A particle of mass 0.02 kg moves simple harmonically with apmplitude 0.12 m and frequency of oscillation 1Hz 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. `(1)/(sqrt2)` time of the maximum
D. Half of the maxmimum

Answer

A

Explanation

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