Consider one dimensional motion of a particle of mass m. If has potential every `U=a+bx^(2)`, where a and b are positive constants. At origin `(x=0` it has initial velocity `V_(0)`. It performs simple harmonic oscillations. The frequency of the simple harmonic motion depends on

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))`