A particle of mass m is located in a one dimensional potential field where potential energy is given by : V(x) = A(1 – cos px), Where A and p are constant. The period of small oscillations of the particle is :

Options

A. `2pi sqrt (( m )/((Ap )))`
B. `2pi sqrt (( m )/(( A p ^(2))))`
C. `2pi sqrt ((m)/( (A)))`
D. `(1)/(2pi) sqrt ((Ap)/(m))`

Answer

B

Explanation

`V = A (1 - cos px )`
`F =- (dV)/(dx)=-Ap sin px`
For small displacement
`sin px = px`
` therefore F = - A p ^(2) x`
`T = 2pi sqrt((m )/(k)) = 2pi sqrt (( m )/( Ap ^(2)))`