transverse wave motion - hashemite university · transverse vs longitudinal waves • transverse...

Transverse Wave Motion

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.

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.

• 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

• 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

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

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

The Wave Equation

2

2

22

21

t

y

cx

y

Tc

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.

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

y

t

ydxTdT

dxdsdxds

x

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

21

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


• 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.

2

2

22

2

1

2

2

2

1

12

2

1

1

1

)(),(

)(),(

)(

t

y

cx

y

xctfct

yxctfc

t

y

xctfx

yxctf

x

y

xctfy


• 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

21

t

y

cx

y

Locus of oscillator displacements

x

Dis

pla

cem

ent

y

a

-a

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

2

2

c

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

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

Equivalent Wave Expressions

where is called wave number.

• Cosine functions are equally valid.

• For both sine and cosine

)(2

sin xctay

)(2sin

x

tay

)(sinc

xtay

)sin( kxtay

ck

2

kxtiaey

The Wave Equation

2

2

22

2

2

2

2

2

2

2

2

2

2

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

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

Particle Velocity

arrows show

the direction

and magnitude

of the particle

velocity

x

yc

t

y

• 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

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

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

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)

• 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 2

since cT

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

titansin

0


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(

0

0

kti

x

tieikT

x

yTeF A

T

c

i

F

ikT

F00

A)(0 kxti

eT

c

i

F

y


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

kxtie

T

cF

yv

ZFv /0

cc

TZ 2

since cT

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


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

1xkti

ieAy

)(

1

1xkti

reBy

)(

2

2xkti

teAy

)(

1

1xkti

ieAy

)(

1

1xkti

reBy

)(

2

2xkti

teAy

find the reflection and

transmission amplitude

coefficients i.e. the relative values

of B1 and A2 with respect to A1

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

triyyy

)(

2

)(

1

)(

1

211xktixktixkti

eAeBeA

0At x )1(Eq211

ABA

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

221111TAkTBkTAk

2

2

1

1

1

1

Ac

TB

c

TA

c

T


These coefficients are independent of

222

2

111

1

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

22

ZZ

Z

A

A

Solving Eqs. (1) and (2)

)(

11

)(

11

)(

1

)(

1

11

11

xktixkti

ri

xktixkti

ri

eBikeAikyyx

eBeAyy

tri

yx

Tyyx

T

)(

22

)(

2

2

2

xkti

t

xkti

t

eAikyx

eAy

1111

0,0At BikAikyyx

txri

220,0At Aiky

xtx

t

tri

yx

Tyyx

T

tx

0,0At

221111AikBikAik

2

2

1

1

1

1

Ac

TB

c

TA

c

T

221111AkBkAk

22111)( AZBAZ

222

2

111

1

Zcc

T

Zcc

T

• 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

22

ZZ

Z

A

A

• 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

Reflection and Transmission of Energy

What 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

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

2

1

2

1

2

1

2

112

1

2

1AZAc

2

1

2

12

21

2

2

1

2

2112

1

2

2

2

2

2

2

1

2

1

2

2

2

22

2

1

2

11

2

1

)(

4)(

2

1

2

1

2

1

2

1

2

1

AZZZ

ZZZZZA

AZBZAcBc

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

1

2

11

2

11

EnergyIncident

Energy Reflected

ZZ

ZZ

A

B

AZ

BZ

2

21

21

2

11

2

224

EnergyIncident

Energy dTransmitte

ZZ

ZZ

AZ

AZ

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


Insertion 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


The impedances Z1 and Z3 of two strings are matched by the

insertion of a length l of a string of impedance Z2


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 dTransmitte

2

11

2

33

AZ

AZ


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

2211xktixktixktixkti

eBeAeBeA

)0at(2211

xBABA

22221111

BikAikTBikAikT

ZcT/ckT /

222111

BAZBAZ


At 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”,6th Edition, pg 122-123 for detail derivation

322

22 AeBeAliklik

33222

22 AZeBeAZliklik

lkrrlkr

r

A

A

2

22

23122

22

13

2

13

2

1

3

sincos1

4


2

1

2

3

13

2

11

2

331

EnergyIncident

Energy dTransmitte

A

A

rAZ

AZ

lkrrlkr

r

2

22

23122

22

13

13

sincos1

4

havewe1sinand0cos,4/chooseweif222

lklkl

23122

2312

13

2

11

2

33when1

4rr

rr

r

AZ

AZ

Standing Waves on a String of Fixed Length

• 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

)()( kxtikxtibeaey

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:

kxaeieeaeytiikxikxti

sin2

02

2

2

yk

x

y

tieba

)(0

Standing Waves

Boundary condition: y = 0 at x = l

Limiting the value of allowed frequencies to:

22

n

n

n

nl

c

l

ncf

nc

l

c

lkl 0sinsin

l

cn

n

kliae

eeae

beae

ti

ikliklti

kltiklti

sin20

0

0)()(

Standing Waves

normal 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

2

nn

l

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

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 aBAnn

22/1

22

we can express this in the form:

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

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:

2

2

1ydx

dxyE2

1

02

1)kinetic(


• 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

x

y

x

y

2

2

2

1

11)()potential(2

1

....112

2

12 2

1

x

y

x

y

...2

)1(1)1(

2

x

nnnxx

n


• For standing waves:

c

xntBtAy

nnnnn

sinsincos

c

x

nnnnnnn

ntBtAy

sincossin

c

x

nnnnc

n nntBtA

x

y

cossincos

dxtBtAEc

xl

nnnnnn

n

2

0

22

2

1sincossin)kinetic(

dxtBtATEc

xl

nnnnc

n

nn

2

0

2

2

1cossincos)potential(

2

2


where m is the mass of the string

= the square of the maximum displacement of the

mode

2cT

)(

)()potentialkinetic(

222

4

1

222

4

1

nnn

nnnn

BAm

BAlE

)(22

nnBA

a

axxaxdx

4

2sin

2sin

2

a

axxaxdx

4

2sin

2cos

2

2

sin0

)/(4

)/(2sin

2

2

0

ldx

l

c

xcx

c

xl

n

nn

2

cos0

)/(4

)/(2sin

2

2

0

ldx

l

c

xcx

c

xl

n

nn

dxtBtAEc

xl

nnnnnn

n

2

0

22

2

1sincossin)kinetic(

dxtBtATEc

xl

nnnnc

n

nn

2

0

2

2

1cossincos)potential(

2

2

2

22

2

1cossin)kinetic(

l

nnnnnntBtAE

22

4

122

4

1)kinetic(

nnnnnBmBlE At any time t:

2

2

2

1sincos)potential(

2

2

l

nnnnc

ntBtATE

n

At any time t: 22

4

122

4

1)potential(

nnnnnAmAlE

2cT

)(

)()potential()kinetic(

222

4

1

222

4

1

nnn

nnnnn

BAm

BAlEE

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)

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

• 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)(

• 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

• 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:

gv

dk

d

k

dk

dvkvkv

dk

d

dk

dv

g

)(

d

dvvv

g

• 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

Standing Waves as Normal Modes of Vibrating String

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

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:

txkAtxynnnn

cossin),(

0sin xkn

Displacement zero (nodes) occur when sine term = 0

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

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)

• 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

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

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

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


where m and n are integers

But:

Fourier amplitude

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.

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

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…..):

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…..):

transverse wave motion - hashemite university · transverse vs longitudinal waves • transverse...

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