Driven oscillation and resonance

Bolt a damped spring-mass oscillator to a periodic force and keep pushing it. Once the transient has died away the motion settles into a steady oscillation at the driving frequency ω_d, with amplitude A = F₀/√((k−mω_d²)²+(cω_d)²) and a phase lag φ = atan2(cω_d, k−mω_d²) behind the force. The headline result is that the amplitude peaks when the driving frequency approaches the natural frequency ω₀ = √(k/m) — that is resonance — and that more damping makes the peak lower and broader. With no driving force the oscillator simply loses its energy to damping and stops. To see what happens without a driving force first, go back to Damped oscillation.

Questions

Why does pushing an oscillator continuously sometimes give a tiny amplitude and sometimes a very large one?

Why does the steady-state amplitude shoot up when the driving frequency approaches the natural frequency ω₀ = √(k/m)?

If you increase the damping coefficient c, does the resonance peak grow or shrink?

Predict

Write down your prediction first: as you sweep the driving frequency ω_d from small to large, does the steady amplitude keep increasing, or rise and then fall? If you double the damping c, does the peak height roughly halve?

Try

Start from the defaults (m = 1 kg, k = 30 N/m, c = 0.8 N·s/m, F₀ = 2 N, ω_d = 5.5 rad/s) and press Play. Watch the displacement–time trace: it starts small, grows, and settles; the dashed lines are the steady-state envelope ±A. Set ω_d to 2, 5.5 and 11 rad/s in turn and note the steady amplitude A and the phase lag φ in the readouts, comparing them with the marker on the response curve. Then set c to 0.4 and 5 N·s/m and compare how sharp the resonance peak is. Finally set F₀ = 0 and watch the trace decay to zero.

Observe

Away from ω₀ the steady amplitude is small: at ω_d = 2 rad/s, A ≈ 0.077 m, and at ω_d = 11 rad/s, A ≈ 0.022 m. At ω_d = 5.5 rad/s, close to ω₀ = √(30/1) ≈ 5.477 rad/s, A ≈ 0.454 m, far larger than at either end. Lowering c from 0.8 to 0.4 roughly doubles the peak to 0.90 m; raising it to 5 flattens the peak to about 0.073 m. The phase lag φ rises monotonically from about 0° at low frequency to about 180° at high frequency, passing through about 90° near resonance. At the start there is also a transient: the amplitude grows from zero and only approaches the steady value after about 2m/c ≈ 2.5 s. With F₀ = 0 there is no steady state at all — the trace decays to rest.

Explain

Where the steady state comes from

Substitute x = A cos(ω_d t − φ) into m x″ + c x′ + k x = F₀ cos(ω_d t). The displacement, velocity and acceleration amplitudes are A, Aω_d and Aω_d². Splitting the result into the component in phase with the drive and the component a quarter cycle behind it leaves k − mω_d² and cω_d, whose combined size is √((k−mω_d²)²+(cω_d)²). Hence A = F₀/√((k−mω_d²)²+(cω_d)²) and φ = atan2(cω_d, k − mω_d²).

Why the peak sits near ω₀ = √(k/m)

The amplitude is largest where the denominator is smallest. Without damping the denominator is |k − mω_d²|, which vanishes at ω_d = √(k/m) = ω₀ and the amplitude diverges: that is the undamped resonance pole. Damping adds the term cω_d, which leaves a non-zero floor near ω₀ and keeps the amplitude finite — at ω_d = ω₀ it equals F₀/(cω₀). The smaller the damping, the lower the floor and the taller and narrower the peak.

Transient, steady state and phase lag

The general solution is the steady solution plus a transient that decays like e^{−(c/2m)t}. Starting from rest the transient cannot be ignored: the amplitude grows over roughly 2m/c before the steady state takes over. The phase lag φ says how far the displacement falls behind the force. Well below resonance φ ≈ 0, with the displacement almost in step with the force; at resonance φ ≈ 90°, with the displacement a quarter cycle behind so that the velocity is in step with the force and the power input is largest; and well above resonance φ → 180°, with the displacement almost opposite to the force.

Damping and the F₀ = 0 case

Damping dissipates mechanical energy through −c v, so with F₀ = 0 the oscillator performs a damped oscillation whose amplitude decays like e^{−(c/2m)t} to rest — there is no steady state. With F₀ > 0 the force keeps feeding energy in, and the steady amplitude is proportional to F₀ and inversely proportional to the damping.

Check

Take m = 1 kg, k = 30 N/m, c = 0.8 N·s/m and F₀ = 2 N. At ω_d = 5.5 rad/s the readouts should give A ≈ 0.454 m and φ ≈ 93°; at ω_d = 2 rad/s, A ≈ 0.077 m and φ ≈ 4°; at ω_d = 11 rad/s, A ≈ 0.022 m and φ ≈ 175°. Setting c = 0.4 N·s/m should roughly double A to 0.90 m, and c = 5 N·s/m should drop it to about 0.073 m. With F₀ = 0 the trace should decay to zero, the readout should show a steady amplitude of 0, and the phase lag should stop showing a number. Use Step to advance 0.10 s at a time and compare the transient with the steady envelope.

Explore further

If the force were not sinusoidal (a periodic square wave, say), could the steady state still be described by a single amplitude? If you coupled the same oscillator to a second direction with a different natural frequency, what would happen? To see how damping drains the energy when no force acts, go back to Damped oscillation; and review where ω₀ = √(k/m) comes from in Spring and period.