EASY · DAMPED AND DRIVEN OSCILLATIONS
AMPLITUDE DECAY AFTER N CYCLES
An underdamped spring–mass system starts from rest at displacement A₀ = 0.20 m. The natural angular frequency is ω₀ = 5 rad/s and the damping coefficient is γ = 0.4 rad/s. What is the amplitude of oscillation after exactly 8 complete damped cycles?
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Step-by-step solution
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Step 1
Find the damped angular frequency ωd of this system.
Hint
The damped frequency is lower than ω₀ because damping slows the oscillation. The discriminant is ω₀² − γ²/4.
Step 2
Calculate the period T_d of one complete damped oscillation.
Step 3
How much total time elapses after n = 8 complete cycles?
Step 4
Write an expression for the oscillation amplitude at time t_n using the exponential envelope.
Solution walkthrough
When a spring is released from displacement A₀ = 0.20 m in a damped system, it does not oscillate at ω₀ — it oscillates at the slightly lower damped frequency ωd = √(ω₀² − γ²/4). Here, with ω₀ = 5 rad/s and γ = 0.4 rad/s, the discriminant is 25 − 0.04 = 24.96, so ωd ≈ 4.996 rad/s — barely changed from ω₀ because γ is small. The period T_d = 2π/ωd ≈ 1.257 s. After 8 full cycles, t = 8 T_d ≈ 10.056 s. The amplitude envelope at that time is A₀ e^(−γt/2) = 0.20 × e^(−0.4 × 10.056 / 2) ≈ 0.20 × e^(−2.011) ≈ 0.20 × 0.134 ≈ 0.027 m. The most common error is writing e^(−γt) without the ½: that would halve the exponent's time constant and make the decay appear four times faster. The ½ arises because the equation of motion is x″ + γx′ + ω₀²x = 0 — the velocity-proportional damping term carries γ, not 2γ, so the amplitude decays at rate γ/2.
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