vibrations and waves 2

1

University of Cape TownDepartment of Physics

PHY2014FVibrations and Waves

Andy BufflerDepartment of PhysicsUniversity of Cape [email protected]

Part 2Coupled oscillatorsNormal modes of continuous systemsThe wave equationFourier analysis

… covering (more or less) French Chapters 2, 5 & 6

2

Problem-solving and homework

Each week you will be given a take-home problem set to complete and hand in for marks ...

In addition to this, you need to work through the following problems in French, in you own time, at home. You will not be asked to hand these in for marks. Get help from you friends, the course tutor, lecturer, ... Do not take shortcuts. Mastering these problems is a fundamental aspect of this course.

The problems associated with Part 2 are:

2-2, 2-3, 2-4, 2-5, 2-6, 5-2, 5-8, 5-9, 6-1, 6-2, 6-6, 6-7, 6-10, 6-11, 6-14

You might find these tougher: 5-4, 5-5, 5-6, 5-7

3

The superposition of periodic motionsTwo superimposed vibrations of equal frequency

1 1 0 1cos( )x A tω φ= +

2 2 0 2cos( )x A tω φ= +combination can be written as

0cos( )x A tω φ= +

Using complex numbers:0 1( )

1 1j tz A e ω φ+=

0 2( )2 2

j tz A e ω φ+= 1 2z z z= +

0 1 2 1( ) ( )1 2

j t jz e A A eω φ φ φ+ −∴ = +

A A2

A1

0 1tω φ+

2 1φ φ−β

2 1φ φ φ= −Phase difference

Then

( )2 2 1sin sinA Aβ φ φ= −and

2 2 21 2 1 2 2 12 cos( )A A A A A φ φ= + + −

Frenchpage 20

4

If we add two sinusoids of slightly different frequency and… we observe “beats”…

x1x2

x1+x2

t

t

Superposed vibrations of slightly different frequency: Beats

2cos tω

1 2cos cost tω ω+1 2cos

2tω ω−

1 2

2beatT π

ω ω=

−

1cos tω

1 2 1 21 2cos cos 2cos cos

2 2t t t tω ω ω ωω ω − + + =

1ω 2ω

Frenchpage 22

5

Combination of two vibrations at right angles

1 1 1cos( )x A tω φ= +

2 2 2cos( )y A tω φ= +???

Write and1 0cos( )x A tω=

Consider case where frequencies are equal and let initial phase difference be

2 0cos( )y A tω φ= +

Case 1 : 0φ = 1 0cos( )x A tω=2 0cos( )y A tω=

2

1

Ay xA

= Rectilinear motion

2φ π= 1 0cos( )x A tω=

2 0 2 0cos( 2) sin( )y A t A tω π ω= + = −Case 2 :

2 2

2 21 2

1x yA A

∴ + = Elliptical path in clockwise direction

Frenchpage 29

φ

6

Case 3 : φ π= 1 0cos( )x A tω= 2

1

Ay xA

= −

Combination of two vibrations at right angles …2

2 0 2 0cos( ) cos( )y A t A tω π ω= + = −

3 2φ π= 1 0cos( )x A tω=

2 0 2 0cos( 3 2) sin( )y A t A tω π ω= + = +Case 4 :

2 2

2 21 2

1x yA A

∴ + = − Elliptical path in anticlockwise direction

4φ π= 1 0cos( )x A tω=2 0cos( 4)y A tω π= +

Case 5 :

Harder to see … use a graphical approach …

7

Superposition of simple harmonic vibrations at right angles with an initial phase difference of 4π

8

Superposition of two perpendicular simple harmonic motions of the same frequency for various initial phase differences.

9

Abbreviated construction for the superposition of vibrations at right angles … see French page 34.

10

Perpendicular motions with different frequencies: Lissajous figures

See French page 35.

