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Ripple tank - strobe frequency

  1. Jul 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Say we are observing plane wave patterns on table. The depth of water is the same everywhere in ripple tank. The frequency of waves is 12 Hz and wave length is L (lamda). We see the wave pattern as stationer while looking at wave pattern through the slots of stroboscope which turning at 2 Hz frequency.

    What is the number of slits if we measure the distance between two wave crest as L/3?

    (A: 18)

    2. Relevant equations

    Fw = n Fs

    Fw: Wave freq
    Fs: Strobe freq
    n: Number of slots

    3. The attempt at a solution

    I could not solve this problem. I can not imagine stationer waves which has smaller wave length.

    Is it possible to see the waves as stationer while slot frequency is bigger then wave frequeny?
     
  2. jcsd
  3. Jul 4, 2015 #2

    TSny

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    If the stroboscope frequency were the same as the wave frequency (12 Hz), then an image of the wave would be projected into the eye every 1/12 of a second. Since the wave would have traveled exactly one wavelength during this time, each image is identical, with the wave crests in each image located at the same place and spaced one wavelength λ apart. If the time between two consecutive images is short enough, then the brain does not perceive any "flicker" and the wave appears to be stationary with the distance between two consecutive crests equal to the wavelength of the traveling wave, λ.

    What do you think you would see if the stroboscope frequency is doubled so that it is twice the wave frequency?
     
    Last edited: Jul 4, 2015
  4. Jul 4, 2015 #3
    Thanks for guide. OK, if the stroboscope (slot) frequency were 24 Hz and the frequency of plane waves were 12 Hz then the waves would be late a half of period. I think would see both crest and troughs of wave alternately.
     
  5. Jul 4, 2015 #4

    TSny

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    I think the problem is meant to be simplified by assuming that you are only looking at wave crests. Thus, assume the light that is illuminating the waves is reflecting off the wave crests and then into your eye. If the slot frequency is 24 Hz, how far apart do you see wavecrests?
     
  6. Jul 4, 2015 #5
    I can not answer this. Because can't imagine. Just a guess, the new wave length would be (λ /2).

    I can not imagine stationary waves for a slot frequency which is higher then the wave frequency. I think, the biggest slot frequency is equal to wave frequency to observe stationary waves. I mean,

    fs = fw
    fs: slot freq.
    fw: wave freq.

    At above equation, fs is maximum. If slot frequency is ½, ⅓, ¼, ... of the fw then I will see stationary waves again. Also, observed wave length will be, 2λ, 3λ, 4λ,... respectively. Because speed of wave does not change (v = λ . f). But as I mentioned, if slot frequency is above the fw then I can not see them stationary anymore, in any manner.

    In my question, observed wave length is λ/3 (distance between two wave crest). In this manner, fs = 3 × fw. Absolutely, if I'm not wrong so far. Anyway, I am calculating but neither can imagine nor think this is possible.
     
  7. Jul 4, 2015 #6

    TSny

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    I don't believe this is correct.

    If fs = 1/2 fw then you would see the same thing as if the slot frequency is equal to fw. To understand this, think about how far the wave will travel during the interval between two slots if fs = 1/2 fw. Draw some pictures of the wave for each slot.

    Then repeat for fs = 1/3 fw.

    Then try fs = 2fw
     
  8. Jul 4, 2015 #7

    TSny

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    Yes, this is correct. Draw pictures of the wave at the slot times to see why.
     
  9. Jul 5, 2015 #8
    You mean, wavelength doesn't change. Does it mean, we don't perceive it as different length too? Honestly I didn't see such a different wavelength but had read it somewhere.

    For fs = 1/2 fw , the wave would advance two wavelengths between successive looks.
    For fs = 1/3 fw , the wave would advance three wavelengths between successive looks.

    For fs = 2 fw , the wave would advance 1/2 wavelengths between successive looks.
     
    Last edited: Jul 5, 2015
  10. Jul 5, 2015 #9
    Which is correct?
    - Impossible to freeze the waves at fs = 3 fw
    - Impossible to observe the wavelength as λ/3
    - both
     
  11. Jul 5, 2015 #10

    TSny

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    Yes, fs = 1/2 fw and fs = 1/3 fw will produce the same stationary pattern with the wave crests λ apart. The fs = 1/3 fw case will have a slower "flicker" than fs = 1/2 fw, but the stationary wave pattern for the two cases will be the same as for fs = fw.

    For the case where fs = 2 fw, you are right that the wave will advance 1/2 wavelength between successive slots. You might think that would mean that you would not see a stationary pattern; rather, you would see an alternation between two stationary patterns where the wave is shifted 1/2 wavelength between the two patterns.

    However, if fs is high enough (I believe about 20 Hz) then you will tend to see both patterns simultaneously due to "persistence of vision" of the eye-brain system. See for example




    So, for fs = 2 fw, you will see the two alternating stationary wave patterns simultaneously and it will look like one stationary wave with wave crests separated by λ/2.
     
    Last edited: Jul 5, 2015
  12. Jul 5, 2015 #11

    TSny

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    In terms of what you see due to persistence of vision, it is possible to freeze the waves at fs = 3 fw such that you see wave crests separated by λ/3.
     
  13. Jul 5, 2015 #12
    Thank you very much. Thanks for your time.

    Those videos are beautiful. We are perceiving the bird and cage at the same place and time. Then we will perceive crest and troughs at the same place in case of fs = 2 fw. Else the wave flicking forth and back at 1/2λ steps?
     
  14. Jul 5, 2015 #13

    TSny

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    Yes, I think so. For fs = 2 fw, the odd numbered slots (1, 3, 5, ...) of the strobe will display the same wave pattern since the wave travels exactly one wavelength during the time between consecutive odd slots. In the figure below, this stationary wave pattern might be the wave shown in blue. The even numbered slots (2, 4, 6,..) will show a stationary wave pattern as shown in black. If the strobe frequency is low, then you would see alternation between the blue stationary wave and the black stationary wave. However, for large enough strobe frequency you would just see both wave patterns simultaneously due to persistence of vision. Thus, you would see a stationary pattern with distance λ/2 between crests.
     

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  15. Jul 5, 2015 #14
    Thank you very very much. Again, thanks for your time. Greatly appreciated.
     
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