Topic 4 transverse wave

Topic 2-1 Transverse Wave Motion

1

UEEP1033 Oscillations and Waves

Topic 4

Transverse Wave Motion


2


Contents• Revision of 1-D wave equation, waves on a

stretched string, polarization• Wave impedance• Reflection and transmission• Impedance matching• Compression waves in a fluid • Waves in 2- and 3-D• Standing waves in a box• Wave groups, group velocity• Dispersion• Waveguides: Cut-off and dispersion in a

confined membrane


3


Definition of Waves

• A wave is a disturbance that moves through a medium without giving the medium, as a whole, any permanent displacement.

• The general name for these waves is progressive wave.• If the disturbance takes place perpendicular to the

direction of propagation of the wave, the wave is called transverse.

• If the disturbance is along the direction of propagation of the wave, it is called longitudinal.


4


Characteristics of Waves

• At any point, the disturbance is a function of time and at any instant, the disturbance is a function of the position of the point.

• In a sound wave, the disturbance is pressure-variation in a medium.

• In the transmission of light in a medium or vacuum, the disturbance is the variation of the strengths of the electric and magnetic fields.

• In a progressive wave motion, it is the disturbance that moves and not the particles of the medium.


5


• To demonstrate wave motion, take the loose end of a long rope which is fixed at the other end quickly up and down

• Crests and troughs of the waves move down the rope

• If the rope is infinity long such waves are called progressive waves

Progressive Waves


6


• If the rope is fixed at both ends, the progressive waves traveling on it are reflected and combined to form standing waves

Standing Waves

The first four harmonics of the standing waves allowed between the two fixed ends of a string


7


Transverse vs Longitudinal Waves

• Transverse wave: the displacements or oscillations in the medium are transverse to the direction of propagation e.g. electromagnetic (EM) waves , waves on strings

• Longitudinal wave: the oscillations are parallel to the direction of wave propagatione.g. sound waves


8


Plane Waves

• Take a plane perpendicular to the direction of wave propagation and all oscillators lying within that plane have a common phase

• Over such a plane, all parameters describing the wave motion remain constant

• The crests and troughs are planes of maximum amplitude of oscillation, which are rad out of phase

• Crest = a plane of maximum positive amplitude• Trough = a plane of maximum negative amplitude


9


The Wave Equation

2

2

22

2 1

t

y

cx

y

T

c 2

+d

T

T

(x +dx, y +dy)

(x , y )

• The wave equation of small element of string of linear density and constant tension T

where

c is the phase or wave velocity.


10


2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

222

tansin small very

sin)sin(

ignored becan thusand small very

1

t

y

Tx

y

t

ydxdx

x

yT

t

ydx

x

y

x

yT

x

yt

ydxTdT

dxdsdxdsx

y

x

y

x

y

dx

ds

x

y

dx

dx

dx

sd

dydxds

xdxx

xdxx

xdxx

x

y

x

ydx

x

y

x

y

x

y

dxx

y

2

2

2

2 1


11


Waves in One Dimension

• Suppose a wave moves along the x-axis with constant velocity c and without any change of shape (i.e. with no dispersion) and the disturbance takes place parallel to the y-axis, then

y (x, t) = f (ct – x) (1)

defines a one-dimensional wave along the positive direction of the x-axis (forward wave)

t

xc


12



• A wave which is the same in all respect but moving in the opposite direction (i.e. along the direction of x decreasing) is given by Eqn. (1) with the sign of v changed:

y (x, t) = f (ct + x) (2)

• This is known as backward wave.


13


2

2

22

2

12

2

2

1

12

2

1

1

1

)(),(

)(),(

)(

t

y

cx

y

xctfct

yxctfc

t

y

xctfx

yxctf

x

y

xctfy


14



• Eqns. (1) and (2) satisfy the second-order partial differential equation:

(3)

• Eqn. (3) is known as the non-dispersive wave equation.

2

2

22

2 1

t

y

cx

y


15


Locus of oscillator displacements

x

Dis

plac

emen

t y

a

-a


16


Solution of Wave equation

• A solution to the wave equation

, where is the oscillation

frequency and

• The wave is moving in the positive x direction.

