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Damped oscillator time for 1/e amplitude

Short answer

With τ=60 s, amplitude a = a0 e^(-t/2τ) so for 1/e, t=2τ=120 s. Energy E = E0 e^(-t/τ) so for 1/e², t=2τ=120 s. Both times are 120 s because energy ∝ amplitude².

If the relaxation time of a damped harmonic oscillator is 60 sec, find the time for which (a) the amplitude falls to 1/e times the initial value and (b) the energy falls to 1/e² times the initial value.

Convention used: for damping force −bv, amplitude a = a0 e^(−bt/2m) and energy E = E0 e^(−bt/m). Relaxation time τ = m/b is the time in which the energy falls to 1/e of its initial value. So a = a0 e^(−t/2τ) and E = E0 e^(−t/τ).

Given τ = 60 s.

(a) Amplitude falls to 1/e

a/a0 = e^(−t/2τ) = e^(−1)
t = 2τ = 2 × 60 = 120 s

(b) Energy falls to 1/e²

E/E0 = e^(−t/τ) = e^(−2)
t = 2τ = 2 × 60 = 120 s

Answer: (a) 120 s (2 min); (b) 120 s.

Both times are the same, as they must be: energy ∝ (amplitude)², so when the amplitude is 1/e the energy is 1/e².

✓ Answer checked by Claude Opus 5.5 (high effort)

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