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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
D
$\frac{2\pi}{3}$
Detailed Solution
Let the displacement of each particle be $x = A\sin\theta$, where $\theta$ is its phase.
When they cross, $x = \frac{A}{2}$ for both: $\frac{A}{2} = A\sin\theta$, so $\sin\theta = \frac{1}{2}$
$\theta = 30^\circ$ or $150^\circ$
Velocity is $v = A\omega\cos\theta$: at $\theta = 30^\circ$ the velocity is positive and at $\theta = 150^\circ$ it is negative.
Since the particles move in opposite directions, one has phase $30^\circ$ and the other $150^\circ$.
Phase difference = $150^\circ - 30^\circ = 120^\circ = \frac{2\pi}{3}$
When they cross, $x = \frac{A}{2}$ for both: $\frac{A}{2} = A\sin\theta$, so $\sin\theta = \frac{1}{2}$
$\theta = 30^\circ$ or $150^\circ$
Velocity is $v = A\omega\cos\theta$: at $\theta = 30^\circ$ the velocity is positive and at $\theta = 150^\circ$ it is negative.
Since the particles move in opposite directions, one has phase $30^\circ$ and the other $150^\circ$.
Phase difference = $150^\circ - 30^\circ = 120^\circ = \frac{2\pi}{3}$