)(2

sin)sin( xctatay

x2

22

c


17


The Wave Equation• At position x = 0, wave equation • Any oscillator to its right at some position x will

be set in motion at some later time t.

• Have a phase lag with respect to the oscillator at x = 0.

• The wavelength is the separation in space between any two oscillators with a phase difference 2 rad.

tay sin

)(2

sin)sin( xctatay


18


The Wave Equation

• The period of oscillation• An observer at any point would be passed by

wavelengths per second.• If the wave is moving to the left the sign is

changed.• Wave moving to right• Wave moving to left

1

c

)(2

sin)sin( xctatay

)(2

sin)sin( xctatay


19


Equivalent Wave Expressions

where is called wave number.• Cosine functions are equally valid.• For both sine and cosine

)(2

sin xctay

)(2sin

xtay

)(sinc

xtay

)sin( kxtay

ck

2

kxtiaey


20


The Wave Equation

2

2

22

2

2

2

2

2

22

2

22

2

1

)sin(),cos(

)sin(),cos(

)sin(

t

y

ct

yk

x

y

x

y

t

x

x

yc

x

y

kt

y

kxtakx

ykxtka

x

y

kxtat

ykxta

t

y

kxtay


21


Wave or Phase velocity• The wave or phase velocity is

• It is the rate at which disturbance moves across the oscillators.

• The oscillator or particle velocity is a simple harmonic velocity

t

xc

t

y

)cos(

)sin(

kxtat

y

kxtay


22


Particle Velocity

arrows show the direction

and magnitude of the particle

velocity

x

yc

t

y


23



• δ is called the phase of y2 relative to y1 and d the path difference:

differencephase2

differencePath

• If δ = 2π, 4π,..., then d = λ, 2λ,..., and we say that the waves are in phase, and y1 = y2.

• If δ = π, 3π,..., then the two waves are exactly out of phase and y1 = – y2.


24


Three Velocities in Wave Motion

1. Particle velocitySimple harmonic velocity of the oscillator about its equilibrium position

2. Wave or phase velocityThe velocity with which planes of equal phase, crests or troughs, progress through the medium

3. Group velocityA number of waves of different frequencies, wavelengths and velocities may be superposed to form a group. Motion of such a pulse would be described by its group velocity


25


• Locus of oscillator displacements in a continuous medium as a wave passes over them travelling in the positive x-direction

• The wavelength is defined as the distance between any two oscillators having a phase difference of 2 rad


26


Wave or Phase Velocity

Wave or Phase Velocity = the rate at which disturbance moves across the oscillators

Wave or Phase Velocity =t

x

Oscillator or Particle Velocity is a simple harmonic velocity

Oscillator or Particle Velocity = t

y


27


Characteristic Impedance of a String

• Any medium through which waves propagate will present an impedance to those waves

• If the medium is lossless, and possesses no resistive or dissipation mechanism, for a string the impedance is determined by inertia and elasticity

• The presence of a loss mechanism will introduce a complex term into the impedance

(the string as a forced oscillator)


28


• The transverse impedance is define as:

Characteristic Impedance of a String

• Characteristic impedance of the string:

(the string as a forced oscillator)

v

FZ

velocitytransverse

forcetransverse

cc

TZ 2since cT


29


Characteristic Impedance of a String(the string as a forced oscillator)

The string as a forced oscillator with a vertical force F0eit driving it at one end

For small :

x

yTTTeF ti tansin0


30



displacement of the progressive waves may be represented exponentially by:

amplitude A may be complex

At the end of the string, where x = 0

)( kxtie Ay

)0(

00

kti

x

ti eikTx

yTeF A

T

c

i

F

ikT

F 00A )(0 kxtieT

c

i

F

y


31



transverse velocity:

velocity amplitude:

transverse impedance:

Characteristic Impedance of the string

Since the velocity c is determined by the inertia and the elasticity, the impedance is also governed by these properties

)(0

kxtieT

cF

yv

ZFv /0

cc

TZ 2since cT


32


Reflection and Transmission


33


Z1 = 1c1

Z2 = 2c2

Reflection and Transmission• Suppose a string consists of two sections smoothly joined at

a point x = 0 with a tension T• Waves on a string of impedance Z1= 1c1 reflected and

transmitted at the boundary x = 0 where the string changes to impedance Z2= 2c2


34



Incident wave:

