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\subsection{Simple Harmonic Motion and Its Governing ODE}
This subsection introduces simple harmonic motion as one-dimensional motion about a stable equilibrium under a linear restoring law.
\dfn{Simple harmonic motion and the equilibrium coordinate}{Let an object move along a line with fixed unit vector $\hat{u}$. Let
\[
\vec{r}(t)=q(t)\hat{u}
\]
denote the object's displacement from a stable equilibrium position, where $q(t)$ is the signed equilibrium coordinate. The motion is called \emph{simple harmonic motion} (SHM) if the net restoring force is proportional to the displacement and points toward equilibrium:
\[
\vec{F}_{\mathrm{net}}=-kq\hat{u}
\]
for some constant $k>0$.
Equivalently, in scalar form along the chosen axis,
\[
F_{\mathrm{net}}=-kq.
\]
The negative sign shows that when $q>0$ the force is negative, and when $q<0$ the force is positive, so the force always points back toward $q=0$.}
\thm{SHM ODE, standard solution, and period relations}{Let an object of mass $m$ move in SHM with equilibrium coordinate $q(t)$ and restoring constant $k>0$. Define
\[
\omega=\sqrt{\frac{k}{m}}.
\]
Then the governing differential equation is
\[
q''+\omega^2 q=0.
\]
Its standard solution may be written as
\[
q(t)=C\cos(\omega t)+D\sin(\omega t),
\]
where $C$ and $D$ are constants set by the initial conditions, or equivalently as
\[
q(t)=A\cos(\omega t+\phi)
\]
for amplitude $A\ge 0$ and phase constant $\phi$.
The period $T$ and frequency $f$ are
\[
T=\frac{2\pi}{\omega},
\qquad
f=\frac{1}{T}=\frac{\omega}{2\pi}.
\]}
\pf{Short derivation from the linear restoring law}{For SHM, the net force along the line of motion is
\[
F_{\mathrm{net}}=-kq.
\]
Newton's second law gives
\[
m\frac{d^2q}{dt^2}=-kq.
\]
Divide by $m$:
\[
\frac{d^2q}{dt^2}+\frac{k}{m}q=0.
\]
If we define
\[
\omega^2=\frac{k}{m},
\]
then the equation becomes
\[
q''+\omega^2 q=0.
\]
The solutions of this constant-coefficient ODE are sinusoidal, so one may write
\[
q(t)=C\cos(\omega t)+D\sin(\omega t).
\]
Because sine and cosine repeat after an angle change of $2\pi$, one full cycle takes time
\[
T=\frac{2\pi}{\omega},
\]
and therefore $f=1/T=\omega/(2\pi)$.}
\ex{Illustrative example}{A particle's equilibrium coordinate satisfies
\[
q''+25q=0.
\]
Identify $\omega$, the period, and the frequency.
Compare this with the SHM form $q''+\omega^2 q=0$. Then
\[
\omega^2=25
\qquad \Rightarrow \qquad
\omega=5.0\,\mathrm{rad/s}.
\]
So the period is
\[
T=\frac{2\pi}{\omega}=\frac{2\pi}{5.0}\,\mathrm{s}=1.26\,\mathrm{s},
\]
and the frequency is
\[
f=\frac{1}{T}=\frac{5.0}{2\pi}\,\mathrm{Hz}=0.796\,\mathrm{Hz}.
\]
Thus this motion is SHM with angular frequency $5.0\,\mathrm{rad/s}$, period $1.26\,\mathrm{s}$, and frequency $0.796\,\mathrm{Hz}$.}
\qs{Worked example}{For one-dimensional SHM about equilibrium, let the equilibrium coordinate be
\[
q(t)=(0.080\,\mathrm{m})\cos\!\left(4\pi t-\tfrac{\pi}{3}\right)
\]
with $t$ in seconds.
Find:
\begin{enumerate}[label=(\alph*)]
\item the amplitude,
\item the angular frequency,
\item the period and frequency,
\item the displacement at $t=0$, and
\item the governing differential equation in the form $q''+\omega^2 q=0$.
\end{enumerate}}
\sol From
\[
q(t)=A\cos(\omega t+\phi),
\]
we identify the amplitude as the coefficient of the cosine and the angular frequency as the coefficient of $t$ inside the cosine.
