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AP EAMCET · PHYSICS · Oscillations

Two particles execute simple harmonic motion along the same straight line with same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distance of \(\frac{1}{\sqrt{2}}\) times the amplitude from their common mean position. The phase difference between the two particles is

  1. A \(30^{\circ}\)
  2. B \(45^{\circ}\)
  3. C \(60^{\circ}\)
  4. D \(90^{\circ}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(90^{\circ}\)

Step-by-step Solution

Detailed explanation

Position of particle \(1, x_1=A \sin \phi_1\) Position of particle 2, \(x_2=A \sin \phi_2\) let particle 1 is moving away from mean position Given, \(x_1=x_2=\frac{A}{\sqrt{2}}\) \(\therefore \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \sin \phi_1 \Rightarrow \phi_1=\frac{\pi}{4}\)…