Q122 / 180 · COMEDK 2023 (Memory Based) · Physics

The phase difference between displacement and velocity of a particle in simple harmonic motion is

  • A
    $\pi \mathrm{rad}$
  • B
    $3 \pi / 2 \mathrm{rad}$
  • C
    zero
  • Correct answer
    D
    $\pi / 2 \mathrm{rad}$

Solution

The displacement of a particle executing SHM is $x=A \sin \omega t$ $\therefore$ Velocity, $v=\frac{d x}{d t}=A \omega \cos \omega t$ $\therefore v=A \omega \sin \left(\omega t+\frac{\pi}{2}\right)$ $\therefore$ Phase difference, $\Delta \phi=\left(\omega t+\frac{\pi}{2}\right)-\omega t=\frac{\pi}{2} \mathrm{rad}$.

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

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