Resonance: the driven, damped oscillator

What's happening

A mass on a spring obeys m x'' + b x' + k x = F₀ cos ωt: the spring pulls back, the damper resists the motion, and a force drives it at angular frequency ω. Started from rest, the motion is the sum of a transient, which oscillates at the system's own frequency and dies away as e^−γt/2 with γ = b/m, and a steady state that follows the drive. The steady state is x = A cos(ωt − φ) with amplitude A = (F₀/m)/√((ω₀² − ω²)² + (γω)²) and a lag φ = tan⁻¹(γω/(ω₀² − ω²)). Well below ω₀ the mass just follows the force; at ω₀ it lags by a quarter cycle and the drive does the most work; far above it the mass moves opposite to the force with a small amplitude. The amplitude peaks a little below ω₀, at ωᵣ = √(ω₀² − γ²/2). The quality factor Q = ω₀/γ measures how sharp the resonance is: the half-power width of the peak is γ, and at resonance the amplitude is about Q times the static stretch F₀/k. In steady state the energy the drive feeds in each cycle equals the energy the damper turns into heat.

A-Level Physics (AQA, OCR, Edexcel): forced oscillations, resonance and damping. First-year university: the driven damped harmonic oscillator, Q factor, bandwidth and phase response.

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Challenge

Predict first: with ω₀ = 4.0 rad/s, m = 1 kg and b = 0.8 kg/s, at what driving frequency is the amplitude largest? How many times bigger than the static stretch F₀/k is it there? Type your frequency in and check.

γ = b/m = 0.8 s⁻¹, so ωᵣ = √(16 − 0.32) = 3.96 rad/s. There A = 1 / (0.8 × √(16 − 0.16)) = 0.314 m, while F₀/k = 1/16 = 0.0625 m: about 5 times bigger, which is Q = ω₀/γ = 5.