--- /dev/null
+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
--- /dev/null
+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.
--- /dev/null
+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 > $@
--- /dev/null
+Construct the atom feed using
+ atomgen -o atom.xml <command> <command options>
+See
+ atomgen --help
+for more information.
+
+For example:
+ atomgen -o atom.xml new --title 'Physics 201' --author 'W. Trevor King' \
+ http://www.physics.drexel.edu/~wking/phys201
+ echo "Changes to the Phys201 website will be noted in this feed." | \
+ atomgen -o atom.xml add -i atom.xml 'Feed purpose' \
+ http://www.physics.drexel.edu/~wking/phys201
+
+
+You can send automatic emails to your students when you publish new
+announcements in the atom feed. The best way I have found to date
+consists of monitoring the posted atom.xml file with rss2email
+ http://rss2email.infogami.com/
+I setup r2e to monitor the feed and email me when there's a change.
+r2e runs every 15 minutes in a cron job
+ 15 * * * * /usr/bin/r2e run
+Then I set up a procmail rule to forward the mail off to the list
+ :0
+ * ^From: "Physics 201, W. Trevor King" <rss2email@null.invalid>$
+ ! `grep -v '^#' $HOME/course/announcements/addresses.txt`
+ # ^--- Make sure those are backticks, BTW (ASCII 96) ---^
--- /dev/null
+# Administrators
+John JJ Smith <john.j.j.smith@drexel.edu>
+Johnny Appleseed <johnny.appleseed@drexel.edu>
+# Students
+Jack <jack@drexel.edu>
+Jill <jill@drexel.edu>
--- /dev/null
+#!/bin/sh
+#
+# Set-up some course-specific details for announcement/script/*.sh
+# Not intended to be used directly
+
+# No trailing slash on the course website, please.
+COURSE_WEBSITE='http://www.physics.drexel.edu/~wking/course'
+COURSE_TITLE='Physics 201'
+ATOM_AUTHOR='W. Trevor King'
--- /dev/null
+#!/bin/sh
+#
+# Create the initial posting for a course. Use carefully, this is
+# just meant to save some typing, and it might not fit into your
+# layout without a bit of tweaking.
+#
+# From the announcements directory
+# usage: script/initial-post.sh
+
+. script/course-details.sh
+
+echo "atomgen -o atom.xml new --title \"$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"
--- /dev/null
+#!/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 <dir> <index> [<note>]
+# where <dir> is one of hwk, lec, rec, lab, exam;
+# <index> is an integer; and <note> 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 <dir> <index> [<note>]"
+ 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"
--- /dev/null
+AddType application/x-httpd-php .shtml
--- /dev/null
+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 $@
--- /dev/null
+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:
+<snip doc/lab/lab1_comment>
+<p>The Interference lab starts on <b>Wed., April 08, 2009</b>.</p>
+<p>
+Wednesday, April 8, Thursday, April 9 and Friday, April 10, 2009:
+EVEN number sections 060, 062, 064, and 66.
+</p>
+<p>
+Wednesday, April 15, Thursday, April 16, and Friday, April 17, 2009:
+ODD number sections 063, 065, 071, and 073.
+</p>
+</snip>
+
+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/).
--- /dev/null
+<!--#include virtual="shared/header.shtml"-->
+
+<?php
+include ("php/atom.php");
+$s=simplexml_load_file('xml/atom.xml');
+printAtom("Announcements", $s, "Feed purpose");
+?>
+
+<!--#include virtual="shared/footer.shtml"-->
--- /dev/null
+<!--#include virtual="shared/header.shtml"-->
+
+<h3>Homeworks</h3>
+
+<?php
+include('php/section_docs.php');
+printSectionDocs('Homework', 'doc/hwk', 'hwk', 20);
+?>
+
+<!--#include virtual="shared/footer.shtml"-->
--- /dev/null
+<!--#include virtual="shared/header.shtml"-->
+
+<br />
+
+<?php
+include('php/people.php');
+$use_pictures = true;
+printPeopleStart($use_picture);
+$s=simplexml_load_file('xml/profs.xml');
+printPeople('Professors', $s, $use_pictures);
+$s=simplexml_load_file('xml/TAs.xml');
+printPeople('Teaching Assistants', $s, $use_pictures);
+printPeopleEnd($use_picture);
+?>
+
+<p style="clear:both;">
+For any questions about the webpage, contact
+<?php include('php/webmaster.php') ?>.
+</p>
+
+<!--#include virtual="shared/footer.shtml"-->
--- /dev/null
+<!--#include virtual="shared/header.shtml"-->
+
+<h3>Exams</h3>
+
+<?php
+include('php/section_docs.php');
+printSectionDocs('Exam', 'doc/exam', 'exam', 20);
+?>
+
+<!--#include virtual="shared/footer.shtml"-->
--- /dev/null
+<!--#include virtual="shared/header.shtml"-->
+
+<h3>Labs</h3>
+
+<ul>
+ <li>All students should be present in Disque-820A - Phys201
+ Laboratory - a few minutes before the start of their scheduled lab.
+ </li>
+ <li>Please note that pre-lab work must be completed before you go to
+ the lab to perform the experiment.
