I need a wave/pressure quibble/wager settled.

I say that the ambient pressure has almost no, or absolutely no, effect on the size and shape of a wave generated by a given motion.

Here is the example just to make things simple:

A 10' x 1' x 1' box 1/2 filled with water, into which a standard brick in dropped from 1" above the surface to generate a wave(actually a bunch of waves).

In order to negate the effects of increased atmospheric density the first test will be in a camber at sea level filled with air, and the second(hi-pressure) test will be in a compression chamber filled with H2 at a pressure where the H2 becomes as dense as sea-level air.

For the purposes of this wager we are discounting any odd differences in the gas's viscosity.

I say the waves will look the same, my opponent says the wave in the pressure tank will be very much muted due to the higher pressure, although he is not sure if it will be 1:1 ratio, or square or cube root, or some other function.


What say yee, and does anyone have the relevant formula?

 

I'm voting with you. Since it's the differential pressures at different depths that determine wave characteristics the absolute pressure should have only a secondary or even tertiary effect, apart from determining the boiling point that is.

 

What you are referring to is basically "damping". In this case Coulomb damping. The fluid above, is the resisting force. Unlike hysteresis damping, with involves elastic restoring forces, such as found in fatigue of structural members.

As such only the amplitude, just as in any normal damping situation, is affected; the period remains the same.

 

So you are incresing the pressure of pure Hydrogen until its density equals that of air at sea level, right? To start with, the meniscus will be different. Also, the water, even though it compresses very little, will have a higher density too. The wave shape should be different. I can take your side for a small fee though.

 

Gonzo, the example of compressed H2 was just to

Gonzo, the example of compressed H2 was just to take Density out of problem as much a practical.

My assertion is that size and shape of the wave is determined by mechanical action of what causes it, independent of surface pressure.

I was going to say "two Ideal Gases with equal different densities" and "non-compressible liquid".


Basically, he thinks the amount of pressure on the surface will make the wave much smaller, due to the pressure itself. I say the wave will be the same as 'exactly' as much pressure is pushing from all directions so the wave doesn't 'know' it is under pressure.

 

The text above equation (4) in the Airy derivation will give you the answer you are after:
http://en.wikipedia.org/wiki/Airy_wave_theory
The relevant formulas are also shown in the table.

There is a small, almost irrelevant, waiver that could come into the discussion. With very small waves, called surface waves rather than gravity waves, the water density influences the propagation. These are tiny waves that you can observe on the crest of smooth gravity waves. Most people will not have observed them. Increasing the pressure will have a slight increase in the water density so the surface waves will be altered minutely.


Rick W

 

I found a photo that gives a good comparison of the size of surface or capillary waves to gravity waves.

The waves generated by the hull are gravity waves. If you look down the face of these waves you will see tiny ripples - almost surface wrinkles. These are the capillary waves. They propagate faster than the gravity waves so are on the leading face of the gravity waves.

These little waves will be affected to a tiny degree by increasing the pressure on the surface but in your example, using hydrogen to achieve the same density as air (say 8 atmospheres), the change in the water density would be a fraction of a percent.

So the gravity waves are not affected by the pressure. The capillary waves, which are minute compared to the gravity waves, are affected minutely by the pressure on the surface.

So you have to decide if a minute amount of a minutely small component of the waves has any significance to your bet.

If you were debating the waves generated by rain drops rather than a brick then the capillary waves would be significant. My view is that they are irrelevant to what you will see with a brick. Here the gravity waves dominate.

Rick W

 

Rick
Again, if you're going to just cut and paste articles try and understand what is written in them.
What is a capillary wave you mentioned above...well, if you bothered to open a text book and read it, rather than just cut and paste articles, you would know that a capillary wave and its existence, is dominated by the surface tension of the fluid in question.
The picture above, those small waves are not capillary waves, they are far to large.

Since for true capillary-gravity waves their mean speed is given by:

u^2 = (Tk/rho) + g/k

which turns out to be about 23cm/sec at the wave length of about 1.7 cm for water and air at room temperature. (T ≈ 75 in c.g.s. units for room temperatures and typical ocean salinities.)
http://www.britannica.com/EBchecked/topic/93834/capillary-wave

Those wave are larger than 1.7cm and as such their restoring force is not the surface tension. So, for longer waves, surface tension is neglected and the waves becomes a 'standard' wave and its associated profile/properties.