Transverse mechanical waves - fundamentals

In summary: I don't get that really. If the water molecules ran downhill this slope, they would run diagonally, not affecting the molecules right below them. So as the slope here becomes less and less steep, the molecules will have less and less energy, and when in equilibrium with the water in the center (i.e. the center of the original depression), no more water molecules will run down the slope (since there's no slope anymore).This is actually two questions that have the same answer.The first question, "how can this hill become higher than the original water level?", sounds like you're trying to figure out where the water gets the energy from to go past its original point of equilibrium. This energy is taken from
  • #1
Mårten
126
1
Hello all!

I'm trying to understand and in detail analyse how transverse mechanical waves work, and have had troubles finding anything on the net, except for the link below. Let's begin with a water wave. The following questions then:

1) Firstly, I wonder if it's possible to find, buy or make some kind of liquid which exhibits a really slow wave speed, preferably something around 1 mm/s. This would have been wonderful to have to make experiments and really see how the wave slowly moves, like it was alive. I suppose it must be a pretty gelly liquid with high viscosity.

2) Someone who knows of an animated computer simulation which shows the various forces which build up the wave crest and the wave trough, and so forth?

3) Why, exactly, is the wave propagating? I've tried to study the following text (see link below). It says that if there is a depression in the water surface, then water on the sides flows down in this depression, building up a small hill (btw, how can this hill become higher than the original water level?). Then, where the water flowed down from the sides, a new depression is formed. But how can this new depression be as deep as the first depression? The water there should level out with the first depression?

http://www.lightandmatter.com/html_books/3vw/ch03/ch03.html#Chapter3 (see figure and 2nd paragraph)
 
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  • #2
Mårten said:
The water there should level out with the first depression?
Right, that would be the point at which the force becomes zero, but because of inertia the medium can't instantly stop moving, not until it goes so far as to build up a force in the opposite direction..
 
  • #3
cesiumfrog said:
Right, that would be the point at which the force becomes zero, but because of inertia the medium can't instantly stop moving, not until it goes so far as to build up a force in the opposite direction..
I don't get that really. If the water molecules ran downhill this slope, they would run diagonally, not affecting the molecules right below them. So as the slope here becomes less and less steep, the molecules will have less and less energy, and when in equilibrium with the water in the center (i.e. the center of the original depression), no more water molecules will run down the slope (since there's no slope anymore).

Also, I've read that the water molecules should run in circles... :confused:
 
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  • #4
Regarding your first question; do you have a video camera? You could record water waves, and then play them back at a lower speed.
Mårten said:
3) Why, exactly, is the wave propagating? I've tried to study the following text (see link below). It says that if there is a depression in the water surface, then water on the sides flows down in this depression, building up a small hill (btw, how can this hill become higher than the original water level?). Then, where the water flowed down from the sides, a new depression is formed. But how can this new depression be as deep as the first depression? The water there should level out with the first depression?

This is actually two questions that have the same answer. Your first question, "how can this hill become higher than the original water level?", sounds like you're trying to figure out where the water gets the energy from to go past its original point of equilibrium. This energy is taken from whatever outside influence caused a depression in the water surface to begin with.

The answer to your second question, "how can the new depression be as deep as the first depression?", depends on the manner in which the waves are being formed. Again, the energy for this wave depression comes from the original event that disturbed the fluid surface in first place. If the surface is disturbed by a short-term event or impact (like a pebble being thrown in), then the second depression cannot be as deep as the first. The water's surface will descend to a point that is below its original level, but above the lowest point in the preceeding trough. If the surface is being disturbed by a continuous and ongoing external force (such as the wind), then this external force can continuously add energy so that it is even possible for the second depression to be deeper than the first.

But, in either case, some event has to put energy into the fluid.
 
  • #5
LURCH said:
Regarding your first question; do you have a video camera? You could record water waves, and then play them back at a lower speed.
No, I haven't... :cry: And it should be a high speed camera I guess. I found this formula here for the speed of a transverse wave, [itex]v = \sqrt{F/\mu}[/itex], and to get v = 0.001 m/s, yo got to have e.g. F = 1 N, and mu = 10^6 kg/m. Hm, that's pretty much...

