From 4a1fc4ce0b17a2d36cb7496bf38269ab2d8e6137 Mon Sep 17 00:00:00 2001 From: William Trevor King Date: Thu, 6 Aug 2009 09:57:00 -0400 Subject: [PATCH] Renamed some of my problems and added two more. --- ...pton-cat.T.tex => problem.compton-cat.tex} | 0 .../wking/problem.lightning-doppler.tex | 58 +++++++ ...o.py => problem.relativistic-limo.limo.py} | 0 ...mo.T.tex => problem.relativistic-limo.tex} | 0 .../wking/problem.sprung-pendulum.tex | 141 ++++++++++++++++++ ...r.jpg => problem.uni-shock.bonbardier.jpg} | Bin ...{problem12.T.tex => problem.uni-shock.tex} | 0 7 files changed, 199 insertions(+) rename latex/problems/wking/{problem28.compton-cat.T.tex => problem.compton-cat.tex} (100%) create mode 100644 latex/problems/wking/problem.lightning-doppler.tex rename latex/problems/wking/{problem09.limo.T.limo.py => problem.relativistic-limo.limo.py} (100%) rename latex/problems/wking/{problem09.limo.T.tex => problem.relativistic-limo.tex} (100%) create mode 100644 latex/problems/wking/problem.sprung-pendulum.tex rename latex/problems/wking/{problem12.T.bombardier.jpg => problem.uni-shock.bonbardier.jpg} (100%) rename latex/problems/wking/{problem12.T.tex => problem.uni-shock.tex} (100%) diff --git a/latex/problems/wking/problem28.compton-cat.T.tex b/latex/problems/wking/problem.compton-cat.tex similarity index 100% rename from latex/problems/wking/problem28.compton-cat.T.tex rename to latex/problems/wking/problem.compton-cat.tex diff --git a/latex/problems/wking/problem.lightning-doppler.tex b/latex/problems/wking/problem.lightning-doppler.tex new file mode 100644 index 0000000..ba5433f --- /dev/null +++ b/latex/problems/wking/problem.lightning-doppler.tex @@ -0,0 +1,58 @@ +\begin{problem} +Lightning strikes a point $1.0\U{km}$ away from where you stand while +the wind blows into your face at $v_w = 50\U{km/h}$. +\Part{a} How much time elapses before you hear the thunder? +\Part{b} If the primary frequency of thunder at the source is +$100\U{Hz}$, what frequency do you hear? +(\url{http://www.lightningsafety.com/nlsi_info/thunder.html}) +\begin{center} +\begin{asy} +import stickfigure; + +pair personPos = (0,0); +pair lightningPos = (2cm,0); +real extraGround = 5mm; +StickFigure s = StickFigure(standsOn=personPos); + + +pair zig = (1mm, 3mm); +path zigzag = (0,0)--zig--(-0.2mm,3.1mm); +path lightning = lightningPos; +pair workingPos; +for(int i=0; i<3; ++i) { + workingPos = point(lightning, length(lightning)); + lightning = lightning -- (shift(workingPos)*zigzag); +} +workingPos = point(lightning, length(lightning)); +lightning = lightning -- (shift(workingPos)*zig); + +draw((personPos-(extraGround,0))--(lightningPos+(extraGround,0))); +s.draw(); +draw(lightning); + +real windXoffset = 5mm; +real windYoffset = 5mm; +draw((lightningPos+(-windXoffset,windYoffset)) + --(personPos+(windXoffset,windYoffset)), Arrow); +label("Wind", (lightningPos+personPos)/2+(0,windYoffset), align=N); +\end{asy} +\end{center} +\end{problem} + +\begin{solution} +\Part{a} +The speed of sound in air at sea level is around $v_s = 343\U{m/s}$. +This speed is relative to the air, so the speed relative to the ground +is $v_s' = v_s+v_w = 393\U{m/s}$. The time it takes the thunder to +reach you is thus +\begin{equation} + \Delta t = \frac{\Delta x}{v_s'} = \frac{1.0\U{km}}{393\U{m/s}} + \approx \ans{2.5\U{s}} +\end{equation} + +\Part{b} +In the reference frame of the wind, the sound made by the moving +thunder source is red-shifted (as the thunder source moves away), but +the sound you hear is blue-shifted (as you move closer), so you here +the thunder at an unshifted \ans{$100\U{Hz}$}. +\end{solution} diff --git a/latex/problems/wking/problem09.limo.T.limo.py b/latex/problems/wking/problem.relativistic-limo.limo.py similarity index 100% rename from latex/problems/wking/problem09.limo.T.limo.py rename to latex/problems/wking/problem.relativistic-limo.limo.py diff --git a/latex/problems/wking/problem09.limo.T.tex b/latex/problems/wking/problem.relativistic-limo.tex similarity index 100% rename from latex/problems/wking/problem09.limo.T.tex rename to latex/problems/wking/problem.relativistic-limo.tex diff --git a/latex/problems/wking/problem.sprung-pendulum.tex b/latex/problems/wking/problem.sprung-pendulum.tex new file mode 100644 index 0000000..51d98ad --- /dev/null +++ b/latex/problems/wking/problem.sprung-pendulum.tex @@ -0,0 +1,141 @@ +\newcommand{\x}{\theta} +\newcommand{\eq}{_\text{eq}} +\newcommand{\xeq}{\x\eq} +\newcommand{\xu}{\x_\text{u}} +\newcommand{\zeq}{z\eq} +\newcommand{\referenceProblem}{Problem 1} + +\begin{problem} +\emph{BONUS PROBLEM}. +Find the angle $\theta$ as a function of time for the left hand bob in +pendulum shown below in terms of the constants shown in the figure +where $d$ is the unstretched length of the spring. +\begin{center} +\begin{asy} +import Mechanics; +real u = 1cm; + +Pendulum pA = makePendulum(angleDeg=-40, length=2u, + angleL="$\theta$", stringL="$r$"); +Pendulum pB = makePendulum(angleDeg=40, length=2u, + angleL="$\theta$", stringL="$r$"); +Spring s = Spring(pFrom=pA.mass.center, pTo=pB.mass.center, + unstretchedLength=abs(pA.mass.center.x)*2, + L="$k$"); +Distance ds = Distance(pFrom=s.pFrom-(0,.7u), + pTo=s.pTo -(0,.7u), + L="$d$"); +s.draw(); +ds.draw(); +pA.draw(drawVertical=true); +pB.draw(); +label("$m$", pA.mass.center); +label("$m$", pB.mass.center); +\end{asy} +\end{center} +\end{problem} + +\begin{solution} +Because the situation is symmetric, the behavior of the left bob is +the same as the behavior of the left half alone +\begin{center} +\begin{asy} +import Mechanics; +real u = 1cm; + +Pendulum p = makePendulum(angleDeg=-40, length=2u, + angleL="$\theta$", stringL="$r$"); +Spring s = Spring(pFrom=p.mass.center, pTo=(0,p.mass.center.y), + unstretchedLength=abs(p.mass.center.x), + L="$2k$"); +Distance ds = Distance(pFrom=s.pFrom-(0,.7u), + pTo=s.pTo -(0,.7u), + L="$d/2$"); +s.draw(); +ds.draw(); +p.draw(drawVertical=true); +label("$m$", p.mass.center); +\end{asy} +\end{center} +where the spring constant doubles, because when the left bob +compresses the spring by $\Delta x$ the right bob does as well, for a +total compression of $2\Delta x$ and a restoring force of $2\Delta x k += \Delta x \cdot 2k$. This is similar to \referenceProblem, except +the