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Damped block-spring oscillation period

Short answer

The damped angular frequency is ω′ = √(k/m − (b/(2m))²). With m=0.750 kg, k=56 N/m, b=0.162 N·s/m, ω′ ≈ 8.640 rad/s, giving period T = 2π/ω′ ≈ 0.727 s.

A 750g block oscillates on the end of a spring whose force constant is k = 56 N/m. The mass moves in a fluid which offers a resistive force F = −bv, where b = 0.162 Ns/m. What is the period of the oscillation?

Given: m = 750 g = 0.750 kg, k = 56 N/m, b = 0.162 N·s/m.

Equation of motion: m x'' + b x' + k x = 0. For weak damping the angular frequency of the oscillation is

ω′ = √(k/m − (b/2m)²)

Steps

  1. k/m = 56 / 0.750 = 74.667 s⁻² (ω0 = 8.641 rad/s).
  2. b/2m = 0.162 / (2 × 0.750) = 0.108 s⁻¹, and (b/2m)² = 0.01166 s⁻².
  3. ω′² = 74.667 − 0.0117 = 74.655, so ω′ = 8.6403 rad/s.
  4. Period:
T = 2π/ω′ = 6.2832 / 8.6403 = 0.7272 s

Compare undamped: T0 = 2π√(m/k) = 0.7271 s. The damping is very small (b² << 4mk since 0.026 << 168), so the period barely changes.

Answer: T ≈ 0.727 s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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