The displacement-time graph of a particle executing SHM is shown in figure. Which of the following statement is/are true ?

Options

A. (a)The velocity is maximum at `t = T//2`
B. (b)The acceleration is maximum at `t = T`
C. (c)The force is zero at `t = 3T//4`
D. (d)The potential energy equals the kinetic energy at `t = T//2`

Answer

B, C, D

Explanation

`:' a = -omega^(2)y` and `t = T`. Y is maximum os accelertion is maximum at `t = Tt`. (`a = -omega^(2)y` and `t = T`)
Also `y = 0` at `t = (3T)/(4)`, so force is zero at `t = (3T)/(4)`
At `t = T/2, v = 0 rArr PE` = oscillation energy