+ </li>
+ <li>All students in a group must actively participate in the lab work.
+ </li>
+ <li>
+ 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
+ <a href="doc/lab/sample_lab_report.pdf">sample lab report</a>.
+ </li>
+ <li>
+ No grade will be given for incomplete pre-lab work or lab reports.
+ </li>
+</ul>
+
+<?php
+include('php/section_docs.php');
+printSectionDocs('Lab', 'doc/lab', 'lab', 20);
+?>
+
+<!--#include virtual="shared/footer.shtml"-->
--- /dev/null
+<!--#include virtual="shared/header.shtml"-->
+
+<h3>Lectures</h3>
+
+<?php
+include('php/section_docs.php');
+printSectionDocs('Lecture', 'doc/lec', 'lec', 20);
+?>
+
+<!--#include virtual="shared/footer.shtml"-->
--- /dev/null
+<!--#include virtual="shared/header.shtml"-->
+
+<h2>Physics 201 - Fundamentals of Physics III</h2>
+<?php include ("php/quarter.php") ?>
+
+<p>
+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.
+</p>
+
+<h4>Resources</h4>
+<ul>
+ <li><a href="announcements.shtml">Announcements</a></li>
+ <li><a href="recitations.shtml">Recitations</a></li>
+ <li><a href="homeworks.shtml">Homeworks</a></li>
+ <li><a href="../contact.shtml">Contact</a></li>
+</ul>
+
+
+<h4>Source code</h4>
+<p>
+For those who are interested, the source code used to generate the
+content of this page is available <a href="source">here</a>.
+</p>
+
+<!--#include virtual="shared/footer.shtml"-->
--- /dev/null
+<?php
+/*
+ * Pretty-print atom feeds in XHTML. For example:
+ * include ('php/atom.xml');
+ * $s=simplexml_load_file('xml/atom.xml');
+ * printAtom("Announcements", $s, "Feed purpose");
+ */
+
+function printContent($xml_data){
+ $xml_parser = xml_parser_create();
+ xml_parse_into_struct($xml_parser, $xml_data, $vals, $index);
+ xml_parser_free($xml_parser);
+ //echo "Index array\n";
+ //print_r($index);
+ //echo "\nVals array\n";
+ //print_r($vals);
+ $content = $vals[1]["value"];
+ echo "<p>\n$content\n</p>\n";
+}
+
+function printAtom($feed_title, $feed, $ignored_title){
+ echo "<h3>$feed_title</h3><p></p>\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 == $ignored_title)
+ continue;
+ echo "<h5><a href=\"$link\">$title</a> $time</h5>\n";
+ printContent($entry->content->asXML());
+ }
+}
+
+?>
--- /dev/null
+<?php
+/*
+ * Pretty-print lists of people from xml data. For example:
+ * include('php/people.php');
+ * $use_pictures = true;
+ * printPeopleStart($use_picture);
+ * $s=simplexml_load_file('xml/profs.xml');
+ * printPeople('Professors', $s, $use_pictures);
+ * $s=simplexml_load_file('xml/TAs.xml');
+ * printPeople('Teaching Assistants', $s, $use_pictures);
+ * printPeopleEnd($use_picture);
+ *
+ * simplexml_load_file idea from 2008 courselist page
+ * http://www.physics.drexel.edu/students/courses/
+ * COPYRIGHT © 2008 DREXEL UNIVERSITY DEPARTMENT OF PHYSICS
+ * Design by Daniel J. Cross
+ */
+function printPeopleStart($use_pictures){
+ if ($use_pictures == false)
+ echo "<table>\n";
+}
+function printPeopleEnd($use_pictures){
+ if ($use_pictures == false)
+ echo "</table>\n";
+}
+function printPeople($title, $people, $use_pictures){
+ if ($use_pictures == true)
+ echo " <h3 style=\"clear: both;\">$title</h3>\n";
+ else
+ echo " <tr><th colspan=\"3\">$title</th></tr>\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 ($use_pictures == true) {
+ echo " <div class=\"person\" style=\"clear: both;\">\n";
+ echo " <img alt=\"$name\" src=\"$picture\" width=\"100px\" height=\"130px\" style=\"float:left;margin-right:10px;border:1px solid black;\"/>\n";
+ echo " <p>\n";
+ echo " <b><a$href>$name</a></b><br/>\n";
+ echo " Email: $email<br/>\n";
+ echo " Office: $office<br/>\n";
+ echo " Hours: $hours<br/>\n";
+ echo " Extension: $extension<br/>\n";
+ echo " </p>\n";
+ echo " </div>\n";
+ } else {
+ echo " <tr><td><a$href>$name</a></td><td>$office</td><td>$email</td></tr>\n";
+ if (strlen($hours) > 0){
+ echo " <tr><td></td><td colspan=\"2\">Office hours: $hours</td></tr>\n";
+ }
+ }
+ }
+}
+?>
--- /dev/null
+<?php
+// return a string representing the current quarter
+echo "Spring 2009";
+?>
--- /dev/null
+<?php
+/*
+ * Pretty-print a list of course section documents (e.g. homework
+ * assignments/solutions, lab prelabs/procedures/reports, ...) from
+ * the contents of a source directory. For example:
+ * include('php/section_docs.php');
+ * printSectionDocs('Recitation', 'doc/rec', 'rec', 20);
+ */
+function printSectionDocs($title, $directory, $startswith, $maxnum){
+ $i = 1;
+ for ($i=$maxnum; $i>=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($comment_file)) {
+ echo "<h4 id=\"s$i\">$title $i</h4>\n";
+ } // 's' b/c id attributes must begin with a letter, not a digit.
