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Q.
A particle of mass $m$ executing $SHM$ with amplitude $A$ and angular frequency $\omega$. The average value of the kinetic energy and potential energy over a period is
Oscillations
Solution:
Let the displacement of the particle executing $SHM$ at any instant of time $t$ from its equilibrium position is given by
$x=A cos \left(\omega t+\phi\right)$
Velocity, $v =\frac{dx}{dt} = -\omega A \,sin\left(\omega t+\phi\right)$
Kinetic energy of the particle is
$ K=\frac{1}{2}mv^{2}$
$= \frac{1}{2}m\omega^{2} A^{2} sin^{2}\left(\omega t+\phi\right)$
Potential energy of the particle is
$ U= \frac{1}{2}m\omega^{2}x^{2}$
$ = \frac{1}{2}m\omega^{2}A^{2} cos^{2} \left(\omega t+\phi\right)$
Average value of kinetic energy over a period is
$ \left(K\right) = \frac{1}{T } \int\limits_{0}^{T} K dt $
$= \frac{1}{T} \int\limits_{0}^{T} \frac{1}{2}m\omega^{2}A^{2} sin^{2}\left(\omega t+\phi\right)dt $
$= \frac{1}{2T} m\omega^{2} A^{2} \int\limits_{0}^{T}\left[\frac{1- cos 2\left(\omega t +\phi\right)}{2}\right]dt $
Since the average value of both a sine and a cosine function for a complete cycle or over a time period $T$ is $0$.
$\therefore \left(K\right)=\frac{1}{4T}m\omega^{2}A^{2}\left[t\right]_{0}^{T}$
$ = \frac{1}{4T} m\omega^{2}A^{2}T$
$ =\frac{1}{4}m\omega^{2}A^{2}$
Average value of potential energy over a period is
$\left(U\right) = \frac{1}{T} \int\limits_{0}^{T} \frac{1}{2}m \omega^{2}A^{2} cos^{2} \left(\omega t +\phi\right)dt$
$ = \frac{1}{2T}m\omega^{2}A^{2} \int\limits_{0}^{T}\left[\frac{1+cos 2\left(\omega t+\phi\right)}{2}\right]dt $
Since the average value of both a sine and a cosine function for a complete cycle or over a time period $T$ is zero.
$\therefore \left(U\right)=\frac{1}{4T} m\omega^{2}A^{2}\left[t\right]_{0}^{T} $
$= \frac{1}{4T} m\omega^{2} A^{2}T $
$ =\frac{1}{4}m\omega^{2}A^{2}$