MCQ
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is
  • A
    $\pi$
  • B
    $2\pi $
  • $0.5\pi $
  • D
    $0$

Answer

Correct option: C.
$0.5\pi $
c
Let $y=A \sin \omega t$

$\frac{d y}{d t}=A \omega \cos \omega t=A \omega \sin \left(\omega t+\frac{\pi}{2}\right)$

Acceleration $=-A \omega^{2} \sin \omega t$

The phase difference between acceleration and velocity is $\pi / 2$

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