Re: Water wave question..

Uncle Ben wrote:
On Jul 2, 5:53 pm, "Spaceman" <space...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
If a 1 centimeter sphere that floats in water
half way normally just sitting there and has a mass of 1gram,
were dropped from 2 centimeters above the water...
What would the wave length from the first peak to the second peak be,
and what would the height of that peak also be?
(anyone know an applet on the web that might do this stuff maybe?)

If you want to make fun of me and my crazy ideas
please do post an answer to the above question.
:)
--
James M Driscoll Jr
Spaceman

Jim, you've asked a difficult question. If we take you seriously
enough to fix the inconsistent data you specified, i.e., choose a mass
that would let your ball float half-way out of the water, it is still
a hard question.

See http://en.wikipedia.org/wiki/Ocean_surface_wave#Science_of_waves

If you paid me $1000, I'd do a bit more work: First, you'd have to
specify the depth of the water. If it is very deep, you could use
equations for deep water waves. You'd also have to determine the
frequency of oscillation of the ball. If you just drop it, the mass
of the ball is significant, but I don't know how to calculate the
frequency with any accuracy.

Then you would know the wavelength = c/f., But wait, the motion of the
ball will not necessarily be sinusoidal; the motion might have a
Fourier spectrum. You would have to consider the difference betwee
wave velocity and group velocity. A wave group, such as a tsunami,
travels at one speed called the group velocity, but little wavelets on
the surface might appear near the pulse, travel over it and disappear
again. They move at the wave velocity.

A big complication the your quiestion is that it concerns the initial
behavior of the water. It might be easier to find the steady state of
the system after a few minutes, But the initial motion of the water is
complicated. Somebody might know how to do it, but I would probably
throw everything into a bathtub and measure what I could.

Maybe this set of complications explains why your correspondents have
been dragging their feet with distractions.
It is true that specifying the mass and the diameter of the balolo is
enough to determine how it floats, and stuffing helium inside it won't
change anything except to add a tiny amount of weight.

But even if you had specified the correct mass that would make the
ball float just the way you say in quiet water. my guess is that
nobody would really attempt to answer your question. It's really hard!

I don't know everything, but I do have a Ph. D. in physics, It's
still a hard question.

I do understand it is hard,
and I am going to give up on it because I goofed on the
ball measurement to begin with and as far as I know the town water
here is most likely not just plain water at all.
The stupid ruler has a blank spot before the 0 and I completely
spaced it.
So I am as stupid as the ruler today.
:(
Sad isn't it.
I need to learn to measure much more carefully and
next time I am getting out the micrometer instead.
Which I should have done in the first place.
:)

--
James M Driscoll Jr
Spaceman


.

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