Question

The `x-t` graph of a particle undergoing simple harmonic motion is shown in figure. Acceleration of particle at `t = 4//3 s` is

A. `(sqrt(3))/(32)pi ^2cm//s^2`
B. `(-pi^2)/(32)cm//s^2`
C. `(pi^2)/(32)cm//s^2`
D. `(-sqrtpi^3)/(32)pi^2cm//s^2`

Answer 1

Correct Answer - D

at time `t=0 s` , the displacement `x=0` , Therefore , the curve is a sinusoidal curve. It represents SHM whose amplitude
`A=1cm and T=8s`.
As, `x=A sin .(2pi)/(T)t`
`therefore` Acceleration, `a=(d^2x)/(dt^2)=-(4pi^2)/(T^2)A sin. (2pi)/(T)t`
At, `t=(4)/(3)s`
Acceleration , `a=-(4pi^2)/(8^2)xx1xxsin.(2pi)/(8)xx(4)/(3)`
`=-(4pi^2)/(64)sin.(pi)/(3)=-(4pi^2)/(64)xx(sqrt(3))/(2)=-(sqrt(3)pi^2)/(32)cm//s^2`