Two Speaker Destructive Interference

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SUMMARY

The discussion centers on the differences in path difference formulas for destructive interference from two speakers, A and B, which are 3.50 m apart and emitting sound at 444 Hz. The path difference for angles closer to speaker A is expressed as dsin(Θ) = (m + 1/4)λ, while for angles closer to speaker B, it is dsin(Θ) = (m + 3/4)λ. This discrepancy arises due to the phase shift of one-fourth of a period between the speakers, necessitating an additional quarter wavelength for complete cancellation at different angles. The speed of sound is given as 340 m/s, and the fundamental equation used is v = fλ.

PREREQUISITES
  • Understanding of wave interference principles
  • Familiarity with the concept of phase shifts in sound waves
  • Knowledge of basic trigonometry related to angles and sine functions
  • Ability to apply the wave equation v = fλ
NEXT STEPS
  • Explore the concept of phase shifts in wave mechanics
  • Study the derivation of path difference formulas for multiple sources
  • Investigate the effects of distance on sound wave interference patterns
  • Learn about the applications of destructive interference in acoustics
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Students studying physics, particularly those focusing on wave mechanics and sound interference, as well as educators seeking to explain complex interference phenomena in classroom settings.

Kharrid
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Homework Statement
Two speakers A and B are 3.50 m apart, and each one is emitting a frequency of 444 Hz. However, because of signal delays in the cables, speaker A is one-fourth of a period ahead of speaker B. For points far from the speakers, find all the angles relative to the centerline at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 m/s.
Relevant Equations
v = f(lambda)
dsin(theta) = (m+ 1/2)(lambda)
I already have the solution in front of me, I am wondering why there is a difference in the formula for path difference. I've attached the problem as well to show the Figure.

What I am struggling to grasp is why the path difference for the angles closer to A is dsin(Θ) = (m+1/4)λ while the path difference for the angles closer to B are dsin(Θ) = (m + 3/4)λ. I know that the phase shift for destructive interference has to be 1/2, but why does it change differently closer to the source with the phase shift and differently farther away from it?

Any help is appreciated!
 

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Kharrid said:
Homework Statement:: Two speakers A and B are 3.50 m apart, and each one is emitting a frequency of 444 Hz. However, because of signal delays in the cables, speaker A is one-fourth of a period ahead of speaker B. For points far from the speakers, find all the angles relative to the centerline at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 m/s.
Homework Equations:: v = f(lambda)
dsin(theta) = (m+ 1/2)(lambda)

I already have the solution in front of me, I am wondering why there is a difference in the formula for path difference. I've attached the problem as well to show the Figure.

What I am struggling to grasp is why the path difference for the angles closer to A is dsin(Θ) = (m+1/4)λ while the path difference for the angles closer to B are dsin(Θ) = (m + 3/4)λ. I know that the phase shift for destructive interference has to be 1/2, but why does it change differently closer to the source with the phase shift and differently farther away from it?

Any help is appreciated!
Closer to A, the path from B is longer, so it puts B further behind. It was already 1/4 behind, so be completely out of phase it needs to be another 1/4 behind, plus whole wavelengths.
Closer to B, the path from A is longer, helping B catch up. To be completely out of phase, the path now needs to be 3/4 longer, etc.
 

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