Simple Harmonic Motion – JEE Advanced Previous Year Questions with Solutions

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When a particle of mass m moves on the x–axis in a potential of the form $\mathrm{V}(\mathrm{x})=\mathrm{k} \mathrm{x}^{2}$, it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{m}{k}}$, 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 $\mathrm{kx}^{2}$ 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 $\mathrm{V}(\mathrm{x})=\alpha \mathrm{x}^{4}(\alpha>0)$ for $|\mathrm{x}|$ near the origin and becomes a constant equal to $\mathrm{V}_{0}$ for $|\mathrm{x}| \geq \mathrm{X}_{0}$ (see figure)