MCQ
Two simple harmonic motions are represented by the equations $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t .$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is
  • A
    $\frac{-\pi}{3}$
  • B
    $\frac{\pi}{6}$
  • $\frac{-\pi}{6}$
  • D
    $\frac{\pi}{3}$

Answer

Correct option: C.
$\frac{-\pi}{6}$
$v_1=\frac{d y_1}{d t}=0.1 \times 100 \pi \cos \left(100 \pi t+\frac{\pi}{3}\right)$
$ v_2=\frac{d y_2}{d t}=-0.1 \pi \sin \pi t=0.1 \pi \cos \left(\pi t+\frac{\pi}{2}\right)$
Phase difference of velocity of first particle with respect to the velocity of $2$ particle at $t=0$ is
$\Delta \phi=\phi_1-\phi_2=\frac{\pi}{3}-\frac{\pi}{2}=-\frac{\pi}{6} .$

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