Damped Harmonic Motion: Under, Critical, and Over

A real oscillator loses energy to friction. The behavior depends on how strong the damping is compared to the natural oscillation. The equation of motion is:

$m\ddot x + c\dot x + kx = 0$

Define the natural frequency $\omega_0 = \sqrt{k/m}$ and damping coefficient $\gamma = c/(2m)$. The character of the solution depends on the dimensionless damping ratio $\zeta = \gamma/\omega_0$:

• $\zeta < 1$ (underdamped): oscillates with decaying envelope. Most common.
• $\zeta = 1$ (critically damped): returns to equilibrium as fast as possible without oscillating. Shock absorbers aim for this.
• $\zeta > 1$ (overdamped): returns slowly to equilibrium without oscillating.

The Three Solutions

Underdamped ($\zeta < 1$):

$x(t) = A\,e^{-\gamma t}\cos(\omega_d\,t + \varphi)$

where $\omega_d = \sqrt{\omega_0^{2} - \gamma^{2}}$ is the damped frequency (slightly less than $\omega_0$).

Critically damped ($\zeta = 1$):

$x(t) = (A + Bt)\,e^{-\omega_0 t}$

Overdamped ($\zeta > 1$):

$x(t) = A\,e^{-(\gamma - \sqrt{\gamma^{2} - \omega_0^{2}})t} + B\,e^{-(\gamma + \sqrt{\gamma^{2} - \omega_0^{2}})t}$

Step-by-Step Solution

Given: Three identical springs with $\omega_0 = 4\;\text{rad/s}$. Compare three damping levels: $\gamma = 0.5$ (light), $\gamma = 4$ (critical), $\gamma = 8$ (heavy).

Find: The behavior of each.

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Step 1 — Compute $\zeta$ for each.

• Light: $\zeta = 0.5/4 = 0.125$ (underdamped)
• Critical: $\zeta = 4/4 = 1.000$ (critically damped)
• Heavy: $\zeta = 8/4 = 2.000$ (overdamped)

Step 2 — Light damping behavior.

$\omega_d = \sqrt{16 - 0.25} \approx 3.969\;\text{rad/s}$ (very close to $\omega_0$).

The mass oscillates many times before stopping, with the amplitude decaying as $e^{-0.5 t}$. Half-life: $\ln 2 / 0.5 \approx 1.386\;\text{s}$.

After 5 seconds, the amplitude has dropped to $e^{-2.5} \approx 0.082$ of the initial value — about 8%.

Step 3 — Critical damping behavior.

The mass returns to equilibrium without overshooting, in the shortest possible time. After about $4/\omega_0 = 1\;\text{s}$, it's effectively at rest. This is what shock absorbers and door closers aim for: fast, smooth, no bounce.

Step 4 — Overdamped behavior.

The mass also returns to equilibrium without oscillating, but slower than critical damping. The system has so much friction that even though it doesn't oscillate, it crawls back to zero.

For our $\gamma = 8$, $\omega_0 = 4$:

$\sqrt{\gamma^{2} - \omega_0^{2}} = \sqrt{64 - 16} = \sqrt{48} \approx 6.928$

So the two decay rates are $8 - 6.928 = 1.072$ and $8 + 6.928 = 14.928$. The slow mode dominates the long-time behavior — relaxation time is $1/1.072 \approx 0.93$ seconds.

Step 5 — Why critical damping is optimal.

If you want a system to return to equilibrium as fast as possible without oscillating, critical damping is the answer. Less damping → it oscillates. More damping → it returns slowly. Exactly $\zeta = 1$ minimizes the settling time.

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Answer:

• Underdamped ($\zeta < 1$): oscillates with shrinking amplitude. The most common case for real oscillators.
• Critically damped ($\zeta = 1$): no oscillation, fastest settling. Used in vehicle suspensions, door closers, gauges.
• Overdamped ($\zeta > 1$): no oscillation, slow settling. Like trying to push a heavy door through molasses.

The visualization shows all three behaviors side by side from the same initial conditions.

Try It

• Adjust the damping with the slider to move smoothly between under-, critical, and over-damped regimes.
• Watch the wave change character as damping crosses $\zeta = 1$.
• The graph shows position vs time for the chosen regime.
• Try $\gamma = 0$ — pure SHM with no decay.