165 lines
4.8 KiB
TeX
165 lines
4.8 KiB
TeX
\subsection{The Simple Pendulum}
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This subsection models a bob of mass on a light string, using angular displacement from the vertical as the natural coordinate.
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\dfn{Simple pendulum and angular coordinate}{Let $m$ denote the bob's mass, let $\ell>0$ denote the string length, let $g$ denote the magnitude of the gravitational field, and let $\theta(t)$ denote the angular displacement from the downward vertical, measured in radians and taken positive in the counterclockwise direction.
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A \emph{simple pendulum} is an idealized system consisting of a point mass $m$ attached to a massless string of fixed length $\ell$, swinging without friction in a uniform gravitational field. The bob moves along a circular arc of radius $\ell$. If $s(t)$ denotes the arc displacement from equilibrium, then
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\[
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s=\ell\theta.
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\]
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The equilibrium position is $\theta=0$.}
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\thm{Exact pendulum equation and small-angle SHM model}{For the simple pendulum above, the exact rotational equation of motion is
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\[
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\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
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\]
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This equation is nonlinear, so the motion is not exactly simple harmonic for arbitrary amplitude.
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If the oscillation remains at small angles so that $|\theta|\ll 1$ radian and $\sin\theta\approx\theta$, then the motion is approximated by
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\[
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\ddot{\theta}+\frac{g}{\ell}\theta=0.
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\]
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Thus the pendulum behaves approximately like SHM with angular frequency
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\[
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\omega=\sqrt{\frac{g}{\ell}},
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\]
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small-angle period
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\[
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T=2\pi\sqrt{\frac{\ell}{g}},
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\]
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and small-angle frequency
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\[
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f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{g}{\ell}}.
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\]}
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\pf{Short derivation from torque and linearization}{About the pivot, the gravitational torque on the bob is restoring, so
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\[
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\tau=-mg\ell\sin\theta.
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\]
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The bob acts like a point mass at distance $\ell$, so its moment of inertia about the pivot is
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\[
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I=m\ell^2.
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\]
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Using rotational Newton's second law, $\sum\tau=I\ddot{\theta}$, gives
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\[
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m\ell^2\ddot{\theta}=-mg\ell\sin\theta.
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\]
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Divide by $m\ell^2$:
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\[
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\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
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\]
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For small oscillations with $|\theta|\ll 1$ radian, use the small-angle approximation $\sin\theta\approx\theta$. Then
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\[
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\ddot{\theta}+\frac{g}{\ell}\theta=0,
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\]
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which is the standard SHM equation with $\omega^2=g/\ell$.}
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\ex{Illustrative example}{A pendulum oscillates through small angles. Its length is changed from $\ell_1=0.50\,\mathrm{m}$ to $\ell_2=2.00\,\mathrm{m}$. How do the period and frequency change?
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For small-angle motion,
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\[
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T=2\pi\sqrt{\frac{\ell}{g}}.
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\]
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Therefore $T\propto\sqrt{\ell}$. Since
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\[
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\frac{\ell_2}{\ell_1}=\frac{2.00}{0.50}=4,
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\]
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the new period is multiplied by
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\[
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\sqrt{4}=2.
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\]
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So the period doubles. Because $f=1/T$, the frequency is cut in half.}
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\qs{Worked AP-style problem}{A simple pendulum has length $\ell=0.90\,\mathrm{m}$. It is pulled aside to a maximum angle $\theta_{\max}=0.10\,\mathrm{rad}$ and released from rest. Take $g=9.8\,\mathrm{m/s^2}$.
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Assume the small-angle model is valid.
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the exact equation of motion and the small-angle approximate equation,
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\item the angular frequency, period, and frequency,
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\item the time required to move from maximum displacement to equilibrium, and
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\item the maximum linear speed of the bob.
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\end{enumerate}}
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\sol Let $\theta(t)$ denote the angular displacement from the downward vertical.
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For part (a), the exact pendulum equation is
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\[
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\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
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\]
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Substitute $g=9.8\,\mathrm{m/s^2}$ and $\ell=0.90\,\mathrm{m}$:
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\[
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\ddot{\theta}+\frac{9.8}{0.90}\sin\theta=0.
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\]
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Thus,
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\[
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\ddot{\theta}+10.9\sin\theta=0
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\]
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to three significant figures.
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Under the small-angle approximation $\sin\theta\approx\theta$, the motion is modeled by
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\[
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\ddot{\theta}+\frac{9.8}{0.90}\theta=0,
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\]
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or
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\[
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\ddot{\theta}+10.9\theta=0.
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\]
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For part (b), compare the small-angle equation with
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\[
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\ddot{\theta}+\omega^2\theta=0.
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\]
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So
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\[
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\omega=\sqrt{\frac{g}{\ell}}=\sqrt{\frac{9.8}{0.90}}=3.30\,\mathrm{rad/s}.
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\]
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Then the period is
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\[
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T=2\pi\sqrt{\frac{\ell}{g}}=\frac{2\pi}{\omega}=\frac{2\pi}{3.30}=1.90\,\mathrm{s}.
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\]
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The frequency is
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\[
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f=\frac{1}{T}=\frac{1}{1.90}=0.526\,\mathrm{Hz}.
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\]
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For part (c), a pendulum in SHM takes one-quarter of a cycle to move from an endpoint to equilibrium. Therefore,
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\[
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t=\frac{T}{4}=\frac{1.90}{4}=0.475\,\mathrm{s}.
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\]
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For part (d), the maximum angular speed in SHM is
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\[
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\dot{\theta}_{\max}=\omega\theta_{\max}.
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\]
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So
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\[
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\dot{\theta}_{\max}=(3.30)(0.10)=0.330\,\mathrm{rad/s}.
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\]
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The bob's linear speed is related by $v=\ell\dot{\theta}$, so the maximum linear speed is
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\[
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v_{\max}=\ell\dot{\theta}_{\max}=(0.90)(0.330)=0.297\,\mathrm{m/s}.
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\]
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Therefore,
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\[
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\ddot{\theta}+10.9\sin\theta=0,
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\qquad
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\ddot{\theta}+10.9\theta=0,
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\]
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\[
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\omega=3.30\,\mathrm{rad/s},
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\qquad
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T=1.90\,\mathrm{s},
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\qquad
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f=0.526\,\mathrm{Hz},
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\]
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and
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\[
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t=0.475\,\mathrm{s},
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\qquad
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v_{\max}=0.297\,\mathrm{m/s}.
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\]
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