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Sound Reflecting of an Obstacle

  1. Apr 25, 2007 #1
    [SOLVED] Sound Reflecting of an Obstacle

    1. The problem statement, all variables and given/known data
    A source of sound waves (wavelength [itex]\lambda[/itex]) is a distance L from a detector. Sound reaches the detector directly, and also by reflecting off an obstacle. The obstacle is equidistant from the source and detector. When the obstacle is a distance d to the right of the line of sight between source and detector, the two waves arrive in phase. How much farther to the right must the obstacle be moved if the two waves are to be out of phase by 1/2 wavelength, so destructive interference occurs? (Assume [itex]\lambda \ll L, d[/itex].)

    2. Relevant equations
    Let c be the distance from the source to the obstacle. The total distance covered by the sound wave bounced off the obstacle, from source to detector, is 2c. Since the waves are in phase, then [itex]L = 2c - n\lambda[/itex] where n is a positive integer.

    3. The attempt at a solution
    I figure, if the two waves are supposed to be out of phase by [itex]1/2 \lambda[/itex], then [itex]L = 2c - n\lambda + 1/2\lambda[/itex] right? So, I have to move the obstacle to the right such that the distance covered by the reflected sound wave, from source to detector, measures [itex]2c + \lambda/2[/itex]. Equivalently, the distance from the source to the obstacle has to measure [itex]c + \lambda/4[/itex].

    If I'm right, the rest is just algebra which I can do. What do you think?
    Last edited: Apr 25, 2007
  2. jcsd
  3. Apr 25, 2007 #2


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    I like your equation here: [itex]L = 2c - n\lambda + 1/2\lambda[/itex]
    Rewritten, it could be:

    [itex]2c - L= (n+\frac{1}{2}) \lambda[/itex]

    read as, the difference in distance between the direct route and reflected route needs to be equal to half the wavelength. I say go for it.
  4. Apr 26, 2007 #3
    Thanks. Calculating c, I get:

    [tex]c = \sqrt{(L/2)^2 + d^2}[/tex]

    Let h be the distance sought by the problem. Then,

    [tex](c + \lambda/4)^2 = (L/2)^2 + (d + h)^2[/tex]

    Solving for h, I get:

    [tex]h = \sqrt{(c + \lambda/4)^2 - (L/2)^2} - d[/tex]

    That looks about right.
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