+ if (is_readable($comment_file)) {
+ readfile($comment_file);
+ }
+ if ($num_readable > 0) {
+ echo "<ul>\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 $front
+
+ echo " <li><a href=\"$url\">$name</a>$mode</li>\n";
+ }
+ echo "</ul>\n";
+ }
+ }
+}
+?>
--- /dev/null
+<?php
+/*
+ simplexml_load_file idea from 2008 courselist page
+ http://www.physics.drexel.edu/students/courses/
+ COPYRIGHT© 2008 DREXEL UNIVERSITY DEPARTMENT OF PHYSICS
+ Design by Daniel J. Cross
+ */
+$s=simplexml_load_file('xml/webmaster.xml');
+$name=$s->name;
+$email=$s->email;
+if (strlen($email) > 0)
+ echo "$name ($email)";
+else
+ echo "$name";
+?>
--- /dev/null
+<!--#include virtual="shared/header.shtml"-->
+
+<h3>Recitations</h3>
+
+<?php
+include('php/section_docs.php');
+printSectionDocs('Recitation', 'doc/rec', 'rec', 20);
+?>
+
+<!--#include virtual="shared/footer.shtml"-->
--- /dev/null
+<br />
+</div> <!--ending the #content div-->
+
+<div id="footer">
+<!-- Footer -->
+ <hr />
+ <p>
+ <a href="./xml/atom.xml">Atom feed</a>
+ </p>
+ <!-- need to make the shtml file executable ($ chmod 744 filename) -->
+ <!-- if XBitHack On, or give footer a .shtml extension. -->
+ <p>
+ Copyright © W. Trevor King. <br />
+ This material is released under the <a href="/~wking/shared/GNU_FDL.txt">GNU FDL</a>.
+ </p>
+ <p>
+ Validate <a href="http://validator.w3.org/check?uri=referer">xhtml</a>
+ | <a href="http://jigsaw.w3.org/css-validator/check?uri=referer">css</a>
+ </p>
+</div>
+
+</body>
+
+</html>
+
--- /dev/null
+<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
+"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
+<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
+
+<!--
+ Copyright (c) 2008 W. Trevor King
+ Permission is granted to copy, distribute and/or modify this document
+ under the terms of the GNU Free Documentation License, Version 1.2
+ or any later version published by the Free Software Foundation;
+ with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts.
+ A copy of the license is included in the section entitled "GNU
+ Free Documentation License".
+-->
+
+<!-- XHTML 1.1 removes the 'name' attribute from 'a' tags,
+ ( http://www.w3.org/TR/xhtml1/#guidelines
+ C.8. Fragment Identifiers )
+ which I imagine makes many browsers confused, so stick with 1.0 for now.
+ <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN"
+ "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> -->
+
+<head>
+
+<meta http-equiv="content-type"
+ content="text/html; charset=utf-8" />
+<meta name="description" content="Physics 201" />
+<meta name="keywords" content="drexel,physics,phys201" />
+<meta name="robots" content="all" />
+<link rel="stylesheet" href="shared/style.css" type="text/css" />
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+<title>Phys201</title>
+</head>
+
+<body>
+<p>
+<a id="top" name="top" />
+</p>
+<div id="header">
+ <h3>Physics 201</h3>
+ <a href="/students/courses/physics-201">Official Homepage</a>
+ |
+ <a href="index.shtml">Homepage</a>
+ |
+ <a href="announcements.shtml">Announcements</a>
+<!-- |
+ <a href="lectures.shtml">Lectures</a> -->
+ |
+ <a href="recitations.shtml">Recitations</a>
+<!-- |
+ <a href="labs.shtml">Labs</a> -->
+ |
+ <a href="homeworks.shtml">Homeworks</a>
+<!-- |
+ <a href="exams.shtml">Exams</a> -->
+ |
+ <a href="../contact.shtml">Contact</a>
+ |
+ <a href="/">Department</a>
+ <hr />
+</div>
+
+<div id="content">
--- /dev/null
+/* 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%;
+}
--- /dev/null
+<?xml version="1.0" encoding="utf-8"?>
+<people>
+ <person>
+ <name>Farnaza Neville Batliwalla</name>
+ <email>fnb23 at drexel dot edu</email>
+ <picture>/students/courses/current/physics-201/faculty/farnaza.jpg</picture>
+ </person>
+ <person>
+ <name>Benjamin Coy</name>
+ <email>btc24 at drexel dot edu</email>
+ <office>Disque 916</office>
+ <extension>1546</extension>
+ <picture>/directory/graduate/small/coy.benjamin.jpg</picture>
+ </person>
+ <person>
+ <name>Edward Damon</name>
+ <email>ead54 at drexel dot edu</email>
+ <office>Disque 705</office>
+ <extension>2732</extension>
+ <picture>/directory/graduate/small/damon.edward.jpg</picture>
+ </person>
+ <person>
+ <name>Nandita Ganesh</name>
+ <email>ng97 at drexel dot edu</email>
+ <picture>/students/courses/current/physics-201/faculty/nandita.jpg</picture>
+ </person>
+ <person>
+ <name>Travis Hoppe</name>
+ <email>travis.aaron.hoppe at drexel dot edu</email>
+ <office>Disque 819A</office>
+ <extension>2057</extension>
+ <url>http://www.physics.drexel.edu/~thoppe</url>
+ <picture>/directory/graduate/small/hoppe.travis.jpg</picture>
+ </person>
