wave motion: revision - university of oxfordwilkinsong/wavemotion...d’alembert solution of wave...

Wave Motion: revision Professor Guy Wilkinson

[email protected] Trinity Term 2014

Introduction Two lectures to remind ourselves of what we learned last term Will restrict discussion to the topics on the syllabus Will re-state main relations/facts/principles you should know. Most of these I will not re-derive (no time) – look back to last term’s notes. Essential information contained in these slides, but some points will be augmented by working through on board. Will illustrate with some typical problems from recent prelims papers. (No guarantee my answers are correct! If working through them we find bugs I shall fix the online version of the slides)

These slides & HT material on http://www.physics.ox.ac.uk/users/wilkinsong

2

Normal modes

4

Normal modes - syllabus

According to the course handbook you should know about the following

Let’s remind ourselves of the essentials, before looking at a few past problems

An illustrative example: coupled pendula

)(

)(

xykl

ymgym

xykl

xmgxm

xy

x y

F F

mg mg

ll

k

First thing to do is to write down the equations of motion:

5

Finding the normal modes

6

Recall there are two good ways to solve these problems:

1. The decoupling method

The equations are decoupled by finding the ‘mode (or normal) coordinates’. If the system is symmetric (i.e. equal masses, spring constants, length of pendula...) then a good bet will be

)(2

1

)(2

1

2

1

yxq

yxq

(the factor of 1/√2 is not necessary for solving the problem, and is often omitted. It’s there to make other expressions, i.e. those for the energy, come out right)

In practice then one adds and subtracts the equations of motion.

If the system is not symmetric then the normal modes will in general be more complicated. In that case I advise a second plan of attack...

2. The matrix method

I’ll remind you of this shortly.

Solving with decoupling method So for the coupled pendula we add and subtract the equations of motion:

yxq 1

1

2

11 qq

to obtain

with l

g2

1

where A1,2 & ϕ1,2 are constants set by inital conditions

)cos( 1111 tAq

7

)( xykl

xmgxm )( xyk

l

ymgym

2

2

22 qq m

k

l

g22

2 with

yxq 2

the solutions of which are clearly

)cos( 2222 tAq

Coupled pendula – the normal modes

)cos( 1111 tAql

g2

1yxq 1

yxq 2)cos( 2222 tAqm

k

l

g22

2

First normal mode: centre-of-mass motion

Second normal mode: relative motion

8

This correlated and anti-correlated motion is given by the form of the normal coordinates and is typical for the simple problems we will encounter.

Coupled pendula: the general solution General solution is a sum of the two normal modes

)cos()cos(

)cos()cos(

222111

222111

tAtAy

tAtAx

The constants A1 ϕ1, A2 and ϕ2 are set by the initial conditions

For example

0)0()0(;)0()0( yxayx

gives

0;0; 121 AaA

x

y

t

t

a

a

T=2π/ω1

9

In this case only the 1st normal mode is excited. (For other possibilities look back at HT lecture notes.)

Solving with matrix method

0

0

dt

ddt

d

2

2

2

2

y

x

m

k

l

g

m

km

k

m

k

l

g

)(

)(

xykl

ymgym

xykl

xmgxm

tieY

X

y

x

Re

0

0

2

2

Y

X

m

k

l

g

m

km

k

m

k

l

g

Expecting an oscillatory solution, so let’s try substituting one in, making use of complex notation

We obtain: eigenvector

equation

X & Y are complex constants

10

Solving with matrix method

0A 0]det[ A

02

2

m

k

l

g

m

km

k

m

k

l

g

So here:

11

We have equation of sort

Non-trivial solution requires matrix has no inverse

Eigenvalue equation m

k

m

k

l

g

2

l

g2

1

m

k

l

g22

2

Substitute these back into the eigenvector equn to find the amplitude ratios, i.e. the relationship between X and Y. Find X=Y and X=-Y, as before.

