Two Superposed Vibrations of Equal Frequency

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SUMMARY

The discussion focuses on calculating the resultant amplitude and phase of a particle subjected to three simple harmonic motions with equal frequency. The amplitudes are 0.25 mm, 0.20 mm, and 0.15 mm, with phase differences of 45 degrees between the first and second components, and 30 degrees between the second and third. The approach involves vector addition of these harmonic motions, starting with the assumption that the first component (0.25 mm) fully displaces the particle. The challenge lies in applying vector addition techniques to three vectors rather than the standard two-vector systems.

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  • Understanding of simple harmonic motion (SHM)
  • Knowledge of vector addition in physics
  • Familiarity with phase differences in wave mechanics
  • Basic trigonometry for calculating resultant vectors
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  • Study vector addition of multiple harmonic motions
  • Learn about phase difference effects in wave interference
  • Explore the concept of resultant amplitude in SHM
  • Review graphical methods for representing harmonic oscillations
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The Problem
A particle is simultaneously subjected to three simple harmonic motions, all of the same frequency and in the x direction. If the amplitudes are 0.25, .20, and 0.15 mm, respectively, and the phase difference between the first and second is 45, and between the second and third is 30, find the amplitude of the resultant displacement and its phase relative to the first (0.25 mm amplitude) component.

I drew a diagram and of the vectors but the equations I have are all for two vectors systems. How do I start this problem off?
 
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Start by assuming the 0.25mm wave is fully dispacing the particle, so its moving it 0.25mm from equilibrium position. the difference between the first and the second is 45 degrees so it will effect it as per the phase its in at that time.
 

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