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Show that for a particle executing SHM, velocity and displacement have a phase difference of $\frac{\pi}{2}$ .

Answers (1)

Let us consider $x = A \sin \omega t$ to be a SHM …….. (i)

Now, $v = \frac{dx}{dt}$

$= A\omega \cos \omega t$

$= A\omega \sin (90^{\circ} + \omega t)$ …….. [since $\sin (90^{\circ} + \theta ) = \cos \theta$ ]

Thus, $v = A\omega \sin \left (\omega t + \left ( \frac{\pi}{2} \right ) \right )$

From (i), we know that,

The phase of displacement = $\omega t$

& from (ii), we know that,

The phase of velocity = $\left (\omega t + \left ( \frac{\pi}{2} \right ) \right )$

Thus, the phase difference = $\omega t + \frac{\pi}{2} - \omega t$

$= \frac{\pi}{2}.$

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infoexpert24

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