equilibrium position is not about $\xeq = 0$, but about $\xeq = +\arcsin(\frac{d\eq}{2r})$, and the spring is always absolutely +horizontal in this case (it was approximately horizontal in +\referenceProblem). Therefor, the procedure is the same with the +small angle approximations replaced by approximations valid around the +new $\xeq$. + +Blowing up the situation around the left bob we have +\begin{center} +\begin{asy} +import Mechanics; +real u = 1cm; + +Pendulum p = makePendulum(angleDeg=-40, length=2u, + angleL="$\theta$"); +Vector fs = Force(p.mass.center, dir=180, mag=1u, L="$F_s$"); +Vector fg = Force(p.mass.center, dir=-90, mag=1u, L="$F_g$"); +Vector dx = Vector(p.mass.center, dir=-40-180, mag=1.5u, L="$x'$"); +Vector dy = Vector(p.mass.center, dir=-40-90, mag=1.5u, L="$y'$"); +Angle xas = Angle(dx.pTip(), p.mass.center, fs.pTip(), L="$\theta$"); +Angle yag = Angle(dy.pTip(), p.mass.center, fg.pTip(), L="$\theta$"); + +xas.draw(); +yag.draw(); +dx.draw(); +dy.draw(); +fs.draw(); +fg.draw(); +p.draw(drawVertical=true); +\end{asy} +\end{center} + +The spring is stretched or compressed by +\begin{equation} + \dd x = r\p[{\sin(\x) - \sin(\xu)}]\;, +\end{equation} +where $\xu$ horizontal displacement of the bob when the spring is +unstretched. + +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{equation} + \sum F_{\tan} = F_s\cos(\x) - F_g\sin(\x) + = -kr\cos(\x)\p[{\sin(\x) - \sin(\xu)}] - mg\sin(\x) +\end{equation} +We also know from Newton's laws that (identically to \referenceProblem) +\begin{equation} + \sum F_{\tan} = ma_{\tan} = m\nderiv{2}{t}{x_{\tan}} + = mr \nderiv{2}{t}{\x} +\end{equation} + +Combining these two formulas for $\sum F$ we have +\begin{equation} + mr\nderiv{2}{t}{\x} = -kr\cos(\x)\p[{\sin(\x) - \sin(\xu)}] - mg\sin(\x) +\end{equation} +To determine the behavior for small deflections, we need to Taylor +expand the right hand side and keep the linear term to find the +effective spring constant. Recall the Taylor expansion of a function +$f(z)$ for $z$ near some $z\eq$ is given by +\begin{equation} + f(z) = f(z\eq) + (z-z\eq) \cdot \p.{\deriv{z}{f}}|_\text{$z\eq$} + \ldots +\end{equation} +So our effective spring constant (valid for small $\dd\x$ around +$\xeq$) is given by +\begin{align} + k' &= \frac{1}{mr}\p.{\deriv{z}{}\p\{{kr\cos(\x)\p[{\sin(\x) - \sin(\xu)}] + mg\sin(\x)}\}}|_\text{$\xeq$} +\end{align} + +Then we can find $\xeq$ by balancing the forces +\begin{align} + 0 = p.{\sum F_{\tan}}|_\text{\xeq} &= -kr\cos(\xeq)\p[{\sin(\xeq) - \sin(\xu)}] - mg\sin(\xeq) \\ + &= -kr\cos(\xeq)\p[{\sin(\xeq) - \sin(\xu)}] - mg\sin(\xeq) \\ + +\end{align} +% WORKING + +\end{solution} diff --git a/latex/problems/wking/problem12.T.bombardier.jpg b/latex/problems/wking/problem.uni-shock.bonbardier.jpg similarity index 100% rename from latex/problems/wking/problem12.T.bombardier.jpg rename to latex/problems/wking/problem.uni-shock.bonbardier.jpg diff --git a/latex/problems/wking/problem12.T.tex b/latex/problems/wking/problem.uni-shock.tex similarity index 100% rename from latex/problems/wking/problem12.T.tex rename to latex/problems/wking/problem.uni-shock.tex -- 2.26.2