Linear Wave Theory: Regular Waves, Study notes of Engineering Physics

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LINEAR WAVE THEORY

PART A

Regular waves

HARALD E. KROGSTAD

AND

ØIVIND A. ARNTSEN

NORWEGIAN UNIVERSITY OF SCIENCE AND TECHNOLOGY

TRONDHEIM

NORWAY

February 2000

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1 INTRODUCTION

These notes give an elementary introduction to linear wave theory. Linear wave theory is the core theory of ocean surface waves used in ocean and coastal engineering and naval architecture. The treatment is kept at a level that should be accessible to first year undergraduate students and does not require more than elementary calculus, probability and statistics.

Part A will cover the linear theory of regular gravity waves on the surface of a fluid, in our case, the surface of water. For gravity waves, gravitation constitutes the restoration force, that is the force that keep the waves going. This applies to waves with wavelengths larger than a few centimeters. For shorter surface waves, capillary forces come into action.

Chapter 2 covers basic wave motion and applies to all kind of waves. In the following chapter we briefly discuss the equations and boundary conditions which lead to water waves. Plane waves are treated in detail and simple superposition is also mentioned. We then proceed to three dimensional waves.

The notes are rather short in the sense that they discuss the equations rather than the applications.

Most of the material covered may be found in standard textbooks on the topic, see the references.

2 BASIC WAVE MOTION

The sine ( or cosine ) function defines what is called a regular wave. In order to specify a

regular wave we need its amplitude , a , its wavelength, λ, its period, T, and in order to be fully

specified. also its propagation direction and phase at a given location and time. All these concepts will be introduced below.

If now t starts to increase, the points [ 0 ( )W defined by [ 0 ( ) /W λ = W / 7 will have the property

that η( [ 0 ( ), )W W = 0 for all t. The point where η is 0, [ 0 , thus moves with velocity λ/T along the

x-axis. The last property stated above shows this in general.

Exercise 2.1: Consider the functions

η 1 = sin(ω W −N[)

and

η 2 = sin(ω W +N[)

where both k and ω are larger than 0. Show that the first represents a wave moving to the right

and the second a wave moving to the left!

An additional angle α in the expression η = D sin(ω W + N[+α) is called a phase term. Show

that a phase term does not affect the wavelength, the period or the propagation direction of the wave.

Exercise 2.2: Show that if we allow ω and k to be negative and arbitrary phase terms to be

included, all functions

D sin(N[ −ωW)

D sin(N[ + ωW +π / 4 )

D cos( ω W − N[ ) + E sin( ωW −N[)

may be written D ’sin( ω ’W − N [’ + α’)for appropriate choices of D’, ω’, N’ and α ’.

The argument of the sine, i.e. ω W − N[+ α, is in general called the phase. The phase is often

denoted by the letter φ (Greek phi ). Since sin( φ + 2 Qπ ) = sin( ),φ Q =L − 1 0 1, , ,L phase

differences of any multiple of 2 do not matter at all. The phase of the point ( [ 1 , W 1 )will be equal to the phase of the point ( [ 2 , W 2 )if

ωW 1 + N[ 1 = ωW 2 +N[ 2

that is, [ [ W W N 7

2 1 2 1

= ω^ =λ

or,

[ [ N

2 =^1 +^ ω^ (^ W^2 −W 1 )

The point [ 2 on the x-axis which moves with velocity ω / N will therefore experience the

same phase for all times. Therefore, the velocity F = ω / N =λ/ 7 is called the phase velocity

associated with the wave.

Let us see what happens if we add two general waves, one travelling to the right and one to the left. We first recall the trigonometric identity

sin( $) + sin( % ) = 2 sin( $^ +^ %^ ) cos($^ −% 2 2

Consider

μ( , )[ W = sin( ω W − N[) + sin( ω W + N[ ) = 2 sin( ω W ) cos( N[)

This function is a product of a sine and a cosine; the first with x as argument and the second with t. Figure 2 shows a plot of the functions for different t’ s.