For part (a),
\[
A=0.080\,\mathrm{m}.
\]
For part (b),
\[
\omega=4\pi\,\mathrm{rad/s}.
\]
For part (c), the period is
\[
T=\frac{2\pi}{\omega}=\frac{2\pi}{4\pi}=0.50\,\mathrm{s}.
\]
Therefore the frequency is
\[
f=\frac{1}{T}=\frac{1}{0.50\,\mathrm{s}}=2.0\,\mathrm{Hz}.
\]
For part (d), substitute $t=0$ into the position function:
\[
q(0)=(0.080)\cos\!\left(-\frac{\pi}{3}\right)\,\mathrm{m}.
\]
Since $\cos(-\pi/3)=\cos(\pi/3)=1/2$,
\[
q(0)=(0.080)\left(\frac12\right)=0.040\,\mathrm{m}.
\]
For part (e), SHM always satisfies
\[
q''+\omega^2 q=0.
\]
Here $\omega=4\pi\,\mathrm{rad/s}$, so
\[
\omega^2=(4\pi)^2=16\pi^2.
\]
Thus the governing ODE is
\[
q''+16\pi^2 q=0.
\]
Therefore,
\[
A=0.080\,\mathrm{m},
\qquad
\omega=4\pi\,\mathrm{rad/s},
\]
\[
T=0.50\,\mathrm{s},
\qquad
f=2.0\,\mathrm{Hz},
\qquad
q(0)=0.040\,\mathrm{m},
\]
and the motion is governed by
\[
q''+16\pi^2 q=0.
\]

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\subsection{The Spring-Mass Oscillator}
This subsection models a mass attached to an ideal spring, using displacement measured from equilibrium so that both horizontal motion and vertical motion about equilibrium take the same mathematical form.
\dfn{Spring-mass oscillator and equilibrium coordinate}{Let $m$ denote the mass of an object and let $k_{\mathrm{eff}}>0$ denote the effective spring constant of the spring system attached to it. Let the motion occur along a line with positive direction given by the unit vector $\hat{u}$.
Define $x(t)$ to be the signed displacement of the mass from its equilibrium position, measured along that line. Then a \emph{spring-mass oscillator} is a system for which the net restoring force is proportional to $x$ and opposite its sign.
For a horizontal spring, $x$ is measured directly from the equilibrium position on the track. For a vertical spring, let $y(t)$ denote the displacement measured from the spring's unstretched length and let $y_{\mathrm{eq}}$ denote the static equilibrium displacement. The equilibrium coordinate is then
\[
x=y-y_{\mathrm{eq}}.
\]
Using $x$ rather than $y$ makes the vertical oscillator look exactly like the horizontal one.}
\thm{Governing equation and period of a spring-mass oscillator}{Let $m$ denote the mass, let $k_{\mathrm{eff}}$ denote the effective spring constant, and let $x(t)$ denote displacement from equilibrium. Then the motion satisfies
\[
m\ddot{x}+k_{\mathrm{eff}}x=0.
\]
Equivalently,
\[
\ddot{x}+\omega^2 x=0,
\qquad
\omega=\sqrt{\frac{k_{\mathrm{eff}}}{m}}.
\]
Thus the motion is simple harmonic. If $T$ denotes the period and $f$ denotes the frequency, then
\[
T=2\pi\sqrt{\frac{m}{k_{\mathrm{eff}}}},
\qquad
f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k_{\mathrm{eff}}}{m}}.
\]
One convenient position model is
\[
x(t)=A\cos(\omega t+\phi),
\]
where $A$ is the amplitude and $\phi$ is a phase constant. Consequently,
\[
v_{\max}=A\omega,
\qquad
a_{\max}=A\omega^2.
\]}
\pf{Why the equation is the same horizontally and vertically}{For horizontal motion, let $x$ denote displacement from equilibrium along $\hat{u}$. The spring force is
\[
\vec{F}_s=-k_{\mathrm{eff}}x\hat{u}.
\]
By Newton's second law,
\[
m\ddot{x}=-k_{\mathrm{eff}}x,
\]
so
\[
m\ddot{x}+k_{\mathrm{eff}}x=0.
