Wave Propagation: Experiences Beyond Equilibrium Position

[scoutfai]

I recall that when in school, teacher said that during a wave propagation, be it the longitudinal wave (like sound wave) or transverse wave (like water surface wave), particle in every position only oscillate along its equilibrium position, and will not experience translational motion. i.e., if the wave propagating to the right, a particle on the surface of water will only bounce up and down but will not moving to the right hand side. Same to the longitudinal wave like a spring compressing and elongating, a section of the spring will go left and right oscillating but will not translate to the right hand side.

However, how come in certain phenomena, such as the tsunami wave, the water wave translate so much and does not stay in equilibrium position?
Another example is the following video of vortex ring (it's in Japanese, just watch the video, ignore the language)
http://www.youtube.com/watch?v=h6LGiPCSx3A&feature=player_embedded
When a displacement is given by the wall of the box to the smoke gas, the smoke gas translate out of the box. Why can't the smoke remains in the box, oscillating in its equilibrium position, but because of the opening, the atmospheric air near to the opening also induced to oscillate, and so the vortex ring will still propagates despite the smoke remains in the box, why can't this phenomena happens? Wave theory fails here?

 

[Antiphon]

A vortex ring is not wave motion. It is a much more general form of motion governed by fluid dynamics.

Water waves are complex but for small amplitudes they obey fairly straitforward wave equations of the type you imagine.

The Tidal Wave (aka Tsunami) also obeys these until the shoreline is too shallow. Then the wave action changes to gross flow. Also, the earthquake is not a small displacement but a large step function with a big "DC" term. This is going to cause big mass flows where the water ends (shore).

 

[Bill_K]

transverse wave (like water surface wave), particle in every position only oscillate along its equilibrium position, and will not experience translational motion. i.e., if the wave propagating to the right, a particle on the surface of water will only bounce up and down but will not moving to the right hand side

Water waves are not transverse, even for water of infinite depth. For a standing wave the particle motion is a straight line, varying from vertical ("transverse") at the crests to horizontal ("longitudinal") at the nodes. In between the motion is slanted, a combination of both. For progressive waves the particle motion is an ellipse.

Water waves are complex but for small amplitudes they obey fairly straitforward wave equations of the type you imagine.

Water waves are totally different. Even small amplitude ones do not obey anything that looks like the usual wave equation. They result from a solution of Laplace's equation ∇²Φ = 0 with a time-dependent boundary condition imposed at the surface. They are dispersive, with ω² = gk tanh(kh), where h is the depth.

 

[AlephZero]

In water surface waves, each particle of water moves in a circle, forwards with the direction of the wave crest, then down, back, and up. The circles have the biggest radius at the surface and get smaller at bigger depths, down to zero radius at the sea bed.

Surface waves are different from many other types of waves because the wave speed depends on the depth of water. As the water gets shallower the speed decreases and (because of conservation of energy) the amplitude increases. The "tidal wave" along the coastline is the final part of a very big circle, when simple surface wave theory would predict the wave speed was zero and the amplitude was infinite.

 

[scoutfai]

A vortex ring is not wave motion. It is a much more general form of motion governed by fluid dynamics.

Water waves are complex but for small amplitudes they obey fairly straitforward wave equations of the type you imagine.

The Tidal Wave (aka Tsunami) also obeys these until the shoreline is too shallow. Then the wave action changes to gross flow. Also, the earthquake is not a small displacement but a large step function with a big "DC" term. This is going to cause big mass flows where the water ends (shore).

For the vortex ring, if instead of a small circular opening on one face of the box, the entire rectangle face of the box is removed. Will vortex ring still be generated (or perhaps vortex rectangle ring now?)?

If the box no longer exist, but just the face (membrane) which gives displacement to the smoke is present. When this face oscillates (like hit by hand, same kind of oscillation when the box still exist), will there be a air pressure wave (i.e. sound wave) exists this time?


As for the tsunami and earthquake, are you saying that wave theory only valid when the displacement is small? I doesn't seem to remember such restriction. Perhaps I forgot.

