A body is executing a linear S.H.M. Its poten | vedclass

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(B) We know that the potential energy $E_P$ of a body in $S.H.M.$ is given by $E_P = \frac{1}{2} Kx^2$,where $K$ is the force constant.For displacemen

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$(\sqrt{E_1}+\sqrt{E_2})^2$

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(B) We know that the potential energy $E_P$ of a body in $S.H.M.$ is given by $E_P = \frac{1}{2} Kx^2$,where $K$ is the force constant.For displacement $x$,$E_1 = \frac{1}{2} Kx^2 \Rightarrow x = \sqrt{\frac{2E_1}{K}}$.For displacement $y$,$E_2 = \frac{1}{2} Ky^2 \Rightarrow y = \sqrt{\frac{2E_2}{K}}$.The potential energy $E$ at displacement $(x+y)$ is given by $E = \frac{1}{2} K(x+y)^2$.Substituting the values of $x$ and $y$:$E = \frac{1}{2} K \left( \sqrt{\frac{2E_1}{K}} + \sqrt{\frac{2E_2}{K}} \right)^2$$E = \frac{1}{2} K \left( \sqrt{\frac{2}{K}} \right)^2 (\sqrt{E_1} + \sqrt{E_2})^2$$E = \frac{1}{2} K \cdot \frac{2}{K} (\sqrt{E_1} + \sqrt{E_2})^2$$E = (\sqrt{E_1} + \sqrt{E_2})^2$.

$A$ body is executing a linear $S.H.M.$ Its potential energies at the displacements $x$ and $y$ are $E_1$ and $E_2$ respectively. Its potential energy at displacement $(x+y)$ will be

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