Fig. 2: The standing wave

Note that the function is always 0 for N[ = Q π , Q= L− 1 0 1, , , L. In this case we therefore do

not have a travelling wave. However, since we still have a periodic behaviour both in x and t, it is customary to call this case a standing wave.

Exercise 2.3 : The phase velocity for light waves is equal to 300 000 km/s. The periods for FM-band broadcasting range from (1/88)·10-6 s to (1/108)·10-6 s. What are the wavelengths of such waves?

Review questions:

1. What is a regular wave?
2. How is the period and the wavelength defined?
3. What is the amplitude of the wave?
4. How do we define the wavenumber and the angular frequency?
5. Why do we call the velocity of the wave the phase velocity, and how can we derive the phase velocity?
6. How can we make up a standing wave?

Note that vectors are written with bold letters. u and w are thus the x - and the z -components of the velocity.

Water is hard to compress, and for our purpose, we will assume that this is impossible, that is, we consider water to be incompressible. In an incompressible fluid, the velocity Y = ( , ,X Y Z ) at each point will satisfy the equation ∂ ∂

X =

[

Y

\

Z

]

called the equation of continuity.

In our case the y -component of the velocity, v, is assumed to be zero, that is, we do not assume any variations across the channel.

If, in addition, the fluid is considered to be irrotational, the velocity may be expressed in terms of a so-called velocity potential Φ such that

X [ Y \ Z ] = ∂Φ ∂ = ∂Φ ∂ = ∂Φ ∂

The concepts "irrotationality" and "velocity potential" are treated in courses in Fluid Mechanics, and also in about every textbook about water waves. If we introduce the velocity potential in the continuity equation ∂ ∂

X =

[

Z

]

we obtain

∂ ∂

2 2

2 2 0

[ ]

(In the general case, we should also add a term ∂ 2 Φ / ∂^2 \ .)

This equation is a very famous partial differential equation called the Laplace equation.

It turns out that Laplace equation is all we have to know about the water motion away from the boundaries and the surface.

The bottom of the channel is not permeable to the water, and therefore the vertical water velocity at the bottom must be zero at all times :

Z [ ] K W

]

( , = − , ) = ∂Φ( ,[ ] K W, )

This constitutes a relation which must hold at the boundary, and it is therefore called a boundary condition. For the moment we assume that the channel very long so that we do not have to bother about conditions and the far ends.

The conditions at the water surface are harder to obtain. It has been observed that the fluid near the surface remains near the surface during the wave motion as long as the motion is smooth. That is, unless the waves break. The first boundary condition at the free surface consists of stating this property in mathematical terms. Consider a part of the surface at two neighbouring times as indicated in Fig. 3.2.

Figure 3.2: Motion of a fluid point on the free surface

The point at ( [ 1 , η( [ 1 , W 1 )) moves with velocity v to ( [ 2 , η( [ 2 , W 2 )) during the time interval

∆W = W 2 −W 1. Thus,

[ W [ W Z W W

[ [ X W W

2 2 1 1 2 1 2 1 2 1

Let us also expand η( [ 1 , W 1 ) in a Taylor series:

η ( [ , W ) η( [ , W ) η( , )( )

[

2 2 =^1 2 + ∂ [^1 W^2 [^2 [ 1

− +L

If this is introduced in first equation above, we obtain

(has to hold at the surface ] =η( , )[ W ).

4. The pressure in the fluid at the free surface is equal to the atmospheric pressure: ∂Φ ∂

W

1 X Z J

(has to hold at the surface ] =η( , )[ W ).

The mathematical problem stated in (1) to (4) is very difficult. No complete solution is known, although we know a lot about special cases.

The next section will treat the case where the magnitude of η( , )[ W is very small compared to

the variations in the x -direction (For a wave we would say that the amplitude is small compared to the wavelength).

