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Homework Help: Loudspeaker interference

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Two loudspeakers emit sound waves along the x axis. Speaker 2 is 2.0 m behind speaker 1. Both loudspeakers are connected to the same signal generator, which is oscillating at 340 Hz, but the wire to speaker 1 passes through a box that delays the signal by 1.47 ms.

    Is the interference along the x axis perfect constructive, perfect destructive or something in between?

    2. Relevant equations

    3. The attempt at a solution

    340 m/s = (340 Hz) * [tex]\lambda[/tex]
    [tex]\lambda[/tex]= 1.0 m

    First, turning to the formula deltaphi= 2pi * (deltax/wavelength)

    plugging in the values delta phi = 2pi*(2.0m/1.0 m) = 4pi

    and so I came up with perfect constructive.

    Is this correct? I am trying to think about how the delay would affect the interference of the waves, so if anyone could offer a good explanation of why the delay does or does not affect the superposition, I would greatly appreciate it.

    Last edited: Sep 7, 2008
  2. jcsd
  3. Sep 7, 2008 #2


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    Homework Helper

    Hi bcjochim07,

    The delay would determine the initial phase difference of the waves as they leave the speaker, which does have to be taken into account.
  4. Sep 13, 2008 #3
    Ok, so the delay is .00147 s, and in .00147 s the undelayed wave from speaker two moves (.00147s)(340m/s)= .50 m.

    .50m/1.0m gives an initial phase difference of half a wavelength, or pi

    delta x then becomes 2m + .50 m = 2.5 m

    then using the formula delta phi= 2pi*(delta x)/lambda + initial delta phi

    delta phi = 2pi* (2.5)/(1.0) + pi = 6 pi. This means that complete constructive interference happens. Is this correct?
  5. Sep 13, 2008 #4


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    I don't believe that is correct. In your equation:

    delta phi= 2pi*(delta x)/lambda + initial delta phi

    you have accounted for the effect of the delay twice: in the (initial delta phi) term, and also by "pretending" that the delay meant that one wave moves 2.5 meters farther than the other. Either way would actually work, but you can only include the effect of the delay once.

    So I would say the term

    2pi*(delta x)/lambda

    accounts for the phase shift due to the actual path length difference, so (delta x) should be 2m. And the initial delay only effects the second term.

    In other words, in your original post you found the phase shift due to the path length difference; now just add the phase shift due to the delay.
  6. Sep 13, 2008 #5
    Oh, of course, I'm not sure why I added .5 to the 2m. So, then the phase shift is 5pi, so the interference is complete destructive
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