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Two speakers driven by the same force

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Two loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 800 Hz.

    (a) What is the phase difference between the two waves when they reach the observer? (Your answer should be between 0 and 2π.)

    (b) What is the frequency closest to 800 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

    2. Relevant equations
    delta r = (phi/2pi)lamda
    frequency = velocity/lamd
    3. The attempt at a solution
    I found the dimensions of the triangle that is formed when the person stands in front of one speaker 4.00m away which is:
    a=4,b=1,c=sqrt(17)
     
  2. jcsd
  3. May 15, 2009 #2

    Redbelly98

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    You'll need to get the difference in the distances to each speaker.
     
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