Reflected wave:

Transmitted wave:

find the reflection and transmission amplitude coefficients

i.e. the relative values of B1 and A2 with respect to A1

)(1

1xktii eAy

)(1

1xktir eBy

)(2

2xktit eAy


35


)(1

1xktii eAy

)(1

1xktir eBy

)(2

2xktit eAy

find the reflection and

transmission amplitude

coefficients i.e. the relative

values of B1 and A2 with respect

to A1


36


Boundary condition No. 1 at the impedance discontinuity at x = 0


1. A geometrical condition that the displacement is the same immediately to the left and right of x = 0 for all time, so that there is no discontinuity of displacement

tri yyy

)(2

)(1

)(1

211 xktixktixkti eAeBeA

0At x )1(Eq211 ABA


37


Boundary condition No. 2 at the impedance discontinuity at x = 0


2. A dynamical condition that there is a continuity of the transverse force T(y/x) at x = 0, and therefore a continuous slope

tri yx

Tyyx

T

at x = 0 for all t

221111 TAkTBkTAk

22

11

11

Ac

TB

c

TA

c

T


38



These coefficients are independent of

2222

1111

and Zcc

TZc

c

T

)2(Eq)( 22111 AZBAZ

Reflection coefficient of amplitude:21

21

1

1

ZZ

ZZ

A

B

Transmission coefficient of amplitude:21

1

1

2 2

ZZ

Z

A

A

Solving Eqs. (1) and (2)


39


)(11

)(11

)(1

)(1

11

11

xktixktiri

xktixktiri

eBikeAikyyx

eBeAyy

tri yx

Tyyx

T

)(22

)(2

2

2

xktit

xktit

eAikyx

eAy

11110,0At BikAikyyx

tx ri

220,0At Aikyx

tx t


40


tri yx

Tyyx

T

tx

0,0At

221111 AikBikAik

22

11

11

Ac

TB

c

TA

c

T

221111 AkBkAk

22111 )( AZBAZ 222

2

1111

Zcc

T

Zcc

T


41


• If Z2 = , B1/A1= 1

incident wave is completely reflected with a phase change of

(conditions that necessary for standing waves to exist)

• If Z2 = 0 (x =0 is a free end of the string)

B1/A1= 1, A2/A1= 2

the flick at the end of a whip or free end string


21

21

1

1

ZZ

ZZ

A

B

21

1

1

2 2

ZZ

Z

A

A


42


• If Z2 = , B1/A1= 1

incident wave is completely reflected with a phase change of (conditions that necessary for standing waves to exist)

• If Z2 = 0

(x =0 is a free end of the string)

B1/A1= 1, A2/A1= 2

the flick at the end of a whip or free end string


43


Reflection and Transmission of EnergyWhat happens to the energy in a wave when it meets a

boundary between two media of different impedance values?(the wave function of transferring energy throughout a medium)

Consider each unit length, mass , of the string as a simple harmonic oscillator of maximum amplitude A

Total energy: = wave frequency

The rate at which energy is being carried along the string:

22

2

1AE

cA22

2

1velocity)(energy


44


Reflection and Transmission of Energy

The rate at which energy leaves the boundary, via the reflected and transmitted waves:

the rate of energy arriving at the boundary x = 0 is the energy arriving with the incident wave:

energy is conserved, and all energy arriving at the boundary in the incident wave leaves the boundary in the reflected and transmitted waves

21

21

21

211 2

1

2

1AZAc

21

212

21

221

22112

12

22

22

21

21

22

222

21

211

2

1

)(

4)(

2

1

2

1

2

1

2

1

2

1

AZZZ

ZZZZZA

AZBZAcBc


45


Reflected and Transmitted Intensity Coefficients

If Z1 = Z2 no energy is reflected and the impedances are said to be matched

2

21

21

2

1

1211

211

EnergyIncident

Energy Reflected

ZZ

ZZ

A

B

AZ

BZ

221

21211

222 4

EnergyIncident

Energy dTransmitte

ZZ

ZZ

AZ

AZ


46


Matching of Impedances

Why Important? • Long distance cables carrying energy must be accurately

matched at all joints to avoid wastage from energy reflection

Example:• The power transfer from any generator is a maximum

when the load matches the generator impedance

• A loudspeaker is matched to the impedance of the power output of an amplifier by choosing then correct turns ratio on the coupling transformer


