\subsection{The Simple Pendulum}

This subsection models a bob of mass on a light string, using angular displacement from the vertical as the natural coordinate.

\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.

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
\[
s=\ell\theta.
\]
The equilibrium position is $\theta=0$.}

\thm{Exact pendulum equation and small-angle SHM model}{For the simple pendulum above, the exact rotational equation of motion is
\[
\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
\]
This equation is nonlinear, so the motion is not exactly simple harmonic for arbitrary amplitude.

If the oscillation remains at small angles so that $|\theta|\ll 1$ radian and $\sin\theta\approx\theta$, then the motion is approximated by
\[
\ddot{\theta}+\frac{g}{\ell}\theta=0.
\]
Thus the pendulum behaves approximately like SHM with angular frequency
\[
\omega=\sqrt{\frac{g}{\ell}},
\]
small-angle period
\[
T=2\pi\sqrt{\frac{\ell}{g}},
\]
and small-angle frequency
\[
f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{g}{\ell}}.
\]}

\nt{The small-angle approximation $\sin\theta\approx\theta$ introduces error that grows with amplitude. For $\theta_{\max}=10^\circ\approx0.17$ rad, the period error is about $0.2\%$. For $\theta_{\max}=30^\circ$, the period error is about $1.7\%$. The exact period is $T_{\mathrm{exact}}=T_0\big(1+\tfrac14\sin^2\frac{\theta_{\max}}{2}+\tfrac{9}{64}\sin^4\frac{\theta_{\max}}{2}+\cdots\big)$, where $T_0=2\pi\sqrt{\ell/g}$. For AP Physics C, the small-angle model is the standard unless stated otherwise.}

\pf{Short derivation from torque and linearization}{About the pivot, the gravitational torque on the bob is restoring, so
\[
\tau=-mg\ell\sin\theta.
\]
The bob acts like a point mass at distance $\ell$, so its moment of inertia about the pivot is
\[
I=m\ell^2.
\]
Using rotational Newton's second law, $\sum\tau=I\ddot{\theta}$, gives
\[
m\ell^2\ddot{\theta}=-mg\ell\sin\theta.
\]
Divide by $m\ell^2$:
\[
\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
\]
For small oscillations with $|\theta|\ll 1$ radian, use the small-angle approximation $\sin\theta\approx\theta$. Then
\[
\ddot{\theta}+\frac{g}{\ell}\theta=0,
\]
which is the standard SHM equation with $\omega^2=g/\ell$.}

\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?

For small-angle motion,
\[
T=2\pi\sqrt{\frac{\ell}{g}}.
\]
Therefore $T\propto\sqrt{\ell}$. Since
\[
\frac{\ell_2}{\ell_1}=\frac{2.00}{0.50}=4,
\]
the new period is multiplied by
\[
\sqrt{4}=2.
\]
So the period doubles. Because $f=1/T$, the frequency is cut in half.}

\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 $g=9.8\,\mathrm{m/s^2}$.

Assume the small-angle model is valid.

Find:
\begin{enumerate}[label=(\alph*)]
\item the exact equation of motion and the small-angle approximate equation,
\item the angular frequency, period, and frequency,
\item the time required to move from maximum displacement to equilibrium, and
\item the maximum linear speed of the bob.
\end{enumerate}}

\sol Let $\theta(t)$ denote the angular displacement from the downward vertical.

For part (a), the exact pendulum equation is
\[
\ddot{\theta}+\frac{g}{\ell}\sin\theta=0.
\]
Substitute $g=9.8\,\mathrm{m/s^2}$ and $\ell=0.90\,\mathrm{m}$:
\[
\ddot{\theta}+\frac{9.8}{0.90}\sin\theta=0.
\]
Thus,
\[
\ddot{\theta}+10.9\sin\theta=0
\]
to three significant figures.

Under the small-angle approximation $\sin\theta\approx\theta$, the motion is modeled by
\[
\ddot{\theta}+\frac{9.8}{0.90}\theta=0,
\]
or
\[
\ddot{\theta}+10.9\theta=0.
\]

For part (b), compare the small-angle equation with
\[
\ddot{\theta}+\omega^2\theta=0.
\]
So
\[
\omega=\sqrt{\frac{g}{\ell}}=\sqrt{\frac{9.8}{0.90}}=3.30\,\mathrm{rad/s}.
\]
Then the period is
\[
T=2\pi\sqrt{\frac{\ell}{g}}=\frac{2\pi}{\omega}=\frac{2\pi}{3.30}=1.90\,\mathrm{s}.
\]
The frequency is
\[
f=\frac{1}{T}=\frac{1}{1.90}=0.526\,\mathrm{Hz}.
\]

For part (c), a pendulum in SHM takes one-quarter of a cycle to move from an endpoint to equilibrium. Therefore,
\[
t=\frac{T}{4}=\frac{1.90}{4}=0.475\,\mathrm{s}.
\]

For part (d), the maximum angular speed in SHM is
\[
\dot{\theta}_{\max}=\omega\theta_{\max}.
\]
So
\[
\dot{\theta}_{\max}=(3.30)(0.10)=0.330\,\mathrm{rad/s}.
\]
The bob's linear speed is related by $v=\ell\dot{\theta}$, so the maximum linear speed is
\[
v_{\max}=\ell\dot{\theta}_{\max}=(0.90)(0.330)=0.297\,\mathrm{m/s}.
\]

Therefore,
\[
\ddot{\theta}+10.9\sin\theta=0,
\qquad
\ddot{\theta}+10.9\theta=0,
\]
\[
\omega=3.30\,\mathrm{rad/s},
\qquad
T=1.90\,\mathrm{s},
\qquad
f=0.526\,\mathrm{Hz},
\]
and
\[
t=0.475\,\mathrm{s},
\qquad
v_{\max}=0.297\,\mathrm{m/s}.
\]
