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……….

Options

A. `(pi A)/(T)`
B. `(3pi^(2) A)/(T)`
C. `(pi A sqrt(3))/(T)`
D. `(pi A sqrt(3))/(2T)`

Answer

C

Explanation

`v= omega sqrt(A^(2)-x^(2))= (2pi)/(T) sqrt(A^(2)-(A^2)/(4))`
`=(2pi)/(T) sqrt((3A^2)/(4))= (pi A sqrt(3))/(T)`.