Driven oscillations and resonance

Because the equation is linear, its general solution is any one particular solution plus the general solution of the homogeneous equation — and the homogeneous equation is exactly the damped oscillator, whose every solution carries the factor e^(−γt/2). That part is the transient: it depends on how the oscillator was released, and it falls by 1/e in a time 2/γ, which is 2Q/ω₀, or about Q/π cycles of the oscillator's own free motion. Take ω₀ = 20 rad s⁻¹ and γ = 2.0 s⁻¹, so Q = 10: that free period is 0.31 s and the decay time is 1.0 s, so ten cycles leave 4.3% of the transient and thirty leave 0.008%. What survives is the particular solution, the steady state. The initial conditions have vanished with the part that died, so the amplitude and the lag measured on a settled oscillator carry no memory of how it was started — they are properties of the driver and the oscillator alone, which is exactly why a resonance curve is reproducible.

A linear operator with constant coefficients turns a sinusoid at ω into another sinusoid at ω, shifted in phase but never in frequency, so try x = A cos(ωt − φ) and see what the equation demands. The three terms are kx = kA cos(ωt − φ), b dx/dt = bωA cos(ωt − φ + π/2), and m d²x/dt² = mω²A cos(ωt − φ + π). Draw them as phasors: one of length kA along the displacement direction, one of length mω²A pointing exactly opposite it, and one of length bωA at right angles to both. Their sum must be the drive phasor, of length F₀. Resolving along and across the displacement gives F₀ cos φ = (k − mω²)A and F₀ sin φ = bωA. Squaring and adding eliminates φ: A = F₀ ÷ √((k − mω²)² + (bω)²), which is (F₀/m) ÷ √((ω₀² − ω²)² + γ²ω²). Dividing instead eliminates A and leaves tan φ = bω ÷ (k − mω²). One triangle has produced both answers.

Below ω₀ the stiffness term k − mω² dominates and is positive, so φ is small: the mass tracks the push almost in step, and A tends to F₀/k, the deflection a steady force would give. This is the stiffness-controlled regime. At ω = ω₀ the stiffness and inertia phasors cancel exactly, the drive has nothing left to balance but damping, and φ = π/2 precisely — for any b, however light. That is the sharpest diagnostic on the curve, because the amplitude peak moves with damping and this crossing does not. Far above ω₀, k − mω² is large and negative, φ approaches π, and A falls as F₀/(mω²): inertia-controlled, the mass barely moves and moves against the push. The numbers in between are unforgiving. Divide the lag formula top and bottom by ω₀² and it becomes tan φ = (ω/Qω₀) ÷ (1 − ω²/ω₀²), so a drive at ω = 0.833 ω₀ with Q = 10 gives 0.0833 ÷ 0.3056 = 0.273 and a lag of only 15.3°, even though ω already sits 17% below resonance.

The amplitude is largest where D(ω) = (ω₀² − ω²)² + γ²ω² is smallest. Treat ω² as the variable: dD/d(ω²) = −2(ω₀² − ω²) + γ², which vanishes at ωₚ² = ω₀² − γ²/2. So the peak is pulled below ω₀, and below the free damped frequency √(ω₀² − γ²/4) as well — three distinct frequencies where the undamped oscillator had one. For light damping the shift is tiny: ωₚ ≈ ω₀[1 − 1/(4Q²)], which at Q = 25 is four parts in ten thousand. Substituting ωₚ back gives a peak height (F₀/m) ÷ (γ√(ω₀² − γ²/4)), exceeding the value at ω₀ by the factor 1/√(1 − 1/(4Q²)), which is 1.0002 at Q = 25. The other end is where it matters. Once γ reaches √2 ω₀, that is Q down to 1/√2 = 0.71, ωₚ² turns negative and there is no peak at all: the curve falls monotonically from F₀/k, which is what a well-damped instrument mount is built to do.

Amplitude is not the only response curve. The driver does work at a rate P = Fv, and with v = −ωA sin(ωt − φ) only the part of the velocity in phase with the force survives a cycle average: ⟨P⟩ = ½ F₀ ω A sin φ. Substituting A and sin φ gives ⟨P⟩ = F₀²γω² ÷ (2m[(ω₀² − ω²)² + γ²ω²]). Divide top and bottom by ω² and the bracket becomes (ω₀²/ω − ω)² + γ², smallest exactly at ω = ω₀ with no shift whatever the damping. There ⟨P⟩ = F₀²/(2b) — for F₀ = 1.5 N and b = 0.50 kg s⁻¹ that is 2.25 W, every watt of it ending in the dashpot. Half power needs (ω₀²/ω − ω)² = γ², or ω₀² − ω² = ±γω, and subtracting the two roots gives a full width Δω = γ exactly. Hence Q = ω₀/Δω: measure a width and you have Q, without ever locating the peak height.

Two assumptions carry everything above. The first is that the restoring force is strictly proportional to displacement. Drive hard enough and it is not: ω₀ starts to depend on amplitude, the peak leans towards higher or lower frequency, the curve becomes multivalued, and the amplitude jumps as the drive is swept up and then back down. A nonlinear oscillator also answers at 2ω and 3ω, which a linear one never does — hearing a harmonic that is not in the drive is the experimental signature. The second is that damping is viscous, proportional to speed; dry friction or turbulent drag gives a peak of different shape and an effective Q that depends on amplitude. Two checks keep you honest. Set F₀ = 0 and the steady state vanishes, leaving the free damped motion. Set γ = 0 away from ω₀ and A = (F₀/m) ÷ |ω₀² − ω²|, the undamped driven result — which diverges at ω₀, a warning that infinite amplitude is a failure of the model, not a prediction.