Lissajous figures forwith various

initial phase differences.2 12ω ω=

φ = 0 4π 3 4π2π π

11

2 1:ω ω1:1

1:2

2:3

1:3

3:4

5:6

4:5

3:5

φ = 0 4π 3 4π2π π

Lissajous figures

12

Coupled oscillators

When we observe two weakly coupled identical oscillators A and B, we see:

t

t

xA

xB

… these functions arise mathematically from the addition of two SHMs of similar frequencies … so what are these two SHMs?

These two modes are known as normal modes … which are states of the system in which all parts of the system oscillate with SHM either in phase or in antiphase.

Frenchpage 121

13

t

t

A B

xA

xB

Coupled oscillators

14

The double mass-spring oscillator

mAxk

mBx k

Individually we know that and A Amx kx= − B Bmx kx= −

For both oscillators: 0km

ω =

Now add a weak coupling force:

mAxk

mBx kck

For mass A: ( )A A c B Amx kx k x x= − + −

or where ,2 20 ( )A A B Ax x x xω= − +Ω −

20

km

ω = 2 ckm

Ω =

15

2 20 ( )A A B Ax x x xω= − +Ω −

The double mass-spring oscillator …2

For mass A:

For mass B: 2 20 ( )B B B Ax x x xω= − −Ω −

… two coupled differential equations … how to solve ?

Adding them:2

202 ( ) ( )A B A B

d x x x xdt

ω+ = − +

Subtract B from A:2

2 202 ( ) ( ) 2 ( )A B A B A B

d x x x x x xdt

ω− = − − − Ω −

Define two new variables:1 A Bq x x= +

2 A Bq x x= −

Then and2

210 12

d q qdt

ω= −2

2 220 22 ( 2 )d q q

dtω= − + Ω

… called “normal coordinates”

16


The two equations are now decoupled … 2

2112 s

d q qdt

ω= −

Write 222

22 fd q qdt

ω= −

2 20sω ω=

2 2 20 2fω ω= + Ω

s = “slow”

f = “fast”

… which have the solutions:1 1cos( )sq C tω φ= +

2 2cos( )fq D tω φ= +

Since and 2 A Bq x x= −

We can write and

1 A Bq x x= +

( )11 22Ax q q= + ( )1

1 22Bx q q= −

17


1 2cos( ) cos( )2 2A s fC Dx t tω φ ω φ= + + +

Then1 2cos( ) cos( )

2 2B s fC Dx t tω φ ω φ= + − +

So xA and xB have been expressed as the sum and difference of two SHMs as expected from observation.… C, D, φ1 and φ2 may be determined from the initial conditions.

… when xA = xB ,then q2 = 0 … there is no contribution from the fast mode and the two masses move in phase … the coupling spring does not change length and has no effect on the motion … 0sω ω=

… when xA = −xB ,then q1 = 0 … there is no contribution from the slow mode … the coupling spring gives an extra force … each mass experiences a force giving ( )2 ck k x− +

2 20 2ω= + Ω

2 2 cf

k km

ω +=

18

The double mass-spring oscillator …5symmetric mode antisymmetric mode mixed mode

19


We now have a system with two natural frequencies, and experimentally find two resonances.

Frequency

Am

plitu

de

20

Pitch and bounce oscillator

dAxk

m

Bx

L

k

Two normal modes (by inspection):

BouncingA Bx x=

Restoring force = 2kx−2

2 2d xm kxdt

∴ = −

Pitching

A Bx x= − AxBx θ

Centre of mass stationary

2bounce

2km

ω =

Iτ θ= −

21 12 12( )kd d mLθ θ= −

2

2

6kdmL

θ θ→ =2

2pitch 2

6k dm L

ω =

Frenchpage 127

21

N = 2

( )1 0 012 sin

2 2 1πω ω ω= =+

( )2 0 022 sin 3

2 2 1πω ω ω= =+

22

N = 4

N = 3

23

N-coupled oscillators

fixed fixedTension T

l

1 2 3 p−1 p p+1 N

… consider transverse displacements that are small.Each bead has mass m

1 2 3 p−1 p p+1 N

1pα −pα

Transverse force on pth particle: 1sin sinp p pF T Tα α−= − +

1 1p p p py y y yT T

l l− +− −

= − +

for small α

y

Frenchpage 136

24

N-coupled oscillators …22

1 12

p p p p pp

d y y y y yF m T T

dt l l− +− −

= = − +

( )2

2 20 0 1 12 2 0p

p p p

d yy y y

dtω ω + −∴ + − − =

where ,20

Tml

ω = p = 1, 2 … N

… a set of N coupled differential equations.

