The potential energy of a particle of mass 'm' situated in a unidimensional potential field varies as `U(x) = U_0 [1- cos((ax)/2)]`, where `U_0` and a are positive constant. The time period of small oscillations of the particle about the mean position-

Options

A. `pisqrt(m/(a^(2) U_(0)))`
B. `2pisqrt((2m)/(a^(2)U_(0)))`
C. `2pisqrt(m/(a^(2)U_(0)))`
D. `4pisqrt(m/(a^(2)U_(0)))`

Answer

D

Explanation

Restoring force `F=(-du)/(dx) =(-d)/(dx) U_(0)[1-cos ((ax)/2)]`
`F(x) =-u_(0) a/2 sin ((ax)/2)`
For small angle `sin((ax)/2) ~~(ax)/2`
`F=-u_(0)(a^(2)x)/4 rArr "acc." =(-u_(0)a^(2)x)/(4m)=-omega^(2) x=((2pi)/T)^(2)xx x`
So, time period `T=4pi sqrt(m/(a^(2)U_(0)))`