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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

`(pi a)/(T)`

B

`(pi 4sqrt(3))/(2T)`

C

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

D

`(pi a sqrt(3))/(T)`

Text Solution

Verified by Experts

The correct Answer is:
D
As, `v = omegasqrt(a^(2) - x^(2))`
`v = omegasqrt(a^(2) - x^(2))`
`= (2pi)/(T)sqrt(a^(2) - (a^(2))/(4)) " "(because x = (a)/(2))`
`= (2pi)/(T).(sqrt(3)a)/(2) = (pi a sqrt(3))/(T)`
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