Question

Two particles execute S.H.M. of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions, each time their displacement is half the amplitude. The phase difference between them is \((\sin 30^\circ = 0 \cdot 5)\)

• A. \(\frac{\pi}{6}\)
• B. \(\frac{5\pi}{6}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{2\pi}{3}\)

Hint:

Key idea: Same displacement + opposite motion $\rightarrow$ symmetric phase points

Answer 1

The Correct Option is D

Concept: For SHM: \[ x = A \sin \theta \]

Step 1: Given displacement. \[ x = \frac{A}{2} \Rightarrow \sin \theta = \frac{1}{2} \Rightarrow \theta = 30^\circ \text{ or } 150^\circ \]

Step 2: Opposite velocities.
Particles moving in opposite directions $\Rightarrow$ phase difference between positions: \[ \Delta \theta = 150^\circ - 30^\circ = 120^\circ \]

Step 3: Convert to radians. \[ 120^\circ = \frac{2\pi}{3} \]

Step 4: Conclusion.
Phase difference = $\frac{2\pi}{3}$ Final Answer: Option (D)