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Forced vibration amplitude and resonance
Short answer
Free vibration occurs at natural frequency without external force; forced vibration is driven by periodic force at driving frequency. Amplitude A = (F0/m)/√((ω0²-p²)²+4γ²p²) peaks at resonance p_r = √(ω0²−2γ²) with A_max = F0/(2mγ√(ω0²−γ²)). Sharpness is measured by quality factor Q = ω0/(2γ).
Distinguish between free and forced vibrations. Give the necessary theory of forced vibration (differential equation), determine the amplitude of oscillation and deduce the condition for resonance amplitude. Explain sharpness of resonance.
Free and forced vibrations
| Free vibration | Forced vibration |
|---|---|
| Body vibrates at its natural frequency after an initial disturbance | Body vibrates at the frequency of an external periodic force |
| No external force after start | Periodic force acts continuously |
| Amplitude falls if damped | Amplitude steady (after transients) |
| Example: pluck a string | Example: tuning fork on a table |
Theory (differential equation)
Mass m, spring constant k, damping force −b v, driving force F0 cos pt:
m x'' + b x' + k x = F0 cos pt
x'' + 2γ x' + ω0² x = f0 cos pt
with 2γ = b/m, ω0² = k/m, f0 = F0/m.
Amplitude
Steady state: x = A cos(pt − φ). Substituting and comparing the coefficients of cos(pt − φ) and sin(pt − φ):
A (ω0² − p²) = f0 cosφ
A (2γ p) = f0 sinφ
A = f0 / √[(ω0² − p²)² + 4γ²p²] , tanφ = 2γp/(ω0² − p²)
Amplitude resonance
A is maximum when D = (ω0² − p²)² + 4γ²p² is minimum. dD/d(p²) = −2(ω0² − p²) + 4γ² = 0:
p_r² = ω0² − 2γ² (p_r < ω0)
A_max = f0 / (2γ √(ω0² − γ²))
Resonance condition: the driving frequency equals √(ω0² − 2γ²), very nearly the natural frequency if damping is small. For γ → 0, A becomes very large at p = ω0.
Sharpness of resonance
It is the rate at which the amplitude falls as p departs from p_r.
A | /\ small damping (sharp)
| / \
| / __ \___ large damping (flat)
|___/_/____\\___ p
p_r
- Small γ gives a tall, narrow peak (sharp); large γ gives a low, broad peak.
- It is measured by the quality factor Q = ω0/2γ = ω0/Δω, where Δω is the half-power bandwidth.
✓ Answer checked by Claude Opus 5.5 (high effort)
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