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 graph
T=8 second , A= 1 cm, `x=A sin omegat=1 sin ((2pi)/8) t`.
`a=-omega^(2)x=-((2pi)/8)^(2)sin((2pi)/8) g cm//s^(2)`
At, `t=4/3` second `a=-((2pi)/8)^(2) sin ((pi)/3) =-(sqrt(3)pi^(2))/32 cm//s^(2)`