+ <person>
+ <name>Trevor King</name>
+ <email>wtk25 at drexel dot edu</email>
+ <office>Disque 927</office>
+ <extension>1818</extension>
+ <url>http://www.physics.drexel.edu/~wking</url>
+ <picture>/directory/graduate/small/king.trevor.jpg</picture>
+ </person>
+ <person>
+ <name>Anitha Manohar</name>
+ <office>Disque 912</office>
+ <email>am627 at drexel dot edu</email>
+ <picture>/students/courses/current/physics-201/faculty/anitha.jpg</picture>
+ </person>
+ <person>
+ <name>Greeshma Manomohan</name>
+ <email>gm322 at drexel dot edu</email>
+ <picture>/students/courses/current/physics-201/faculty/greeshma.jpg</picture>
+ </person>
+ <person>
+ <name>Ryan Michaluk</name>
+ <email>rmm622 at drexel dot edu</email>
+ <office>Disque 916</office>
+ <extension>1546</extension>
+ <picture>/directory/graduate/small/michaluk.ryan.jpg</picture>
+ </person>
+ <person>
+ <name>Pavithra Ramakrishnan</name>
+ <office>Disque 915</office>
+ <extension>2739</extension>
+ <email>pr323 at drexel.edu</email>
+ <picture>/students/courses/current/physics-201/faculty/pavithra.jpg</picture>
+ </person>
+ <person>
+ <name>John Schreck</name>
+ <email>jss74@drexel.edu</email>
+ <office>Disque 705</office>
+ <extension>2732</extension>
+ <picture>/directory/graduate/small/schreck.john.jpg</picture>
+ </person>
+</people>
--- /dev/null
+#!/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)
--- /dev/null
+#!/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)
--- /dev/null
+<?xml version="1.0" encoding="utf-8"?>
+<people>
+ <person>
+ <name>T. S. Venkataraman</name>
+ <office>Disque 912</office>
+ <hours>Open Door Policy Every Term</hours>
+ <email>venkat at drexel dot edu</email>
+ <extension>2721</extension>
+ <url>/directory/faculty/homepage.html?name=Venkat</url>
+ <picture>/students/courses/current/physics-201/faculty/venkatMATE.bmp</picture>
+ </person>
+</people>
--- /dev/null
+<?xml version="1.0" encoding="utf-8"?>
+<person>
+ <name>Trevor King</name>
+ <email></email>
+</person>
--- /dev/null
+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
--- /dev/null
+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.
--- /dev/null
+# 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
--- /dev/null
+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 > $@
--- /dev/null
+\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}
--- /dev/null
+../../problems/problem12.V1.tex
\ No newline at end of file
--- /dev/null
+../../problems/problem12.V2.tex
\ No newline at end of file
--- /dev/null
+../../problems/problem12.V3.tex
\ No newline at end of file
--- /dev/null
+\documentclass[letterpaper, 10pt]{article}
+
+\PassOptionsToPackage{nosolutions}{problempack}
+
+\input{all_problems}
--- /dev/null
+\documentclass[letterpaper, 10pt]{article}
+
+\PassOptionsToPackage{solutions}{problempack}
+
+\input{all_problems}
--- /dev/null
+\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}
--- /dev/null
+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.
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
+
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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 $\ans{0.54<v/c<1}$, where the upper bound is
+because no massive observer can ever reach the speed of light.
+\end{solution}
--- /dev/null
+\begin{problem*}{9.V13}
+A proton (rest energy $938\U{MeV}$) moving with velocity $v$ has a
+total energy of $1173\U{MeV}$. \Part{a} What is the proton's velocity
+(in multiples of $c$ and in $m/s$)? \Part{b} What is the proton's
+linear momentum (in MeV/$c$)?
+\end{problem*} % Prof. Venkat lecture problem #13
+
+\begin{solution}
+\Part{a}
+Using our results from Problem 9.35,
+\begin{align}
+ \gamma &= \frac{E}{m_0 c^2} = \frac{1173}{938} = 1.25 \\
+ v &= \sqrt{1-\frac{1}{\gamma^2}} c = 0.600c = \ans{180\U{Mm/s}}
+\end{align}
+
+\Part{b}
+\begin{equation}
+ p = m v = \gamma m_0 v = 1.25\cdot 938\U{MeV/$c^2$} \cdot 0.600c
+ = \ans{704\U{MeV/$c$}}
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{9.V17}
+A particle of rest mass $1.2\U{MeV/$c^2$}$ and kinetic energy
+$2.0\U{MeV}$ collides with a stationary particle of rest mass
+$2.4\U{MeV/$c^2$}$. After the collision the particles stick
+together. \Part{a} What is the speed of the first particle before the
+collision? \Part{b} What is the total energy of the first particle
+before the collision? \Part{c} What is the total initial momentum of
+the system? \Part{d} What is the rest mass of the system after the
+collison? \Part{e} What is the total kinetic energy after the
+collision? \Part{f} What conclusions can you draw from your ansers in
+parts \Part{d} and \Part{e}?
