Damping & Resonance

Add damping and a driving force, and trace the resonance curve

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Drag the block to pull it aside, then let go. The views zoom automatically to fit the motion.

Position x(t), last 10 sxdashed: envelope ±A₀e^(−γt)
Driving force

System

Underdamped (γ < ω₀)

Oscillates at ω₁ = √(ω₀² − γ²) inside an envelope that shrinks like e^(−γt).

ω₀ = √(k/m)
5.00 rad/s
γ = b/2m
0.50 s⁻¹
Critical b = 2√(km)
10.00 N·s/m
Q = ω₀/2γ
5.00
ω₁ = √(ω₀² − γ²)
4.97 rad/s
Decay time 1/γ
2.00 s

Resonance curve

Steady-state amplitude as a multiple of the static stretch F₀/k, against ω_d (rad/s). The red dot is the current drive.

10003691215ω₀

Phase lag δ

How far the motion trails the driving force. It passes 90° exactly at ω₀.

180°90°0°03691215ω₀

Damping: where the energy goes

A real spring does not oscillate forever. Air resistance, internal friction or a shock absorber add a force that opposes the velocity, F = −bv. The equation of motion becomes m x'' + b x' + k x = 0. Dividing by m, two rates control everything: the natural frequency ω₀ = √(k/m) and the damping rate γ = b/2m.

When γ is smaller than ω₀ the mass still oscillates, slightly slower than ω₀, inside an envelope that shrinks like e^(−γt). Energy, which is proportional to amplitude squared, decays like e^(−2γt). Increase b until γ equals ω₀ and the oscillation disappears: this is critical damping, the quickest way back to rest without overshooting, which is what car suspensions and door closers aim for. Beyond that the system is overdamped and creeps back slowly.

Driving and resonance

Now push the mass with a periodic force F₀ cos(ωdt). At first the motion is a mix of the driven response and the natural, decaying oscillation (the transient). After a few decay times 1/γ the transient is gone and the mass oscillates at the driving frequency, not its own, with a fixed amplitude A and a fixed phase lag δ behind the force. This is the steady state.

Drive slowly and the spring simply follows the force: A ≈ F₀/k and the motion is in phase. Drive very fast and the mass cannot keep up: A falls towards zero and the motion is half a cycle out of phase. In between, near ω₀, the force pushes in step with the velocity, energy pours in on every cycle, and the amplitude peaks. That is resonance. The lighter the damping, the taller and narrower the peak.

The quality factor Q = ω₀/2γ measures how sharp the resonance is. At resonance the amplitude is about Q times the static stretch F₀/k, the peak's width at 1/√2 of its height is about ω₀/Q, and a free oscillator rings for roughly Q/π cycles before its energy drops to e−2 of the start.

Key formulas

Equation of motion

m x'' + b x' + k x = F₀ cos(ωdt)

b
damping coefficient (N·s/m)
F₀, ω_d
driving amplitude (N) and angular frequency (rad/s)

Natural frequency, damping rate, Q

ω₀ = √(k/m),   γ = b/(2m),   Q = ω₀/(2γ) = √(km)/b

Free underdamped motion

x(t) = A₀ e−γt cos(ω₁t − φ),   ω₁ = √(ω₀² − γ²)

Steady-state amplitude and phase

A(ωd) = (F₀/m) / √((ω₀² − ωd²)² + (2γωd)²)
tan δ = 2γωd / (ω₀² − ωd²)

The peak is not exactly at ω₀

The amplitude is largest at ωr = √(ω₀² − 2γ²), a little below ω₀. For light damping (large Q) the difference is tiny. The phase lag, however, is exactly 90° at ωd = ω₀ for any damping, because there the force is in step with the velocity.

The three damping regimes

RegimeConditionBehaviour
Underdampedγ < ω₀ (b < 2√(km))Oscillates at ω₁ = √(ω₀² − γ²) while the amplitude decays as e^(−γt)
Critically dampedγ = ω₀ (b = 2√(km))Fastest return to equilibrium with no overshoot
Overdampedγ > ω₀ (b > 2√(km))Slow, non-oscillating return; the slow rate is γ − √(γ² − ω₀²)

How to use this visualizer

  1. 1
    With the driving force off, drag the block aside and let go. The dashed curves show the envelope ±A₀e^(−γt) that the peaks follow.
  2. 2
    Increase b and watch the regime label change. Press Critical to set b = 2√(km) and see the fastest non-oscillating return.
  3. 3
    Turn the driving force on and press Reset to start from rest. The transient dies out and the motion settles onto the dashed steady-state amplitude.
  4. 4
    Sweep ω_d across ω₀. The red dot moves along the resonance curve, and the phase lag passes through 90°.
  5. 5
    Lower b to raise Q and sharpen the peak. Set b = 0 and drive near (not at) ω₀ to see beats.

Practice problems

Problem 1: Classify the damping

A 2.0 kg mass on a spring with k = 50 N/m has damping b = 4.0 N·s/m. Find ω₀, γ, the damping regime, Q, and the time for the amplitude to halve when it oscillates freely.
Show solution

ω₀ = √(50 / 2.0) = 5.0 rad/s

γ = b / 2m = 4.0 / 4.0 = 1.0 s⁻¹

γ < ω₀, so the motion is underdamped (critical b would be 2√(km) = 20 N·s/m)

Q = ω₀ / 2γ = 5.0 / 2.0 = 2.5

Amplitude halves when e^(−γt) = ½: t = ln 2 / γ = 0.693 s

Underdamped, Q = 2.5, half-life of amplitude ≈ 0.69 s

Problem 2: Amplitude at resonance

A 1.0 kg mass on a spring with k = 100 N/m and b = 0.50 N·s/m is driven by F₀ = 2.0 N. Find the steady-state amplitude and phase lag when ω_d = ω₀, and when ω_d = 5.0 rad/s.
Show solution

ω₀ = √(100 / 1.0) = 10 rad/s, γ = 0.50 / 2.0 = 0.25 s⁻¹, Q = 10 / 0.5 = 20

At ω_d = ω₀ the first term vanishes: A = (F₀/m) / (2γω₀) = 2.0 / (2 × 0.25 × 10) = 0.40 m

Check: Q × F₀/k = 20 × 0.020 m = 0.40 m. Phase lag δ = 90°.

At ω_d = 5.0: ω₀² − ω_d² = 100 − 25 = 75, 2γω_d = 2.5

A = 2.0 / √(75² + 2.5²) = 2.0 / 75.04 = 0.0267 m

tan δ = 2.5 / 75 → δ = 1.9°

0.40 m at resonance (δ = 90°); 2.7 cm at 5 rad/s (δ ≈ 1.9°)

Discussion

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