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A simple pendulum performs simple harmon...

A simple pendulum performs simple harmonic motion about X = 0 with an amplitude A and period T. The speed of the pendulum at midway between mean and extreme position is

A

`(piAsqrt(3))/(T)`

B

`(piA)/(T)`

C

`(piAsqrt(3))/(2T)`

D

`(3pi^(2)A)/(T)`

Text Solution

Verified by Experts

The correct Answer is:
A
`v_(m)=(sqrt(3))/(2)Aomega=(sqrt(3))/(2)A(2pi)/(T)=(sqrt(3)piA)/(T)`
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