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Wave interference, finding amplitude in certain situations

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity is minimum when speaker 2 is 10 cm behind speaker 1. The intensity increases as speaker 2a is moved forward and first reaches maximum, with amplitude 2 a, when it is 30 cm in front of speaker 1.

    What is the amplitude of the sound if the speakers are placed side by side?

    2. Relevant equations

    ΔΦ = 2π[(Δx)/λ] = 2mπ , m = 0,1,2,3......

    3. The attempt at a solution

    I found the following:

    λ = 80cm
    Phase difference = ¾π (0.75π) =~ 2.36radians

    I tried drawing it but I am not getting the correct answer, and I don't know what to do...
     
  2. jcsd
  3. Sep 8, 2007 #2

    learningphysics

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    You have the amplitude... so try to write the wave equation for the two waves... Use the equation:

    y = A cos (kx) for speaker 1

    y = A cos (kx + phi) for speaker 2

    you know k = 2pi/wavelength. you know A = 80. Try to use the fact that when speaker two is moved 30m in front of speaker 1, then there is constructive interference... hint a motion of 30m, is a shift in the positive x direction.

    Once you know the equations for both waves, add them and find the amplitude for the sum.
     
  4. Nov 24, 2007 #3
    Excuse me but why would A= 80?
     
  5. Nov 24, 2007 #4
    wouldnt x =80
     
  6. Nov 25, 2007 #5

    learningphysics

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    I'm sorry. A is not 80... amplitude is unknown variable a. I should have written wavelength is 80cm.

    I'm getting a phase difference of -0.7854 radians

    Using these and adding

    y = a cos (kx) to

    y = a cos (kx + phi)

    where phi = -0.7854 and k = 2pi/0.80 = 7.854

    you can get the amplitude of the sum in terms of a... you can either use trig identites to get the amplitude of the sum... or use phasors...
     
    Last edited: Nov 25, 2007
  7. Nov 25, 2007 #6
    what about x?

    y = a cos ( (2pi/.80)(0)

    +

    y = a cos ( (2pi/.80)(0.30) -0.7854)

    is that correct?
     
    Last edited: Nov 25, 2007
  8. Nov 25, 2007 #7

    learningphysics

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    No. don't substitute in an x.

    2pi/.80 = 7.854

    You need the function:

    y = acos(7.854x) + acos(7.854x - 0.7854)

    don't substitute in an x... you need to find the amplitude of this function... you can write this sum of two cosines as a single cosine...

    try to use this identity. cosA + cosB = 2*(cos[(A+B)/2])*(cos[(A-B)/2]), where A = 7.854x and B = 7.854x - 0.7854

    you should get something of the form:
    y = (unknownamplitude)*cos(7.854x + unknownphase)
     
  9. Jan 28, 2008 #8
    What is The amplitude of the sound if the speakers are placed side by side?

    A = 2*a*cos([tex]\Delta\Phi[/tex]/ 2)

    The amplitude has its max value A=2a if cos([tex]\Delta\Phi[/tex]/2) = +/- 1.
    This is not the case so take a out.

    2*cos([tex]\Delta\Phi[/tex]/2)=.765

    Something a friend told me that's a little clearer to me and perhaps others.
     
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