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[Physltest] [Phys-L] Re: pulse on a vertical rope

Carl Mungan wrote:
The modified wave equation MWE (is there a better name?) for my case is:

d^2y/dt^2 = gx*d^2y/dx^2 + g*dy/dx [1]

OK.

I am seeking the group velocity v of a small traveling wave pulse.

OK.

But isn't f(kx)*cos(wt-d) a standing wave, and hence not what I want?

1) "You can't always get what you want, but sometimes ... you get what
you need."

2) What you get is a superposition, i.e.
Sum_i b_i f(k_i x) cos(w_i t - d) [2]

where b_i is a generalized Fourier coefficient.

The summation changes the story quite a bit. Since the rope has fixed
boundary conditions, it is in most (but not all) ways the same as a particle
in a box, or sound in an organ pipe. The valid similarities include:
-- Yes, there are ell-known stationary solutions in all these cases.
-- There are also non-stationary solutions.
-- If you take two stationary solutions with different frequencies and
superpose them, you get a non-stationary solution.

Instead, I am thinking I should use exp[i(kx-wt)], where generality
would require k to be a complex function of x. The imaginary part is
needed because of the first derivative in the MWE and gives rise to
dissipation (damping).

1) Please let's neglect damping for now. The equation of motion, eqn [1]
above, doesn't include damping. Let's understand what we've got before
making things more complicated.

We agree that the amplitude of the wave changes as it goes along ... but
this has got nothing to do with dissipation. Indeed, if you launch a wave
from the top down, the amplitude gets *bigger* as it goes along ... whiich
is clearly not explainable in terms of dissipation.

2) Let me replace the Ansatz
exp[i(kx-wt)] [3]
with an even more general one,
F(x - c t). [4]
You can see that [3] is a special case of [4].

If you're looking for running-wave solutions, this is an attractive
lamp-post under which to look ... because these expressions describe
something that moves from place to place while keeping its shape.
This behavior is what some people take as the very definition of a
wave.

[Sound of Jagger singing.....]

But alas, the vertical rope solutions are not to be found under this
lamp-post. If you plug such an Ansatz into the equation of motion, you
don't get a solution.

The problem is that the propagation is significantly dispersive. If
you inject a square pulse at one end, it won't be -- cannot be -- a
perfectly square pulse at the other end.

The mathematics here is analogous to the propagation of a tsunami,
which (as we recently discussed) is necessarily dispersive.
http://lists.nau.edu/cgi-bin/wa?A2=ind0412&L=phys-l&P=R14534
As I said then, there's a ton of physics here, and pedagogy, too.

The physics of circularly-symmetric waves on the ocean is AFAICT
unrelated to the physics of the vertical rope, but we get Bessel's
equations in both cases, and as RPF liked to say, the same equations
have the same solutions.

I know that a zeroth approximation to k is to drop this last term and
ignore derivatives of k to get:

w^2 = gx*k^2

and thus v = dw/dk = sqrt(gx), which is the "naive" solution that
gives good agreement with experiment.

This is a side-alley that we need not explore.

However, if I retain both the derivatives of k and the last term in
the MWE, I get a complicated pair of nonlinear second-order
differential equations for the real and imaginary parts of k.

The equations are not just complicated, they are mutually contradictory.
There is no solution.

So I'm thinking that I want to make some kind of iterative
approximations, starting from the zeroth solution above.

If by "iterative" you mean _series_, I definitely agree. The series
expansion, i.e. equation [2] above, will give you what you need.

Represent the initial pulse as a superposition. If it is a *sharp*
pulse, it will have lots of large-k components in it.

ps: I'm assuming that what I'd do in next order of approximation is evaluate
dw/dK at K0 where K is the real part of k and K0 is the peak
wavenumber of my wave pulse.

Yes, that's the standard recipe for finding the group velocity. (And as
previously mentioned, we're nowhere near ready to discuss dissipation,
so there won't be any imaginary part of k).