Review questions:

1. What is the equation of continuity?
2. What is the velocity potential?
3. Which equation must the velocity potential satisfy? What is it called?
4. Which condition must hold at the bottom of the channel?
5. What is a boundary condition?
6. What is the physical content of the kinematic boundary condition?
7. How can we state the kinematic boundary condition in mathematical terms?
8. What is the physical content of the dynamic boundary condition?

4. SMALL AMPLITUDE WAVES

The equations stated in the previous section are much too complicated to be solved in full generality. We are going to linearize the equation and the boundary condition, and in order to do so, we shall apply a useful technique called dimensional analysis and scaling. Actually, in most textbooks, the linearization is treated very briefly.

Assume that the typical length scale for variations in the x -direction is L (for ocean waves, L could typically be of order 100m which we write as O (100m) ). Assume further that the time scale is T. (This could be a typical wave period which for ocean waves would be around 8s.)

The amplitude is of the order A, that is, η = 2 ( $). The two physical parameters in our

problem are

h = the mean water depth

and

g = the acceleration of gravity.

(It turns out that water density and viscosity, which did not occur in our equations anyway, are of virtually no significance.)

From the five quantities

L, T, A, h and g

we may form three dimensionless combinations :

1

2

3

2

K

J

(There are other possible combinations but these turn out to be the most convenient).

The small amplitude gravity waves case is when π 1 << 1 , that is, when A << L, and gravity is

essential, that is, π 3 = 2 ( ) 1.

The appropriate water velocity scale follows from the vertical motion of the surface. Thus the scale for | v | is A/T. Consider the kinematic condition,

∂ ∂

W

X

[

Z.

The first term is O ( A/T ), the right hand side is of the same order, whereas the second term is

X [

η = ( ) = ( ) × ( )

Since A/L is supposed to be much smaller than 1, we may neglect the second term and use the simplified kinematic condition ∂ ∂

W

Z

in the present case.

We then consider the dynamic condition:

( , , ) [ W = − ( , ).

W

0 J η [ W

We are now finally ready for attacking the linearized equations :

∂ ∂

2 2

2 2 0

Φ ( , , ) [ ] W Φ( , , ) ,

[

[ ] W

]

K ] η , (1)

]

( ,[ ] K W, ) 0 , (2)

W

( , )[ W Z [( , , ) 0 W , (3)

W

( , , )[ 0 W J η( , ). [ W (4)

We are primarily looking for solutions that are regular waves so let us first see whether (1) may have such solutions. For a given z , we thus assume that Φ has the form

Φ( , , )[ ] W = $ ]( ) sin( ωW − N[+φ 0 )

where k , ω and φ 0 are unknowns and A is an amplitude which we assume is dependent of z (It

is conceivable that A should depend on z ). If this function is inserted into (1), we easily obtain

− N $ ]^2 ( ) + $ ’’( ) sin(] ωW − N[+ φ 0 )= 0

Shall this be fulfilled for all x and t , we must have that the term in the bracket vanishes completely. This leads to an second order ordinary linear differential equation for A which has the general solution

$ ]( ) = & 1 cosh(N] +& 2 )

Equation (2) requires ∂Φ ∂

]

[ ] K W G$

G]

( , , ) (] K ) sin( ωW N[ φ 0 ) 0 ,

which means that $ ’( ] = −K )= 0. But $ ’( )] = N& 1 sinh( N] +& 2 )which vanishes at ] = − Kif & 2 = NK. Thus, a possible solution which satisfies both (1) and (2) is

Φ( , , )[ ] W = & 1 cosh( N ]( + K)) sin( ωW − N[+φ 0 ).

