![IIT-JEE 2011 Paper 2 Offline Physics - Simple Harmonic Motion Question 18 English](https://imagex.cdn.examgoal.net/ZFyH3fE4pDjAZu1Ia/fyx7AtLnyyulRwQlvd1GYyzE9sBKf/ufM381hIDIhu1GE3ouJsk5/image.png)
When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, 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) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).
If the total energy of the particle is E, it will perform periodic motion only if
When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to $$\sqrt {{m \over k}} $$, 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) = $$\alpha$$x4 ($$\alpha$$ > 0) for | x | near the origin and becomes a constant equal to V0 for (see figure).
For periodic motion of small amplitude A, the time period T of this particle is proportional to