The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle of `t=4//3s` is

Options

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

Answer

D

Explanation

Let `x=Asin(omegat+phi)`
At `t=0`, `x=0`, `phi=0`
`T=8s`, `omega=(2pi)/(T)=(2pi)/(8)=pi/4 rad//s`, `a=1cm`
`x=A sin omegat=1xxsin((pi)/(4))t`
At `t=4/3s`, `x=sin((pi)/(4))*4/3=sqrt3/2`
`a=-omega^2x=-(pi/4)^2sqrt3/2=-(sqrt3pi^2)/(32)cm//s^2`