+\end{problem*} % Prof. Venkat lecture problem #17
+
+\begin{solution}
+\Part{a}
+Using our results from Problem 9.35,
+\begin{align}
+ \gamma_{i,1} &= \frac{K}{m_{0,1} c^2} + 1 = \frac{2.0}{1.2} + 1 = 2.67 \\
+ v_{i,1} &= \sqrt{1-\frac{1}{\gamma_{i,1}^2}} c = \ans{0.927c}
+\end{align}
+
+\Part{b}
+\begin{equation}
+ E_{i,1} = \gamma m_{0,1} c^2 = 2.67 \cdot 1.2\U{MeV} = \ans{3.20\U{MeV}}
+\end{equation}
+
+\Part{c}
+\begin{equation}
+ p_i = m_{i,1}v_{i,1} = \gamma_{i,1} m_0 v_{i,1}
+ = 2.67 \cdot 1.2\U{MeV/$c^2$} \cdot 0.927c
+ = \ans{2.97\U{MeV/$c$}}
+\end{equation}
+
+\Part{d}
+Conserving momentum
+\begin{equation}
+ p_i = p_f = \gamma_f m_{0,f} v_f = 2.97\U{MeV/$c$}
+\end{equation}
+Conserving energy
+\begin{equation}
+ E_i = (1.2+2.0+2.4)\U{MeV} = 5.6\U{MeV} = E_f = \gamma_f m_{0,f} c^2
+\end{equation}
+Using the relativistic energy-momentum relationship (Eqn.~9.22)
+\begin{align}
+ E_f^2 &= p_f^2 c^2 + m_{0,f}^2 c^4 \\
+ m_{0,f} c^2 &= \sqrt{E_f^2 - p_f^2 c^2}
+ = \sqrt{(5.6\U{MeV})^2 - (2.97\U{MeV})^2}
+ = 4.75\U{MeV} \\
+ m_{0,f} &= \ans{4.75\U{MeV/$c^2$}}
+\end{align}
+
+``But wait!'' you might say, ``In our mechanics class, we learned that
+you only conserve momentum for inelastic collisions. You conserve
+both momentum and energy for elastic collisions.'' If so, your memory
+is tricking you. For elastic collisions (and not for inelastic
+collisions), you conserve \emph{kinetic energy}. Note that we don't
+conserve kinetic energy (see \Part{f}), but the total energy, which is
+always conserved in isolated systems.
+
+\Part{e}
+The kinetic energy is the increase in energy over the rest energy
+\begin{equation}
+ K_f = E_f - m_{0,f} c^2 = 5.6\U{MeV} - 4.75\U{MeV} = \ans{0.852\U{MeV}}
+\end{equation}
+
+\Part{f}
+$2.0-0.852=1.15\U{MeV}$ of initial kinetic energy is converted to rest
+energy during the collision. The increased rest energy shows that the
+internal energy of the composite partical has increased. For example,
+if the particles were lumps of clay, they would be warmer after the
+collision, and this increased thermal energy would show up as an
+increased rest mass. Or if the particles were oppositely charged, the
+increased electric potential energy would show up in the increased
+rest mass.
+\end{solution}
--- /dev/null
+\begin{problem}
+The $S'$ frame moves along the $x$ axis of the $S$ frame with a
+constant velocity $v$. Observers in $S$ see an explosion at $t_1=0$
+at $x_1=1080\U{m}$. They see a second explosion $8.0\U{$\mu$s}$ later
+at $x_2=3000\U{m}$. In $S'$, the explosions occur at the same $x'$.
+\Part{a} Use the Lorentz transforms to determine the velocity (in
+terms of $c$) of the $S'$ frame with respect to the $S$ frame.
+\Part{b} At what $x'$ (in m) do the two explosions occur in the $S'$
+frame?
+\Part{c} Instead, is it possible for the two explosions to occur
+\emph{simultaneously} in the $S'$ frame? If so, how fast should the
+$S'$ frame be moving relative to the $S$ frame?
