Physics - SHM Question with Solution | TestHub

We know, Potential Energy ${\mathrm{E}}_{\mathrm{P}}=\frac{1}{2}{\mathrm{Kx}}^2$ $\therefore {\mathrm{E}}_1=\frac{1}{2}{\mathrm{Kx}}^2\Rightarrow \mathrm{x}=\sqrt{\frac{2{\mathrm{E}}_1}{\mathrm{K}}}$... (i) and ${\mathrm{E}}_2=\frac{1}{2}{\mathrm{Ky}}^2\Rightarrow \mathrm{y}=\sqrt{\frac{2{\mathrm{E}}_2}{\mathrm{K}}}$... (ii) Given, total displacement $=(x+y)$ $\therefore$ Potential energy at displacement $(x+y)$, $\mathrm{E}$ is $\frac{1}{2}\mathrm{K}(\mathrm{x}+\mathrm{y}{)}^2$ $\begin{array}{rl}& =\frac{1}{2}K{\left(\sqrt{\frac{2E_1}{K}}+\sqrt{\frac{2E_2}{K}}\right)}^2\\ & =\frac{1}{2}K\left[\frac{2E_1}{K}+\frac{2E_2}{K}+2\left(\sqrt{\frac{E_1}{K}}\right)\left(\sqrt{\frac{2E_2}{K}}\right)\right]\\ & ={\left(\sqrt{E_1}+\sqrt{E_2}\right)}^2=\left(E_1+E_2+2\sqrt{E_1{E}_2}\right)\end{array}$