Damped Oscillations

Introduction

Damping removes energy from oscillating systems. Linear damping force F = -bv leads to three regimes: underdamped (oscillating decay), critically damped (fastest return), and overdamped (slow return).

Equation of Motion

mẠ+ bẋ + kx = 0 or Ạ+ 2βẋ + ω₀²x = 0 where β = b/(2m) is damping constant, ω₀ = √(k/m) is natural frequency.

Underdamped Regime (β < ω₀)

Solution: x(t) = Ae^(-βt)cos(ω't + φ) where ω' = √(ω₀² - β²) is damped frequency. Amplitude decays exponentially. Decay time τ = 1/β. Quality factor Q = ω₀/(2β) measures oscillation persistence. Many oscillators are underdamped.

Critically Damped (β = ω₀)

Solution: x(t) = (A + Bt)e^(-βt). Fastest return to equilibrium without oscillation. Important for instruments (galvanometers, door closers) where quick settling without overshoot is desired.

Overdamped (β > ω₀)

Solution: x(t) = Ae^(-β₁t) + Be^(-β₂t) where β₁₂ = β ± √(β² - ω₀²). Slow exponential return, slower than critical damping. No oscillation. Occurs in viscous fluids or heavy damping.

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Introduction

Damping removes energy from oscillating systems. Linear damping force F = -bv leads to three regimes: underdamped (oscillating decay), critically damped (fastest return), and overdamped (slow return).

Equation of Motion

mẠ+ bẋ + kx = 0 or Ạ+ 2βẋ + ω₀²x = 0 where β = b/(2m) is damping constant, ω₀ = √(k/m) is natural frequency.

Underdamped Regime (β < ω₀)

Solution: x(t) = Ae^(-βt)cos(ω't + φ) where ω' = √(ω₀² - β²) is damped frequency. Amplitude decays exponentially. Decay time τ = 1/β. Quality factor Q = ω₀/(2β) measures oscillation persistence. Many oscillators are underdamped.

Critically Damped (β = ω₀)

Solution: x(t) = (A + Bt)e^(-βt). Fastest return to equilibrium without oscillation. Important for instruments (galvanometers, door closers) where quick settling without overshoot is desired.

Overdamped (β > ω₀)

Solution: x(t) = Ae^(-β₁t) + Be^(-β₂t) where β₁₂ = β ± √(β² - ω₀²). Slow exponential return, slower than critical damping. No oscillation. Occurs in viscous fluids or heavy damping.

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Energy Dissipation

Power dissipated by damping: P = F_damping·v = -bv². Energy decays as E = E₀e^(-2βt). Time for energy to decay to 1/e: τ_E = 1/(2β) = τ/2 where τ is amplitude decay time.

Solved Example: Damped Oscillator Analysis

A 1 kg mass on spring (k = 100 N/m) has damping coefficient b = 2 kg/s. Initially displaced 0.1 m and released. Find: (a) Oscillation regime, (b) Damped period if applicable, (c) Amplitude after 10 cycles. Solution: ω₀ = √(k/m) = √100 = 10 rad/s. β = b/(2m) = 2/2 = 1 s⻹. (a) β < ω₀ (1 < 10), so underdamped. (b) ω' = √(ω₀² - β²) = √(100 - 1) = 9.95 rad/s. T' = 2π/ω' = 2π/9.95 = 0.631 s. Compare to undamped T = 2π/10 = 0.628 s. Damping slightly increases period. (c) Amplitude decays as A(t) = A₀e^(-βt). Time for 10 cycles: t = 10T' = 6.31 s. A = 0.1 × e^(-1×6.31) = 0.1 × e^(-6.31) = 0.1 × 0.0018 = 1.8×10^-4 m = 0.18 mm. After 10 cycles, amplitude reduced from 10 cm to 0.18 mm - about 1/500 of original. Quality factor Q = ω₀/(2β) = 10/2 = 5. Moderately damped system.

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