47


Matching of ImpedancesInsertion of a coupling element

between two mismatched impedances

Remark: when a smooth joint exists between two strings of different impedances, energy will be reflected at the boundary

Goal: to eliminate energy reflection and match the impedances

Require to match the impedances Z1 = 1c1 and Z3 = 3c3 by

the smooth insertion of a string of length l and impedance

Z2 = 2c2

Our problem is to find the values of l and Z2


48


Matching of ImpedancesThe impedances Z1 and Z3 of two strings are matched by the

insertion of a length l of a string of impedance Z2


49



we seek to make the ratio

Boundary conditions: y and T(y/x) are continuous across the junctions

x = 0 and x = l

1EnergyIncident

Energy dTransmitte211

233

AZ

AZ


50



Between Z1 and Z2 the continuity of y gives:

Continuity of T(y/x) gives

Dividing the above equation by and remember

At x = 0

)(2

)(2

)(1

)(1

2211 xktixktixktixkti eBeAeBeA

)0at(2211 xBABA

22221111 BikAikTBikAikT

ZcT/ckT /

222111 BAZBAZ


51


Matching of ImpedancesAt x = l

Continuity of T(y/x) gives:

Continuity of y gives:

From the four boundary equations, solve for the ratio A3/A1

Refer to the H.J. Pain, “The Physics of Vibrations and Waves”,6 th Edition, pg 122-123 for detail derivation

32222 AeBeA liklik

3322222 AZeBeAZ liklik

lkrrlkr

r

A

A

222

2312222

13

213

2

1

3

sincos1

4


52



21

23

13211

233 1

EnergyIncident

Energy dTransmitte

A

A

rAZ

AZ

lkrrlkr

r

222

2312222

13

13

sincos1

4

havewe1sinand0cos,4/chooseweif 222 lklkl

231222312

13211

233 when1

4rr

rr

r

AZ

AZ


53


Standing Waves on a String of Fixed Length


54


• A string of fixed length l with both ends rigidly clamped

• Consider wave with an amplitude a traveling in the positive x-direction and an amplitude b traveling in the negative x-direction

• The displacement on the string at any point is given by:

Standing Waves

with the boundary condition that y = 0 at x = 0 and x = l

)()( kxtikxti beaey


55



56


Standing Waves

Boundary condition: y = 0 at x = 0

A wave in either direction meeting the infinite impedance at either end is completely reflected with a phase change in amplitude

a = b

An expression of y which satisfies the standing wave time dependent form of the wave equation:

kxaeieeaey tiikxikxti sin2

022

2

ykx

y

tieba )(0


57


Standing Waves

Boundary condition: y = 0 at x = l

Limiting the value of allowed frequencies to:

22n

nn

nl

c

l

ncf

nc

l

c

lkl 0sinsin

l

cnn

kliae

eeae

beae

ti

ikliklti

kltiklti

sin20

0

0 )()(


58


Standing Wavesnormal frequencies or modes of vibration:

Such allowed frequencies define the length of the string as an exact number of half wavelengths

(Fundamental mode)

The first four harmonics, n =1, 2, 3, 4 of the standing waves allowed between the two fixed ends of a string

l

xn

c

xn

sinsin

2nn

l


59



60


Standing Waves• For n > 1, there will be a number of positions along the

string where the displacement is always zero called nodes or nodal point

These points occur where

there are (n1) positions equally spaced along the string in the nth harmonic where the displacement is always zero

• Standing waves arise when a single mode is excited and the incident and reflected waves are superposed

• If the amplitudes of these progressive waves are equal and opposite (resulting from complete reflection), nodal points will exist

