Phase Difference of Two Particles in SHM

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SUMMARY

Two particles executing simple harmonic motion (SHM) with the same amplitude and frequency along parallel lines exhibit a phase difference of π/3 radians. When their displacement reaches half the amplitude (a/2), the sine function indicates that the corresponding angles are π/6 and 5π/6. Therefore, the phase difference is calculated as the difference between these angles, confirming the established relationship in SHM.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM)
  • Knowledge of trigonometric functions and their properties
  • Familiarity with phase difference concepts in wave mechanics
  • Basic mathematical skills for solving equations
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  • Explore the mathematical derivation of phase difference in SHM
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Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?
 
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You haven't shown any attempt. But I'm still giving you a lot of hint. The rest is upto you.

For a particle executing shm, if x=a/2, then a/2=a*sin(wt) => sin(wt) = ½ = sin pi/6 or sin(pi-pi/6). Can you find the phase diff now?
 

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