Prof. Alejandro Ucan-Puc
By the end of this session, you can:
$$m x'' + c x' + k x = f(t)$$
| Term | Mechanical meaning | Circuit meaning |
|---|---|---|
| $m x''$ | inertia | $Lq''$ (inductance) |
| $c x'$ | damping | $Rq'$ (resistance) |
| $kx$ | spring restoring force | $q/C$ (capacitor voltage) |
| $f(t)$ | applied force | $E(t)$ (applied voltage) |
| 0–10 | Modeling: Newton + Hooke | 10–28 | Undamped free motion |
|---|---|---|---|
| 28–33 | Qualitative reasoning | 33–55 | Damped free motion |
| 55–60 | Break / retrieval quiz | 60–78 | Forced motion and resonance |
| 78–95 | Series RLC circuits | 95–100 | Wrap-up / exit ticket |
Convention: measure displacement $x$ from equilibrium, with downward positive.
If the spring's static extension is $s$, Newton's law gives
$$m x'' = mg - k(s+x) - c x' + f(t).$$
At equilibrium, $mg=ks$. Cancel the weight and static spring force:
$$m x''+c x'+kx=f(t).$$
Set $c=0$ and $f(t)=0$. Then
$$x''+\omega^2x=0, \qquad \omega^2=\frac{k}{m}.$$
$$x(t)=c_1\cos(\omega t)+c_2\sin(\omega t)=A\sin(\omega t+\phi).$$
A 24-lb weight stretches a spring 4 in. It is pulled 4 in below equilibrium and released from rest. Find its motion.
$$\omega^2=\frac{k}{m}=96, \qquad \omega=4\sqrt{6}\ \text{rad/s}.$$
Applying the initial conditions gives
$$x(t)=\frac{1}{3}\cos(4\sqrt{6}\,t)\ \text{ft}.$$
Its amplitude is $1/3$ ft and its period is $T=\pi/(2\sqrt{6})$ s. With no damping, the motion continues indefinitely.
A mass of $1/2$ slug is attached to a spring with $k=36$ lb/ft. It is displaced 2 in below equilibrium and released from rest.
Compare your sign and units with a partner before revealing the result.
$$\omega=\sqrt{36/(1/2)}=6\sqrt{2}\ \text{rad/s}, \qquad T=\frac{\pi}{3\sqrt{2}}\ \text{s}.$$
The initial displacement is $2$ in $=1/6$ ft, so
$$x(t)=\frac{1}{6}\cos(6\sqrt{2}\,t)\ \text{ft}.$$
Suppose the motion satisfies
$$x''+\frac{1}{4}e^{-0.1t}x=0.$$
As $t\to\infty$, the restoring-force coefficient tends to zero.
Predict: Why should we not expect sustained oscillation? What equation describes the long-time trend?
For large $t$, $e^{-0.1t}/4\approx0$, so the equation approaches $x''=0$.
$$x(t)\approx C_1t+C_2.$$
The restoring force weakens away; it cannot keep pulling the mass back strongly enough for continuing oscillations. The solution may keep stretching rather than settling into a periodic motion.
Habit to build: inspect the coefficients and limiting equation before reaching for a solution method.
$$m x''+\beta x'+kx=0, \qquad mr^2+\beta r+k=0.$$
| Discriminant | Motion | Response shape |
|---|---|---|
| $\beta^2-4mk>0$ | Overdamped | Two real roots; no oscillation |
| $\beta^2-4mk=0$ | Critically damped | Repeated root; fastest non-oscillatory return |
| $\beta^2-4mk<0$ | Underdamped | Decaying oscillation |
Sketch each case: mark equilibrium, whether it crosses, and how the amplitude changes over time.
A 4-lb weight stretches a spring 2 ft. The damping constant is $\beta=1$. The mass is pulled 1 ft above equilibrium and launched downward at 12 ft/s.
With a repeated root, $x=(c_1+c_2t)e^{-4t}$. From $x(0)=-1$, $c_1=-1$.
$$x'(0)=c_2-4c_1=12 \quad\Longrightarrow\quad c_2=8.$$
$$x(t)=(-1+8t)e^{-4t}\ \text{ft}.$$
This is critically damped: it crosses equilibrium once, then returns toward equilibrium without oscillating.
Equilibrium occurs when $x=0$:
$$-1+8t=0 \quad\Longrightarrow\quad t=\frac18\ \text{s after release}.$$
Velocity is $x'=(12-32t)e^{-4t}$. Its zero is at $t=3/8$ s, where
$$x(3/8)=2e^{-3/2}\ \text{ft}.$$
$3/8$ s is measured from release. The time after passing equilibrium is $3/8-1/8=1/4$ s.
Without notes, write down:
Take a short break, then compare answers with a partner.
For sinusoidal forcing,
$$m x''+c x'+kx=F_0\cos(\gamma t), \qquad x=x_c+x_p.$$
Solve the particular-response part of
$$x''+4x=\cos(2t).$$
The natural frequency is $2$, the same as the forcing frequency. The usual cosine/sine trial duplicates a homogeneous solution, so include a factor of $t$:
$$x_p(t)=\frac{t}{4}\sin(2t).$$
The resonant particular solution has an amplitude envelope proportional to $t$: in the ideal undamped model, the amplitude grows without bound.
With damping, the long-time amplitude stays bounded; it reaches a peak near the natural frequency.
Check: In a forced problem, name which part eventually fades and which part remains while the input continues.
Let $q(t)$ be charge and $i(t)=q'(t)$ current. Kirchhoff's voltage law gives
$$Lq''+Rq'+\frac{1}{C}q=E(t).$$
| Mechanical | Electrical |
|---|---|
| mass $m$ | inductance $L$ |
| damping $c$ | resistance $R$ |
| spring constant $k$ | $1/C$ |
| displacement $x$ | charge $q$ |
| velocity $x'$ | current $i=q'$ |
| force $f(t)$ | voltage $E(t)$ |
Take $L=1$ H, $R=2\ \Omega$, $C=1/5$ F, and a constant source $E=10$ V, with $q(0)=0$ and $i(0)=0$.
$$q''+2q'+5q=10.$$
The characteristic roots are $-1\pm2i$ and the constant particular solution is $q_p=2$ C. Thus
$$q(t)=2-e^{-t}(2\cos2t+\sin2t)\ \text{C}.$$
Differentiate charge to obtain
$$i(t)=q'(t)=5e^{-t}\sin2t\ \text{A}.$$
Before calculating, vote: over-, critically, or underdamped?
One model, three readings: mechanical free motion, mechanical forced motion, and electrical charge response.