0sinsin

l

xn

c

xn

),.....,3,2,1,0( nrrl

xn

0sin xkn

rxkn


61


Standing Waves

the complete expression for the displacement of the nth harmonic is given by:

c

xtBtAy n

nnnnn

sinsincos

c

xtitiay n

nnn

sinsincos)(2

where the amplitude of the nth mode is given by aBA nn 22/122

we can express this in the form:


62


Standing Wave Ratio• If a progressive wave system is partially reflected from a

boundary, let the amplitude reflection coefficient B1/A1 = r, for r < 1

• The maximum amplitude at reinforcement is (A1 + B1), the minimum amplitude (A1 B1)

• The ratio of the maximum to minimum amplitudes is called standing wave ratio (SWR)

• Reflection coefficient:

r

r

BA

BA

1

1SWR

11

11

1SWR

1SWR

1

1

A

Br


63


Energy in Each Normal Mode of a Vibrating String

• A vibrating string possesses both kinetic and potential energy

• Kinetic energy of an element of length dx and linear density

• Total kinetic energy:

221 ydx

dxyE 21

021)kinetic(


64



• Potential energy = the work done by thee tension T in extending an element of length dx to a new length ds when the string is vibrating

neglect higher powers of y/x

dxT

dxTdxdsTE

xy

xy

2

2

2

1

11)()potential(21

....112

21

2 21

xy

xy

...2

)1(1)1( 2

x

nnnxx n


65


Energy in Each Normal Mode of a Vibrating String• For standing waves:

cxntBtAy nnnnn

sinsincos

cx

nnnnnnnntBtAy sincossin

cx

nnnncn nn tBtAx

y

cossincos

dxtBtAE cxl

nnnnnnn 2

0

2221 sincossin)kinetic(

dxtBtATE cxl

nnnncnnn 2

0

221 cossincos)potential( 2

2


66



where m is the mass of the string

= the square of the maximum displacement of the mode

2cT

)(

)()potentialkinetic(222

41

22241

nnn

nnnn

BAm

BAlE

)( 22nn BA


67


a

axxaxdx

4

2sin

2sin 2 a

axxaxdx

4

2sin

2cos2

2

sin0)/(4

)/(2sin2

2

0

ldx

l

cxcx

cxl

n

nn

2

cos0)/(4

)/(2sin2

2

0

ldx

l

cxcx

cxl

n

nn

dxtBtAE cxl

nnnnnnn 2

0

2221 sincossin)kinetic(

dxtBtATE cxl

nnnncnnn 2

0

221 cossincos)potential( 2

2


68


222

21 cossin)kinetic( l

nnnnnn tBtAE

224122

41)kinetic( nnnnn BmBlE At any time t:

22

21 sincos)potential( 2

2l

nnnncn tBtATE n

At any time t: 224122

41)potential( nnnnn AmAlE

2cT

)(

)()potential()kinetic(222

41

22241

nnn

nnnnn

BAm

BAlEE


69


Wave Groups and Group Velocity

• Waves to occur as a mixture of a number or group of component frequencies

e.g. white light is composed of visible wavelength spectrum of 400 nm to 700 nm

• The behavior of such a group leads to the group velocity

dispersion causes the spatial separation of a white light into components of different wavelength (different colour)


70


Superposition of two waves of almost equal frequencies

• A group consists of two components of equal amplitude a but frequencies 1 and 2 which differ by a small amount.

• Their displacements:

• Superposition of amplitude and phase:

)cos()cos( 222111 xktayxktay

2

)(

2

)(cos

2

)(

2

)(cos2 21212121

21

xkktxkktayyy

a wave system with a frequency (1+ 2)/2 which is very close to the frequency of either component but with a maximum amplitude of 2a, modulated in space and time by a very slowly varying envelope of frequency (1 2)/2 and wave number (k1 k2)/2


71




72


• The velocity of the new wave is


so that the component frequencies and their superposition, or group will travel with the same velocity, the profile of their combination in Figure 5.11 remaining constant

)/()( 2121 kk

ckk 2211 //If the phase velocities , gives

ckk

kkc

kk

21

21

21

21 )(


73


• For the two frequency components have different phase velocities so that 1/k1 2/k2


The superposition of the two waves will no longer remain constant and the group profile will change with time

