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

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Question

(C) The kinetic energy $(K.E.)$ and potential energy $(U)$ of a simple harmonic oscillator are given by:$K.E. = \frac{1}{2} k(A^2 - x^2)$$U = \frac{1}

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Kinetic energy is maximum,potential energy is minimum

Vedclass · Answer

(C) The kinetic energy $(K.E.)$ and potential energy $(U)$ of a simple harmonic oscillator are given by:$K.E. = \frac{1}{2} k(A^2 - x^2)$$U = \frac{1}{2} k x^2$At the mean position,the displacement $x = 0$.Substituting $x = 0$ into the equations:$K.E. = \frac{1}{2} k A^2$ (which is the maximum value)$U = \frac{1}{2} k (0)^2 = 0$ (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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