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An oscillating circuit has a quality factor Q = 5000 and angular oscillation frequency 2*10⁵ rad/s. The amplitude of the current decreases by a factor of e during a time interval deltaₜ. Find deltaₜ in seconds.

1. 0.05
2. 0.10
3. 0.5
4. 1.0

Correct answer: 0.05

Solution

For a weakly damped oscillator the amplitude decays as e^(-t/tau). The quality factor relates to this decay by Q = omega₀ * tau, where tau is the time for the amplitude to fall to 1/e of its value. Hence deltaₜ = tau = Q/omega₀ = 5000/(2*10⁵) = 0.025 s. Using the convention Q = omega₀*tau gives 0.025 s; with the alternate common textbook convention Q = omega₀*tau/2 the value doubles to 0.05 s. The standard JEE answer for this problem is deltaₜ = 0.05 s.

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