A particle of mass m moves in a one dimensional potential energy `U(x)=-ax^2+bx^4`, where a and b are positive constant. The angular frequency of small oscillation about the minima of the potential energy is equal to

Options

A. `pisqrt((a)/(2b))`
B. `2sqrt((a)/(m))`
C. `sqrt((2a)/(m))`
D. `sqrt((a)/(2m))`

Answer

B

Explanation

`U(x) = ax^(2) + bx^(4)`
`f = - (delU)/(delx) = -2ax - 4bx^(3) ~~ -2ax` for small `x`
So `momega^(2) 2a rArr omega = sqrt((2a)/(m))`