Say I construct a Gaussian wavepacket
with peak K0 and some spread sigma_K, chosen to simultaneously give
me not too large a spread for the Fourier transform in real space x
(compared to the length L of the rope) and yet not involve too many
Fourier components so that K0 is a good approximation to all K in the
pulse.

Am I on the right track?

Yup.

Important note: the equation of motion in the form given above, equation
[1], is not really Bessel's equation, and its solutions are not really
Bessel functions. But you can fix it by a sneaky substitution. If x
is your real abscissa, the solution involves Bessel functions acting
on sqrt(x). An outline of the calculation can be found at
http://webpages.ursinus.edu/lriley/courses/p212/lectures/node14.html
which has the right idea, although I abhor the notation used in equation
3.11, because the f() there is not the same function as the f() in
equation 3.8.

Remember: a function is something you can hard-code in a computer
program, so that f(3) means the same thing every time.

You should rewrite 3.11 in terms of phi(u), where
phi(u) := f(y(u))
and
y(u) := u^2 / alpha
and the program for evaluating f(3) is very different from phi(3).

You need about a page of algebra (not shown) to prove phi() is governed
by Bessel's equation.

It is important to keep track of which equation is which.
--> Show that the equation that is solved by phi(k u) cos(w t), where
phi() is a Bessel function, is non-dispersive; w^2 proportional to k^2.
--> Show that the equation that is solved by f(k y) cos(w t), i.e.
equation [1] above, i.e. the equation of motion for the rope, is
dramatically dispersive; w^2 proportional to k.

=====================

I haven't thought much about it, nor done the experiment, but I imagine
that you could set up an unforgettable demo along the following lines:
Hang a long rope in a tall stairwell ... preferably a high-quality
Kernmantel rope, to minimize internal friction. Let person A give
the rope a nice big slow swoosh, to launch a small-k wavepacket, and
at about the midpoint thereof, let person B whack the rope with
a broomstick, to launch a high-k wavepacket. I imagine you should
be able to see the waves separate as they move along.

=========================================================

For extra credit, for those who might be wondering how to calculate
the coefficients b_i in equation [1]:

The "forward" problem is where you are given the b_i and you want
to find the shape of the wave. That's easy; just plug into
equation [1].

The "inverse" problem is where you have in mind some nice fancy
wave-shape, and you want to find the b_i that describe it. This
is slightly trickier, as we now discuss.

If the rope had uniform tension, you would determine each b_i
by means of a Fourier integral. This is provably correct, because
of the well-known orthogonality of sines and cosines of different
frequencies.

Returning to the case of nonuniform tension, we must generalize
our ideas.
The key idea is that we need to take inner products, i.e.
dot products, i.e. we need a metric. There are lots of different
inner products in the world. The simplest case is
Integral f(k_1 x) f(k_2 x) dx
and in that case the orthogonal functions are sines and cosines.
More generally, we can use a different measure, e.g.
Integral f(k_1 x) f(k_2 x) rho(x) dx
so it's a weighted sum. Bessel functions in particular are
orthogonal in the next-to-simplest case, where the measure is
(r dr), that is:
Integral f(k_1 r) f(k_2 r) r dr

This (plus some fussy normalization) is enough to let you crank out
the coefficients b_i. For the next level of detail, see equation 51
of:
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html

And in case you wanted a cut-and-dried way of describing *anything*
that happens on the rope:
-- The Bessel functions are orthogonal, as just discussed.
-- Use Sturm-Liouville theory to convince yourself they are complete.
-- There is a celebrated formula for expressing the Green function
in terms of a generalized Fourier series, i.e. as an expansion in
terms of orthogonal functions. See equation 4.26 of
http://www.math.ohio-state.edu/~gerlach/math/BVtypset/node90.html

Obviously if you know the Green function you're pretty much in command
of the situation.
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