Sound waves propagation problem

  • Thread starter qban88
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  • #1
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Homework Statement


A loudspeaker at the origin emits sound waves on a day when the speed of sound is 340 m/s. A crest of the wave simultaneously passes listeners at the{xy} coordinates (43 ,0) and (0,33).What are the lowest two possible frequencies of the sound?

Homework Equations


v(sound)= frequency * wavelength

The Attempt at a Solution


My assumption is the wavelength must be a number that when multiplied by two different integers the result is 33 and 43. I need to get something like the greatest common factor between 33 and 43 but it is a non-integer and I don't know what to do. Moreover, the distance between the 2 crests (10 m) has to be also the product of an integer and the wavelength.
I am trying to find the system of equations to solve for the maximum wavelength that would give me the lowest frequency and then calculate the next frequency. Any help would be greatly appreciated.
Yes, this is a homework problem and it is past due already but I spent almost 3 hours trying to figure it out and by know I just want to find out how is done. Thanks
 

Answers and Replies

  • #2
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Group 1 is at (33,0) and group 2 is at (0,43).

[tex] x^2 + y^2 = d^2 [/tex]

[tex] d = 54.2 m [/tex]

[tex] f = \frac{v}{\lambda} [/tex]

Since both groups experience a peak you know that they are an even integer of wavelength apart where [tex]\lambda n = d[/tex]
 
Last edited:
  • #3
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Thanks for the response. I see what you say but I can't find any way to find the second equation in order to find n . The way I was seeing it is that the crest of the wave that hits the points does not belong to the same wave but to consecutive waves in the same train wave. Otherwise, the speed of propagation in y would be less than the speed in x which is not the case. The longest wavelength using your suggestion would be 54.2/2 = 27.1 which is not compatible with the data that there is a crest at 33 and another at 43. I am putting the two points in the same line since I assumed the speed of propagation is the same in every direction so what matters is the distance from the origin if I am not mistaken. I tried to make the wavelength equal to 1 m (the greatest common divisor) which would give me the lowest frequency and the wavelength equal to 0.5 which would give me the second lowest frequency but that wasn't the answer.

43/lambda = integer 1 33/lambda = integer 2 It seems to me that actually the longest wavelength is a non integer greater than 1.
Again, thanks for the input.
 
Last edited:

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