Consider yet again the 1D vibrating string, this time without a viscous medium but with spring ‘anchors’ – similar to a vibrating beam sitting on an elastic foundation

T

T

ψ

x

Elastic springs: providing a stiffness per unit length of σ

Conservative system: No energy loss

For an element of length δx, the stiffness produces an extra transverse force of ( )txx ,ψσδ−

* Fading away

Evanescent* waves

Note that there is no energy loss here (unlike the viscous damping considered earlier)

Reworking the wave equation, it is straightforward to show that the equilibrium equation now becomes

ψσδψδψµδ xx

xTt

x −∂∂

=∂∂

2

2

2

2

orψωψψ 2

2

22

2

2

cxc

t−

∂∂

=∂∂

whereµTc = and

µσω =c

As usual , we investigate a forward travelling wave solution:

( )kxtjeAV −= ω

2222ckc ωω +=This time we find the

Dispersion Relation

So the wave number isc

k c22 ωω −

±=

If ω > ωc k real: is what we have looked at before

If ω < ωc k imaginary: this is the new situation jk α±=

We then have: ( ) tjxjxtj eAeeAtx ωααωψ ±± ==),(

Evidently, the solution +αx is physically unacceptable: the amplitude would increase with position

tjxeAetx ωαψ −=),( An evanescent wave

Phase (ωt) is now independent of position, x; like a standing wave but with exponential shape

Phase & Group Velocity of an evanescent wave2222

ckc ωω +=

222

1 ⎟⎠⎞

⎜⎝⎛−

=−

==

ωωωω

ωω

cc

pcc

kv

ck c

22 ωω −±=

Recall the dispersion equation:

Phase velocity:

We have

Group velocity:

kcdkd 222 =

ωω

212

1 ⎟⎠⎞⎜

⎝⎛−=⎟

⎠⎞

⎜⎝⎛⋅⋅===

−

ωωω

ωω c

g ck

cckcdkdv

Note that:

0,2 == cgp cvv ω if dispersive-non is and usual, As

Water WavesIs complex because of surface tension, gravity, the constraint of the ocean bed, … We might expect displacements of the form

y

xψx

ψy

)sin()(),,(

)cos()(),,(

x

y

kxtyAtxy

kxtyAtxy

x

y

−=

−=

ωψ

ωψ

h: the sea depth

Incompressible

Irrotational (no viscosity)

122

=⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞

⎜⎝⎛

x

x

y

yAA

ψψNote that:

Surface y=0

Dispersion Relation for water waves

To derive the dispersion relation requires that we apply Bernoulli’s theorem, which states that the total energy per unit mass has the same value at each point along a given streamline (the path followed by a particle in steady-state flow. We will NOT derive the dispersion relation here (there are three slides at the end that give an outline of how to do it if you are interested).

Dispersion relation relates the circular frequency to the wave number k. This is not straightforward for water waves!!!

2ω

khkgk tanh32 ⎟

⎠⎞⎜

⎝⎛ += ρ

σω

Gravity termSurface Tension term

The dispersion relation for water waves:

Generally, we can ignore the surface tension term – and will do so in a minute. The surface tension term is significant only for short wavelengths - which we call ripplesk/2πλ =

At 20oC, σ = 0.073Nm-1, for water, and ρ = 1000 kgm-3 so the critical wavelength is

Waves much shorter than this will be dominated by surface tension

λ = 17 mm

Ripples

The two terms in brackets are equal when

ρσ 3kgk =

21

2 ⎟⎠⎞⎜

⎝⎛= gρσπλ

sinceλπ2

=k

khkgk tanh32 ⎟

⎠⎞⎜

⎝⎛ += ρ

σωRecall thedispersion relation

Neglect the gravity term, and assume kh >> 1 (true for all but the shallowest puddles) 1tanh ≈kh

21

3⎟⎠⎞⎜

⎝⎛≈ ρσω k

The Phase Velocity of the ripples is

21

21

2 ⎟⎠⎞⎜

⎝⎛=⎟

⎠⎞⎜

⎝⎛≈= ρλ

πσρ

σω kk

vp

- shortest waves travel fastest

The Group Velocity is

pg vkdkdv

23

23 2

1

=⎟⎠⎞⎜

⎝⎛≈= ρσω

Since vg > vp, individual ripples appear to travel backwards through the parent group as it propagates

Any waves whose wavelength are significantly longer than 17mm can be considered gravity waves, and we ignore the surface tension term

