A particle of mass `10gm` is placed in a potential field given by `V = (50x^(2) + 100)J//kg`. The frequency of oscilltion in `cycle//sec` is

Options

A. `(10)/(pi)`
B. (5)/(pi)`
C. `(100)/(pi)`
D. `(50)/(pi)`

Answer

B

Explanation

Potential energy `U = mV`
`rArr U = (50 x^(2) + 100) 10^(-2)`
`F = -(dU)/(dx)=-(100x)10^(-2)`
`rArr m omega^(2)x=-(100 xx10^(-2))10^(-2)`
`10 xx 10^(-3) omega^(2)x=100 xx 10^(-2)x`
`rArr omega^(2) =100,omega =10`
`rArr f =(omega)/(2pi)=(10)/(2 pi)=(5)/(pi)`.