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. `sqrt2pi`

Answer

C

Explanation

`F=-(dU)/(dx)=-8sin2x`
For small oscillation, `sin2x=2x`
i.e.,
Since `amu-x`, the oscillation are simple harmonic in nature.
`T=pisqrt(|(x)/(a)|)=2pisqrt((1)/(16))=(pi)/(2)s`