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