C.4 · IB Physics
Resonance
Drive an oscillator at different frequencies and see how its response peaks near its natural frequency. Damping lowers and broadens the peak, and the response lags further behind the driver as you pass through resonance.
Predict first
Start from the default settings (f₀ = 50 Hz, γ = 6 s⁻¹, f = 50 Hz). If you halve the damping γ to 3 s⁻¹, what happens to the amplitude ratio A ÷ A₀ at f = 50 Hz?
At f = f₀ the formula gives A ÷ A₀ = f₀² ÷ (γf₀) = f₀ ÷ γ, so halving γ doubles the response (8.33 to 16.7).
Check it in the model: Move the Damping, γ slider from 6 to 3 s⁻¹ and compare the Amplitude ratio A ÷ A₀ readout.
response A ÷ A₀ chosen frequency · A₀ is the response at very low frequency
tan φ = γf ÷ (f₀² − f²)
Q = f₀ ÷ γ
Try this
- Start at f₀ = 50 Hz, γ = 6 s⁻¹ and f = 50 Hz. Lower γ to 2 s⁻¹ and compare the height and the width of the peak.
- Sweep the driving frequency from 10 Hz to 90 Hz with the phase curve on. At what frequency is the lag 90°, and how does it change with γ?
- Set f₀ = 60 Hz, γ = 4 s⁻¹ and f = 30 Hz. Is the response large or small, and how do the two traces compare?
Exam tip: maximum amplitude occurs when the driving frequency is close to the natural frequency; light damping gives a tall, narrow peak and heavy damping a low, broad one, with the peak moving to slightly lower frequency.