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When a particle of mass `m` moves on the x-axis in potential of the from `V(x) = kx^2` it performs simple harmonic motion. The corresponding time period is proportional to `m, k` as can be seen easily using dimensional analysis. 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 to infinity. Consider a particle of mass `m` moving on the x-axis. Its potential energy is `V(x) = prop x^4(prop gt 0) fro |x|` near the origin and becomes a constant equal to `V_0 for |x| ge X_0` (see Fig. 8.12).

If the total energy of the particle is `E`. it will perform periodic motion only if.

A

`A sqrt((m)/(prop))`

B

`(1)/(A) sqrt((m)/(prop))`

C

`A sqrt((prop)/(m))`

D

`A sqrt((2 prop)/(m))`

Text Solution

Verified by Experts

The correct Answer is:
B
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))`.
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