# Tsunami Wave

by mysearch
Tags: tsunami, wave
 PF Gold P: 513 Hi, I was wondering if anybody is in a position to resolve some questions about tsunami waves as they relate to the general physics of mechanical waves. I will briefly try to outline the issues: 1. The energy of a mechanical wave, i.e. one dependent on the physical interaction of the particles in the propagating medium is normally said to be proportional to the square of the amplitude of the wave. 2. Wikipedia has a page on ocean waves that gives a generalised solution of a wave’s propagation velocity: http://en.wikipedia.org/wiki/Ocean_surface_wave. However, this solution can be simplified for discussion by considering the following deep and shallow depth solutions: [1] $$v = \sqrt { \frac {g\lambda}{2 \pi} }$$ deep water [2] $$v = \sqrt { gd}$$ shallow water 3. Now descriptions of tsunami waves suggest that the amplitude of the wave is relatively small, e.g. 1 metre, but the wavelength is very long, e.g. 10-100km, in deep water. As such, equation [1] would suggest that these waves also move very fast. 4. However, if tsunami waves are classified as mechanical waves, they seem to conflict with the energy of the wave being proportional to the square of the amplitude.I have some thoughts of why this might be the case, but would like to get some general feedback on the specific issue associated with energy in bullet (4) above: 5. At some level, the energy of the wave seems to be related to the volume of water displaced by the crest and trough of the wave. In the case of tsunami waves, the energy-volume is still very large, even though the amplitude is small, because the wavelength is so long. 6. As the tsunami wave approaches shallow water, equation [2] would restrict the propagation velocity is a function of the depth of the water, not its wavelength. 7. Does the amplitude of the wave grow to compensate, i.e. is this some form of conservation of energy? 8. While I assume that some wave energy must be lost to friction in shallow water, is it correct to assume, in this specific case, that the wave energy is still characterised by the volume of water displaced, i.e. is this representative of potential energy? 9. Finally, does the propagation velocity of a mechanical wave reflect some sort of notion of the kinetic energy associated with the wave? I am asking this question because a SHM wave model cycles between potential and kinetic energy maximaWould appreciate any clarification of any of the issues raised. Thanks
 PF Gold P: 513 By way of an addendum to my original post, I am trying to consider a tsunami wave in the context of a 2-dimensional surface wave model, which radiates outwards from some central event. Clearly, this event has to inject the necessary energy into the system to trigger the wave in the first place, e.g. o Seismic event, ocean floor sinks o Meteor strike Is correct to assume that some portion of the energy of the originating event is transported away in the form of the tsunami wave, analogous to a large pebble in a large pond? Now, in the pebble model, the initial energy is also being distributed over an ever-increasing circumference, such that the amplitude at any point is proportional to: [1] $$A \propto \frac{1}{\sqrt{radius}}$$ However, the issue of interest is whether a tsunami wave corresponds to this general model of a 2D surface wave in terms of the energy being proportional to the square of the amplitude of the wave?
 Sci Advisor P: 2,506 In response to points 7 and 8; the growth of wave amplitude si usually described as being a result of a focussing of the wave energy, so I would say yes, this is a sort of conservation of energy, and the volume of water displaced tries to remain constant by trading off wavelength for amplitude (wave height).
HW Helper
P: 8,961

## Tsunami Wave

7, 8 explain why a tsunami 'rears up' when it hits the coast.
I assume that if there was an island with no shallow water - just a steep volcanic peak in mid ocean (like La Palma) - then the tsunami would flow past with only a temporary small rise in sea level?
 P: 745 Some quick thoughts that may require confirmation: 1. I believe this is true of a wave expanding with a spherical wave front. However, a tsunami is a surface wave and therefore has a cylindrical wavefront. This means that it loses energy proportional to the radius only, i.e. not the square of the radius, such that it does not attenuate as much due to geometric spreading as a spherical wave would. 2. Are those equations for the group or the phase velocity? The main package of energy travels at the group velocity so really this is the velocity that concerns us. The good thing is that the group velocity is slower than the phase velocity, so if you got it wrong and calculated the time the wave was gonna hit you based on the phase velocity, you'd be the "right side" of wrong, if you catch my drift. As the wave approaches shallower water, the group velocity of the front of the wave slows down but the back is still moving quicker so that the wave length is squished. As a direct consequence of this the amplitude of the wave increases simply by conservation of energy.
HW Helper
P: 8,961
 Quote by billiards However, a tsunami is a surface wave and therefore has a cylindrical wavefront. This means that it loses energy proportional to the radius only, i.e. not the square of the radius, such that it does not attenuate as much due to geometric spreading as a spherical wave would.
It could be even worse. If the Tsunami is caused by a landslip and that is long compared to the wavelength then it could be a sort of colimated beam

