Spring-Mass Systems and Harmonic Oscillators

The canonical second-order physical ODE

Newton's second law applied to a mass on a spring with a dashpot (viscous damper) gives $m \, x'' + c \, x' + k \, x = 0.$

Forces in play:

• $-k\,x$: spring restoring force (Hooke's law).
• $-c\,x'$: drag force proportional to velocity (dashpot / viscous damping).
• $m\,x''$: inertia.

This is the textbook damped harmonic oscillator — one of the most-studied equations in physics and engineering.

The given system

• Mass: $m = 2$ kg.
• Stiffness: $k = 50$ N/m.
• Damping: $c = 4$ N·s/m.

So $2 x'' + 4 x' + 50 x = 0$.

Step-by-step

Step 1 — Normalise to standard form.

Divide by $m = 2$: $x'' + 2 x' + 25 x = 0.$

Step 2 — Characteristic equation (see #181).

$r^{2} + 2 r + 25 = 0$

Discriminant $\Delta = 4 - 100 = -96 < 0$ → complex conjugate roots: $r = \frac{-2 \pm \sqrt{-96}}{2} = -1 \pm i \sqrt{24} = -1 \pm 2\sqrt{6}\, i.$

So $\alpha = -1$, $\beta = 2\sqrt{6} \approx 4.899$.

Step 3 — Identify the regime.

$\Delta < 0$ ⇒ under-damped. Oscillates while decaying.

The critical damping would be $c_{\text{crit}} = 2\sqrt{m k} = 2\sqrt{2 \cdot 50} = 20 \text{ N·s/m}.$

Our $c = 4$ is far below 20, so we're well into the under-damped regime.

Step 4 — General solution.

$\boxed{\, x(t) = e^{-t}\bigl(C_1 \cos(2\sqrt{6}\, t) + C_2 \sin(2\sqrt{6}\, t)\bigr) \,}$

Natural frequency (un-damped): $\omega_0 = \sqrt{k/m} = \sqrt{25} = 5$ rad/s. Damped frequency: $\omega_d = \sqrt{\omega_0^2 - \gamma^2} = \sqrt{25 - 1} = 2\sqrt{6} \approx 4.899$ rad/s. Decay rate: $\gamma = c / (2m) = 1$ s⁻¹. Time constant (envelope): $1/\gamma = 1$ s.

Step 5 — Apply initial conditions.

Say $x(0) = 0.1$ m (displaced 10 cm) and $x'(0) = 0$ (released from rest).

$x(0) = C_1 = 0.1$. $x'(0) = -C_1 + \beta C_2 = 0 \implies C_2 = C_1 / \beta = 0.1 / (2\sqrt{6}) \approx 0.0204$.

$x(t) \approx e^{-t}\bigl(0.1 \cos(2\sqrt{6}\, t) + 0.0204 \sin(2\sqrt{6}\, t)\bigr).$

Verification (structural)

Both $e^{-t} \cos(2\sqrt{6}\, t)$ and $e^{-t} \sin(2\sqrt{6}\, t)$ solve $x'' + 2x' + 25x = 0$ because $r = -1 \pm 2\sqrt{6}i$ satisfy the characteristic equation. Any linear combination does too (superposition — see #200).

The three damping regimes, geometrically

Let $\gamma = c/(2m)$ (decay rate) and $\omega_0 = \sqrt{k/m}$ (natural frequency). Then:

• Under-damped ($\gamma < \omega_0$): oscillates inside an envelope $\pm A e^{-\gamma t}$. Bouncy. (Our case.)
• Critically damped ($\gamma = \omega_0$): $(C_1 + C_2 t) e^{-\gamma t}$. Returns fastest, no overshoot.
• Over-damped ($\gamma > \omega_0$): two decaying exponentials with different rates. Sluggish return, no oscillation.

Shock absorbers in cars are tuned near critical damping — any less and the car bounces; any more and it wallows.

Energy and the envelope

Total mechanical energy $E(t) = \tfrac{1}{2} m x'^2 + \tfrac{1}{2} k x^2$ decays exponentially: $E(t) = E(0) \, e^{-2\gamma t} \times \text{(oscillation)}.$ Damping dissipates energy at the rate $c \, x'^2$ per unit time. Not conserved — the dashpot converts it to heat.

Quality factor Q

A dimensionless measure of "how bouncy" an oscillator is: $Q = \frac{\omega_0}{2 \gamma} = \frac{\sqrt{m k}}{c}.$

• $Q \gg 1$: very under-damped, rings many cycles before decaying. Tuning forks, crystals, high-Q LC circuits.
• $Q \approx 0.5$: critically damped.
• $Q < 0.5$: over-damped.

Our system: $Q = \sqrt{100}/4 = 10/4 = 2.5$. Moderately under-damped — you'd hear a few dying bounces.

Where this equation shows up

• Mechanical: car suspension, building response to wind, pendulums, bridges, watch escapements.
• Electrical: RLC circuit (#203) — literally the same ODE with $Q \leftrightarrow x$, $L \leftrightarrow m$, $R \leftrightarrow c$, $1/C \leftrightarrow k$.
• Acoustic: speaker cone dynamics, room resonance.
• Biological: heart rhythms (to first order), swinging limbs during walking.

Master this one ODE and you have a mental model for half of undergraduate physics.

Common mistakes

• Forgetting to divide by $m$. The characteristic equation is $r^2 + (c/m) r + (k/m) = 0$ — easy to accidentally use $r^2 + c r + k$.
• Confusing natural and damped frequency. $\omega_0 = \sqrt{k/m}$ is what you'd see with no damping; $\omega_d = \sqrt{\omega_0^2 - \gamma^2}$ is what you actually observe (slightly slower).
• Sign of the damping force. Damping opposes motion, so the term has sign $+c x'$ on the left side (or $-c x'$ as a force on the right).
• Reading the amplitude off the wrong thing. The envelope is $e^{-\gamma t}$, where $\gamma = c/(2m)$, not $c/m$.

Try it in the visualization

Draw the mass sliding on a track, attached to a spring (coil drawing) and dashpot (cylinder). Animate $x(t)$ as the mass moves. Overlay the $x(t)$ curve with its $\pm e^{-\gamma t}$ envelope. Slide $c$ from $0$ to $30$ to watch the three regimes morph: pure oscillation → under-damped → critically damped → over-damped.