Illustrate these methods with a couple of past exam questions

12

• TT 2008, Q11 • TT 2011, Q8

We have reminded ourselves about:

Let’s apply these techniques to

Trinity Term 2009, Q11

13

14


)2(

)2(2

122

211

xxkxm

xxkxm

Equations of motion:

Different masses involved, so recommend matrix method.

tieX

X

x

x

2

1

2

1Re

0

0

22

2

1

2

2

X

X

m

k

m

km

k

m

k

Usual ansatz

yields eigenvector equation

which gives

2

3

2

32

m

k

2

3

2

32

m

kand

with the X1/X2 amplitude ratios obtained by substituting ω2 results back into eigenvector equation

31

1

2

1

X

X

13

1

2

1

X

X

15


So general solution is

)cos(13

)cos(31

1

t

At

Ax

)cos()cos(2 tAtAx

Use initial conditions to fix A+,- and Φ+,- . It is the requirement that the masses start at rest that determines Φ+,- = 0, while the initial displacements fix A+,-.

)cos()cos(3

02 tt

xx

)cos(

13

1)cos(

31

1

3

01 tt

xx

16


17


)(22

)(2

122

211

xxkxm

xxkxm

Different masses involved, so recommend matrix method.


tieX

X

x

x

2

1

2

1Re

Usual ansatz

yields eigenvector equation

0

022

2

1

2

2

X

X

m

k

m

km

k

m

k

Solving this gives in the case mk /3 21 2XX

The other solution, with is . This is non-oscillatory ! 021 XX

tmkA

x

tmkAx

)/3(cos2

)/3(cos

2

1

vtBx

vtBx

2

1So have and

with A,B,Φ and v constants to be determined from initial conditions

18


tm

krtmk

rx

tm

krtmk

rx

3

20)/3(cos

20

3

20)/3(cos

10

2

1

i.e. anticorrelated oscillations superimposed upon spinning motion

Velocity of mass 2m is zero when 2

3

3

k

mt

Other topics

19

• Reminder about energy • Driving term – TT2012, Q10

Energy of system with normal modes

kxyyxm

k

l

gmyxmU

)(

2

1)(

2

1 2222

)(2

1

)(2

1

21

21

qqy

qqx

This a bit opaque. How does it look in terms of normal coordinates?

Recall (here with normalisation factors included):

and l

g2

1

m

k

l

g22

2

This gives a much more transparent expression, and without cross-terms:

So total energy of system = sum of energies in each excited mode

2

2

2

2

2

2

2

1

2

1

2

12

1

2

1

2

1

2

1qmqmqmqmU

energy in mode 1 energy in mode 2

20

Energy will include kinetic and potential contributions, with each coordinate featuring. So returning to the case of the two coupled pendula:

21


22



2

2

01

2

01 3 yyy

)(23 2

02

2

0

2

012 thyyy

These are inhomogeneous, due to presence of h(t) term. To find the normal modes we must find the complementary functions, i.e. solve homogenous case:

mLT 2/2

0 with

2

2

01

2

01 3 yyy

2

2

0

2

012 3 yyy

Symmetric system, so this is easily done by the decoupling method

with and with pp 2

02 qq 2

0421 yyp 21 yyq

02 p 02 q

23


Finding steady state amplitude means finding the particular integral

)Re()]4)(2/[(2 22

0

22

00

4

01

tiehy

)Re()]4)(2/[()3(2 22

0

22

0

22

00

4

02

tiehy

when and 4/01 hy 02 )4/3( hy 0

when 03

)Re(2 01

tiehy 02 y

2h0

h0

L

2L L

and

Waves

24

25

Waves - syllabus

According to the course handbook you should know about the following

Let’s remind ourselves of the essentials, before looking at a few past problems

26

Waves - syllabus

Focus first on the below topics

and illustrate by looking at Long Vacation 2011, Q9

Waves on a stretched string Consider a segment of string of linear density ρ stretched under tension T

x

T

y

xx

T

,

small

27

2

2

2

2

t

y

Tx

y

Key point – means that no net horizontal force at first order, only vertical force, and we can make approximations such as sinδθ ≈ δθ

For full derivation see HT lecture notes or text book, e.g. French

2

2

22

2 1

t

y

cx

y

Tc

This is the wave equation, which (anticipating solution) we can write

with

d’Alembert solution of wave equation

28

2

2

22

2 1

t

y

cx

y

ctxv

ctxu

y is a function of x and t. Define new variables so that y is now a function of u and v

02

vu

y)()(),( vgufvuy With chain rule

we can show

)()(),( ctxgctxftxy

So general solution of wave equation is

Here f and g are any functions of (x-ct) and (x+ct), determined by initial conditions. A common scenario is that they are sinusoidal.

Note that describes forward-going wave and )( ctxf

)( ctxg describes a backwards-going one !

with c the (phase) velocity of the wave

Sinusoidal waves

)()(),( ctxgctxftxy

A very common functional dependence for f and g...