Normal mode solutions: sinp py A tω=

Substitute to obtain N simultaneous equations

( ) ( )2 2 20 0 1 12 0p p pA A Aω ω ω + −− + − + =

or2 2

1 1 020

2p p

p

A AA

ω ωω

+ −+ − +=

25


From observation of physical systems we expect sinusoidal “shape functions” of the form sinpA C pθ=

Substitute into 2 2

1 1 020

2p p

p

A AA

ω ωω

+ −+ − +=

And apply boundary conditions and0 0A = 1 0NA + =

and

n = 1, 2, 3, … N (modes)

( )02 sin2 1n

nNπω ω=+

… find that 1

nNπθ =+

There are N modes:sin sin sin

1pn pn n n npny A t C tN

πω ω= =+

26

02ωnω

0 1 2 3 N+1 n

( )02 sin2 1n

nNπω ω=+


For small N:

27

For n << N :( ) ( )

sin2 1 2 1

n nN Nπ π

=+ +

( )0

022 1 1n

n nN N

πωπω ω = = + + then

i.e. n nω ∝ for n << N

In many systems of interest N is very large… and we are only interested in the lowest frequency modes.

02ωnω

0 n << N N+1 n

linear region


28

N coupled oscillators have N normal modes and hence N resonances

response

ω


29

Continuous systems

30

Continuous systems

Consider a string stretched between two rigid supports …

x = 0 x = Ltension T

String has mass m and mass per unit length m Lµ =

Suppose that the string is disturbed in some way:

y

x

The displacement y is a function of x and t : ( , )y x t

31

T

Tθ

θ θ+ ∆

y

x xx x+ ∆

Consider the forces on a small length of string …

Restrict to small amplitude disturbances … then is small andθ

cos 1θ = sin tan yx

θ θ θ ∂= = =

∂The tension T is uniform throughout the string.Net horizontal force is zero: cos( ) cos 0T Tθ θ θ+ ∆ − =

Vertical force: sin( ) sinF T Tθ θ θ= + ∆ −

Then x x xy yF T Tx x+∆

∂ ∂= −

∂ ∂

Normal modes of a stretched string

Frenchpage 162

32

x x xy yF T Tx x+∆

∂ ∂= −

∂ ∂

Use( ) ( )dg g x x g x

dx x+ ∆ −

=∆

Then2

2

yF T xx∂

= ∆∂

or ( )2 2

2 2

y yx T xt x

µ ∂ ∂∆ = ∆

∂ ∂

giving2 2

2 2

y yx T t

µ∂ ∂ = ∂ ∂

Check: has the dimensions

Tµ21 v

Then isthe speed at which a wave propagates along the string … see later

v T µ=

Write2 2

2 2 2

1y yx v t∂ ∂

=∂ ∂

One dimensional wave equation

Normal modes of a stretched string …2

µ: mass per unit length

33

Look the standing wave (normal mode) solutions … Normal mode: all parts of the system move in SHM at the same frequency …

Write: ( , ) ( )cosy x t f x tω=

( )f x is the “shape” function … substitute into wave equation2 2

2 2

( , ) ( ) cosy x t d f x tx dx

ω∂=

∂

( )2

22

( , ) ( ) cosy x t f x tt

ω ω∂= −

∂

22

2 2

( ) 1cos ( )cosd f x t f x tdx v

ω ω ω= −

which must be true for all t

then2 2

2 2 ( )d f f xdx v

ω= −


34

2 2

2 2 ( )d f f xdx v

ω= −

… which has the same form as the eq. of SHM:2

202

d x xdt

ω= −

… has general solution: 0sin( )x A tω φ= +

Thus we must have: ( ) sinf x A xvω = +Φ

Apply boundary conditions: y = 0 at x = 0 and x = L

(0) 0f∴ = and ( ) 0f L =

x = 0, f =0 : 0 sin 0Avω = +Φ

i.e.

