The graph shows the variation of displacement of a particle executing `S.H.M` with time. We inference form this graph that :-

Options

A. the force is zero at time `(3T)/(4)`
B. the velocity is maximum at time `T/2`
C. the acceleration is maximum at time `T`
D. the P.E. is equal to half of total energy at time `T/2`

Answer

B

Explanation

For (A) : at `t = (3T)/(4)`, particle at extreme position
`:' a = -omega^(2)x :' F ne 0`
For (B) at `t = T//2`, particle at mean position
`(t = T//2)`
`v = omegaA`
For (C) : at `t = T`, particle at mean position
`rArr a = -omega^(2)x = 0`
For (D) : at `t = t//2`, particle at mean position
`(t= T//2)`
so `x = 0 rArr U = 1/2 kx^(2) = 0`