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Wavelength of Sound

  1. Dec 22, 2017 #1
    • Member advised to use the formatting template for all homework help requests
    1. The problem statement, all variables and given/known data
    Hello, If one can see the image I posted, the question that follows is this: Use the information provided to obtain a value for the wavelength of sound emitted. The signal is at maximum intensity at X, and minimum at Y.

    http://uploads.im/7wJOq.jpg
    upload_2017-12-22_10-19-42.png

    2. Relevant equations
    Now I know that the wavelength equation is v = f • λ, but in this case I don't have the frequency nor speed, so I have to use the lengths given to find the wavelength.

    I know in Young's double slit experiment, the wavelength is: {Fringe Separation x Slit separation / Source to screen distance}.
    3. The attempt at a solution
    I believe after observing this diagram, the solution will be similar, as there is source to screen distance involved, but there are no fringes or slits as this is sound we are dealing with. I now know that the path difference is the difference in distance from the sources to the point, in this case Y, which will be 0.12m.
    But I am still stuck, I don't know how to calculate the wavelength even using the phase difference formula because for that the wavelength is needed in the first place.
     
    Last edited: Dec 22, 2017
  2. jcsd
  3. Dec 22, 2017 #2

    scottdave

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    There should be more information given. What is happening at point Y? Is this where the signal intensity is cancelled out? or max intensity?
    What would you know about possible values of wavelength if the signals cancelled out each other at this location? Think about the differences in distance.
     
  4. Dec 22, 2017 #3

    scottdave

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  5. Dec 22, 2017 #4

    scottdave

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    If they meet at that point 180° (½ wavelength) out of phase, then the sound intensity will be a minimum. Other possibilities are 3/2 wavelength, 5/2, etc. Use what you know about the speed of sound to see if these make sense as viable frequencies.

    Do you see why 3/2 and 5/2 work as well (mathematically, maybe not practically though)?
     
  6. Dec 22, 2017 #5
    Is it that they work well because, as you add 180 degrees each time, the improper fractions such as 3/2 and 5/2 will show places of destructive interference? where as 2/2 or 4/2 would be constructive?
     
  7. Dec 23, 2017 #6
    Edit: I figured it out, the interference at Y is a half shift of the wavelength, which is given by the formula, R1-R equals n/2 and so on, and the wavelength of the wave is 0.24m, and from there i worked out the frequency, but the question now asks to find what distance the speakers are apart, can anyone help?
     
  8. Dec 27, 2017 #7

    PeterO

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    The diagram looks an awful lot like a right angled triangle - have you tried Pythagoras?
     
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