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A simple pendulum performs simple harmon...

A simple pendulum performs simple harmonic motion about `x=0` with an amplitude a ans time period T. The speed of the pendulum at `x = (a)/(2)` will be

A

`(piasqrt(3))/(T)`

B

`(piasqrt(3))/(2T)`

C

`(pia)/(T)`

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
`v=omega sqrt(A^(2)-x^(2))`
`=(2pi)/(T)sqrt(a^(2)-(a^(2))/(T))=(pia sqrt(3))/(T)`
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