When a particle of mass m moves on the x-axis in a potential of the form `V(x) =kx^(2)` it performs simple harmonic motion. The correspondubing time period is proprtional to `sqrtm/h`, as can be seen easily using dimensional analusis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of `x=0` in a way different from `kx^(2)` and its total energy is such that the particle does not escape toin finity. Consider a particle of mass m moving on the x-axis. Its potential energy is `V(x)=ax^(4)(agt0)` for |x| neat the origin and becomes a constant equal to `V_(0)` for |x|impliesX_(0)` (see figure).
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For periodic motion of small amplitude A,the time period (T) of thes particle is proportional to.

Options

A. `A sqrt((m)/(prop))`
B. `(1)/(A) sqrt((m)/(prop))`
C. `A sqrt((prop)/(m))`
D. `A sqrt((2 prop)/(m))`

Answer

B

Explanation

Dimension of `alpha` can be found as
`[alpha] = ML^(-2) T^(-2)`
Only option (b) has dimenstion of time
Alternatively :
From conservation of energy :
`(1)/(2)m((dx)/(dt))^(2) + alpha x^(4) = alpha A^(4)`
`((dx)/(dt))^(2) = (2 alpha)/(m) (A^(4)-x^(4))`
`int dt = sqrt((m)/(2 alpha)) underset(0) overset(A) int (dx)/(sqrt(A^(4) -x^(4)))`
`rArr t = (1)/(A) sqrt((m)/(2 alpha)) underset(0) overset(1) int (du)/(sqrt(1-u^(4)))` [Substitute `x = Au`]
`rArr t prop (1)/(A) sqrt((m)/(2 alpha))`.