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Phase Change and soundwaves

  1. Dec 7, 2005 #1
    Hmm, more questions on phase change.

    I was looking at this question:

    "Waves from a radio station have a wavelength of 319 m. They travel by two paths to a home receiver 14.2 km from the transmitter. One path is a direct path, and the other is by reflection from a mountain directly behind the home receiver. What is the minimum distance from the mountain to the receiver that produces destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)
    Answer: 7.98e+01 m"

    Now, it looks like the answer, to me, implies that the path difference is lambda/4. Why would it be lambda/4 and not lambda/2? Lambda/2 gives complete destructive interference.
     
  2. jcsd
  3. Dec 7, 2005 #2

    Astronuc

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    But then there is the distance of the mountain from the transmitter, which would perhaps also be distance of [itex]\lambda[/itex]/4, so over the two distances, the total distance would be [itex]\lambda[/itex]/2.
     
  4. Dec 10, 2005 #3

    Ouabache

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    I think Astronuc is onto something, but it may not be quite that simple.
    It seems that the receiver (rcvr) is 14.2km from the transmitter (xmtr). When they say the mountain is behind the rcvr, I sense they imply that the xmtr is in front of the rcvr. The rcvr is 14200/319 ≈ 44 1/2 [itex]\lambda[/itex] (wavelengths) away from the xmtr. The mountain could be 1/4 [itex]\lambda[/itex] behind the rcvr, so the the total path reflected off the mountain would be 44 1/2 + 1/4 + 1/4 = 45[itex]\lambda[/itex]. The difference between direct and reflected path would then be 1/2 [itex]\lambda[/itex] just as in Astronuc's example.
     
    Last edited: Dec 10, 2005
  5. Dec 11, 2005 #4

    lightgrav

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    The number of wavelengths from transmitter to receiver is UNimportant .
    These are traveling waves, not standing waves ; only the difference between wavelengths along each path determines the interference condition.

    (I would point out that their presumption of no E-field flip is unrealistic.)
     
  6. Dec 14, 2005 #5

    Ouabache

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    The question only refers to the minimum distance from the mountain to the receiver that produces destructive interference at the receiver.

    Astronuc and I both used examples where the minimum reflected wave arrives at the rcvr [itex]1/2 \lambda [/itex] out of phase with a direct wave. The concept works when you picture the mountain along the same path as the xmtr and rcvr. Then the [itex]1/4 \lambda [/itex] distance between the mountain and rcvr makes sense. I agree, we need to make an assumption that the reflected wave does not change polarization.
     
    Last edited: Dec 14, 2005
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