MCQ
Two particles executing $S.H.M.$ of same frequency, meet at $x=+A / 2$, while moving in opposite directions. Phase difference between the particles is .........
- A$\frac{\pi}{6}$
- B$\frac{\pi}{3}$
- C$\frac{5 \pi}{6}$
- ✓$\frac{2 \pi}{3}$
✓
Answer
Correct option: D.
$\frac{2 \pi}{3}$
d
(d)
(d)
$x=A \sin \omega t$
When displacement
$x=\frac{A}{2}$
$\frac{A}{2}=A \sin (\omega t+\phi)$
$\sin ^{-1} \frac{1}{2}=\omega t+\phi$
$\omega t+\phi=30^{\circ} \text { or } 150^{\circ}$
When particles are in opposite direction at one lime phase is $30^{\circ}$ and at the other $150^{\circ}$. So phase difference is $120^{\circ}$.
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