LURCH said:
But, in either case, some event has to put energy into the fluid.
Okey, thank's for a good explanation there regarding input of energy. I'm getting closer. Still, I cannot see the whole picture. I'll try to give a new, maybe more pure example.

Imagine a water surface (a glass of water may be good for experiment) where you carefully put down the end of a knife, just a millimetre or so, and let the knife be in that position. As soon as the knife touches the water surface, a wave pulse is seen emerging from where the knife touched the water surface. Imagine now the water surface close up to the knife, just after impact. There the water must be a little bit higher than the rest of the water, because the water where the knife now is, must go somewhere. Now, this higher water would like to come in equilibrium with the water outside, so the higher water flows outwards to the lower water.

Now, this had been perfectly okey for me, if what happened here had been a sort of pulse, where the higher water flowed outward, but as soon as that water was as high as the water outside no more flow would occur. So the pulse would not be like a hill going outwards from the impact place, more like a step function, that slowly decreased the height of the step as it moved further away from the impact. But now it is a hill - how come? And, as pointed out before, I cannot see that this higher water has momentum that makes it go below the equilibrium level, since the flow seems to be diagonal. And, it's not even diagonal, they say it's circular.

(Maybe, what I'm ultimately out for is the micro mechanics that's behind this circular motion of the water molecules.)
 
  • #6
Mårten said:
Hello all!

Marten, I can do little more than share your curiosity!


"1) Firstly, I wonder if it's possible to find, buy or make some kind of liquid which exhibits a really slow wave speed, preferably something around 1 mm/s. I suppose it must be a pretty gelly liquid with high viscosity."


I don't know anything physical that would transverse at + 1 mm/s. But those executive and kid's toys featuring waves are made w/ mineral spirits and dyed water. Both are a low vis fluids. Waves in those mediums giddy up! Now place these two immiscibles together and the wave speed at the interface is dramatically s-l-o-w.
 
  • #7
Mårten said:
If the water molecules ran downhill this slope, they would run diagonally, not affecting the molecules right below them.

The water molecules in a surface water wave do not run diagonally down the slope of the wave. Each molecule oscillates vertically. Neighboring molecules oscillate slightly out of step with each other, i.e. with slightly different phases, although still with the same frequency.
 
  • #8
jtbell said:
The water molecules in a surface water wave do not run diagonally down the slope of the wave. Each molecule oscillates vertically.
Well, if the depth is finite, the water parcels do take http://en.wikipedia.org/wiki/Airy_wave_theory" paths. But I thought the OP's confusion might be better cleared up first with reference to simpler waves, such as of a rope.
 
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  • #9
jtbell said:
The water molecules in a surface water wave do not run diagonally down the slope of the wave. Each molecule oscillates vertically. Neighboring molecules oscillate slightly out of step with each other, i.e. with slightly different phases, although still with the same frequency.
Okey, let's think of this vertical oscillation then. And my example above with the water that is higher just close up to the knifepoint that was just now inserted into the water surface. Now, what force drives these molecules vertically down?

I found this animation of water waves. It's nice, but I don't get the mechanics driving the points. http://www.kettering.edu/~drussell/Demos/waves/water.gif

Btw, I suppose the water we are talking about here has a constant density, and that there is no kind of compression of the water anywhere. Correct me if this is an erroneous assumption.

cesiumfrog said:
Well, if the depth is finite, the water parcels do take http://en.wikipedia.org/wiki/Airy_wave_theory" paths. But I thought the OP's confusion might be better cleared up first with reference to simpler waves, such as of a rope.
I have indeed problems understanding the mechanics of a wave in a rope as well, but I thought I will take that issue later...
 
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  • #10
cesiumfrog said:
Well, if the depth is finite, the water parcels do take http://en.wikipedia.org/wiki/Airy_wave_theory" paths. But I thought the OP's confusion might be better cleared up first with reference to simpler waves, such as of a rope.

You're right, I remember that now from somewhere. I agree that a stretched string or rope would be a better system to start with. I'm sure many classical mechanics textbooks derive the wave equation for a stretched string. I've seen it, but I don't remember the details and I don't have any of those books here at home.

For surface waves in deep water you probably have to deal with the surface tension, and I suspect that in some approximation you can consider the molecules as moving in very narrow vertical ellipses, i.e. practically in a vertical straight line. This would be for a propagating wave, relatively far from the source.

