Physics - W.E.P. Question with Solution | TestHub

Given, $V=k{x}^3$ Total energy $(E)=$ Maximum potential energy $=k{a}^3$ From conservation of energy $\begin{array}{rl}\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2+k{x}^2& =k{a}^3\\ \Rightarrow \frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2& =k\left(a^3-x^3\right)\\ \Rightarrow \frac{dx}{dt}& =\sqrt{\frac{2k}{m}\left(a^3-x^3\right)}\\ \frac{dx}{\sqrt{a^3-x^3}}& =\sqrt{\frac{2k}{m}}dt\end{array}$ or Integrating ${\int }_0^a\frac{dx}{\sqrt{a^3-x^3}}=\sqrt{\frac{2k}{m}}{\int }_0^{T/4}dt$ or ${\int }_0^a\frac{dx}{\sqrt{a^3-x^3}}=\sqrt{\frac{2k}{m}}\frac{T}{4}$ Let $x=a\sin \theta \Rightarrow dx=a\cos \theta d\theta$ $\therefore {\int }_0^{\pi /2}\frac{a\cos \theta d\theta }{\sqrt{a^3-(a\sin \theta {)}^3}}=\sqrt{\frac{2k}{m}}\frac{T}{4}$