 2. Are those equations for the group or the phase velocity? The main package of energy travels at the group velocity so really this is the velocity that concerns us. The good thing is that the group velocity is slower than the phase velocity, so if you got it wrong and calculated the time the wave was gonna hit you based on the phase velocity, you'd be the "right side" of wrong, if you catch my drift.
Isn't a Tsunami a soliton?

 As the wave approaches shallower water, the group velocity of the front of the wave slows down but the back is still moving quicker so that the wave length is squished. As a direct consequence of this the amplitude of the wave increases simply by conservation of energy.
Or just conservation of mass. It's really a single wave - a hump of water 1m high and 100km long, it's that extra few cubic km of water arriving that does the damage
P: 2,506
 Quote by mgb_phys 7, 8 explain why a tsunami 'rears up' when it hits the coast. I assume that if there was an island with no shallow water - just a steep volcanic peak in mid ocean (like La Palma) - then the tsunami would flow past with only a temporary small rise in sea level?
I can't say for certain, but I believe that is incorrect. The tsunami travels across the deep ocean as a long wavelength with low amplitude, then the leading edge encounters a cliff face. Rather than slowing down, like when approaching a normal beach, the front of the wave abruptly stops. I believe this will cause the remainder of the wave to mount up much more suddenly than it does when approaching land through shallow water.
PF Gold
P: 513
Thanks for all the useful points raised. A few quick comments in response:
 Quote by LURCH In response to points 7 and 8; the growth of wave amplitude si usually described as being a result of a focussing of the wave energy, so I would say yes, this is a sort of conservation of energy, and the volume of water displaced tries to remain constant by trading off wavelength for amplitude (wave height).
I think that in the context of this Earth bound system, the conservation of energy would apply, but becomes very difficult to resolve due to factors such as ‘focussing and seabed drag etc, especially in shallow water. This was why I was focusing on the deep water solution.
 Quote by mgb_phys I assume that if there was an island with no shallow water - just a steep volcanic peak in mid ocean (like La Palma) - then the tsunami would flow past with only a temporary small rise in sea level?
I tend to agree, but I think it depends on what scale you want to look at this problem. See post #7 for a slightly different take.
 Quote by billiards 1. I believe this is true of a wave expanding with a spherical wave front. However, a tsunami is a surface wave and therefore has a cylindrical wavefront. This means that it loses energy proportional to the radius only, i.e. not the square of the radius, such that it does not attenuate as much due to geometric spreading as a spherical wave would.
Thanks, I made a typo, the equation should have been with respect to amplitude, i.e. the energy is the square of the amplitude, but the energy is distributed over the circumference of the wavefront at radius [r]:

$$A^2 = E/2 \pi r$$
$$A \propto 1/\sqrt{r}$$ I corrected post2 to avoid confusion
 Quote by billiards 2. Are those equations for the group or the phase velocity? The main package of energy travels at the group velocity so really this is the velocity that concerns us. The good thing is that the group velocity is slower than the phase velocity, so if you got it wrong and calculated the time the wave was gonna hit you based on the phase velocity, you'd be the "right side" of wrong, if you catch my drift.
According to the Wikipedia reference, it’s the phase speed, which make sense because the Tsumani seems to correspond to a fundamental wave of long wavelength that would quickly leave behind any secondary modulated waves of shorter wavelength associated with a group velocity.
 Quote by billiards As the wave approaches shallower water, the group velocity of the front of the wave slows down but the back is still moving quicker so that the wave length is squished. As a direct consequence of this the amplitude of the wave increases simply by conservation of energy.
I think this is a good practical description, i.e. the front of wave slows first and as the back of the wave catches up, it causes the wave height to increase. The actual helight is determined by other physics, e.g. individual waves break when their wave height H is larger than 0.8 times the water depth h
 Quote by mgb_phys Isn't a Tsunami a soliton?
Thanks. I hadn’t heard this term before. http://en.wikipedia.org/wiki/Soliton

