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

Answer

D

Explanation

In general displacement of a particle performing S.H.M. is given by :
`x A sin (omegat + delta) " "…(i)`
Here in this case particle at t = 0 is crossing origin in positive direction hence `delta = 0`.
time period of motion : T= 8s
`omega = (2pi)/(T) = (2pi)/(8) = (pi)/(4)`
Substituting in equation (i) we get
`x = A sin ((pi)/(4)t)`
Here amplitude A = 1 cm
`rArr x = sin ((pi)/(4)t)` , where x is measured in cm.
`rArr (dx)/(dt) = (pi)/(4) cos ((pi)/(4)t)`
`rArr a = (d^(2)x)/(dt^(2)) = (pi^(2))/(16) sin ((pi)/(4)t)`
Substituting t = 4/3 s
`a = -(pi^(2))/(16)sin ((pi)/(4) xx (4)/(3)) = -(pi^(2))/(16) sin(pi//3)`
`rArr = -(pi^(2)sqrt(3))/(32) cm//s^(2)`