0 sinA Lvω =

0Φ =

x = L, f =0 : L nvω π=i.e. n = 1,2,3,…


35

nn v

Lπω =Write n = 1,2,3,…

( ) sin sinn nx n v n xf x A Av L L

π π = =

Therefore

shape function, or “eigenfunction”

x = 0 x = Ln = 1

n = 3

n = 2

n = 5

n = 4

( )1( ) sinf x A x Lπ=

( )2( ) sin 2f x A x Lπ=

( )3( ) sin 3f x A x Lπ=

( )4( ) sin 4f x A x Lπ=

( )5( ) sin 5f x A x Lπ=

1 v Lω π=

3 3 v Lω π=

2 2 v Lω π=

5 5 v Lω π=

4 4 v Lω π=


36

Normal modes of a stretched string

n = 1

n = 2

n = 3

n = 4

38

Full solution for our standing waves:

( , ) sin cosn nn xy x t A t

Lπ ω =

n

n vLπω =

The mode frequencies are evenly spaced: 1n nω ω=

nω

0 1 2 3 n

3ω

2ω

1ω

(recall the beaded string)

This continuum approach breaks down as the wavelength approaches atomic dimensions… also if there is any “stiffness” in the spring which adds an additional restoring force which is more pronounced in the high frequency modes.


39

All motions of the system can be made up from the superposition of normal modes

1( , ) sin cos( )n n n

n

n xy x t A tLπ ω φ

∞

=

= +

∑

with nn v

Lπω =

Note that the phase angle is back since the modes may not be in phase with each other.


40

Whispering galleries

… best example is the inside dome of St. Paul’s cathedral.

If you whisper just inside the dome, then an observer close to you can hear the whisper coming from the opposite direction … it has travelled right round the inside of the dome.

41

Longitudinal vibrations of a rodx x x+ ∆

x xξ ξ+ + ∆ + ∆x ξ+1F

2F

section of massive rod

section is displaced and stretched by an unbalanced force

Average stress =xξ∆∆

Average strain = Yxξ∆∆

Y : Young’s modulus

stress at = (stress at x) + x x+ ∆(stress) x

x∂

∆∂

Frenchpage 170

42

If the cross sectional area of the rod is α

1F Yxξα ∂

=∂

2

2 2F Y Y xx xξ ξα α∂ ∂

= + ∆∂ ∂

and

2

1 2 2F F Y x maxξα ∂

− = ∆ =∂2 2

2 2Y x xx tξ ξα ρα∂ ∂

∴ ∆ = ∆∂ ∂

or2 2

2 2x Y tξ ρ ξ∂ ∂=

∂ ∂

2 2

2 2 2

1x v tξ ξ∂ ∂

∴ =∂ ∂

Yvρ

=

Longitudinal vibrations of a rod …2

43

Apply boundary conditions: one end fixed and the other free

x = L :

x = 0 : i.e. 0Φ =

n = 1,2,3,…

( , ) ( )cosx t f x tξ ω=Look for solutions of the type:

