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concepts/mechanics/u7/m7-1-shm.tex

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