7.3 Applications

Big idea

The previous two sections built machinery. This one shows why it was worth building: one second-order linear equation with constant coefficients describes a mass bouncing on a spring, a door closing on a damper, a car settling after a bump, and the charge sloshing in a circuit. Those systems share no parts. They share an equation.

The equation comes out of Newton’s second law. A mass on a spring feels a restoring force proportional to its displacement from equilibrium, a resisting force roughly proportional to its speed, and possibly an external push. Setting mass times acceleration equal to the sum of those forces produces $mx'' + cx' + kx = F(t)$, whose three coefficients are inertia, resistance and stiffness.

The qualitative behavior is then decided by algebra you already did. The discriminant $c^2 - 4mk$ of $mr^2 + cr + k = 0$ splits the motion into three regimes. Too much damping and the mass creeps back to equilibrium without crossing it; too little and it oscillates inside a shrinking envelope. The boundary between them is the fastest return without overshoot, which is what a shock absorber is designed for.

The electrical version is the same statement with different letters. Inductance resists changes in current the way mass resists changes in velocity, resistance dissipates energy the way friction does, and a capacitor stores charge the way a spring stores displacement. With $L$, $R$ and $1/C$ in the slots held by $m$, $c$ and $k$, every result about vibrating masses becomes one about circuits.

Decoder

For a damped mass on a spring, the sign of $c^2 - 4mk$ decides everything: positive gives decay with no oscillation, zero gives the fastest decay without oscillation, and negative gives oscillation inside a decaying envelope.

The comparison is damping against the product of inertia and stiffness. Heavy damping wins outright and the motion is a sum of two decaying exponentials, which can cross zero at most once. Light damping loses, complex roots appear, and the imaginary part becomes a frequency.

The word “envelope” carries the underdamped case: the solution is $e^{-ct/(2m)}$ times a combination of sine and cosine, so the exponential sets the ceiling and floor while the trigonometric factor fills the space between. Each swing is smaller than the last, so the motion is not periodic, though its zero crossings stay evenly spaced.

The classic mistake is to read “critically damped” as “no motion”. The mass still moves and can still overshoot once; what critical damping rules out is repeated oscillation.

Definitions and results

The spring-mass equation. Measure $x(t)$ from equilibrium, positive downward. Hooke’s law gives a restoring force $-kx$, damping contributes $-cx'$, and an external force is $F(t)$. Newton’s second law reads $mx'' = -kx - cx' + F(t)$, or

$$ mx'' + cx' + kx = F(t) $$

Gravity does not appear: the equilibrium stretch $s$ satisfies $mg = ks$, so measuring from equilibrium cancels the weight.

Free undamped motion. With $c = 0$ and $F = 0$ the equation is $x'' + \omega^2 x = 0$ where $\omega = \sqrt{k/m}$. The roots are $\pm i\omega$, so $x(t) = c_1 \cos \omega t + c_2 \sin \omega t$: simple harmonic motion with period $T = 2\pi/\omega$, which depends only on $m$ and $k$, never on how far you pulled the mass.

Amplitude and phase. The same motion is one shifted cosine, $x(t) = A\cos(\omega t - \phi)$, with $A = \sqrt{c_1^2 + c_2^2}$, $\cos\phi = c_1/A$ and $\sin\phi = c_2/A$; the cosine difference identity recovers $c_1\cos\omega t + c_2 \sin\omega t$. The amplitude is the farthest the mass gets from equilibrium, the phase where in the cycle it starts.

The damping discriminant. With $c > 0$ the characteristic equation $mr^2 + cr + k = 0$ has roots $r = \left(-c \pm \sqrt{c^2 - 4mk}\right)/(2m)$. Both have negative real part when $m$, $c$, $k$ are positive, so free damped motion always decays.

Overdamped and critically damped motion. If $c^2 - 4mk > 0$ the roots are real, distinct and negative, and $x = c_1 e^{r_1 t} + c_2 e^{r_2 t}$. If $c^2 - 4mk = 0$ the single root is $r = -c/(2m)$ and $x = (c_1 + c_2 t)e^{rt}$, where the factor of $t$ grows but the exponential wins. Neither form oscillates, so the mass crosses equilibrium at most once. Critical damping is the least damping that avoids oscillation, the design target for shock absorbers and door closers.

Underdamped motion. If $c^2 - 4mk < 0$ the roots are $-c/(2m) \pm i\mu$ with $\mu = \sqrt{4mk - c^2}\,/(2m)$, and

$$ x(t) = e^{-ct/(2m)}\left(c_1 \cos \mu t + c_2 \sin \mu t\right) $$

The damped frequency $\mu$ is below the undamped $\omega$: friction slows the oscillation as well as shrinking it.

Forced vibration and resonance. With $F(t) = F_0 \cos \gamma t$ and no damping, undetermined coefficients gives a bounded particular solution provided $\gamma \ne \omega$. When $\gamma = \omega$ the trial form collides with the complementary solution, so it must be multiplied by $t$, and the amplitude grows without bound: that is resonance. With damping the growth is capped and the transient decays, leaving a steady-state oscillation at the driving frequency.

