In a simple harmonic oscillator,at the mean p | vedclass

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(C) In $S.H.M.$,the kinetic energy of a particle at any displacement $x$ is given by:$K.E. = \frac{1}{2} m \omega^{2} (a^{2} - x^{2})$The potential en

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Kinetic energy is maximum,potential energy is minimum

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(C) In $S.H.M.$,the kinetic energy of a particle at any displacement $x$ is given by:$K.E. = \frac{1}{2} m \omega^{2} (a^{2} - x^{2})$The potential energy of a particle at any displacement $x$ is given by:$P.E. = \frac{1}{2} m \omega^{2} x^{2}$where $a$ is the amplitude of the particle and $x$ is the displacement from the mean position.At the mean position,the displacement $x = 0$.Substituting $x = 0$ into the equations:$K.E. = \frac{1}{2} m \omega^{2} a^{2}$ (which is the maximum value).$P.E. = \frac{1}{2} m \omega^{2} (0)^{2} = 0$ (which is the minimum value).Therefore,at the mean position,kinetic energy is maximum and potential energy is minimum.

In a simple harmonic oscillator,at the mean position

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