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Sound emitted from the origin

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data
    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 coordinates (38,0) and (0,31).

    What are the lowest two possible frequencies of the sound?

    Honestly, I am most clueless on how to do this.
    2. Relevant equations
    I want to say that if the crest passes over each individual 38 and 31 m at the same, that has something to do with the wavelength[tex]\lambda[/tex].


    3. The attempt at a solution
    I know that v=[tex]\lambda[/tex]f, But I'm unsure how you would use 38 and 31 m to figure out frequencies that would apply for each of them. Any starting point would be appreciated.
     
  2. jcsd
  3. Jan 19, 2010 #2

    kuruman

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    What is the definition of wavelength? It is the distance from ... to ... ?
     
  4. Jan 19, 2010 #3
    The distance between one peak to another. I still don't see how you would use the two different lengths. Are each of the lengths representative of the two different frequencies?
     
  5. Jan 19, 2010 #4

    kuruman

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    The origin emits spherical waves that expand in all directions. An emitted wavefront, say a crest, moves away from the origin at the speed of sound. Do the peaks mentioned in the problem belong to the same wavefront or to different wavefronts? What do you think?
     
  6. Jan 19, 2010 #5
    Well, if they arrive at different points at the same time, they would have to be different wave fronts, because they're different distances. Also, the problem says the crest of the wave passes both listeners simultaneously.
     
  7. Jan 19, 2010 #6

    kuruman

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    So if they belong to different wavefronts, what could the possible wavelengths be?
     
  8. Jan 19, 2010 #7
    That's where I'm scratching my head. It can't be 38 and 31 m, because that's ridiculously low. It would have to be something that divides evenly into both. Or the fact that 7 m is separated between the two is also significant.

    Ah ha! So, since the difference between them is 7 m, that would mean that that would be a possible wavelength along with 14 m. Then you divide 340 by 7 and 14. That gives you 49 and 97. Again, thanks, man. Incredibly simple, but it took talking it out to understand it.
     
  9. Jan 19, 2010 #8
    Think it over one more time. You've found that the distance between two crests (You don't know if they're consecutive or not) is 7 meters.

    And you've said yourself that the distance between two consecutive crests is defined as the wavelength of the wave.

    Using this information, what formulation can you find for the frequency? Note that there's an infinite number of possible frequencies!

    Note that you don't know the phase at the source at that moment in time. You don't know how many wavelengths the source has 'sent out', it could be 3, 4, or 4.669.
     
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