A body is performing simple harmonic motion with amplitude `a` and time period `T` variation of its acceleration `(f)` with time`(t)` is shown in figure If at time `t` velocity of the body is `v` which of the following graph is correct?

Options

A.
B.
C.
D.

Answer

D

Explanation

Since acceleration of the particle at initial moment is maximum possible and is negative, therefore, the particle is at right extreme position at this moment .
When the particle is released, it starts to move to the left. It means, velocity starts to increase from zero initial value to negative value and its magnitude becomes maximum possible at mean position (at `t=(T)/(4)`). It means at `t=(T)/(4)`, kinetic energy is equal to maximum possible.
At `t=(T)/(2)`, the particle comes to instantaneous rest at left extreme position. It means at `t=(T)/(2)`, `v` is equal to zero. Hence kinetic energy is equal to zero. At `t=(3T)/(4)`, particle comes back to mean position and now moves to the right. Therefore, velocity is positive and has maximum possible magnitude. Therefore kinetic energy is maximum possible.
At `t=T`, particle comes back to initial position (extreme right position). Velocity and kinetic energy become equal to zero.