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 propagation
e.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
tansinsmallvery
sin)sin(
ignoredbecanthusandsmallvery
1
t
y
Tx
y
t
y
dxdx
x
y
T
t
y
dx
x
y
x
y
T
x
y
t
y
dxTdT
dxdsdxds
x
y
x
y
x
y
dx
ds
x
y
dx
dx
dx
sd
dydxds
xdxx




















































































































xdxx
xdxx
x
y
x
y
dx
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
x
c




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
1
2
2
2
1
12
2
1
1
1
)(),(
)(),(
)(
t
y
cx
y
xctfc
t
y
xctfc
t
y
xctf
x
y
xctf
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
Displacementy
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




2
2

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




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




19
Equivalent Wave Expressions
where is called wave number.
• Cosine functions are equally valid.
• For both sine and cosine
)(
2
sin xctay 


)(2sin


x
tay 
)(sin
c
x
tay  
)sin( kxtay  
c
k




2
 kxti
aey 
 

20
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
y
c
x
y
kt
y
kxtak
x
y
kxtka
x
y
kxta
t
y
kxta
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
x
c



t
y


)cos(
)sin(
kxta
t
y
kxtay







22
Particle Velocity
arrows show
the direction
and magnitude
of the particle
velocity
x
y
c
t
y






23
• δ is called the phase of y2 relative to y1 and d the path
difference:
differencephase
2
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 velocity
Simple harmonic velocity of the oscillator about its equilibrium
position
2. Wave or phase velocity
The velocity with which planes of equal phase, crests or troughs,
progress through the medium
3. Group velocity
A 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 the string:
v
F
Z 
velocitytransverse
forcetransverse
c
c
T
Z   2
since cT 

29
The string as a forced oscillator with a vertical force F0eit driving it at one end
For small : 








x
y
TTTeF 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
)( kxti
e 
 Ay
)0(
0
0











 kti
x
ti
eikT
x
y
TeF A








T
c
i
F
ikT
F 00
A )(0 kxti
e
T
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
kxti
e
T
c
F 






 yv 
ZFv /0
c
c
T
Z   2
since cT 

32
Reflection and Transmission

33
Z1 = 1c1
Z2 = 2c2
• 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
1xkti
i eAy 

)(
1
1xkti
r eBy 

)(
2
2xkti
t eAy 


35
)(
1
1xkti
i eAy 

)(
1
1xkti
r eBy 

)(
2
2xkti
t 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 y
x
Tyy
x
T






at x = 0 for all t
221111 TAkTBkTAk 
2
2
1
1
1
1
A
c
T
B
c
T
A
c
T


38
 These coefficients are independent of 
222
2
111
1
and Zc
c
T
Zc
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
xktixkti
ri
xktixkti
ri
eBikeAikyy
x
eBeAyy






  tri y
x
Tyy
x
T





)(
22
)(
2
2
2
xkti
t
xkti
t
eAiky
x
eAy






  11110,0At BikAikyy
x
tx ri 



220,0At Aiky
x
tx t 




40
  tri y
x
Tyy
x
T
tx





 0,0At
221111 AikBikAik 
2
2
1
1
1
1
A
c
T
B
c
T
A
c
T

221111 AkBkAk 
22111 )( AZBAZ 
222
2
111
1
Zc
c
T
Zc
c
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 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
1
AE 
cA22
2
1
velocity)(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
2
1
2
1
2
1
2
11
2
1
2
1
AZAc 
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
AZ
ZZ
ZZZZZ
A
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
1
2
11
2
11
EnergyIncident
EnergyReflected















ZZ
ZZ
A
B
AZ
BZ
 2
21
21
2
11
2
22 4
EnergyIncident
EnergydTransmitte
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
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

48
The 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
1
EnergyIncident
EnergydTransmitte
2
11
2
33

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
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
AeBeA liklik

  33222
22
AZeBeAZ liklik

    lkrrlkr
r
A
A
2
22
23122
22
13
2
13
2
1
3
sincos1
4








52
2
1
2
3
13
2
11
2
33 1
EnergyIncident
EnergydTransmitte
A
A
rAZ
AZ

    lkrrlkr
r
2
22
23122
22
13
13
sincos1
4


havewe1sinand0cos,4/chooseweif 222  lklkl
  23122
2312
13
2
11
2
33
when1
4
rr
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 

02
2
2



yk
x
y
ti
eba 
 )(0

57
Standing Waves
Boundary condition: y = 0 at x = l
Limiting the value of allowed frequencies to:
22
n
n
n
n
l
c
l
nc
f








 n
c
l
c
l
kl 0sinsin
l
cn
n


 
 kliae
eeae
beae
ti
ikliklti
kltiklti
sin20
0
0 )()(







58
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



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 (n1) 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( nrr
l
xn



0sin xkn
 rxkn

61
Standing Waves
the complete expression for the displacement of the nth
harmonic is given by:
 
c
x
tBtAy n
nnnnn

 sinsincos
 
c
x
titiay n
nnn

 sinsincos)(2
where the amplitude of the nth mode is given by   aBA nn 2
2/122

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
1
SWR
11
11
1SWR
1SWR
1
1



A
B
r

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:
2
2
1 ydx 
dxyE 2
1
02
1)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
x
y
x
y



















 
2
2
2
1
11)()potential(
2
1
    ....11
2
2
1
2 2
1




  



x
y
x
y
...
2
)1(
1)1( 2


 x
nn
nxx n

65
• For standing waves:
  c
xn
tBtAy nnnnn

 sinsincos
  c
x
nnnnnnn
n
tBtAy 
 sincossin
  c
x
nnnnc
n nn
tBtA
x
y 



cossincos
  dxtBtAE c
xl
nnnnnn
n
 2
0
22
2
1
sincossin)kinetic(
  dxtBtATE c
xl
nnnncn
nn 
 2
0
2
2
1
cossincos)potential( 2
2

66
where m is the mass of the string
= the square of the maximum displacement of the
mode
2
cT 
)(
)()potentialkinetic(
222
4
1
222
4
1
nnn
nnnn
BAm
BAlE


)( 22
nn BA 

67
a
axx
axdx
4
2sin
2
sin2
 a
axx
axdx
4
2sin
2
cos2

  2
sin
0)/(4
)/(2sin
2
2
0
l
dx
l
c
xcx
c
xl
n
nn
 


  2
cos
0)/(4
)/(2sin
2
2
0
l
dx
l
c
xcx
c
xl
n
nn
 


  dxtBtAE c
xl
nnnnnn
n
 2
0
22
2
1
sincossin)kinetic(
  dxtBtATE c
xl
nnnncn
nn 
 2
0
2
2
1
cossincos)potential( 2
2

68
  2
22
2
1
cossin)kinetic( l
nnnnnn tBtAE 
22
4
122
4
1
)kinetic( nnnnn BmBlE At any time t:
  2
2
2
1
sincos)potential( 2
2
l
nnnncn tBtATE n


At any time t: 22
4
122
4
1
)potential( nnnnn AmAlE 
2
cT 
)(
)()potential()kinetic(
222
4
1
222
4
1
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
xkktxkkt
ayyy
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
c
kk
kk
c
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
21
velocityGroup

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:
gv
dk
d
k





dk
dv
kvkv
dk
d
dk
d
vg 

 )(


d
dv
vvg

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
x
Axy sin)0,( 11 




 

L
x
Axy
2
sin)0,( 22 




 

L
x
Axy
3
sin)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

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Topic 4 transverse wave