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A particle of mass m moves in a one dime...

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

A

`pisqrt((a)/(2b))`

B

`2sqrt((a)/(m))`

C

`sqrt((2a)/(m))`

D

`sqrt((a)/(2m))`

Text Solution

Verified by Experts

The correct Answer is:
B
`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))`
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