(0, ) 0tξ =

0F Yxξα ∂

= =∂

( ) sinf x A xvω = +Φ

where

… then cos 0Lvω =

or ( )12L n

vω π= −

( ) ( )1 12 2

n

n v n YL Lπ π

ωρ

− −= =The natural angular frequencies

Longitudinal vibrations of a rod …3

44

x = 0 x = L

n = 1

n = 3

n = 2

n = 5

n = 4

1 2Y

Lπω

ρ=

232

YLπω

ρ=

352

YLπω

ρ=

45n = 1 n = 2 n = 3

Normal modes for different boundary conditions

Simply supported

Clamped one end

Free both ends

Clamped both ends

46

The elasticity of a gasl

A , pρBulk modulus: dpK V

dV= −

Kinetic theory of gases: Pressure 2 21rms rms3 3

mp v vAl

ρ= =

If then21rms2kE mv=

23

kEpA l

=

Now move piston so as to compress the gas… work done on gas: kW pA l E∆ = − ∆ = ∆

Then ( )2

2 2 2 5( )3 3 3 3

kk

E l l lp E pA l p pA l A l A l l∆ ∆ ∆ ∆

∆ = − ∆ = − ∆ − = −

adiabatic53

pK V pV∆

= − =∆

giving

and 1.667K pv

ρ ρ= =

Frenchpage 176

47

Sound waves in pipes

A sound wave consists of a series of compressions and rarefactions of the supporting medium (gas, liquid, solid)

In this wave individual molecules move longitudinally with SHM. Thus a pressure maximum represents regions in which the molecules have approached from both sides, receding from the pressure minima.

wave propagation

Frenchpage 174

48

Longitudinal wave on a spring

49

Pressure pp0

Flow velocity u

0

x

x

x

x

0 :t =

2 :t T=p

u

Standing sound waves in pipes

50

Standing sound waves in pipes …2

Consider a sound wave in a pipe. At the closed end the flow velocity is zero (velocity node, pressure antinode). At the open end the gas is in contact with the atmosphere, i.e. p = p0 (pressure node and velocity antinode).

p

u

p0

0

pressure node

velocity antinode

pressure antinode

velocity node

Open end Closed end

51

Pipe closed at both ends

Pipe open at both ends

Pipe open at one end

nn v

Lπω =

2 2n nvL

fλ

= =

2nvfL

=

( ) ( )2 1 2 14 4

n n vLf

λ− −= =

( )2 14

n vf

L−

=

( )2 12n

n vLπ

ω−

=

Standing sound waves in pipes …3

52

Sound

Audible sound is usually a longitudinal compression wave in air to which the eardrum responds.

Velocity of sound (at NTP) ~ 330 m s-1

By considering the transport of energy by a compression wave, can show that 2 2 22 mP f Avsπ ρ=

… where A is cross sectional area of air column and sm is maximum displacement of air particle in longitudinal wave

Then intensity = 2 2 22 mP f vsA

π ρ= unit: W m-2

53

The human ear detects sound from ~10-12 W m-2 to ~1 W m-2

… use a logarithmic scale for I :

100

10log II

β

=

decibels

where I0 = “reference intensity” = 10-12 W m-2

Intensity level or “loudness”:

Sound …2

54

Musical sounds

Waveforms from real musical instruments are complex, and may contain multiple harmonics, different phases, vibrato, ...

Pitch is the characteristic of a sound which allows sounds to be ordered on a scale from high to low (!?). For a pure tone, pitch is determined mainly by the frequency, although sound level may also change the pitch. Pitch is a subjective sensation and is a subject in “psychoacoustics”.

The basic unit in most musical scales is the octave. Notes judged an octave apart have frequencies nearly (not exactly) in the ratio 2:1. Western music normally divides the octave into 12 intervals called semitones ... which are given note names (A through G with sharps and flats) and designated on musical scales.

55

Timbre is used to denote “tone quality” or “tone colour” of a sound and may be understood as that attribute of auditory sensation whereby a listener can judge that two sounds are dissimilar using any criteria other than pitch, loudness or duration. Timbre depends primarily on the spectrum of the stimulus, but also on the waveform, sound pressure and temporal characteristics of the stimulus.