Again, thanks for the comments, but my main issue is still to try and get some physical understanding of the general assertion of wave mechanics that the energy of a mechanical wave is the square of its amplitude. However, when I looked at a tsunami wave, purely a practical example, the energy also seems dependent on the wavelength. I am making this statement because it seemed to me that the energy of this specific wave was reflected in the mass of the water displaced, which seems to be a function of not only the amplitude, but also the wavelength?
PF Gold
P: 2,283
 Quote by mysearch 1. The energy of a mechanical wave, i.e. one dependent on the physical interaction of the particles in the propagating medium is normally said to be proportional to the square of the amplitude of the wave. 2. Wikipedia has a page on ocean waves that gives a generalised solution of a wave’s propagation velocity: http://en.wikipedia.org/wiki/Ocean_surface_wave. However, this solution can be simplified for discussion by considering the following deep and shallow depth solutions: [1] $$v = \sqrt { \frac {g\lambda}{2 \pi} }$$ deep water [2] $$v = \sqrt { gd}$$ shallow water 3. Now descriptions of tsunami waves suggest that the amplitude of the wave is relatively small, e.g. 1 metre, but the wavelength is very long, e.g. 10-100km, in deep water. As such, equation [1] would suggest that these waves also move very fast. 4. However, if tsunami waves are classified as mechanical waves, they seem to conflict with the energy of the wave being proportional to the square of the amplitude.I have some thoughts of why this might be the case, but would like to get some general feedback on the specific issue associated with energy in bullet (4) above: 5. At some level, the energy of the wave seems to be related to the volume of water displaced by the crest and trough of the wave. In the case of tsunami waves, the energy-volume is still very large, even though the amplitude is small, because the wavelength is so long. 6. As the tsunami wave approaches shallow water, equation [2] would restrict the propagation velocity is a function of the depth of the water, not its wavelength. 7. Does the amplitude of the wave grow to compensate, i.e. is this some form of conservation of energy? 8. While I assume that some wave energy must be lost to friction in shallow water, is it correct to assume, in this specific case, that the wave energy is still characterised by the volume of water displaced, i.e. is this representative of potential energy? 9. Finally, does the propagation velocity of a mechanical wave reflect some sort of notion of the kinetic energy associated with the wave? I am asking this question because a SHM wave model cycles between potential and kinetic energy maximaWould appreciate any clarification of any of the issues raised. Thanks
Hi mysearch,

The energy is generally proportional to the square of the amplitude and the wavelength or $E \propto \lambda a^2$. Since tsunamis have ridiculously long wavelengths it should seem reasonable to compare them. However, remember that ocean waves are highly non-linear by nature and that simplified models are only estimates for how they behave.

Nonetheless tsunamis are a wave phenomenon and will therefore follow certain physical laws defined for all waves. Tsunamis behave physically as shallow-water waves (i.e. non-dispersive) so their phase velocity = group velocity = $\sqrt{gd}$. Of course this is just an approximation mind you. Plainly the velocity is tremendous for deepwater waves. This large velocity also indicates that they have an enormous amount of energy even though they have a small amplitude.

The rate of energy loss is typically proportional to the inverse of the wavelength. Thus tsunamis dissipate very little energy. In the absence of bottom friction, like in deep water, the wave maintains its large amount of energy. As the tsunami approaches the coast it will begin to slow and experience a run-up creating large surges.

To give a better understanding of how much energy is in the tsunami, it may be helpful to look at how much energy it takes to create it (and hence transferred to the wave). The energy in a tsunami wave can be defined mathematically as:

$$E = \frac{1}{2}\rho g \lambda La^2$$

By inspection we see that the energy is indeed a function of the wavelength and the amplitude as suspected.

Hope this helps.

CS
PF Gold
P: 513
Hi Stewartcs,

Thank you very much, that was extremely helpful. I had been struggling to find any useful references in this area. However, could I just ask for some additional clarifications on a few of the points you made?

$$E = \frac{1}{2}\rho g \lambda La^2$$

First, could you clarify the nature of the parameters [rho] and [L] as I am interested in whether they reflect the mass of water displaced in the originating event, which caused the wave. See my following diatribe for details.

