A simple pendulum performs 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. `(pia)/(T)`
B. `(3pi^(2)a)/(T)`
C. `(pi a sqrt(3))/(T)`
D. `(pi a sqrt(3))/(2T)`.

Answer

C

Explanation

Speed `v=omega sqrt(a^(2)-x^(2)),x=(a)/(2)`
`:." "v=omega sqrt(a^(2)-(a^(2))/(4))=omega sqrt((3a^(2))/(4))`
`=(2pi)/(T)(a sqrt(3))/(2)=(pi a sqrt(3))/(T)`
Correct choice is ( c ).