A simple pendulum performs simple harmonic motion about.x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a2 will be

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πa32T

πaT

3π2aT

πa3T

answer is D.

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Detailed Solution

v = dydt = Aω cos ωt = Aω1-sin2ωt= ωA2-y2Here, y = a2∴ v = ωa2-a24 = ω3a24 = 2πTa32 = πa3T

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