Gravity Waves in Deep Water

khkgk tanh32 ⎟

⎠⎞⎜

⎝⎛ += ρ

σω

1tanh ≈kh

As we noted earlier, any wave much larger than 17 mm is a gravity wave – surface tension is negligible, and the dispersion relationship reduces to

( ) 21

gk≈ω

For a typical example of ocean swell (λ ~ 100m so vp ~ 10ms-1)

Typical example of a deep-water wave is ocean swell

Recall the dispersion equation for water waves

In this case also, kh >> 1, because h is huge so

21

21

2⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛≈=

πλω g

kg

kvp

Phase velocity

pg vkg

dkdv

21

21 2

1

=⎟⎠⎞

⎜⎝⎛≈=

ω

Group velocity

)sin()(),,(

)cos()(),,(

x

y

kxtyAtxy

kxtyAtxy

x

y

−=

−=

ωψ

ωψ

122

=⎟⎠⎞⎜

⎝⎛+⎟

⎠⎞

⎜⎝⎛

x

x

y

yAA

ψψ

Water motion for deep waves

In general:

From which we noted:

It follows that all the water particles move in circles

For deep waves, it is reasonable to suppose that since there is no obviously preferred direction for the depth of the wave (x,y)

AyAyA xy == )()(

Shallow water wavesA typical example is sea-shore waves.

If the beach shelves smoothly, h gradually decreases, so the approaching waves slow down, so their amplitude increases (conservation of energy) until they break – surf.

khkgk tanh32 ⎟

⎠⎞⎜

⎝⎛ += ρ

σωRecall once more the dispersion relation:

In this particular case, kh can be assumed to be small, so:

( )3

31tanh khkhkh −≈

Or approximated further: 22 ghk=ω

From this it is easily seen that: ghvv gp ≈≈

and so shallow water waves are approximately non-dispersive

TsunamiThe first recorded tsunami was in 479BC, and up to the present day there have been approximately 300.

Highest recorded was 75m high in Alaska (1964) following an earthquake in Prince William Sound.

Highest speed to date is 490mph.

Aleutian Islands 1946, waves 6.1m high

Indonesia/Thailand tsunami 2004

Following a 9.0 magnitude earthquake off the coast of Sumatra, a massive tsunami and tremors struck Indonesia and southern Thailand Lanka on 26th December 2004, killing over 104,000 people in Indonesia and over 5,000 in Thailand.

Images from the web site: http://www.disasterscharter.org/disasters/CALLID_079_e.html

Khao Lak, Phuket Island, one year pre-disaster

Tourist hotels

Forest plantations

December 29th, Ikonos satellite image

Tsunami

Produced by submarine earthquake, leading to waves of massive wavelength, but small amplitude (~1m).

In fact, the wavelength λ is so larget that the product kh = 2πh/λ << 1 even in the Pacific ocean. That is, a tsunami can be considered a ‘shallow water’ wave. So:

ghvv gp ≈≈

In an ocean of depth 4km, say,

As the wave approaches the shore, h decreases sharply, so the amplitude increases massively.

Worse, as it is a ‘shallow’ water wave, it is nearly non-dispersive, so the energy is concentrated into a few massive wave crests

-13 ms20010104 =⋅⋅== ghvp

The End

Outline derivation of the dispersion relation for water waves

Bernoulli’s theorem states that the total energy per unit mass has the same value at each point along a given streamline (the path followed by a particle in steady-state flow

VvpW ++= 2

21

ρ

PressureFluid speed Potential energy per

unit mass

We now apply this along a streamline at the surface of the water

For this course, just accept this as a given!! Evidently, there’s no point solving a PDE unless the equation makes sense for the physics/engineering of the situation to hand.

For the potential energy, V we assume that only gravity is involved:

ψgV =

2

2

xpp a ∂

∂−=

ψσ

The velocity, v, on the surface y=0

),0(),0(),( 222 xvxvxyv xy +=

Pressure depends on surface tension, s (N/m), and atmospheric pressure, pa

Surface tension gives a pressure difference of σ x radius of curvature

In Bernoulli’s theorem, we have the term

ψρσ

ρ ⎟⎟⎠

⎞⎜⎜⎝

⎛++= gkpW a 2

And this must be independent of x

Gravity & surface tension

khkgk tanh32 ⎟

⎠⎞⎜

⎝⎛ += ρ

σω

Gravity termSurface Tension term

Generally, we can ignore the surface tension term – and will do so in a minute. The surface tension term is significant only for short wavelengths - ripples

This equation is called the Dispersion Relation for water waves

Adding in the boundary conditions (incompressible, irrotational, zero displacement of the ocean bed, …) we (eventually) find* that

*you have to take my word for it…k/2πλ =