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

B

Explanation

Here `U=4(1-cos2x)J`
`therefore F=-(dU)/(dx)=-8sin2x`
Acceleration, `a=(F)/(m)=-8sin2x" "(because m=1kg)`
For small oscillations, `sin2x~~2x therefore a=-16x" "...(i)`
Since `aprop-x`, the oscillations are SHM in nature.
In SHM, `a=-omega^(2)x" "...(ii)`
Comparing (i) and (ii), we get `omega^(2)=16oromega=sqrt(16)`
`therefore` Time period, `T=(2pi)/(omega)=(2pi)/(sqrt(16))=(pi)/(2)s`