+\end{problem} % Prof. Venkat homework problem 4.3
+
+\begin{solution}
+\Part{a}
+From the Lorentz transformations
+\begin{align}
+ x_1' &= \gamma \p({x_1 - v t_1}) = 0 \\
+ x_2' &= \gamma \p({x_2 - v t_2}) = 0 \\
+ \Delta x' &= x_2'-x_1' = \gamma \p({(x_2-x_1 - v(t_2-t_1)})
+ = \gamma \p({\Delta x - v \Delta t}) = 0 \\
+ 0 &= \Delta x - v \Delta t \\
+ v &= \frac{\Delta x}{\Delta t} = \frac{3000-1080}{8.0}\U{m/$\mu$s}
+ = \ans{0.801c}
+\end{align}
+
+\Part{b}
+Since $x_1'=x_2'$, we only need to find $x_1'$
+\begin{align}
+ \gamma &= \frac{1}{\sqrt{1-v^2/c^2}}
+ = \frac{1}{\sqrt{1-0.801^2}} = 1.67 \\
+ x_1' &= \gamma(x_1 - vt_1) = \gamma x_1
+ = \ans{1800\U{m}}
+\end{align}
+
+\Part{c}
+From the Lorentz transformations
+\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}\cdot8.0\U{$\mu$s}}{(3000-1080)\U{m}}c
+ = 1.25c
+\end{align}
+Nothing can go faster than the speed of light, so it is not possible
+for these two events to happen simultaneously.
+
+The fact that these events occur at the same position in one frame
+means that the distance between the events is \emph{timelike} (it's
+possible for a particle moving slower than the speed of light to be
+present at two timelike events.) For timelike events, the concepts of
+an \emph{earlier} and \emph{later} event are well defined, since the
+earlier event will come first in \emph{any} reference frame. For
+example, your birth and death events are \emph{timelike}, since you
+move slower than the speed of light and will be present at both.
+Therefore, there is no reference frame in which you die before you are
+born (which would be very confusing).
+
+If you would have to travel constantly at the speed of light to be
+present at two events, the distance between those two events is
+\emph{lightlike}. In this case, the time-ordering of the two events
+is still well defined.
+
+If you would have to travel faster than the speed of light to be
+present at two events, the distance between those two events is
+\emph{spacelike}. In this case, the time-ordering of the two events
+depends on your reference frame, but the world doesn't explode,
+because nothing that happens at one event can send information
+affecting the other.
+\end{solution}
--- /dev/null
+\begin{problem}
+\Part{a} How long (in Earth time) does it take for a plane to circle
+the Earth at low altitude (average radius $6,370\U{km}$) going
+600\U{mph}? \Part{b} How much less time passes according to a clock
+on the plane?
+
+For the purpose of this problem, you may assume the Earth is
+stationary and not rotating.
+\end{problem}
+
+\begin{solution}
+\Part{a}
+\begin{align}
+ 600\U{mph} \cdot \frac{1.6\U{km}}{1\U{mile}}\cdot\frac{1\U{hour}}{3600\U{s}}
+ &= 267\U{m/s} \\
+ v\Delta t &= \Delta x = 2 \pi r \\
+ \Delta t &= \frac{2 \pi r}{v}
+ = \frac{2 \pi \cdot 6.37\E{6}\U{m}}{267\U{m/s}}
+ = \ans{150\U{ks}} = \ans{41.7\U{hours}}
+\end{align}
+
+\Part{b}
+This is pretty much a twin-paradox, with the clock on the
+constantly-accelerating plane running slower than the clock on Earth.
+\begin{align}
+ \Delta t' &= \frac{\Delta t}{\gamma} = \sqrt{1-\p({\frac{v}{c}})^2} \Delta t
+ = (1-3.95\E{-13}) \Delta t \\
+ \Delta t - \Delta t' &= 3.95\E{-13} \Delta t = \ans{59.3\U{ns}}
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem}
+Returning to Earth in his rocket ship at $0.6c$, Jack calls Jill (who
+is on Earth) to let her know he's almost home. He sets up his radio
+transmitter to broadcast at $500\U{MHz}$. \Part{a} In Jill's frame,
+how much time passes beween two succesive maxima of the voltage across
+Jack's antenna? \Part{b} What is the frequency of the signal as it
+reaches Jill?
+\end{problem}
+
+\begin{solution}
+\Part{a}
+Jack's carrier wave has a period of
+\begin{equation}
+ T = \frac{1}{f} = 2.00\U{ns} \;.
+\end{equation}
+Jill thinks Jack's clock is time-dilated (i.e. runs slower), so the
+time between voltage maxima in Jill's frame is
+\begin{equation}
+ \Delta t' = \gamma T = \frac{1}{\sqrt{1-\p({\frac{v}{c}})^2}} T
+ = 1.25 \cdot T = \ans{2.50\U{ns}} \;.
+\end{equation}
+Note that this time between antenna voltage maxima is \emph{not} the
+period of the light Jill recieves, because Jack is moving towards her
+(this would be the period she recieved if Jack were moving
+perpendicular to her).
+
+\Part{b}
+During the time $\Delta t'$ between successive voltage maxima, Jack
+came $v\Delta t'$ closer to Earth while the earlier maximum came
+$c\Delta t'$ closer. The wavelength of the signal Jill recieves is
+thus
+\begin{equation}
+ \lambda' = (c-v)\Delta t' = 0.4 c \Delta t' = 0.300\U{m} \;,
+\end{equation}
+and the frequency is
+\begin{align}
+ c T' &= \lambda' \\
+ f' &= \frac{1}{T'} = \frac{c}{\lambda'} = \ans{1.00\U{GHz}} \;.
