Interference of rock music at a concert

AI Thread Summary
At an outdoor rock concert, two loudspeakers are positioned 2.4 meters apart, with a listener standing 17.7 meters from one and 20.7 meters from the other. The technician tests frequencies from 20Hz to 30,000Hz, prompting a discussion on sound interference. The listener notes that due to the differing distances, sound waves will arrive out of phase, leading to constructive or destructive interference. To find the lowest frequency for minimum signal, it is essential to understand that the path length difference must be an odd multiple of half-wavelengths for cancellation to occur. Thus, calculating how many wavelengths fit into the distance difference at 20Hz is a critical step in solving the problem.
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Homework Statement


Two loudspeakers at an outdoor rock concert are located 2.4 meters apart. You are standing 17.7 meters from one of the speakers and 20.7 from the other. During a sound check, the technician sends the exact same frequency to both speakers while you listen. The technician starts at 20Hz and slowly increases it to 30,000Hz.

What is the lowest frequency where you will hear a minimum signal ?

2. Equations

y(t) = Acos(wt) + Acos(wt)

3. Attempt at the solution

I am not sure how to go about this one. But here is my thoughts on it. Since you are at a different differences from each of the speakers, the waves will arrive out of phase and will interfere constructively or destructively. I am thinking that this is related to beats. Any help?
 
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It isn't related to the beats, but the path length has to be an odd multiple of a half-wavelength in order for there to be a minimum. That's because such a path length difference ensures crests overlap with troughs and cancel.
 


So would I just find how much of the 20Hz signal would fit on each distance?
 
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