1/R_1 & 0 \\ + 1 & 0 & R_3/R_1 & \big| & 0 & 0 & 1/R_1 + \end{pmatrix} + \rightarrow \\ + \begin{pmatrix} + 1 & -1 & -1 & \big| & -1 & 0 & 0 \\ + 0 & R_2/R_1 +1 & 1 & \big| & 1 & 1/R_1 & 0 \\ + 0 & 1 & R_3/R_1 +1 & \big| & 1 & 0 & 1/R_1 + \end{pmatrix} + &\rightarrow + \begin{pmatrix} + 1 & -1 & -1 & \big| & -1 & 0 & 0 \\ + 0 & 1 & \frac{R_1}{R_2 +R_1} & \big| & \frac{R_1}{R_2 +R_1} & \frac{1}{R_2+R_1} & 0 \\ + 0 & 1 & \frac{R_3+R_1}{R_1} & \big| & 1 & 0 & 1/R_1 + \end{pmatrix} + \rightarrow \\ + \begin{pmatrix} + 1 & -1 & -1 & \big| & -1 & 0 & 0 \\ + 0 & 1 & \frac{R_1}{R_2 +R_1} & \big| & \frac{R_1}{R_2 +R_1} & \frac{1}{R_2+R_1} & 0 \\ + 0 & 0 & \frac{R_3+R_1}{R_1} - \frac{R_1}{R_2+R_1} & \big| & 1-\frac{R_1}{R_2 + R_1} & \frac{-1}{R_2+R_1} & 1/R_1 + \end{pmatrix} + &\rightarrow + \begin{pmatrix} + 1 & -1 & -1 & \big| & -1 & 0 & 0 \\ + 0 & 1 & \frac{R_1}{R_2 +R_1} & \big| & \frac{R_1}{R_2 +R_1} & \frac{1}{R_2+R_1} & 0 \\ + 0 & 0 & \frac{R_3R_2 + R_2R_1+ R_1R_3+ R_1^2-R_1^2}{R_1(R_2+R_1)} & \big| & \frac{R_2}{R_2 + R_1} & \frac{-1}{R_2+R_1} & 1/R_1 + \end{pmatrix} + \rightarrow \\ + \begin{pmatrix} + 1 & -1 & -1 & \big| & -1 & 0 & 0 \\ + 0 & 1 & \frac{R_1}{R_2 +R_1} & \big| & \frac{R_1}{R_2 +R_1} & \frac{1}{R_2+R_1} & 0 \\ + 0 & 0 & \frac{A}{R_1(R_2+R_1)} & \big| & \frac{R_2}{R_2 + R_1} & \frac{-1}{R_2+R_1} & 1/R_1 + \end{pmatrix} + &\rightarrow + \begin{pmatrix} + 1 & -1 & -1 & \big| & -1 & 0 & 0 \\ + 0 & 1 & \frac{R_1}{R_2 +R_1} & \big| & \frac{R_1}{R_2 +R_1} & \frac{1}{R_2+R_1} & 0 \\ + 0 & 0 & 1 & \big| & \frac{R_1R_2}{A} & \frac{-R_1}{A} & \frac{R_1+R_2}{A} + \end{pmatrix} + \rightarrow \\ + \begin{pmatrix} + 1 & -1 & -1 & \big| & -1 & 0 & 0 \\ + 0 & 1 & 0 & \big| & \frac{R_1}{R_2 +R_1}\left(1-\frac{R_1R_2}{A}\right) & \frac{R_1}{R_2+R_1}\left(\frac{1}{R_1}+\frac{R_3}{A}\right) & \frac{-R_1}{A} \\ + 0 & 0 & 1 & \big| & \frac{R_1R_2}{A} & \frac{-R_1}{A} & \frac{R_1+R_2}{A} + \end{pmatrix} + &\rightarrow + \begin{pmatrix} + 1 & 0 & 0 & \big| & -0.2308 & 0.0385 & 0.0577 \\ + 0 & 1 & 0 & \big| & 0.3077 & 0.1154 & -0.0769 \\ + 0 & 0 & 1 & \big| & 0.4615 & -0.0769 & 0.1346 + \end{pmatrix} + \rightarrow \\ +\end{align} +Where $A \equiv R_1 R_2 + R_2 R_3 + R_3 R_1$. diff --git a/latex/problems/problem21.25.tex b/latex/problems/problem21.25.tex new file mode 100644 index 0000000..3252231 --- /dev/null +++ b/latex/problems/problem21.25.tex @@ -0,0 +1,63 @@ +\begin{problem*}{21.25} +A battery has an emf of $\epsilon = 15.0\U{V}$. The terminal voltage +of the battery is $V_t = 11.6\U{V}$ when it is delivering $P = +20.0\U{W}$ of power to an external load resistor $R$. +\Part{a} What is the value of $R$? +\Part{b} What is the internal resistance $r$ of the battery? +\end{problem*} % problem 21.25 + +\begin{solution} +\begin{center} +\begin{empfile}[1] +\begin{emp}(0,0) + path P; + pair N[]; + numeric dx, ddy, f; + dx := 2cm; + ddy := 3pt; + f := 2.5; + % draw the battery branch components + battery.B(origin, 90, "\epsilon", ""); + resistor.B(B.B.p, normal, 90, "r", ""); + N[0] := B.B.n; + N[1] := R.B.r; + % draw the resistor branch components + centerto.A(N[0], N[1], dx, res); + resistor.A(A, normal, 90, "R", ""); + N[2] := (xpart R.A.l, ypart N[0]); + N[3] := (xpart N[2], ypart N[1]); + % draw the currents + centreof.i(N[1], N[3], cur); + current.A(c.i, phi.i, "I", ""); + % draw the wires + wire(N[0], R.A.l, rlsq); + wire(N[1], R.A.r, rlsq); + % draw the nodes + draw N[0] withpen pencircle scaled 3pt; + draw N[1] withpen pencircle scaled 3pt; + % box the battery + P := (N[0]+(-dx/f,ddy))--(N[0]+(dx/f,ddy))--(N[1]-(-dx/f,ddy))--(N[1]-(dx/f,ddy))--cycle; + draw P dashed evenly; + % label the battery + ctext.lft(N[0]+(-dx/f,ddy), N[1]-(dx/f,ddy), "$V_t$", witharrow); +\end{emp} +\end{empfile} +\end{center} +\Part{a} +Using the power absorbed by $R$ +\begin{align} + P &= IV_t = \frac{V_t^2}{R} \\ + R &= \frac{V_t^2}{P} = \frac{(11.6\U{V})^2}{20.0\U{W}} = \ans{6.75\Omega} +\end{align} + +\Part{b} +The current through the entire ciruit is given by +\begin{equation} + I = \frac{V_t}{R} = \frac{P}{V_t} = 1.72\U{A} +\end{equation} +So the internal resistance is given by +\begin{align} + \epsilon - V_t &= I r \\ + r &= \frac{\epsilon - V_t}{I} = \frac{3.4\U{V}}{1.72\U{A}} = \ans{1.97\Omega} +\end{align} +\end{solution} diff --git a/latex/problems/problem21.27.tex b/latex/problems/problem21.27.tex new file mode 100644 index 0000000..4dd4ea1 --- /dev/null +++ b/latex/problems/problem21.27.tex @@ -0,0 +1,62 @@ +\begin{problem*}{21.27} +\Part{a} Find the equivalent resistance between points $a$ and $b$ in +Figure P21.27. +\Part{b} A potential difference of $V = 34.0\U{V}$ is applied between +points $a$ and $b$. Calculate the current in each resistor. +\begin{center} +\begin{empfile}[5] +\begin{emp}(0,0) + input makecirc; % circuit drawing functions + initlatex(""); + numeric dx, dy; + dx := 1cm; + dy := 1cm; + dotlabel.lft("a", origin); + resistor.A(origin, normal, 0, "R_1", "4.00\ohm"); + resistor.B((0,dy)+R.A.r, normal, 0, "R_2", "7.00\ohm"); + resistor.C((0,-dy)+R.A.r, normal, 0, "R_3", "10.0\ohm"); + resistor.D((xpart(R.C.r),0), normal, 0, "R_4", "9.00\ohm"); + wire(R.A.r, R.B.l,nsq); + wire(R.A.r, R.C.l,nsq); + draw R.A.r withpen pencircle scaled 3pt; + wire(R.D.l, R.B.r,nsq); + wire(R.D.l, R.C.r,nsq); + draw R.D.l withpen pencircle scaled 3pt; + dotlabel.rt("b", R.D.r); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 21.27 + +\begin{solution} +%(Numbering from right to left and top to bottom, +% $R_1 = 4.00\U{$\Omega$}$, +% $R_2 = 7.00\U{$\Omega$}$, +% $R_3 = 10.0\U{$\Omega$}$, and +% $R_4 = 9.00\U{$\Omega$}$.) + +\Part{a} +First, we calculate the equivalent resistance to the two resistors in parallel +\begin{equation} + R_p = \left(\frac{1}{R_2} + \frac{1}{R_1}\right)^{-1} = 4.12\U{$\Omega$} +\end{equation} +Then we calculate the equivalent resistance of the three resistors in series +\begin{equation} + R_{ab} = R_1 + R_p + R_4 = \ans{17.1\U{$\Omega$}} +\end{equation} + +\Part{b} +Now applying $V = IR$ to the equivalent system +\begin{equation} + I_{ab} = I_1 = I_4 = I_p = \frac{V}{R_{ab}} = \ans{1.99\U{A}} +\end{equation} +From which we can compute the voltage across the parallel resistors +\begin{equation} + V_p = I_p R_p = 8.18\U{V} +\end{equation} +Giving us currents of +\begin{align} + I_2 &= \frac{V_p}{R_2} = \ans{ 1.17\U{A}} \\ + I_3 &= \frac{V_p}{R_3} = \ans{ 0.818\U{A}} +\end{align} +\end{solution} diff --git a/latex/problems/problem21.29.tex b/latex/problems/problem21.29.tex new file mode 100644 index 0000000..96b7092 --- /dev/null +++ b/latex/problems/problem21.29.tex @@ -0,0 +1,99 @@ +\begin{problem*}{21.29} +Consider the circuit shown in Figure P21.29. +Find \Part{a} the current in the $R_1 = 20\Omega$ resistor and + \Part{b} the potential difference between points $a$ and $b$. +\begin{center} +\begin{empfile}[2] +\begin{emp}(0,0) + pair N[]; + numeric dx, ddx, dy; + dx := 4cm; + ddx := 1cm; + dy := 1.5 cm; + % draw the top branch components + resistor.T(origin, normal, 0, "R_3", "10.0\ohm"); + battery.T(R.T.r, 0, "V", "25.0 V"); + N[0] := R.T.l; + N[1] := B.T.p; + % draw the top-middle branch components + centerto.t(N[0], N[1], -dy, res); + resistor.t(t, normal, 0, "R_4", "10.0\ohm"); + N[2] := (xpart N[0], ypart R.t.l); + N[3] := (xpart N[1], ypart R.t.r); + % draw the bottom-middle branch components + centerto.b(N[0], N[1], -2dy, res); + resistor.b(b, normal, 0, "R_5", "5.00\ohm"); + N[4] := (xpart N[0], ypart R.b.l); + N[5] := (xpart N[1], ypart R.b.r); + % draw the bottom branch components + N[6] := N[2] - (ddx,0); + N[7] := N[3] + (ddx,0); + N[8] := N[6] - (0,2dy); + N[9] := N[7] - (0,2dy); + centreof.L(N[8], N[6], res); + resistor.L(c.L, normal, phi.L, "\mbox{5.00\ohm}", "$R_2$"); + centreof.R(N[9], N[7], res); + resistor.R(c.R, normal, phi.R, "R_1", "20.0\ohm"); + % draw the currents + %centreof.T(R.T.r, B.T.n, cur); + %current.T(c.T, phi.T, "", ""); + % draw the wires + wire(R.t.l, N[0], rlsq); + wire(R.t.r, N[1], rlsq); + wire(R.b.l, N[2], rlsq); + wire(R.b.r, N[3], rlsq); + wire(N[2], R.L.r, rlsq); + wire(N[3], R.R.r, rlsq); + wire(N[8], R.L.l, nsq); + wire(N[9], R.R.l, nsq); + wire(N[8], N[9], nsq); + % draw the nodes + draw N[2] withpen pencircle scaled 3pt; + draw N[3] withpen pencircle scaled 3pt; + draw (N[2]+N[6])/2 withpen pencircle scaled 3pt; + draw (N[3]+N[7])/2 withpen pencircle scaled 3pt; + % label the connection points + puttext.top("$a$", (N[2]+N[6])/2); + puttext.top("$b$", (N[3]+N[7])/2); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} + +\begin{solution} +Label the voltage $V = 25.0\U{V}$ and the resistances (clockwise from +$b$) $R_1 = 20.0\Omega$, $R_2 = 5.00\Omega$, $R_3 = 10.0\Omega$, $R_4 += 10.0\Omega$, and $R_5 = 5.00\Omega$. + +Computing some equivalent resistance of $R_1$ and $R_2$ in series we +have +\begin{equation} + R_s = R_1 + R_2 = 25.0\Omega +\end{equation} +Computing the equivalent resistance of $R_s$, $R_4$, and $R_5$ in +parallel we have +\begin{equation} + R_p = \left(\frac{1}{R_4} + \frac{1}{R_5} + \frac{1}{R_s}\right)^{-1} + = 2.94\Omega +\end{equation} +And the equivalent resistance of the entire setup is +\begin{equation} + R_e = R_p + R_3 = 12.94\Omega +\end{equation} + +The total current is then (from Ohm's law) +\begin{equation} + I_e = \frac{V}{R_e} = 1.93\U{A} +\end{equation} +And the voltage from $a$ to $b$ is +\begin{equation} + V_{ab} = I_e R_p = \ans{5.68\U{V}} +\end{equation} +Which is what they were looking for in \Part{b}. + +The current through the branch with $R_1$ and $R_2$ is then +\begin{equation} + I_s = \frac{V_{ab}}{R_s} = \ans{227\U{mA}} +\end{equation} +Which is what they were looking for in \Part{a}. +\end{solution} diff --git a/latex/problems/problem21.30.tex b/latex/problems/problem21.30.tex new file mode 100644 index 0000000..9c699ec --- /dev/null +++ b/latex/problems/problem21.30.tex @@ -0,0 +1,68 @@ +\begin{problem*}{21.30} +Three $R = 100\U{$\Omega$}$ resistors are connected as shown in Figure +P21.30. The maximum power that can safely be delivered to any one +resistor is $P_{max} = 25.0\U{W}$. +\Part{a} What is the maximum voltage that van be applies to the +terminals $a$ and $b$? +\Part{b} For the voltage determined in \Part{a}, what is the power +delivered to each resistor? +What is the total power delivered? +\begin{center} +\begin{empfile}[6] +\begin{emp}(0,0) + input makecirc; % circuit drawing functions + initlatex(""); + numeric dx, dy; + dx := 1cm; + dy := 1cm; + dotlabel.lft("a", origin); + resistor.A(origin, normal, 0, "R_1", "100\ohm"); + resistor.B((0,dy)+R.A.r, normal, 0, "R_2", "100\ohm"); + resistor.C((0,-dy)+R.A.r, normal, 0, "R_3", "100\ohm"); + wire(R.A.r, R.B.l,nsq); + wire(R.A.r, R.C.l,nsq); + draw R.A.r withpen pencircle scaled 3pt; + wire((xpart(R.B.r),0), R.B.r,nsq); + wire((xpart(R.B.r),0), R.C.r,nsq); + wire((xpart(R.B.r),0), (xpart(R.B.r)+dx,0),nsq); + draw (xpart(R.B.r)+dx,0) withpen pencircle scaled 3pt; + dotlabel.rt("b", (xpart(R.B.r)+dx,0)); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 21.30 + +\begin{solution} +\Part{a} +The current through the entire setup $I_{ab}$ all goes through $R_1$, +so $I_{ab} = I_1$. Then it splits 50/50, so $I_{ab} = 2I_2 = 2I_3$. +($R_1$ and $R_2$ each get half the current going through $R_1$). +Because it gets the most current, the maximum current $I_{ab}$ is when +the power $P_1$ absorbed by $R_1$ is $P_{max}$. +\begin{align} + P_{max} &= \frac{V_1^2}{R_1} \\ + V_1 &= \sqrt{R_1 P_{max}} = 50\U{V} +\end{align} +So $I_1 = I_{ab} = V_1/R_1 = 0.500\U{A}$. + +The equivalent resistance of the two parallel resistors is +\begin{equation} + R_p = \left( \frac{1}{R_1} + \frac{1}{R_2}\right)^{-1} = 50\U{$\Omega$} +\end{equation} +So the voltage drop over them is $V_p = I_{ab} R_p = 25.0\U{V}$. + +Adding the two voltages together +\begin{equation} + V_{ab} = V_1 + V_p = \ans{75.0\U{V}} +\end{equation} + +\Part{b} +The power absorbed by the other two resistors is then +\begin{equation} + P_2 = P_3 = I_2 V_p = 0.250\U{A} \cdot 25.0\U{V} = \ans{6.25\U{W}} \;, +\end{equation} +and the total power delivered is +\begin{equation} + P = P_1 + P_2 + P_3 = (25 + 2\cdot6.25)\U{W} = \ans{37.5\U{W}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem21.31.tex b/latex/problems/problem21.31.tex new file mode 100644 index 0000000..525f0c7 --- /dev/null +++ b/latex/problems/problem21.31.tex @@ -0,0 +1,85 @@ +\begin{problem*}{21.31} +Calculate the power delivered to each resistor in the circuit shown in +Figure P21.31. +\begin{center} +\begin{empfile}[3] +\begin{emp}(0,0) + pair N[]; + numeric ddy, dx; + ddy := 6pt; + dx := 3cm; + % draw the left branch components + battery.L(origin, 90, "\mbox{18.0 V}", "$V$"); + N[0] := B.L.n-(0,ddy); + N[1] := B.L.p+(0,ddy); + % draw the horizontal components + N[2] := N[0]+(dx,0); + N[3] := N[1]+(dx,0); + centreof.B(N[0], N[2], res); + resistor.B(c.B, normal, phi.B, "R_4", "4.00\ohm"); + centreof.T(N[1], N[3], res); + resistor.T(c.T, normal, phi.T, "R_1", "2.00\ohm"); + % draw the middle branch components + centreof.M(N[2], N[3], res); + resistor.M(c.M, normal, phi.M, "R_2", "3.00\ohm"); + % draw the right branch components + N[4] := 2N[2]-N[0]; + N[5] := 2N[3]-N[1]; + centreof.R(N[4], N[5], res); + resistor.R(c.R, normal, phi.R, "R_3", "1.00\ohm"); + % draw the currents + %centreof.T(R.T.r, B.T.n, cur); + %current.T(c.T, phi.T, "", ""); + % draw the wires + wire(B.L.p, R.T.l, udsq); + wire(B.L.n, R.B.l, udsq); + wire(R.T.r, N[3], nsq); + wire(R.B.r, N[2], nsq); + wire(R.M.r, N[3], nsq); + wire(R.M.l, N[2], nsq); + wire(R.R.r, N[3], udsq); + wire(R.R.l, N[2], udsq); + % draw the nodes + draw N[2] withpen pencircle scaled 3pt; + draw N[3] withpen pencircle scaled 3pt; +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 21.31 + +\begin{solution} +Label the voltage $V = 18.0\U{V}$ and the resistors (starting in the +upper left) $R_1 = 2.00\Omega$, $R_2 = 3.00\Omega$, $R_3 = +1.00\Omega$, and $R_4 = 4.00\Omega$. + +To find the total current through the circuit, we compute its +equivalent resistance. First for the two resistors in parallel +\begin{equation} + R_p = \left(\frac{1}{R_2} + \frac{1}{R_3}\right){-1} = 0.750\Omega + %1/3 + 1 = 4/3 +\end{equation} +And then for the complete circuit +\begin{equation} + R_c = R_1 + R_p + R_4 = 6.75\Omega +\end{equation} + +Using Ohm's law to calculate the total current $I_c = I_1 = I_4$ we have +\begin{equation} + I_c = \frac{V}{R_c} = 18/6.75 = 2.67\U{A} +\end{equation} +And the powers dissipated through $R_1$ and $R_4$ are +\begin{align} + P_1 &= I_1 V_1 = I_c^2 R_1 = \ans{14.2\U{W}} \\ + P_4 &= I_c^2 R_4 = \ans{28.4\U{W}} +\end{align} + +The voltage across $R_p$ is given by +\begin{equation} + V_p = I_c R_p = 2\U{V} +\end{equation} +And the powers dissipated through $R_2$ and $R_3$ are +\begin{align} + P_2 &= I_2 V_2 = \frac{V_p^2}{R_2} = \ans{1.33\U{W}} \\ + P_3 &= \frac{V_p^2}{R_3} = \ans{4\U{W}} +\end{align} +\end{solution} diff --git a/latex/problems/problem21.32.tex b/latex/problems/problem21.32.tex new file mode 100644 index 0000000..ab1fe7a --- /dev/null +++ b/latex/problems/problem21.32.tex @@ -0,0 +1,120 @@ +\begin{problem*}{21.32} +Four resistors are connected to a battery as shown in Figure P21.32. +The current in the battery is $I$, the battery emf is $\epsilon$, and +the resistor values are $R_1 = R$, $R_2 = 2R$, $R_3 = 4R$, and $R_4 = +3R$. +\Part{a} Rank the resistors according to the potential difference +across them, form largest to smallest. Note any cases of equal +potential difference. +\Part{b} Determine the potential difference across each resistor in +terms of $\epsilon$. +\Part{c} Rank the resistors according to the current in them, from +largest to smallest. Note any cases of equal current. +\Part{d} Determine the current in each resistor in terms of $I$. +\Part{e} If $R_3$ is increased, what happens to the current in each of +the resistors? +\Part{f} In the limit that $R_3 \rightarrow \infty$, what are the new +values of the current in each resistor in terms of $I$, the original +current in the battery? +\begin{center} +\begin{empfile}[7] +\begin{emp}(0,0) + input makecirc; % circuit drawing functions + initlatex(""); + pair N[]; + numeric dx, dy; + dx := 2cm; + dy := 1cm; + battery.B(origin, 90, "\mathcal{E}", ""); + resistor.A(B.B.p, normal, 90, "R_1", "$R$"); + N[0] := B.B.n + (dx,0); + N[1] := R.A.r + (dx,0); + centerto.A(R.A.l, R.A.r, dx, res); + resistor.D(A, normal, 90, "R_4", "$3R$"); + resistor.B(N[1]+(dx/2,0), normal, 0, "R_2", "$2R$"); + resistor.C(N[0]+(dx/2,0), normal, 0, "R_3", "$4R$"); + centreof.B(R.A.r, N[1], cur); + current.A(c.B, phi.B, "I", ""); + centreof.C(R.B.r, R.C.r, cur); + current.B(c.C, phi.C, "I_2", ""); + centreof.D(R.D.l, N[0], cur); + current.C(c.D, phi.D, "I_3", ""); + wire(R.A.r, R.D.r,rlsq); + wire(B.B.n, R.D.l,rlsq); + wire(N[1], R.B.l,rlsq); + wire(N[0], R.C.l,rlsq); + wire(R.B.r, R.C.r,nsq); + draw N[0] withpen pencircle scaled 3pt; + draw N[1] withpen pencircle scaled 3pt; +\end{emp} +\end{empfile} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +$R_2$ and $R_3$ both have $I_2$ passing through them, so from Ohm's +law we know $V_2=I_2 R_2 < V_3=I_2 R_3$, because $R_2 < R_4$. $R_4$ +and the equivalent resistance $R_s=R_2+R_3$ are in parallel, so they +have the same voltage across them. Because $V_4=V_s=V_2+V_4$, the +voltage $V_4$ across $R_4$ is greater than either $V_2$ or $V_3$. +Finally, the equivalent resistance of $R_4$ and $R_s$ in parallel is +given by +$$ + R_p=\p({\frac{1}{R_4}+\frac{1}{R_s}})^{-1} + =\p({\frac{1}{3R}+\frac{1}{6R}})^{-1} + =3R\p({1+\frac{1}{2}})^{-1} + =3R\cdot\frac{2}{3} + =2R \;, +$$ +so $R_p > R_1$. Since both $R_p$ and $R_1$ have $I$ going through +them, and $V_p=V_4=V_s > V_1$. We still need to place $V_1$ relative +to $V_2$ and $V_3$, so we use the formula for voltage across series +resistors +$$ + I=\frac{V_A}{R_A}=\frac{V_B}{R_B} \;. +$$ +$R_p=2R_1$, so $V_1=V_p/2$, and $R_3=2R_2$, so $V_3=2V_2$. +$V_3 + V_2=V_p$, so $V_3=2/3\cdot V_p$ and $V_2=V_p/3$. +The final ranking is therefore $\ans{V_4=V_p > V_3=2/3\cdot V_p > V_1=V_p/2 > V_2=V_p/3}$. + +\Part{b} +We've done most of the work in \Part{a}. +$$ + \mathcal{E}=V_1 + V_p=\frac{3 V_p}{2} \;, +$$ +so +\begin{align} + V_4 &= V_p =\ans{\frac{2\mathcal{E}}{3}} \\ + V_1 &= \frac{V_p}{2}=\ans{\frac{\mathcal{E}}{3}} \\ + V_2 &= \frac{V_p}{3}=\ans{\frac{2\mathcal{E}}{9}} \\ + V_3 &= \frac{2V_p}{3}=\ans{\frac{4\mathcal{E}}{9}} +\end{align} + +\Part{c} +$I=I_2 + I_3$, and all our currents are positive as we've labled them, +so $I$ is greater than $I_2$ and $I_3$. $R_4=3R < R_s=6R$, so $I_3 > +I_2$. The final ranking is therefore $I > I_3 > I_2$, with $I_2$ +passing through both $R_2$ and $R_3$. + +\Part{d} +To be quantitative about \Part{c}, we can use Ohm's law for each current: +\begin{align} + I &= \frac{V_1}{R_1}=\frac{\mathcal{E}}{3}\cdot\frac{1}{R}=\frac{\mathcal{E}}{3R} \\ + I_3 &= \frac{V_p}{R_4}=\frac{2\mathcal{E}}{3}\cdot\frac{1}{3R}=\frac{2}{3}\cdot\frac{\mathcal{E}}{3R}=\ans{\frac{2I}{3}}\\ + I_2 &= \frac{V_p}{R_s}=\frac{2\mathcal{E}}{3}\cdot\frac{1}{6R}=\frac{1}{3}\cdot\frac{\mathcal{R}}{3R}=\ans{\frac{I}{3}}\;. +\end{align} +We see that $I=I_2+I_3$, as it should by Kirchhoff's junction rule. + +\Part{e} +If $R_3$ increases, $R_s$ increases and $R_p$ increases, so $I_2$ and +$I$ decrease. The change in $I_3$ is a balance of increased flow +relative to $I_2$ and decreased overall $I$. We see that less current +through $I$ drops $V_1$, but $V_1+V_4=\mathcal{E}$ which doesn't +change, so $V_4$ increases, so $I_3$ increases. + +\Part{f} +As $R_3 \rightarrow \infty$, $I_2$ is choked off entirely, so $I=I_3$. +So $I$ flows through $R_1$ and $R_4$, and nothing flows through $R_2$ +and $R_3$. +\end{solution} diff --git a/latex/problems/problem21.35.tex b/latex/problems/problem21.35.tex new file mode 100644 index 0000000..12caf89 --- /dev/null +++ b/latex/problems/problem21.35.tex @@ -0,0 +1,165 @@ +\begin{problem*}{21.35} +Determine the current in each branch of the circuit shown in Figure P21.35. +\begin{center} +\begin{empfile}[1] +\begin{emp}(0,0) + pair N[]; + numeric dx, dy; + dx := 3cm; + dy := 1cm; + % draw the middle branch components + battery.B(origin, 90, "\epsilon_2", "4 V"); + resistor.B(B.B.p, normal, 90, "", "1.00\ohm"); + resistor.b(R.B.r, normal, 90, "", "5.00\ohm"); + N[0] := B.B.n + (0,0); + N[1] := R.b.r + (0,0); + % draw the left branch components + centerto.A(N[0], N[1], -dx, res); + resistor.A(A, normal, 90, "R_1", "8.00\ohm"); + N[2] := (xpart R.A.l, ypart N[0]); + N[3] := (xpart N[2], ypart N[1]); + % draw the right branch components + N[6] := N[1] + (dx,0); + resistor.c(N[6], normal, -180, "", "3.00\ohm"); + N[4] := (xpart N[6], ypart N[0]); + N[5] := (N[4]+N[6])/2; % midpoint of right branch + centreof.c(N[4], N[5], bat); + battery.C(c.c, phi.c, "\epsilon_3", "12 V"); + centreof.C(N[5], N[6], res); + resistor.C(c.C, normal, phi.C, "", "1.00\ohm"); + % draw the currents + centreof.i(N[3], R.A.r, cur); + current.A(c.i, phi.i, "I_1", ""); + centreof.I(R.B.l, R.b.r, cur); + current.B(c.I, phi.I, "I_2", ""); + centreof.j(B.C.p, R.C.l, cur); + current.C(c.j, phi.j, "I_3", ""); + % draw the wires + wire(N[0], R.A.l, rlsq); + wire(N[1], R.A.r, rlsq); + wire(N[1], R.c.r, nsq); + wire(N[6], R.C.r, nsq); + wire(R.C.l, B.C.p, nsq); + wire(N[0], B.C.n, rlsq); + % draw the nodes + draw N[0] withpen pencircle scaled 3pt; + draw N[1] withpen pencircle scaled 3pt; +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 21.35 + +\begin{solution} +Let $I_1$ be the current on the left branch (going down), $I_2$ be the +current on the middle branch (going up), and $I_3$ be the current on +the right branch (going up). From Kirchhoff's junction rule, we know. +\begin{equation} + I_1 = I_2 + I_3 +\end{equation} + +Let $\epsilon_2 = 4.00\U{V}$ be the voltage across the middle battery, +and $\epsilon_3 = 12.0\U{V}$ be the voltage across the right battery. + +Using our knowledge of series resistors, we find +\begin{align} + R_1 &= 8.00\Omega \\ + R_2 &= 5.00\Omega + 1.00\Omega = 6.00\Omega \\ + R_3 &= 3.00\Omega + 1.00\Omega = 4.00\Omega +\end{align} +We can use Ohm's law to find the voltage drops across them in the +direction of their current. + +Now using Kirchhoff's loop rule on the left-center and left-right loops +respectively we have +\begin{align} + 0 &= \epsilon_2 - I_2 R_2 - I_1 R_1 \\ + 0 &= \epsilon_3 - I_3 R_3 - I_1 R_1 +\end{align} + +So we have our three equations relating our unknown currents. If +you're comfortable with linear algebra (take a look at my linear +algebra intro if you want to get comefortable), you can express these +as a matrix +\begin{equation} + \begin{pmatrix} + 0 \\ + \epsilon_2 \\ + \epsilon_3 + \end{pmatrix} + = + \begin{pmatrix} + -1 & 1 & 1 \\ + R_1 & R_2 & 0 \\ + R_1 & 0 & R_3 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{equation} + +Inverting the 3x3 matrix, we get +\begin{equation} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} + = \begin{pmatrix} + -1 & 1 & 1 \\ + 8.00\Omega & 6.00\Omega & 0 \\ + 8.00\Omega & 0 & 4.00\Omega + \end{pmatrix}^{-1} + \begin{pmatrix} + 0 \\ + \epsilon_2 \\ + \epsilon_3 + \end{pmatrix} + = + \begin{pmatrix} + -0.2308 & 0.0385 & 0.0577 \\ + 0.3077 & 0.1154 & -0.0769 \\ + 0.4615 & -0.0769 & 0.1346 + \end{pmatrix} + \begin{pmatrix} + 0 \\ + 4.00 \\ + 12.0 + \end{pmatrix} + = + \ans{ + \begin{pmatrix} + 0.8462 \\ + -0.4615 \\ + 1.3077 + \end{pmatrix} + \U{A} + } +\end{equation} +Where $I_2 < 0$ indicates that current actually flows in the opposite +direction to what we expected. + +If you're not comfortable with linear algebra, you can solve the +equations using your method of choice. If no methods make sense to +you, come talk to me or get someone else to teach you one. If you +want to double check your algebra, I work the solution out +symbolically in my linear algebra introduction in traditional equation +format as well as in matrix format. + +The benifit of the linear algebra is that most graphing calculators +can do the matrix inversion for you. On the TI-89, you can do +\begin{align} +&[-1,1,1;8,6,0;8,0,4] \rightarrow A \\ +&[0;4;12] \rightarrow I \\ +&A^{-1}*I \\ +& \qquad\qquad + \begin{pmatrix} + 0.8462 \\ + -0.4615 \\ + 1.3077 + \end{pmatrix} +\end{align} +(I don't have a TI-89, so if this is wrong, let me know\ldots, see my +linear algebra introduction for TI-83+ rules). +\end{solution} diff --git a/latex/problems/problem21.38.tex b/latex/problems/problem21.38.tex new file mode 100644 index 0000000..cec9b8b --- /dev/null +++ b/latex/problems/problem21.38.tex @@ -0,0 +1,263 @@ +\begin{problem*}{21.38} +The following equations describe an electric circuit: +\begin{align} + -(220\Omega)I_1 + 5.80\U{V} - (370\Omega)I_2 &= 0 \label{eq.2_L1}\\ + (370\Omega)I_2 + (150\Omega)I_3 - 3.10\U{V} &= 0 \label{eq.2_L2}\\ + I_1 + I_3 - I_2 &= 0 \label{eq.2_J} +\end{align} +\Part{a} Draw a diagram of the circuit. +\Part{b} Calculate the unknowns and identify the physical meaning of +each unknown. +\end{problem*} % problem 21.38 + +\begin{solution} +\Part{a} +Looking at the three equations, we see that the only unknowns are +$I_1$, $I_2$, and $I_3$. That looks like a circuit with current in +three branches. +\begin{center} +\begin{empfile}[2a] +\begin{emp}(0,0) + numeric dx, dy; + dx := .5cm; + dy := .5cm; + % draw dashed branches (with single CCW spiral) + draw (0,0)--(0,dy)--(-dx,dy)--(-dx,0)--(dx,0)--(dx,dy)--(0,dy) dashed evenly; + % draw the nodes + draw (0,0) withpen pencircle scaled 3pt; + draw (0,dy) withpen pencircle scaled 3pt; +\end{emp} +\end{empfile} +\end{center} +By looking at Eqn.