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A body is executing S.H.M. Its potential energy is ' $\mathrm{P}_1$ ' and ' $\mathrm{P}_2$ ' at displacements ' $x$ ' and ' $y$ ' respectively. The potential energy at displacement $(\mathrm{x}+\mathrm{y})$ is
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Verified Answer
The correct answer is:
$\sqrt{\mathrm{p}_1}+\sqrt{\mathrm{p}_2}=\sqrt{\mathrm{p}}$
At displacement $\mathrm{x}$
At displacement y
$\mathrm{E}_1=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2$
From equations
(1), Or (2) and (3) it follows that
$\sqrt{\mathrm{E}}=\sqrt{\mathrm{E}_1}+\sqrt{\mathrm{E}_2}$
Or $\mathrm{E}=\mathrm{E}_1+\mathrm{E}_2+2 \sqrt{\mathrm{E}_1 \mathrm{E}_2}$
At displacement y
$\mathrm{E}_1=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2$
From equations
(1), Or (2) and (3) it follows that
$\sqrt{\mathrm{E}}=\sqrt{\mathrm{E}_1}+\sqrt{\mathrm{E}_2}$
Or $\mathrm{E}=\mathrm{E}_1+\mathrm{E}_2+2 \sqrt{\mathrm{E}_1 \mathrm{E}_2}$
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