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

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 the graph it is clear that the amplitude is `1 cm` and the period is `8 second`. Therefore the equation for the (S.H.M.) is
`x = a sin((2 pi)/(T)xx t = 1 sin ((2pi)/(8)) t = sin ((pi)/(4))t`
The velocity (v) of the particle at any instant of time 't' is
`v= ( d x)/(d t) = (d)/(d t)[sin((pi)/(4))t] = (pi)/(4) cos ((pi)/(4)) t`
The acceleration of the particle is
`(d^2 x)/(d t^2) = -((pi)/(4))^2 sin ((pi)/(4))t`
At `t = (4)/(3) s` we get
`(d^2 x)/(d t^2) = -((pi)/(4))^2 si n (pi)/(4) xx (4)/(3) = (-pi^2)/(16) si n(pi)/(3)`
= ` (-sqrt((3 pi^2)))/(32) cm//s^2`.