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Homework Help: Amplitude of Sound Waves from Two Sources at a Point

  1. Apr 21, 2007 #1
    [SOLVED] Amplitude of Sound Waves from Two Sources at a Point

    Problem. Two sources, A and B, emit sound waves, in phase, each of wavelength [itex]\lambda[/itex] and amplitude [itex]D_M[/itex]. Consider a point P that is a distance [itex]r_A[/itex] from A and [itex]r_B[/itex] from B. Show that if [itex]r_A[/itex] and [itex]r_B[/itex] are nearly equal ([itex]r_A - r_B \ll r_A[/itex]). then the amplitude varies approximately with position as

    [tex]\frac{2D_M}{r_A} \, \cos \frac{\pi}{\lambda} (r_A - r_B)[/tex]

    Let D(x, t) be the function that describes the displacement of the sound waves at some time t and a distance x from the source. I figure that the displacement at point P must be [itex]D(r_A, t) + D(r_B, t)[/itex] right? One thing I'm noticing is that the expression for the amplitude given in the problem statement does not vary with time. What gives?
  2. jcsd
  3. Apr 24, 2007 #2
    If I look at the situation when t = 0, I get that the displacement at P is

    [tex]D_M \sin kr_A + D_M \sin kr_B = 2D_M \sin \frac{\pi}{\lambda} (r_A + r_B) \cos \frac{\pi}{\lambda} (r_A - r_B)[/tex]

    How in the world does the sine expression simplify to 1/rA?
  4. Apr 24, 2007 #3


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    This problem describes the double slit interference pattern at a point near the central maximum. The signal might vary with time, but per definition the amplitude is the maximum (minimum) strength of the resulting signal.
    Last edited: Apr 24, 2007
  5. Apr 24, 2007 #4
    Interesting. However, I'm still puzzled as why the sine term simplifies to 1/rA.
  6. Apr 25, 2007 #5


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    I can get close :

    The phase difference, [tex]\delta[/tex], between the two waves at the point P will be small under the stated conditions. Since the amplitude of the resultant at P is proportional to


    which under the stated conditions reduces to


    The phase difference at the point P is given by

    [tex]\delta = k \Delta r[/tex]

    with the wave number [tex]k = \frac{2 \pi}{\lambda}[/tex]. We therefore have that

    [tex]\cos(\frac{\pi \Delta r}{\lambda})[/tex]

    for the above mentioned.

    I am very suspicious about the [tex]\frac{1}{r}[/tex] term in the given solution, since the dimension is then incorrect.
    Last edited: Apr 25, 2007
  7. Apr 25, 2007 #6
    How do you know that? Where did you get that expression from?

    You're right. Perhaps it is incorrect. I think we can safely ignore the 1/rA term.
  8. Apr 25, 2007 #7
    It just hit me. The problem wants the maximum displacement, i.e. the amplitude at point P. The displacement at P is given by:

    [tex]2D_M \cos \frac{\pi}{\lambda} (r_A - r_B) \sin (\pi / \lambda (r_A + r_B) + \omega t)[/tex]

    and the maximum displacement or amplitude is just

    [tex]2D_M \cos \frac{\pi}{\lambda} (r_A - r_B)[/tex]

    Last edited: Apr 25, 2007
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