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The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is :
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Verified Answer
The correct answer is:
$0.5 \pi$
Let $y=A \sin \omega t=A \omega^2 \sin (\omega t+\pi)$
$\begin{aligned}
& \therefore \quad \frac{d y}{d t}=A \omega \cos \omega t \\
& \therefore \quad=A \omega \sin \left(\omega t+\frac{\pi}{2}\right) \\
& \text { Acceleration }=-A \omega^2 \cos \omega t=A \omega^2 \\
& \sin (\omega t+\pi) \\
& \therefore \text { Phase difference }=\pi-\frac{\pi}{2}
\end{aligned}$
$\begin{aligned}
& \therefore \quad \frac{d y}{d t}=A \omega \cos \omega t \\
& \therefore \quad=A \omega \sin \left(\omega t+\frac{\pi}{2}\right) \\
& \text { Acceleration }=-A \omega^2 \cos \omega t=A \omega^2 \\
& \sin (\omega t+\pi) \\
& \therefore \text { Phase difference }=\pi-\frac{\pi}{2}
\end{aligned}$
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