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Constructive Interference of Sound

  1. Dec 13, 2009 #1
    Hi,

    Two speakers, A and B, are located at x = +0.5 m and x = -0.5 m. A 680 Hz signal is sent to both speakers. You then walk around the origin, x = 0, in a circle of radius 5.0 m. v_sound = 340 m/s

    If you walk once around the complete circle, how many intensity maxima do you hear?


    This is what I did: m*wavelength = path difference
    Since the maximum path difference is 1.0 m and the wavelength is 0.5 metres, m can only be 0 or 1 or 2. This corresponds to 6 points on the circle (2 points per value of m).
    Is this the correct way of doing this question?

    The correct answer is 8, not 6, and I don't know where the extra 2 points come from!

    Please help me. Thank you very much!
     
  2. jcsd
  3. Dec 13, 2009 #2
    If you are walking around a circle at 5m doesn't the path to each source depend on sine and cosine relationships, or some other variable geometry? In other words, I am asking, is the path difference to be considered as the distance of the listener to each source, and not the distance between the two sources?
     
  4. Dec 13, 2009 #3
    The path difference is supposed to be the difference in distance that sound waves from each source must travel to get to you. So, for example, if you are 3 metres from a certain source and there is another source 1 metre directly behind the first source, the path difference would be 4-3=1 metre. I hope that makes sense.
    I know for sure that the maximum path difference is 1 m because I had to determine that in another part of the question and the answer key agrees with me on that part.
     
  5. Dec 13, 2009 #4
    Perhaps I have the entire concept wrong and there's some other mysterious formula that I am supposed to use...?
     
  6. Dec 13, 2009 #5
    Are the two speakers being driven in-phase or out-of-phase (polarity reversed)?
    Bob S
     
  7. Dec 13, 2009 #6
    I think we're supposed to assume that they're emitting sound waves in phase and that the only phase difference arises from the path difference.
     
  8. Dec 13, 2009 #7

    atyy

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    Maybe there are 4 points corresponding to m=1? By circular symmetry?
     
  9. Dec 13, 2009 #8
    How do I find the four points?
     
  10. Dec 13, 2009 #9

    atyy

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    Draw the speakers and the circle. Draw the points where m=0 (no path difference), and the points where m=2 (maximal path difference). Presumably the m=1 points are between those.

    To start, let's use x and y axes. I would guess that the two m=0 points are at (x=0, y=5) and (x=0, y=-5). Do you agree?
     
  11. Dec 14, 2009 #10
    Yes, and the points where m=2 are at (5,0) and (-5,0)?
     
  12. Dec 14, 2009 #11

    atyy

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    Yes. So if you try to place only two m=1 points between your m=0 and m=2 points, will you be able to respect the symmetry of the situation?
     
  13. Dec 14, 2009 #12

    atyy

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    A bit more explicit, assume that an m=1 point lies on the circle between an m=0 and an m=2 point. Show by symmetry that there are 3 other points with the same path difference.
     
  14. Dec 14, 2009 #13
    I see that by fitting two m=1 points between each m=0 and m=2 points, I would get a total of 8 intensity maxima positions, but I don't understand how we can assume that there are TWO instead of one m=1 points between an m=0 and an m=2 point.
     
  15. Dec 14, 2009 #14
    Maybe I'm missing something, but I don't think that is necessary.

    Start the circle on the positive x axis, speakers are situated at -0.5 and +0.5 also on the x axis so you are in an m=2 mode. As I walk around the circle clockwise there is an m=0 maxima at 90 degrees. Presumably I have walked thruough an m=1 max on the way,. If you continue the circle tour there will be 7 maxima before retrurning to the starting point which will be eight.
     
    Last edited: Dec 14, 2009
  16. Dec 14, 2009 #15

    rl.bhat

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    If O is the center of AB, OY the axis and P is the position of the observer, four position of P for m = 1 makes angles pi/4, 3pi/4, 5pi/4 and 7pi/4 with respect to Y-axis. So you get four points for maximum.
     
  17. Dec 14, 2009 #16

    atyy

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    It's a good question how many m=1 points lie between each pair of m=0 and m=2 points. But start by assuming that there is only one m=1 point between each pair, and ask, "how many quadrants to a circle are there?"
     
  18. Dec 14, 2009 #17
    There are four quadrants in a circle, so I would say there should only be four m=1 points, that is, one m=1 point between each m=0 and m=2 points. I don't understand why there should be TWO m=1 points in each of the four quadrants of the circle.
     
  19. Dec 14, 2009 #18

    atyy

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    2+2+4=?
     
  20. Dec 14, 2009 #19
    Eight! I forgot to add those initial 4 points. Haha thanks so much for helping me out and for the quick replies!
     
  21. Dec 14, 2009 #20

    atyy

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    :smile: Actually, to be more careful, we should show using trigonometry and algebra that there is only one m=1 point between each pair of m=0 and m=2 points. But perhaps that can be left for another time.
     
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