ees report

O N T H E M AT H E M AT I C A L M O D E L L I N G O F S TA N D I N G WAV E SI N TA U T W I R E S

james pollard

Solutions to the One Dimensional Wave Equation in Stiff Wires Held Under Tension

July 2015 – version 0.1

[ June 28, 2015 at 22:39 – classicthesis version 0.1 ]

James Pollard: On the Mathematical Modelling of Standing Waves in TautWires, Solutions to the One Dimensional Wave Equation in Stiff WiresHeld Under Tension, c© July 2015


A B S T R A C T

Short summary of the contents. . .

iii


C O N T E N T S

i waves 1

1 an introduction to waves 3

1.1 The Definition of a Wave 3

1.2 The Wave Equation 6

2 solutions to the wave equation 9

2.1 General Solution 9

2.2 The First Case - Simple Waves 10

2.3 The Second Case - More Complicated Waves 10

2.4 The Third Case (Standing Wave): F(x) = G(x), F(x) 6=0 10

ii the showcase 11

3 math test chapter 13

3.1 Some Formulas 13

3.2 Various Mathematical Examples 14

iii appendix 15

a appendix test 17

a.1 Appendix Section Test 17

a.2 Another Appendix Section Test 18

bibliography 19

v


L I S T O F F I G U R E S

vi


L I S T O F TA B L E S

Table 1 Autem usu id 18

vii


L I S T I N G S

Listing 1 A floating example 18

viii


A C R O N Y M S

DRY Don’t Repeat Yourself

API Application Programming Interface

UML Unified Modeling Language

ix


Part I

WAV E S

It would be remiss of the author not to introduce what ismeant by a wave when writing on the mathematical mod-elling of waves. In this section we explore the one dimen-sional wave equation, it’s derivation, and it’s solutions.


1A N I N T R O D U C T I O N T O WAV E S

1.1 the definition of a wave

It is very hard to pin down a general, non-mathematical definitionof a wave; it might even be easier to skip straight to a mathematicaldefinition, but it is worthwhile considering what is meant by the term‘wave’, as it aids in the linking of abstract mathematical formulæ tophysical phenomena.

A wave can be defined simply as an oscillation that varies in spaceand time. By this we mean that if we consider the wave at a singlepoint in time, if we move in one direction in space1 some value willoscillate. In the same way, if we consider the wave at a single point inspace, as we move in one direction through time the same value willoscillate. The two oscillations do not have to be identical.

A wave has a time period, T , the ‘distance’ in the time dimensionbetween the same point on two subsequent oscillations, a wavelength,λ, the distance in the spatial dimension between the same point ontwo subsequent oscillations, and an amplitude, A, the maximum dis-placement from the undisturbed equilibrium position. We will alsodefine the frequency, f, as the number of oscillations which occur persecond, and so this is just the reciprocal of the time period. It is im-portant to note that the oscillation does not have to be of a spatialnature, such as the height of a point at a given spatial co-ordinate: itcould be the temperature, or the magnitude of the electric field, orsomething else entirely.

We can attempt to define this mathematically by coming up with anattempt at an equation for a wave. If we have some periodic function,F, and y is the property which oscillates, then:

y(x, t) = F(kx−ωt) (1.1)

where:x is the space co-ordinate

t is the time co-ordinate

k is the wave number2

ω is the angular frequency3

No one part of this equation is ‘the wave’. Rather the whole equationdefines a wave.

1 We shall only be considering waves in one dimension, as a wire may be consideredto be one dimensional for the purposes of mathematical modelling.

3 The function F may not necessarily have period 2π, so the usual definition of wavenumber may not hold

3 Strictly, this should be relative to 2π, however it will be relative to the period of F.

3


4 an introduction to waves

Through the use of Fourier Series, we may define any periodicfunction as a (potentialy infinite) sum of sinusoidal and cosinusoidalwaves4, and so we will simplify our definition considerably if we letF(x) = sin(x), knowing that when we want to produce a more compli-cated wave, we are just summing together several sinusoidal wavesto form a new solution.5

Our new and simplified equation is just:

y(x, t) = A sin(kx−ωt+φ) (1.2)

with the same definitions as before.6 This time, however, we have in-cluded φ, a phase constant, to allow for the starting conditions of thewave to be set. When we define a function we do not know anythingabout, such as F, we have no need for this phase constant, as we canimagine that F(x) = P(x+φ), where P is some other function. Oncewe define F(x) = sin(x) we must introduce a phase constant.

We shall now try link this equation to our physical understandingof a wave, previously outlined. To begin with we shall try to workout the amplitude. This is exceedingly simple: we know that as A isa constant, the maximum will occur when sin(...) is at it’s maximum,which is 1, and so the amplitude will be A · 1, or just A.

