The acceleration a of a particle undergoing SHM is shown in the figure. Which of the labelled points corresponds to the particle being at `-x_("max")`

Options

A. `(sqrt(3))/(32) pi^2 cm//s^2`
B. `(-pi^2)/(32) cm//s^2`
C. `(pi^2)/(32) cm//s^2`
D. `-(sqrt(3))/(32) pi^2 cm//s^2`

Answer

D

Explanation

The given motion is represented by
`(omega =(2pi)/(T)=(2pi)/(8)=(pi)/(4))`
`x=1 sin((pi)/(4)t),(d^(2)x)/(dt^(2))=(-pi^(2))/(16) sin(pi//4)t`
At `t = 4//3 s,(d^(2)x)/(dt^(2))=-(sqrt(3))/(32) pi^(2) cm//s^(2)`.