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Wave energy density

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  1. May 12, 2015 #1
    1. The problem statement, all variables and given/known data
    In deep water, it can be shown that ##\omega = \sqrt{gk/2}##

    Orcas (killer whales) have been filmed generating waves in deep water to knock seals off ice floes. Assume that an orca can generate a wavepacket which is simply half a sine wave (i.e. ##\sin(kx)## from ##kx =0## to ##kx= \pi##). Show that the kinetic energy density in this wave can be written: $$U = \frac{1}{32} \rho g \frac{A \lambda}{\pi^2}$$

    2. Relevant equations


    3. The attempt at a solution
    After several attempts, I seem to have noticed a discrepancy. I would assume that energy density has the units ##\text{Jm}^{-3}## or ##\text{kgm}^{-1} \text{s}^{-2}.## The equation has the units ## \text{kgm}^{-3} \text{ms}^{-2} \text{mm }## or ##\text{kgms}^{-2}##, assuming A has the unit of metre.

    Are my assumptions wrong?
     
  2. jcsd
  3. May 12, 2015 #2

    BvU

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    No idea. What are all these variables ? And how are they linked in your relevant equations ?
     
  4. May 13, 2015 #3
    ##\rho## is the water density, ##g## the acceleration due to gravity, ##k## is the wavenumber, so ##\frac{2 \pi}{\lambda}##, where ##\lambda## is the wavelength. I forgot to add that ##A## is the wave amplitude.

    I guess I'm too used to assuming symbols are 'standardised'...
     
  5. May 13, 2015 #4

    BvU

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    Standardization is a good thing -- it allows people to communicate at a certain level. Specialists at specialist level and such. Problems arise when non-initiated folks have to make too many assumptions about what they think they pick up. PF is somewhat of a mixed bag in that you can have real specialists in your subject running to your rescue, but you can also have the bad luck that some good-willing, curious and interested outsider tries to put in his two cents while not being restricted by any knowledge about the subject at hand, just gut feeling and a decent education (if you're not too unlucky :smile:).

    In this case I didn't see any quick responses to your thread in spite of a decent number of views, and figured I might help out by asking for clarification.


    And I was intrigued by your ##
    U = \tfrac{1}{32} \rho g \frac{A \lambda}{\pi^2}## dimension: of U presumably ##
    \text{kg m}^{-1} \;\text{s}^{-2}## a.k.a. pressure , but ##[\rho \;g \;A \; \lambda] = \text{kg m}^{-3} \ \text m \; \text{s}^{-2} \ \text m\ \text m\ = \text{kg} \; \text{s}^{-2}## and not ## \text{kg m} \; \text{s}^{-2}## as you found.

    Furthermore, your "deep water" addition had me wondering what kind of waves this is about. Amplitude in m is displacement and that seems strange. But amplitude of pressure doesn't fit the dimensions either.

    And finally, I couldn't find your equation, not even here (where they let A stand for Area ... again, no dimensional fit). Their (2.80) ##\rho v^2## does pass the dimensional check...
     
    Last edited: May 13, 2015
  6. May 13, 2015 #5
    It appears that in this case it's energy per unit area, analogous to energy density on a string being per unit length.
     
  7. May 13, 2015 #6

    BvU

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    Ah, good. ##\text{kg} \; \text{s}^{-2}## it is, then. So with your equation dimensions corrected, the expression passes and your assumptions hold !

    But we're still no further than twinklestar in 2012 ...

    I'm intrigued by this tidbit of animal kingdom know-how. The physics, I mean: they generate this wave, but apparently it can propagate and stay focused enough to tilt an ice floe ? Wow !

    [edit] seen the video. They hunt in teams. Even more impressive. And it's not what I would call deep water waves at all (the water is deep allright, but the waves are surface waves -- goes to show you need to communicate what's 'standard' in your communication...:wink:).
     
    Last edited: May 13, 2015
  8. May 14, 2015 #7
    I will show what the result of my previous attempts were. So first I will find the mass of the water. If its density is ##\rho##, the mass is $$ \rho l \int_{0}^{\frac{ \pi}{k}} A \sin(kx) = \frac{2Al \rho}{k},$$

    where ##l## is the water width.

    The kinetic energy per volume is given by ##\frac{1}{2} \rho A^2 \omega^2 \cos^2(kx) ## ( I think), but to find the total kinetic energy and then divide it by the length and width is what I'm unsure about. My initial assumption was that each small volume of the wave has some kinetic energy so we must multiply by the height and width and then integrate over the length. Is it instead that the kinetic energy above is the energy of the thin 'column' rather than just one small volume of it of it ( so we replace ##\rho## by by mass per area) , if that makes sense?
     
    Last edited: May 14, 2015
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