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A simple pendulum executing S.H.M. has p...

A simple pendulum executing S.H.M. has period T and amplitude A. Its speed, when at a distance `(A)/(4)` is :

A

`(piAsqrt(15))/(2T)`

B

`(piA sqrt(15))/(T)`

C

`(piA)/(2T)`

D

`(2piA)/(T)`.

Text Solution

Verified by Experts

The correct Answer is:
A
`v=omega sqrt(A^(2)-y^(2))=(2pi)/(T) sqrt(A^(2)-(A^(2))/(16))`
`=(2pi)/(T)sqrt((15)/(16)A^(2))=(pi sqrt(15)A)/(2T)`
Hence correct choice is (a).
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