\]
For vertical motion, choose downward as positive. Let $y$ denote the downward displacement from the unstretched length, and let $y_{\mathrm{eq}}$ denote the equilibrium value. At equilibrium,
\[
mg-k_{\mathrm{eff}}y_{\mathrm{eq}}=0.
\]
Now write the actual position as $y=y_{\mathrm{eq}}+x$, where $x$ is displacement from equilibrium. Then
\[
m\ddot{y}=mg-k_{\mathrm{eff}}y=mg-k_{\mathrm{eff}}(y_{\mathrm{eq}}+x).
\]
Since $mg-k_{\mathrm{eff}}y_{\mathrm{eq}}=0$ and $\ddot{y}=\ddot{x}$, this becomes
\[
m\ddot{x}=-k_{\mathrm{eff}}x.
\]
So in either viewpoint,
\[
m\ddot{x}+k_{\mathrm{eff}}x=0.
\]
Comparing with $\ddot{x}+\omega^2x=0$ gives $\omega=\sqrt{k_{\mathrm{eff}}/m}$, and then $T=2\pi/\omega$ and $f=1/T$.}
\ex{Illustrative example}{A block of mass $m=0.40\,\mathrm{kg}$ oscillates on a frictionless horizontal surface attached to a spring system with effective spring constant $k_{\mathrm{eff}}=100\,\mathrm{N/m}$. Find the angular frequency, period, and frequency.
Use
\[
\omega=\sqrt{\frac{k_{\mathrm{eff}}}{m}}=\sqrt{\frac{100}{0.40}}=\sqrt{250}=15.8\,\mathrm{rad/s}.
\]
Then
\[
T=2\pi\sqrt{\frac{m}{k_{\mathrm{eff}}}}=2\pi\sqrt{\frac{0.40}{100}}=0.397\,\mathrm{s},
\]
and
\[
f=\frac{1}{T}=\frac{1}{0.397}=2.52\,\mathrm{Hz}.
\]
So the oscillator has angular frequency $15.8\,\mathrm{rad/s}$, period $0.397\,\mathrm{s}$, and frequency $2.52\,\mathrm{Hz}$.}
\qs{Worked AP-style problem}{A mass $m=0.60\,\mathrm{kg}$ hangs from a vertical spring with spring constant $k=150\,\mathrm{N/m}$. Choose downward as positive. Let $y$ denote the mass's downward displacement from the spring's unstretched length, let $y_{\mathrm{eq}}$ denote the equilibrium displacement, and let
\[
x=y-y_{\mathrm{eq}}
\]
denote displacement from equilibrium.
The mass is pulled downward $0.080\,\mathrm{m}$ from equilibrium and released from rest.
Find:
\begin{enumerate}[label=(\alph*)]
\item the equilibrium displacement $y_{\mathrm{eq}}$,
\item the differential equation for $x(t)$ together with the angular frequency and period, and
\item the maximum speed of the mass.
\end{enumerate}}
\sol At static equilibrium, the acceleration is zero, so the net force is zero:
\[
mg-ky_{\mathrm{eq}}=0.
\]
Therefore,
\[
y_{\mathrm{eq}}=\frac{mg}{k}=\frac{(0.60\,\mathrm{kg})(9.8\,\mathrm{m/s^2})}{150\,\mathrm{N/m}}=0.0392\,\mathrm{m}.
\]
So the equilibrium position is $3.92\times 10^{-2}\,\mathrm{m}$ below the unstretched length.
Now write the motion in terms of displacement from equilibrium:
\[
x=y-y_{\mathrm{eq}}.
\]
The net force is
\[
F_{\mathrm{net}}=mg-ky=mg-k(y_{\mathrm{eq}}+x).
\]
Using $mg-ky_{\mathrm{eq}}=0$, this becomes
\[
F_{\mathrm{net}}=-kx.
\]
Apply Newton's second law:
\[
m\ddot{x}=-kx.
\]
Hence the differential equation is
\[
0.60\,\ddot{x}+150x=0,
\]
or equivalently,
\[
\ddot{x}+250x=0.
\]
Therefore,
\[
\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{150}{0.60}}=\sqrt{250}=15.8\,\mathrm{rad/s}.