~\ref{eq.2_J} and identifying it with Kirchhoff's +junction rule on junction $A$, we can get current directions. +\begin{center} +\begin{empfile}[2b] +\begin{emp}(0,0) + numeric dx, dy; + dx := 1cm; + dy := 1cm; + % draw the nodes + draw (0,dy) withpen pencircle scaled 3pt; + draw (0,0) withpen pencircle scaled 3pt; + puttext.bot("$A$", (0,0)); + % draw dashed branches (with single CCW spiral) + draw (0,0)--(0,dy)--(-dx,dy)--(-dx,0)--(dx,0)--(dx,dy)--(0,dy) dashed evenly; + centreof.i((-dx,0), (0,0), cur); + current.A(c.i, phi.i, "", "$I_1$"); + centreof.I((0,0), (0,dy), cur); + current.B(c.I, phi.I, "I_2", ""); + centreof.j((dx,0), (0,0), cur); + current.C(c.j, phi.j, "I_3", ""); +\end{emp} +\end{empfile} +\end{center} +Eqn.~\ref{eq.2_L1} looks like a Kirchhoff's loop rule involving only +branches 1 and 2. The first term $-(220\Omega)I_1$ looks like a +$V=IR$ resistor drop in the direction of the current on branch 1, so +let's add a $220\Omega$ resistor to branch 1. Because the voltage +drops in our loop equation, we must be moving in the direction of the +current. Continuing through the Eqn.~\ref{eq.2_L1}, we see a constant +voltage increase, which looks like we crossed a battery from the +negative to positive side, so we'll add that onto branch 1 too. +Finally, there is a $-(370\Omega)I_2$ drop which looks like crossing a +resistor in the direction of the current on branch 2, so let's add a +$370\Omega$ resistor to branch 2. +\begin{center} +\begin{empfile}[2c] +\begin{emp}(0,0) + pair N[]; + numeric dx; + dx := 2cm; + % draw the left branch components + resistor.A(origin, normal, -90, "", "220\ohm"); + battery.A(R.A.r, -90, "", "5.80 V"); + N[0] := B.A.p; + N[1] := R.A.l; + N[2] := N[0] + (dx,0); + centreof.i(N[0], N[2], cur); + current.A(c.i, phi.i, "", "$I_1$"); + % draw the middle branch components + N[3] := (xpart N[2], ypart N[1]); + centerto.B(R.A.r, R.A.l, dx, res); + resistor.B(B, normal, 90, "", "370\ohm"); + centreof.I(N[2], R.B.l, cur); + current.B(c.I, phi.I, "I_2", ""); + % draw the right branch components + N[4] := N[2]+(dx,0); + N[5] := (xpart N[4], ypart N[3]); + centreof.j(N[4], N[2], cur); + current.C(c.j, phi.j, "I_3", ""); + % draw the wires + wire(N[0], N[2], rlsq); + wire(N[2], R.B.l, rlsq); + wire(N[1], R.B.r, nsq); + draw N[2]--N[4]--N[5]--N[3] dashed evenly; + % draw the nodes + draw N[2] withpen pencircle scaled 3pt; + draw N[3] withpen pencircle scaled 3pt; + % draw the loop direction + imesh((N[0]+N[3])/2, ypart (N[1]-N[0])/4, dx/4, ccw, 90, ""); +\end{emp} +\end{empfile} +\end{center} + +Eqn.~\ref{eq.2_L2} looks like another Kirchhoff's loop rule, this time +involving only branches 2 and 3. The first term $-(370\Omega)I_2$ +looks like a resistor gain \emph{against} the direction of the current +on branch 2. We already have a $370\Omega$ resistor to branch 2, so +this term just tells us we're moving upstream against $I_2$. +Continuing through the Eqn.~\ref{eq.2_L2}, we see another voltage +\emph{gain} $(150\Omega)I_3$. If we're moving upstream on $I_2$, +we'll also be moving upstream on $I_3$, so this voltage gain must be a +$150\Omega$ resistor on branch 3. The last term is a constant votage +\emph{drop}, which looks like we crossed a battery from the positive +to negative side, so we'll add that onto branch 3 too. +\begin{center} +\begin{empfile}[2d] +\begin{emp}(0,0) + pair N[]; + numeric dx; + dx := 2cm; + % draw the left branch components + resistor.A(origin, normal, -90, "", "220\ohm"); + battery.A(R.A.r, -90, "", "5.80 V"); + N[0] := B.A.p; + N[1] := R.A.l; + N[2] := N[0] + (dx,0); + centreof.i(N[0], N[2], cur); + current.A(c.i, phi.i, "", "$I_1$"); + % draw the middle branch components + N[3] := (xpart N[2], ypart N[1]); + centerto.B(R.A.r, R.A.l, dx, res); + resistor.B(B, normal, 90, "", "370\ohm"); + centreof.I(N[2], R.B.l, cur); + current.B(c.I, phi.I, "I_2", ""); + % draw the right branch components + N[4] := N[2]+(dx,0); + N[5] := (xpart N[4], ypart N[3]); + centreof.j(N[4], N[2], cur); + current.C(c.j, phi.j, "I_3", ""); + resistor.C(N[4], normal, 90, "", "150\ohm"); + battery.C(N[5], -90, "\mbox{3.10 V}", ""); + % draw the wires + wire(N[0], N[2], rlsq); + wire(N[2], R.B.l, rlsq); + wire(N[1], R.B.r, nsq); + wire(N[2], N[4], nsq); + wire(N[3], N[5], nsq); + % draw the nodes + draw N[2] withpen pencircle scaled 3pt; + draw N[3] withpen pencircle scaled 3pt; + % draw the loop direction + imesh((N[2]+N[5])/2, ypart (N[3]-N[2])/4, dx/4, ccw, 90, ""); +\end{emp} +\end{empfile} +\end{center} + +\Part{b} +Solve using your method of choice. With linear algebra: +\begin{equation} + \begin{pmatrix} + 5.80\U{V} \\ + 3.10\U{V} \\ + 0 + \end{pmatrix} + = + \begin{pmatrix} + 220\Omega & 370\Omega & 0 \\ + 0 & 370\Omega & 150\Omega \\ + 1 & -1 & 1 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{equation} + +Inverting the 3x3 matrix, + we get +\begin{equation} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} + = + \begin{pmatrix} + 0.0031 & -0.0022 & 0.3267 \\ + 0.0009 & 0.0013 & -0.1942 \\ + -0.0022 & 0.0035 & 0.4791 + \end{pmatrix}^{-1} + \begin{pmatrix} + 5.80\U{V} \\ + 3.10\U{V} \\ + 0 + \end{pmatrix} + = + \ans{ + \begin{pmatrix} + 11.0 \\ + 9.13 \\ + -1.87 + \end{pmatrix} + \U{mA} + } +\end{equation} + +With regular algebra, we can save ourselves a bit of work by noticing +that this problem is the same as the one we just did (35)! Well, now +we have a battery on the first branch and none on the second, and the +batteries are facing down\ldots If we flip the picture over and swap +the first and second branches\ldots +\begin{center} +\begin{empfile}[2e] +\begin{emp}(0,0) + pair N[]; + numeric dx; + dx := 2cm; + % draw the left branch components (now middle) + resistor.A(origin, normal, 90, "", "220\ohm"); + battery.A(R.A.r, 90, "", "5.80 V"); + N[0] := B.A.p; + N[1] := R.A.l; + N[2] := N[0] - (dx,0); + centreof.i(N[1], N[0], cur); + current.A(c.i, phi.i, "", "$I_1$"); + % draw the middle branch components (now left) + N[3] := (xpart N[2], ypart N[1]); % (now bottom) + centerto.B(R.A.r, R.A.l, -dx, res); + resistor.B(B, normal, 90, "", "370\ohm"); + centreof.I(N[2], R.B.r, cur); + current.B(c.I, phi.I, "I_2", ""); + % draw the right branch components + N[4] := N[0]+(dx,0); + N[5] := (xpart N[4], ypart N[3]); % (now bottom) + centreof.j(N[5], N[4], cur); + current.C(c.j, phi.j, "I_3", ""); + resistor.C(N[4], normal, -90, "\mbox{150\ohm}", ""); + battery.C(N[5], 90, "", "3.10 V"); + % draw the wires + wire(N[0], N[2], rlsq); + wire(N[1], R.B.l, rlsq); + wire(N[2], R.B.r, nsq); + wire(N[2], N[4], nsq); + wire(N[3], N[5], nsq); + % draw the nodes + draw N[0] withpen pencircle scaled 3pt; + draw N[1] withpen pencircle scaled 3pt; +\end{emp} +\end{empfile} +\end{center} +Alright, now it looks like the figure in Problem 35, except that the +things labeled $X_1$ and $X_2$ are reversed. We can take our analytic +solution to 35 (see the linear algebra notes) and exchange $1 +\leftrightarrow 2$ giving + +\begin{align} + \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_1}{R_1}}{\frac{R_2}{R_3}+\frac{R_2}{R_1}+1} &= I_2 = \ans{9.13\U{mA}} \\ + \frac{\epsilon_1}{R_1} - \frac{1}{R_1}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_1}{R_1}}{\frac{1}{R_3}+\frac{1}{R_1}+\frac{1}{R_2}} &= I_1 = \ans{11.0\U{mA}} \\ + \frac{\epsilon_3}{R_3} - \frac{1}{R_3}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_1}{R_1}}{\frac{1}{R_3}+\frac{1}{R_1}+\frac{1}{R_2}} &= I_3 = \ans{-1.87\U{mA}} +\end{align} +\end{solution} diff --git a/latex/problems/problem21.40.tex b/latex/problems/problem21.40.tex new file mode 100644 index 0000000..4a94c5f --- /dev/null +++ b/latex/problems/problem21.40.tex @@ -0,0 +1,84 @@ +\begin{problem*}{21.40} +A dead battery is charged by connecting it to the live battery of +another car with jumper cables (Fig.~P21.40). Determine the current +in the starter and in the dead battery. +\end{problem*} % problem 21.40 + +\begin{solution} +Let $V_L = 12\U{V}$ and $R_L = 0.01\Omega$ be the parameters of the +live battery, $V_D = 10\U{V}$ and $R_D = 1.00\Omega$ be the parameters +of the dead battery, and $R_S = 0.06\Omega$ be the resistance of the +starter. Let $I_L$ be the current going upward in the left branch, +$I_D$ be the current going upward in the middle branch, and $I_S$ be +the current going downward in the right branch. + +Applying Kirchhoff's junction rule to the top node, we have +\begin{equation} + I_L + I_D - I_S = 0 +\end{equation} + +Applying Kirchhoff's loop rule to the outer and right loops +respectively, we have +\begin{align} + V_L - I_L R_L - I_S R_S &= 0 \\ + V_D - I_D R_D - I_S R_S &= 0 +\end{align} + +Solving these using linear algebra (or your method of choice) +\begin{align} + \begin{pmatrix} + 0 \\ + V_L \\ + V_D + \end{pmatrix} + &= + \begin{pmatrix} + 1 & 1 & -1 \\ + R_L & 0 & R_S \\ + 0 & R_D & R_S + \end{pmatrix} + \begin{pmatrix} + I_L \\ + I_D \\ + I_S + \end{pmatrix} + = + \begin{pmatrix} + 1 & 1 & -1 \\ + 0.01 & 0 & 0.06 \\ + 0 & 1.00 & 0.06 + \end{pmatrix} + \begin{pmatrix} + I_L \\ + I_D \\ + I_S + \end{pmatrix} \\ + \begin{pmatrix} + I_L \\ + I_D \\ + I_S + \end{pmatrix} + &= + \begin{pmatrix} + 0.850 & 15.0 & -0.850 \\ + 0.0085 & -0.850 & 0.992 \\ + -0.142 & 14.2 & 0.142 + \end{pmatrix} + \begin{pmatrix} + 0 \\ + 12 \\ + 10 + \end{pmatrix} + = + \ans{ + \begin{pmatrix} + 172 \\ + -0.283 \\ + 171 + \end{pmatrix} + \U{A} + } +\end{align} +Where $I_2 < 0$ indicates that the current in the middle branch +actually flows downward, recharging the dead battery. +\end{solution} diff --git a/latex/problems/problem21.42.tex b/latex/problems/problem21.42.tex new file mode 100644 index 0000000..c9e0340 --- /dev/null +++ b/latex/problems/problem21.42.tex @@ -0,0 +1,114 @@ +\begin{problem*}{21.42} +A $C = 2.00\U{$\mu$F}$ capacitor with an intial charge of +$Q=5.10\U{$\mu$C}$ is discharged through an $R=1.30\Omega$ resistor. +\Part{a} Calculate the current in the resistor $t_a = 9.00\U{$\mu$s}$ +after the resistor is connected across the terminals of the capacitor. +\Part{b} What charge remains on the capacitor after $t_b = 8.00\U{$\mu$s}$? +\Part{c} What is the maximum current in the resistor? +\end{problem*} % problem 21.42 + +\begin{solution} +\Part{a} +The current through the entire circuit follows +\begin{equation} + I = \frac{Q}{RC}e^{-t/RC} \label{eq.3_I} +\end{equation} +So +\begin{equation} + I(t_a) = \frac{5.10\E{-6}\U{C}}{1.30\Omega\cdot2.00\E{-6}\U{F}} e^{\frac{-9.00\E{-6}\U{s}}{1.30\Omega\cdot2.00\E{-6}\U{F}}} + = \ans{61.6\U{mA}} +\end{equation} + +\Part{b} +The charge on the capacitor follows +\begin{equation} + q = Qe^{-t/RC} \label{eq.3_q} +\end{equation} +So +\begin{equation} + I(t_a) = 5.10\E{-6}\U{C} e^{\frac{-8.00\E{-6}\U{s}}{1.30\Omega\cdot2.00\E{-6}\U{F}}} + = \ans{235\U{nC}} +\end{equation} + +\Part{c} +Plugging $t=0$ into our equation from \Part{a} we have +\begin{equation} + I_{max} = \frac{Q}{RC} = \ans{1.96\U{A}} +\end{equation} + +For those who are interested the derivation of Eqns.~\ref{eq.3_I} and +\ref{eq.3_q} is pretty straightforward. Consider the circuit +\begin{center} +\begin{empfile}[3] +\begin{emp}(0,0) + pair N[]; + numeric Lres; + % calculate the length of a resistor + centreof.R((1,0), (-1,0), res); + Lres := 2*(xpart c.R); + % define the nodes + N[0] := (0,0); + N[1] := (Lres,0); + N[2] := (Lres,Lres); + N[3] := (0,Lres); + resistor.R(N[1], normal, 90, "R", ""); + centreof.I(N[3], N[2], cur); + current.I(c.I, phi.I, "I", ""); + centreof.C(N[0], N[3], cap); + capacitor.C(c.C, normal, phi.C, "C", ""); + wire(N[0], N[1], nsq); + wire(N[1], R.R.l, nsq); + wire(R.R.r, N[2], nsq); + wire(N[2], N[3], nsq); + wire(N[3], C.C.r, nsq); + wire(C.C.l, N[0], nsq); + labeloffset := 4pt; + puttext.urt("$q$", (0, Lres/2)); + puttext.lrt("$-q$", (0, Lres/2)); +\end{emp} +\end{empfile} +\end{center} +The current $I = -dq/dt$. + +(Note: In the book it gives $I \equiv dQ/dt$ [Eqn.~21.2], but that +$dQ$ is the charge passing through a given cross section. Our $q$ is +the charge on the top capacitor plate. As charge leaves the top +capacitor plate ($dq/dt < 0$), it passes through a point in the wire +in the direction we've specified for $I$ ($I > 0$), so $I = -dq/dt$. +This is the point that tripped me up in Wednesday's recitation, right +after I had warned about equating symbols without thinking about what +they ment :p) + +Using Kirchhoff's loop rule and the definition of capacitance $Q = CV$ +and resistance $V = IR$, we have +\begin{align} + \Delta V &= +\frac{q}{C} - IR = 0 \\ + \frac{q}{C} &= IR = -\frac{dq}{dt}R \\ + \frac{-dt}{RC} &= \frac{dq}{q} \;. +\end{align} +Integrating both sides we have +\begin{equation} + \int \frac{-dt}{RC} = \frac{-1}{RC} \int dt = \frac{-1}{RC} (t+A) + = \int \frac{dq}{q} = \ln(q) \;, +\end{equation} +where $A$ is some constant of integration because we were taking +indefinite integrals. We want a function for $q(t)$, so we take $e$ +to the power of both sides +\begin{equation} + e^{-(t+A)/RC} = e^{-t/RC - A/RC} = A'e^{-t/RC} + = e^{\ln(q)} = q \;, +\end{equation} +where $A' \equiv e^{-A/RC}$ is just another way of thinking about our +arbitrary integration constant $A$. Looking at our initial condition, +$q(t=0) = Q$, the initial charge on the capacitor, and comparing with +our equation we have +\begin{equation} + q(t=0) = A' e^{-0/RC} = A' = Q \;, +\end{equation} +so we can replace $A'$ with $Q$ to get Eqn.~\ref{eq.3_q}. + +Eqn.~\ref{eq.3_I} follows from Eqn.~\ref{eq.3_q} using our $I = -dq/dt$: +\begin{equation} + I = -\frac{d}{dt}\p({Q e^{-t/RC}}) = -Q \frac{d}{dt}e^{-t/RC}) = -Q \frac{-1}{RC} e^{-t/RC} = \frac{Q}{RC} e^{-t/RC} +\end{equation} +\end{solution} diff --git a/latex/problems/problem21.45.tex b/latex/problems/problem21.45.tex new file mode 100644 index 0000000..449baf8 --- /dev/null +++ b/latex/problems/problem21.45.tex @@ -0,0 +1,112 @@ +\begin{problem*}{21.45} +The circuit in Figure P21.45 has been connected for a long time. +\Part{a} What is the voltage $V_c$ across the capacitor? +\Part{b} If the battery is disconnected, how long does it take the +capacitor to discharge to $V_c'=1/10\cdot V$? +\begin{center} +\begin{empfile}[4] +\begin{emp}(0,0) + pair N[]; + numeric dx, Lres; + dx := 3cm; + % calculate the length of a resistor + centreof.R((1,0), (-1,0), res); + Lres := 2*(xpart c.R); + % draw the bridge + N[0] := origin; + N[3] := N[0] - (0,Lres*sqrt(2)); + N[1] := N[0] - ((1,1)*Lres/sqrt(2)); + N[2] := N[0] + ((1,-1)*Lres/sqrt(2)); + resistor.A(N[1], normal, 45, "R_1=1.00\ohm", ""); + resistor.B(N[3], normal, 135, "R_2=4.00\ohm", ""); + resistor.C(N[3], normal, 45, "", "$R_3=2.00\ohm$"); + resistor.D(N[2], normal, 135, "", "$R_4=8.00\ohm$"); + centreof.C(N[1], N[2], cap); + capacitor.C(c.C, normal, phi.C, "C", ""); + %centerto.B(N[3], N[1], -dx, bat); % I don't know why this way doesn't work, but it appears to not fully define B + %battery.B(B, 90, "\mathcal{E}", "10.0 V"); + centreof.B(N[3]-(dx,0), N[0]-(dx,0), bat); + battery.B(c.B, phi.B, "V", "10.0 V"); + % draw the wires + wire(N[1], C.C.l, nsq); + wire(N[2], C.C.r, nsq); + wireU(N[0], B.B.p, 3pt, udsq); + wireU(N[3], B.B.n, -3pt, udsq); + % draw the nodes + draw N[0] withpen pencircle scaled 3pt; + draw N[1] withpen pencircle scaled 3pt; + draw N[2] withpen pencircle scaled 3pt; + draw N[3] withpen pencircle scaled 3pt; +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 21.45 + +\begin{solution} +Labeling the resistors counterclockwise from the upper left we have + $R_1 = 1.00\Omega$, + $R_2 = 4.00\Omega$, + $R_3 = 2.00\Omega$, and + $R_4 = 8.00\Omega$. +Let $V = 10.0\U{V}$ be the voltage on the battery + and $C = 1.00\U{$\mu$F}$ be the capacitance of the capacitor. + +\Part{a} +Because the system has been running for a long time, the system must +be close to equilibrium. Therefore, the current through the capacitor +must be zero (otherwise the voltage across the capacitor would be +changing, and you wouldn't be at equilibrium). The resistor bridge +then reduces to two parallel circuits, and we can apply Ohm's law to +determine $V_c$ + +Starting with the left side of the bridge (calling the current $I_L$), +\begin{align} + V &= I_L (R_1 + R_2) & + I_L &= \frac{V}{R_1 + R_2} +\end{align} +And on the right calling the current $I_R$ +\begin{equation} + I_R = \frac{V}{R_3 + R_4} +\end{equation} + +So using Ohm's law to compute the voltage across the capacitor, we +call the voltage on the bottom wire $0$ and have the voltage $V_L$ on the left at +\begin{equation} + V_L = I_L R_2 = \frac{V R_2}{R_1+R_2} = 8\U{V} +\end{equation} +And the voltage $V_R$ to the right of the capacitor is +\begin{equation} + V_R = I_R R_3 = \frac{V R_3}{R_3+R_4} = 2\U{V} +\end{equation} +So the voltage across the capacitor is +\begin{equation} + V_c = V_L - V_R = \ans{6\U{V}} +\end{equation} + +\Part{b} +Once we remove the battery, we see that the capacitor discharges +through two paths in parallel, $R_1 \rightarrow R_4$ and $R_2 +\rightarrow R_3$. +The eqivalent resistances of these two parallel branches (top and +bottom) are +\begin{align} + R_T &= R_1 + R_4 & + R_B &= R_2 + R_3 +\end{align} +So the total equivalent resistance is +\begin{equation} + R = \left(\frac{1}{R_T}+\frac{1}{R_B}\right)^{-1} + = 3.60\Omega +\end{equation} + +The voltage of a discharging capacitor depends on time according to +\begin{equation} + V_c' = V_c e^{-t/RC} +\end{equation} +So using $V_c' = V_c/10$ we have +\begin{align} + 10 &= \frac{V_c}{V_c'} = \frac{V_c}{V_c e^{-t/RC}} = e^{t/RC} \\ + \ln(10) &= \frac{t}{RC} \\ + t &= RC\ln(10) = \ans{8.29\U{$\mu$s}} +\end{align} +\ens{solution} diff --git a/latex/problems/problem21.46.tex b/latex/problems/problem21.46.tex new file mode 100644 index 0000000..4401ce5 --- /dev/null +++ b/latex/problems/problem21.46.tex @@ -0,0 +1,39 @@ +\begin{problem*}{21.46} +A $C=10.0\U{$\mu$F}$ capacitor is sharged by a $\epsilon= 10.0\U{V}$ +battery through a resistance $R$. The capacitor reaches a potential +difference of $V_C(t_f)=4.00\U{V}$ at the instant $t_f = 3.00\U{s}$ +after the charging begins. Find $R$. +\end{problem*} % problem 12.46 + +\begin{solution} +Applying Kirchhoff's loop rule, +\begin{align} + \epsilon - V_C - V_R &= \epsilon - \frac{q}{C} - IR = 0 \\ + R \frac{dq}{dt} &= \epsilon - \frac{q}{C} \\ + \frac{dq}{dt} &= \frac{C\epsilon - q}{RC} \\ + \frac{dq}{C\epsilon - q} &= \frac{dt}{RC} \\ + \int \frac{dq}{C\epsilon -q} &= \int \frac{dt}{RC} \\ + -\ln (C\epsilon-q) &= t/RC + A \\ + C\epsilon - q &= Be^{-t/RC} \\ + q &= C\epsilon - Be^{-t/RC} \\ + q &= C\epsilon(1-e^{-t/RC}) +\end{align} +Where $A$ is a constant of integration, + $B = e^{-A}$ is another way of writing that constant, + $C\epsilon = Q$ (because as $t \rightarrow \infty$, $q \rightarrow Q$), + and $B = Q = C\epsilon$ (because at $t=0$, $q=0$). + +Note: The book derives this on pages 709-710 if you want more details. + +Now applying this to our particular problem, +\begin{align} + V_C(t_f) &= \frac{q(t_f)}{C} + = \frac{C\epsilon}{C}(1-e^{-t_f/RC}) \\ + \frac{V_C(t_f)}{\epsilon} &= 1 - e^{-t_f/RC} \\ + e^{-t_f/RC} &= 1 - \frac{V_C(t_f)}{\epsilon} \\ + \frac{-t_f}{RC} &= \ln\left(1 - \frac{V_C(t_f)}{\epsilon}\right) \\ + R &= \frac{-t_f}{C \ln(1 - V_C(t_f)/\epsilon)} + = \frac{-3.00\U{s}}{10.0\E{-6}\U{F}\cdot\ln(1-4.00\U{V}/10.0\U{V})} + = \ans{587\U{k$\Omega$}} +\end{align} +\end{solution} diff --git a/latex/problems/problem21.53.tex b/latex/problems/problem21.53.tex new file mode 100644 index 0000000..6ab5f88 --- /dev/null +++ b/latex/problems/problem21.53.tex @@ -0,0 +1,29 @@ +\begin{problem*}{21.53} +An electric heater is rated at $P_H = 1500\U{W}$, + a toaster at $P_T = 750\U{W}$, and + an electric grill at $P_G = 1000\U{W}$. +The three appliances are connected to a common $V = 120\U{V}$ + household circuit. +\Part{a} How much current does each draw? +\Part{b} Is a circuit with a $V_{max} = 25.0\U{A}$ circuit breaker + sufficient in this situation? + Explain your answer. +\end{problem*} % problem 21.53 + +\begin{solution} +\Part{a} +Using $P=IV$ we have +\begin{align} + I_H &= \frac{P_H}{V} = \ans{12.5\U{A}} & + I_T &= \frac{P_T}{V} = \ans{6.25\U{A}} & + I_G &= \frac{P_G}{V} = \ans{8.33\U{A}} +\end{align} + +\Part{b} +If all the appliances are running together, the circuit draws +\begin{equation} + I = I_H + I_T + I_G = 27.1\U{A} +\end{equation} +So you will be fine with a $25\U{A}$ breaker unless you plan to run +all three at the same time. +\end{solution} diff --git a/latex/problems/problem21.55.tex b/latex/problems/problem21.55.tex new file mode 100644 index 0000000..bc5545a --- /dev/null +++ b/latex/problems/problem21.55.tex @@ -0,0 +1,17 @@ +\begin{problem*}{21.55} +Four $V = 1.50\U{V}$ AA batteries in series are used to power a +transistor radio. If the batteries can move a charge of $\Delta Q = +240\U{C}$, how long will they last if the radio has a resistance of $R += 200\Omega$? +\end{problem*} % problem 21.55 + +\begin{solution} +Using Kirchhoff's loop rule, +\begin{align} + V + V + V + V - IR &= 0 \\ + I &= \frac{4V}{R} = \frac{\Delta q}{\Delta t} \\ + \Delta t &= \frac{\Delta q R}{4V} + = \frac{240\U{C}\cdot200\Omega}{4\cdot1.50\U{V}} + = 8000\U{s} = \ans{2.22\U{hours}} +\end{align} +\end{solution} diff --git a/latex/problems/problem21.58.tex b/latex/problems/problem21.58.tex new file mode 100644 index 0000000..c33745d --- /dev/null +++ b/latex/problems/problem21.58.tex @@ -0,0 +1,78 @@ +\begin{problem*}{21.58 +A battery with emf $\epsilon$ is used to charge a capacitor $C$ +through a resistor $R$ as shown in Figure 21.25. +Show that half the energy supplied by the battery appears as internal +energy in the resistor and that half is stored in the capacitor. +\begin{center} +\begin{empfile}[6] +\begin{emp}(0,0) + pair N[]; + numeric dx, dy, Lres; + % calculate the length of a resistor + centreof.R((1,0), (-1,0), res); + Lres := 2*(xpart c.R); + dy := 1cm; + dx := 2.5*Lres; + N[0] := origin; + N[1] := N[0] + (dx/2,0); + N[2] := N[0] + (dx,0); + N[3] := N[0] + (0,dy); + N[4] := N[3] + (dx/2,0); + N[5] := N[3] + (dx,0); + % draw the top branch + centreof.R(N[3], N[4], res); + resistor.R(c.R, normal, phi.R, "R", ""); + centreof.C(N[4], N[5], cap); + capacitor.C(c.C, normal, phi.C, "C", ""); + % draw the bottom branch + centreof.S(N[0], N[1], swt); + switch.S(c.S, NO, phi.S, "", "Switch"); + centreof.B(N[2], N[1], bat); + battery.B(c.B,phi.B, "\epsilon", ""); + % draw the wires + wire(N[3], R.R.l, nsq); + wire(R.R.r, C.C.l, nsq); + wire(C.C.r, N[5], nsq); + wire(N[0], N[3], nsq); + wire(N[2], N[5], nsq); + wire(N[0], st.S.l, nsq); + wire(st.S.r, B.B.p, nsq); + wire(B.B.n, N[2], nsq); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} + +\begin{solution} +%The capacitor charges to a final charge of $Q=C\epsilon$, at which +%point no more current flows through the system. +The total current through the system is given by +\begin{equation} + I = I_o e^{-t/RC} = \frac{\epsilon}{R}e^{-t/RC} +\end{equation} +Which allows us to compute the energy put out by the battery. +Power is the time derivative of energy so +\begin{equation} + E_b = \int_0^\infty P\cdot dt + = \int_0^\infty I\epsilon\cdot dt + = \frac{\epsilon^2}{R}\int_0^\infty e^{-t/RC}\cdot dt + = \frac{\epsilon^2}{R}\left. -RC e^{-t/RC}\right|_0^\infty + = \frac{\epsilon^2}{R}\left(0 - (-RC e^0)\right) + = C\epsilon^2 +\end{align} +Similarly for the energy absorbed by the resistor +\begin{align} + E_r = \int_0^\infty P\cdot dt + = \int_0^\infty I^2 R\epsilon\cdot dt + = \frac{\epsilon^2}{R}\int_0^\infty e^{-2t/RC}\cdot dt + = \frac{\epsilon^2}{R}\left. \frac{-RC}{2} e^{-2t/RC}\right|_0^\infty + = \frac{E_b}{2} = \frac{1}{2}C\epsilon^2 +\end{align} +And we already know the energy stored in a capacitor with a voltage +$\epsilon$ is +\begin{equation} + E_c = \frac{1}{2}C\epsilon^2 = \frac{E_b}{2} +\end{equation} +So the battery energy splits evenly between the capacitor and the +resistor, and we're done. +\end{solution} diff --git a/latex/problems/problem22.01.tex b/latex/problems/problem22.01.tex new file mode 100644 index 0000000..f48d948 --- /dev/null +++ b/latex/problems/problem22.01.tex @@ -0,0 +1,88 @@ +\begin{problem*}{22.1} +Determine the initial direction of the deflection of charged particles +as they enter magnetic fields as shown in Figure P22.1. +\begin{center} +\Part{a} +\empaddtoprelude{% + pair p; + numeric Dx, Dy, ddx, dx, dy, nx, ny; + Dx := 2cm; + Dy := 2cm; + ddx := .4cm; + nx := 4; + ny := 4; + dx := Dx/nx; + dy := Dy/ny; +} +\begin{empfile}[1a] +\begin{emp}(0,0) + for i=1 upto nx : + for j=1 upto ny : + p := draw_Bfletch((dx*(i-.5),dy*(j-.5))); + endfor; + endfor; + draw origin--(Dx,0)--(Dx,Dy)--(0,Dy)--cycle dashed evenly; + p := draw_velocity((-2ddx,Dy/2),(-ddx,Dy/2),2ddx); + draw_pcharge((-ddx,Dy/2), 6pt); +\end{emp} +\end{empfile} +\hspace{1cm} +\Part{b} +\begin{empfile}[1b] +\begin{emp}(0,0) + for i=1 upto nx : + p := draw_Bfield((dx*(i-.5),0), (dx*(i-.5),.5dy), (ny-1)*dy); + endfor; + draw origin--(Dx,0)--(Dx,Dy)--(0,Dy)--cycle dashed evenly; + p := draw_velocity((Dx+2ddx,Dy/2),(Dx+ddx,Dy/2),2ddx); + draw_ncharge((Dx+ddx,Dy/2), 6pt); +\end{emp} +\end{empfile} +\hspace{1cm} +\Part{c} +\begin{empfile}[1c] +\begin{emp}(0,0) + for j=1 upto ny : + p := draw_Bfield((0,dy*(j-.5)), (.5dx,dy*(j-.5)), (nx-1)*dx); + endfor; + draw origin--(Dx,0)--(Dx,Dy)--(0,Dy)--cycle dashed evenly; + p := draw_velocity((Dx+2ddx,Dy/2),(Dx+ddx,Dy/2),2ddx); + draw_pcharge((Dx+ddx,Dy/2), 6pt); +\end{emp} +\end{empfile} +\hspace{1cm} +\Part{d} +\begin{empfile}[1d] +\begin{emp}(0,0) + for j=1 upto ny : + p := draw_Bfield((dx/2,dy*(j-.5))-dir(45), (dx/2,dy*(j-.5)), (ny-j)*sqrt(2)*dy); + endfor; + for i=1 upto nx : + p := draw_Bfield((dx*(i-.5),dy/2)-dir(45), (dx*(i-.5),dy/2), (nx-i)*sqrt(2)*dx); + endfor; + draw origin--(Dx,0)--(Dx,Dy)--(0,Dy)--cycle dashed evenly; + draw (Dx/2,-ddx)--(Dx/2,Dy) dashed withdots scaled .3; + label.urt(btex $45^\circ$ etex, draw_langle((Dx/2,Dy),(Dx/2,Dy/2),(Dx,Dy), dx/2)); + p := draw_velocity((Dx/2,-2ddx),(Dx/2,-ddx),2ddx); + draw_pcharge((Dx/2,0-ddx), 6pt); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} + +\begin{solution} +Using our right hand rule for the cross product, and + $\vect{F}_B = q \vect{v} \times \vect{B}$ + +\Part{a} +Force is up. + +\Part{b} +Force is out of the page. + +\Part{c} +No force. + +\Part{d} +Force is into the page. +\end{solution} diff --git a/latex/problems/problem22.03.tex b/latex/problems/problem22.03.tex new file mode 100644 index 0000000..417da3e --- /dev/null +++ b/latex/problems/problem22.03.tex @@ -0,0 +1,24 @@ +\begin{problem*}{22.3} +A proton travels with a speed of $v = 3.00\E{6}\U{m/s}$ at an angle of +$\theta = 37.0\dg$ with the direction of a magnetic field of $B = +0.300\U{T}$ in the $+y$ direction. What are \Part{a} the magnitude of +the magnetic force on the proton and \Part{b} its acceleration? +\end{problem*} % problem 22.3 + +\begin{solution} +We'll pick the \ihat\ direction so that \vect{v} has a positive $x$-component. + +\Part{a} +\begin{equation} + F_B = q v B \sin\theta = (1.60\E{-19}\U{C})\cdot(3.00\E{6}\U{m/s})\cdot(0.300\U{T})\cdot\sin37.0\dg + = \ans{8.67\E{-14}\U{N}} +\end{equation} +and the direction of the force is in the \khat\ direction. + +\Part{b} +Using $\vect{F} = m \vect{a}$ we have +\begin{equation} + \vect{a} = \frac{\vect{F}}{m} = \frac{8.67\E{-14}\U{N}}{1.67\E{-27}\U{kg}} + = \ans{5.19\E{13}\U{m/s$^2$}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem22.04.tex b/latex/problems/problem22.04.tex new file mode 100644 index 0000000..e361fee --- /dev/null +++ b/latex/problems/problem22.04.tex @@ -0,0 +1,25 @@ +\begin{problem*}{22.4} +An electron is accelerated through $V = 2400\U{V}$ from rest and then +enters a uniform $B = 1.70\U{T}$ magnetic field. What are \Part{a} +the maximum and \Part{b} the minimum values of the magnetic force this +charge can experience? +\end{problem*} % problem 22.4 + +\begin{solution} +First we compute the electron's velocity $v$ upon entering the field. +Conserving energy +\begin{align} + qV &= \frac{1}{2} m v^2 \\ + v &= \sqrt{\frac{2qV}{m}} + = 29.0\U{Mm/s} +\end{align} + +The magnetic force is given by $\vect{F} = q\vect{v}\times\vect{B}$, +so it is maximized when \vect{B} is perpendicular to \vect{v}, at +which point +\begin{equation} + F = qvB = \ans{7.90\U{pN}} +\end{equation} +The force is minimized then \vect{B} is parallel (or anti-parallel) to +\vect{v}, at which point $\ans{F = 0}$. +\end{solution} diff --git a/latex/problems/problem22.06.tex b/latex/problems/problem22.06.tex new file mode 100644 index 0000000..14432df --- /dev/null +++ b/latex/problems/problem22.06.tex @@ -0,0 +1,23 @@ +\begin{problem*}{22.6} +A proton moves with a velocity of $\vect{v} = (2\ihat -4\jhat ++\khat)\U{m/s}$ in a region in which the magnetic field is $\vect{B} = +(\ihat + 2\jhat -3\khat)\U{T}$. What is the magnitude of the magnetic +force this charge experiences? +\end{problem*} % problem 22.6 + +\begin{solution} +\begin{align} + \vect{F} &= q \vect{v}\times\vect{B} + = q \left| + \begin{matrix} + \ihat & \jhat & \khat \\ + 2 & -4 & 1 \\ + 1 & 2 & -3 + \end{matrix} + \right| + = q [\ihat(12-2) -\jhat(-6-1) +\khat(4-(-4))] + = q (10\ihat+7\jhat+8\khat) \\ + |\vect{F}| &= q \sqrt{10^2+7^2+8^2} + = \ans{2.34\E{-18}\U{N}} +\end{align} +\end{solution} diff --git a/latex/problems/problem22.08.tex b/latex/problems/problem22.08.tex new file mode 100644 index 0000000..2b726a0 --- /dev/null +++ b/latex/problems/problem22.08.tex @@ -0,0 +1,35 @@ +\begin{problem*}{22.8} +An electron moves in a circular path perpendicular to a constant +magnetic field of magnitude $B=1.00\U{mT}$. The angular momentum of +the electron about the center of the circle is $L=4.00\E{-25}\U{Js}$. +Determine + \Part{a} the radius of the circular path and + \Part{b} the speed of the electron. +\end{problem*} % problem 22.8 + +\begin{solution} +Angular momentum is defined as +\begin{equation} + \vect{L} = \vect{r}\times\vect{p} = m\vect{r}\times\vect{v} \;, +\end{equation} +which for circular orbits reduces to +\begin{equation} + L = mrv \;, +\end{equation} +because \vect{r} and \vect{v} are perpendicular. + +We also have +\begin{align} + F_c &= qvB = m\frac{v^2}{r} \\ + qBr &= mv \;, +\end{align} +which combined with the angular momentum formula give two equations +with two unknowns. + +Solving for the unknowns +\begin{align} + L &= qBr^2 \\ + r &= \sqrt{\frac{L}{qB}} = \ans{0.0500\U{m}} \\ + v &= \frac{L}{mr} = \ans{4.79\U{km/s}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem22.10.tex b/latex/problems/problem22.10.tex new file mode 100644 index 0000000..32c9fb3 --- /dev/null +++ b/latex/problems/problem22.10.tex @@ -0,0 +1,27 @@ +\begin{problem*}{22.10} +A velocity selector consists of electric and magnetic fields described +by the expressions $\vect{E} = E\khat$ and $\vect{B} = B\jhat$, with +$B = 15.0\U{mT}$. Find the value of $E$ such that a $K = 750\U{eV}$ +electron moving in the \ihat\ direction is undeflected. +\end{problem*} % problem 22.10 + +\begin{solution} +The force from the magnetic field is in the $-\khat$ direction (right +hand rule), so the sign of $E$ must be negative (to push the electron +in the \khat\ direction). + +The velocity of the electron is given by +\begin{align} + K &= \frac{1}{2} m v^2 \\ + v &= \sqrt{\frac{2K}{m}} + = \sqrt{\frac{2\cdot(750\cdot1.60\E{-19}\U{J})}{9.11\E{-31}\U{kg}}} + = 16.2\U{Mm/s} +\end{align} + +Balancing the magnitudes of the two forces +\begin{align} + F_e = qE &= F_B = qvB \\ + E &= vB = \ans{243\U{kV/m}} +\end{align} +So $\vect{E} = -243\khat\U{kV/m}$. +\end{solution} diff --git a/latex/problems/problem22.12.tex b/latex/problems/problem22.12.tex new file mode 100644 index 0000000..137cb73 --- /dev/null +++ b/latex/problems/problem22.12.tex @@ -0,0 +1,58 @@ +\begin{problem*}{22.12} +A cyclotron designed to accelerate protons has an outer radius of $R = +0.350\U{m}$. The protons are emitted nearly at rest from a source at +the center and are accelerated through $V = 600\U{V}$ each time they +cross the gap between the dees. The dees are between the poles of an +electromagnet where the field is $B = 0.800\U{T}$. +\Part{a} Find the cyclotron frequency $f$. +\Part{b} Find the speed $v_e$ at which the protons exit the cyclotron and +\Part{c} their kinetic energy $K$. +\Part{d} How many revolutions $N$ does a proton make in the cyclotron? +\Part{e} For what time $\Delta t$ interval does one proton accelerate? +\end{problem*} % problem 22.12 + +\begin{solution} +\Part{a} +Protons with velocities $v$ in a constant magntic field will move in +circles of radius $r$ in the plane perpendicular to the magnetic +field. The centerward acceleration is given by +\begin{align} + F_c = m \frac{v^2}{r} &= F_B = qvB \\ + v &= \frac{qrB}{m} +\end{align} +Their velocity can also be related to their period $T=1/f$ by +\begin{equation} + v = \frac{dr}{dt} = \frac{2\pi r}{T} = 2\pi r f +\end{equation} +So +\begin{align} + 2\pi r f &= \frac{qrB}{m} \\ + f &= \frac{qB}{2\pi m} = 12.2\U{MHz} +\end{align} +This is the frequency of revolution for a proton \emph{anywhere} +inside the cyclotron. + +\Part{b} +Using our $v(r)$ equation from \Part{a} when $r = R$, we have +\begin{equation} + v_e = \frac{qRB}{m} = \frac{(1.60\E{-19}\U{C})\cdot(0.350\U{m})\cdot(0.800\U{T})}{1.69\E{-27}\U{kg}} = \ans{26.8\U{Mm/s}} +\end{equation} + +\Part{c} +\begin{equation} + K = \frac{1}{2} m v^2 = \frac{(qrB)^2}{2m} = \ans{6.01\E{-13}\U{J}} +\end{equation} + +\Part{d} +The kinetic energy $K$ is built up from $2N$ passes through $V$ (twice +per revolution). +\begin{equation} + N = \frac{K}{2qV} = \frac{6.01\E{-13}\U{J}}{2\cdot(1.60\E{-19})\cdot(600\U{V})} + = \ans{3130} +\end{equation} + +\Part{e} +\begin{equation} + \Delta t = T N = N/f = \ans{257\U{$\mu$s}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem22.15.tex b/latex/problems/problem22.15.tex new file mode 100644 index 0000000..7c67167 --- /dev/null +++ b/latex/problems/problem22.15.tex @@ -0,0 +1,33 @@ +\begin{problem*}{22.15} +A wire carries a steady current of $A = 2.40\U{A}$. A straight +section of the wire is $l = 0.750\U{m}$ long and lies in the +\ihat\ direction within a uniform magbnetic field, $\vect{B} = +1.60\khat\U{T}$. What is the magnetic force on the section of wire? +\end{problem*} % problem 22.15 + +\begin{solution} +We can find the magnetic force on a wire using $\vect{F} = +q\vect{v}\times\vect{B}$. Consider an infinitesimal bit of wire of +length $d\vect{s}$, a current $I = dq/dt$ means that $dq$ will move +through this bit of wire in time $dt$. So the force on the bit of +wire is +\begin{equation} + \vect{dF} = dq \frac{\vect{ds}}{dt} \times \vect{B} + = \frac{dq}{dt} \vect{ds} \times \vect{B} + = I \vect{ds} \times \vect{B} +\end{equation} +If the wire is straight, we can integrate easily to find the total +force on the whole segment +\begin{equation} + \vect{F} = \int_0^l \vect{dF} = \int_0^l I \vect{ds} \times \vect{B} + = I B \sin\theta \int_0^l ds = I l B \sin\theta + = I \vect{l} \times \vect{B} +\end{equation} + +Plugging in for our specific case we get a force in the $-\jhat$ +direction from the right-hand-rule, with a magnitude of +\begin{equation} + F = IlB = (2.40\U{A})\cdot(0.750\U{m})\cdot(1.60\U{T}) = 2.88\U{N} +\end{equation} +So $\ans{\vect{F} = -2.88\jhat\U{N}}$ +\end{solution} diff --git a/latex/problems/problem22.16.tex b/latex/problems/problem22.16.tex new file mode 100644 index 0000000..776ae27 --- /dev/null +++ b/latex/problems/problem22.16.tex @@ -0,0 +1,25 @@ +\begin{problem*}{22.16} +A wire $l = 2.80\U{m}$ in length carries a current of $I = 5.00\U{A}$ +in a region where a uniform magnetic field has a magnitude of $B = +0.390\U{T}$. Calculate the magnitude of the magnetic force on the +wire assuming that the angle between the magnetic field and the +current is + \Part{a} $\theta_a = 60.0\dg$, + \Part{b} $\theta_b = 90.0\dg$, and + \Part{c} $\theta_c = 120\dg$. +\end{problem*} % problem 22.16 + +\begin{solution} +Using our formula for the force on a wire due to a uniform field we +have +\begin{align} + \vect{F} &= I\vect{l}\times\vect{B} \\ + F &= IlB\sin\theta \;, +\end{align} +so just pluggging in +\begin{align} + F_a &= IlB\sin\theta_a = \ans{4.73\U{N}}\\ + F_b &= IlB\sin\theta_b = \ans{5.46\U{N}}\\ + F_c &= IlB\sin\theta_c = \ans{4.73\U{N}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem22.21.tex b/latex/problems/problem22.21.tex new file mode 100644 index 0000000..bd01b35 --- /dev/null +++ b/latex/problems/problem22.21.tex @@ -0,0 +1,70 @@ +\begin{problem7}{22.21} +A rectangular coil consists of $N=100$ closely wrapped turns and has +dimensions $a = 0.400\U{m}$ and $b = 0.300\U{m}$. The coil is hinged +along the $y$ axis, and its plane makes an angle $\theta = 30.0\dg$ +with the $x$ axis (Fig.~P22.21). What is the magnitude of the torque +exerted on the coil by a uniform magnetic field $B = 0.800\U{T}$ +directed along the x axis whwn the current is $I=1.20\U{A}$ in the +direction shown? What is the expected direction of motion of the +coil? +\begin{center} +\begin{empfile}[7] +\begin{emp}(0,0) + pair p[]; + numeric dirz, dirl, Dx, Dy, Dz, Dlxz, Dly, ddy; + dirz := -160; + dirl := -20; + Dx := 2cm; + Dy := 2cm; + Dz := 1cm; + Dlxz := 1.5cm; + Dly := 1.5cm; + ddy := 5pt; + % draw axes + drawarrow origin--(Dx,0) withcolor black withpen pencircle scaled 0pt; + label.bot(btex x etex, (Dx,0)); + drawarrow origin--(0,Dy) withcolor black withpen pencircle scaled 0pt; + label.lft(btex y etex, (0,Dy)); + drawarrow origin--(Dz*dir(dirz)) withcolor black withpen pencircle scaled 0pt; + label.lft(btex z etex, (Dz*dir(dirz))); + % draw box + p[0] := origin; + p[1] := Dlxz*dir(dirl); + p[2] := p[1]+(0,Dly); + p[3] := p[0]+(0,Dly); + draw p[0]--p[1]--p[2]--p[3]--cycle withcolor (1,.2,0) withpen pencircle scaled 2pt; + label.lft(btex $a = 0.400\mbox{ m}$ etex, draw_length(p[3],p[0],3pt)); + label.bot(btex $b = 0.300\mbox{ m}$ etex, draw_length(p[0],p[1],3pt)); + label.rt(btex $30^\circ$ etex, draw_langle((Dx,0),p[0],p[1],Dlxz/2.2)); + drawarrow (p[3]+0.2(p[2]-p[3])+(0,ddy))--(p[2]-0.2(p[2]-p[3])+(0,ddy)) withcolor (1,.2,1) withpen pencircle scaled 1pt; + label.urt(btex $I = 1.20\mbox{ A}$ etex, (p[2]+p[3])/2+(0,ddy)); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 22.21 + +\begin{solution} +Using our formula for force on a wire segment $\vect{F} = +I\vect{l}\times\vect{B}$, and recalling that torque is defined $\tau = +\vect{r}\times\vect{F}$, we can use the right hand rule to find the +direction of motion. + +The torque from the portion of the coil lying on the $y$ axis is zero, +because the lever arm is zero ($r$ in the torque equation). The +torque from the top portion is also zero, because the force is in the +\jhat\ direction and the coil is not free to rotate in that direction. +Similarly the torque from the bottom portion is zero, because the +force is in the $-\jhat$ direction. All the torque comes from the +force on the outer leg, giving a force in the \khat\ direction. So +\ans{we expect the angle $\theta$ to increase}. + +To find the magnitude of the torque, we simply plug in +\begin{equation} + \tau = \vect{r}\times\vect{F} = bF\cos\theta = b\cos\theta\cdot(NIaB) + = IabB\cos\theta + = \ans{9.98\U{J}} +\end{equation} +Where we multiplied the force from a single wire by $N$ because there +are $N$ wraps, and took $\cos\theta$ to get the perpendicular force +because $\theta$ is the complement of the angle between $r$ and $F$. +\end{solution} diff --git a/latex/problems/problem22.33.tex b/latex/problems/problem22.33.tex new file mode 100644 index 0000000..9999e25 --- /dev/null +++ b/latex/problems/problem22.33.tex @@ -0,0 +1,38 @@ +\begin{problem*}{22.33} +In studies of the possibility of migrating birds using the Earth's +magnetic field for navigation, birds have been fitted with coils as +``caps'' and ``collars'' as shown in Figure P22.33. +\Part{a} If the identical coils have radii of $r=1.20\U{cm}$ and are +$d=2.20\U{cm}$ apart, with $N=50$ turns of wire apiece, what current +should they both carry to produce a magnetic field of +$B=4.50\E{-5}\U{T}$ halfway between them? +\Part{b} If the resistance of each coild is $R=210\Omega$, what +voltage should the battery supplying each coil have? +\Part{c} What power is delivered to each coil? +\end{problem*} % problem 22.33 + +\begin{solution} +\Part{a} +From page 745 we have the magnetic field along the axis of symmetry of +a loop of wire as +\begin{equation} + \vect{B} = \frac{\mu_0 I r^2}{2(x^2 + r^2)^{3/2}}\ihat +\end{equation} +All of our $2N$ loops are equidistant from the center, with $x = d/2$ so +\begin{align} + B &= \frac{\mu_0 I r^2}{2(x^2 + r^2)^{3/2}}\cdot(2N) \\ + I &= \frac{(x^2+r^2)^{3/2} B}{\mu_0 N r^2} + = \ans{21.4\U{mA}} +\end{align} + +\Part{b} +Using Ohm's law +\begin{equation} + V = IR = \ans{4.51\U{V}} +\end{equation} + +\Part{c} +\begin{equation} + P = IV = \ans{96.7\U{mW}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem22.34.tex b/latex/problems/problem22.34.tex new file mode 100644 index 0000000..e106c71 --- /dev/null +++ b/latex/problems/problem22.34.tex @@ -0,0 +1,60 @@ +\begin{problem*}{22.34} +Two long, parallel conductors, separated by $r = 10.0\U{cm}$, carry +current in the same direction. The first wire carries current $I_1 = +5.00\U{A}$, and the second carries $I_2 = 8.00\U{A}$. +\Part{a} What is the magnitude of the magnetic field $B_1$ created by +$I_1$ at the location of $I_2$? +\Part{b} What is the force per unit length exerted by $I_1$ on $I_2$? +\Part{c} What is the magnitude of the magnetic field $B_2$ created by +$I_2$ at the location of $I_1$? +\Part{d} What is the force per unit length exerted by $I_2$ on $I_1$? +\end{problem*} % problem 22.34 + +\begin{solution} +\Part{a} +From Ampere's law, the $B$ field generated by a long, thin current is +\begin{equation} + B = \frac{\mu_0 I}{2 \pi r} +\end{equation} +Plugging in $I_1$, we have +\begin{equation} + B_1 = \frac{\mu_0 I_1}{2 \pi r} = \ans{10.0\U{$\mu$T}} \label{eqn.34_b1} +\end{equation} +This $B$ field depends on your distance from $I_1$, but because the +wires are parallel, the $B$ field from $I_1$ is constant along $I_2$ +We can use the right hand rule to determine that $\vect{B}_1$ is +perpendicular to both $I_1$ and $r$. + +\Part{b} +From $F_B = q\vect{v}\times\vect{B}$ we have the force on a current +carrying wire in a uniform magnetic field as +\begin{equation} + F_B = I\vect{l}\times\vect{B} +\end{equation} + +Combining these two equations, we have the force per unit length of +$I_1$ on $I_2$ as +\begin{equation} + F_{B12}/l = I_2 B_1 = \frac{\mu_0 I_1 I_2}{2 \pi r} = \ans{80.0\U{$\mu$N}} \label{eqn.34_fb12} +\end{equation} +where there is no $\sin\theta$ term in the cross product, because +$B_1$ is perpendicular to $I_2$. +By drawing the situation and doing some right hand rules, you can +convince yourself that this force is \emph{attractive}. + +\Part{c} +Because the situation in \Part{c} is identical to \Part{a} with $I_1 +\leftrightarrow I_2$, we simply relabel eqn.