The average kinetic energy in one time period | vedclass

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(C) The displacement of a particle in $SHM$ is given by $x = A \sin(\omega t)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.The kinet

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$\frac{1}{4} m \omega^2 A^2$

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(C) The displacement of a particle in $SHM$ is given by $x = A \sin(\omega t)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.The kinetic energy $(KE)$ is given by $KE = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.The average kinetic energy over one time period $T$ is given by $\langle KE angle = \frac{1}{T} \int_0^T KE \, dt$.Substituting the expression for $KE$: $\langle KE angle = \frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \, dt$.Using the identity $\cos^2(\omega t) = \frac{1 + \cos(2\omega t)}{2}$,the integral becomes:$\langle KE angle = \frac{m A^2 \omega^2}{2T} \int_0^T \frac{1 + \cos(2\omega t)}{2} \, dt$.Since the integral of $\cos(2\omega t)$ over a full period $T$ is $0$,we get:$\langle KE angle = \frac{m A^2 \omega^2}{4T} \int_0^T 1 \, dt = \frac{m A^2 \omega^2}{4T} 	imes T = \frac{1}{4} m \omega^2 A^2$.

The average kinetic energy in one time period of a $SHM$ is :-

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