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A simple pendulum perform simple harmoni...

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

A

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

B

`(piA)/(T)`

C

`(piAsqrt3)/(2T)`

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
As `v=omegasqrt(A^(2)-x^(2))`
at `X=A/2`
`v=omega sqrt(A^(2)-((A)/(2))^(2))implies v=(2pi)/(T)*(sqrt3)/(2)A=(sqrt(3)piA)/(T)`.
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