\ \ref{eqn.34_b1}. +\begin{equation} + B_2 = \frac{\mu_0 I_2}{2 \pi r} = \ans{16.0\U{$\mu$T}} +\end{equation} + +\Part{d} +Eqn.\ \ref{eqn.34_fb12} is identical under the relabeling, so we have +another attractive force at the same magnitude +\begin{equation} + F_{B21}/l = \ans{80\U{$\mu$N}} +\end{equation} +as we would expect from Newton's third law (for every action there is +an equal and opposite reaction). +\end{solution} diff --git a/latex/problems/problem22.37.tex b/latex/problems/problem22.37.tex new file mode 100644 index 0000000..d61d521 --- /dev/null +++ b/latex/problems/problem22.37.tex @@ -0,0 +1,68 @@ +\begin{problem*}{22.37} +Four long, parallel conductors carry equal currents of $I = +5.00\U{A}$. Figure P22.37 is an end view of the conductors. The +current direction is into the page at points $A$ and $B$ and out of +the page at points $C$ and $D$. Calculate the magnitude and direction +of the magnetic field at point $P$, located at the center of the +square of edge length $a=0.200\U{m}$. +\begin{center} +\begin{empfile}[2] +\begin{emp}(0cm, 0cm) + pair p; + numeric r; + r := 1cm; + dotlabel.bot(btex $P$ etex, origin); + draw (-r,r)--(-r,-r)--(r,-r)--(r,r)--cycle dashed evenly; + p := draw_Ifletch((-r, r)); + p := draw_Ifletch((-r,-r)); + p := draw_Itip( ( r,-r)); + p := draw_Itip( ( r, r)); + label.bot(btex 0.200\mbox{ m} etex, (0,-r)); + label.rt(btex 0.200\mbox{ m} etex, (r,0)); + labeloffset := 4pt; + label.lft(btex $A$ etex, (-r, r)); + label.lft(btex $B$ etex, (-r,-r)); + label.rt( btex $D$ etex, ( r,-r)); + label.rt( btex $C$ etex, ( r, r)); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 22.37 + +\begin{solution} +First, let us pick a coordinate system by choosing unit vectors. +Let \ihat\ be down and to the left, + \jhat\ be down and to the right, and + \khat\ be straight down. + +Using the right-hand rule, we determine the direction of the magnetic +field at $P$ generated by each wire to be +\begin{align} + \widehat{B_A} &= \ihat \\ + \widehat{B_B} &= \jhat \\ + \widehat{B_C} &= \ihat \\ + \widehat{B_D} &= \jhat +\end{align} + +The magnitude of each $B$ is given by +\begin{equation} + B = \frac{\mu_0 I}{2 \pi r} +\end{equation} +And since the currents have the same magnitude, and each corner is +equidistant from the square center, each magnetic field contribution +will have the same magnitude. The distance $r$ is given by +\begin{equation} + r = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} + = \frac{a}{\sqrt{2}} +\end{equation} + +We still have to add our vector fields, which gives +\begin{equation} + \vect{B}_P = \vect{B}_A + \vect{B}_B + \vect{B}_C + \vect{B}_D + = 2B(\ihat + \jhat) + = 2\frac{\mu_0 I}{2 \pi r}\cdot\sqrt{2}\khat + = \frac{\sqrt{2}\mu_0 I}{\pi r}\khat + = \frac{2\mu_0 I}{\pi a}\khat + = \ans{20\U{$\mu$T}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem22.39.tex b/latex/problems/problem22.39.tex new file mode 100644 index 0000000..3d1f297 --- /dev/null +++ b/latex/problems/problem22.39.tex @@ -0,0 +1,31 @@ +\begin{problem*}{22.39} +A packed bundle of $N = 100$ long, straight, insulated wires forms a +cylinder of radius $R = 0.500\U{cm}$. +\Part{a} If each wire carries $I = 2.00\U{A}$, what are the magnitude +and direction of the magnetic force per unit length acting on a wire +located $r = 0.200\U{cm}$ from the center of the bundle? +\Part{b} Would a wire on the outer edge of the bundle experience a +force greater or smaller than the value calculated in \Part{a}. +\end{problem*} % problem 22.39 + +\begin{solution} +\Part{a} +Drawing an Amperian loop at a radius $r$, the number of enclosed wires is +\begin{equation} + N_{enc} = N \frac{\pi r^2}{\pi R^2} = N \frac{r^2}{R^2} \;. +\end{equation} +So the magnetic field is +\begin{equation} + B = \frac{\mu_0 I N r^2/R^2}{2 \pi r} = \frac{\mu_0 I N r}{2 \pi R^2} + = 3.17\E{-3}\U{T} \;, +\end{equation} +and the force per unit length is +\begin{equation} + \frac{F}{l} = IB = \frac{\mu_0 I^2 N r}{2 \pi R^2} = \ans{6.34\U{mN/m}} \;. +\end{equation} + +\Part{b} +From the previous equation, you can see the force increases linearly +with $r$ (so long as we stay inside the wire), so the force will be +\ans{greater} at the outer edge. +\end{solution} diff --git a/latex/problems/problem22.43.tex b/latex/problems/problem22.43.tex new file mode 100644 index 0000000..7434069 --- /dev/null +++ b/latex/problems/problem22.43.tex @@ -0,0 +1,22 @@ +\begin{problem*}{22.43} +Niobium metal becomes superconducting when cooled below 9K. Its +superconductivity is destroyed when the surface $B$ field exceeds +$B_{max} = 0.100\U{T}$. Determine the maximum current in a +$d=2.00\U{mm}$ diameter niobium wire can carry and remain +superconducting, in the absence of any external $B$ field. +\end{problem*} % problem 22.43 + +\begin{solution} +For long, cylindrical wires, the magnetic field a distance $r$ from +the center of the wire is +\begin{equation} + B = \frac{\mu_0 I}{2 \pi r} +\end{equation} +As long as you are outside the wire. + +Therefore, the magnetic field at the surface is maximized when +\begin{align} + B_{max} &= \frac{\mu_0 I_{max}}{2 \pi r} \\ + I_{max} &= (2 \pi r B_{max})/\mu_0 = \ans{500\U{A}} +\end{align} +\end{solution} diff --git a/latex/problems/problem22.48.tex b/latex/problems/problem22.48.tex new file mode 100644 index 0000000..f0ec3e1 --- /dev/null +++ b/latex/problems/problem22.48.tex @@ -0,0 +1,47 @@ +\begin{problem*}{22.48} +In Bohr's 1913 model of the hydrogen atom, the electron is in a +circular orbit of radius $r = 5.29\E{-11}\U{m}$, and its speed is $v = +2.19\E{6}\U{m/s}$. +\Part{a} What is the magnitude of the magnetic moment \vect{\mu} due to the +electron's motion? +\Part{b} If the electron moves in a horizontal circle, +counterclockwise as seen from above, what is the direction of \vect{\mu}? +\end{problem*} % problem 22.48 + +\begin{solution} +\Part{a} +The magnetic moment is defined on page 742 as +\begin{equation} + \vect{\mu} = I \vect{A} +\end{equation} + +The area swept out by our electron is just +\begin{equation} + A =\pi r^2 +\end{equation} + +The current is the amount of charge circling the nucleus in a unit +time. +Because +\begin{equation} + \Delta x = v \Delta t +\end{equation} +The time $\tau$ taken for an entire circuit is +\begin{equation} + \tau = \frac{\Delta x}{v} = \frac{2 \pi r}{v} +\end{equation} +The current is then given by +\begin{equation} + I = \frac{\Delta q}{\Delta t} = \frac{q_e v}{2 \pi r} +\end{equation} + +Plugging $I$ and $A$ into our moment equation +\begin{equation} + \mu = \frac{q_e v}{2 \pi r}\cdot \pi r^2 = (q_e v r)/2 = + \ans{9.27\E{-24}\U{A m$^2$}} +\end{equation} + +The direction of the current is opposite the direction of the electron +(because the electron has negative charge), so the direction of +\vect{\mu} is down. +\end{solution} diff --git a/latex/problems/problem22.56.tex b/latex/problems/problem22.56.tex new file mode 100644 index 0000000..55da363 --- /dev/null +++ b/latex/problems/problem22.56.tex @@ -0,0 +1,21 @@ +\begin{problem*}{22.56} +An $m = 0.200\U{kg}$ metal rod carrying a current of $I = 10.0\U{A}$ +glides on two horizontal rails $l = 0.500\U{m}$ apart. +What vertical magnetic field is required to keep the rod moving at a +contant speed if the coefficient of kinetic friction between the rod +and rails is $\mu_k = 0.100$? +\end{problem*} % problem 22.56 + +\begin{solution} +Balancing the forces on the rod in the \jhat\ direction +\begin{align} + N - mg &= 0 \\ + N &= mg \;, +\end{align} +and in the \ihat\ direction +\begin{align} + IlB - \mu_k N &= 0 \\ + B &= \frac{\mu_k m g}{Il} = \ans{ 39.2\U{mT}} \;, +\end{align} +where we are ignoring the magnetic field generated by the current in the rails. +\end{solution} diff --git a/latex/problems/problem22.57.tex b/latex/problems/problem22.57.tex new file mode 100644 index 0000000..31ec4fe --- /dev/null +++ b/latex/problems/problem22.57.tex @@ -0,0 +1,40 @@ +\begin{problem*}{22.57} +A positive charge $q = 3.20\E{-19}\U{C}$ moves with a velocity +$\vect{v} = (2\ihat + 3\jhat - \khat)\U{m/s}$ through a region where +both a uniform magnetic field and a uniform electric field exist. +\Part{a} Calculate the total force $F$ on the moving charge (in +unit-vector notation), taking $\vect{B} = (2\ihat + 4\jhat ++\khat)\U{T}$ and $\vect{E} = (4\ihat - \jhat - 2\khat)\U{V/m}$. +\Part{b} What angle $\theta$ does the force vector \vect{F} make with \ihat? +\end{problem*} % problem 22.57 + +\begin{solution} +\Part{a} +From Chapter 19, +\begin{equation} + \vect{F}_E = q \vect{E} = q(4\ihat - \jhat - 2\khat)\U{N/C} +\end{equation} + +From this chapter +\begin{equation} + \vect{F}_B = q\vect{v}\times\vect{B} + = q \begin{vmatrix} \ihat & \jhat & \khat \\ + 2 & 3 & -1 \\ + 2 & 4 & 1 \end{vmatrix} + = q [(3+4)\ihat - (2+2)\jhat + (8-6)\khat] + = q (7\ihat - 4\jhat + 2\khat)\U{N/C} +\end{equation} + +So the total force is given by +\begin{equation} + \vect{F} = \vect{F}_E + \vect{F}_B + = q [(4+7)\ihat + (-1-4)\jhat + (-2+2)\khat]\U{N/C} + = q (11\ihat - 5\jhat)\U{N/C} + = \ans{(35.2\ihat - 16.0\jhat)\E{-19}\U{N}} +\end{equation} + +\Part{b} +\begin{equation} + \theta = \arctan\left(\frac{-5}{11}\right) = \ans{-24.4\dg} +\end{equation} +\end{solution} diff --git a/latex/problems/problem22.58.tex b/latex/problems/problem22.58.tex new file mode 100644 index 0000000..54d2599 --- /dev/null +++ b/latex/problems/problem22.58.tex @@ -0,0 +1,78 @@ +\begin{problem*}{22.58} +Protons having a kinetic energy of $K=5.00\U{MeV}$ are moving in the +\ihat\ direction and enter a magnetic field $B = 0.050\khat\U{T}$ +directed out of the plane of the page and extending from $x=0$ to +$x=1.00\U{m}$ as shown in Figure P22.58. +\Part{a} Calculate the $y$ component of the protons' momentum as they +leave the magnetic field. +\Part{b} Find the angle $\alpha$ between the initial velocity vector of +the proton beam, and the velocity vector after the beam emerges from +the field. +Ignore relativistic effects and note that $1\U{eV} = 1.60\E{-19}\U{J}$. +\begin{center} +\begin{empfile}[6] +\begin{emp}(0,0) + pair p; + numeric Dx, Dy, ddx, dx, dy, nx, ny; + Dx := 3cm; + Dy := 2cm; + ddx := .4cm; + nx := 6; + ny := 4; + dx := Dx/(nx); + dy := Dy/(ny); + for i=1 upto nx : + for j=1 upto ny : + p := draw_Btip((dx*(i-.5),dy*(j-.5))); + endfor; + endfor; + draw origin--(Dx,0)--(Dx,Dy)--(0,Dy)--cycle dashed evenly; + drawarrow (.9ddx,Dy/2){right} .. (Dx+ddx,Dy/4) withcolor red; + p := draw_velocity((-2ddx,Dy/2),(-ddx,Dy/2),2ddx); + draw_pcharge((-ddx,Dy/2), 6pt); + draw_ijhats((-Dy/2, Dy/4), 0, Dy/3); +\end{emp} +\end{empfile} +\end{center} +\end{problem*} % problem 22.58 + +\begin{solution} +\Part{b} +As in our cyclotron problem (Recitation 7, Problem 12), we know +\begin{align} + F_c &= m\frac{v^2}{r} = qvB \\ + mv &= qrB +\end{align} +And +\begin{align} + K &= \frac{1}{2}mv^2 \\ + v &= \sqrt{\frac{2K}{m}} = 30.9\U{Mm/s} +\end{align} +So the radius of the circular arc our protons make in the constant +magnetic field region is +\begin{equation} + r = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2K}{m}} + = \frac{1}{qB}\sqrt{2Km} + = \ans{6.47\U{m}} +\end{equation} +Drawing out the center of the circle the beam would make and doing a +bit of geometry, we see that +\begin{equation} + \alpha = \arcsin\left(\frac{\Delta x}{r}\right) + = \ans{8.90\dg} +\end{equation} + +\Part{a} +Because the \emph{speed} of the particles doesn't change because of a +magnetic field's perpendicular force, we can find the protons' speed +in the $y$ direction on exiting by +\begin{equation} + v_y = v \sin(\alpha) +\end{equation} +So the $y$ momentum is +\begin{equation} + p_y = m v_y = m v \sin(\alpha) = m v \frac{\Delta x}{r} + = \frac{ m v \Delta x }{(mv)/(qB)} = qB\Delta x + = \ans{8.00\E{-21}\U{kg m/s}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem23.01.tex b/latex/problems/problem23.01.tex new file mode 100644 index 0000000..3280163 --- /dev/null +++ b/latex/problems/problem23.01.tex @@ -0,0 +1,24 @@ +\begin{problem*}{23.1} +A flat loop of wire consisting of a single turn of cross-sectional +area $A=8.00\U{cm$^2$}$ is perpendicular to a magnetic field that +increases uniformly in magnitude from $B_i = 0.500\U{T}$ to $B_f = +2.50\U{T}$ in $1.00\U{s}$. What is the resulting induced current if +the loop has a resistance of $R = 2.00\Omega$. +\end{problem*} % problem 23.1 + +\begin{solution} +By Faraday's law +\begin{equation} + \varepsilon = - \frac{d\Phi_B}{dt} + = - \frac{(2.0\U{T})\cdot(8.00\E{-4}\U{m$^2$})}{1.00\U{s}} + = - 1.6\U{mV} \;. +\end{equation} + +By Ohm's law +\begin{align} + \varepsilon &= V = IR \\ + I &= \frac{\varepsilon}{R} = \frac{-1.6\U{mV}}{2.00\Omega} + = \ans{-0.80\U{mA}} \;. +\end{align} +\end{solution} + diff --git a/latex/problems/problem23.02.tex b/latex/problems/problem23.02.tex new file mode 100644 index 0000000..df069ac --- /dev/null +++ b/latex/problems/problem23.02.tex @@ -0,0 +1,23 @@ +\begin{problem*}{23.2} +An $N = 25$ turn circular coil of wire has diameter $d = 1.00\U{m}$. +It is placed with it's axis along the direction of the Earth's +magnetic field of $B = 50.0\U{$\mu$T}$, and then in $t = 0.200\U{s}$ +it is flipped 180\dg. An average emf of what magnitude is generated +in the coil? +\end{problem*} % problem 23.2 + +\begin{solution} +The flux before the flip is +\begin{equation} + \Phi_{Bi} = AB = N \pi r^2 B \;, +\end{equation} +and the flux after the flip is +\begin{equation} + \Phi_{Bf} = -AB = -N \pi r^2 B \;. +\end{equation} + +From Ampere's law +\begin{equation} + \varepsilon = - \frac{d\Phi_B}{dt} = 2 N \pi r^2 B/dt = \ans{9.82\U{mV}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem23.06.tex b/latex/problems/problem23.06.tex new file mode 100644 index 0000000..6e6618a --- /dev/null +++ b/latex/problems/problem23.06.tex @@ -0,0 +1,32 @@ +\begin{problem*}{23.6} +A coil of $N=15$ turns and radius $R=10.0\U{cm}$ surrounds a long +solenoid of radius $r=2.00\U{cm}$ and $n=1.00\E{3}\U{turns/m}$ +(Fig.~P23.6). The current in the solenoid changes as +$I=(5.00\U{A})\sin(120t)$. Find the induced emf in the $15$ turn coil +as a function of time. +\end{problem*} % problem 23.6 + +\begin{solution} +Because the solenoid is long, we can pretend it is infinite, so all +the magnetic field is contained inside the solenoid, and there is no +magnetic field outside (see page 751). + +The field inside the solenoid is given by +\begin{equation} + B = \mu_0 n I \;, +\end{equation} +so the flux through the large coil is +\begin{equation} + \Phi_B = \int BdA = N \pi r^2 B = N \pi r^2 \mu_0 n I \;. +\end{equation} + +The induced emf is then +\begin{equation} + \varepsilon = -\frac{d\Phi_B}{dt} = - \pi\mu_0 n N r^2 \frac{dI}{dt} + = -\pi\mu_0 n N r^2 (5.00\U{A}\cdot120\U{Hz})\cos(120t) + = \ans{-14.2\cos(t \cdot 120\U{rad/s})\U{mV}} \;, +\end{equation} +where we are assuming that the units on 120 are rad/s, otherwise we'd +have to convert them to rad/s to make the units work out on the +coefficient. +\end{solution} diff --git a/latex/problems/problem23.07.tex b/latex/problems/problem23.07.tex new file mode 100644 index 0000000..57fa317 --- /dev/null +++ b/latex/problems/problem23.07.tex @@ -0,0 +1,18 @@ +\begin{problem*}{23.7} +An $N=30$ turn circular coil of radius $r = 4.00\U{cm}$ and resistance +$R = 1.00\Omega$ is placed in a magnetic field directed perpendicular +to the plane of the coil. The magnitude of the magnetic field varies +with time according to $B = 0.0100t + 0.0400t^2$, where $t$ is in +seconds and $B$ is in Tesla. Calculate the induced emf in the coil +at $t= 5.00\U{s}$. +\end{problem*} % problem 23.7 + +\begin{solution} +The magnetic flux through the loop is +\begin{align} + \Phi_B &= AB = N \cdot \pi r^2 \cdot B \\ + \varepsilon &= - \frac{d\Phi_B}{dt} = - 30 \cdot \pi r^2 \cdot \frac{dB}{dt} + = -30 \cdot \pi (0.0400\U{m})^2 \cdot (0.100 + 0.800t)\U{T/s} + = \ans{61.8\U{mV}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem23.10.tex b/latex/problems/problem23.10.tex new file mode 100644 index 0000000..2ce4c19 --- /dev/null +++ b/latex/problems/problem23.10.tex @@ -0,0 +1,37 @@ +\begin{problem*}{23.10} +A piece of insulated wire is shaped into a figure eight as shown in +Figure P23.10. The radius of the upper circle is $r_s = 5.00\U{cm}$ +and that of the lower circle is $r_b = 9.00\U{cm}$. The wire has a +uniform resistance per unit length of $\lambda = 3.00\U{$\Omega$/m}$. +A uniform magnetic field is applied perpendicular to the plane of the +two circles, in the direction shown. The magnetic field is increasing +at a constant rate of $dB/dt = 2.00\U{T/s}$. Find the magnitude and +direction of the induced current in the wire. +\end{problem*} % problem 23.10 + +\begin{solution} +Pick a direction for the current to be counterclockwise in the bottom +loop (so clockwise in the top). Thus, the area vector of the top loop +is antiparallel to \vect{B} and that of the bottom loop is parallel to +\vect{B}. The magnetic flux is then +\begin{equation} + \Phi_B = \vect{A}\cdot\vect{B} = (\pi r_s^2 - \pi r_b^2)B \;. +\end{equation} + +Using Ampere's law +\begin{equation} + \varepsilon = - \frac{d\Phi_B}{dt} = \pi(r_b^2 - r_s^2)\frac{dB}{dt} \;. +\end{equation} + +The resistance of the entire figure eight is +\begin{equation} + R = \lambda (2 \pi r_s + 2 \pi r_b) \;. +\end{equation} + +Plugging that into Ohm's law yields +\begin{align} + \varepsilon &= V = I R \\ + I &= \frac{(r_b^2 - r_s^2) \frac{dB}{dt}}{2 \lambda (r_s + r_b)} + = \ans{25.2\U{mA}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem23.12.tex b/latex/problems/problem23.12.tex new file mode 100644 index 0000000..7a3f495 --- /dev/null +++ b/latex/problems/problem23.12.tex @@ -0,0 +1,21 @@ +\begin{problem*}{23.12} +Consider the arrangement shown in Figure P23.12. Assume that $R = +6.00\Omega$, $l = 1.20\U{m}$, and a uniform $B=2.50\U{T}$ magnetic +field is directed into the page. At what speed should the bar be +moved to produce a current of $0.500\U{A}$ in the resistor. +\end{problem*} + +\begin{solution} +This problem is almost identical to the recitation Problem 13. +Copying the induced current formula: +\begin{equation} + I = \frac{\varepsilon}{R} = \frac{-lvB}{R} \;, +\end{equation} +where the $-$ sign indicates the current is counterclockwise (out of the +page), so current flows upward through the bar. + +We can solve this equation for $v$, yeilding +\begin{equation} + v = \frac{IR}{lB} = \ans{1.00\U{m/s}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem23.13.tex b/latex/problems/problem23.13.tex new file mode 100644 index 0000000..8d9bff2 --- /dev/null +++ b/latex/problems/problem23.13.tex @@ -0,0 +1,56 @@ +\begin{problem*}{23.13} +Figure P23.12 shows a top view of a bar that can slide without +friction. The resistor is $R = 6.00\Omega$, and a $B = 2.50\U{T}$ +magnetic field is directed perpendicularly downward, into the paper. +Let $l = 1.20\U{m}$. +\Part{a} Calculate the applied force required to move the bar to the +right at a constant speed $v = 2.00\U{m/s}$. +\Part{b} At what rate is energy delivered to the resistor? +\end{problem*} % problem 23.13 + +\begin{solution} +\Part{a} +Let $x$ be the width of the enclosed loop. The magnetic flux is then +\begin{equation} + \Phi_B = AB = xlB +\end{equation} +So the induced emf is +\begin{equation} + \varepsilon = -\frac{d\Phi_B}{dt} = -lB \frac{dx}{dt} = -lvB +\end{equation} +So the induced current is +\begin{equation} + I = \frac{\varepsilon}{R} = \frac{-lvB}{R} +\end{equation} +The $-$ sign indicates the current is counterclockwise (out of the +page), so current flows upward through the bar, so the magnetic force +on the bar is to the left, so our applied force must be \ans{to the right}. + +The work begin done by the applied force is +\begin{equation} + W = F \cdot dx \;. +\end{equation} +So the power input from the force is +\begin{equation} + P_F = \frac{W}{dt} = F \frac{dx}{dt} = Fv \;. +\end{equation} + +All of this power must be dissipated by the resistor, so the current is +\begin{align} + P &= I^2 R \\ + I &= \sqrt{\frac{P}{R}} = \sqrt{\frac{Fv}{R}} \;. +\end{align} + +We combine both current equations to yield +\begin{align} + \frac{-lvB}{R} &= \sqrt{\frac{Fv}{R}} \\ + (lvB)^2 &= RFv \\ + F &= \frac{v(lB)^2}{R} = \ans{3.00\U{N}} \;. +\end{align} + +\Part{b} +Going back and plugging in $F$, +\begin{equation} + P_F = Fv = \ans{6.00\U{W}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem23.22.tex b/latex/problems/problem23.22.tex new file mode 100644 index 0000000..df18844 --- /dev/null +++ b/latex/problems/problem23.22.tex @@ -0,0 +1,38 @@ +\begin{problem*}{23.22} +A rectangular coil with resistance $R$ has $N$ turns, each of length +$l$ and width $w$ as shown in Figure P23.22. The coil moves in a +uniform magnetic field \vect{B} with constant velocity $v$. What are +the magnitude and direction of the total magnetic force on the coild +as it + \Part{a} enters, + \Part{b} moves within, and + \Part{c} leaves + the magnetic field. +\end{problem*} % problem 23.22 + +\begin{solution} +\Part{a} +As in Problem 13, $d\Phi_B/dt = wBv$, so the induced current is +\begin{align} + I = \frac{\varepsilon}{R} = \frac{-d\Phi_B/dt}{R} = \frac{-wvBN}{R} \;, +\end{align} +where the $-$ sign indicates it is counterclockwise (against the +changing flux direction). The force on the leading wires is +\begin{equation} + \vect{F} = I\vect{l}\times\vect{B} = -I\cdot Nw\cdot B\ihat + = \ans{\frac{-v(wBN)^2}{R}\ihat} \;. +\end{equation} + +\Part{b} +Once the coil is inside the magnetic field, the flux becomes constant, +so there is no induced emf driving a current, and thus \ans{no net + force} on the coil. + +\Part{c} +The situation here is the inverse of that in \Part{a}, so the induced +emf is clockwise, but the current through the portion of loop in the +magnetic field is \emph{still up}, so the force is unchanged. +\begin{equation} + \vect{F} = \ans{\frac{-v(wBN)^2}{R}\ihat} +\end{equation} +\end{solution} diff --git a/latex/problems/problem23.53.tex b/latex/problems/problem23.53.tex new file mode 100644 index 0000000..0bc23c2 --- /dev/null +++ b/latex/problems/problem23.53.tex @@ -0,0 +1,46 @@ +\begin{problem*}{23.53} +A particle with a mass of $m = 2.00\E{-16}\U{kg}$ and a charge of $q = +30.0\U{nC}$ starts from rest, is accelerated by a strong electric +field, and is fired from a small source inside a region of uniform +constant magnetic field $B = 0.600\U{T}$. The velocity of the +particle is perpendicular to the field. The circular orbit of the +particle encloses a magnetic flux of $\Phi_B = 15.0\U{$\mu$Wb}$. +\Part{a} Calculate the speed of the particle. +\Part{b} Calculate the potential difference through which the particle +accelerated inside the source. +\end{problem*} % problem 23.53 + +\begin{solution} +\Part{a} +For particles circling in a uniform, perpendicular magnetic field, +\begin{align} + F_c &= m \frac{v^2}{r} = qvB \\ + mv &= qrB +\end{align} + +Letting $\tau$ be the period, from $\Delta x = v \Delta t$ we have +\begin{equation} + \tau = \frac{2 \pi r}{v} = \frac{2 \pi r m}{qrB} = \frac{2 \pi m}{qB} + = 69.8\U{ns} +\end{equation} +The inverse of our cyclotron frequency from Recitation 7. + +The flux and magnetic field give us radius by +\begin{align} + \Phi_B &= AB = \pi r^2 B \\ + r &= \sqrt{\frac{\Phi_B}{\pi B}} = \ans{2.82\U{mm}} +\end{align} + +So the speed is given by +\begin{equation} + v = \frac{2 \pi r}{\tau} = \frac{qB}{2 \pi m}\sqrt{\frac{\Phi_B}{\pi B}} + = \ans{254\U{km/s}} +\end{equation} + +\Part{b} +Conserving energy +\begin{align} + K &= \frac{1}{2}mv^2 = q\Delta V \\ + \Delta V &= \frac{mv^2}{2q} = \ans{215\U{V}} +\end{align} +\end{solution} diff --git a/latex/problems/problem23.64.tex b/latex/problems/problem23.64.tex new file mode 100644 index 0000000..c87a321 --- /dev/null +++ b/latex/problems/problem23.64.tex @@ -0,0 +1,30 @@ +\begin{problem*}{23.64} +A novel method of storing energy has been proposed. A huge, +underground, superconducting coil, $d = 1.00\U{km}$ in diameter, would +be fabricated. It would carry a maximum current of $I=50.0\U{kA}$ +through each winding of an $N = 150$ turn Nb$_3$Sn solenoid. +\Part{a} If the inductance of this huge coil were $L = 50.0\U{H}$, +what would be the total energy stored? +\Part{b} What would be the compressive force per meter length acting +between two adjacent windings $r = 0.250\U{m}$ apart? +\end{problem*} % problem 23.64 + +\begin{solution} +\Part{a} +\begin{equation} + U_L = \frac{1}{2} L I^2 = \ans{62.5\E{10}\U{J}} +\end{equation} + +\Part{b} +Because the radius of the loop is so much larger than the spacing +between windings, we can ignore the curvature of the wires and treat +them as infinitely long and parallel. +Then the magnetic field of one at the location of it's neighbor is +\begin{equation} + B = \frac{\mu_0 I}{2 \pi r} +\end{equation} +And the force per unit length is +\begin{equation} + F/l = IB = \frac{\mu_0 I^2}{2 \pi r} = \ans{2000\U{N/m}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem24.07.tex b/latex/problems/problem24.07.tex new file mode 100644 index 0000000..4112ccb --- /dev/null +++ b/latex/problems/problem24.07.tex @@ -0,0 +1,52 @@ +\begin{problem*}{24.7} +Figure 24.3 shows a plane electromagnetic sinosoidal wave propogating +in the $x$ direction. Suppose the wavelength is $50.0\U{m}$ and the +electric field vibrates in the $xy$ plane with an amplitude of +$22.0\U{V/m}$. Calculate \Part{a} the frequency of the wave +and \Part{b} the magnitude and direction of \vect{B} when the electric +field has its maximum value in the negative $y$ direction. \Part{c} +Write an expression for the \vect{B} with the correct unit vector, +with numerical values for $B_\text{max}$, $k$, and $\omega$, and with +its magnidude in the form +\begin{equation} + B = B_\text{max}\cos(kx-\omega t) +\end{equation} +\end{problem*} % problem 24.7 + +\begin{solution} +\Part{a} +This is just a units conversion +\begin{equation} + f = \frac{c}{\lambda} = \frac{3.00\E{8}\U{m/s}}{50.0\U{m/cycle}} + = 6.00\E{6}\U{cycles/s} = \ans{6.00\U{MHz}} +\end{equation} + +\Part{b} +The magnitude of $B$ in an electromagnetic plane wave is given by +$B=E/c$. The direction of the wave's motion is given by the Poynting +vector $\vect{S}=\vect{E}\times\vect{B}$. Using the right-hand-rule +for the cross product, we see that when \vect{E} is in the $-\jhat$ +direction and \vect{S} is in the \ihat\ direction, \vect{B} must be in +the $-\khat$ direction. Putting this together +\begin{equation} + \vect{B_0} = \frac{-E_0}{c}\khat = \ans{-73.3\U{nT}\cdot\khat} +\end{equation} + +\Part{c} +Because it is a sinusoidal wave moving in the \ihat\ direction, we know +$B$ must look something like +\begin{equation} + \vect{B} = \vect{B_0} \cos(kx - \omega t + \phi) \;. +\end{equation} +We already found \vect{B_0} in \Part{b}, and we don't have any phase +information, so we can drop $\phi$. That leaves +\begin{align} + k &= \frac{2\pi}{\lambda} = 0.126\U{rad/m} \\ + \omega &= 2\pi f = 3.77\E{7}\U{rad/s} +\end{align} +so +\begin{equation} + \vect{B} = \ans{-73.3\U{nT} \cdot + \cos(0.126\U{rad/m}\cdot x - 3.77\E{7}\U{rad/s}\cdot t) \khat} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem24.08.tex b/latex/problems/problem24.08.tex new file mode 100644 index 0000000..d3d54fa --- /dev/null +++ b/latex/problems/problem24.08.tex @@ -0,0 +1,34 @@ +\begin{problem*}{24.8} +In SI units, the electric field in an electromagnetic wave is described by +\begin{equation} + E_y = 100\sin(1.00\E{7}x - \omega t) +\end{equation} +Find \Part{a} the amplitude of the corresponding magnetic field +oscillations, \Part{b} the wavelength $\lambda$, and \Part{c} the +frequency $f$. +\end{problem*} % problem 24.8 + +\begin{solution} +\Part{a} +The amplitude is the magnitude of the oscillation, which just comes +from the prefactor outside the trig function. In this case, +$A=\ans{100\U{V/m}}$ + +\Part{b} +By comparing with the standard form of sinusoidal waves +\begin{equation} + Y = A \sin(kx - \omega t) \;, +\end{equation} +we see that the wavenumber $k=1.00\E{7}\U{rad/m}$. Converting radians +to cycles and inverting yields +\begin{equation} + \lambda = \frac{2\pi\U{rad/cycle}}{k} = 628\U{nm/cycle} +\end{equation} + +\Part{c} +Once we know the length of a cycle, and how fast the wave is moving, we can find out how many of them occur in a second +\begin{equation} + f = \frac{c}{\lambda} = \frac{3.00\U{m/s}}{628\U{nm/cycle}} + = 477\E{12}\U{cycles/s} = \ans{477\U{THz}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem24.09.tex b/latex/problems/problem24.09.tex new file mode 100644 index 0000000..bd71e29 --- /dev/null +++ b/latex/problems/problem24.09.tex @@ -0,0 +1,43 @@ +\newcommand{\Em}{E_\text{max}} +\newcommand{\Bm}{B_\text{max}} +\newcommand{\ctrig}{\cos(kx-\omega t)} +\newcommand{\strig}{\sin(kx-\omega t)} +\begin{problem*}{24.9} +Verify by substitution that the following equations are solutions to +Equations 24.15 and 24.16 respectively: +\begin{align} + E &= \Em\ctrig \\ + B &= \Bm\ctrig +\end{align} +\begin{align*} + \npderiv{2}{x}{E} &= \epsilon_0\mu_0 \npderiv{2}{t}{E} \tag{24.15} \\ + \npderiv{2}{x}{B} &= \epsilon_0\mu_0 \npderiv{2}{t}{B} \tag{24.16} +\end{align*} +\end{problem*} % problem 24.9 + +\begin{solution} +This is just an excercise in partial derivatives. +\begin{align} + \pderiv{x}{E} &= -\Em\strig\cdot k \\ + \npderiv{2}{x}{E} &= -\Em k\ctrig\cdot k = -k^2 E\\ + \pderiv{t}{E} &= -\Em\strig\cdot (-\omega) \\ + \npderiv{2}{t}{E} &= \Em\omega\ctrig\cdot (-\omega) = -\omega^2 E \\ + \frac{k\U{rad/m}}{\omega\U{rad/s}} &= \frac{1}{c} + = \sqrt{\epsilon_0\mu_0} \label{eq.c_to_e_mu} \\ + \npderiv{2}{x}{E} &= \frac{k^2}{\omega^2} \npderiv{2}{t}{E} + = \epsilon_0\mu_0 \npderiv{2}{t}{E} +\end{align} +which is what we set out to show. Note that we used Equation 24.17 in +Equation \ref{eq.c_to_e_mu}. The situation for $B$ is exactly the +same with the replacement $E\rightarrow B$. +\begin{align} + \pderiv{x}{B} &= -\Bm\strig\cdot k \\ + \npderiv{2}{x}{B} &= -\Bm k\ctrig\cdot k = -k^2 B\\ + \pderiv{t}{B} &= -\Bm\strig\cdot (-\omega) \\ + \npderiv{2}{t}{B} &= \Bm\omega\ctrig\cdot (-\omega) = -\omega^2 B \\ + \frac{k\U{rad/m}}{\omega\U{rad/s}} &= \frac{1}{c} + = \sqrt{\epsilon_0\mu_0} \\ + \npderiv{2}{x}{B} &= \frac{k^2}{\omega^2} \npderiv{2}{t}{B} + = \epsilon_0\mu_0 \npderiv{2}{t}{B} +\end{align} +\end{solution} diff --git a/latex/problems/problem24.18.T.tex b/latex/problems/problem24.18.T.tex new file mode 100644 index 0000000..b181441 --- /dev/null +++ b/latex/problems/problem24.18.T.tex @@ -0,0 +1,25 @@ +\begin{problem} +\Part{a} Caluculate the inductance of an LC circuit that oscillates at +$60\U{Hz}$ when the capacitance is $5.00\U{$\mu$F}$. \Part{b} A +resistor is inserted into the LC loop shown in Figure 24.8. Give a +qualitative description of current oscillation in the new circuit. +\end{problem} % based on P24.18 + +\begin{solution} +\Part{a} +The oscillation frequency of an LC circuit is given in Equation 24.24 +(with the derivation in the preceeding few equations) +\begin{equation} + f_0 = \frac{1}{2\pi\sqrt{LC}} \;. +\end{equation} +So +\begin{align} + LC &= \frac{1}{(2\pi f_0)^2} \\ + L &= \frac{1}{C (2\pi f_0)^2} = \ans{1.41\U{H}} +\end{align} + +\Part{b} +When current passes through the resistor some electrical energy is +converted into heat, so the LRC circuit would act as a damped harmonic +oscillator (see Section 12.6 for more on damped oscillations). +\end{solution} diff --git a/latex/problems/problem24.22.tex b/latex/problems/problem24.22.tex new file mode 100644 index 0000000..16069e6 --- /dev/null +++ b/latex/problems/problem24.22.tex @@ -0,0 +1,28 @@ +\begin{problem*}{24.22} +An AM radio station broadcasts isotropically (equally in all +directions) with an average power of $4.00\U{kW}$. A dipole recieving +antenna $65.0\U{cm}$ long is at a location $4.00\U{miles}$ from the +transmitter. Compute the amplitude of the emf that is induced by this +signal between the ends of the recieving antenna. +\end{problem*} % problem 24.22 + +\begin{solution} +To find the signal intensity at our antenna, we note that the power +broadcast from the station is spread out over a sphere of radius +$R=4.00\U{miles}$. The average intensity is then +\begin{equation} + I = S_\text{avg} = \frac{P}{A} = \frac{P}{4\pi R^2} + = \frac{4.00\E{3}\U{W}}{4\pi(4.00\U{miles}\cdot 1.609\E{3}\U{m/mile})^2} + = 7.68\U{$\mu$W/m$^2$} \;. +\end{equation} +From Equation 24.27, we see +\begin{align} + S_\text{avg} &= \frac{E_\text{max}^2}{2\mu_0 c} \tag{24.27} \\ + E_\text{max} &= \sqrt{2\mu_0 c S_\text{avg}} = 76.1\U{mV/m} +\end{align} +The total voltage difference produced across our length $L=65.0\U{cm}$ +antenna is then +\begin{equation} + \Delta V = LE_\text{max} = \ans{49.4\U{mV}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem24.25.tex b/latex/problems/problem24.25.tex new file mode 100644 index 0000000..dffb1c5 --- /dev/null +++ b/latex/problems/problem24.25.tex @@ -0,0 +1,47 @@ +\begin{problem*}{24.25} +The filament of an incandescent lamp has a $150\U{\Ohm}$ resistance +and carries a direct current of $1.00\U{A}$. The filament is +$8.00\U{cm}$ long and $0.900\U{mm}$ in radius. \Part{a} Calculate +thte Poynting vector at the surface of the filament, associated with +the static electric field producing the current and the curret's +static magnetic field. \Part{b} Find the magnitude of the static +electric and magnetic fields at the surface of the filament. +\end{problem*} % problem 24.25 + +\begin{solution} +\Part{a} +The hot resistor will be radiating heat, and none of the electric or +magnetic fields change with time, so we expect a constant Poynting +vector of magnitude +\begin{equation} + S = \frac{P}{A} = \frac{I^2R}{2\pi r L} + = \ans{332\U{kW/m$^2$}} \;. +\end{equation} +This Poynting vector will always point away from the wire (in the +direction the radiation is going). + +\Part{b} +The electric field is given by Ohm's law. +\begin{align} + V &= I R \\ + E &= \frac{V}{L} = \frac{IR}{L} + = \frac{1.00\U{A}\cdot150\U{\Ohm}}{8.00\E{-2}\U{m}} + = \ans{1875\U{V/m}} \;. +\end{align} +The magnetic field from a long, straight wire is +\begin{equation} + B = \ans{\frac{I}{2\pi r}} \;, +\end{equation} +so the magnetic field at the surface of the wire is +\begin{equation} + B = 177\U{T} +\end{equation} +The electric field is along the wire, and the magnetic field is +perpendicular to the current, so the Poynting vector points directly +out (perpendicular to the wire's surface) and has a magnitude +\begin{equation} + S = EB\sin(90\dg) = EB = \frac{IR}{L}\cdot\frac{I}{2\pi r} + = \frac{I^2R}{2\pi r L} \;, +\end{equation} +which is the same expression we found in \Part{a}. +\end{solution} diff --git a/latex/problems/problem24.39.T.tex b/latex/problems/problem24.39.T.tex new file mode 100644 index 0000000..8c8e6f6 --- /dev/null +++ b/latex/problems/problem24.39.T.tex @@ -0,0 +1,26 @@ +\begin{problem} +You're listening to WKDU (transmitted from Van Rensselaer Hall) on +91.7fm while watching a basketball game $400\U{m}$ away at the DAC. How +many wavelengths are between you and the transmitter? FM channel +names give the carrier frequency in MHz. +% Building from http://en.wikipedia.org/wiki/WKDU +% Lat, Long: +39° 57' 36.00" N, 75° 11' 27.00" W, from +% http://www.fcc.gov/fcc-bin/fmq?list=0&facid=17596 +% +% DAC Lat, Long: 39° 57' 22.99" N, 75° 11' 26.51" W, from +% http://en.wikipedia.org/wiki/Daskalakis_Athletic_Center +% +% Distance calculated following the Vincenty algorithm according to +% http://en.wikipedia.org/wiki/Great-circle_distance#The_geographical_formula +\end{problem} % based on P24.39 + +\begin{solution} +The wavelength of the signal is +\begin{equation} + \lambda = \Delta x = v \Delta t = c T = \frac{c}{f} = 3.27\U{m} +\end{equation} +So the number of wavelengths in $400\U{m}$ is +\begin{equation} + N = \frac{L}{\lambda} = \frac{400}{3.27} = \ans{122} +\end{equation} +\end{solution} diff --git a/latex/problems/problem24.55.T.tex b/latex/problems/problem24.55.T.tex new file mode 100644 index 0000000..6f5eea0 --- /dev/null +++ b/latex/problems/problem24.55.T.tex @@ -0,0 +1,41 @@ +\begin{problem} +Assume that the intensity of solar radiation incident on the cloudtops +of the Earth is $1370\U{W/m$^2$}$. \Part{a} Calculate the total power +radiated by the Sun. \Part{b} Determine the maximum values of the +electric and magnetic fields in the sunlight at their source in the +photosphere (\url{http://en.wikipedia.org/wiki/Sun#Photosphere}) a +distance of $6.96\E{8}\U{m}$ from the center of the sun. +\end{problem} % based on P24.55 + +\begin{solution} +\Part{a} +The intensity at Earth is the total power spread over a sphere of +radius $r_E = 150\E{9}\U{m}$ (the Earth-Sun distance). +\begin{align} + \frac{P}{4\pi r_E^2} &= I_E \\ + P &= 4\pi r_E^2 I_E = 4\pi (150\E{9}\U{m})^2 \cdot 1370\U{W/m$^2$} + = \ans{3.87\E{26}\U{W}} \;. +\end{align} + +\Part{b} +The intensity in the photosphere is +\begin{equation} + I_P = \frac{P}{4\pi r_P^2} = \frac{4\pi r_E^2 I_E}{4\pi r_P^2} + = I_E\p({\frac{r_E}{r_P}})^2 = 63.6\U{MW/m$^2$} \;. +\end{equation} +The intensity of light is related to the peak electric field by +Equation 24.27 +\begin{equation} + I = \avg{S} = \frac{1}{2\mu_0 c} E_\text{max}^2 \;. +\end{equation} +So +\begin{equation} + E_\text{max} = \sqrt{2\mu_0 c I} + = \sqrt{2 \cdot 4\pi\E{-7} \cdot 3\E{8} \cdot 63.6\E{6}} + = \ans{219\U{kV/m}} \;. +\end{equation} +In a light wave, $E_\text{max} = cB_\text{max}$ so +\begin{equation} + B_\text{max} = \frac{E_\text{max}}{c} = \ans{730\U{$\mu$T}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem24.57.T.antenna.jpg b/latex/problems/problem24.57.T.antenna.jpg new file mode 100644 index 0000000..7c3eb4b Binary files /dev/null and b/latex/problems/problem24.57.T.antenna.jpg differ diff --git a/latex/problems/problem24.57.T.tex b/latex/problems/problem24.57.T.tex new file mode 100644 index 0000000..535f3b9 --- /dev/null +++ b/latex/problems/problem24.57.T.tex @@ -0,0 +1,72 @@ +\begin{problem} +Set-top loop antennas are sometimes used to pick up UHF TV broadcast +with a carrier frequency $f$ and peak electric field at the antenna of +$E_\text{max}$. The changing magnetic flux in the antenna loop +produces an emf which matches the broadcast signal. \Part{a} Using +Faraday's law, derive an expression for the amplitude of the emf in a +single-turn circular loop of radius $r$, if $r$ is much less than the +broadcast wavelength. \Part{b} If the TV station is due East of your +house and the electric field oscillates vertically, how would you +orient the antenna for best reception? +\begin{center} +\includegraphics[width=1.25in]{antenna} % http://www.jascoproducts.com/products/pc/viewPrd.asp?idproduct=96&IDCategory=19 +\end{center} +\end{problem} % based on P24.57 + +\begin{solution} +\Part{a} +Faraday's law is given in Equation 24.6 +\begin{equation} + \oint \vect{E}\cdot \dd \vect{s} = -\deriv{t}{\Phi_B} \;. +\end{equation} +The left-hand side of Faraday's law is the induced emf. Working on +the right-hand side and noting that $\Phi_B = \vect{A}\cdot\vect{B} = +AB\cos(\theta)$, we see +\begin{align} + \oint \vect{E}\cdot \dd \vect{s} + &= -A\cos(\theta)\deriv{t}{B} + = -\pi r^2\cos(\theta) \deriv{t}{B} + = -\pi r^2\cos(\theta) \deriv{t}{}\p({B_\text{max} \sin(kx-wt)}) + = \pi r^2\cos(\theta) \omega B_\text{max} \cos(kx-wt) \;. +\end{align} +So the amplitude of the emf is +\begin{equation} + A_\text{emp} = \pi r^2 \omega B_\text{max} \cos(\theta) + = 2 \pi^2 r^2 f B_\text{max} \cos(\theta) + = \ans{2 \pi^2 r^2 f \frac{E_\text{max}}{c} \cos(\theta)} \;. +\end{equation} + +\Part{b} +\begin{center} +It is best to have the antennal oriented in the plane of the electric +field (red), so that the magnetic field (perpendicular to the page in +my drawing) creates the most flux through the antenna loop. +\begin{asy} +import Mechanics; +real u = 6cm; + +real r = 0.1; // m +real f = 300e6; // Hz +real v = 3e8; // m/s +real L = v/f; // m, wavelength +real xstart = 8r; +real mag = 16r; + +Vector x = Vector(center=(xstart*u,0), mag=mag*u, dir=180, L="signal"); +path a = scale(r*u)*unitcircle; // antenna + +draw(a); +x.draw(); +label("station", (xstart*u,0), S); + +real amp = r/2; +int n = 200; +real dx = mag/n; +path pE = (xstart*u,0); // Left speaker output wave +for (int i=1; i<=n; ++i) { + pE = pE..(xstart-i*dx, amp*sin(2pi*dx*i/L))*u; +} +draw(pE, red); +\end{asy} +\end{center} +\end{solution} diff --git a/latex/problems/problem27.02.T.tex b/latex/problems/problem27.02.T.tex new file mode 100644 index 0000000..729fa5b --- /dev/null +++ b/latex/problems/problem27.02.T.tex @@ -0,0 +1,35 @@ +\begin{problem} +$11500\U{Hz}$ sound waves moving at $350\U{m/s}$ hit a wall with two +slits $50.0\U{cm}$ appart. +\Part{a} At what angle is the first minimum located? +\Part{b} At what angle is the first minimum located if the sound is replaced +by $10.0\U{GHz}$ microwaves? +\end{problem} % Based on Problem 27.2 + +\begin{solution} +The first minimum occurs when the waves are $180\dg$ out of phase (due +to a path length difference of $\lambda/2$). +\begin{align} + d \sin(\theta_\text{min}) &= \p({n+\frac{1}{2}}) \lambda \qquad n = 0,1,2,\ldots \\ + \theta_\text{min} &\approx \frac{\p({n+\frac{1}{2}}) \lambda}{d} \qquad n = 0,1,2,\ldots +\end{align} +So the first minimum ($n=0$) is located at +\begin{equation} + \theta_\text{min} = \arcsin\p({\frac{\lambda}{2d}}) \;. +\end{equation} + +\Part{a} +For $f = 1150\U{Hz}$ sound, $\lambda = vT = v/f = 3.04\U{cm}$ and +\begin{equation} + \theta_\text{min} = \arcsin\p({\frac{3.04\U{cm}}{2\cdot 50.0\U{cm}}}) + = \ans{30.4\U{mrad}} = \ans{1.74\dg} \;. +\end{equation} + +\Part{b} +For $f = 10.0\U{Hz}$ light, $\lambda = c/f = 3.00\U{cm}$ (pretty much +the same wavelength as the sound in \Part{a}) and +\begin{equation} + \theta_\text{min} = \arcsin\p({\frac{3.00\U{cm}}{2\cdot 50.0\U{cm}}}) + = \ans{30.0\U{mrad}} = \ans{1.72\dg} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem27.07.T.tex b/latex/problems/problem27.07.T.tex new file mode 100644 index 0000000..f3e74ef --- /dev/null +++ b/latex/problems/problem27.07.T.tex @@ -0,0 +1,27 @@ +\begin{problem} +Two slits are separated by $1\U{mm}$. A beam of $440\U{nm}$ light +incident on the slits produces an interference pattern. \Part{a} +Determine the number of intensity maxima observed in the angular range +$-20.0\dg < \theta < 20.0\dg$. \Part{b} If the light is shifted to a +longer wavelength, will the number of maxima in that range increase or +decrease? +\end{problem} % based on Problem 27.7 + +\begin{solution} +\Part{a} +Following the path-length argument outlined in the previous problem +for interference minima, we have maxima located at +\begin{align} + d \sin(\theta_\text{max}) &= n \lambda \qquad n = 0,1,2,\ldots \\ + n_{20\dg} &= \frac{d \sin(\theta)}{\lambda} + = \frac{1\E{-3}\U{m} \cdot \sin(20\dg)}{440\E{-9}\U{m}} = 777.32 \;. +\end{align} +So the number of maxima in the specified range is +\begin{equation} + 2 \cdot \floor(n_{20\dg}) + 1 = \ans{1555} +\end{equation} + +\Part{b} +A longer wavelength will increase the spacing between the interference +maxima, \ans{decreasing} the total number in the specified angle range. +\end{solution} diff --git a/latex/problems/problem27.15.tex b/latex/problems/problem27.15.tex new file mode 100644 index 0000000..301f497 --- /dev/null +++ b/latex/problems/problem27.15.tex @@ -0,0 +1,40 @@ +\begin{problem} +An oil film ($n = 1.5$) floats on the surface of a bowl of water. The +film is illuminated by a white light placed directly above the bowl. +Red light at $\lambda = 650\U{nm}$ is the most strongly reflected +color. +\Part{a} How thick is the oil? +\Part{b} What color is most strongly transmitted? % warning, students don't like ultraviolet as a color. +\end{problem} % Based on Problem 27.15 + +\begin{solution} +\Part{a} +Constructive intereference on reflection follows Equation 27.11 +\begin{equation} + 2nt = \p({m+\frac{1}{2}})\lambda \qquad m = 0,1,2,\ldots +\end{equation} +Where the $+\frac{1}{2}$ is due to the $180\dg$ phase change from a +reflection off the interface in the direction of increasing index of +refraction (for the path that bounces off the air-oil interface). + +The strength of the reflected color decreases as $m$ increases, due to +some light being absorbed by the oil and decoherence from scattering. +Since red light is the most strongly reflected, $m = 0$, and the +thickness of the film is +\begin{equation} + t = \frac{\lambda}{4n} = \ans{108\U{nm}} +\end{equation} + +\Part{b} +Constructive interference on transmission occurs when +\begin{equation} + 2nt = m\lambda \qquad m = 1,2,3,\ldots +\end{equation} + +Because of absorbtion, the strongest transmission will be for $m=1$, so the +most strongly transmitted color is +\begin{equation} + \lambda_t = 2nt = \ans{325\U{nm}} +\end{equation} +which is in the ultraviolet. +\end{solution} diff --git a/latex/problems/problem27.V1.tex b/latex/problems/problem27.V1.tex new file mode 100644 index 0000000..741bb22 --- /dev/null +++ b/latex/problems/problem27.V1.tex @@ -0,0 +1,89 @@ +\begin{problem} +A double slit interference pattern is produced by two slits each of +width $0.35\U{mm}$ and slit separation $4.2\U{mm}$ on a screen placed +parallel to the slits at a distance $1.5\U{m}$ from the slits. The +slits are illuminated by light of wavelength $580\U{nm}$. Given: +$Y_m=m(D\lambda/d)$ \Part{a} What is the angular separation (in +degrees) between the central fringe and the 8th bright +fringe? \Part{b} What is the fringe width (in $\mu$m) of the pattern +on the screen? \Part{c} Assuming that we can see a large number of +fringes on the screen, which bright fringes between the central fringe +and the 40th bright fringe will be missing on the screen? Show +appropriate calculations and give reasons to substantiate your +answer. \Part{d} If however, a thin mica sheet of index of refraction +$n=1.58$ and thickness $t$ covers one of the slits, the central point +on the screen is now occupied by the $29th$ bright fringe. What is the +thickness $t$ (in $\mu$m) of the mica sheet? +\end{problem} + +\begin{solution} +\Part{a} +Bright fringes appear when the pathlengths differ by integer numbers +of wavelengths. +\begin{align} + d\sin(\theta) &= m \lambda \label{eq.doubleslit_max_angle} \\ + \theta &= \arcsin\p({\frac{m\lambda}{d}}) \\ + \theta_0 &= \arcsin(0) = 0\dg \\ + \theta_8 &= \arcsin\p({\frac{8\lambda}{d}}) + = \arcsin\p({\frac{8\cdot580\E{-9}\U{m}}{4.2\E{-3}\U{m}}}) + = 0.0633\dg \\ + \Delta \theta &= \theta_8 - \theta_0 = \ans{0.0633\dg} +\end{align} + +\Part{b} +To find the fringe width we take the small-angle approximation of +Eqn.