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Q.
A particle executing $SHM$. The phase difference between acceleration and displacement is
Oscillations
Solution:
Let the displacement of a particle executing simple harmonic motion at any instant $t$ is, $x = A cos \omega t$
Velocity, $v=\frac{dx}{dt} = \frac{d}{dt}\left(A cos \omega t\right) = -A\omega sin \omega t$
Acceleration, $a = \frac{dv}{dt} = -A \omega^{2} cos \omega t $
$= A\omega^{2} cos \left(\omega t +\pi\right) $
Phase of displacement, $\phi_{1} = \omega t $
Phase of acceleration, $\phi_{2} = \omega t +\pi $
$ \therefore $ Phase difference $= \phi_{2} -\phi_{1} $
$= \left(\omega t + \pi\right)-\omega t = \pi$