Question:

Two particles execute SHM with same amplitude and frequency. When displacement is half amplitude, find phase difference.

Show Hint

Same displacement + opposite motion $\Rightarrow$ take supplementary angles.
Updated On: Apr 23, 2026
  • \(30^\circ\)
  • \(60^\circ\)
  • \(90^\circ\)
  • \(120^\circ\)
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The Correct Option is D

Solution and Explanation

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

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

Step 2:
Opposite directions
Particles have same displacement but opposite velocities. \[ \text{Phase difference} = 150^\circ - 30^\circ = 120^\circ \] Conclusion: \[ 120^\circ \]
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