What Distance from Speaker A Will a Microphone Detect Minimum Sound Intensity?

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Homework Help Overview

The problem involves two loudspeakers emitting sound waves in phase, with a focus on determining the distance from one speaker at which a microphone detects minimum sound intensity due to destructive interference. The context includes concepts from wave physics and sound propagation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of Pythagoras' theorem and the calculation of wavelengths. Questions are raised about the phase difference at various distances and how to determine the conditions for destructive interference.

Discussion Status

Participants are exploring the relationship between path lengths from the speakers to the microphone and the resulting phase differences. Some guidance has been offered regarding the calculation of wavelengths and the conditions for destructive interference, but there is no explicit consensus on the next steps.

Contextual Notes

There is mention of using the speed of sound and frequency to calculate wavelength, along with the assumption that the speakers are in phase. The discussion reflects uncertainty about the initial approach and the calculations involved.

patrickmoloney
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Homework Statement



Two loudspeakers A and B of equal power are separated by a distance of 1.4 m. Both the speakers emit sound waves in phase and of frequency 450 Hz. If a microphone were to be moved from A in a direction perpendicular to AB, at what distances from A will it detect a minimum of sound intensity? ( Take the velocity of sound in air to be 330 m/s )


Homework Equations



c=fλ,

The Attempt at a Solution



I have no idea where to even start. I tried using Pythagoras' theorem and could go no further.
 
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Pythagoras' theorem is certainly relevant.
At a point distance x from A along that line, how far is it from B?
How many wavelengths is it from each? What will the phase difference be at that point?
 
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Distance is sqrt(1.96)+x². How do I calculate the wavelength at each point. lambda=c/f = 330/ 450 = 0.7333 which is all I know. I also know that destructive interference occurs at lamda/2
 
patrickmoloney said:
Distance is sqrt(1.96)+x².
That's not what you meant to write, I hope.
How do I calculate the wavelength at each point.
I did not suggest calculating a wavelength at each point. There is only one wavelength, which you have calculated. I said to calculate the number of wavelengths represented by each of the two distances. But the alternative below may be simpler.
lambda=c/f = 330/ 450 = 0.7333 which is all I know. I also know that destructive interference occurs at lamda/2
For sources in phase, same frequency, completely destructive interference occurs when the difference in the path lengths is lambda/2. What, as a function of x, is the difference in the path lengths? How many wavelengths is that?
 

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