The displacement of a particle of mass `3g` executing simple harmonic motion is given by `x =3sin (0.2t)` in `SI` units. The kinetic energy of the particle at a point which is at a displacement equal to `1//3` of its amplitude from its mean position is

Options

A. `12 xx 10^(-3) J`
B. `25 xx 10^(-3) J`
C. `0.48 xx 10^(-3) J`
D. `0.24 xx 10^(-3) J`

Answer

C

Explanation

`K = (1)/(2) m omega^(2) (A^(2) - x^(2))`
`= (1)/(2) (3 xx 10^(-3)) (0.2)^(2) (3^(2) - ((3)/(3))^(2))`
`= 0.48 xx 10^(-3)`J