+\end{align}
+
+If we wanted to leave the whole problem in symbolic notation, we have
+\begin{align}
+ f' &= \frac{c}{\lambda'} = \frac{c}{(c-v)\Delta t'}
+ = \frac{1}{(1-\frac{v}{c})\gamma T}
+ = \frac{\sqrt{1-\p({\frac{v}{c}})^2}}{1-\frac{v}{c}} f
+ = \frac{\sqrt{\p({1+\frac{v}{c}})\p({1-\frac{v}{c}})}}{1-\frac{v}{c}} f
+ = \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}} f \;,
+\end{align}
+which is the source of the text's Equation 24.22.
+\end{solution}
--- /dev/null
+from pylab import *
+from matplotlib.patches import Polygon
+import sys
+
+filename = None
+if len(sys.argv) >= 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)
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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 $A \equiv R_1 R_2 + R_2 R_3 + R_3 R_1$.
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
+
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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<Ticks.length; ++i) {
+ if (Ticks[i] == x)
+ return xLabels[i];
+ }
+ return "Unknown";
+}
+draw(graph(lorentzian,0,2X),red);
+xaxis(0,LeftTicks(ticklabel, Ticks=Ticks));
+yaxis("intensity",0);
+
+pair[] dxs = {(X-FWHM/2, 0), (X+FWHM/2, 0)};
+pair[] leftBar = {(X-FWHM/2, 0), (X-FWHM/2, lorentzian(X))};
+pair[] rightBar = {(X+FWHM/2, 0), (X+FWHM/2, lorentzian(X))};
+pair[] fwhmBar = {(X-FWHM/2, lorentzian(X)/2), (X+FWHM/2, lorentzian(X)/2)};
+draw(graph(dxs), green);
+draw(graph(leftBar), dotted);
+draw(graph(rightBar), dotted);
+draw(graph(fwhmBar), blue);
+label("$\Delta \lambda$", align=N, (X, lorentzian(X)));
+\end{asy}
+\end{center}
+\end{problem} % similar to Problem 28.34.
+
+\begin{solution}
+\Part{a}
+From the Rydberg formula
+\begin{align}
+ \frac{1}{\lambda} &= R_H\p({\frac{1}{n_1^2} - \frac{1}{n_2^2}}) \\
+ f &= \frac{c}{\lambda} = c R_H\p({\frac{1}{n_1^2} - \frac{1}{n_2^2}})
+ = 3\E{8}\U{m/s} \cdot 1.097\E{7}\U{1/m} \p({1 - \frac{1}{4}})
+ = \ans{2.47\E{15}\U{Hz}}
+\end{align}
+
+\Part{b}
+Using the energy/time uncertainty principle,
+\begin{align}
+ \Delta E \Delta t &\ge \frac{\hbar}{2} \\
+ \Delta E &\ge \frac{\hbar}{2\Delta t} \approx \frac{\hbar}{2 \tau}
+ = 3.3\E{-26}\U{J} = 2.1\E{-7}\U{eV} \\
+ \frac{\Delta E}{E} &= \frac{\Delta E}{hf} \ge \ans{2.0\E{-8}}
+\end{align}
+
+\Part{bonus}
+We can see find the natural linewidth $\Delta \lambda$ through a
+propagation-of-errors approach
+\begin{align}
+ \lambda &= \frac{hc}{E} = \frac{c}{f} = 121\U{nm} \\
+ \partial \lambda &= -\frac{hc}{E^2} \partial E
+ = -\lambda \frac{\partial E}{E} \;.
+\end{align}
+Since $\Delta E \ll E$, this relationship doesn't change much over the
+span of $E \pm \Delta E/2$, so
+\begin{equation}
+ \Delta \lambda \approx -\lambda \frac{\Delta E}{E}
+ \ge \ans{2.4\U{fm}} \;,
+\end{equation}
+where we dropped the sign because we only care about the range $\Delta
+\lambda$, not whether increasing $E$ increases or decreases $\lambda$.
+
+Note that the propagation-of-errors approach would \emph{not} work for
+Problem 1, since $\partial K/\partial p (p=0) = 0$. Therefore,
+assuming the slope to be constant over the range $\Delta p$ would give
+$\Delta K = 0$. Instead, we used a quasi-graphical approach that took
+advantage of our understanding of $K(p) = p^2/2m$.
+\end{solution}
--- /dev/null
+\begin{problem}
+An electron that has an energy of approximately $6\U{eV}$ moves
+between rigid walls $1.00\U{nm}$ apart. Find \Part{a} the quantum
+number $n$ for the energy state that the electron occupies
+and \Part{b} the precise energy of the electron.
+\end{problem} % based on 28.38
+
+\begin{solution}
+\Part{a}
+The allowed energy levels for a particle in a box are (Equation 28.30)
+\begin{align}
+ E_n &= \frac{h^2 n^2}{8mL^2} \;.
+\end{align}
+For an electron ($m=9.11\E{-31}\U{kg}$) in a box of length
+$L=1\U{nm}$, this works out to
+\begin{align}
+ E_1 &= 0.377\U{eV} \\
+ E_2 &= 1.51\U{eV} \\
+ E_3 &= 3.39\U{eV} \\
+ E_4 &= 6.02\U{eV} \\
+ E_5 &= 9.41\U{eV}
+\end{align}
+So the electron is in the $\ans{n=4}$ state.