By way of background, my interest is not directly with tsunami waves, I just happen to come across a description of them, which I couldn’t resolve in terms of general wave mechanics, i.e. the general statement that energy (E) is proportional to the square of the amplitude (A). I haven’t seen any mention of wavelength in standard text up until now. I was also looking into mechanical wave as opposed to EM wave, because of the different correlation to energy, i.e. Planck equation E=hf. Clearly, EM are also capable of propagate through a vacuum, whereas mechanical waves depend on ‘vibrations’ of neighbouring particles within the physical propagating media.
 Quote by stewartcs The energy is generally proportional to the square of the amplitude and the wavelength or $E \propto \lambda a^2$. Since tsunamis have ridiculously long wavelengths it should seem reasonable to compare them. However, remember that ocean waves are highly non-linear by nature and that simplified models are only estimates for how they behave.
I think I understand the caveats concerning linearity, but is your statement concerning energy being related to amplitude and wavelength generally true for all mechanical waves? I ask because this seems to be of fundamental importance to anybody trying to gain some physical understanding of the nature of mechanical waves. For example, it seemed to me, although I admit this is just speculation, that a mechanical wave starts by acquiring energy from outside the wave system, which it then propagates through a medium. If I use the pebble-in-the-pond analogy, an initial central wave is created, e.g. meteor strike or seismic shift in the seabed, which then radiates outwards. However, given that there is no net propagation of water, only wave energy, does this energy correlate to the potential energy of the source?

What I mean by this is that the original event caused a large displacement of water, which had mass (m) and subject to gravity [g] due to its height (A) and width ($$\lambda$$). Therefore, if this displacement disappears, i.e. the water returns to a flat surface, it would seem logical to assume that the wave carried away most of this potential energy. Of course, in the case of a radiating circular wave, the initial central energy of the water is dissipated across the increasing radius of the circumference wave.

Again, I am raising these points, because I am interested in trying to understand the interplay of potential and kinetic energy in mechanical waves. In a SHM model, the wave function persists because energy is converted from potential energy to kinetic energy and back again, assuming no loss to fiction etc. However, it seems that, in this case, the wave is propagating the original potential energy, although local particles within the propagating media are subject to kinetic movement, i.e. energy exchange.

I know I am pushing my luck, but I would also like to understand what determines the frequency/wavelength of a mechanical wave in the first place. For example, in the case of an EM wave in vacuum, energy defines the frequency, the vacuum media defines the propagation velocity [c] and therefore the wavelength is determined by virtue of $$c= f * \lambda$$. Is there any analogous logic to mechanical waves?

Anyway, again many thanks for all your previous input.
PF Gold
P: 2,283
 Quote by mysearch Hi Stewartcs, Thank you very much, that was extremely helpful. I had been struggling to find any useful references in this area. However, could I just ask for some additional clarifications on a few of the points you made? $$E = \frac{1}{2}\rho g \lambda La^2$$ First, could you clarify the nature of the parameters [rho] and [L] as I am interested in whether they reflect the mass of water displaced in the originating event, which caused the wave. See my following diatribe for details. By way of background, my interest is not directly with tsunami waves, I just happen to come across a description of them, which I couldn’t resolve in terms of general wave mechanics, i.e. the general statement that energy (E) is proportional to the square of the amplitude (A). I haven’t seen any mention of wavelength in standard text up until now. I was also looking into mechanical wave as opposed to EM wave, because of the different correlation to energy, i.e. Planck equation E=hf. Clearly, EM are also capable of propagate through a vacuum, whereas mechanical waves depend on ‘vibrations’ of neighbouring particles within the physical propagating media. I think I understand the caveats concerning linearity, but is your statement concerning energy being related to amplitude and wavelength generally true for all mechanical waves? I ask because this seems to be of fundamental importance to anybody trying to gain some physical understanding of the nature of mechanical waves. For example, it seemed to me, although I admit this is just speculation, that a mechanical wave starts by acquiring energy from outside the wave system, which it then propagates through a medium. If I use the pebble-in-the-pond analogy, an initial central wave is created, e.g. meteor strike or seismic shift in the seabed, which then radiates outwards. However, given that there is no net propagation of water, only wave energy, does this energy correlate to the potential energy of the source? What I mean by this is that the original event caused a large displacement of water, which had mass (m) and subject to gravity [g] due to its height (A) and width ($$\lambda$$). Therefore, if this displacement disappears, i.e. the water returns to a flat surface, it would seem logical to assume that the wave carried away most of this potential energy. Of course, in the case of a radiating circular wave, the initial central energy of the water is dissipated across the increasing radius of the circumference wave. Again, I am raising these points, because I am interested in trying to understand the interplay of potential and kinetic energy in mechanical waves. In a SHM model, the wave function persists because energy is converted from potential energy to kinetic energy and back again, assuming no loss to fiction etc. However, it seems that, in this case, the wave is propagating the original potential energy, although local particles within the propagating media are subject to kinetic movement, i.e. energy exchange. I know I am pushing my luck, but I would also like to understand what determines the frequency/wavelength of a mechanical wave in the first place. For example, in the case of an EM wave in vacuum, energy defines the frequency, the vacuum media defines the propagation velocity [c] and therefore the wavelength is determined by virtue of $$c= f * \lambda$$. Is there any analogous logic to mechanical waves? Anyway, again many thanks for all your previous input.
The parameters $\rho$ and L are representative of the mass of the wave. $\rho$ is the water density and L describes part of the volume of the water ($\lambda$ and the amplitude describe the rest). The initial seismic event causes a transfer of energy to the water by displacing it upward. This energy, for the most part, is carried along all the way to the coastline with little dissipation as previously noted in the form of a wave.

