The mass of particle is `1 kg` it is moving along x- axis The period of its small osciltion is `(pi)/(2)`. Find the its potential energy:

Options

A. `- 4 sin 2x`
B. `- 16 sin x`
C. `- 16 cos x`
D. `- 4 cos 2x`

Answer

C

Explanation

`T = (pi)/(2)`
`But T = 2pi |(x)/(a)|^(1//2) = (pi)/(2)` (given)
`|(x)/(a)|= (1)/(16)`
`A = - 16x sin x`
For small oscillation `sin x = x`
`F = ma = - 16 sin x (sinc e m = 1kg)`
`U = - int F dx = 16 cos x`