...is sinusoidal. In this case it is usual to write:

)cos()cos(),( tkxBtkxAtxy

or Asin(kx-ωt) ... etc (choice doesn’t matter, unless we are comparing one wave with another and then relative phases become important)

with k and ω (and A and B) constants

• speed of wave

• frequency

where ω is angular frequency

• wavelength where k is the wave-number (or wave-vector if also used to indicate direction of wave)

kc /

2//1 Tf

k/2

29

Notation choices

)cos(),( tkxAtxy Sinusoidal solution (writing here, for compactness, only the forward-going solution)

Using the relationships between k,ω, λ & c this can be expressed in many forms

)](cos[),( ctxkAtxy

Also note that sometimes it is convenient to write

)cos(),( kxtAtxy

A very frequent approach is to use complex notation (we already made use of this when analysing normal modes, and you will have seen it in circuit analysis)

)](exp[Re),( tkxiAtxy

or if it’s important to pick out sine function. Note that often the ‘Re’ or ‘Im’ is implicit, and it gets omitted in discussion.

)](exp[Im),( tkxiAtxy

30

Energy and impedance for travelling wave on string

31

Energy stored in a mechanical wave

Integrate kinetic energy and potential energy densities over an integer number of wavelengths to show they contribute equally and give

2

2

1

d

ddensity KE

t

y

x

K

2

2

1

d

ddensity PE

x

yT

x

U

nA 22

2

1Total energy in n wavelengths =

t

yx

yT

v

FZ

y

y

so with

)sin(),( tkxAtxy )(

T

c

TTkZ

The characteristic impedance Z is defined as the applied driving force acting in the y-direction divided by the velocity of the string in the y-direction

Characteristic impedance

(Note sign on driving force which ensures Z positive for forward wave!)

Power flow = kTA 2

2

1

Long Vacation 2011, Q9

32

Waves at boundaries

33

This question a good opportunity to remind ourselves what happens to waves at boundaries, so let’s answer it in a more general way than is being asked by first allowing strings to be different (e.g. different densities)

x=0

ρ1

ρ2

x→

M

So we must allow for reflected and transmitted waves. To satisfy boundary conditions all waves must have same frequencies, but their velocity and wave-vector will depend on which string they are on

)](exp[Re 1xktiA

)](exp['Re 1xktiA )](exp[''Re 2xktiA

Incident

Reflected Transmitted

Complex notation convenient for these problems. Pay attention to signs!

34

Long Vacation 2011, Q9 Write down boundary conditions

M

x=0 θ1 θ2

T T

1. Equation of motion for mass

021

0

2

2or1

2

)sinsin(

x

x

TTt

yM

x

y

11

11

tan

tansin

x

y

22

22

tan

tansin

0

12

0

2

2or1

2

xxx

y

x

yT

t

yM

2. Strings stay stuck to mass

),0(),0( 21 tyty

Apply to expressions for incident, reflected and transmitted waves

')('' 11

2

2 TAikTAikmTikA

''' AAA

(1)→

(2)→

35

Long Vacation 2011, Q9

Evaluate r ≡ A’/A and t ≡ A’’/A

mikkT

Tk

A

At

2

21

1

)(

2''

mikkT

imkkT

A

Ar

2

21

2

21

)(

)('

Specialising to case where strings are identical and so k1=k2 =k gives

ip

ip

mikT

imr

12 2

2

ipmikT

kTt

1

1

2

22 kT

mp

2

2with and

(Aside: one is often asked about transmitted and reflected energy.

So make sure you remember wave power ) 2)Amplitude(2

1kT

Phase of r = )(tan2

1 p • m low, phase is –π/2

• m high, phase is -π

but this case is slightly artificial, as no reflection in case m=0!

36

Waves - syllabus Now let’s consider

with reference to questions: TT 2009, Q6; TT 2010, Q6; TT 2012, Q8

Wave packets and group velocity

37

A single wave cannot transmit information. To do that we need a wave packet. Any wave packet can be formed from a sum of single waves. Simplest example:

Sum together two waves which differ by 2δω and 2δk in angular frequency and wave-number, respectively:

txkkAy

txkkAy

)()(sin

)()(sin

2

1

)sin()cos(221 tkxtxkAyyy

y1

y2

y

Describes envelope – so envelope moves with velocity and indeed

)sin()cos(2 tkxtxkA

Group velocity while phase velocity k

vgd

d

kvp

k

Dispersion

38

Dispersion is when there is not a linear relationship between ω and k.

