Driven Oscillations & Resonance

Add a periodic push to the damped oscillator and the equation m d²x/dt² + b dx/dt + kx = F₀ cos ωt stays linear, so every solution is the sum of two pieces. One is the free damped motion of the previous lesson: it rings at ω′, its size is set by how the motion started, and its envelope decays as exp(−t/τ) with τ = 2m/b. The other is a particular solution that oscillates at the drive frequency ω, with an amplitude and lag fixed by m, b, k, F₀ and ω alone. For the 0.20 kg mass with b = 0.40 kg s⁻¹, τ = 1.0 s: driven at ω₀ from rest, its amplitude closes the gap to the steady value by a factor e every second, reaching 95% of it after 3 s, and after 5 s the transient is below 1% of its starting size. Until then the two pieces add, and if ω differs from ω′ their sum swells and fades at the difference frequency, beats that die with the transient. Everything that follows is about the piece that stays.

Try x = A cos(ωt − δ) and write θ = ωt − δ. Then kx = kA cos θ, m d²x/dt² = −mω²A cos θ and b dx/dt = −bωA sin θ. Expand the drive about the same angle: F₀ cos ωt = F₀ cos(θ + δ) = F₀ cos δ cos θ − F₀ sin δ sin θ. The equation must hold at every instant, and cos θ and sin θ are independent functions of time, so their coefficients match separately: (k − mω²)A = F₀ cos δ and bωA = F₀ sin δ. Those are the two legs of a right triangle whose hypotenuse is F₀, the picture drawn in the figure. Squaring and adding gives A = F₀/√((k − mω²)² + (bω)²); dividing gives tan δ = bω/(k − mω²). The across leg is set by the net stiffness, the spring constant less the mω² needed to accelerate the mass, and it changes sign at ω₀. The up leg is drag, which is never negative, so δ runs from 0 to π and never beyond.

The denominator names which term controls the motion. Take the 0.20 kg mass on the 80 N m⁻¹ spring with b = 0.40 kg s⁻¹, pushed with F₀ = 0.40 N, whose static stretch is F₀/k = 5.0 mm. At ω = 10 rad s⁻¹ the net stiffness is 80 − 0.20 × 100 = 60 N m⁻¹ against a drag term of only 4.0 N m⁻¹, so A = 0.40/√(60² + 4.0²) = 6.7 mm and δ = 3.8°: the spring rules, and the mass moves almost in step with the push. At ω₀ = 20 rad s⁻¹ the net stiffness is zero, spring and inertia cancelling exactly, and drag alone limits the motion: A = 0.40/(0.40 × 20) = 50 mm with δ = 90°. At ω = 40 rad s⁻¹ the net stiffness is 80 − 320 = −240 N m⁻¹, so inertia wins: A = 1.7 mm, a third of the static stretch, and δ = 176°, the mass moving against the push. The lags at ω₀/2 and 2ω₀ sit the same 3.8° from 0° and 180°. That is general: δ at ω₀²/ω equals 180° minus δ at ω, for any Q.

The drive delivers power P = Fv. The velocity v = −ωA sin(ωt − δ) leads the displacement by a quarter cycle, so averaging F₀ cos ωt × v over a cycle gives P̄ = ½ F₀ωA sin δ. With bωA = F₀ sin δ this becomes P̄ = ½ bω²A², exactly the mean rate at which drag removes energy: in steady state every joule supplied is dissipated. Written as P̄ = (F₀²/2b) sin²δ, it peaks where δ = 90°, at ω₀ whatever b is. For the running example the peak is 0.16/0.80 = 0.20 W; at 10 rad s⁻¹ the drive delivers only sin²(3.8°) = 0.44% of it, although the amplitude there is 13% of the resonant value. Power halves where δ = 45° or 135°, where |k − mω²| = bω: for the example 19.0 and 21.0 rad s⁻¹, a band b/m = 2.0 rad s⁻¹ wide whose geometric mean is ω₀. The displacement peak sits slightly lower, at ω₀√(1 − 1/(2Q²)) = 19.95 rad s⁻¹; only at small Q do the power peak, the displacement peak and ω′ visibly separate.

With Q = mω₀/b, three results that look separate become one. Height: at ω₀, A = F₀/(bω₀) = Q × F₀/k, so a matched drive multiplies the static stretch by Q; for the example, 10 × 5.0 mm = 50 mm. Width: the half-power band is Δω = b/m = ω₀/Q. Time: the transient, and with it the build-up from rest, decays with τ = 2m/b = 2Q/ω₀, which is Q/π periods. Multiply width by time and Q cancels, Δω × τ = 2: a sharp resonance is necessarily a slow one. A 440 Hz tuning fork with Q ≈ 1000, a representative value, responds strongly only within a band 0.44 Hz wide and takes τ ≈ 0.72 s to ring up or down. Q also counts energy: in steady state at ω₀, Q = 2π × (energy stored) ÷ (energy dissipated per cycle), so the example oscillator, with Q = 10, loses and is handed back 63% of its stored energy every cycle.

Resonance lets a small periodic force produce a large, frequency-selective motion, and that cuts both ways. Tuning forks, quartz clock crystals and ultrasonic cleaners are driven at ω₀ on purpose, so that a weak drive gives a strong response in a narrow band. The same arithmetic is a hazard when the drive is unwanted. A pump running near the ω₀ of its mounting shakes up to Q times harder than its static load suggests, and a washing-machine drum crosses the resonance of its suspension on every spin-up, where a quick crossing helps because the motion needs a time of order τ to build. The formula names the cures: detune by changing k or m so the operating frequency sits far from ω₀; add damping, which lowers the peak in proportion to Q and widens the band as ω₀/Q; or run well above ω₀, where A falls as F₀/(mω²). London's Millennium Bridge, which swayed under crowds when it opened in 2000, was fixed with dampers. Not every famous failure fits the model: the 1940 Tacoma Narrows collapse was aeroelastic flutter, a wind force that depends on the deck's own motion.