Now we shall try to find the wavelength. As the wavelength ispurely a spatial phenomenon, we shall assume that the time co-ordinateremains constant, and only allow the space co-ordinate to vary. Tomake the calculations easier, we shall assume that t is fixed at 0, al-though it would not actually matter if it were something else, as longas it is constant. We shall also assume that the phase is fixed at 0.Again this is purely to make the calculations easier. This simplifiesout equation to just:

y(x) = A sin(kx) (1.3)

We want to find the distance between the same point on two subse-quent oscillations. If we let the first point be x0, and the second pointbe x0 + λ, then:

y(x0) = y(x0 + λ) (1.4)

as to be the same point, they must have the same value. Hence:

sin(kx0) = sin(k(x0 + λ))

4 Here, by wave is meant any oscillating function, not a wave as is the subject of ourdiscussion.

5 This principle of superposition only holds for a linear system.6 This time the wavenumber and angular frequency will be relative to 2π.


1.1 the definition of a wave 5

as sine has a period of 2π:

kx0 + 2π = kx0 + kλ

2π = kλ

λ =

∣∣∣∣2πk∣∣∣∣ (1.5)

We can use the same technique to find the time period. By lettingx and φ be fixed at zero, as this time we are concerned with a timebased property, we can see that:

T =

∣∣∣∣2πω∣∣∣∣ (1.6)

Trivially:

f =∣∣∣ω2π

∣∣∣ (1.7)

At this point it might be useful to calculate the velocity of the wave.This is quite simple to do, as we can just use the relation s = vt. Weknow that over one time period, one complete oscillation will occur,and we know that two points on subsequent oscillations are separatedby the wavelength. One complete oscillation means that the wave has‘cycled’ through once, and is at the same point on the next oscillation,so over one time period the wave will have moved the distance of onewavelength. We can substitute these into the equation, and we get:

s = vt

λ = vT(=v

f)

2π

k= v · 2π

ω

v =2π

k· ω2π

v =ω

k(= fλ) (1.8)

One point we have neglected to consider is the direction of motionof the wave. We can treat velocity as a signed value rather than a vec-tor per se as we are dealing purely in one dimension. Let us see whathappens if we plot y(x, t) = sin(x− t). All we are looking for hereis the direction of motion, so we can remove any unnecessary coef-ficients and constants. As we can see, a negative t coefficient resultsin motion towards the positive direction, hence resulting in a positivevelocity. As our equation has a ‘built in’ minus sign, a positive valueof ω results in a positive velocity and so we don’t need to alter thesigns in our velocity equation.


6 an introduction to waves

1.2 the wave equation

We have succeeded in producing a mathematical equation that de-fines some sort of wave - it has a clearly defined amplitude, wave-length and time period, and we have said that we can use FourierSeries to produce a general oscillation from a series of sinusoidal os-cillations.

However, although our equation does define some sort of wave,do all waves fit this definition? The answer is no; we need a morefundamental mathematical definition.

In order to do this, we will consider our first equation, and calculateits second partial derivatives with respect to space:

y(x, t) = F(kx−ωt+φ)

∂

∂x(F(kx−ωt+φ)) = kF ′(kx−ωt+φ)

∂

∂x

(kF ′(kx−ωt+φ)

)= k2F ′′(kx−ωt+φ)

∂2y

∂x2= k2F ′′(kx−ωt+φ) (1.9)

and time:

y(x, t) = F(kx−ωt+φ)

∂

∂t(F(kx−ωt+φ)) = −ωF ′(kx−ωt+φ)

∂

∂t

(−ωF ′(kx−ωt+φ)

)= ω2F ′′(kx−ωt+φ)

∂2y

∂t2= ω2F ′′(kx−ωt+φ) (1.10)

By taking out a factor of F ′′(kx−ωt+φ) we can see that:

1

k2· ∂

2y

∂x2=

1

ω2· ∂

2y

∂t2

ω2

k2· ∂

2y

∂x2=∂2y

∂t2

(1.11)

and as v = ωk :

v2 · ∂2y

∂x2=∂2y

∂t2

∂2y

∂t2= v2 · ∂

2y

∂x2(1.12)


1.2 the wave equation 7

This final result is known as the Wave Equation7. As we derived itfrom our first equation, we know that our first equation is a solutionto this equation. This equation defines every wave possible; any equa-tion for a wave will be a particular solution to this equation and sowhenever we come up with what we think is an equation for a wave,we can check whether it is indeed a wave by seeing if it is a solutionto the equation.

7 Specifically it is the one dimensional wave equation.


2S O L U T I O N S T O T H E WAV E E Q U AT I O N

2.1 general solution

We start with the equation we previously derived:

∂2y

∂t2= v2 · ∂

2y

∂x2(2.1)

This is a linear, second order partial differential equation. If wewant to find the true general form of a wave, we must solve thisequation1.

The general solution is as follows:

y(x, t) = F(x− vt) +G(x+ vt) (2.2)

This two terms can be thought of as two waves: note that thesewill both have the same magnitude of velocity, but they will travelin opposite directions, as within each function, the t coefficient is ofopposite sign.

At first sight, it might look as though our original simplified guesswas off a different form, however using the relation v = ω

k , we canrewrite our original guess as follows:

y(x, t) = A sin(k(x− vt) +φ) (2.3)

Having done this, it is clear that our original guess is of the formof a solution, where:F(x) = A sin(kx+φ)

G(x) = 0

Now that we have a general solution, shall temporarily move awayfrom the mathematical viewpoint, and see what actually happenswhen we define F and G. There are three special cases, each of whichwe shall consider:

1. F(x) 6= G(x),G(x) = 0

2. F(x) 6= G(x), F(x) 6= 0,G(x) 6= 0

3. F(x) = G(x), F(x) 6= 0

1 We are not actually going to solve this: this is a purely mathematical venture; it isfar more interesting to examine specific solutions to the equation, which is what weshall do instead.