\]
The period is
\[
T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{0.60}{150}}=0.397\,\mathrm{s}.
\]
Because the mass is released from rest $0.080\,\mathrm{m}$ from equilibrium, the amplitude is
\[
A=0.080\,\mathrm{m}.
\]
For simple harmonic motion,
\[
v_{\max}=A\omega.
\]
So
\[
v_{\max}=(0.080)(15.8)=1.26\,\mathrm{m/s}.
\]
Thus,
\[
y_{\mathrm{eq}}=0.0392\,\mathrm{m},
\qquad
\ddot{x}+250x=0,
\qquad
\omega=15.8\,\mathrm{rad/s},
\qquad
T=0.397\,\mathrm{s},
\]
and the maximum speed is
\[
v_{\max}=1.26\,\mathrm{m/s}.
\]

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\subsection{Energy in Simple Harmonic Motion}
This subsection uses energy to describe how a frictionless spring-mass oscillator trades energy between motion and spring deformation.
\dfn{Kinetic, potential, and total energy in spring SHM}{Consider a block of mass $m$ attached to an ideal spring of spring constant $k$ and moving frictionlessly along the $x$-axis. Let $x$ denote the signed displacement from equilibrium, let $\vec{v}=\dot{x}\hat{\imath}$ denote the block's velocity, let $v=|\vec{v}|=|\dot{x}|$ denote its speed, and let $A>0$ denote the amplitude of the motion.
The kinetic energy is
\[
K=\tfrac12 mv^2=\tfrac12 m\dot{x}^2.
\]
The spring potential energy is
\[
U_s=\tfrac12 kx^2.
\]
The total mechanical energy is
\[
E=K+U_s=\tfrac12 m\dot{x}^2+\tfrac12 kx^2.
\]}
\thm{Conserved-energy relation for SHM}{For the frictionless spring-mass oscillator above, the total mechanical energy is constant:
\[
E=\tfrac12 m\dot{x}^2+\tfrac12 kx^2=\tfrac12 kA^2.
\]
Therefore, at any displacement $x$,
\[
\dot{x}^2=\frac{k}{m}\left(A^2-x^2\right),
\qquad
v=\sqrt{\frac{k}{m}\left(A^2-x^2\right)}.
\]
In particular, the maximum speed occurs at equilibrium $x=0$:
\[
v_{\max}=\sqrt{\frac{k}{m}}\,A.
\]}
\nt{At the turning points $x=\pm A$, the block reverses direction, so $v=0$, $K=0$, and all the mechanical energy is spring potential energy:
\[
U_s=E=\tfrac12 kA^2.
\]
At equilibrium $x=0$, the spring is neither stretched nor compressed, so $U_s=0$ and all the energy is kinetic:
\[
K=E=\tfrac12 kA^2.
\]
Thus SHM continually swaps energy between kinetic and potential forms. If $x(t)=A\cos(\omega t+\phi)$, then $U_s\propto \cos^2(\omega t+\phi)$ and $K\propto \sin^2(\omega t+\phi)$, so the two energy curves are out of phase and each repeats twice during one full oscillation.}
\pf{Short derivation from conservation of mechanical energy}{For a frictionless spring-mass system, the only horizontal interaction is the spring force
\[
\vec{F}_s=-kx\hat{\imath},
\]
which is conservative. Therefore the mechanical energy $E=K+U_s$ is constant. Using the spring potential-energy function,
\[
U_s=\tfrac12 kx^2,
\]
the total energy at any instant is
\[
E=\tfrac12 m\dot{x}^2+\tfrac12 kx^2.
\]
At a turning point, $x=\pm A$ and $\dot{x}=0$, so
\[
E=\tfrac12 kA^2.
\]
Equating the two expressions for $E$ gives
\[
\tfrac12 m\dot{x}^2+\tfrac12 kx^2=\tfrac12 kA^2.
\]
Solving for $\dot{x}^2$ yields
\[
\dot{x}^2=\frac{k}{m}\left(A^2-x^2\right),
\]
and taking the positive square root gives the speed formula for the magnitude $v=|\dot{x}|$.}
\qs{Worked example}{A block of mass $m=0.40\,\mathrm{kg}$ is attached to an ideal horizontal spring of spring constant $k=160\,\mathrm{N/m}$ and oscillates frictionlessly with amplitude $A=0.10\,\mathrm{m}$. At one instant, the block is at displacement $x=+0.060\,\mathrm{m}$ from equilibrium.
