\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
\[
v_{\max}=1.26\,\mathrm{m/s}.
\]