Note that the direction of the tsunami wave is mainly orthogonal to the direction of the seismic fault line.

Generally speaking the energy of a mechanical wave is typically not dependent on the wavelength, rather only proportional to the square of amplitude. However, I presumed you were only talking about tsunamis which behave slightly differently than a normal mechanical wave (although they are very similar as described above). The total energy initially transferred to the tsunami wave was given above as E, which is dependent on the wavelength.

Hope this helps.

CS
PF Gold
P: 513
Thanks for the clarification of the equation. The dependency on the mass of the wave makes sense, if the source of the energy being dissipated by the wave is proportional to the potential energy of the event that caused the wave.
 Quote by stewartcs Note that the direction of the tsunami wave is mainly orthogonal to the direction of the seismic fault line.
Does this preclude a tsunami being modelled on the pebble-in-the-pond analogy, i.e. a meteor in the middle of the Pacific causes a radially expanding tsunami?
 Quote by stewartcs Generally speaking the energy of a mechanical wave is typically not dependent on the wavelength, rather only proportional to the square of amplitude.
Actually, this is still my central issue. If mechanical waves have a potential energy source, then I don’t understand how amplitude alone is really representative. I accept that, as in the case of tsunami waves, the energy of a mechanical wave may well always be proportional to the square of the amplitude, but this seems to be only part of the equation.

$$v = \sqrt{\frac{T}{\mu}}$$

For example, I found the propagation formula above for a wave on a string, where [T] is tension and [mu] is linear mass, although I have not had much luck in finding any direct energy related equations for a string wave. However, just guessing, I would have thought that the energy of this type of wave would also be a function of the amplitude and wavelength linked to the linear mass of the string?

One final point of interest, while there are many sources that specify various equations for the propagation velocity [v] and the fundamental equation $$v= f \lambda$$, there seems to be little discussion of how to determine either the frequency or wavelength of a mechanical wave from this product relationship. I am assuming the type of media affects the outcome?
PF Gold
P: 2,283
 Quote by mysearch Does this preclude a tsunami being modelled on the pebble-in-the-pond analogy, i.e. a meteor in the middle of the Pacific causes a radially expanding tsunami?
No, not for a meteor strike.

 Quote by mysearch Actually, this is still my central issue. If mechanical waves have a potential energy source, then I don’t understand how amplitude alone is really representative. I accept that, as in the case of tsunami waves, the energy of a mechanical wave may well always be proportional to the square of the amplitude, but this seems to be only part of the equation. $$v = \sqrt{\frac{T}{\mu}}$$ For example, I found the propagation formula above for a wave on a string, where [T] is tension and [mu] is linear mass, although I have not had much luck in finding any direct energy related equations for a string wave. However, just guessing, I would have thought that the energy of this type of wave would also be a function of the amplitude and wavelength linked to the linear mass of the string?
Perhaps this will help:

