Driven oscillators

A driven oscillator selects amplitude and phase

An external periodic force can maintain motion after transients decay. The steady response depends on the relation between drive frequency and the system’s natural frequency, not just on the input force magnitude.

Substitute a sinusoidal trial response

For m x¨ + c x˙ + kx = F₀ cosωt, use complex notation x = Re[Xe^{iωt}]. Derivatives multiply X by iω and −ω², giving X = F₀/[k − mω² + i cω]. Its magnitude is A = F₀/√[(k − mω²)² + (cω)²], while the complex denominator supplies phase lag.

Let m = 1 kg, k = 100 N/m, c = 4 N·s/m, F₀ = 10 N. At ω = 5 rad/s, A = 10/√[(75)² + 20²] = 0.131 m; at ω = 10 rad/s, A = 10/40 = 0.25 m.

Infinite resonance belongs to the undamped ideal

If c = 0 and ω = ω₀, the steady formula diverges because energy is added every cycle. Any real loss limits amplitude, and nonlinearities eventually shift or cap the response.

Work it: Calculate A and phase at ω = 8, 10, and 12 rad/s for the parameters above. Identify the frequency with the largest response and explain why it is near, but not exactly, ω₀.