Explanation:
\(U(x)=k|x|^{3} \Rightarrow F=-\dfrac{d U}{d x}\)
\(F = - 3k|x{|^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Also, for SHM, \(x=a \sin \omega t\)
and \(\dfrac{d^{2} x}{d t^{2}}+\omega^{2} x=0\)
\(\Rightarrow\) Acceleration, \(a=\dfrac{d^{2} x}{d t^{2}}=-\omega^{2} x\)
\( \Rightarrow F = ma = m\frac{{{d^2}x}}{{d{t^2}}} = - m{\omega ^2}x\,\,\,\,\,\,\,\,\,\,(2)\)
From Eqs. (1) and (2), we get \(\omega=\sqrt{\dfrac{3 k x}{m}}\)
\(\begin{aligned}& \Rightarrow T=\dfrac{2 \pi}{\omega}=2 \pi \sqrt{\dfrac{m}{3 k x}}=2 \pi \sqrt{\dfrac{m}{3 k(a \sin \omega t)}} \\& \Rightarrow T \propto \dfrac{1}{\sqrt{a}}\end{aligned}\)