--- /dev/null
+\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}
--- /dev/null
+\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}
projects / course.git / commitdiff
