A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $\mathrm{U}\left(\mathrm{x}\right) = \mathrm{k}{\left[\mathrm{x}\right]}^3,$ where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

$\mathrm{U} = \mathrm{k}|\mathrm{x}{|}^3 \Rightarrow \mathrm{F} = -\frac{\mathrm{dU}}{\mathrm{dx}} = -3\mathrm{k}|\mathrm{x}{|}^2$----------------(i)

Also, for SHM $\mathrm{x} = \mathrm{a} \sin \mathrm{\omega t} \mathrm{and} \frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}+{\mathrm{\omega }}^2\mathrm{x}= 0$

$\Rightarrow \mathrm{acceleration} = \frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2} = -{\mathrm{\omega }}^2\mathrm{x} \Rightarrow \mathrm{F} = \mathrm{ma}$

$= \mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2} = -{\mathrm{m\omega }}^2\mathrm{x}----\left(\mathrm{ii}\right)$

From equation (i) and (ii) we get $\mathrm{\omega } = \sqrt{\frac{3\mathrm{kx}}{\mathrm{m}}}$

$\Rightarrow \mathrm{T} = \frac{2\mathrm{\pi }}{\mathrm{\omega }} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{3\mathrm{kx}}} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{3\mathrm{k}(\mathrm{a} \sin \mathrm{\omega t})}} \Rightarrow \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{a}}}$