A particle of mass 0.02 kg moves simple harmonically with amplitude 0.12 m and frequency of oscillation 1 Hz. At t=0, the particle is at x=0.24 m (the mean position) then the acceleration of the particle at t=0.5 s is

Options

A. zero
B. maximum
C. T time of the maximum
D. Half of the maximum

Answer

A

Explanation

`m=0.05 kg , A = 0.12 m, v=1 Hz`
At `t=0, x_(0)=0.24" " :.x=x_(0)+A sin omega t`
`rArr x=0.24 +(0.12) sin (2pi) t " " :.a= -(0.12)(2pi)^(2) sin(2pi)t" " :. "At t"=0.5s`,
`a= -(0.12) (4pi^(2)) sin(2pi xx 0.5) rArr a=0`