Spring-Mass-Damper — Free Vibration

A mass on a spring with a damper. Pull it aside, let it go, watch it settle. The damping ratio ζ decides whether it rings, dies smoothly, or crawls back.

Visualization

Input

m

Mass

\mathrm{kg}

k

Spring stiffness

\mathrm{N/m}

c

Damping coefficient

\mathrm{N\cdot s/m}

x_0

Initial displacement

\mathrm{m}

Show the time-response plot (engineering view)

The simplest interesting oscillator

A mass on a spring with a damper is a one-line ODE that captures more than its share of physics. Newton's second law on the mass gives:

m\,\ddot{x} + c\,\dot{x} + k\,x = 0

The spring pulls the mass back toward equilibrium (the $kx$ term). The damper resists velocity (the $c\dot{x}$ term). Source: OpenStax University Physics §15.5 Eq. 15.23; Wikipedia — Harmonic oscillator.

Two derived numbers determine everything about the response:

\omega_n = \sqrt{\frac{k}{m}} \qquad \zeta = \frac{c}{2\sqrt{m k}}

$\omega_n$ is the natural frequency — how fast it would ring with no damping. $\zeta$ is the damping ratio — how much the damper kills the oscillation, expressed as a fraction of the critical amount.

Three regimes

The character of the response depends only on $\zeta$ :

$\zeta < 1$ — underdamped. The mass oscillates at the slightly slower damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$ , with the amplitude decaying exponentially inside an envelope $\pm x_0\,e^{-\zeta\omega_n t}$ . A small $\zeta$ rings for a long time before settling.
$\zeta = 1$ — critically damped. The fastest possible return to equilibrium without overshoot (OpenStax §15.5).
$\zeta > 1$ — overdamped. No oscillation; the mass crawls back smoothly. Big slow shock absorbers, or a finger pushed through honey.

Critical damping happens when the damper exactly matches $c_{\text{crit}} = 2\sqrt{m k}$ . Equivalently: when the discriminant of the characteristic equation vanishes, i.e. when $b = \sqrt{4mk}$ in the OpenStax notation.

Reading the plot

For the underdamped case, the visualization above shows two things at once: the actual response $x(t)$ in solid blue, and the decay envelope $\pm x_0\,e^{-\zeta\omega_n t}$ as the faint dashed curves. The peaks of the oscillation always touch the envelope — that's how damping bleeds energy out of the system.

Slide $c$ from zero (perfect harmonic motion, ringing forever) up through critical (the "best" damping for a quick settle) and into overdamped (sluggish return). Watch the period stretch from $T_n$ to $T_d$ as $\zeta$ grows, then disappear entirely once $\zeta \ge 1$ .

Show your work

Undamped natural angular frequency: $\omega_n \;=\; \sqrt{\dfrac{k}{m}} \;=\; \sqrt{\dfrac{\textcolor{#dc4124}{\left(100\right)}}{\textcolor{#dc4124}{\left(1\right)}}} \;=\; 10\;\mathrm{rad/s}$
Critical damping coefficient: $c_{\text{crit}} \;=\; 2\sqrt{m k} \;=\; 2\sqrt{\textcolor{#dc4124}{\left(1\right)} \textcolor{#dc4124}{\left(100\right)}} \;=\; 20\;\mathrm{N\cdot s/m}$
Damping ratio: $\zeta \;=\; \dfrac{c}{2\sqrt{m k}} \;=\; \dfrac{\textcolor{#dc4124}{\left(2\right)}}{2\sqrt{\textcolor{#dc4124}{\left(1\right)} \textcolor{#dc4124}{\left(100\right)}}} \;=\; 0.1$
Damped natural angular frequency (zero if ζ ≥ 1): $\omega_d \;=\; \omega_n\sqrt{\,1 - \zeta^{2}\,} \;=\; 9.95\;\mathrm{rad/s}$
Undamped natural period: $T_n \;=\; \dfrac{2\pi}{\omega_n} \;=\; 0.6283\;\mathrm{s}$
Damped period (NaN if ζ ≥ 1): $T_d \;=\; \dfrac{2\pi}{\omega_d} \;=\; 0.6315\;\mathrm{s}$
Decay coefficient (envelope rate): $\gamma \;=\; \dfrac{c}{2 m} = \zeta\,\omega_n \;=\; \dfrac{\textcolor{#dc4124}{\left(2\right)}}{2 \textcolor{#dc4124}{\left(1\right)}} = \zeta\,\omega_n \;=\; 1\;\mathrm{1/s}$

Worked Example

For m = 1 kg, k = 100 N/m, c = 2 N·s/m: ω_n = √(100/1) = 10 rad/s; ζ = 2/(2√100) = 0.10; ω_d = 10·√(1 − 0.01) = 9.9499 rad/s. These are direct evaluations of the formulas in the references.

Quantity	Expected	Computed	Δ
omega_n	10	10	0
zeta	0.1	0.1	0
omega_d	9.95	9.95	4.371 \times 10^{-6}
c_crit	20	20	0

Source: OpenStax §15.5 / Wikipedia: Harmonic oscillator — formulas applied to chosen inputs

Worked Example

Critical damping is defined by c² = 4mk, or equivalently c = 2·√(mk). For m = 2 kg, k = 50 N/m, the critical damping coefficient is c_crit = 2·√(100) = 20 N·s/m, giving ζ = 1 exactly when c = 20. Setting c at the critical value returns ζ = 1, ω_d = 0.

Quantity	Expected	Computed
c_crit	20	20
zeta	1	1
omega_d	0	0

Source: OpenStax §15.5 — critical damping condition b = √(4mk)

Sources

[1] OpenStax. University Physics Volume 1 — §15.5 Damped Oscillations, eq. m·d²x/dt² + b·dx/dt + kx = 0 (Eq. 15.23); ω₀ = √(k/m) (Eq. 15.25); ω = √(ω₀² − (b/2m)²) (Eq. 15.26); critical b = √(4mk); underdamped solution x(t) = A₀·exp(−bt/2m)·cos(ωt + φ) (Eq. 15.24) (openstax.org)
[2] Wikipedia — Harmonic oscillator (mass-spring-damper), eq. d²x/dt² + 2ζω₀(dx/dt) + ω₀²x = 0; ω₀ = √(k/m); ζ = c/(2√(mk)); ω₁ = ω₀√(1−ζ²); three regimes: underdamped (ζ<1), critically damped (ζ=1), overdamped (ζ>1) (en.wikipedia.org)

Assumptions

Linear (Hookean) spring: restoring force = −kx (OpenStax §15.5).
Linear viscous damping: damping force = −c·ẋ (OpenStax §15.5).
Free vibration — no external forcing function (OpenStax §15.5 Eq. 15.23 sets the right-hand side to zero).
Constant m, k, c throughout the motion (small displacements, idealised dashpot).
Initial conditions x(0) = x₀ and ẋ(0) = 0 (released from rest at the displaced position).