The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in meters. The period of small oscillation (in sec) is

Options

A. `2pi`
B. `pi`
C. `(pi)/(2)`
D. `sqrt(2)pi`

Answer

C

Explanation

`U=4(1cos2x) J`
`F=-(dU)/(dx)=-8 sin 2x`
for small oscillation `sin 2x ~~2x`
F=-16x
Comparing whih =-kx
k=16
`nomega^(2)=16`
`omega=sqrt((16)/(m))`
`T=2pisqrt((m)/(16))`
`T=2pisqrt((1)/(16))=(pi)/(2)sec `