Two Speaker Destructive Interference

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The discussion focuses on the differences in path difference formulas for destructive interference in a two-speaker setup. For angles closer to speaker A, the path difference 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 variation arises because speaker A is initially ahead by a quarter wavelength, and the longer path from B to points near A requires an additional quarter wavelength to achieve complete destructive interference. Conversely, the longer path from A to points near B allows speaker B to catch up, necessitating a total path difference of three-quarters of a wavelength for cancellation. Understanding these phase shifts is crucial for analyzing sound interference patterns.
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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