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
读数
发生了什么
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
考试大纲
- A-Level Physics AQA, OCR, Edexcelforced oscillations, resonance and damping
- First-year universitythe driven damped harmonic oscillator, Q factor, bandwidth and phase response
挑战
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.
系统地学
GCSE Combined and Separate Science 中讲这个主题的课：先看分步讲解，再做考试风格的题目，每次数字都不同。 课程为英文。
常见问题
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.