The displacement vs time graph of a particle undergoing S.H>M. Is shown below . The acceleration of the particle at `t=2//3s` is

Options

A. `(sqrt3pi^(2))/(8)cm//s^(2)`
B. `(pi^(2))/(8)cm//s^(2)`
C. `(-sqrt3pi^(2))/(8)cm//(s^(2))`
D. `(-pi^(2))/(80cm//s^(2))`

Answer

C

Explanation

From the x -t graph,
T = 4 s
`omega = (2pi)/(T) = (2pi)/(4) = (pi)/(2)` rad
`t = (2)/(3)s`
`omegat = (pi)/(2) xx (2)/(3) = (pi)/(3) rad`
`|a| = omega^(2)A sin omegat `
` = ((pi)/(2))^(2) xx 1 xx sin(pi)/(3)`
` = (pi^(2))/(4) xx (sqrt(3))/(2) = (sqrt(3))/(8)pi^(2) cm//s^(2)`