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Power supplied to a string by a rod

Short answer

Average power = ½ μ v ω² A². v = √(15/0.002) = 86.6 m/s, ω = 2π×12 = 75.4 rad/s, A = 1.5 mm → P_avg ≈ 1.11×10⁻³ W (1.11 mW).

A rod vibrating at 12 Hz generates harmonic waves with amplitude of 1.5mm in a string of linear mass density 2 gm/m. If the tension in the string is 15N, what is the average power supplied by the source?

Given: f = 12 Hz, amplitude A = 1.5 mm = 1.5 × 10⁻³ m, μ = 2 g/m = 2 × 10⁻³ kg/m, T = 15 N.

Formulas

v = √(T/μ) ,   ω = 2πf ,   P_avg = ½ μ v ω² A²

Steps

  1. Wave speed: v = √(15 / 2×10⁻³) = √7500 = 86.6 m/s.
  2. ω = 2π × 12 = 75.40 rad/s; ω² = 5684.9 rad²/s².
  3. A² = (1.5×10⁻³)² = 2.25 × 10⁻⁶ m².
  4. Power:
P_avg = ½ × 2×10⁻³ × 86.6 × 5684.9 × 2.25×10⁻⁶
      = 1.0×10⁻³ × 86.6 × 0.012791
      = 1.11 × 10⁻³ W

Answer: average power supplied ≈ 1.1 × 10⁻³ W (1.11 mW).

✓ Answer checked by Claude Opus 5.5 (high effort)

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