It remains to be seen whether (3) and (4) can be satisfied. Equation (4) actually gives an

expression for η since

cosh( ( )) cos( )

[ W

J W

[ ] W

J

& N ] K W N[

1 0

But (3) must also hold, that is,

∂ ∂

W

[ W

J

( , ) & cosh( NK ) sin( W N[ )

2 1 0

must be equal to

Z [ ] W ] ( , = , ) = ∂Φ( ,[ ] , )W N& sinh(NK ) sin( W N[ ) ∂

Shall this last condition be true for all x - and t -s, we must have

J

cosh( NK ) =N sinh( NK)

or

ω 2 = JN tanh( KN)

This is an equation which says that ω and k can not be chosen at will. For a given N ≠ 0, only

the two frequencies, ω and - ω which satisfies the equation are allowed. The equation is called

a dispersion relation. The dispersion relation tells us how the frequency and the wavenumber are connected.

If we now let φ 0 = −π / 2 and set

J

& 1 cosh( NK ) =D

we recover the familiar running regular wave for η:

η( , )[ W = D sin( ω W − N[).

For the potential Φ we obtain:

Φ( , , ) cosh( ( )) sin( / )

cosh( ) cosh( ( )) cos( )

cosh( ( )) cosh( )

cos( )

[ ] W & N ] K W N[

DJ

NK

N ] K W N[

DJ N ] K

NK

W N[

$#^

= +^ −

The equations for η, Φ and the dispersion relation represent the core of linear wave theory.

For small values of the argument

tanh( ) (^ (^ ))^ (^ (^ )) ( ( )) ( ( ))

[ H^ H ( )

H H

[ 2 [ [ 2 [

2 [ 2 [

[ 2 [

[ [ = (^) [ [

= +^ +^ −^ −^ +

− −

(^2 )

Moreover,

tanh( )[ H^ H H H

[ [ = (^) [ [ [

− − →∞ 1.

We recall that h is the water depth, and N = 2 π / λ where λ is the wavelength. Thus,

NK = 2 πK / λ. If kh is small, then h << λ, that is, the water depth is much smaller than the

wavelength. This corresponds to shallow water. Conversely, if kh is large, this corresponds to deep water. Let us consider the dispersion relation in these particular cases.

Shallow water:

Now NK << 1 and tanh( NK) may be replaced by kh. Thus, ω 2 = JN NK⋅ , or

ω = ±( JK) 1 2/^ N.

Deep water:

In this case, we set tanh( NK) = 1 and

ω = ± JN

Note there is a wide range of water depths which are neither shallow nor deep for a given wavelength. The rules of thumb are:

• Use the deep water expression when K >λ / 2.
• Use the shallow water expression when K <λ / 20

Exercise 5.1 : When tanh( x ) > 0.99, we may replace tanh( x ) by 1 for practical calculations. Find, by means of your calculator, a value [ 0 such that tanh( x ) > 0.99 when [ 0 < x. Verify that

we may use ω 2 = JN when K >λ / 2. Find the maximum relative error in ω if we use the

shallow water expression when K <λ / 20.

Exercise 5.2 : Determine the wavelength of a wave with period 10s when the water depth is a) 2000m , b) 1m. (Answ. (a) 156m, (b) 31.3m)

Exercise 5.3 Show that in deep water, the wavelength, λ , in metres of a wave with period T

seconds is 1.56 T.

The phase velocity, cp , of a regular wave was defined as

F N

S 7

We recall that this was, e.g. the speed of the top (crest) of the wave as it moves along. From the dispersion relation we obtain the following expression for the phase velocity

F (^) S =J^ NK

tanh( ).

Let us see what this amount to in deep and shallow water. In shallow water, we obtain the

somewhat surprising answer that the phase velocity is independent of both ω and k:

F

N

JK N

N

S =^ ω≈^ ⋅^ = JK

However, the velocity is now dependent of the depth, h. Waves for which the phase velocity is constant are called non-dispersive (Light waves in vacuum and regular sound waves in air are also non-dispersive).

For deep water we obtain

F N J

J JN

N

J

S N

= ω^ ≈ ω = = = ω 2 / ω

In deep water, the speed increases with increasing wave period and wavelength. The graph

below is copied from the book of K.F. Bowden. Note that as long h < λ / 10, the waves move

with constant speed.