Interference of rock music at a concert

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SUMMARY

The discussion focuses on sound interference at an outdoor rock concert involving two loudspeakers positioned 2.4 meters apart. The listener is located 17.7 meters from one speaker and 20.7 meters from the other. The technician tests frequencies from 20Hz to 30,000Hz, and the key takeaway is that a minimum signal occurs when the path length difference between the speakers is an odd multiple of half-wavelengths. This principle of destructive interference is critical for understanding sound wave behavior in this context.

PREREQUISITES
  • Understanding of sound wave properties and interference
  • Knowledge of wave equations, specifically y(t) = Acos(wt) + Acos(wt)
  • Familiarity with frequency and wavelength relationships
  • Basic concepts of phase difference in wave mechanics
NEXT STEPS
  • Study the principles of wave interference and how it applies to sound
  • Learn about calculating wavelength from frequency using the formula λ = v/f
  • Explore the concept of beats and its relation to sound waves
  • Investigate practical applications of sound interference in concert acoustics
USEFUL FOR

Acoustics engineers, sound technicians, physics students, and anyone interested in the principles of sound wave interference at live events.

Bryon
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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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