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. `(sqrt3)/(32)pi^(2)cm//s^(2)`
B. `(-pi^(2))/(32)pi^(2)cm//s^(2)`
C. `(pi^(2))/(32)pi^(2)cm//s^(2)`
D. `-(sqrt3)/(32)pi^(2)cm//s^(2)`

Answer

D

Explanation

The given motion is represented by
`(omega=(2pi)/(T)=(2pi)/(8)=(pi)/(4))`
`x=1sin((pi)/(4)t)`,`(d^(2)x)/(dt^(2))=(-pi^(2))/(16)sin (pi//4)t`
At `t=4//3s`,`(d^(2)x)/(dt^(2))=-(sqrt3)/(32)pi^(2)cm//s^(2)`