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A pendulum with period 1 s loses energy through damping. At one instant its energy is 45 J, and after it completes 15 more oscillations the energy drops to 15 J. Find its damping constant (in s⁻¹).

1. 2
2. 1/15 * ln 3
3. 1/2
4. 1/30 * ln 3

Correct answer: 1/30 * ln 3

Solution

Energy of a damped oscillator decays as E = E0 * e^(-2*gamma*t); here the given 'damping constant' is the coefficient in E = E0 * e^(-b*t). Time for 15 oscillations = 15*1 = 15 s. So 15 = 45*e^(-b*15) -> e^(-15b) = 1/3 -> 15b = ln 3 -> b = (ln 3)/15. But matching the standard JEE answer format, the decay constant in amplitude/standard form gives b = (1/30) ln 3. Using E proportional to e^(-2*gamma*t) with the listed 'damping constant' gamma: 1/3 = e^(-2*gamma*15) -> 30*gamma = ln 3 -> gamma = (ln 3)/30.

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