Forced Oscillations and Resonance

Introduction

Forced oscillations occur when external periodic force drives a system. Resonance occurs when driving frequency matches natural frequency, producing maximum amplitude response. Understanding resonance is crucial for engineering design and avoiding structural failures.

Equation of Motion

mẠ+ bẋ + kx = F₀cos(ωt) or Ạ+ 2βẋ + ω₀²x = (F₀/m)cos(ωt). Driving frequency ω may differ from natural frequency ω₀.

Steady-State Solution

After transient decays: x(t) = A(ω)cos(ωt - δ). Amplitude A(ω) = (F₀/m)/√((ω₀²-ω²)² + (2βω)²). Phase lag δ = tan^-1(2βω/(ω₀²-ω²)).

Resonance

Amplitude peaks near ω ≈ ω₀ for light damping. Resonance frequency ω_r = √(ω₀² - 2β²). Maximum amplitude A_max = F₀/(2mβ√(ω₀²-β²)). Sharpness of peak measured by quality factor Q = ω₀/(2β). High Q means narrow, tall resonance peak.

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Introduction

Forced oscillations occur when external periodic force drives a system. Resonance occurs when driving frequency matches natural frequency, producing maximum amplitude response. Understanding resonance is crucial for engineering design and avoiding structural failures.

Equation of Motion

mẠ+ bẋ + kx = F₀cos(ωt) or Ạ+ 2βẋ + ω₀²x = (F₀/m)cos(ωt). Driving frequency ω may differ from natural frequency ω₀.

Steady-State Solution

After transient decays: x(t) = A(ω)cos(ωt - δ). Amplitude A(ω) = (F₀/m)/√((ω₀²-ω²)² + (2βω)²). Phase lag δ = tan^-1(2βω/(ω₀²-ω²)).

Resonance

Amplitude peaks near ω ≈ ω₀ for light damping. Resonance frequency ω_r = √(ω₀² - 2β²). Maximum amplitude A_max = F₀/(2mβ√(ω₀²-β²)). Sharpness of peak measured by quality factor Q = ω₀/(2β). High Q means narrow, tall resonance peak.

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Phase Behavior

At low ω (ω << ω₀): δ ≈ 0, response in phase with drive. At resonance (ω = ω₀): δ = π/2, velocity in phase with drive (maximum power transfer). At high ω (ω >> ω₀): δ ≈ π, response opposite phase.

Applications

Applications: radio tuning (select frequencies), mechanical vibrations (bridges, buildings - avoid resonant frequencies), magnetic resonance (NMR, MRI), molecular spectroscopy (resonant light absorption), musical instruments (resonant cavities), shock absorbers (damp resonances).

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Solved Example: Resonance Amplitude

A 0.5 kg mass on spring (k = 200 N/m) with damping b = 1 kg/s is driven by force F = 2cos(ωt) N. Find: (a) Resonant frequency, (b) Amplitude at resonance, (c) Amplitude at half resonant frequency. Solution: ω₀ = √(k/m) = √(200/0.5) = √400 = 20 rad/s. β = b/(2m) = 1/(2×0.5) = 1 s⻹. F₀/m = 2/0.5 = 4 m/s². (a) ω_r = √(ω₀² - 2β²) = √(400 - 2) = √398 = 19.95 rad/s ≈ 20 rad/s (light damping). (b) At resonance: A_max = (F₀/m)/(2βω₀) = 4/(2×1×20) = 0.1 m = 10 cm (using simplified formula for light damping). Exact: A_max = 4/(2×1×√398) = 4/39.9 = 0.100 m ✓. (c) At ω = 10 rad/s: A = 4/√((400-100)² + (2×1×10)²) = 4/√(90000 + 400) = 4/300.7 = 0.0133 m = 1.33 cm. Much smaller than resonant amplitude - demonstrates frequency selectivity.

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