Q116 / 200 · NEET 2007 · Physics

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

  • A
    $\pi$
  • B
    $0.70 \pi$
  • C
    zero
  • Correct answer
    D
    $0.5 \pi$

Solution

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

Hence, the correct answer is option: 'D'.

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