~\ref{eq.doubleslit_max_angle} +\begin{align} + Y_m &= D\tan(\theta) \approx \frac{D}{d}d\sin(\theta) + = \frac{D}{d} m\lambda \\ + \Delta y &= y_1 - y_0 = \frac{D\lambda}{d} + = \frac{1.5\U{m}\cdot580\E{-9}\U{m}}{4.2\E{-3}\U{m}} + = \ans{207\U{$\mu$m}} +\end{align} + +\Part{c} +Up to now we have ignored the larger envelope caused by the +single-slit interference inside each slit. That has interference +minima when the slit can be broken up into an integer number of +self-annihilating regions. The difference in path length between rays +from the top and bottom of a self-annihilating region should be +$\lambda$ (so that the ray from the top cancels the ray from the +middle, the ray slightly below the top cancels the ray slightly below +the middle, etc.) With this idea and a little trig we find length +along the slit of a self-annihilating region should be given by +\begin{align} + x\sin(\theta) &= \lambda & + x &= \frac{\lambda}{\sin(\theta)} +\end{align} +So there are interference minima when we can fit an integer number of +these regions into $a$: +\begin{align} + a &= nx = n \frac{\lambda}{\sin(\theta)} \\ + a\sin(\theta) &= n\lambda +\end{align} +for any non-zero, integer $n$. The single-slit interference effect +kills a double-slit maximum when they occur at the same angle +\begin{align} + \sin(\theta) &= \frac{n}{a}\lambda = \frac{m}{d}\lambda \\ + \frac{n}{a} &= \frac{m}{d} \\ + \frac{m}{n} &= \frac{d}{a} = \frac{4.2\U{mm}}{0.35\U{mm}} = 12 +\end{align} +So the $m(n=1)=\ans{12^\text{th}}$, $m(n=2)=\ans{24^\text{th}}$, and +$m(n=3)=\ans{36^\text{th}}$ fringes are missing. + +\Part{d} +While passing through the mica (almost perpendicularly), the one ray +travels an `extra' $nt-t=(n-1)t$ wavelengths (compared to it's +neighbor passing through air). Because the central point on the +screen is the $29^\text{th}$ fringe, and the $0^\text{th}$ fringe +occurs where the effective path lengths are equal, the mica-ray must +have traveled an extra $29\lambda$ of effective distance. Therefore +\begin{align} + 29\lambda &= (n-1)t \\ + t &= \frac{29\lambda}{n-1} + = \frac{29\cdot580\E{-9}\U{m}}{0.58} = \ans{29.0\U{$\mu$m}} +\end{align} + +By `effective' distances, I mean `measured in wavelengths traveled', +as opposed to `measured in meters'. +\end{solution} diff --git a/latex/problems/problem28.02.T.tex b/latex/problems/problem28.02.T.tex new file mode 100644 index 0000000..e5ebbe0 --- /dev/null +++ b/latex/problems/problem28.02.T.tex @@ -0,0 +1,56 @@ +\begin{problem} +In Homework 3, you showed that the sun emits $3.87\E{26}\U{W}$ of +power. \Part{a} Use Stefan's law to calculate the surface temperature +at the photosphere ($r = 6.96\E{8}\U{m}$). \Part{b} Estimate the +power needed to produce the same spectrum with an incandescent light +bulb (i.e. to heat the bulb to the same temperature). Model the light +bulb's tungsten filament as a cylinder $58.0\U{cm}$ long and +$45\U{$\mu$m}$ in diameter with emissivity of $0.45$. +% emissivity from http://www.pyrometry.com/farassociates_tungstenfilaments.pdf +% model the bulb as a cylinder 65\U{mm} bulb diameter. (48 kW at 2000 K) +\Part{c} Tungsten melts at $3695\U{K}$, so we cannot actually operate the +filament at the same temperature as the sun. What temperature is the +filament from \Part{b} if we only radiate at $60\U{W}$? +% filament geometry and tungsten melting temperature from +% http://en.wikipedia.org/wiki/Incandescent_light_bulb + +Because of their long length, tungsten filaments are usually coiled +twice. See \url{http://en.wikipedia.org/wiki/Electrical_filament} for +some nice pictures. + +This problem is similar to P28.2 and P28.55. +\end{problem} % based on problem 28.2 + +\begin{solution} +\Part{a} +\begin{align} + P &= \sigma A e T^4 \\ + T &= \p({\frac{P}{\sigma A e}})^{1/4} + = \p({\frac{P}{\sigma 4 \pi r^2 e}})^{1/4} + = \p({\frac{3.87\E{26}\U{W}}{5.6696\E{-8}\U{W/m$^2$K$^4$} \cdot 4 \pi (6.96\E{8}\U{m})^2 \cdot 1}})^{1/4} + = \ans{5790\U{K}} \;, +\end{align} +where we treat the sun as a black body ($e = 1$). + +\Part{b} +Let the subscript w denote the tungsten filament and s denote the sun. +\begin{align} + \frac{P_\text{w}}{P_\text{s}} + &= \frac{\sigma A_\text{w} e_\text{w} T^4}{\sigma A_\text{s} e_\text{s} T^4} + = \frac{L_\text{w} 2 \pi r_\text{w} e_\text{w}}{4 \pi r_\text{s}^2} \\ + P_\text{w} + &= \frac{L_\text{w} r_\text{w} e_\text{w}}{2 r_\text{s}^2} P_\text{s} + = \frac{0.58\U{m} \cdot \frac{45\E{-6}\U{m}}{2} \cdot 0.45}{2 \cdot (6.96\E{8}\U{m})^2} \cdot 3.87\E{26} + = \ans{2.35\U{kW}} +\end{align} + +\Part{c} +\begin{equation} + T = \p({\frac{P}{\sigma A e}})^{1/4} + = \p({\frac{P}{\sigma L 2 \pi r e}})^{1/4} + = \p({\frac{60\U{W}}{5.6696\E{-8}\U{W/m$^2$K$^4$} \cdot 0.58\U{m} + \cdot 2 \pi \cdot \frac{45\E{-6}\U{m}}{2} \cdot 0.45}})^{1/4} + = \ans{2310\U{K}} +\end{equation} + +\end{solution} diff --git a/latex/problems/problem28.04.tex b/latex/problems/problem28.04.tex new file mode 100644 index 0000000..4010ba6 --- /dev/null +++ b/latex/problems/problem28.04.tex @@ -0,0 +1,19 @@ +\begin{problem*}{28.4} +Calculate the energy, in electron volts, of a photon whose frequency +is \Part{a} $620\U{THz}$, \Part{b} $3.10\U{GHz}$, and \Part{c} +$46.0\U{MHz}$. \Part{d} Determine the corresponding wavelengths for +these photons and state the classification of each on the +electromagnetic spectrum. +\end{problem*} % problem 28.4 + +\begin{solution} +\begin{center} +\begin{tabular}{r r r r} + & \Part{a} & \Part{b} & \Part{c} \\ + $E = hf$ & $2.56\U{eV}$ & $12.8\U{$\mu$eV}$ & $190\U{neV}$ \\ + $\lambda=c/f$ & $484\U{nm}$ & $9.67\U{cm}$ & $6.51\U{m}$ \\ + Class & Visible (blue) & Microwave & Radio +\end{tabular} +\end{center} +See Figure 24.12 (page 823) in the text for a classification spectrum. +\end{solution} diff --git a/latex/problems/problem28.06.tex b/latex/problems/problem28.06.tex new file mode 100644 index 0000000..910f1be --- /dev/null +++ b/latex/problems/problem28.06.tex @@ -0,0 +1,22 @@ +\begin{problem*}{28.6} +The average threshold of dark-adapted (scotopic) vision is +$4.00\E{-11}\U{W/m$^2$}$ at a central wavelength of $500\U{nm}$. If +light having this intensity and wavelength enters the eye and the +pupil is open to its maximum diameter of $8.50\U{mm}$, how many +photons per second enter the eye? +\end{problem*} % problem 28.6 + +\begin{solution} +The total power into the eye is +\begin{equation} + P = IA = \pi r^2 I = \frac{\pi d^2 I}{4} = 2.27\U{fW} \;, +\end{equation} +and the energy per photon is +\begin{equation} + E = hf = \frac{hc}{\lambda} = 2.48\U{eV} = 3.97\E{-19}\U{J} \;, +\end{equation} +so the number of photons entering per second is +\begin{equation} + \Phi_p = \frac{P}{E} = \ans{5710} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem28.07.T.tex b/latex/problems/problem28.07.T.tex new file mode 100644 index 0000000..0904115 --- /dev/null +++ b/latex/problems/problem28.07.T.tex @@ -0,0 +1,18 @@ +\begin{problem} +A simple pendulum as a length of $2.00\U{m}$ and a mass of +$3.00\U{kg}$. The amplitude of oscillation is $5.00\U{cm}$. Assuming +that energy is quantized, calculate the quantum number of the pendulum. +\end{problem} % based on problem 28.7 + +\begin{solution} +Using Planck's assumption that energy is quantized (Equation 28.2) +\begin{align} + E_n = nhf &= mgh_\text{max} = mgL (1-\cos \theta_\text{max}) + = mgL \p({1 - \sqrt{1 - (a/L)^2}}) + = 18.4\U{mJ} \\ + f &= \frac{1}{2\pi}\sqrt{\frac{g}{L}} = 352\U{mHz} \\ + n &= \frac{E_n}{hf} + = \frac{18.4\U{mJ}}{6.63\E{-34}\U{J$\cdot$s}\cdot352\U{mHz}} + = \ans{7.87\E{31}} +\end{align} +\end{solution} diff --git a/latex/problems/problem28.09.T.tex b/latex/problems/problem28.09.T.tex new file mode 100644 index 0000000..af0193b --- /dev/null +++ b/latex/problems/problem28.09.T.tex @@ -0,0 +1,33 @@ +\begin{problem} +Sodium has a work function of $2.46\U{eV}$. \Part{a} Find the +cutoff wavelength and cutoff frequency for the photoelectic +effect. \Part{b} What is the stopping potential if the incident light +has a wavelength of $440\U{nm}$? +\end{problem} % based on problem 28.9 + +\begin{solution} +\Part{a} +The cutoff wavelength is the wavelength where the incoming light has +barely enough energy to free an electron, i.e. all of the photon's +energy goes into overcoming the work function barrier. +\begin{align} + hf &= \phi \\ + f &= \frac{\phi}{h} + = \frac{2.46\U{eV}\cdot 1.60\E{-19}\U{J/eV}}{6.63\E{-34}\U{J$\cdot$s}} + = 592\U{THz} \\ + \lambda f &= c \\ + \lambda &= \frac{c}{f} = \ans{507\U{nm}} +\end{align} + +\Part{b} +The photon brings in $hf$, but much of that energy goes to overcoming +the work function barrier. The left over energy $hf-\phi$ becomes the +electron's kinetic energy. The stopping potential is the voltage +change which matches that kinetic energy. +\begin{align} + K_\text{max} &= hf - \phi = h\frac{c}{\lambda} - \phi + = 5.98\E{-20}\U{J} = 0.358\U{eV} \\ + \Delta V_S &= K_\text{max}/e = \ans{0.374\U{V}} +\end{align} + +\end{solution} diff --git a/latex/problems/problem28.09.tex b/latex/problems/problem28.09.tex new file mode 100644 index 0000000..8e98c14 --- /dev/null +++ b/latex/problems/problem28.09.tex @@ -0,0 +1,32 @@ +\begin{problem*}{28.9} +Molybdenum has a work function of $4.20\U{eV}$. \Part{a} Find the +cutoff wavelength and cutoff frequency for the photoelectic +effect. \Part{b} What is the stopping potential if the incident light +has a wavelength of $180\U{nm}$? +\end{problem*} % problem 28.9 + +\begin{solution} +\Part{a} +The cutoff wavelength is the wavelength where the incoming light has +barely enough energy to free an electron, i.e. all of the photon's +energy goes into overcoming the work function barrier. +\begin{align} + hf &= \phi \\ + f &= \frac{\phi}{h} + = \frac{4.20\U{eV}\cdot 1.60\E{-19}\U{J/eV}}{6.63\E{-34}\U{J$\cdot$s}} + = \ans{1.01\U{PHz}} \\ + \lambda f &= c \\ + \lambda &= \frac{c}{f} = \ans{296\U{nm}} +\end{align} + +\Part{b} +The photon brings in $hf$, but much of that energy goes to overcoming +the work function barrier. The left over energy $hf-\phi$ becomes the +electron's kinetic energy. The stopping potential is the voltage +change which matches that kinetic energy. +\begin{align} + K_\text{max} &= hf - \phi = h\frac{c}{\lambda} - \phi + = 4.30\E{-19}\U{J} = 2.69\U{eV} \\ + \Delta V_S &= K_\text{max}/e = \ans{2.69\U{V}} +\end{align} +\end{solution} diff --git a/latex/problems/problem28.10.tex b/latex/problems/problem28.10.tex new file mode 100644 index 0000000..50cafdf --- /dev/null +++ b/latex/problems/problem28.10.tex @@ -0,0 +1,27 @@ +\begin{problem*}{28.10} +Electrons are ejected from a metallic surface with speeds ranging up +to $4.60\E{5}\U{m/s}$ when light with a wavelength of $625\U{nm}$ is +used. \Part{a} What is the work function of the surface? \Part{b} +What is the cutoff frequency for this surface? +\end{problem*} % problem 28.10 + +\begin{solution} +\Part{a} +Conserving energy +\begin{align} + hf &= \frac{hc}{\lambda} = \phi + K_\text{max} + = \phi + \frac{1}{2} m_e v_\text{max}^2 \\ + \phi &= \frac{hc}{\lambda} - \frac{1}{2} m_e v_\text{max}^2 + = \frac{6.63\E{-34}\U{J$\cdot$s}\cdot3.00\E{8}\U{m/s}}{625\E{-9}\U{m}} + - \frac{1}{2}\cdot 9.11\E{-31}\U{kg}\cdot(4.60\E{5}\U{m/s})^2 + = \ans{2.21\E{-19}\U{J}} = \ans{1.38\U{eV}} +\end{align} + +\Part{b} +Light at the cutoff frequency only barely supplies enough energy to +overcome the work function. +\begin{align} + hf_\text{cut} &= \phi \\ + f_\text{cut} &= \frac{\phi}{h} = \ans{334\U{THz}} \\ +\end{align} +\end{solution} diff --git a/latex/problems/problem28.13.tex b/latex/problems/problem28.13.tex new file mode 100644 index 0000000..8d8a431 --- /dev/null +++ b/latex/problems/problem28.13.tex @@ -0,0 +1,23 @@ +\begin{problem*}{28.13} +An isolated copper sphere of radius $5.00\U{cm}$, initially uncharged, +is illuminated by ultraviolet light of wavelength $200\U{nm}$. What +charge will the photoelectric effect induce on the sphere? The work +function for copper is $4.70\U{eV}$. +\end{problem*} % problem 28.13 + +\begin{solution} +As light lands on the sphere, electrons are blasted off into oblivion. +As they leave, the sphere accumulates positive charge, so it takes a +bit more energy to pull the next electrons away. Eventually the +system reaches equilibrium when the initial kinetic energy of electron +blasting off is not quite enough for it to escape the electro-static +attraction to the positive sphere. In math +\begin{align} + K_\text{max} &= hf - \phi = \frac{hc}{\lambda} - \phi + = 2.40\E{-19}\U{J} = 1.50\U{eV} \\ + &= k_e \frac{Qq}{R} \\ + Q &= \frac{K_\text{max}R}{k_e q} + = \frac{2.40\E{-19}\U{J}}{8.99\E{9}\U{N$\cdot$m$^2$/C$^2$}\cdot 1.60\E{-19}\U{C}} + = \ans{8.34\U{pC}} = \ans{52.1\E{6}\U{electrons}} +\end{align} +\end{solution} diff --git a/latex/problems/problem28.14.tex b/latex/problems/problem28.14.tex new file mode 100644 index 0000000..bc1032d --- /dev/null +++ b/latex/problems/problem28.14.tex @@ -0,0 +1,25 @@ +\begin{problem*}{28.14} +Calculate the energy and momentum of a photon of wavelength $700\U{nm}$. +\end{problem*} % problem 28.14 + +\begin{solution} +You should be familiar with these equations by now (after our time +with the Bohr atom and relativity). +The energy is given by (Eq's 28.3 and 11.15) +\begin{equation} + E = \frac{hc}{\lambda} = \frac{1240\U{eV$\cdot$nm}}{700\U{nm}} + = \ans{1.77\U{eV}} \;. +\end{equation} +For momentum you can either use the relativistic energy-momentum +equation (Eq.~9.22) +\begin{align} + E^2 &= p^2c^2 + m_0^2c^4 \\ + E_\text{photon} &= p c \\ + p &= \frac{E}{c} = \ans{1.77\U{eV/$c$}} = \ans{9.46\E{-28}\U{kg$\cdot$m/s}} +\end{align} +or the de Broglie formula (Eq.~28.10) +\begin{align} + p &= \frac{h}{\lambda} = \frac{6.63\E{-34}\U{J$\cdot$s}}{700\U{nm}} + = \ans{9.46\E{-28}\U{kg$\cdot$m/s}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem28.15.tex b/latex/problems/problem28.15.tex new file mode 100644 index 0000000..e033770 --- /dev/null +++ b/latex/problems/problem28.15.tex @@ -0,0 +1,39 @@ +\begin{problem*}{28.15} +X-rays having an energy of $300\U{keV}$ undergo Compton scattering +from a target. The scattered rays are detected at $37.0\dg$ relative +to the incident rays. Find \Part{a} the Compton shift at this +angle, \Part{b} the energy of the scattered x-ray, and \Part{c} the +energy of the recoiling electron. +\end{problem*} % problem 28.15 + +\begin{solution} +\Part{a} +From the Compton shift equation (Eq.~28.8) +\begin{align} + \lambda' - \lambda_0 &= \frac{h}{m_e c} (1-\cos\theta) \\ + \Delta \lambda &= 2.43\U{pm}(1-\cos 37.0\dg) + = \ans{489\U{fm}} \;. +\end{align} + +\Part{b} +The wavelength of the incoming photon was +\begin{equation} + \lambda_0 = \frac{hc}{E_0} = \frac{1240\U{eV$\cdot$nm}}{300\U{keV}} + = 4.13\U{pm} \;. +\end{equation} +The scattered wavelength is thus +\begin{equation} + \lambda' = \lambda_0 + \Delta \lambda = (4.13+0.489)\U{pm} = 4.62\U{pm} \;, +\end{equation} +and the energy of the scattered photon is +\begin{equation} + E' = \frac{hc}{\lambda'} = \frac{1240\U{eV$\cdot$nm}}{4.62\U{pm}} + = \ans{268\U{keV}} \;. +\end{equation} + +\Part{c} +All the energy lost by the photon must go into the recoiling electron so +\begin{equation} + E_e = E_0 - E' = (300-268)\U{keV} = \ans{31.7\U{keV}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/problem28.16.tex b/latex/problems/problem28.16.tex new file mode 100644 index 0000000..62ef19b --- /dev/null +++ b/latex/problems/problem28.16.tex @@ -0,0 +1,28 @@ +\begin{problem*}{28.16} +A $0.110\U{nm}$ photon collides with a stationary electron. After +the collision, the electron moves forward and the photon recoils +backward. Find the momentum and the kinetic energy of the electron. +\end{problem*} % problem 28.16 + +\begin{solution} +The photon scatters by $180\dg$, so from the Compton shift equation +\begin{equation} + \lambda' = \lambda_0 + \frac{h}{m_e c}(1-\cos 180\dg) + = \lambda_0 + \frac{2h}{m_e c} + = (110+4.85)\U{pm} = 115\U{pm} \;. +\end{equation} +The kinetic energy of the electron is given by the change in photon +energy (just like problem 28.15). +\begin{equation} + K = E_0 - E' = \frac{hc}{\lambda_0} - \frac{hc}{\lambda'} + = 11.3\U{keV} - 10.8\U{keV} + = \ans{476\U{eV}} \;. +\end{equation} +We conserve momentum to find the electron's momentum, using +$p_\text{photon}=E/c$. +\begin{align} + p_i &= \frac{E_0}{c} = p_f = p_e - \frac{E'}{c} \\ + p_e &= \frac{E_0+E'}{c} = (11.3+10.8)\U{keV/$c$} = \ans{22.1\U{keV/$c$}} + = \ans{1.18\E{-23}\U{kg$\cdot$m/s}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem28.25.T.tex b/latex/problems/problem28.25.T.tex new file mode 100644 index 0000000..3a95f90 --- /dev/null +++ b/latex/problems/problem28.25.T.tex @@ -0,0 +1,66 @@ +\begin{problem} +The resolving power of a microscope depends on the wavelength of light +used. If you want to ``see'' an atom, you must resolve features on +the order of $0.1\U{\AA}$. \Part{a} If you use electrons (in an electron +microscope), what minimum kinetic energy would they require? \Part{b} +If you used photons (in a light microscope), what minimum kinetic +energy would they require? +\end{problem} % Problem 28.25 + +\begin{solution} +\Part{a} +This is a de Broglie problem, since the resultion of the microscope is +on the order of the wavelength of the probe photon or electron. We +use de Broglie's formula to relate the particle's wavelength to it's +momentum $\lambda p = h$ (Equation 28.10). + +First, assume our electrons have non-relativistic speeds and we can +use $p = m_0v$ (as opposed to the relativistic Equation 9.15 $p = +\gamma m_0 v$). +\begin{align} + \lambda &= \frac{h}{p} = \frac{h}{m_0v} \\ + v &= \frac{h}{m_0\lambda} \\ + K &= \frac{1}{2} m_0 v^2 = \frac{h^2}{2m_0\lambda^2} \;. +\end{align} +We can resolve features on the order of a wavelength, so let's set +$\lambda = 0.1\U{\AA}$. +\begin{equation} + K = \frac{6.626\E{-34}\U{J$\cdot$s$^2$}}{2\cdot 9.11\E{-31}\U{kg} \cdot (1\E{-11}\U{m})^2} + = \ans{2.41\E{-15}\U{J}} = \ans{15.1\U{keV}} \;. +\end{equation} +We assumed that the electrons were non-relativistic, so we check our +calculated speed +\begin{equation} + v = \frac{h}{m_0\lambda} = 0.727\E{8}\U{m/s} = 0.242 c \;. +\end{equation} +This is on the border of the relativistic behavior, so we can go back +and redo the calculation relativistically. +\begin{align} + p &= \frac{h}{\lambda} = 6.63\E{-23}\U{kg$\cdot$m/s$^2$} \\ + E^2 &= p^2 c^2 + m_0^2 c^4 \\ + K &= E - E_\text{rest} = E - m_0 c^2 = \sqrt{p^2 c^2 + m_0^2 c^4} - m_0 c^2 + = \ans{2.38\E{-15}\U{J}} = \ans{14.8\U{keV}} \;. +\end{align} +Which is pretty close to our non-relativistic answer. The $E^2$ +formula is simply a rephrasing of $E = \gamma m_0 c^2$ in terms of +momentum. The derivation is sketched out in the text around Equation +9.22. + +If you wanted to get really fancy, you could use the formula for the +resolution of a circular-aperature microscope (Equation 27.15) +\begin{align} + \theta_\text{min} &= 1.22\frac{\lambda}{D} \\ + \Delta x_\text{min} &= L\tan(\theta_\text{min}) +\end{align} +to determine the required wavelength of light, but you'd have to make +guesses about the diameter of the aperature $D$ and the distance +between the aperature and the speciment $L$. + +\Part{b} +For light, we use the relativistic formula with $m_0 = 0$ +\begin{equation} + E = K = p c = \frac{hc}{\lambda} = \frac{1240\U{eV$\cdot$nm}}{0.01\U{nm}} + = \ans{124\U{keV}} = \ans{1.98\E{-14}\U{J}} \;, +\end{equation} +around 8 times larger than the energy needed using electrons. +\end{solution} diff --git a/latex/problems/problem28.25.tex b/latex/problems/problem28.25.tex new file mode 100644 index 0000000..61564d3 --- /dev/null +++ b/latex/problems/problem28.25.tex @@ -0,0 +1,64 @@ +\begin{problem} +The resolving power of a microscope depends on the wavelength of light +used. If one wished to ``see'' an atom, a resolution of approximately +$1.00\E{-11}\U{\m}$ would be required. \Part{a} If electrons are used +(in an electron microscope), what minimum kinetic energy is required for the electrons? \Part{b} If photons are used, what minimum photon energy is needed to obtain the required resolution? +\end{problem} % Problem 28.25 + +\begin{solution} +\Part{a} +This is a de Broglie problem, since the resultion of the microscope is +on the order of the wavelength of the probe photon or electron. We +use de Broglie's formula to relate the particle's wavelength to it's +momentum $\lambda p = h$ (Equation 28.10). + +First, assume our electrons have non-relativistic speeds and we can +use $p = m_0v$ (as opposed to the relativistic Equation 9.15 $p = +\gamma m_0 v$). +\begin{align} + \lambda &= \frac{h}{p} = \frac{h}{m_0v} \\ + v &= \frac{h}{m_0\lambda} \\ + K &= \frac{1}{2} m_0 v^2 = \frac{h^2}{2m_0\lambda^2} \;. +\end{align} +We