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Destructive vs Constructive interference

  1. Mar 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Two loudspeakers, A and B are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 694 hz. You are standing between the speakers, along the line connecting them and are at a point of constructive interference. How far must you walk toward speaker B to reach the first point of destructive interference?

    2. Relevant equations
    v=lambda/f


    3. The attempt at a solution
    I don't really know how to approach this, besides obviously solving for wavelength, .494 m . After this, i tried drawing pictures to relate things but didn't really know what I was doing. All I know is that constructive interference occurs in whole number integers of wavelength or something like that and destructive in n+ 1/2
     
  2. jcsd
  3. Mar 29, 2016 #2

    blue_leaf77

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    If you stand on the connecting line between the speakers, you will be exposed to two counterpropagating waves - what you will hear is a standing wave. Modelling the two waves as ##y_1(x,t) = A \sin(kx-\omega t)## and ##y_2(x,t) = A \sin(kx+\omega t)##, how does the superposition ##y_1+y_2## look like?
     
  4. Mar 29, 2016 #3
    First, it helps to understand constructive interference which occurs when the the peaks of the 2 waves occur at the same time, the 2 waves are "in phase" with each other. In the adverse, destructive interference occurs when the peaks of one wave occur at the same time as the valleys in the other wave, the 2 waves are "out of phase" with each other.

    Consider now, 2 speakers placed beside each other, emitting the same frequency wave and in the same direction. If you stand in front of the pair you will notice that the sound is louder than it would be from only one of the speakers, because the peaks and valleys are in phase, as above. If we move one speaker further away by a distance of 1/2 of the frequency, the peaks and valleys will be out of phase and there will be a noticeable decrease in volume and potentially no sound will be heard at all. If we move the speaker further back to equal the distance of the length of the frequency, the 2 waves will again be in phase and the sound will again be louder.

    Now, with your example of having the 2 identical speakers facing each other, standing at the middle point will again have the 2 waves in phase with each other. When we move closer to one speaker, we are also moving away from the other speaker by the same amount. If our goal is to stand at a point that is equal in length to 1/2 of the frequency to achieve destructive interference, we need to move 1/4 of the length toward one speaker which will also put us 1/4 of the length away from the other, which will give us the 1/2 length we require.

    Hth. :-)
     
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