Find:
\begin{enumerate}[label=(\alph*)]
\item the total mechanical energy of the oscillator,
\item the spring potential energy and kinetic energy at $x=+0.060\,\mathrm{m}$,
\item the speed of the block at that displacement, and
\item the maximum speed and where it occurs.
\end{enumerate}}
\sol Let $E$ denote the total mechanical energy. Because the motion is frictionless,
\[
E=\tfrac12 kA^2.
\]
Substitute $k=160\,\mathrm{N/m}$ and $A=0.10\,\mathrm{m}$:
\[
E=\tfrac12 (160)(0.10)^2=80(0.010)=0.80\,\mathrm{J}.
\]
So the oscillator's total mechanical energy is
\[
0.80\,\mathrm{J}.
\]
At $x=+0.060\,\mathrm{m}$, the spring potential energy is
\[
U_s=\tfrac12 kx^2=\tfrac12 (160)(0.060)^2.
\]
Since $(0.060)^2=0.0036$,
\[
U_s=80(0.0036)=0.288\,\mathrm{J}.
\]
Then the kinetic energy is
\[
K=E-U_s=0.80-0.288=0.512\,\mathrm{J}.
\]
Now use kinetic energy to find the speed:
\[
K=\tfrac12 mv^2.
\]
So
\[
0.512=\tfrac12 (0.40)v^2=0.20v^2.
\]
Thus,
\[
v^2=\frac{0.512}{0.20}=2.56,
\qquad
v=1.60\,\mathrm{m/s}.
\]
For the maximum speed, use the equilibrium position $x=0$, where all the energy is kinetic:
\[
\tfrac12 mv_{\max}^2=E=0.80\,\mathrm{J}.
\]
Therefore,
\[
0.20v_{\max}^2=0.80,
\qquad
v_{\max}^2=4.0,
\qquad
v_{\max}=2.0\,\mathrm{m/s}.
\]
Equivalently,
\[
v_{\max}=\sqrt{\frac{k}{m}}\,A=\sqrt{\frac{160}{0.40}}(0.10)=20(0.10)=2.0\,\mathrm{m/s}.
\]
Therefore,
\[
E=0.80\,\mathrm{J},
\qquad
U_s=0.288\,\mathrm{J},
\qquad
K=0.512\,\mathrm{J},
\]
\[
v=1.60\,\mathrm{m/s},
\qquad
v_{\max}=2.0\,\mathrm{m/s}\text{ at }x=0.
\]

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\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
\[
f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{g}{\ell}}.
\]}
\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
\[
t=0.475\,\mathrm{s},
\qquad
v_{\max}=0.297\,\mathrm{m/s}.
\]

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\subsection{Physical Pendulum and Small-Angle Linearization}
This subsection models the small oscillations of a rigid body that swings about a fixed pivot under gravity.
\dfn{Physical pendulum, pivot-to-CM distance, and angular coordinate}{Let a rigid body of mass $m$ swing in a vertical plane about a fixed pivot point $O$. Let $C$ denote the center of mass of the body, let
\[
d=OC
\]
denote the distance from the pivot to the center of mass, and let $\theta(t)$ denote the angular displacement from the stable vertical equilibrium position.
Such a system is called a \emph{physical pendulum}. Unlike a simple pendulum, the body's mass is distributed throughout the rigid object, so its rotational inertia must be included in the dynamics.}
\thm{Exact torque equation and small-angle SHM model}{Let $m$ denote the mass of the rigid body, let $d$ denote the distance from the pivot to the center of mass, let $I$ denote the moment of inertia of the body about the pivot, let $g$ denote the magnitude of the gravitational field, and let $\theta(t)$ denote the angular displacement from stable equilibrium.
Then the exact rotational equation of motion is
\[
I\ddot{\theta}=-mgd\sin\theta,
\]
or equivalently,
\[
I\ddot{\theta}+mgd\sin\theta=0.
\]
For small angular displacements, use the linearization $\sin\theta\approx\theta$ to obtain
\[
I\ddot{\theta}+mgd\,\theta=0.
