Quick Sound Question Homework: Lowest & Next Two Freqs at 3m & 3.8m

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SUMMARY

The discussion focuses on calculating the lowest frequency and the subsequent frequencies that cause destructive interference between two loudspeakers positioned 2.5 m apart, with a listener standing 3.0 m from one speaker and 3.8 m from the other. The lowest frequency determined is 214.375 Hz, with the next two frequencies calculated as 643.125 Hz and 1071.875 Hz, which correspond to three and five times the fundamental frequency, respectively. The reasoning behind these frequencies relates to the path difference and wavelength adjustments necessary for achieving destructive interference.

PREREQUISITES
  • Understanding of sound wave properties, including frequency and wavelength
  • Knowledge of destructive interference principles in wave physics
  • Familiarity with the equation v = λf, where v is velocity, λ is wavelength, and f is frequency
  • Basic concepts of standing waves and their relationship to frequency in acoustics
NEXT STEPS
  • Study the principles of wave interference and its applications in acoustics
  • Learn about the relationship between frequency and wavelength in sound waves
  • Explore the concept of standing waves in different mediums, such as half-open pipes
  • Investigate the mathematical derivation of frequencies for destructive interference in various configurations
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics, acoustics, and sound engineering. This discussion is also beneficial for educators teaching concepts of interference and frequency calculations.

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Homework Statement



Two loudspeakers are 2.5 m apart. A person stands 3.0 m from one speaker and 3.8 m from the other. Assume the air temperature is 20°C.

(a) What is the lowest frequency at which destructive interference will occur at this point?

(b) Calculate the next two frequencies that also result in destructive interference at this point.

second lowest frequency?

third lowest frequency?

Homework Equations



v=lambda*f

v = velocity (m/s)
lambda = wavelength (m)
f = frequency (Hz)

The Attempt at a Solution



Alright, so part A understand. The difference between the distances would be half the wavelength, double it, plug it into the equation, find out that

Part A: f = 214.375 Hz.

Answers to part B: 643.125 Hz (triple original answer) and 1071.875 (5x the original answer)

Now what I don't get is why the next two frequencies are triple/5x my original answer instead of double/triple. I thought that an overtone would be where the next frequency would be, which I thought was double/triple/etc. the fundamental frequency.

Any help is greatly appreciated!
 
Last edited:
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Shakas said:
Now what I don't get is why the next two frequencies are triple/5x my original answer instead of double/triple. I thought that an overtone would be where the next frequency would be, which I thought was double/triple/etc. the fundamental frequency.

Any help is greatly appreciated!

For the lowest frequency, the listener is standing at a point where there is one half-wavelength path difference between the two sources. If you double the frequency, what happens to the wavelength of the emitted sound? What would the path difference be now in terms of the new wavelength? Would that give you destructive or constructive interference?

[EDIT: It occurs to me that this is similar to the situation for the question of the wavelengths that lead to standing waves in a half-open pipe of a specified length...]
 
Last edited:
Oh, I think I got it. If I were to double the frequency, then the wavelength would be halved. But since the guy is already standing at half the wave length, he would hear constructive interference. So to get destructive interference he has to be at 3/2 the wave length. Then you have to multiply the wavelength by 2/3. Same deal with the next frequency.

Thanks for the help!
 

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