9


10 solutions to the wave equation

2.2 the first case - simple waves

This is the most simple case.

2.3 the second case - more complicated waves

2.4 the third case (standing wave): F(x) = G(x) , F(x) 6= 0

This is perhaps the particularly interesting case. We shall devote themost time to it as it is the case that applies to waves in strings, as weshall see in a later chapter.


Part II

T H E S H O W C A S E

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3M AT H T E S T C H A P T E R

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3.1 some formulas

Due to the statistical nature of ionisation energy loss, large fluctu-ations can occur in the amount of energy deposited by a particletraversing an absorber element1. Continuous processes such as multi-ple scattering and energy loss play a relevant role in the longitudinaland lateral development of electromagnetic and hadronic showers,and in the case of sampling calorimeters the measured resolutioncan be significantly affected by such fluctuations in their active lay-ers. The description of ionisation fluctuations is characterised by thesignificance parameter κ, which is proportional to the ratio of meanenergy loss to the maximum allowed energy transfer in a single colli-sion with an atomic electron: You might get

unexpected resultsusing math inchapter or sectionheads. Consider thepdfspacing option.

κ =ξ

Emax(3.1)

Emax is the maximum transferable energy in a single collision with anatomic electron.

Emax =2meβ

2γ2

1+ 2γme/mx + (me/mx)2

,

where γ = E/mx, E is energy and mx the mass of the incident par-ticle, β2 = 1− 1/γ2 and me is the electron mass. ξ comes from theRutherford scattering cross section and is defined as:

ξ =2πz2e4NAvZρδx

meβ2c2A= 153.4

z2

β2

Z

Aρδx keV,

where

1 Examples taken from Walter Schmidt’s great gallery:http://home.vrweb.de/~was/mathfonts.html

13


http://home.vrweb.de/~was/mathfonts.html

14 math test chapter

z charge of the incident particle

NAv Avogadro’s number

Z atomic number of the material

A atomic weight of the material

ρ density

δx thickness of the materialκ measures the contribution of the collisions with energy transfer

close to Emax. For a given absorber, κ tends towards large values if δxis large and/or if β is small. Likewise, κ tends towards zero if δx issmall and/or if β approaches 1.

The value of κ distinguishes two regimes which occur in the de-scription of ionisation fluctuations:

1. A large number of collisions involving the loss of all or most ofthe incident particle energy during the traversal of an absorber.

As the total energy transfer is composed of a multitude of smallenergy losses, we can apply the central limit theorem and de-scribe the fluctuations by a Gaussian distribution. This case isapplicable to non-relativistic particles and is described by theinequality κ > 10 (i. e., when the mean energy loss in the ab-sorber is greater than the maximum energy transfer in a singlecollision).

2. Particles traversing thin counters and incident electrons underany conditions.

The relevant inequalities and distributions are 0.01 < κ < 10,Vavilov distribution, and κ < 0.01, Landau distribution.

3.2 various mathematical examples

If n > 2, the identity

t[u1, . . . ,un] = t[t[u1, . . . ,un1

], t[u2, . . . ,un]]

defines t[u1, . . . ,un] recursively, and it can be shown that the alterna-tive definition

t[u1, . . . ,un] = t[t[u1,u2], . . . , t[un−1,un]

]gives the same result.


Part III

A P P E N D I X


AA P P E N D I X T E S T

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Etiam ac leo a risus tristique nonummy. Donec dignissim tinciduntnulla. Vestibulum rhoncus molestie odio. Sed lobortis, justo et pretiumlobortis, mauris turpis condimentum augue, nec ultricies nibh arcupretium enim. Nunc purus neque, placerat id, imperdiet sed, pel-lentesque nec, nisl. Vestibulum imperdiet neque non sem accumsanlaoreet. In hac habitasse platea dictumst. Etiam condimentum facili-sis libero. Suspendisse in elit quis nisl aliquam dapibus. Pellentesqueauctor sapien. Sed egestas sapien nec lectus. Pellentesque vel dui velneque bibendum viverra. Aliquam porttitor nisl nec pede. Proin mat-tis libero vel turpis. Donec rutrum mauris et libero. Proin euismodporta felis. Nam lobortis, metus quis elementum commodo, nunc lec-tus elementum mauris, eget vulputate ligula tellus eu neque. Viva-mus eu dolor.

a.1 appendix section test

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17


18 appendix test

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fastidii ea ius germano demonstratea

suscipit instructior titulo personas

quaestio philosophia facto demonstrated

Table 1: Autem usu id.

Listing 1: A floating example

1 for i:=maxint to 0 do

begin

{ do nothing }

end;

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a.2 another appendix section test

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D E C L A R AT I O N

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Manchester, July 2015

James Pollard, June 28, 2015


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