"The amount of energy carried by a wave is related to the amplitude of the wave. A high energy wave is characterized by a high amplitude; a low energy wave is characterized by a low amplitude. The amplitude of a wave refers to the maximum amount of displacement of a particle on the medium from its rest position. The logic underlying the energy-amplitude relationship is as follows: If a slinky is stretched out in a horizontal direction and a transverse pulse is introduced into the slinky, the first coil is given an initial amount of displacement. The displacement is due to the force applied by the person upon the coil to displace it a given amount from rest. The more energy that the person puts into the pulse, the more work which he/she will do upon the first coil. The more work which is done upon the first coil, the more displacement which is given to it. The more displacement which is given to the first coil, the more amplitude which it will have. So in the end, the amplitude of a transverse pulse is related to the energy which that pulse transports through the medium. Putting a lot of energy into a transverse pulse will not effect the wavelength, the frequency or the speed of the pulse. The energy imparted to a pulse will only effect the amplitude of that pulse"

Source: http://www.glenbrook.k12.il.us/gbssc...es/u10l2c.html

 Quote by mysearch One final point of interest, while there are many sources that specify various equations for the propagation velocity [v] and the fundamental equation $$v= f \lambda$$, there seems to be little discussion of how to determine either the frequency or wavelength of a mechanical wave from this product relationship. I am assuming the type of media affects the outcome?
IIRC, the source of the energy that creates the wave and the medium (if any) in which it propagates determine these.

Hope this helps.

CS
PF Gold
P: 513
CS,
Really appreciate the help. I am trying to read deeper into the issues in the background, but it is useful to use this thread to resolve some the issues that initially puzzled me.
 Quote by stewartcs ”…Putting a lot of energy into a transverse pulse will not effect the wavelength, the frequency or the speed of the pulse. The energy imparted to a pulse will only effect the amplitude of that pulse….”
This seems a categorical statement, which I need to consider. However, one of the interesting things about wave mechanics is applying the theory to real-world situations.

For example, a weight on a spring acting as lossless SHM pendulum can be described in terms of a sine wave. The initial vertical displacement defines the potential energy being put into the system: $$1/2kA^2$$, which reduces to zero as the mass springs back through the equilibrium point, having been totally converted to kinetic energy at this specific point. In this example, the square of the amplitude of the wave seems to unambiguously reflect the potential energy at any position without reference to wavelength. This example can also calculate the angular frequency ($$w=\sqrt{k/m}$$) as function of 2 physical attributes of the media, i.e. elasticity and mass, which does indeed suggests that the media defines frequency, at least, in this case. However, there doesn’t appear to be any real propagation velocity, as the wave seems to be essentially a function of time only.

In the case of a surface wave, things don’t seem so straightforward. Presumably, these waves are more like a pulse than sine wave, although I am assuming a Fourier series would allow the pulse to be modelled as fundamental sine wave plus its harmonic components? It also seems difficult to associate the potential energy of this type of wave to a specific point, as per the example above, because this energy seems to associated with the pulse/wave as a whole, i.e. height (A) and width ($$\lambda$$) etc. The issues of any conversion between potential and kinetic forms also seems less obvious, but presumably is required in any real world propagation process? Finally, while we have touched on several equations that specify a propagation velocity for this type of wave, I haven’t come across any equation that actually allow the frequency/wavelength to be calculated as a function of the media ,as per the SHM, possibly it is too complex to generalise?

