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. `Asqrt((2alpha)/(m))`

Answer

B

Explanation

Dimension `alpha` can be found as `[alpha]=ML^(-2)T^(-2)`
Only option (B) has dimension of time
Alternatively:
From conservation of energy: `(1)/(2)m((dx)/(dt))^(2)+ax^(4)=alphaA^(4)`, `((dx)/(dt))^(2)=(2a)/(m)(A^(4)-x^(4)) `
`overset(T//4)underset(0)intdt=sqrt((m)/(2a))overset(A)underset(0)int(dx)/(sqrt(A^(4)-x^(4)))`
implies `(T)/(4)=(1)/(A)sqrt((m)/(2a))overset(1)underset(0)int(dx)/(sqrt(1-u^(4)))` [Substitudex=Au]
implies `Tprop(1)/(A)sqrt((m)/(a))`