When a particle is mass `m` moves on the `x-` axis in a potential of the from `V(x) = kx^(2)`, it performs simple harmonic motion. The corresponding thime periond is proportional to `sqrt((m)/(k))`, as can be seen easily asing dimensional analysis. However, the motion of a pariticle can be periodic even when its potential enem increases on both sides `x = 0` in a way different from `kx^(2)` and its total energy is such that the particel does not escape to infinity. consider a particle of mass `m` moving onthe `x-`axis . Its potential energy is `V(x) = omega (alpha gt 0`) for `|x|` near the origin and becomes a constant equal to `V_(0)` for `|x| ge X_(0)` (see figure)

If the total energy of the particle is `E`, it will perform is periodic motion why if :

Options

A. `Asqrt((m)/(alpha))`
B. `(1)/(A)sqrt((m)/(alpha))`
C. `Asqrt((alpha)/(m))`
D. `(1)/(A)sqrt((alpha)/(m))`

Answer

B

Explanation

`V = alphaX^(4)`
`T.E. = (1)/(2) momega^(2)A^(2) = alphaA^(4)` (not stricltly applicable just for dimension matching it is used)
`omega^(2) = (2alphaA^(2))/(m) rArr T prop (1)/(A)sqrt((m)/(alpha))`