Anyway, many thanks for all the help.
 PF Gold P: 513 Hi, I was wondering if anybody was in a position to answer any of the additional questions about tsunami waves below. In post #9, the following equation was cited for the energy associated with a tsunami wave: $$E = \frac{1}{2}\rho g (\lambda L)a^2$$ http://www.scribd.com/doc/4633126/Fi...ics-of-Tsunami I tracked down a few papers that reference this equation and it appears to be based on a model of a wave with a volume of water referred to a waterberg`. However, it seems that the essence of this equation can be reduced to the classical equation for potential energy: $$E_p = mgh = \rho (x*y*z) * g * y/2 = \frac{1}{2} \rho g (x z) y^2$$ Where mass [m] is the volume [x,y,z] times density and [y/2] is the average height. Note: these equations still reflect that the energy (E) of the wave is proportional to the square of its amplitude, but the huge energy of a tsunami is really a reflection of the net volume of the wave based on its very long wavelength. However, this analysis appears to be predicated on the fact that most tsunami waves are caused by seismic events, where the wavefront generated is essentially aligned and orthogonal to the direction of a seismic fault line. These papers also suggest a number of assumptions that I would like to raise for clarification: 1. While tsunami waves exist in deep water, because the wavelength can be much greater than the sea depth, they are often modelled using the shallow water equations cited in post #1. 2. While deep water is said to be dispersive, shallow water appears to be modelled as non-dispersive. Given the assumption in bullet (1) does it follow that tsunami waves should be modelled as non-dispersive? 3. In a non-dispersive medium, the equation $$[v=f \lambda]$$ holds for all frequencies, as long as the characteristics of medium do not change, i.e. the phase velocity of all waves of different frequency is the same? 4. Can a tsunami be modelled as a Fourier series of different frequency components? 5. Would these frequencies travel at the same phase velocity based on bullet (3)?It is said that due to their long wavelength, tsunami waves dissipate very little energy. However, I have attached a diagram that is more representative of a circular expanding wave, which by definition would dissipate its energy over an ever-increasing circumference – see post #2/#8. 6. Is this a valid model for a tsunami wave? 7. While the initial energy would still be proportional to the volume of water displaced; would it make sense to model this example on the volume of a cone? 8. While the propagation of a wave involves a local conversion of potential energy to kinetic energy, i.e. initial wave height into a rate of change of amplitude, is it true to say that only potential energy is being transported by the wave? 9. With reference to the attached diagram; what would determine the very long wavelength generally associated with tsunami waves?Appreciate that all these questions may be excessive, but would appreciate any insights or clarification of any of the issues. Thanks Attached Thumbnails
PF Gold
P: 2,283
 Quote by mysearch 1. While tsunami waves exist in deep water, because the wavelength can be much greater than the sea depth, they are often modelled using the shallow water equations cited in post #1. 2. While deep water is said to be dispersive, shallow water appears to be modelled as non-dispersive. Given the assumption in bullet (1) does it follow that tsunami waves should be modelled as non-dispersive?
Yes, they should be modeled as non-dispersive.

Deep-water approximation refers to the depth of the water divided by the wavelength. Tsunamis have wavelengths of hundreds of kilometers, so they are shallow water waves.

However, most shallow water and deep water waves are somewhat dispersive. Tsunamis are just slightly dispersive so they are most accurately modeled as non-dispersive waves even though they may exist in deepwater. Only solitons are non-dispersive.

If you are looking for a Tsunami model, check out the NOAA Tsunami site. They give some details (IIRC) about how they do it.

Hope this helps.

CS
PF Gold
P: 513
 Quote by stewartcs Yes, they should be modeled as non-dispersive… Tsunamis are just slightly dispersive so they are most accurately modeled as non-dispersive waves even though they may exist in deepwater…. If you are looking for a Tsunami model, check out the NOAA Tsunami site.
Thanks. Approximating to non-dispersive will allow a simplification of the frequency spread by propagation velocity. Based on the shallow water model, can the propagation equation still be equated to frequency * wavelength, e.g.

$$v = \sqrt{gd} = f * \lambda$$?

I will assume that the circular model is possible for tsunamis and therefore the energy in successive radiating circular waves must fall as stated. However, is the ultra long wavelength primarily established, in most practical cases, by the seabed lifting/falling over a very large area?

I will take this displacement of a large volume of water as the initial potential energy input into the wave system, which is radiated away by successive circular waves. If I assume no net radial movement of water, the kinetic energy would seem to only exist in the localised rise and fall of the wave. However, it would seem that the energy of any section of the wave, in motion, at any radius must be the sum of this kinetic energy plus the potential energy associated with the amplitude of the wave and its wavelength. As such, the total energy being transported by the wave exists in both kinetic and potential forms.

Given that the kinetic energy exists as a function of the potential energy, i.e. wave amplitude, it will also be proportional to the square of the amplitude aggegated over the wavelength.
PF Gold
P: 2,283
 Quote by mysearch Thanks. Approximating to non-dispersive will allow a simplification of the frequency spread by propagation velocity. Based on the shallow water model, can the propagation equation still be equated to frequency * wavelength, e.g. $$v = \sqrt{gd} = f * \lambda$$?
Yes. In fact the approximate definitions for the phase speed are found from that relation.

Starting with the relations:

$$c = f \lambda$$

or

$$c = \frac{\omega}{k}$$

where,
c is the phase speed
k is the wave number

and

$$\omega^2 = gk^2d$$, which is the shallow water dispersion relation

We get:

$$c = \sqrt{gd}$$

CS

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