Dispersive medium = medium in which the phase velocity is frequency dependent

(i.e. /k not constant)

kkk

21

21velocityGroup


74


• If a group contain a number of components of frequencies which are nearly equal the original, expression for the group velocity is written:


Since = kv (v is the phase velocity)

group velocity:

gvdk

d

k

dk

dvkvkv

dk

d

dk

dvg

)(

d

dvvvg


75


• A non-dispersive medium where /k is constant, so that vg = v, for instance free space behaviour towards light waves

• A normal dispersion relation, vg < v

• An anomalous dispersion relation, vg > v


76


Standing Waves as Normal Modes of Vibrating String


77


Characteristic of a Normal Mode

• all the masses move in SHM at the same frequency

• normal modes are completely independent of each other

• general motion of the system is a superposition of the normal modes

• All of these properties of normal modes are shared by standing waves on a vibrating string

• all the particles of the string perform SHM with the same frequency

• the standing waves are the normal modes of the vibrating string

Standing Waves as Normal Modes


78


Superposition of Normal Modes

the expression for the n-th normal mode of a vibrating string of length L

the motion of the string will be a superposition of normal modes given by:


79


txkAtxy nnnn cossin),(

0sin xkn

Displacement zero (nodes) occur when sine term = 0

,....)2,1,0( nnxkn


80


Example: superposition of the 3rd normal mode with a relative amplitude of 1.0 and the 13th normal mode with a relative amplitude of 0.5

3rd harmonic y3(x, 0) of a string at t = 0

13th harmonic y13(x, 0) of a string at t = 0

The superposition of the two harmonics to give the resultant shape of the string at t = 0

(a)

(b)

(c)


81


• To excite the two normal modes in this way, we would somehow have to constrain the shape of the string as in (c) and then release it at time t = 0

• It is impractical to do this and in practice we pluck a string to cause it to vibrate

• Example the string is displaced a distance d at one quarter of its length

• Initially, the string has a triangular shape and this shape clearly does not match any of the shapes of the normal modes


82


it is possible to reproduce this triangular shape by adding together the normal modes of the string with appropriate amplitudes

The first three excited normal modes of the string: y1(x, 0), y2(x, 0) and y3(x, 0)

L

xAxy sin)0,( 11

L

xAxy

2sin)0,( 22

L

xAxy

3sin)0,( 33

• Even using just the first three normal modes we get a surprisingly good fit to the triangular shape

• By adding more normal modes, we would achieve even better agreement, especially with respect to the sharp corner

The superposition of the first three normal modes gives a good reproduction of the initial triangular shape of the string except for the sharp corner


83


When we pluck a string we excite many of its normal modes and the subsequent motion of the string is given by the superposition of these normal modes according to equation

Amplitudes of Normal Modes

The initial shape of the string f (x), i.e. at t = 0 is given by


84


The expansion of the above equation is known as a Fourier series and the amplitudes A1, A2, . . . as Fourier coefficients or Fourier amplitude

Any shape f (x) of the string with fixed end points [f (0) = f (L) = 0] can be written as a superposition of these sine functions with appropriate values for the coefficients A1, A2, . . . , i.e. in the form


Fourier series


85



where m and n are integers

But:

Fourier amplitude


86


Example

A string of length L is displaced at its mid-point by a distance d and released at t = 0, as shown in figure below. Find the first three normal modes that are excited and their amplitudes in terms of the initial displacement d.


87


Let the shape of the string at time t = 0 by the function y = f (x)

Solution:

Inspection of figure shows that:

To cope with the ‘kink’ in f (x) at x = L/2, we split the integral in the Fourier amplitude equation (An in slide-9) into two parts, so that


88


Substituting for f (x) over the appropriate ranges of x, the right-hand side of this equation becomes:

Useful formula for the indefinite integrals

The final result is

Solution (continued…..):


89


An = 0 for even values of n: we only excite those modes that have odd values of n, since modes with even n have a node at the mid-point of the string and so will not be excited

the amplitudes An of these normal modes:

frequencies given by:

Solution (continued…..):

Topic 4 transverse wave

Science

waves waves

transverse wave motion

waves topic

waves plane waves

waves characteristics

waves definition of

wave velocity

progressive wave motion