Two particles are oscillating along two close parallel straight l… - Vidyadip

MCQ

Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

A
$\pi$
B
$\frac{\pi}{6}$
C
$0$
✓
$\frac{2 \pi}{3}$

✓

Answer

Correct option: D.

$\frac{2 \pi}{3}$

d
(d) $y = a\sin (\omega t + {\phi _0})$. According to the question

$y = \frac{a}{2}$

$ \Rightarrow \frac{a}{2} = a\sin (\omega \,t + {\phi _0})$

$ \Rightarrow (\omega \,t + {\phi _0}) = \phi = \frac{\pi }{6}$ or $\frac{{5\pi }}{6}$
Physical meaning of $\phi = \frac{\pi }{6}$ : Particle is at point $P$ and it is going towards B
Physical meaning of $\phi = \frac{{5\pi }}{6}$ : Particle is at point $P$ and it is going towards $O$
So phase difference $\Delta \phi = \frac{{5\pi }}{6} - \frac{\pi }{6} = \frac{{2\pi }}{3} = 120^\circ $

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