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

From graph
`T = 8` second, `A = 1 cm, x = A sinomegat. = 1sin(2pi)/(8)t`.
`a = -omega^(2)x = -((2pi)/(8))^(2) sin((2pi)/(8))t cm//s^(2)`
At, `t = (4)/(3)` second, `a = -((2pi)/(8))^(2) sin"(pi)/(3) = - (sqrt(3) pi^(2))/(32) cm//s^(2)`