To take Mårten's example with the knife point touching the surface of the water to start the wave, the water next to the knife initially rises because of capillary action (I think that's what it's called) on the surface of the knife. This is what makes water rise by itself a bit in a narrow tube that you dip vertically into the water. The molecules next to the knife, in turn pull up their neighbors which then fall back a bit, and those neighbors pull up their neighbors, etc. I've never seen this "initial" process analyzed mathematically.
 
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  • #11
Mårten said:
Okey, let's think of this vertical oscillation then. And my example above with the water that is higher just close up to the knifepoint that was just now inserted into the water surface. Now, what force drives these molecules vertically down?
Btw, I know of course there is gravity. But since the water is higher here, there is also a normal force from the underlying molecules, so the total force should be zero downwards (assuming that the water cannot be compressed). So the molecules ought not to go straight downwards. Like a steep pile of sand that you construct - the grains of sand will fall down more diagonally, not vertically, since there are other grains of sand in the way, just below them.

jtbell said:
I'm sure many classical mechanics textbooks derive the wave equation for a stretched string.
They do, but mine doesn't explain properly how the forces come about, at least so I can understand it...

jtbell said:
To take Mårten's example with the knife point touching the surface of the water to start the wave, the water next to the knife initially rises because of capillary action (I think that's what it's called) on the surface of the knife. This is what makes water rise by itself a bit in a narrow tube that you dip vertically into the water. The molecules next to the knife, in turn pull up their neighbors which then fall back a bit, and those neighbors pull up their neighbors, etc. I've never seen this "initial" process analyzed mathematically.
I've been thinking of that, but I was thinking that you could disregard that fact. If you have dishwashing liquid instead (which I think doesn't have much of surface tension), and put down the knife point into the liquid, you will still have a wave, at least if you put down the knife point not too slowly. That was what I would expect at least...

Edit: Some reading makes me conclude that we cannot at all disregard surface tension and capillary effects, since what we are talking about here is actually called capillary waves, see:
http://en.wikipedia.org/wiki/Capillary_wave
 
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  • #12
I've now started a new thread, dealing with the maybe more easily understood mechanics of the wave in a rope - see here: https://www.physicsforums.com/showthread.php?t=274913

I've now understood that this area is actually extremely complicated - even Feynman admits that... :yuck: (see link above in my previous post).

Still, I insist in that there ought to be some form of conceptual answer, not too much contaminated by ugly formulas, to why the water molecules behave like circles, like they do in this figure below, and why they ultimately spread their motion to neigboring molecules, so that the wave propagates, which is the main thing I haven't understood yet.

http://www.kettering.edu/~drussell/Demos/waves/water.gif
 
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What are transverse mechanical waves?

Transverse mechanical waves are a type of wave that travels through a medium by oscillating perpendicular to the direction of the wave's motion. This means that the particles in the medium move up and down or side to side as the wave passes through.

What is the difference between transverse and longitudinal waves?

The main difference between transverse and longitudinal waves is the direction in which the particles of the medium move as the wave passes through. In transverse waves, the particles move perpendicular to the direction of the wave's motion, while in longitudinal waves, the particles move parallel to the direction of the wave's motion.

What are some examples of transverse mechanical waves?

Some examples of transverse mechanical waves include water waves, electromagnetic waves, and seismic S waves. In water waves, the particles of water move up and down as the wave passes through, while in electromagnetic waves, the electric and magnetic fields oscillate perpendicular to the direction of the wave's motion. Seismic S waves are a type of earthquake wave that causes particles in the ground to move side to side as the wave passes through.

How do transverse mechanical waves travel?

Transverse mechanical waves travel by transferring energy from one particle to the next as they pass through a medium. The particles do not travel with the wave but instead oscillate in place, causing neighboring particles to also oscillate and transfer the wave's energy.

What factors affect the speed of transverse mechanical waves?

The speed of transverse mechanical waves is affected by the properties of the medium through which they are traveling, such as density, elasticity, and temperature. The speed is also influenced by the frequency and wavelength of the wave. In general, waves travel faster in denser and more elastic mediums, and at higher frequencies and shorter wavelengths.

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