1

Model 3: Tidal wavesNote: there are a number of different types of wave:1)       Ripples (wavelength ~ between 1 and 2cm; governed by surface tension, fluid density ρ, and gravity g.2)       Surface waves (wavelength shorter than fluid depth h)3)       Tidal waves (wavelength > fluid depth h)

Also called tsunamis

Can be very big. Highest recorded tsunami occurred in Lituya Bay, Alaska. This was the result of a massive landslide that fell into the bay on July 9, 1958. The resulting wave surged up the slope on the other side of the narrow bay to a height of 518 m.

What usually causes them?

An offshore earthquake that moves the ocean bottom in the vertical direction. Waves propagate toward coast growing higher as the water becomes shallow.

2003/ 1

2

Imagine an array of impermeable panels in a narrow canal:

3

A wave enters from the right. Panels at the left move forward and rise.

4

The wave proceeds:

5

Cont.

6

Cont.

7

Cont.

8

Step 1: Model the problem and find an equation

Keep it simple (1 dimensional). Assume the wave is travelling up a canal of width b. The fluid is water so we can consider it to be incompressible. Imagine that we place two “floating” impermeable panels at x and x+Δx (Note: these are free to move in the x direction):

h

x x+x

figure 13: Side view of canal with single wave travelling in positive x direction.

9

The volume of water V1 between x and x+Δx is given

by:

xhbxxxhbV 1

10

A moment later the wave has progressed forward in x. The water has shifted a distance u from x so the first panel is now

located at x+u. The second panel is located at x+Δx+u+Δx(∂u/∂x):

x+u x+x+u+ x(∂u/∂x)

h+vh

The water level has risen an average level of v between the two panels. The volume of water between the panels can now be written:

x

uxxbvhux

x

uxuxxbvhV2

11

The expression u(x)+ x(∂u/∂x) is an estimator of the value of u(x+x)

u(x)

x x+x

u(x)

u(x)+ x(∂u/∂x) u(x+x)

∂u/∂x

12

Imagine impermeable panels

x+u x+xx x+x+u+ x(∂u/∂x)

u u+ x(∂u/∂x)

13

The fluid is incompressible so:

21 VV

Thus

x

uxxbvhxhb

We can expand the right hand side:

x

uxvxv

x

uxhxhb

x

uxxbvhxhb

14

Divide by b and simplify:

vx

uh

0

x

uxvxv

x

uxhxhb

x

uxxbvhxhb

Very small

xvx

uxh0

x

uhv

rearrange

(21)

15

Now consider the forces that act on the water:

 Consider a horizontal slab of water between panels located distance y above the bottom:

y

x x+x

h+vxh+vx+xy

16

The pressure at position x

and distance y above the

bottom is: yvhgP xyx ,

At location x+∆x the pressure at y is:

yvhgP xxyxx ,

x

y

h+vx-y h+vx+x-y

17

The force at each end equals the local pressure times the cross-sectional area by :

y

h+vx-y h+vx+x-y

ybyvhgF xyx , ybyvhgF xxyxx ,

From inspection we see that the net horizontal force F = -Parea = -P by= ybvvgybPPF xxxyxyxx ,,

18

From Newton we have:Force = mass × acceleration

y

h+vx-y h+vx+x-y

ybyvhgF xyx , ybyvhgF xxyxx ,

2

2

t

uybx

2

2

t

uybxybvvg xxx

Force = mass × acceleration

19

Simplify this expression:

2

2

t

uybxybvvg xxx

2

2

t

uxvvg xxx

Divide by Δx:

2

2

t

u

x

vvg xxx

Let Δx→0:

2

2

t

u

x

vg

(22)

20

Recall eqn. 21:

x

uhv

Differentiate both sides of eqn. 21 by x:

(21)

2

2

x

uh

x

v

(23)

Substitute eqn. 23 into eqn. 21:

2

2

2

2

t

u

x

ugh

2

22

2

2

x

uc

t

u

Rearrange

The now familiar wave equation:

With wavespeed is c: ghc (25)

21

Calculate tsunami speeds:

Average depth of ocean is approximately 4,500m. Therefore c= 210m/s or 756 km/hr. In 1000 m of water the speed reduces to 356 km/hr. A tsunami travels at relatively slow speeds in shallow water and fast speeds in deep water.

Sea

Continental shelf

Auckland

22

Standing waves in a fountainA periodic disturbance to water in a narrow channel can set up a

standing wave phenomenon. I witnessed this at a fountain in Auckland city. This was an example of a so-called self-excited

vibration.

Water jet

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The water jet was uncovered by the water wave once per cycle. The periodicity was fixed by the frequency of the tidal wave within the fountain. Once per cycle the water jet fed energy into the

vibration.

24

Step 1: The equationWe use the 1 dim wave equation:

2

22

2

2

x

uc

t

u

Step 2: The general solutionWe can reuse the solution (eqn. 8):

)cos()sin()cos(sin, tcDtcCxBxAtxu

25

Step 3: Boundary conditionsFor a standing tidal wave the water movement u at

each end boundary is zero:

x=0At x=0 u(0,t) = 0

0)cos()sin()()0(,0 tcDtcCBtTXtu

Thus B= 0

26

At x=L u(L,t) = 0

0)cos()sin(sin)()(, tcDtcCLAtTLXtLu

So sin(L) = 0

  or

L = nπ n= 1,2,3,….

L

ctnD

L

ctnC

L

xnAtxu nnnn

cossinsin,

L

ctnb

L

ctna

L

xntxu nnn

cossinsin,

27

1st mode

L

gh

L

c 1

L

ghf

21

L

ctb

L

cta

L

xtxu

cossinsin, 111