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Sound problem: Destructive Interference

  1. Dec 4, 2006 #1
    The drawing shows a loudspeaker A and point C, where a listeer is positioned. A second loudspeaker B is located somewhere to the right of A. Both speakers vibrate in phase and are playing a 68.6-Hz tone. The speed of sound is 343 m/s. What is the closest to speaker A that speaker B can be located, so that the listener hears no sound?

    Picture: [​IMG]


    First, I connected C and B to make a triangle. I then bisect the triangle with a line straight down from C to make 2 triangles (with the extra point being D). I call AB line = x and CD line = y. Next, I set up the destructive interference where the point in which line AB has to be 1 wavelength and line CB is half wavelength. So, the difference between AC and CB has to be

    CB - CA = (n*wavelength)/2

    for smallest distance, n would have to be 1.

    with this, I list all the known and unknowns which are

    AD = AC*cos(60)
    DB = AB-AD -> x-cos(60)
    CD = AC*sin(60)

    CB = sqrt(sin^2(60)+ (x-cos^2(60)))

    So then:

    wavelength/2 = sqrt(sin^2(60)+cos^2(60) + x^2 - 2*cos(60x))

    I solve to get:

    x^2 - (2*cos(60)x) + (1 + wavelength^2/4)


    use quadratic equation to get

    x = 2.845 and x = -1.845

    The correct answer is 3.89 m.

    So, I don't know if I'm missing something here. I hope you guys can help. Thanks.
     
  2. jcsd
  3. Dec 5, 2006 #2

    OlderDan

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    Science Advisor
    Homework Helper

    I can't say I followed all of your trig, but it looks to me like you said

    CB = sqrt(sin^2(60)+ (x-cos^2(60))) = wavelength/2 = sqrt(sin^2(60)+cos^2(60) + x^2 - 2*cos(60x))

    and that is not true

    CB = CA + wavelength/2

    You might consider using the Law of Sines to do this problem. CB is easy to find and you know angle CAB = 60°. Finding angle CBA and then angle ACB is easier with LoS than what you did.
     
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