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A body is executing a linear S.H.M. Its potential energies at the displacement ' $x$ ' and ' $y$ ' are ' $E_1$ ' and ' $\mathrm{E}_2$ ' respectively. Its potential energy at displacement $(\mathrm{x}+\mathrm{y})$ will be
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Verified Answer
The correct answer is:
$\left(\sqrt{\mathrm{E}_1}+{\sqrt{\mathrm{E}_2}}\right)^2$
We know,
Potential Energy $\mathrm{E}_{\mathrm{P}}=\frac{1}{2} \mathrm{Kx}^2$
$\therefore \quad \mathrm{E}_1=\frac{1}{2} \mathrm{Kx}^2 \Rightarrow \mathrm{x}=\sqrt{\frac{2 \mathrm{E}_1}{\mathrm{~K}}}$... (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 $=(x+y)$
$\therefore \quad$ Potential energy at displacement $(x+y)$,
$\mathrm{E}$ is $\frac{1}{2} \mathrm{~K}(\mathrm{x}+\mathrm{y})^2$
$\begin{aligned}
& =\frac{1}{2} K\left(\sqrt{\frac{2 E_1}{K}}+\sqrt{\frac{2 E_2}{K}}\right)^2 \\
& =\frac{1}{2} K\left[\frac{2 E_1}{K}+\frac{2 E_2}{K}+2\left(\sqrt{\frac{E_1}{K}}\right)\left(\sqrt{\frac{2 E_2}{K}}\right)\right] \\
& =\left(\sqrt{E_1}+\sqrt{E_2}\right)^2=\left(E_1+E_2+2 \sqrt{E_1 E_2}\right)
\end{aligned}$
Potential Energy $\mathrm{E}_{\mathrm{P}}=\frac{1}{2} \mathrm{Kx}^2$
$\therefore \quad \mathrm{E}_1=\frac{1}{2} \mathrm{Kx}^2 \Rightarrow \mathrm{x}=\sqrt{\frac{2 \mathrm{E}_1}{\mathrm{~K}}}$... (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 $=(x+y)$
$\therefore \quad$ Potential energy at displacement $(x+y)$,
$\mathrm{E}$ is $\frac{1}{2} \mathrm{~K}(\mathrm{x}+\mathrm{y})^2$
$\begin{aligned}
& =\frac{1}{2} K\left(\sqrt{\frac{2 E_1}{K}}+\sqrt{\frac{2 E_2}{K}}\right)^2 \\
& =\frac{1}{2} K\left[\frac{2 E_1}{K}+\frac{2 E_2}{K}+2\left(\sqrt{\frac{E_1}{K}}\right)\left(\sqrt{\frac{2 E_2}{K}}\right)\right] \\
& =\left(\sqrt{E_1}+\sqrt{E_2}\right)^2=\left(E_1+E_2+2 \sqrt{E_1 E_2}\right)
\end{aligned}$
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