Musical sounds ...2

One subjective rating scale for timbre (von Bismarck, 1974)

dullcompact

full

colourful

sharpscattered

empty

colourless

56

Two dimensional systems

… the membrane has mass per unit area , and a surface tension S which gives a force SΔl perpendicular to a length Δl in the surface …

y

x

ΔyΔx

The forces on the shaded portion are …

SΔx

SΔx

SΔySΔy

Consider an elastic membrane clamped at its edges …

Frenchpage 181

σ

57

If the membrane is displaced from the z = 0 plane then a cross section through the shaded area shows:

θ

θ θ+ ∆

z

x xx x+ ∆

SΔy

SΔy

… looks exactly like the case of the stretched string.2

2

zS y xx∂

∆ ∆∂

The transverse force on the element will be

And if we looked at a cross section perpendicular to this … the transverse force will be 2

2

zS x yy∂

∆ ∆∂

Two dimensional systems …2

58

The mass of the element is .x yσ∆ ∆

Thus2 2 2

2 2 2

z z zS y x S x y x yx y t

σ∂ ∂ ∂∆ ∆ + ∆ ∆ = ∆ ∆

∂ ∂ ∂

or2 2 2

2 2 2

z z zx y S t

σ∂ ∂ ∂ + = ∂ ∂ ∂

… a two dimensional wave equation

… with the wave velocity Svσ

=


59

( , , ) ( ) ( )cosz x y t f x g y tω=2 2

2 2 ( )cosz d f g y tx dx

ω∂=

∂

( )2

22 ( ) ( ) cosz f x g y t

tω ω∂

= −∂

Look for normal mode solutions of the form:

2 2

2 2 ( )cosz d g f x ty dy

ω∂=

∂

2 2

2 2( )cos ( )cosd f d gg y t f x tdx dy

ω ω+ =

2

2 ( ) ( )cosf x g y tvω ω−

2 2 2

2 2 2

1 1d f d gf dx g dy v

ω+ = −i.e.

In a similar fashion to the 1D case, find …

1

1( ) sinnx

n xf x ALπ

=

and2

2( ) sinny

n yg y BLπ

=


60

…then1 2 1 2

1 2,( , , ) sin sin cosn n n n

x y

n x n yz x y t C tL Lπ π ω

=

1 2

22

1 2,n n

x y

n v n vL Lπ πω

= +

where the normal mode frequencies are

e.g. for a membrane having sides 1.05L and 0.95L

then1 2

2 21 2

, 1.05 0.95n nn nv

Lπω = +


61

up

down1,1

2,1 2,2

3,1 3,2

Normal modes of a rectangular membrane

62

Normal modes of a circular membrane

1,0 2,0 3,0

1,1 2,1 2,2

63

Modes of vibration of a 38 cm cymbal. The first 6 modes resemble those of a flat plate ... but after that the resonances tend to be combinations of two or more modes.

64

Normal modes of a circular drum

65

Chladni plates

66

Soap films

67

Holographic interferograms of the top and bottom plates of a violin at several resonances.

68

Holographic interferograms of a classical guitar top plate at several resonances.

69

Holographic interferograms showing the vibrations of a 0.3 mm thick trombone driven acoustically at 240 and 630 Hz.

70

Time-average hologram interferograms of inextensional modes in a C5 handbell

71

one point per normal mode

1 2

2 21 2

, 2 2n nn v n vf

L L = +

Normal modes of a square membrane

12vL

22vL

32vL

42vL

01

2vL

22vL

32vL

42vL

52vL

0

0 1 2 3 4 5

0

1

2

3

4

4,3f

1n

2n

Normal modes having the same frequency are said to be degenerate

area per point =2

2vL

72

Normal modes of a square membrane … for large n1 and n2

1n

2ndf

f

area =

2

2vL

area per mode =

14 (2 )f dfπ

Number of modes with frequencies between f and (f + df) =

214

2(2 ) Lf dfv

π 2

2

2 L f dfv

π=

73

Three dimensional systems

Consider some quantity Ψ which depends on x , y , z and t , e.g. the density of air in a room.

