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 $(KE)$ and potential energy $(PE)$ at a displacement $y$ are given by:$KE = \frac{1}{2} m \omega

Vedclass · Accepted Answer

$A/\sqrt{2}$

Vedclass · Answer

(C) For a particle executing $SHM$,the kinetic energy $(KE)$ and potential energy $(PE)$ at a displacement $y$ are given by:$KE = \frac{1}{2} m \omega^{2}(A^{2} - y^{2})$$PE = \frac{1}{2} m \omega^{2} y^{2}$Given that $KE = PE$,we equate the two expressions:$\frac{1}{2} m \omega^{2}(A^{2} - y^{2}) = \frac{1}{2} m \omega^{2} y^{2}$$A^{2} - y^{2} = y^{2}$$A^{2} = 2y^{2}$$y^{2} = \frac{A^{2}}{2}$$y = \pm \frac{A}{\sqrt{2}}$Thus,the displacement is $A/\sqrt{2}$.

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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