Q .     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 corresponding time period is proportional to `sqrt(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 kx2 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) = alpha x^4 (alpha> 0)` for |x| near the origin and becomes a constant equal to `V_0` for `|x| >= X_0`
For periodic motion of small amplitude A, the time period T this particle can be

JEE 2010 ADVANCED Paper 1
A

`A\sqrt{\frac{m}{\alpha}}`

B

`\frac{1}{A}\sqrt{\frac{m}{\alpha}}`

C

`A\sqrt{\frac{\alpha}{m}}`

D

`A\sqrt{\frac{2\alpha}{m}}`

0
Success Rate: NAN %
Sr. No.SubjectTagCorrectRecorded Answer