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Calculating total energy carried by a ripple

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A two-dimensional water wave spreads in circular ripples. Show that the amplitude [itex]A[/itex] at a distance [itex]R[/itex] from the initial disturbance is proportional to [itex]1/\sqrt{R}[/itex] Hint: Consider the energy carried by one outward moving ripple

    2. Relevant equations

    Kinetic energy carried by a pulse in a one-dimensional string:

    [tex]\frac{1}{4} \mu \omega^2 A^2 \lambda [/tex]

    where [itex]\mu[/itex] is the mass per unit length, [itex] \omega [/itex] is the angular frequency of the wave, [itex]\lambda[/itex] is the wavelength

    3. The attempt at a solution

    To attempt this, I copied the derivation in the book for the 1-D string. They did the kinetic energy part, but not the potential energy part, since it comes out exactly the same at the kinetic energy part. I wish they had, because I don't understand how they did it, namely where the gravitational constant go? Anyway, I'm going to put what I did for the kinetic energy part here and I would appreciate any comments as to its correctness. I'm lost on how to do the potential energy part, so please give me some hints on that too if you can!

    We use the equation derived for the transverse velocity in the 1D case:

    [tex] -\omega A \cos(kx-\omega t) [/tex]

    Consider one ripple, with the origin at the center. This equation can be applied along a cross section of the ripple (think of a ray emanating from the center, we apply the equation along that line). Since it's a circle, we'll use polar coordinates. Instead of [itex] x [/itex], the velocity now depends on [itex] r [/itex]. We also take [itex] t [/itex] to be 0 since we can start time whenever we want (at least I think this is the reason they do this). Now we integrate. It must be a double integral, since it's 2-dimensional. Also, it doesn't depend on [itex] \theta [/itex], so we get

    [tex] \frac{1}{2}\mu \omega^2 A^2\int_0^{2\pi} \int_{R-\lambda/2}^{R+\lambda/2} \cos^2(kr) r dr d\theta [/tex]

    where we have used the fact that the kinetic energy is given by [itex] \frac{1}{2} m v^2 [/itex].

    I'll skip the details of the integration since this post is already kind of long, but this is what I got:

    [tex] \frac{1}{8} \mu \omega^2A^2R\lambda [/tex]

    I'm tempted to say this is right, since after adding the potential energy part (which I'm guessing is similar to this answer) and solving for [itex] A[/itex], we would indeed get that it is proportional to [itex] 1/\sqrt{R} [/itex]. Plus that fact that the constant went from [itex] 1/4 [/itex] to [itex] 1/8 [/itex] between the 1D and 2D case is also encouraging, but I'm not sure why I think that, just seems to make sense!

    Anyway, is this right? I'm not sure I'm right in just plugging in [itex] r[/itex] for [itex] x[/itex] when using polar coordinates, shouldn't it be [itex] x = r\cos(\theta) [/itex]? Also, how in the heck do you do the potential energy part? If potential energy is [itex] mgh [/itex], where does the [itex] g [/itex] end up going? Is there another equation for potential energy I don't know about?

    Well, congrats if you made it this far. I think this is a really interesting problem and I'm probably making more difficult than it is, so I'd appreciate any useful insight! Thanks!
  2. jcsd
  3. Apr 8, 2013 #2
    One can trivially show that the energy in any "elementary" oscillator, of which any wave is made of, is proportional to the square of its amplitude. This is because any wave is made of harmonic oscillations, described by ## A \sin \phi ##, where ## \phi ## is the phase of the wave. Thus the maximum kinetic energy of the oscillator is proportional to ## A^2 ##, which must then be true for the entire wave.

    Now think about the number of those elementary oscillators as the wave spread. Is the number constant? Or does it depend on the radius somehow?
  4. Apr 8, 2013 #3
    I kind of see what you're saying, but i don't see how the radius comes into it. In the 1D case, it didn't depend on the distance from the source.
  5. Apr 8, 2013 #4
    Assume that oscillators have some fixed length. How many are excited by the wave at any given time as it propagates?
  6. Apr 9, 2013 #5


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    voko is suggesting you carve each ripple into narrow sections - cutting radially - using the same width at all radii. So as you go further out there are more sections. The total energy is the same in each complete ripple.
    This works ok in terms of PE, but to conclude the KE works the same way I think you need to use properties of SHM (to which voko alluded).
  7. Apr 9, 2013 #6
    Ok, so since the motion is described by something like [itex] A\sin(\phi) [/itex] the velocity for one radial section is something like [itex] A\cos(\phi) [/itex], and so the kinetic energy is proportional to [itex] A^2 [/itex]. So by conservation of energy, the energy is always prportional to [itex] A^2 [/itex].

    Next, since the ripple grows bigger as it travels away from the disturbance, the number of radial sections grows, so the total energy would be proportional to [itex] 2\pi R A^2[/itex]. Is this what you mean? So is my derivation above wrong?
  8. Apr 9, 2013 #7


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    No, I think your derivation is fine, but voko and I are suggesting you could have avoided the integration by just thinking about conservation of energy.
  9. Apr 10, 2013 #8
    THanks guys! By the way, do you guys have any idea about how to derive the formula for the potential energy in the case of the 1D string? (It comes out the same as the KE, but I can't figure out why)
  10. Apr 10, 2013 #9


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    SHM always, AFAIK, involves the trading of energy back and forth between two modes. When you talk of the amount of energy in one mode, you really mean the difference between the min and max energy in that mode. By work conservation, this is necessarily the same for both modes.
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