The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is :

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Aicharya · Accepted Answer

Let $y=A \sin \omega t=A \omega^2 \sin (\omega t+\pi)$ $\begin{aligned} & 	herefore \quad \frac{d y}{d t}=A \omega \cos \omega t \ & 	herefore \quad=A \omega \sin \left(\omega t+\frac{\pi}{2}ight) \ & 	ext {Acceleration }=-A \omega^2 \cos \omega t=A \omega^2 \sin (\omega t+\pi) \ & 	herefore 	ext {Phase difference }=\pi-\frac{\pi}{2} \end{aligned}$ ^

The phase difference between the instantaneous velocity and acceleration of a particle executing simple…

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