The phase of a particle executing simple harm | vedclass

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Question

(D) For a particle executing $S.H.M.$,the displacement is given by $y = a \sin(\omega t + \phi)$.Assuming the initial phase $\phi = 0$,the displacemen

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

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(D) For a particle executing $S.H.M.$,the displacement is given by $y = a \sin(\omega t + \phi)$.Assuming the initial phase $\phi = 0$,the displacement is $y = a \sin(\omega t)$.At phase $\omega t = \frac{\pi}{2}$,the displacement $y = a \sin(\frac{\pi}{2}) = a$,which is the maximum displacement.The acceleration of the particle is given by $A = -\omega^2 y = -\omega^2 a \sin(\omega t)$.At phase $\omega t = \frac{\pi}{2}$,the acceleration $A = -\omega^2 a \sin(\frac{\pi}{2}) = -\omega^2 a$.The magnitude of acceleration is $|A| = \omega^2 a$,which is the maximum acceleration.Therefore,at a phase of $\frac{\pi}{2}$,the particle has both maximum displacement and maximum acceleration.

The phase of a particle executing simple harmonic motion is $\frac{\pi}{2}$ when it has:

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