Resonance: the driven, damped oscillator
Drive a damped spring at different frequencies. Near its natural frequency the response grows sharply, and the damping sets how tall and how narrow that peak is.
Displacement and drive vs time
Readouts
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.
- Equation of motionm x'' + b x' + k x = F₀ cos ωt
- AmplitudeA = (F₀/m) / √((ω₀² − ω²)² + (γω)²)
- Phase lagtan φ = γω / (ω₀² − ω²)
- Resonant frequencyωᵣ = √(ω₀² − γ²/2)
- Q factor and bandwidthQ = ω₀ / γ, Δω ≈ γ
On the syllabus
- A-Level Physics AQA, OCR, Edexcelforced oscillations, resonance and damping
- First-year universitythe driven damped harmonic oscillator, Q factor, bandwidth and phase response
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.
Learn this properly
Lessons from GCSE Combined and Separate Science on this topic: worked steps, then exam-style questions with new numbers every time.
- Transverse and longitudinal wavesWaves and the electromagnetic spectrumFree
- Springs, elasticity and the force-extension practicalForces, work, elasticity, moments and pressurePlus
See the whole courseThe full course, topic tests and mock papers come with Plus. See Plus
FAQ
How do I find the resonance frequency of a damped oscillator?
Work out γ = b/m, then ωᵣ = √(ω₀² − γ²/2). A 1 kg mass with ω₀ = 4 rad/s and b = 0.8 kg/s has γ = 0.8 s⁻¹, so ωᵣ = √(16 − 0.32) = 3.96 rad/s and Q = ω₀/γ = 5. If γ² ≥ 2ω₀² there is no peak at all. Pick Resonance frequency under Solve for to see each step.
Why is the resonant frequency lower than the natural frequency?
Damping shifts the peak of the amplitude curve down to ωᵣ = √(ω₀² − γ²/2). For light damping the shift is tiny; with heavy damping (γ² > 2ω₀²) there is no peak at all and the amplitude just falls as the frequency rises.
What is the Q factor?
Q = ω₀/γ, the natural frequency divided by the damping rate. A high Q means a tall, narrow resonance peak and an oscillator that rings for about Q/π cycles before dying away. A tuning fork has Q of about 1000; a car suspension about 1.
Why does the phase lag go from 0 to 180°?
At low frequency the spring dominates and the mass moves with the force. At ω₀ the spring and inertia cancel, so only the damper resists: the velocity is in step with the force and the displacement lags by 90°. At high frequency inertia dominates and the mass moves opposite to the force.
What are transients?
When the drive starts, the motion includes a part at the system's own frequency that decays as e−γt/2. After a few times 2/γ only the steady state at the drive frequency is left. That is why the measured amplitude settles onto the analytic curve.