Two consequences:

1. Phase velocity, ω/k, depends on ω and k. e.g. Light in medium m has refractive index n and velocity cm, where

That’s why a prism splits light.

2. Group velocity ≠ phase velocity

nccm /

kvg

d

dIf and it follows that

kvp

dk

dvkvv

p

pg

d

dvvv

p

pg

d

dn

nn

cvg 1and and

(more important that you can derive these, rather than learn them!)

Dispersion question: TT 2009, Q6

39

(1)

4222222 4 DBAc 2

2

2 )2(

d

dv

c

DA

and (1) → so

kk d

d

d

d

d

d

Now and 2

2

d

d

kk

hence result

Standing waves: TT 2010, Q6

40

0x Lx

tkxA

tkxAtkxAtxy

cossin2

)sin()sin(),(

Sum right- and left-going travelling waves of same amplitude and frequency:

Standing waves: every point on the string moves with a certain time dependence (cosωt), but the amplitude depends on its position along the string (sinkx)

factorised spatial & temporal dependence

...

41

Standing waves: TT 2010, Q6 ...

0x Lx

Ends of string being fixed determine boundary conditions: 0),(),0( tLyty

angular frequency of normal modes

tkxAtxy cossin2),( nLkn L

Tnn

and

and putting in numbers, with n=1, gives T=71 N

Wave equation revisited – solving by separation of variables

42

2

2

22

2 1

t

y

cx

y

We have already solved the wave equation using the d’Alembert approach

Can also be solved by looking for solutions which have the ‘separated’ form

)()(),( tTxXtxy

i.e. that factorise into functions that are separate functions of x and t. This is just the situation that applies to standing waves !

T

T

cX

X 2

1

set each side equal to some separation constant –k2

kxBkxAxX sincos)( cktEcktDtT sincos)(

with A,B,D and E constants defined by initial conditions

This yields and

Standing waves: TT 2012 Q8

43

44

Standing waves: TT 2012 Q8 Let’s rearrange question so we can discuss the relevant topics more clearly

Going from general solution to specific solution through applying initial conditions and monitoring subsequent evolution with time

Energy of system

45

Standing waves: TT 2012 Q8 Know from separation of variables that a solution to wave equation is

)sincos)(sincos(),( kctDkctCkxBkxAtxy

and we also have four boundary conditions:

1. String initially at rest, i.e. for all x

2. y(0,t)=0

3. y(L,t)=0 where n any integer. This is the eigenvalue eqn. and discretises k. Each value of n corresponds to a normal mode.

4. Form of initial displacement involves normal modes 1 and 3. From these we fix coefficient of mode 1 to be 1 and 3 to be -2/3, and all others 0.

0/ ty 0D

0 A

nkL

L

ct

L

x

L

ct

L

xtxy

3cos

3sin

3

2cossin),( hence

This first returns to initial displacement when cLt /2

Energy of standing waves

46

Normal mode n for our string, with given boundary conditions:

L

ctn

L

xnFtxy nn

cossin),(

Calculate kinetic energy, Kn, and potential energy, Un, for each mode

xt

yK

L

nn d

2

1

0

2

x

x

yTU

L

nn d

2

1

0

2

L

Tnn

4

22

nnnnn

LFUKE

Evaluate and sum with

What about the case when several normal modes are excited (as in question)?

1n

nEENote that all cross-terms have vanished due to the orthogonality of sines

mnxL

xm

L

xnL

withdsinsin0

i.e. all these terms are zero

So total energy is weighted sum of all the excited normal modes

47

Standing waves: TT 2012 Q8

xx

tyTtU

L

d)0(

2

1)0(

0

2

Initial energy of string is all in PE

L

TtU

4

5)0(

2

1n

nEEThis result makes sense as it equals total energy of system as calculated from

4

22

nnn

LFE

and

Kinetic energy of each standing wave, i.e. kinetic energy of each mode

xt

yK

L

nn d

2

1

0

2

L

ctncn

LAK nn

222sin)(

4

where An is the amplitude coefficient for mode n (here A1=1 and A3=-2/3 and others =0)

wave motion: revision - university of oxfordwilkinsong/wavemotion...d’alembert solution of wave...

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