 

[Bill_K]

In water surface waves, each particle of water moves in a circle, forwards with the direction of the wave crest, then down, back, and up. The circles have the biggest radius at the surface and get smaller at bigger depths, down to zero radius at the sea bed.

Sorry, this is wrong. As stated above, the particle motion is an ellipse:

δx = Ak/ω cosh k(y+h) cos (kx + ωt)
δy = Ak/ω sinh k(y+h) sin (kx + ωt)

And at the bottom, y = -h, the motion does not go to zero, it becomes a horizontal line.

 

[scoutfai]

Water waves are not transverse, even for water of infinite depth. For a standing wave the particle motion is a straight line, varying from vertical ("transverse") at the crests to horizontal ("longitudinal") at the nodes. In between the motion is slanted, a combination of both. For progressive waves the particle motion is an ellipse.

Water waves are totally different. Even small amplitude ones do not obey anything that looks like the usual wave equation. They result from a solution of Laplace's equation ∇²Φ = 0 with a time-dependent boundary condition imposed at the surface. They are dispersive, with ω² = gk tanh(kh), where h is the depth.

"water wave is not transverse wave" All right this is new to me. Perhaps in school time the claim that water surface wave is a transverse wave is an approximation (if so, under what condition this approximation becomes good?)

Nevertheless, will the particle of such a water surface wave (the one you claim) still oscillates in relative to its equilibrium position? Or, they will eventually experience translational displacement?

 

[scoutfai]

In water surface waves, each particle of water moves in a circle, forwards with the direction of the wave crest, then down, back, and up. The circles have the biggest radius at the surface and get smaller at bigger depths, down to zero radius at the sea bed.

Surface waves are different from many other types of waves because the wave speed depends on the depth of water. As the water gets shallower the speed decreases and (because of conservation of energy) the amplitude increases. The "tidal wave" along the coastline is the final part of a very big circle, when simple surface wave theory would predict the wave speed was zero and the amplitude was infinite.

Hence, the simple surface wave theory must be wrong or inadequate to explain tidal wave, right? Because by observation we knew that when tidal wave strikes, the amplitude is huge but not infinite and the speed is definitely not zero.

Your explanation on the particle of water surface wave moving in circle seems very abstract for me to imagine, have you came across any website that has a decent graphic simulation of it?

 

[scoutfai]

Sorry, this is wrong. As stated above, the particle motion is an ellipse:

δx = Ak/ω cosh k(y+h) cos (kx + ωt)
δy = Ak/ω sinh k(y+h) sin (kx + ωt)

And at the bottom, y = -h, the motion does not go to zero, it becomes a horizontal line.

You claim that the water surface wave is not a transverse wave. But, how would you explain the phenomena where a stone is dropped onto the surface of a static water surface, upon contact, a circular wave propagating outward? If a very light particle indicator (such as a tiny piece of polystyrene) is placed at a distance away from the center, when the wave arrives at there, the indicator will only bounce up and down, and no nett translational displacement?

 

[olivermsun]

But actually you do notice the flotsam moving back-and-forth as well as up-and-down.

 

[seonshrestha]

why does mechanical transverse waves travel only in solid medium such as rope? and longitudinal mechanical waves travel in solid , liquid and gas?

 

[olivermsun]

Transverse waves in a solid work because, by definition, solids can support various stresses that fluids do not. If you stretch a rubber bar, it wants to resist and rebound -- it resists tension stress. You can also imagine deforming the rubber bar by translating (not compressing or stretching) the ends back and forth. The bar resists the shear stress as well. A bar shaped blob of water would just flow and relieve either kind of stress.

An important example of a transverse wave is the S-wave in seismology (also known as the "Secondary" wave although I always remember it as "shear"!) .

The rope is a little simpler because the shearing stress is not particularly important, but the longitudinal movement is still resisted by tension in the rope.

Longitudinal waves, on the other hand, can be supported by pressure changes alone, so they work in solid, liquid, and gas.

Of course, longitudinal and transverse waves are not the only possible waves in these materials. The most familiar examples of waves for many of us are surface gravity waves, which are neither!

 

[seonshrestha]

thanx a lot sir.