Natural Science Forum / Physics / General Physics / July 2008

On Jul 3, 3:14 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
wrote:

It is a better-framed question, Jim, since you have defined the
frequency of motion of the system.

But it is still a hard problem.  I replied on your other thread at
message 65, although the numbers change when somebody sticks one in
above mine.  I explained a bit about why the question is harde.  The
fellow who posted just above me on that thread has additional good
information.

Uncle Ben

On Jul 3, 3:14 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
wrote:

It is a better-framed question, Jim, since you have defined the
frequency of motion of the system.

But it is still a hard problem.  I replied on your other thread at
message 65, although the numbers change when somebody sticks one in
above mine.  I explained a bit about why the question is hard.  The
fellow who posted just above me on that thread has additional good
information.

If you want to make it an easy question, you may actualy get an
answer.  Just say that the water is very deep and the motion continues
a long time, and what you want is the wavength of the circular wave
pattern that develops. You give up on finding the height of the waves,
which will diminish anyway as the waves get farther from the ball.

Uncle Ben

As Uncle Ben said, it's better but still difficult.

The period of 1 second will set the frequency of the waves
(at least in steady state).  So f = 1 cycle per second, or
f = 1 Hz, and period is T = 1 sec.  

Let's check to see if we're in a shallow water regime.  That
is, check to see if the depth is less than 1/20 of the
wavelength.  If that is true, then the wave speed will be
given by sqrt(g*d) where d is the depth:

 v = sqrt(g*d) = sqrt(981 cm/sec^2 * 6 cm) = 76.7 cm/sec

The wavelength will then be

 L = v/f = 76.7 cm

L/20 is 3.84 cm, which is less than the depth of 6 cm, so we're
close to, but not in the (simpler) shallow water regime.  Rats.

Start again, assuming deep water regime.  Now the velocity is
given by:

 v = g*T/(2*pi) = 1.56 m/s
 L = 1.56 m

For deep water we want to have d > L/2.  Rats again.  So we're
in between the deep and shallow regimes and must resort to the
full machinery for calculation:

 v = sqrt(g*L/(2*pi) * tanh(2*pi*d/L))

 Letting L = v/f this becomes

 v = g/(2*pi*f)*tanh(2*pi*f*d/v)

The hyperbolic tangent function including v in its argument
makes this hard to solve symbolically for v, so I'll resort to
a quick and dirty numerical solution.  I get:

 v = 73.6 cm/sec   (close to the shallow water value as expected)
 L = 73.6 cm

So there's wavelength part of your question answered at least.
The waves will have nearly 74 centimeters between peaks.

On Jul 3, 2:14 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
wrote:

Let's try a deep water wave model first:
speed of wave = (9.8 m/s/s) x (1 s) / (2pi) = 1.6 m/s, suggesting a
wavelength of 1.6 m. The depth of the tank is less than 1/20th of the
wavelength, and so the deep water model doesn't work so well, and we
should switch to the shallow water model:
speed of wave = sqrt (9.8 m/s/s x 1/2 (0.01 m)) = 0.22 m/s, which
suggests a wavelength of 22 cm. The real experimental answer will be a
little higher than this model suggests, since we're in a transition
zone between shallow and deep water treatments.

Now, how do you have the foggiest idea whether this is right or not?

PD

On Jul 3, 11:14 am, "Spaceman" <space...@yourclockmalfunctioned.duh>
wrote:

[...]

The question is why should we teach you anything after you spending
year after year sh.tting on the only people who can help you?

On Jul 3, 3:14 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
wrote:

Jim, several people has introduced equations for the velocity of water
waves and have shown that it depends on the depth of the water.  When
the water getsw deep enough, that doesn't matter so much, but when the
water gets shallow, you will find the the speed of the wave slows down
as the water gets shallower.

That gives a neat explanation as to why waves at the seashore "break"
as they approach.  The crests of the wave are in deeper water than the
troughs of the wave.  So the crests go faster than the troughs. That
is why they outrun the troughs and the wave breaks.

The reason that waves in shallow water go slower than those in deep
water is that the action of the wave down below the surface involves
bits of water that move in verticle circles of a size that is related
to the wavelength.  When the water is too shallow, this motion is
impeded, and the wave slows down.

Note that the water doesn't go with the wave except for short
distances, in which the water goes back and forth, staying in one
vicinity.

These are just some interesting facts about water waves.  They make an
interesting field of study.

Ben