A body is executing S.H.M. Its potential ener | vedclass

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(D) The potential energy $P$ of a body executing $S$.$H$.$M$. at displacement $x$ is given by $P = \frac{1}{2} k x^2$,where $k = m \omega^2$.Given,$P_

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$\sqrt{P_1}+\sqrt{P_2}=\sqrt{P}$

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(D) The potential energy $P$ of a body executing $S$.$H$.$M$. at displacement $x$ is given by $P = \frac{1}{2} k x^2$,where $k = m \omega^2$.Given,$P_1 = \frac{1}{2} k x^2 \implies \sqrt{P_1} = x \sqrt{\frac{k}{2}}$  ---$(1)$And $P_2 = \frac{1}{2} k y^2 \implies \sqrt{P_2} = y \sqrt{\frac{k}{2}}$  ---$(2)$Let the potential energy at displacement $(x+y)$ be $P$. Then,$P = \frac{1}{2} k (x+y)^2$Taking the square root of both sides:$\sqrt{P} = (x+y) \sqrt{\frac{k}{2}}$$\sqrt{P} = x \sqrt{\frac{k}{2}} + y \sqrt{\frac{k}{2}}$Substituting equations $(1)$ and $(2)$ into this expression:$\sqrt{P} = \sqrt{P_1} + \sqrt{P_2}$

$A$ body is executing $S$.$H$.$M$. Its potential energy is '$P_1$' and '$P_2$' at displacements '$x$' and '$y$' respectively. The potential energy at displacement $(x+y)$ is

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