The RLC circuit. In a loop with inductance $L$, resistance $R$, capacitance $C$ and voltage $E(t)$, the charge $q(t)$ satisfies

$$ Lq'' + Rq' + \frac{1}{C}q = E(t) $$

with current $i = q'$. Term by term: $L$ with $m$, $R$ with $c$, $1/C$ with $k$, $q$ with $x$, $i$ with velocity, $E(t)$ with the driving force. The discriminant becomes $R^2 - 4L/C$, so the same test names the regime.

Worked examples

A mass in free vibration

A $2$ kg mass hangs on an undamped spring with $k = 18$ N/m. It starts $1$ m below equilibrium moving upward at $3$ m/s, so $2x'' + 18x = 0$, that is $x'' + 9x = 0$ with $\omega = 3$ rad/s.

From $x = c_1 \cos 3t + c_2 \sin 3t$ and $x(0) = 1$, $c_1 = 1$. Since $x' = -3c_1 \sin 3t + 3c_2 \cos 3t$ and upward is negative, $x'(0) = 3c_2 = -3$ gives $c_2 = -1$ and $x = \cos 3t - \sin 3t$. Then $A = \sqrt2$, with $\cos\phi = 1/\sqrt2$ and $\sin\phi = -1/\sqrt2$, so $\phi = -\pi/4$ and

$$ x(t) = \sqrt{2}\cos\left(3t + \frac{\pi}{4}\right) $$

The period is $2\pi/3$ seconds and the mass stays within $\sqrt2$ m of equilibrium. Check at $t = 0$: $\sqrt2 \cos(\pi/4) = 1$, the initial position.

One system, three dampers

Take $m = 1$ and $k = 4$. The discriminant is $c^2 - 16$, so the boundary sits at $c = 4$.

With $c = 5$: $r^2 + 5r + 4 = 0$ gives $r = -1, -4$, so $x = c_1 e^{-t} + c_2 e^{-4t}$, overdamped.

With $c = 4$: $r^2 + 4r + 4 = (r+2)^2$ gives the double root $-2$, so $x = (c_1 + c_2 t)e^{-2t}$, critically damped.

With $c = 2$: $r^2 + 2r + 4 = 0$ gives $r = -1 \pm i\sqrt3$, so $x = e^{-t}(c_1 \cos \sqrt3 t + c_2 \sin \sqrt3 t)$, underdamped. From $x(0) = 1$, $c_1 = 1$; differentiating, $x'(0) = -c_1 + \sqrt3 c_2 = -1$ gives $c_2 = 0$ and $x = e^{-t}\cos\sqrt3 t$. The mass crosses equilibrium repeatedly while the envelope $e^{-t}$ closes in.

Driving at the natural frequency

Solve $x'' + 9x = 6\cos 3t$ with $x(0) = 0$ and $x'(0) = 0$.

The natural and driving frequencies are both $3$, so $\cos 3t$ already solves the homogeneous equation. Multiply the trial form by $t$: with $x_p = t(A\cos 3t + B\sin 3t)$, two differentiations give $x_p'' + 9x_p = -6A\sin 3t + 6B\cos 3t$, so $A = 0$, $B = 1$ and $x_p = t\sin 3t$. Both initial conditions then force $c_1 = c_2 = 0$, leaving $x = t\sin 3t$, an oscillation whose amplitude grows linearly forever. No real spring does this; the model breaks first.

The same equation as a circuit

A series circuit has $L = 1$ H, $R = 4$ ohms and $C = 1/20$ F, with no applied voltage. The capacitor starts uncharged and the initial current is $8$ A, so with $1/C = 20$ the problem is $q'' + 4q' + 20q = 0$, $q(0) = 0$, $q'(0) = 8$.

The discriminant $R^2 - 4L/C = 16 - 80 = -64$ is negative, so the circuit is underdamped: the charge should oscillate while decaying. The roots are $(-4 \pm 8i)/2 = -2 \pm 4i$, so $q = e^{-2t}(c_1 \cos 4t + c_2 \sin 4t)$. From $q(0) = 0$, $c_1 = 0$, and $q'(0) = 4c_2 = 8$, giving

$$ q(t) = 2e^{-2t}\sin 4t, \qquad i(t) = q'(t) = e^{-2t}(8\cos 4t - 4\sin 4t) $$

Check the current at $t = 0$: $8 - 0 = 8$ A, as given. The mechanical system $x'' + 4x' + 20x = 0$ has the identical solution, with the resistor as damper, the inductor as mass and the capacitor as spring.

Practice

Start undamped. Build the equation from $m$ and $k$, solve it, and convert the answer to amplitude, period and phase.

Then add damping. Compute $c^2 - 4mk$, name the regime, and only then write the solution in the matching form.

Last, the electrical version. Set up $Lq'' + Rq' + q/C = E(t)$, solve for the charge, then differentiate for the current.

Quiz

Six items on free vibration, the damping regimes, resonance, and RLC circuits.