# Two Speaker Destructive Interference

• Kharrid
In summary, the problem involves two speakers emitting a frequency of 444 Hz with a one-fourth period delay between them, causing destructive interference at certain angles. The path difference for angles closer to speaker A is dsin(Θ) = (m+1/4)λ while for angles closer to B it is dsin(Θ) = (m + 3/4)λ. This is due to the longer path from speaker B to points closer to A, causing it to be further behind and requiring an additional 1/4 wavelength to be completely out of phase. The opposite is true for points closer to B.

#### Kharrid

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.