\]
Therefore the motion is approximately simple harmonic with angular frequency
\[
\omega=\sqrt{\frac{mgd}{I}}
\]
and period
\[
T=2\pi\sqrt{\frac{I}{mgd}}.
\]
A simple pendulum is the special case in which all the mass is concentrated a distance $L$ from the pivot, so $I=mL^2$ and $d=L$.}
\pf{Short derivation from torque and linearization}{The weight $m\vec{g}$ acts at the center of mass. When the body is displaced by angle $\theta$, the gravitational torque about the pivot is restoring, so
\[
\tau=-mgd\sin\theta.
\]
For rotation about a fixed axis, Newton's second law for rotation gives
\[
\sum \tau=I\ddot{\theta}.
\]
Hence,
\[
I\ddot{\theta}=-mgd\sin\theta,
\]
which is the exact equation.
If the oscillations are small, then $\sin\theta\approx\theta$, so the equation becomes
\[
I\ddot{\theta}+mgd\,\theta=0.
\]
Divide by $I$ to get
\[
\ddot{\theta}+\frac{mgd}{I}\theta=0.
\]
Comparing with the SHM form $q''+\omega^2 q=0$ shows that
\[
\omega^2=\frac{mgd}{I},
\qquad
\omega=\sqrt{\frac{mgd}{I}}.
\]
Therefore,
\[
T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{mgd}}.
\]}
\ex{Illustrative example}{Show that the simple pendulum is a special case of the physical pendulum formula.
For a point mass $m$ at distance $L$ from the pivot,
\[
I=mL^2,
\qquad
d=L.
\]
Substitute into the physical-pendulum period formula:
\[
T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{mL^2}{mgL}}=2\pi\sqrt{\frac{L}{g}}.
\]
This is exactly the small-angle period of a simple pendulum.}
\qs{Worked AP-style problem}{A uniform rod of mass $m=1.50\,\mathrm{kg}$ and length $L=0.90\,\mathrm{m}$ is pivoted about one end and allowed to swing in a vertical plane. Let $\theta(t)$ denote the angular displacement from the stable vertical equilibrium position. Assume the oscillations are small.
Find:
\begin{enumerate}[label=(\alph*)]
\item the pivot-to-center-of-mass distance $d$ and the rod's moment of inertia $I$ about the pivot,
\item the small-angle differential equation for $\theta(t)$, and
\item the period of oscillation.
\end{enumerate}}
\sol For a uniform rod pivoted about one end, the center of mass is at the midpoint, so
\[
d=\frac{L}{2}=\frac{0.90\,\mathrm{m}}{2}=0.45\,\mathrm{m}.
\]
The moment of inertia of a uniform rod about one end is
\[
I=\frac{1}{3}mL^2.
\]
Substitute the given values:
\[
I=\frac{1}{3}(1.50)(0.90)^2\,\mathrm{kg\cdot m^2}.
\]
Since $(0.90)^2=0.81$,
\[
I=\frac{1}{3}(1.50)(0.81)=0.405\,\mathrm{kg\cdot m^2}.
\]
For small oscillations, a physical pendulum satisfies
\[
I\ddot{\theta}+mgd\,\theta=0.
\]
Now compute $mgd$:
\[
mgd=(1.50)(9.8)(0.45)=6.615.
\]
So the differential equation is
\[
0.405\,\ddot{\theta}+6.615\,\theta=0.
\]
Divide by $0.405$:
\[
\ddot{\theta}+16.3\,\theta=0.
\]
Thus,
\[
\omega=\sqrt{16.3}=4.04\,\mathrm{rad/s}.
\]
The period is
\[
T=\frac{2\pi}{\omega}=\frac{2\pi}{4.04}=1.56\,\mathrm{s}.
\]
Equivalently, using the period formula directly,
\[
T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{0.405}{6.615}}=1.56\,\mathrm{s}.
\]
Therefore,
\[
d=0.45\,\mathrm{m},
\qquad
I=0.405\,\mathrm{kg\cdot m^2},
\]
and the small-angle motion is governed by
\[
\ddot{\theta}+16.3\,\theta=0,
\qquad
T=1.56\,\mathrm{s}.
\]