In three dimensions:2 2 2 2

2 2 2 2 2

1x y z v t

∂ Ψ ∂ Ψ ∂ Ψ ∂ Ψ+ + =

∂ ∂ ∂ ∂

which can be written:2

22 2

1v t

∂ Ψ∇ Ψ =

∂

The solutions for a rectangular enclosure:

1 2 3

22 2

31 2, ,n n n

x y z

n vn v n vL L L

ππ πω

= + +

…and for a cube:1 2 3

2 2 2, , 1 2 3n n n

v n n nLπω = + +

Frenchpage 188

74

How many modes are there with frequencies in the rangef and (f + df) … ?Set up an imaginary cubic lattice with spacing 2v L

2n

1n

3n

df

f

… and consider positive frequencies only.

Volume of shell = 218 (4 )f dfπ

3

2vL

Volume per mode =

Number of modes with frequencies between f and (f + df) =

321

82(4 ) Lf dfv

π

3 2

3

4 L f dfv

π=

Three dimensional systems …2

75

Number of modes with frequencies between f and (f + df)

2

3

4 V f dfv

π=

… holds for any volume V … provided its dimensions are much greater than the wavelengths involved.… need to multiply by a factor of 2 when dealing with electromagnetic radiation (2 polarization states) …

“Ultraviolet catastrophe” for blackbody radiation …

Equipartition theorem: in thermal equilibrium each mode has an average energy in each of its two energy storesHence, energy density of radiation in a cavity:

12 Bk T

( )2

123

4 2 2V f dfdf kTc

πµ =

or2

3

8 f kTcπµ =

f

µexperiment

!?

Three dimensional systems …3

76

Planck was able to show, effectively by assuming that energy was emitted an absorbed in quanta of energy hf , that the average energy of a cavity mode was not kTbut

1hf kT

hfe −

where Planck’s constant h = 6.67 10-34 J K-1

Then2

3

8 1hf kT

f df hfdfc e

πµ =−

… which agrees extremely well with experiment.

energy density

no. of modes in range f to f +df

average energy per

mode

Planck’s law

78

Introduction to Fourier methods

We return to our claim that any physically observed shape function of a stretched string can be made up from normal mode shape functions.

x

( )f x

i.e.1

( ) sinnn

n xf x BLπ∞

=

=

∑… a surprising claim … ? … first find nB

… multiply both sides by

and integrate over the range x = 0 to x = L

1sin n xLπ

1 1

10 0

( )sin sin sinL L

nn

n x n x n xf x dx B dxL L Lπ π π∞

=

=

∑∫ ∫

Frenchpage 189

79

1 1

10 0

( )sin sin sinL L

nn


=

=

∑∫ ∫

Fourier methods …2

If the functions are well behaved, then we can re-order things:

1 1

10 0

( )sin sin sinL L

nn


=

=

∑∫ ∫

[n1 is a particular integer and n can have any value between 1 and .] ∞

Integral on rhs:

( ) ( )1 11

0 0

1sin sin cos cos2

L L n n x n n xn x n x dx dxL L L L

π ππ π − + = −

∫ ∫

80


Both (n1 + n) and (n1 − n) are integers, so the functions

on the interval x = 0 to Lmust look like… from which it is evident that

( )1cosn n x

Lπ ±

( )1

0

cos 0L n n x

dxL

π ±=

∫

Except for the special case when n1 and n are equal … then … ( )1cos 1n n x

Lπ −

=

and( )1

0

cosL n n x

dx LL

π −=

∫

81


Thus all the terms in the summation are zero, except for the single case when n1 = n i.e.

1 2nLB=

( ) ( )1

1 11

0 0

1( )sin cos cos2

L L

n

n n x n n xn xf x dx B dxL L L

π ππ − + = −

∫ ∫

1

1

0

2 ( )sinL

nn xB f x dx

L Lπ =

∫i.e.

We have found the value of the coefficient for some particular value of n1 … the same recipe must work for any value, so we can write:

0

2 ( )sinL

nn xB f x dx

L Lπ =

∫

82


The important property we have used is that the functions

and

are “orthogonal over the interval x = 0 to x = L.”

i.e.

