The idea: push at the right rate

A mass on a spring has its own rate, the natural angular frequency ω₀ = √(k/m). Here m = 1 kg and k = 16 N/m, so ω₀ = 4 rad/s. Air or oil slows it down with a damping rate γ. Now push it with a force F₀·cos ωt that goes back and forth at a rate ω of our choice. After a while the mass swings at the driver's rate ω, not its own, with a steady amplitude A that depends hugely on how close ω is to ω₀. That big swing near ω₀ is resonance.

Steady amplitude A divided by F₀/k against ω/ω₀ for Q = 10, 4, 2, 1 and 0.5: the Q = 10 curve peaks at 10 near ω = ω₀, Q = 0.5 has no peak; below, the phase lag δ rises from 0° through 90° at ω₀ to 180°, steeply for high Q
Low damping (high Q) gives a tall, narrow peak at ω₀; every curve lags 90° at ω₀.

The equation and its steady answer

Newton's law for the mass, with the spring force −kx, a drag force −2mγ·v and the drive:

m·x″ + 2mγ·x′ + k·x = F₀·cos ωt        or   x″ + 2γx′ + ω₀²x = (F₀/m)·cos ωt

On the page the drive is the far end of the spring: a driver plate that moves d·cos ωt with d = F₀/k = 6.25 cm. The spring then pulls with k·(d·cos ωt − x), which is the same equation. Try a steady answer x = A·cos(ωt − δ) and you get

A(ω) = (F₀/m) / √((ω₀² − ω²)² + (2γω)²)
tan δ = 2γω / (ω₀² − ω²)        δ from 0° to 180°

A is the steady-state amplitude, δ the phase lag: the mass reaches its peak a fraction δ/360° of a cycle after the driver does. The two arrows in the phasor circle on the canvas turn at ω, δ apart; the shadow of each arrow on the axis is the plate's and the mass's position.

Three regimes

drivewho winsAδon the page (γ = 0.5)
ω ≪ ω₀the spring≈ F₀/k, the static stretch≈ 0°, in phaseDemo 1: ω = 0.25·ω₀, A ≈ 6.65 cm, δ ≈ 3.8°
ω = ω₀nobody: spring and inertia cancel, damping is leftQ·F₀/k90°Demo 2: A = 25 cm, δ = 90°
ω ≫ ω₀the mass's inertia≈ F₀/(mω²), tiny→ 180°, oppositeDemo 3: ω = 3·ω₀, A ≈ 0.78 cm, δ ≈ 174.6°

At resonance the lag of 90° means the mass moves fastest exactly when the push is strongest, so the force does positive work in every part of the cycle. Below and above ω₀ the force pushes against the motion for part of each cycle and gives some energy back.

Three phasor diagrams. Below resonance the mass arrow lies on the drive arrow, δ ≈ 0°, A ≈ F₀/k. At resonance it is a quarter turn behind, δ = 90°, A = Q·F₀/k. Above resonance it points the opposite way, δ ≈ 180°, A small
The mass arrow trails the drive arrow by 0°, 90° or nearly 180°, depending on whether ω is below, at or above ω₀.

Q, peak height and width

The quality factor Q = ω₀/(2γ) sums up the damping. For small damping:

  • the peak is Q times the static stretch: A_max ≈ Q·F₀/k;
  • the peak is 2γ wide: between the two ω where A drops to A_max/√2 (half the power), Δω ≈ 2γ = ω₀/Q;
  • left alone, the oscillation rings for about Q cycles before its amplitude falls to e^(−π) ≈ 1/23.

The exact peak sits a bit below ω₀, at ω_r = √(ω₀² − 2γ²). With heavy damping, γ ≥ ω₀/√2 (Q ≤ 0.71), there is no peak at all: A is biggest at ω = 0 and falls from there.

γ (1/s)QA at ω₀width Δω (green bracket)demo
0.21062.5 cm≈ 0.4 rad/sDemo 4: tall, narrow
0.5425 cm≈ 1 rad/sthe start
1212.5 cm≈ 2 rad/s
216.25 cmtoo wide
40.53.125 cmno peakDemo 5: less than F₀/k

Changing γ keeps the old curve in grey so you can compare. Demo 7 walks ω across the peak at Q = 4: the orange dot climbs from 8.2 cm to 25 cm and falls to 2.1 cm while δ goes from 9.5° to 170.5°.

From rest: transient + steady state

The page starts the mass at rest, so x(0) = 0 and v(0) = 0. The steady answer alone does not fit that start, so a free, damped oscillation is added, the transient:

x(t) = A·cos(ωt − δ) + e^(−γt)·(C₁·cos ω_d t + C₂·sin ω_d t),   ω_d = √(ω₀² − γ²)

C₁ and C₂ are chosen so that x(0) = 0 and x′(0) = 0. The transient swings at the natural rate and dies as e^(−γt), the same envelope as on Damped Oscillation; after a few 1/γ only the steady swing is left. At resonance the swing simply grows toward A. Near resonance the two rates are close and the sum beats: the two parts drift in and out of step every 2π/|ω_d − ω| (Demo 6: ω = 3.6 rad/s, ω_d ≈ 3.995 rad/s, beats about every 15.9 s, fading as the transient dies).

Resonance in the real world

  • A child on a swing: small pushes, timed at the swing's own rate, build a big swing.
  • A wine glass rings for a long time when tapped (high Q). A singer who holds its note loud enough can make the swing so big that the glass breaks.
  • Bridges: the Millennium Bridge in London (2000) swayed sideways when walkers fell into step with it; it was fixed by adding dampers, which lower Q. The Tacoma Narrows collapse (1940) was wind flutter, a related but self-excited effect rather than a simple periodic push.
  • Tuned mass dampers in tall buildings are a second, damped oscillator tuned to the building's ω₀ that soaks up the swing.
  • Radio tuning: an LC circuit obeys the same equation; a high Q lets one station through and rejects its neighbours.

See also

Without damping or a drive this is the Spring Oscillator, x = A·cos(ωt + φ) with ω = √(k/m). The transient on this page is the decaying wave of Damped Oscillation.

What the page leaves out

Overdamped transients (γ > ω₀; the page stops at critical damping γ = ω₀ = 4), drives that are not a pure cosine (any periodic push is a sum of cosines, and each gets its own A and δ), the power the driver feeds in (biggest exactly at ω₀), springs that stop obeying Hooke's law at big swings (nonlinear resonance and jumps), parametric resonance (pumping a swing by standing up and squatting), and coupled oscillators such as a tuned mass damper. The mass, spring and push stay fixed at 1 kg, 16 N/m and 1 N.