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Decay time of waves in a swimming pool

  1. Aug 10, 2015 #1
    Hello,
    Let's say I have an ideal cylindrical pool (rigid, vertical walls) of diameter d, with water of depth h. Take the ratio of d:h to be around 5:1 - 10:1. If I press down on the surface with a plunger of width w, I cause a standing wave resembling a single-node Bessel function, like a vibrating membrane with non-fixed edges. Taking into account the viscosity and surface tension, what are the expressions for (and how are they obtained)
    1. wave decay time
    2. wave amplitude
    I've searched all over, but mostly get expressions for the surface heights obtained via slosh analysis and nothing for the decay time of the wave. Thanks!
     
  2. jcsd
  3. Aug 16, 2015 #2
    Maybe the following paper will be useful in finding the decay of the standing surface waves caused by viscosity:-
    http://sites.apam.columbia.edu/courses/apph4200x/Behroozi_Viscous Gravity Waves.pdf
    As for the height of the wave, I presume the volume of water in a half wave will equal the displacement of the plunger.
     
  4. Aug 16, 2015 #3

    Bystander

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    Can you write shear rate as a function of (r,w)?
     
  5. Aug 29, 2015 #4
    Thanks for responding! Sorry it took so long to respond as I have had to set the project aside for a short time. I'll be working on it here and there as my time permits.

    tech99: I haven't had a chance to read the paper in depth, but it looks promising. As for the wave height, I believe that nonlinearities will not permit such a simple relationship between the displacement and water volume, especially near resonance. I say this also because the wave amplitude visibly grows to its maximum after some time of being constantly driven at the resonant frequency. That is, the wave heights differ at the beginning and at resonance while having the same displacement (or so it appears).

    Bystander: I am not sure about that. My expertise is not in fluids, so I need to look into exactly how the shear rate is defined. When I figure that out, I will let you know.

    Thanks again!
     
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