1sin n xLπ

sin n xLπ

1

0

sin sinL n x n x dx

L Lπ π =

∫10 if n n≠

1 if 2L n n=

Read French pages 195-6

83


The most general case (where there can be nodal or antinodal boundary conditions at x = 0 and x = L) is

0

1( ) cos sin

2 n nn

A n x n xf x A BL Lπ π∞

=

= + +

∑

where

0

2 ( )sinL

nn xB f x dx

L Lπ =

∫

0

2 ( )cosL

nn xA f x dx

L Lπ =

∫

84

Fourier methods …7One of the most commonly encountered uses of Fourier methods is the representation of periodic functions of time in terms of sine and cosine functions …

Put 2Tπ

Ω =

This is the lowest frequency in … clearly there are higher frequencies … by the same method as before, write …

( )f t

0

1

2 2( ) cos sin2 n n

n

A nt ntf t A BT Tπ π∞

=

= + +

∑0

1cos sin

2 n nn

A A n t B n t∞

=

= + Ω + Ω∑

where and ( )0

2 ( )sinT

nB f t n t dtT

= Ω∫( )0

2 ( )cosT

nA f t n t dtT

= Ω∫

T( )f t

t

85

Waveforms of ...

a flute

a clarinet

an oboe

a saxophone

86

Fast Fourier transform experiments, 10 March 2008

87


88


89


90


91

Odd functions

An odd periodic function ( ) ( )f t f t− = −

where 2 2T t T− < <

t tt

( )f t ( )f t ( )f t

…can be expressed as a sum of sine functions

0( ) sinn

nf t B n t

∞

=

= Ω∑ 2Tπ

Ω =

( )0

2 ( )sinT

nB f t n t dtT

= Ω∫

92

Even functions

An even periodic function ( ) ( )f t f t− = +

where 2 2T t T− < <

t tt

( )f t ( )f t ( )f t

…can be expressed as a sum of cosine functions2Tπ

Ω =0

1( ) cos

2 nn

Af t A n t∞

=

= + Ω∑

( )0

2 ( )cosT

nA f t n t dtT

= Ω∫

93

Fourier methods … Example

Find Fourier coefficients for the case:( )f t

t

1

-1

This is an odd function: ( ) ( )f t f t− = −1

( ) sinnn

f t B n t∞

=

∴ = Ω∑

( )0

2 ( )sinT

nB f t n t dtT

= Ω∫

( ) ( )2

0 2

2 2(1)sin ( 1)sinT T

T

n t dt n t dtT T

= Ω + − Ω∫ ∫

( ) ( )2

0 2

2 1 2 1(1) cos ( 1) cosT T

Tn t n t

T n T n = − Ω + − − Ω Ω Ω

94

Fourier methods … Example cont.2Tπ

Ω =

[ ] [ ]1 1cos cos0 cos 2 cosnB n n nn n

π π ππ π

∴ = − − + −

For even n: cos 2 cos 1n nπ π= = even 0nB∴ =

For odd n: and cos 1nπ = − cos 2 1nπ = +

odd 1 1 4( 1 1) (1 1)nB

n n nπ π π∴ = − − − + + =

4 1 1( ) sin sin3 sin5 ...3 5

f t t t tπ ∴ = Ω + Ω + Ω +

95

Fourier sums … Example 3

2 terms

8 terms

4 terms

200 terms

50 terms

20 terms2T

2T−

96


T2 terms

4 terms

3 terms

50 terms

20 terms

8 terms0

97


T2 terms

8 terms

4 terms

200 terms

50 terms

20 terms0

98

t

t

t

t

t

t

f

f

f

f

f

f

Time domain Frequency spectrum

1f

1 1 1 1 3 5 7f f f f

17 f

1 1 1 1 3 5 7f f f f

Fourier transforms

vibrations and waves 2

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