The K.E. and P.E. of a particle executing SHM | vedclass

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Question

(C) For a particle executing $SHM$,the kinetic energy $(KE)$ at displacement $y$ is given by $KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$.The poten

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$A/\sqrt{2}$

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(C) For a particle executing $SHM$,the kinetic energy $(KE)$ at displacement $y$ is given by $KE = \frac{1}{2} m \omega^{2} (A^{2} - y^{2})$.The potential energy $(PE)$ at displacement $y$ is given by $PE = \frac{1}{2} m \omega^{2} y^{2}$.Given that $KE = PE$,we equate the two expressions:$\frac{1}{2} m \omega^{2} (A^{2} - y^{2}) = \frac{1}{2} m \omega^{2} y^{2}$.Canceling the common terms $\frac{1}{2} m \omega^{2}$ from both sides,we get:$A^{2} - y^{2} = y^{2}$.Rearranging the terms:$A^{2} = 2y^{2}$.Solving for $y$:$y^{2} = \frac{A^{2}}{2} \implies y = \frac{A}{\sqrt{2}}$.

The $K.E.$ and $P.E.$ of a particle executing $SHM$ with amplitude $A$ will be equal when its displacement is-

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