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Q.
A particle executing $SHM$ with time period $T$ and amplitude $A$. The mean velocity of the particle averaged over quarter oscillation is
Oscillations
Solution:
Let the displacement of the particle executing $SHM$ at any instant $t$ is , $x = A sin \omega t$
Velocity, $v = \frac{dx}{dt} = \frac{d}{dt} A sin \omega t = A w cos \omega t$
The mean velocity of the particle averaged over quarter oscillation is
$(v)_{0\to{T/4}}=\frac{ \int\limits_{0}^{T/4} v dt }{\int\limits_{0}^{T/4} dt} =\frac{ \int\limits_{0}^{T/4}A\omega cos\omega t dt}{T/4}$
$ \frac{\frac{A\omega}{\omega}\left[sin \omega t\right]_{0}^{T/4}}{T/4} $
$= \frac{A\left[sin \omega t\right]_{0}^{T/4}}{T/4} = \frac{4A}{T}$