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Grade 11 Physics; Doppler Effect.

  1. Oct 2, 2007 #1
    Hi, and thank you for your time to read this.

    I'm having a difficult time figuring out the answer to this question;
    Q: A porpoise swims toward a school of fish and emits a high pitched sound.
    The porpoise listens for the echo of its own squeak.

    I have to predict whether the observer will experience an increase or a decrease
    in the observed frequency of the sound.

    Here's my opinion.

    A: Since the distance to be covered by each wave is decreased by the motion
    of the source of the sound, the number of waves that strike the porpoise's eardrum
    in a second will increase and be interpreted as a higher frequency.

    Could anyone correct me if I'm wrong? or not going the right way?

    Thanks again!
  2. jcsd
  3. Oct 2, 2007 #2


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    OK, that is correct as it stands. But you probably ought to argue your point in terms of what happens on each leg of the sound wave's travel. How does the frequency reaching the fish compare with the frequency the porpoise "squeaked"? How do the frequency the porpoise hears compare with the frequency reflecting off the fish? What causes the change on each leg?

    (Incidentally, a similar comparison is made by the software interpreting the radio wave signals bounced off approaching motor vehicles by a highway patrolman's "radar gun", telling the good officer the speed of the vehicle's travel [as I've occasionally found to my cost...].)
  4. Oct 2, 2007 #3
    During the sound wave's travel, each second the wavelength of sound wave decreases, meaning that the porpoise is experiencing stronger echo.

    That's the farthest I could think of right now... :(
    does this also have to do with amplification of sound? as the sound wave reflects off the fish?
  5. Oct 2, 2007 #4


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    We're just considering the frequency or wavelength of the sound here. The intensity will continue to diminish (the wave gets weaker) as it travels, following an inverse square law each way; the returning sound (or radio) wave actually diminishes in strength according to an inverse-fourth power law of the total round-trip distance ( 1/(x^2) times 1/(x^2) because of the two legs of the round trip).

    Your explanation of the wave having a higher frequency (or shorter wavelength) because of the motion of the porpoise toward the fish is correct for the outbound wave. That gives the frequency that reflects off the fish. But the frequency the porpoise hears is higher still. Why would that be?
  6. Oct 2, 2007 #5

    sigh... I'm sorry I just cannot think beyond the fact that the reflected frequency is still high because of the decreasing distance between the porpoise and the fish.
  7. Oct 2, 2007 #6


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    OK, I'm not sure how far you've gone into the Doppler Effect in your class. The change is related to the motion toward the fish. The distinction I'm making is that two things can cause the increase in frequency:

    on the outbound leg, the sound is being emitted by the moving porpoise, so the wavelength of the sound wave itself is what is being compressed; the frequency of the sound travelling toward the fish is higher than what the porpoise itself "squeaked";

    on the return leg, that higher frequency is unchanged -- a portion of the sound wave simply bounces off the fish and returns toward the porpoise; however, since the porpoise is traveling into the returning sound, it passes through the "peaks and troughs" (actually, increases and decreases in the water's density caused by the pressure of the sound wave) more often than the porpoise would if it were stationary; so it hears the frequency of the wave as being increased still more.

    On the way out, we have a "moving source (porpoise) - stationary observer (fish)" situation; on the wave's return, we have a "stationary source (fish reflecting sound) - moving observer (porpoise)" situation. The change in the frequency is (nearly) the same on each leg, so the total frequency change is (about) double what would be expected from the porpoise's speed toward the fish alone. [An exact calculation shows that the shift in frequency is very close to but not exactly doubled.]

    Maybe this is more discussion than you wanted...
  8. Oct 2, 2007 #7
    Thank you, you have been a great help

    I'm taking this course through Adult School and it's all independent learning,
    the booklet I have has lack of information for a complete understanding and I
    should have been more responsible for finding resources but science is just a panic to me :(

    Once again, I really appreciate your help
  9. Oct 2, 2007 #8


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    I understand the situation. I've sort of given you the introductory college physics description of the Effect (the theory goes a bit deeper still, but that's mostly refinements in the description). If you live somewhere with a good-sized library, you might want to have a look at the physics textbooks they have. One tries out texts the way one tries out shoes. You find ones that fit you the best, but they may fit you differently on another day. Not every textbook is uniformly good on every topic (I oughta know, I've looked at a bunch of 'em by now...), so you may need to look around a bit when you are learning different topics.
  10. Oct 2, 2007 #9
    Thank you very much for your advice :)
    I will visit a local library and take a look for reference
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