can resolve features on the order of a wavelength, so let's set +$\lambda = 0.1\U{\AA}$. +\begin{equation} + K = \frac{6.626\E{-34}\U{J$\cdot$s$^2$}}{2\cdot 9.11\E{-31}\U{kg} \cdot (1\E{-11}\U{m})^2} + = \ans{2.41\E{-15}\U{J}} = \ans{15.1\U{keV}} \;. +\end{equation} +We assumed that the electrons were non-relativistic, so we check our +calculated speed +\begin{equation} + v = \frac{h}{m_0\lambda} = 0.727\E{8}\U{m/s} = 0.242 c \;. +\end{equation} +This is on the border of the relativistic behavior, so we can go back +and redo the calculation relativistically. +\begin{align} + p &= \frac{h}{\lambda} = 6.63\E{-23}\U{kg$\cdot$m/s$^2$} \\ + E^2 &= p^2 c^2 + m_0^2 c^4 \\ + K &= E - E_\text{rest} = E - m_0 c^2 = \sqrt{p^2 c^2 + m_0^2 c^4} - m_0 c^2 + = \ans{2.38\E{-15}\U{J}} = \ans{14.8\U{keV}} \;. +\end{align} +Which is pretty close to our non-relativistic answer. The $E^2$ +formula is simply a rephrasing of $E = \gamma m_0 c^2$ in terms of +momentum. The derivation is sketched out in the text around Equation +9.22. + +If you wanted to get really fancy, you could use the formula for the +resolution of a circular-aperature microscope (Equation 27.15) +\begin{align} + \theta_\text{min} &= 1.22\frac{\lambda}{D} \\ + \Delta x_\text{min} &= L\tan(\theta_\text{min}) +\end{align} +to determine the required wavelength of light, but you'd have to make +guesses about the diameter of the aperature $D$ and the distance +between the aperature and the speciment $L$. + +\Part{b} +For light, we use the relativistic formula with $m_0 = 0$ +\begin{equation} + E = K = p c = \frac{hc}{\lambda} = \frac{1240\U{eV$\cdot$nm}}{0.01\U{nm}} + = \ans{124\U{keV}} = \ans{1.98\E{-14}\U{J}} \;, +\end{equation} +around 8 times larger than the energy needed using electrons. +\end{solution} diff --git a/latex/problems/problem28.31.T.tex b/latex/problems/problem28.31.T.tex new file mode 100644 index 0000000..e822518 --- /dev/null +++ b/latex/problems/problem28.31.T.tex @@ -0,0 +1,77 @@ +\begin{problem} +It would take a huge potential energy barrier to confine an electron +to the nucleus of an atom (diameter $d \approx 10\U{fm}$). \Part{a} +Use the Heisenberg uncertainty principle to find the momentum +uncertainty of such a bound electron. \Part{b} Use the monemtum +uncertainty from \Part{a} to find the minimum binding energy $U$. +Note that the total energy $E = K+U < 0$ for a bound particle. You +may use the non-relativistic form of kinetic energy even though it's +not particularly valid for this situation. +%In recitation Problem 28.XX we found the energy of a proton bound to +%the nucleus (modeled as a $10.0\U{fm}$ box) by using the de Broglie +%wavelength $\lambda = h/p$ and the non-relativistic kinetic energy +%$K=p^2/2m$. You can use that approach to ... +\end{problem} % starts out along the lines of Problem 28.31. + +\begin{solution} +\Part{a} +The uncertainty principle gives a lower bound on our momentum +uncertainty $\Delta p$ +\begin{align} + \Delta x \Delta p &\ge \frac{\hbar}{2} \\ + \Delta p &\ge \frac{\hbar}{2 \Delta x} = \frac{\hbar}{2 d} + = \ans{5.27\E{-21}\U{kg$\cdot$m/s}} \;. +\end{align} + +\Part{b} +The momentum uncertainty puts a lower bound on the kinetic energy +uncertainty $\Delta K$. +\begin{center} +\begin{asy} +import graph; + +real u=1cm; +real dp=2; +real k=1; +real m=1; +real pMax=2; +real width=4u; // because I can't get ScaleX to work... + +real dK=(dp/2)**2/(2m); +real K(real p) { return p**2/(2m); } + +size(width,0); +scale(Linear, Linear); +xlimits(-pMax,pMax); +ylimits(0,pMax**2/(2m)); +xaxis("$p$"); +yaxis("$K$",0); + +real scalex(real x) { return x/pMax*(width/2); } + +pair[] dps = {(-dp/2,0),(dp/2,0)}; +pair[] dKs = {(dp/2,0),(dp/2,dK)}; + +draw(graph(K,-pMax,pMax),red); +draw(graph(K,-dp/2,dp/2),blue); +draw(graph(dps), green); +draw(graph(dKs), green); +label("$\Delta p$", align=S, Scale((0,0))); +label("$\Delta K$", align=E, Scale((dp/2,dK/2))); +\end{asy} +\end{center} +\begin{equation} + \Delta K = \frac{(\Delta p/2)^2}{2 m} = \frac{\hbar^2}{32 d^2 m} +\end{equation} +The binding potential must be at least this deep, or the electron +would occasionally have enough kinetic energy to escape. +\begin{equation} + \Delta U < -\Delta K \approx -\frac{\hbar^2}{32 d^2 m} + = -3.8\U{pJ} = -24\U{MeV} \;. +\end{equation} +So the energy barrier is $\ans{24\U{MeV}}$, which is much greater than +the electron's rest mass energy of $511\U{keV}$, so the electron would +be extremely relativistic. This is one reason why light particles +such as electrons do not collapse into the nucleus of the atom, +despite the electromagnetic attraction to the protons. +\end{solution} diff --git a/latex/problems/problem28.34.T.tex b/latex/problems/problem28.34.T.tex new file mode 100644 index 0000000..9ba5838 --- /dev/null +++ b/latex/problems/problem28.34.T.tex @@ -0,0 +1,100 @@ +\begin{problem} +The time/energy Heisenberg uncertainty principle is the source of an +natural linewidth $\Delta \lambda$ in photons emitted from atoms when +electrons change orbitals. \Part{a} Calculate the frequency of light +emitted in the $n_2 \rightarrow n_1$ transition for +Hydrogen. \Part{b} Assuming that transition has an average lifetime +of $\tau = 1.6\U{ns}$, estimate the relative uncertainty in the energy +of the emitted photon. +%rate from +%http://www.springerlink.com/content/n15u027840428w14/ +%see also +%http://en.wikipedia.org/wiki/Spontaneous_emission#Rate_of_spontaneous_emission +\begin{center} +\begin{asy} +import graph; + +real u=1cm; +real dx=0.3u; +real X=10; +real H=5; +real FWHM=1; + +real lorentzian(real x) { + return H*(FWHM/2)/((x-X)**2 + (FWHM/2)**2); +} + +size(4u,0); +scale(Linear, Linear); +xlimits(0, 2X); +ylimits(0, lorentzian(X)); + +real[] Ticks = {X}; +string[] xLabels = {"$\lambda_0$"}; +string ticklabel(real x) { + int i; + for (i=0; i L$. +%so the normalization condition, Equation 28.23, reduces to $\int_0^L |\psi|^2 dx = 1$. +\end{problem} % problem 28.46 + +\begin{solution} +\begin{equation} + 1 = \int_{-\infty}^{\infty} \psi dx \psi^* = \int_0^L \psi^2 dx \;, +\end{equation} +where we used the fact that $\psi(x) = 0$ for $x < 0$ and $x > L$ to +reduce the range of integration, and the fact that $\psi$ is real to +reduce $\psi\psi^2 = \psi^2$. So we must integrate +\begin{equation} + 1 = \int_0^L \psi^2 dx = \int_0^L A^2 \sin^2\p({\frac{n\pi x}{L}}) dx + = A^2 \int_0^{\pi} \sin^2(nu) \frac{Ldu}{\pi} + = \frac{LA^2}{\pi} \int_0^{\pi} \sin^2(nu) du \;, +\end{equation} +where $u \equiv \pi x/L$, so $dx = Ldu/\pi$. + +There are two possible approaches. The easiest way is to use +symmetry. We're integrating over multiples of half wavelengths of +$\sin$ ($\lambda = 2\pi/n$ so $\pi = n\lambda/2$), so we're +integrating through full wavelengths of $\sin^2$. Over a full +wavelength, the averages $<\sin^2> = <\cos^2> = 1/2$ since $\sin^2 + +\cos^2 = 1$. so +\begin{align} + 1 &= \frac{LA^2}{\pi} \int_0^\pi \sin^2(nu) du + = \frac{LA^2}{\pi} \cdot \pi \frac{1}{2} = \frac{LA^2}{2} \\ + A &= \ans{\sqrt{\frac{2}{L}}} \;. +\end{align} + +The slightly harder way is to use the double-angle identity +$\sin^2(\theta) = [1-\cos(2\theta)]/2$. +\begin{align} + 1 &= \frac{LA^2}{\pi} \int_0^\pi \sin^2(nu) du + = \frac{LA^2}{2\pi} \int_0^\pi [1-\cos(2nu)] du + = \frac{LA^2}{2\pi} \cdot \p[{\int_0^\pi du - \int_0^\pi \cos(2nu) du}] \\ + &= \frac{LA^2}{2\pi} \cdot \p[{u + \frac{1}{2n}\sin(2nu)}]_0^\pi + = \frac{LA^2}{2\pi} \cdot \pi + = \frac{LA^2}{2} \\ + A &= \ans{\sqrt{\frac{2}{L}}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/problem28.56.millikan.ppm b/latex/problems/problem28.56.millikan.ppm new file mode 100644 index 0000000..9f92590 --- /dev/null +++ b/latex/problems/problem28.56.millikan.ppm @@ -0,0 +1,61 @@ +P6 +480 283 +255 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+ + 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+ + 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ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿþþþÿÿÿÿÿÿþþþþþþÿÿÿÿÿÿþþþüüüÿÿÿÿÿÿýýýüüüÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿýýýÿÿÿÿÿÿÿÿÿÿÿÿûûûNNN<< (1200-500)/(442-93)*(Xp-93) + 500 +# Yp -> (2.5-0.5)/(64-215)*(Yp-215) + 0.5 +550.143 0.44702 +688.539 1.00331 +740.688 1.18874 +818.911 1.58609 +959.312 2.0894 +1181.95 3.00331 diff --git a/latex/problems/problem28.56.tex b/latex/problems/problem28.56.tex new file mode 100644 index 0000000..32ab154 --- /dev/null +++ b/latex/problems/problem28.56.tex @@ -0,0 +1,69 @@ +\begin{problem*}{28.56} +Figure P28.56 shows the stopping potential versus the incident photon +frequency for the photoelectric effect for sodupm. Use the graph to +find \Part{a} the work function, \Part{b} the ratio $h/e$, +and \Part{c} the cutoff wavelength. The data are taken from +R.A.~Millikan, \emph{Physical Review} 7:362 (1916). +\begin{center} +\begin{asy} +import graph; +size(5cm, 4cm, IgnoreAspect); + +real fmin = 400; +real fmax = 1200; + +pair[] points={(550,0.827),(689,1.20),(741,1.39), + (818,1.79),(959,2.29),(1182,3.20)}; // (THz, V) + +real fit(real f) +{ + real work_fn = 1.62; // electron-volts + real h = 6.56e-34; // joule-seconds + real E = h*f*1e12 / 1.6e-19; // electron-volts + real Vs = E-work_fn; + return Vs; +} + +draw(graph(fit, fmin, fmax), red); +draw(graph(points), p=invisible, marker=marker(scale(1pt)*unitcircle, blue)); + + +pen thin=linewidth(0.5*linewidth())+grey; +xaxis("$f$ (THz)",BottomTop, + LeftTicks(begin=false,end=false,extend=true,ptick=thin)); +yaxis("$\Delta V_s$ (V)",LeftRight, + RightTicks(begin=false,end=false,extend=true,ptick=thin)); +\end{asy} +\end{center} +\end{problem*} % problem 28.56 + +\begin{solution} +\Part{a} +The fit line passes nearby the points $(400\U{THz},0\U{V})$ and +$(1.20\U{PHz},3.3\U{V})$. In point-slope form, the fit line is then +\begin{align} + \Delta V_s-0\U{V} &= \frac{(3.3-0)\U{V}}{(1200-400)\U{THz}} (f-400\U{THz}) \\ + \Delta V_s &= 4.125\text{mV/THz}\cdot (f-400\U{THz}) +\end{align} +The work function is the inverse $y$-intercept, so +\begin{equation} + \phi = -\Delta V_s(f=0) = -4.125\text{mV/THz}\cdot(-400\U{THz}) + = \ans{1.65\U{eV}} +\end{equation} + +\Part{b} +The theoretical form for the fit line is +\begin{align} + e\Delta V_s &= hf - \phi \\ + \Delta V_s &= \frac{h}{e}f - \phi +\end{align} +so $\frac{h}{e} = \ans{4.12\U{mV/THz}} = \ans{4.12\U{pV/s}}$. + +\Part{c} +The cutoff wavelength is given by the work function and conservation +of energy. +\begin{align} + hf_\text{cut} &= \frac{hc}{\lambda_\text{cut}} = \phi \\ + \lambda_\text{cut} &= \frac{hc}{\phi} = \ans{751\U{nm}} +\end{align} +\end{solution} diff --git a/latex/problems/problem28.57.tex b/latex/problems/problem28.57.tex new file mode 100644 index 0000000..c3f374a --- /dev/null +++ b/latex/problems/problem28.57.tex @@ -0,0 +1,75 @@ +\begin{problem*}{28.57} +The following table shows data obtained in a photoelectric +experiment. \Part{a} Using these data, make a graph similar to Active +Figure 28.9 that plots as a straight line. From the graph, +determine \Part{b} an experimental value for Planck's constant (in +joule-seconds) and \Part{c} the work function (in electron volts) for +the surface. (Two significant figures for each answer are +sufficient.) +\begin{center} +\begin{tabular}{r r} +Wavelength (nm) & Maximum Kinetic Energy of Photoelectrons (eV) \\ +$588$ & $0.67$ \\ +$505$ & $0.98$ \\ +$445$ & $1.35$ \\ +$399$ & $1.63$ +\end{tabular} +\end{center} +\end{problem*} % problem 28.57 + +\begin{solution} +\Part{a} +We can convert the wavelengths to frequencies and graph them +\begin{center} +\begin{tabular}{r r} +$\lambda$ (nm) & $f=c/\lambda$ (THz) \\ +$588$ & $510$ \\ +$505$ & $594$ \\ +$445$ & $674$ \\ +$399$ & $751$ +\end{tabular} \\ +\begin{asy} +import graph; +import stats; +size(8cm, 5cm, IgnoreAspect); + +real c = 299792458.0; + +real fmin = 0; +real fmax = 800; + +real[] x = {588, 505, 445, 399}; // nm +real[] y = {0.67, 0.98, 1.35, 1.63}; // eV + +// convert wavelength in nm to freq in THz +int i; +for (i=0; iU$. Classically, we would +expect all the particles to continue on, although with reduced speed. +According to quantum mechanics, a fraction of the particles are +reflected at the barrier. Prove that the reflection coefficient $R$ +for this case is +\begin{equation} + R = \frac{(k_1-k_2)^2}{(k_1+k_2)^2} \;, +\end{equation} +where $k_1=2\pi/\lambda_1$ and $k_2=2\pi/\lambda_2$ are the wave +numbers for the incident and transmitted particles. Proceed as +follows. Impose the boundary conditions $\psi_1=\psi_2$ and +$d\psi_1/dx = d\psi_2/dx$ at $x=0$ to find the relationships between +$B$ and $A$. Then evaluate $R=B^2/A^2$. + +Assume the wave function $\psi_1 = Ae^{ik_1x}+Be^{-ik_1x}$ satisfies +the Schr\"odinger equation in region 1, for $x<0$. Also assume that +$\psi_2 = Ce^{ik_2x}$ satisfies the Schr\"odinger equation in region +2, for $x>0$. These assumptions will be derived in the posted +solutions in case you are interested, but they are pretty +straightforward. +\begin{center} +\begin{asy} +import Mechanics; + +real u=.8cm; +real U=1u; +real E=1.2U; +real bonusU=0.3u; +real xmax=2u; +real xpos=-u; + +path pU = (-xmax,-bonusU)--(-xmax,0)--(0,0)--(0,U)--(xmax,U)--(xmax,-bonusU)--cycle; +draw(pU, black); +fill(pU, blue+4white); +Vector v = Velocity(center=(xpos, E)); +label("incoming particle", v.center, W); +v.draw(); +dot(v.center); +Distance dE = Distance(pFrom=(xpos, 0), pTo=(xpos, E), L="E"); +dE.draw(rotateLabel=false); +Distance dU = Distance(pFrom=(-xpos, 0), pTo=(-xpos, U), L="U"); +dU.draw(rotateLabel=false); +\end{asy} +\end{center} +\end{problem} % problem 28.62 + +\begin{solution} +Show that $\psi_1$ satisfies the Schr\"odinger equation +\begin{equation} + -\frac{\hbar}{2m}\frac{d^2\psi}{dx^2} = (E-U)\psi +\end{equation} +in region 1. +\begin{align} + \frac{d\psi_1}{dx} &= ik_1Ae^{ik_1x}-ik_1Be^{-ik_1x} \\ + \frac{d\psi_1^2}{dx^2} &= i^2k_1^2Ae^{ik_1x}+i^2k_1^2Be^{-ik_1x} = -k_1^2\psi \;. +\end{align} +So the Schr\"odinger equation is satisfied if +\begin{align} + k_1^2 \frac{\hbar}{2m} &= E-U_1 = E \\ + k_1 &= \frac{\sqrt{2mE}}{\hbar} +\end{align} + +Show that $\psi_2$ satisfies the Schr\"odinger equation in region 2. +\begin{align} + \frac{d\psi_2}{dx} &= ik_2Ce^{ik_2x} \\ + \frac{d\psi_2^2}{dx^2} &= i^2k_2^2Ce^{ik_2x} = -k_2^2\psi \;. +\end{align} +So the Schr\"odinger equation is satisfied if +\begin{align} + k_2^2 \frac{\hbar}{2m} &= E-U_2 = E-U \\ + k_2 &= \frac{\sqrt{2m(E-U)}}{\hbar} +\end{align} + +Imposing the continuous $\psi$ boundary condition +\begin{equation} + \psi_1(x=0) = \psi_2(x=0) \qquad \rightarrow \qquad + A+B = C +\end{equation} + +Imposing the smooth $\psi$ boundary condition +\begin{equation} + \frac{d\psi_1}{dx}(x=0) = \frac{d\psi_2}{dx}(x=0) \qquad \rightarrow \qquad + ik_1A-ik_1B = ik_2C \qquad \rightarrow \qquad \frac{k_1}{k_2}(A-B) = C +\end{equation} + +Putting this together to find the reflection coefficient +\begin{align} + A+B &= C = \frac{k_1}{k_2}(A-B) \\ + k_2B + k_1B &= k_1A - k_2A \\ + B &= \frac{k_1-k_2}{k_1+k_2}A \\ + \frac{B}{A} &= \frac{k_1-k_2}{k_1+k_2} \\ + R &= \frac{B^2}{A^2} = \p({\frac{B}{A}})^2 = \frac{(k_1-k_2)^2}{(k_1+k_2)^2} \;, +\end{align} +which is what we set out to show. +\end{solution} diff --git a/latex/problems/problem28.V1.tex b/latex/problems/problem28.V1.tex new file mode 100644 index 0000000..2287ca2 --- /dev/null +++ b/latex/problems/problem28.V1.tex @@ -0,0 +1,35 @@ +\begin{problem} +Light of intensity $1.0\U{$\mu$W/cm$^2$}$ falls on a clean metal +surface $2.5\U{cm$^2$}$ in area. Assume that the surface reflects +$95\%$ of the incident light and that only $4\%$ of the absorbed energy +lies above the threshold frequency of the spectrum. \Part{a} How much +energy (in eV) is available for photoelectric effect in this case? +\Part{b} Assume that the photons in the region above the threshold +frequency have an effective wavelength of $250\U{nm}$, how many +electrons will be emitted per second? \Part{c} Find the stopping +potential for sodium ($\phi=2.3\U{eV}$) if the photoelectrons are +produced by the light of wavelength $\lambda=250\U{nm}$. +\end{problem} + +\begin{solution} +\Part{a} +\begin{align} + P_\text{total} &= IA = 2.5\U{$\mu$W} \\ + P_\text{absorbed} &= 0.05\cdot P_\text{total} = 125\U{nW} \\ + P_\text{thresh} &= 0.04\cdot P_\text{absorbed} = 5.00\U{nW} + = \ans{31.2\U{GeV/s}} +\end{align} + +\Part{b} +\begin{align} + E_\text{photon} &= hf = \frac{hc}{\lambda} = 4.96\U{eV} \\ + \Phi &= \frac{P_\text{thresh}}{E_\text{photon}} + = \ans{6.30\E{9}\U{photons/second}} +\end{align} + +\Part{c} +\begin{equation} + \Delta V_s = \frac{E_\text{photon} - \phi}{e} = \frac{(4.96-2.3)\U{eV}}{e} + = \ans{2.7\U{V}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem28.V2.tex b/latex/problems/problem28.V2.tex new file mode 100644 index 0000000..010872b --- /dev/null +++ b/latex/problems/problem28.V2.tex @@ -0,0 +1,36 @@ +\begin{problem} +This problem is one of estimating the time lag (expected classically +but not observed) for the photoelectric effect. Assume that a point +source of light gives $3\U{Watt}=3\U{Joule/sec}$ of light +energy. \Part{a} Assuming uniform radiation in all directions, find +the light intensity in eV/m$^2\cdot$sec at a distance of $2\U{m}$ from +the point source. \Part{b} Assuming some reasonable size for an atom, +find the energy/time incident on the atom for this intensity. \Part{c} +If the work function is $1.6\U{eV}$, how long does it take for this +much energy to be absorbed, assuming that all the energy hitting the +atom is absorbed? +\end{problem} + +\begin{solution} +\Part{a} +The intensity $R=2\U{m}$ from the source is +\begin{equation} + I = \frac{P}{4\pi R^2} = 0.0597\U{W/m$^2$} + = \ans{3.73\E{17}\U{eV/m$^2$}} \;. +\end{equation} + +\Part{b} +The atom presents a cross-sectional surface area of +\begin{equation} + A_a = \pi r^2 \;, +\end{equation} +with $r\approx1\U{\AA}$ ($r_\text{Hydrodgen}=0.529\U{\AA}$). So +\begin{equation} + P_a = IA_a = \ans{11.7\U{meV/s}} +\end{equation} + +\Part{c} +\begin{equation} + \Delta t = \frac{\phi}{P_a} = \ans{137\U{s}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem28.V3.tex b/latex/problems/problem28.V3.tex new file mode 100644 index 0000000..3abb410 --- /dev/null +++ b/latex/problems/problem28.V3.tex @@ -0,0 +1,40 @@ +\begin{problem} +Ultra-violet light of wavelength $\lambda$ incident on an emitter +surface gives rise to photoelectrons with maximum kinetic energy +$2.00\U{eV}$ whereas, for a wavelength $3\lambda/4$, the maximum +kinetic energy of emitted photoelectrons from the same surface is +$3.47\U{eV}$. \Part{a} Determine the wavelength $\lambda$ (in nm) of +these ultraviolet light waves incident on the emitter +surface. \Part{b} What are the photoelectric work function $\phi$ (in +eV) and the threshold wavelength $\lambda_\text{threshold}$ (in nm) of the +photons for photoelectron emission from this emitter surface. \Part{c} If +the wavelength of incident violet light is $400\U{nm}$, find the maximum +kinetic energy (in eV) of emitted photoelectrons. +\end{problem} + +\begin{solution} +\Part{a} +Balancing energy in both cases +\begin{align} + \phi &= \frac{hc}{\lambda} - K_\text{max} = \frac{4hc}{3\lambda} - K_\text{max}' \\ + K_\text{max}' - K_\text{max} &= \frac{1}{\lambda}(\frac{4}{3}hc-hc) \\ + \lambda &= \frac{hc}{3(K_\text{max}' - K_\text{max})} + = \ans{281\U{nm}} \\ + \frac{3}{4}\lambda = \ans{211\U{nm}} +\end{align} + +\Part{b} +\begin{align} + \phi &= \frac{hc}{\lambda} - K_\text{max} + = \ans{2.41\U{eV}} \\ + \phi &= \frac{hc}{\lambda_\text{threshold}} \\ + \lambda_\text{threshold} &= \frac{hc}{\phi} + = \ans{514\U{nm}} \\ +\end{align} + +\Part{c} +\begin{equation} + K_\text{max} = \frac{hc}{\lambda} - \phi + = \ans{0.690\U{eV}} +\end{equation} +\end{solution} diff --git a/latex/problems/problem28.compton-cat.T.tex b/latex/problems/problem28.compton-cat.T.tex new file mode 100644 index 0000000..346e1dc --- /dev/null +++ b/latex/problems/problem28.compton-cat.T.tex @@ -0,0 +1,19 @@ +\begin{problem} +You are working in the radiology department at a vetrinary hospital. The cat you are trying to X-ray refuses to hold still so you volunteer to stand at the table and hold the cat down during the exposure. If X-rays of wavelength $\lambda = 24\U{pm}$ enter the cat from above, what is the wavelength of the photons that enter your head after scattering through a $135\dg$ angle? +\end{problem} + +\begin{solution} +Compton effect +\begin{align} + \lambda' - \lambda_0 &= \frac{h}{m_e c}(1-\cos\theta) \\ + \lambda' &= \lambda_0 + \frac{h}{m_e c}(1-\cos\theta) + = \lambda_0 + \frac{hc}{m_e c^2}(1-\cos\theta) + = 24\U{pm} + \frac{1240\U{eV$\cdot$nm}}{511\U{keV}}(1-\cos 135\dg) + = \ans{28.1\U{pm}} \;. +\end{align} +These are still quite penatrative X-rays, which is why radiologists +try to hold down cats with tape and sandbags when taking radiographs. +Determining the intensity of the scattered beam is more complicated +and would involve some sort of volume integral over the atoms in the +cat and the scattering cross section per atom. +\end{solution} diff --git a/latex/problems/problem28.thornton_and_rex.eqn5.23.T.tex b/latex/problems/problem28.thornton_and_rex.eqn5.23.T.tex new file mode 100644 index 0000000..3a2a4c3 --- /dev/null +++ b/latex/problems/problem28.thornton_and_rex.eqn5.23.T.tex @@ -0,0 +1,77 @@ +\begin{problem} +In the particle-wave duality, localized particles are modeled as wave +packets, with both a group speed and a phase speed. Between Equations +28.13 and 28.16, the text shows that the group speed $v_g$ of a wave +function $\psi$ is the same as the particle speed $u$. Treat the +particle as a non-relativistic de Broglie wave, and use $v_p = \lambda +f$ to show that the phase speed $v_p = u/2 \ne u = v_g$. +\end{problem} % based on Thornton and Rex's 3rd edition Equation 5.34. + +\begin{solution} +Here's a picture of the wave packet, just as a reminder of what we're +talking about. +\begin{center} +\begin{asy} +import graph; + +real u=1cm; +real dp=2; +real f=440; +real v=343; +real plength=5; // packet length scale + +real xMax=4plength; +real width=6u; + +real lambda = v/f; +real k = 2pi/lambda; + +real envelope(real x) { return exp(-(x/plength)**2); } +real amplitude(real x) { return envelope(x) * cos(k*x); } + +size(width,width/2,IgnoreAspect); +scale(Linear, Linear); +xlimits(-xMax,xMax); +ylimits(-1.2,1.2); +xaxis("$x$"); +//yaxis("$y$"); + +draw(graph(envelope,-xMax,xMax, n=200, operator ..),blue); +draw(graph(amplitude,-xMax,xMax, n=800, operator ..),red); + +real vlength = 8lambda; +real vxpos = 3.625lambda; +pair[] vgs = {(vxpos, envelope(vxpos)), (vxpos+vlength, envelope(vxpos))}; +pair[] vps = {(vxpos, amplitude(vxpos)), (vxpos+vlength, amplitude(vxpos))}; + +real scalex(real x) { return x/xMax*(width/2); } + +draw("$v_g$", graph(vgs), N, Arrow); +draw("$v_p$", graph(vps), S, Arrow); +\end{asy} +\end{center} +We start with the given definition of the phase velocity +\begin{equation} + v_p = \lambda f \;. +\end{equation} +DeBroglie tells us (Equations 28.10 and 28.11) +\begin{align} + \lambda &= \frac{h}{p} \\ + E &= hf \;, +\end{align} +so +\begin{equation} + v_p = \lambda f = \frac{hf}{p} = \frac{E}{p} \;. +\end{equation} +Then we use the non-relativistic kinetic energy and momentum +\begin{align} + E &= \frac{p^2}{2m} \\ + p &= mu \\ + v_p &= \frac{E}{p} = \frac{\frac{p^2}{2m}}{p} = \frac{p}{2m} + = \frac{mu}{2m} = \frac{u}{2} \;. +\end{align} + +As a side note, I have no idea why the text uses $u$ instead of $v$ +for Equation 28.16, but I though I should stick with it here to avoid +adding to the confusion. +\end{solution} diff --git a/latex/problems/problem28.xray.T.tex b/latex/problems/problem28.xray.T.tex new file mode 100644 index 0000000..dd32614 --- /dev/null +++ b/latex/problems/problem28.xray.T.tex @@ -0,0 +1,38 @@ +\begin{problem} +X-rays generation (e.g. for medical imaging) can be modeled as an +inverse photoelectric effect (basically the regular photoelectric +effect played backwards in time). An electron beam is fired into an +anode, which absorbs the electrons and emits radiation (the X-rays). +If the electrons are accelerated by $50\U{kV}$ towards a Tungsten +surface ($\phi = 4.5\U{eV}$), find the wavelength of the emitted +photons predicted by this model. +\begin{center} +\includegraphics[width=2in]{xray_tube} % http://en.wikipedia.org/wiki/File:Roentgen-Roehre.svg +\end{center} + +\emph{HINT}. Follow the energy flow through the system. Ignore anode +heating. +\end{problem} % based on Urbanc's X-ray production lecture notes + +\begin{solution} +The kinetic energy of the incoming electrons (due to the accelerating +voltage) is $e\Delta V = 50\U{keV}$. They gain an additional +$\phi=4.5\U{eV}$ of energy while ``dropping into'' the lower energy +bound states of the metal, but this energy is much less than the energy +from the accelerating voltage, so we'll ignore it. + +Assuming all the energy from the electons is converted into photons +(since we're ignoring anode heating), we get +\begin{align} + E &= hf = \frac{hc}{\lambda} \\ + \lambda &= \frac{hc}{E} = \frac{hc}{e\Delta V} + = \frac{1240\U{eV$\cdot$nm}}{5\U{keV}} + = \ans{0.0248\U{nm}} +\end{align} + +The production of X-rays in tubes is more accurately modeled as a +combination of X-ray flourescence and bremsstrahlung. Our ``inverse +photoelectric effect'' prediction is the gives the shortest wavelength +(highest frequency) of the emitted photons. For more details, see +\url{http://en.wikipedia.org/wiki/X-ray#Medical_physics}. +\end{solution} diff --git a/latex/problems/problem28.xray.T.xray_tube.png b/latex/problems/problem28.xray.T.xray_tube.png new file mode 100644 index 0000000..245f63d Binary files /dev/null and b/latex/problems/problem28.xray.T.xray_tube.png differ diff --git a/latex/problems/question09.11.T.tex b/latex/problems/question09.11.T.tex new file mode 100644 index 0000000..396febd --- /dev/null +++ b/latex/problems/question09.11.T.tex @@ -0,0 +1,48 @@ +\begin{problem} +By what fraction does the mass of a $m=10\U{g}$, $k=500\U{N/m}$ spring +increase when it is compressed by $1\U{cm}$? +\begin{center} +\begin{asy} +import Mechanics; +real u = .5cm; + +Spring Su = Spring(pFrom=(0,0), pTo=(4u,0), k=500, L="$m$"); +Spring Sc = Spring(pFrom=(0,-2u), pTo=(3u,-2u), k=500, L="$m'$"); +Distance d = Distance(pFrom=(4u,-2u), pTo=(3u,-2u), scale=u, L=rotate(90)*Label("1 cm")); +Su.draw(); +Sc.draw(); +d.draw(); +\end{asy} +\end{center} +\end{problem} % Based on Q9.11 + +\begin{solution} +Compression increases the potential energy of the spring by +\begin{equation} + \Delta U = \frac{1}{2} k \Delta x^2 + = \frac{1}{2} \cdot 500\U{N/m} \cdot \p({0.01\U{m}})^2 + = 25.0\U{mJ} \;. +\end{equation} +From Einstein's mass-energy equivalence, increasing the spring's +energy must also increase its mass, since mass and energy are two ways +of talking about the same stuff. +\begin{align} + \Delta E &= \Delta m c^2 \\ + \Delta m &= \frac{\Delta E}{c^2} = \frac{\Delta U}{c^2} + = \frac{0.025\U{J}}{(3\E{8}\U{m/s})^2} + = 2.78\E{-19}\U{kg} \;. +\end{align} +So the fractional mass increase is +\begin{equation} + \frac{\Delta m}{m} = \frac{2.78\E{-19}\U{kg}}{0.010\U{kg}} + = \ans{2.78\E{-17}} \;. +\end{equation} + +This mass difference is quite small, which is why it took so long to +come up with the $E=mc^2$ idea. Notice, though, that the mass +difference is equal to the mass of 16 billion protons (at +$1.67\E{-27}\U{kg}$ a pop). Nuclear reactions achieve their high +energies through small energy changes for an \emph{enourmous} number +of nuclei (on the order of Avogadro's number $N_A = +6.022\E{23}\U{particles/mole}$) +\end{solution} diff --git a/latex/problems/question12.07.T.tex b/latex/problems/question12.07.T.tex new file mode 100644 index 0000000..f24e2dc --- /dev/null +++ b/latex/problems/question12.07.T.tex @@ -0,0 +1,41 @@ +\begin{problem} +A frictionless block-spring system oscillates with amplitude $A$. If +the mass of the block is doubled without changing the amplitude, +\Part{a} does the total energy change? +\Part{b} does the frequency of oscillation change? +\begin{center} +\begin{asy} +import Mechanics; +real u = 1cm; + +Surface table = Surface(pFrom=(0,0), pTo=(2u,0)); +Surface wall = Surface(pFrom=(0,0.5u), pTo=(0,0), thickness=0); + +real blockside = 0.3u; +pair blockpos = (1.5u,blockside/2); +path block = shift(blockpos-blockside*(.5,.5))*scale(blockside)*unitsquare; +Spring s = Spring(pFrom=(0,blockpos.y), pTo=blockpos, width=0.6blockside); + +table.draw(); +wall.draw(); +s.draw(); +filldraw(block); +\end{asy} +\end{center} +\end{problem} % Derived from Ch. 12, Question 7. + +\begin{solution} +\Part{a} +The total energy is equal to the spring potential energy at maximum +extension (when the kinetic energy is zero), so $E_\text{T} = +\frac{1}{2}kA^2$. Neither the spring constant, nor the amplitude +changed, so \ans{the total energy is unaffected}. + +\Part{b} +The heavier mass will move more slowly under the influnce of the same +spring, so the frequency is smaller for the bigger mass. Quantitatively +\begin{equation} + f = \frac{1}{2\pi}\omega = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\;, +\end{equation} +so doubling the mass reduces the frequency to $\ans{f' = \frac{f}{\sqrt{2}}}$. +\end{solution} diff --git a/latex/problems/section11.05.T.tex b/latex/problems/section11.05.T.tex new file mode 100644 index 0000000..a51ca84 --- /dev/null +++ b/latex/problems/section11.05.T.tex @@ -0,0 +1,54 @@ +\begin{problem} +\emph{BONUS PROBLEM}. Use Bohr's assumptions in Section 11.5 to +derive a formula for the allowed energy levels in singly ionized +helium (He$^+$). +\end{problem} % based on the Bohr model derivation of the Hydrogen + % energy levels in Section 11.5. + +\begin{solution} +Following the derivation of hydrogen energy levels in the book, we +have our analog for Equation 11.19 (total energy of the electron) +\begin{equation} + E = K + U_e = \frac{1}{2} m_e v^2 - k_e \frac{Ze^2}{r} \label{eq.bohr_energy} +\end{equation} +where $Z$ is the atomic number of our element (1 for H, 2 for He). + +Applying Newton's second law and Coulomb's law to the electron's +circular motion, +\begin{equation} + \frac{m_e v^2}{r} = F = \frac{k_e Ze^2}{r^2} \label{eq.bohr_newton} \;, +\end{equation} +so the kinetic energy of the electron is (our Equation 11.20 analog) +\begin{equation} + K = \frac{1}{2} m_e v^2 = \frac{k_e Ze^2}{2r} = \frac{U_e}{2} \;. +\end{equation} + +Plugging this expression for $K$ into eq.~\ref{eq.bohr_energy} +\begin{equation} + E = -\frac{k_e Ze^2}{2r} +\end{equation} + +We can find the allowed value for $r$ by substituting angular momentum +conservation $m_e v r = n \hbar$ into eq.~\ref{eq.bohr_newton} +\begin{align} + m_e v^2 &= \frac{k_e Ze^2}{r} \\ + m_e \p({\frac{n\hbar}{m_e r}})^2 &= \frac{k_e Ze^2}{r} \\ + n^2 \hbar^2 &= k_e Z e^2 r m_e \\ + r = \frac{n^2 \hbar^2}{m_e k_e Z e^2} \;, +\end{align} +which is our analog to Equation 11.22. + +Plugging this expression for $r$ into our electron energy forumula +\begin{align} + E_n &= -\frac{m_e k_e^2 Z^2 e^4}{2 n^2 \hbar^2} = Z^2 \cdot E_{n\text{H}} \\ + E_{n,\text{He}} &= 2^2 E_{n,\text{H}} = \ans{-\frac{54.42\U{eV}}{n^2}} +\end{align} +which is our analog to Equation 11.25. + +Note that the only difference between this derivation and the book's +hydrogen derivation is the replacement $k_e e^2 \rightarrow k_e Ze^2$ +in Coulomb's law. This is also the only place the constant $e$ comes +into the derivation. Simply matching and replacing $e^2$ with $Ze^2$ +in Equations 11.23 and 11.24 would produce the correct answer without +following every step of the derivation. +\end{solution} diff --git a/latex/rec/Makefile b/latex/rec/Makefile new file mode 100644 index 0000000..c110949 --- /dev/null +++ b/latex/rec/Makefile @@ -0,0 +1,21 @@ +# give numbers for assigned recitations +REC_NUMS = +# give numbers for recitations whose solutions should be posted +# (don't install source until the solutions should be published) +SOLN_NUMS = + +INSTALL_DIR := $(INSTALL_DIR)/doc/rec +export INSTALL_DIR + +install : + @for i in $(REC_NUMS:%=rec%); do \ + echo "make install-probs in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install-probs); done + @for i in $(SOLN_NUMS:%=rec%); do \ + echo "make install-solns in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install-solns); done + +clean : + @for i in $(REC_NUMS:%=rec%); do \ + echo "make clean in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) clean); done diff --git a/latex/rec/rec1/Makefile b/latex/rec/rec1/Makefile new file mode 100644 index 0000000..7d9a65a --- /dev/null +++ b/latex/rec/rec1/Makefile @@ -0,0 +1,41 @@ +THIS_DIR = $(shell basename $(PWD)) +RECITATION_NUMBER = $(THIS_DIR:rec%=%) +SOURCE_FILES = all_problems.tex probs.tex sols.tex problem[0-9].tex +OTHER_FILES = Makefile +DIST_FILES = $(SOURCE_FILES) $(OTHER_FILES) +DIST_FILE = $(THIS_DIR)_source.tar.gz +DIST_DIR = rec + +all : sols.pdf probs.pdf + +view : all + xpdf probs.pdf & + xpdf sols.pdf & + +%.pdf : %.tex $(SOURCE_FILES) + pdflatex $(patsubst %.tex,%,$<) + asy $(patsubst %.tex,%,$<) + pdflatex $(patsubst %.tex,%,$<) + +semi-clean : + rm -f *.log *.aux *.out *.thm *.toc *.pre *_[0-9]_.tex *.asy + +clean : semi-clean + rm -f *.pdf $(DIST_FILE) $(DIST_DIR) install* + +$(DIST_FILE) : $(DIST_FILES) + mkdir $(DIST_DIR) + cp -Lrp $^ $(DIST_DIR) + tar -chozf $@ $(DIST_DIR) + rm -rf $(DIST_DIR) + +install : install-probs install-solns + +install-probs : probs.pdf + scp -p $< $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR)/rec$(RECITATION_NUMBER)_problems.pdf + @date > $@ + +install-solns : sols.pdf $(DIST_FILE) + scp -p $< $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR)/rec$(RECITATION_NUMBER)_solutions.pdf + scp -p $(DIST_FILE) $(INSTALL_USER)@$(INSTALL_HOST):$(SOURCE_DIR) + @date > $@ diff --git a/latex/rec/rec1/all_problems.tex b/latex/rec/rec1/all_problems.tex new file mode 100644 index 0000000..f314bd4 --- /dev/null +++ b/latex/rec/rec1/all_problems.tex @@ -0,0 +1,18 @@ +\usepackage[author={W. Trevor King}, + coursetitle={Physics 201}, + classtitle={Recitation 1}, + subheading={Chapter 12}]{problempack} +\usepackage[inline]{asymptote} +\usepackage{wtk_cmmds} + +\begin{document} + +\maketitle + +\input{problem1} +\input{problem2} +\input{problem3} +\input{problem4} +\input{problem5} + +\end{document} diff --git a/latex/rec/rec1/problem1.tex b/latex/rec/rec1/problem1.tex new file mode 120000 index 0000000..9d042e9 --- /dev/null +++ b/latex/rec/rec1/problem1.tex @@ -0,0 +1 @@ +../../problems/problem12.02.tex \ No newline at end of file diff --git a/latex/rec/rec1/problem2.tex b/latex/rec/rec1/problem2.tex new file mode 120000 index 0000000..fad3569 --- /dev/null +++ b/latex/rec/rec1/problem2.tex @@ -0,0 +1 @@ +../../problems/problem12.05.tex \ No newline at end of file diff --git a/latex/rec/rec1/problem3.tex b/latex/rec/rec1/problem3.tex new file mode 120000 index 0000000..9ad2afd --- /dev/null +++ b/latex/rec/rec1/problem3.tex @@ -0,0 +1 @@ +../../problems/problem12.12.tex \ No newline at end of file diff --git a/latex/rec/rec1/problem4.tex b/latex/rec/rec1/problem4.tex new file mode 120000 index 0000000..1964d5c --- /dev/null +++ b/latex/rec/rec1/problem4.tex @@ -0,0 +1 @@ +../../problems/problem12.15.tex \ No newline at end of file diff --git a/latex/rec/rec1/problem5.tex b/latex/rec/rec1/problem5.tex new file mode 120000 index 0000000..64a05d8 --- /dev/null +++ b/latex/rec/rec1/problem5.tex @@ -0,0 +1 @@ +../../problems/problem12.18.tex \ No newline at end of file diff --git a/latex/rec/rec1/probs.tex b/latex/rec/rec1/probs.tex new file mode 100644 index 0000000..fa21ac1 --- /dev/null +++ b/latex/rec/rec1/probs.tex @@ -0,0 +1,5 @@ +\documentclass[letterpaper, 10pt]{article} + +\PassOptionsToPackage{nosolutions}{problempack} + +\input{all_problems} diff --git a/latex/rec/rec1/sols.tex b/latex/rec/rec1/sols.tex new file mode 100644 index 0000000..1596819 --- /dev/null +++ b/latex/rec/rec1/sols.tex @@ -0,0 +1,5 @@ +\documentclass[letterpaper, 10pt]{article} + +\PassOptionsToPackage{solutions,loose}{problempack} + +\input{all_problems} diff --git a/latex/syllabus/Makefile b/latex/syllabus/Makefile new file mode 100644 index 0000000..4ebddb1 --- /dev/null +++ b/latex/syllabus/Makefile @@ -0,0 +1,17 @@ +PDFS = syllabus.pdf +INSTALL_DIR := $(INSTALL_DIR)/doc + +all : $(PDFS) + +.SECONDEXPANSION: +$(PDFS) : $$(patsubst %.pdf,%.tex,$$@) + pdflatex $< + pdflatex $< + +clean : + rm -rf *.log *.aux *.out *.pdf install + +install : $(PDFS) + scp -p $^ $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR) + scp -p $(^:%.pdf=%.tex) $(INSTALL_USER)@$(INSTALL_HOST):$(SOURCE_DIR) + @date > $@ diff --git a/pdf/Makefile b/pdf/Makefile new file mode 100644 index 0000000..866110e --- /dev/null +++ b/pdf/Makefile @@ -0,0 +1,11 @@ +SUBDIRS = lab lec exam + +install : + @for i in $(SUBDIRS); do \ + echo "make install in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install); done + +clean : + @for i in $(SUBDIRS); do \ + echo "make clean in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) clean); done diff --git a/pdf/README b/pdf/README new file mode 100644 index 0000000..940822b --- /dev/null +++ b/pdf/README @@ -0,0 +1 @@ +This directory is for PDFs for which we do not have LaTeX source. diff --git a/pdf/exam/Makefile b/pdf/exam/Makefile new file mode 100644 index 0000000..639c020 --- /dev/null +++ b/pdf/exam/Makefile @@ -0,0 +1,16 @@ +# give numbers for the exams +EXAM_NUMS = +OTHER_DIRS = + +INSTALL_DIR := $(INSTALL_DIR)/doc/exam +export INSTALL_DIR + +install : + @for i in $(EXAM_NUMS:%=exam%) $(OTHER_DIRS); do \ + echo "make install in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install); done + +clean : + @for i in $(EXAM_NUMS:%=exam%) $(OTHER_DIRS); do \ + echo "make clean in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) clean); done diff --git a/pdf/exam/README b/pdf/exam/README new file mode 100644 index 0000000..159d497 --- /dev/null +++ b/pdf/exam/README @@ -0,0 +1 @@ +See ../lab/README and ../lec/README. diff --git a/pdf/exam/exam1/Makefile b/pdf/exam/exam1/Makefile new file mode 100644 index 0000000..54ac527 --- /dev/null +++ b/pdf/exam/exam1/Makefile @@ -0,0 +1,36 @@ +THIS_DIR = $(shell basename $(PWD)) +EXAM_NUMBER = $(THIS_DIR:exam%=%) +SOURCE_FILES = $(shell echo *.jpg) +OTHER_FILES = Makefile +DIST_FILES = $(SOURCE_FILES) $(OTHER_FILES) +DIST_FILE = $(THIS_DIR)_source.tar.gz +DIST_DIR = hwk + +all : out.pdf + +view : all + pdfinfo out.pdf + xpdf out.pdf & + +%.pdf : %.jpg + convert $< $@ + +out.pdf : $(SOURCE_FILES:%.jpg=%.pdf) + gs -dBATCH -dNOPAUSE -q -sDEVICE=pdfwrite -sOutputFile=$@ $^ + +semi-clean : + rm -f *.log + +clean : semi-clean + rm -f *.pdf $(DIST_FILE) $(DIST_DIR) install* + +$(DIST_FILE) : $(DIST_FILES) + mkdir $(DIST_DIR) + cp -rp $^ $(DIST_DIR) + tar -chozf $@ $(DIST_DIR) + rm -rf $(DIST_DIR) + +install : out.pdf $(DIST_FILE) + scp -p $< $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR)/exam$(EXAM_NUMBER)_solutions.pdf + scp -p $(DIST_FILE) $(INSTALL_USER)@$(INSTALL_HOST):$(SOURCE_DIR) + @date > $@ diff --git a/pdf/lab/Makefile b/pdf/lab/Makefile new file mode 100644 index 0000000..6af16c0 --- /dev/null +++ b/pdf/lab/Makefile @@ -0,0 +1,16 @@ +# give numbers for the labs +LAB_NUMS = +OTHER_DIRS = sample-lab + +INSTALL_DIR := $(INSTALL_DIR)/doc/lab +export INSTALL_DIR + +install : + @for i in $(LAB_NUMS:%=lab%) $(OTHER_DIRS); do \ + echo "make install in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install); done + +clean : + @for i in $(LAB_NUMS:%=lab%) $(OTHER_DIRS); do \ + echo "make clean in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) clean); done diff --git a/pdf/lab/README b/pdf/lab/README new file mode 100644 index 0000000..2e28c31 --- /dev/null +++ b/pdf/lab/README @@ -0,0 +1,26 @@ +Facilitate posting of source-less lab PDFs. Usage example: + +Place Lab 1 PDFs into the lab1 subdirectory + $ ls lab1 + Makefile prelab.pdf procedure.pdf report.pdf + +Check that a reasonable PDF is generated + $ cd lab1 + $ make && xpdf out.pdf + +Add lab1 to the to-be-posted list by adding a `1' to `LAB_NUMS' in +./Makefile, so that line looks like + LAB_NUMS = 1 + +Future course-wide makes should publish the Lab 1 PDFs and source to +the appropriate locations in the course website. + +To add Lab 2 PDFs, simply create a lab2 subdirectory, copy over the +lab1 Makefile, and repeat the above procedure using `2's instead of +`1's. + + +The framework also supports bonus directories such as sample-lab by +adding them to the `OTHER_DIRS' variable in Makefile, e.g. + OTHER_DIRS = sample-lab +See the sample-lab subdirectory for more details. diff --git a/pdf/lab/lab1/Makefile b/pdf/lab/lab1/Makefile new file mode 100644 index 0000000..48fbee8 --- /dev/null +++ b/pdf/lab/lab1/Makefile @@ -0,0 +1,30 @@ +THIS_DIR = $(shell basename $(PWD)) +LAB_NUMBER = $(THIS_DIR:lab%=%) +SOURCE_FILES = +PDFS = prelab.pdf procedure.pdf report.pdf +OTHER_FILES = Makefile $(PDFS) +DIST_FILES = $(SOURCE_FILES) $(OTHER_FILES) +DIST_FILE = $(THIS_DIR)_source.tar.gz +DIST_DIR = lab + +clean : + rm -rf $(DIST_FILE) $(DIST_DIR) install* + +$(DIST_FILE) : $(DIST_FILES) + mkdir $(DIST_DIR) + cp -rp $^ $(DIST_DIR) + tar -chozf $@ $(DIST_DIR) + rm -rf $(DIST_DIR) + +install : install-pdfs install-source + +install-pdfs : $(PDFS) + @for i in $(PDFS); do \ + echo " install $(THIS_DIR)/$$i as lab$(LAB_NUMBER)_$$i..."; \ + scp -p $$i $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR)/lab$(LAB_NUMBER)_$$i; \ + done + @date > $@ + +install-source : $(DIST_FILE) + scp -p $(DIST_FILE) $(INSTALL_USER)@$(INSTALL_HOST):$(SOURCE_DIR) + @date > $@ diff --git a/pdf/lab/sample-lab/Makefile b/pdf/lab/sample-lab/Makefile new file mode 100644 index 0000000..34a84f3 --- /dev/null +++ b/pdf/lab/sample-lab/Makefile @@ -0,0 +1,29 @@ +THIS_DIR = $(shell basename $(PWD)) +SOURCE_FILES = +PDFS = report.pdf +OTHER_FILES = Makefile $(PDFS) +DIST_FILES = $(SOURCE_FILES) $(OTHER_FILES) +DIST_FILE = $(THIS_DIR)_source.tar.gz +DIST_DIR = lab + +clean : + rm -rf $(DIST_FILE) $(DIST_DIR) install* + +$(DIST_FILE) : $(DIST_FILES) + mkdir $(DIST_DIR) + cp -rp $^ $(DIST_DIR) + tar -chozf $@ $(DIST_DIR) + rm -rf $(DIST_DIR) + +install : install-pdfs install-source + +install-pdfs : $(PDFS) + @for i in $(PDFS); do \ + echo " install $(THIS_DIR)/$$i as $(THIS_DIR)_$$i..."; \ + scp -p $$i $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR)/$(THIS_DIR)_$$i; \ + done + @date > $@ + +install-source : $(DIST_FILE) + scp -p $(DIST_FILE) $(INSTALL_USER)@$(INSTALL_HOST):$(SOURCE_DIR) + @date > $@ diff --git a/pdf/lab/sample-lab/report.pdf b/pdf/lab/sample-lab/report.pdf new file mode 100644 index 0000000..73ac722 Binary files /dev/null and b/pdf/lab/sample-lab/report.pdf differ diff --git a/pdf/lec/Makefile b/pdf/lec/Makefile new file mode 100644 index 0000000..40559ab --- /dev/null +++ b/pdf/lec/Makefile @@ -0,0 +1,16 @@ +# give numbers for the lecs +LEC_NUMS = +OTHER_DIRS = + +INSTALL_DIR := $(INSTALL_DIR)/doc/lec +export INSTALL_DIR + +install : + @for i in $(LEC_NUMS:%=lec%) $(OTHER_DIRS); do \ + echo "make install in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) install); done + +clean : + @for i in $(LEC_NUMS:%=lec%) $(OTHER_DIRS); do \ + echo "make clean in $$i..."; \ + (cd $$i; $(MAKE) $(MFLAGS) clean); done diff --git a/pdf/lec/README b/pdf/lec/README new file mode 100644 index 0000000..e19e2a2 --- /dev/null +++ b/pdf/lec/README @@ -0,0 +1,25 @@ +Facilitate posting of scanned lecture notes. Usage example: + +Scan Lecture 1 notes page-by-page into the lec1 subdirectory + $ ls lec1 + Makefile scan1.png scan2.png ... + +If your scans are JPEGs instead of PNGs, you will need to make +approptiate changes to lec1/Makefile. + $ sed -i 's/[.]png/.jpg/' lec1/Makefile + +Check that a reasonable PDF is generated + $ cd lec1 + $ make && xpdf out.pdf + +Add lec1 to the to-be-posted list by adding a `1' to `LEC_NUMS' in +./Makefile, so that line looks like + LEC_NUMS = 1 + +Future course-wide makes should publish the Lecture 1 PDF and source +to the appropriate locations in the course website. + +To add Lecture 2 notes, simply create a lec2 subdirectory, copy over +the lec1 Makefile, and repeat the above procedure using `2's instead +of `1's. + diff --git a/pdf/lec/lec1/Makefile b/pdf/lec/lec1/Makefile new file mode 100644 index 0000000..df751ea --- /dev/null +++ b/pdf/lec/lec1/Makefile @@ -0,0 +1,36 @@ +THIS_DIR = $(shell basename $(PWD)) +LECTURE_NUMBER = $(THIS_DIR:lec%=%) +SOURCE_FILES = $(shell echo *.png) +OTHER_FILES = Makefile +DIST_FILES = $(SOURCE_FILES) $(OTHER_FILES) +DIST_FILE = $(THIS_DIR)_source.tar.gz +DIST_DIR = hwk + +all : out.pdf + +view : all + pdfinfo out.pdf + xpdf out.pdf & + +%.pdf : %.png + convert $< $@ + +out.pdf : $(SOURCE_FILES:%.png=%.pdf) + gs -dBATCH -dNOPAUSE -q -sDEVICE=pdfwrite -sOutputFile=$@ $^ + +semi-clean : + rm -f *.log + +clean : semi-clean + rm -f *.pdf $(DIST_FILE) $(DIST_DIR) install* + +$(DIST_FILE) : $(DIST_FILES) + mkdir $(DIST_DIR) + cp -rp $^ $(DIST_DIR) + tar -chozf $@ $(DIST_DIR) + rm -rf $(DIST_DIR) + +install : out.pdf $(DIST_FILE) + scp -p $< $(INSTALL_USER)@$(INSTALL_HOST):$(INSTALL_DIR)/lec$(LECTURE_NUMBER)_slides.pdf + scp -p $(DIST_FILE) $(INSTALL_USER)@$(INSTALL_HOST):$(SOURCE_DIR) + @date > $@