+
+\Part{b}
+The precise energy is $E_4 = \ans{6.02\U{eV}}$.
+\end{solution}
--- /dev/null
+\begin{problem}
+The wave function for a particle confined to moving in a
+one-dimensional box is
+\begin{equation}
+ \psi(x) = A \sin\p({\frac{n\pi x}{L}})
+\end{equation}
+Use the normalization condition to show that
+\begin{equation}
+ A = \sqrt{\frac{2}{L}}
+\end{equation}
+\emph{HINT}: Because the box length is $L$, the wave function is zero
+for $x < 0$ and for $x > L$.
+%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}
--- /dev/null
+P6
+480 283
+255
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+\ 1\ 1\ 1\0\0\0\0\0\0\f\f\f\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\ 2\ 2\ 2\0\0\0\0\0\0\ 3\ 3\ 3\0\0\0\ 2\ 2\ 2\0\0\0\0\0\0\0\0\0\ 5\ 5\ 5\b\b\b\ 2\ 2\ 2\0\0\0\ 1\ 1\ 1\0\0\0\ 1\ 1\ 1\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\ 2\ 2\ 2\0\0\0\0\0\0\ 2\ 2\ 2\0\0\0\ 1\ 1\ 1\0\0\0\0\0\0\ 1\ 1\ 1\ 2\ 2\ 2\0\0\0\0\0\0\0\0\0\ 1\ 1\ 1\ 3\ 3\ 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1\ 1\ 1\ f\ f\ f\ 1\ 1\ 1\ 1\ 1\ 1\0\0\0\11\11\11\ e\ e\ e\ 5\ 5\ 5\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\ 1\ 1\ 1\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\ 1\ 1\ 1\0\0\0\0\0\0\ 1\ 1\ 1\ 1\ 1\ 1\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\ 5\ 5\ 5\0\0\0\0\0\0\ 3\ 3\ 3\0\0\0\0\0\0\ 1\ 1\ 1\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\ 1\ 1\ 1\0\0\0\0\0\0\ 1\ 1\ 1\ 1\ 1\ 1\0\0\0\ 1\ 1\ 1\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\ 5\ 5\ 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1\ 1\ 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+
+
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NNN\ f\ f\ 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+
+
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e\ e\ 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e\ e\ 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e\ e\ 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2\ 2\ 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5\ 5\ 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***ØØØ®®®\ 4\ 4\ 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\ No newline at end of file
--- /dev/null
+# Xp -> (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
--- /dev/null
+\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}
--- /dev/null
+\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; i<x.length; ++i) {
+ x[i] = c/x[i]/1e3;
+}
+
+linefit L=leastsquares(x, y);
+write("work function");
+write(L.b);
+write("h/e");
+write(L.m);
+
+real fit(real f)
+{
+ return L.b + L.m*f;
+}
+
+draw(graph(fit, fmin, fmax), red);
+draw(graph(x, y), 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("$K_{max}$ (V)",LeftRight,
+ RightTicks(begin=false,end=false,extend=true,ptick=thin));
+\end{asy}
+\end{center}
+
+The least-squares fit yields parameters \Part{c}
+$\phi=\ans{1.4\U{eV}}$ and $h/e = 4.04\U{mV/THz}$. So \Part{b}
+$h=\ans{6.5\E{-34}\U{J$\cdot$s}}$.
+\end{solution}
--- /dev/null
+\begin{problem}
+\emph{BONUS PROBLEM}. Particles incident from the left are confronted
+with a step in potential energy. Located at $x=0$, the step has a
+height $U$. The particles have energy $E>U$. 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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+# 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
--- /dev/null
+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 > $@
--- /dev/null
+\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}
--- /dev/null
+../../problems/problem12.02.tex
\ No newline at end of file
--- /dev/null
+../../problems/problem12.05.tex
\ No newline at end of file
--- /dev/null
+../../problems/problem12.12.tex
\ No newline at end of file
--- /dev/null
+../../problems/problem12.15.tex
\ No newline at end of file
--- /dev/null
+../../problems/problem12.18.tex
\ No newline at end of file
--- /dev/null
+\documentclass[letterpaper, 10pt]{article}
+
+\PassOptionsToPackage{nosolutions}{problempack}
+
+\input{all_problems}
--- /dev/null
+\documentclass[letterpaper, 10pt]{article}
+
+\PassOptionsToPackage{solutions,loose}{problempack}
+
+\input{all_problems}
--- /dev/null
+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 > $@
--- /dev/null
+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
--- /dev/null
+This directory is for PDFs for which we do not have LaTeX source.
--- /dev/null
+# 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
--- /dev/null
+See ../lab/README and ../lec/README.
--- /dev/null
+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 > $@
--- /dev/null
+# 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
--- /dev/null
+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.
--- /dev/null
+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 > $@
--- /dev/null
+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 > $@
--- /dev/null
+# 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
--- /dev/null
+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.
+
--- /dev/null
+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 > $@
projects / course.git / commitdiff
