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Question about stationary waves

  1. May 8, 2009 #1
    I have a question about stationary waves. Anti-nodes are where waves are in phase and nodes are where the waves are out of phase, right? But don't the waves have to be in phase for a stationary wave to be produced (so there wouldn't be any nodes)? Or do they only have to be coherent?
  2. jcsd
  3. May 9, 2009 #2
    In order to create a standing, or stationary wave you have to cause an oscillation at a constant frequency at a multiple of the resonant frequency of the medium, this constant frequency is how the stationary wave is created. if a wave is sent down a string, then another sent down the other side of the string out of phase with the first, then when they overlapped, you would see no wave in the string because they are out of phase and therefore cancel each other off. The time it takes for the pulsed wave to move along the string is the resonant frequency of the string. When the string oscillates at a multiple of its resonant frequency, then the reflected waves match up with the emitted waves which causes the amplitudes to positively add which is what causes the antinodes, or peaks in the wave. The nodes are not actually points where the emitted wave cancels with the reflected wave, but rather the points where both waves have an amplitude of zero, and since 0+0=0, then the amplitude at the nodes is zero.
  4. May 9, 2009 #3
    So a stationary wave is produced when both waves are in phase right? Anti-nodes are where both waves are at the peak and nodes are where both waves have 0 amplitude. Is that correct?
  5. May 10, 2009 #4
    Also, theoretically, do the two waves have to be exactly in phase to produce a stationary wave?

    And are anti-nodes in phase and nodes out of phase or are they both in-phase (otherwise we wouldn't get a stationary wave)

    Finally, why is the distance between two adjacent nodes or anti-nodes half of the wavelength? Is it because
    http://www.csounds.com/ezine/winter1999/beginner/sine.gif [Broken]
    if we took the horizontal distance between the two peaks at y=1 and y=-1(probably easier to look at with y = |sinx| because they are adjacent) it is pi which is half of the "wavelength" (2 pi). Would a stationary wave with this look like y = sinx and y = - sinx on the same axes?
    Last edited by a moderator: May 4, 2017
  6. May 10, 2009 #5


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    Since the 2 waves are travelling in opposite directions, they will be in phase at some locations, and out of phase at other locations.


    Moving half a wavelength means changing the phase of 1 wave by π radians, and the other wave by . The net effect is a relative phase change of 2π, so the same condition (in-phase or out-of-phase) always holds at half a wavelength away.
  7. May 10, 2009 #6
    No, they have the same phase at the anti-nodes and opposite phase (180 degrees diference) at the nodes. The phase difference for a given point should be constant (in time) in order to have a stationary wave.
  8. May 14, 2009 #7
    I don't get it. Don't the waves need a constant phase relationship for a stationary wave to be formed? So they are in phase and this is the constant relationship, so why will they ever be out of phase? Aren't all the points in phase? Although I guess it makes sense because they are both traveling in opposite directions, so some points will be in phase and some out of phase.

    Also, when we have double slit diffraction are the minima on the interference pattern where the two waves are out of phase? But again, aren't the two waves always in phase because they are essentially the same wave through each slit? So how would the waves ever cancel each other out if they are exactly in phase?

    Thank you for any help.
    Last edited: May 14, 2009
  9. May 14, 2009 #8


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    You can't simply say that two waves are "in phase" or "out of phase", without specifying a location. Let's look at an example:

    Wave#1 = cos(x+t)
    Wave#2 = cos(x-t)​

    We'll look at these waves at two locations.

    First, at x=0 we have

    Wave#1 = cos(t)
    Wave#2 = cos(-t) = cos(t) = Wave#1

    So the waves are in phase at x = 0

    Secondly, look at the waves when x=π/2:

    Wave#1 = cos(π/2+t) = -sin(t)
    Wave#2 = cos(π/2-t) = sin(t) = -Wave#1
    These same waves are out of phase at x = π/2
  10. May 15, 2009 #9

    Andy Resnick

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    I wonder if the confusion is driven by nomenclature. The *relative* phase between the two waves must be a constant everywhere (and at all times) in order to get stationary interference.
  11. May 15, 2009 #10
    At all times but not everywhere.
    If it is the same everywhere you'll have no pattern.
    As Redbelly has shown already, the phase difference at x=0 is zero and it will be zero at any time.
    At x=pi/2 the phase difference is 180 degrees and it will be 180 degrees at any time.

    However the phase difference at x=0 is not the same as the phase difference at pi/2.
  12. May 15, 2009 #11
    Thank you. So what does "constant phase difference" mean?
  13. May 15, 2009 #12


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    Can you provide some context? How have you seen/heard that phrase used?

    It could mean that, at a given location, the phase difference is constant for all time. As in the example I posted in Post #8.
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