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sluo
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Homework Statement
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
Homework 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
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!