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Dispersion relation for a surface wave of a pool of water

  1. Dec 23, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-12-23_11-20-28.png upload_2017-12-23_11-20-46.png

    2. Relevant equations


    3. The attempt at a solution

    ## v = \frac { \omega } k ##

    ## \omega = \sqrt{ kg \tanh (k) } ##


    I have no idea to guess the graph.

    I put g = 9.8 and tried to calculate ## \omega ## for different values of k.

    ## \omega (0 ) = 0,

    \omega (30) = 17.15

    ~~\omega (20) = 14 ##

    But, these values don’t match any of the graph.

    So, what to do now?
     
    Last edited: Dec 23, 2017
  2. jcsd
  3. Dec 23, 2017 #2

    Charles Link

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    You missed a factor of ## h ## in ## \tanh (kh) ##. (Editing: Oh I missed the part ## h=1##). One problem with trying to rely on numbers on the axes here is they don't label their units, so you don't know what number to use for ## g ##. ## \\ ## A hint that should help: What is the functional behavior of ## \tanh(k) ## for large ## k ##? Thereby, what does ## \omega ## vs. ## k ## look like for large ## k ##? ## \\ ## Also, what is the slope for very small ## k ##? Is it some constant or is it zero? (Hint: Expand ## \sinh(x) ## and ## \cosh(x) ## in Taylor series for small ##x ##). ## \\ ## You could also try using ## g=980 ## cm/sec^2 , but for this problem, I think that leads you to an incorrect answer.
     
    Last edited: Dec 23, 2017
  4. Dec 23, 2017 #3
    I didn’t know the graph of ## \tanh ( k ) ## and I don’t want to look for it right now. 1st I want to solve this question with the knowledge I have and then I will look to the graph.

    ## \tanh ( k) = \frac { e^k - e^{-k} } { e^k + e^{-k} } ## .....(.1)

    As k tends to infinity, ## \tanh ( k) ## tends to 1.

    So, for very large k, ## \omega \approx \sqrt{ kg} ## .....(2)

    So, as k goes to infinity, ##\omega## should go to infinity . .....(3)

    (b) and (d) fulfils condition (3).


    ## \tanh ( k ) = k \frac { 1 + \frac { k^2}{3!} +…}{ 1 + \frac { k^2}{2!} + …} ## .....(4)

    So, for small k, ## \tanh ( k) \approx k ## .....(5)

    ## \omega \approx k\sqrt{ g} ## .....(6)

    So, ##\omega## goes from ##\sqrt{ g}## k for small k to ##\sqrt {gk} ## for large k. This is shown by (b).

    So, the correct option is (b).

    Is this correct?
     
  5. Dec 23, 2017 #4

    Charles Link

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    Correct=I would pick (b) also. A couple of things to observe: For large ## k ##, ## \omega=\sqrt{g} \sqrt{k} ## so instead of looking like a straight line, the second derivative will be negative. ## \\ ## Also for small ## k ##, ## d \omega /dk=\sqrt{g} ##, rather than 0. ## \\ ## (d) comes close to working in some ways, but the second derivative in (d) is slightly positive for large ## k ##.
     
  6. Dec 23, 2017 #5
    Thanks.
    Is there any insight article in PF, which helps in understanding how to guess a graph without plotting it exactly?
     
  7. Dec 23, 2017 #6

    Charles Link

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    Not that I know of. I thought of one additional feature though that helps to show (b) is correct: For large ## k ## (in this case even 18 is large), ## \omega(36)/\omega(18) ## needs to be approximately ## \sqrt{36/18}=\sqrt{2} ##. This would rule out (c).
     
  8. Dec 24, 2017 #7
    That's a nice idea to get rid of g.
    Then I can have one ratio for small k and another ratio for large k and see which graph meets this ratio. If I got one meeting this ratio, then surely this will be the required graph.
    I see that tanh (k) ≈ k is a good approximation for k < 0.5,
    tanh (k) ≈ 1 is a good approximation for k > 10.

    Now, small k approximation is not useful here due to scale of the graphs. So, one can use large k approximation to eliminate some options.
    Thanks for it.



    upload_2017-12-24_18-42-10.png
     

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