Q.
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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a
πa32T
b
πaT
c
3π2aT
d
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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