At what position in simple harmonic motion ar | vedclass

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(C) The kinetic energy $(K)$ of a particle in simple harmonic motion is given by $K = \frac{1}{2}m\omega^2(a^2 - y^2)$.The potential energy $(U)$ is g

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$a/\sqrt{2}$

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(C) The kinetic energy $(K)$ of a particle in simple harmonic motion is given by $K = \frac{1}{2}m\omega^2(a^2 - y^2)$.The potential energy $(U)$ is given by $U = \frac{1}{2}m\omega^2y^2$.Given that the kinetic energy equals the potential energy $(K = U)$:$\frac{1}{2}m\omega^2(a^2 - y^2) = \frac{1}{2}m\omega^2y^2$Canceling the common terms $\frac{1}{2}m\omega^2$ from both sides:$a^2 - y^2 = y^2$$a^2 = 2y^2$$y^2 = \frac{a^2}{2}$$y = \pm \frac{a}{\sqrt{2}}$Therefore,the kinetic and potential energies are equal at the position $y = a/\sqrt{2}$.

At what position in simple harmonic motion are the kinetic energy and potential energy equal?

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