\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
\[
q''+16\pi^2 q=0.
\]
