Lecture 4

Waves

♦ Phase velocity and group velocity. Dispersion.

♦ Energy and transport of energy by a wave

Information transmission

Plane wave does not convey information - must be modulated

e.g.

y = Asin kx !"t( ),

kx !"t #"T / 2

kx !"t >"T / 2= 0,

Rate if information transfer : velocity of envelope … “GROUP VELOCITY”

To construct this modulated wave need a wave packet made up of an

infinite number of frequencies - Fourier series (2nd year topic)

y(x,t) = Dncos k

nx !"

nt( )

n=1

N

#

c.f. speed of individual waves of definite frequency … “PHASE VELOCITY” v=!

n

kn

"

#$%

&'

Group and phase velocity

Group velocity, g....... g =L2

t1

y(x,t) = Dncos k

nx !"

nt( )

n=1

N

#

Phase velocity, v..... e.g. v =L1

t1

(Since many frequencies need not be the same

as the group velocity)

y(x,t) = Dncos k

nx !"

nt( )

n=1

N

#

Non-dispersive medium

vn= v

y(x,t) = Dncos k

nx ! vt( )

n=1

N

" # y(x ! vt)

i.e. g=v

Dispersive medium

vn! v

i.e. g ! v

! "n= k

nv !

#"

#k$"

k= v

e.g Light through glass prism

Different colours ! different angles, because the

refractive index µ (= c/v) depends on colour (!),

i.e.v depends on !.

Dispersion: variation of wave speed with frequency(or wave number)

Simple example : superposition of two waves

y

1= Asin k + !k( )x " # + !#( )t$

%&'

y2= Asin k ! "k( )x ! # ! "#( )t$

%&'

y = y1+ y

2= 2Acos !k.x " !# .t( )sin kx "#t( )

!k << k, !" <<"

Phase velocity v !!

kGroup velocity g =

!"

!k (velocity of envelope)

•

• Superposition of travelling waves y1(x, t) and y2(x, t):

y1 = A sin[(k + δk)x− (ω + δω)t] , y2 = A sin[(k − δk)x− (ω − δω)t]

• Using sinα+ sinβ = 2 sin[(α+ β)/2] cos[(α− β)/2], the resultant wave is

y = y1 + y2 = 2A sin(kx− ωt) cos(δk x− δω t)

♠ 1st factor sin varies with frequency ω and wave number k,

i.e., close to the original waves y1 and y2,

and corresponding speed v = ω/k (phase velocity).

♠ 2nd factor cos varies much more slowly,

with frequency δω and wave number δk

⇒ amplitude modulation, moving at speed vg = δω/δk (group velocity).

The modulating envelope encloses a group of short waves.

For δω , δk → 0 , vg =dω

dk

To construct a finite pulse need a superposition of an

infinite number of waves of different frequencies

If distribution is peaked around ! , k

Phase velocity v !!

kGroup velocity g =

d!

dk (velocity of envelope)

In a dispersive medium pulse spreads out

… but mean position moves with group velocity

N .B. g ! c

In a non-dispersive medium pulse maintains its shape d!

dk=!

k= constant

because of dispersion, i.e., dv/dk nonzerodistinction between g and v arises

There are many equivalent expressions for the group velocity:

g =

d!

dk

! = vk g = v + kdv

dk

k = 2! / " g = v ! "

dv

d"

v = c / µ g =

c

µ1+

!µ

dµ

d!"#$

%&'

In terms of wavelength !' in vacuum

g ! v 1!1

1+v

"#d "#dv

$

%

&&&

'

(

)))

�" = "c

v

f =c

�"=v

"

• Waves travelling through a dispersive medium:

group velocity vg =dω

dk=

d(vk)

dk= v + k

dv

dk

= v +2π

λ

dv

dλ

(

−λ2

2π

)

= v − λdv

dλ

EXAMPLE

• Suppose the dispersion relation v = v(λ) is given by

v2 = c2 + λ2ω2

0

Then 2v dv = 2λ dλ ω2

0, i.e.

dv

dλ=

λω2

0

v

So vg = v − λdv

dλ= v − λ2ω2

0

v, i.e. vg = v − v2 − c2

v=

c2

v=⇒ vgv = c2

product of phase and group velocities equals c2

Example: waves in deep water

v ∝√λ , i.e. v =

C√k

with C constant

Since v = ω/k, then ω/k = C/√k

⇒ ω = C√k

Therefore g =dω

dk=

C

2√k

=1

2v

• group velocity is half the phase velocity:

component wave crests run rapidly through the group,

first growing in amplitude and then disappearing

Energy of vibrating string

y = Asin kx !"t( )

KE of section = 1

2!"x

#y

#t

$%&

'()

2

=1

2!A2* 2

cos2 kx +*t( )"x

Kinetic Energy

KE =1

2!A

2" 2cos

2kx '#"t( )

x

x+ l$

% dx ' =1

2!A

2" 2 &l$2

(integer l wavelengths)

Energy of vibrating string

y = Asin kx !"t( )

KE of section = 1

2!"x

#y

#t

$%&

'()

2

=1

2!A2* 2

cos2 kx +*t( )"x

Kinetic Energy

Potential Energy

PE in stretched string element = T ! l " !x( ) = T!x 1+! y

!x

#$%

&'(

2

"1

#

$%%

&

'(()

1

2TA2k 2 cos2 kx "*t( )!x

v =! / k = T / " ! Tk2= "# 2 ! PE=KE! (Example of virial theorem)

KE =1

2!A

2" 2cos

2kx '#"t( )

x

x+ l$

% dx ' =1

2!A

2" 2 &l$2

KE / l! =1

4"A2# 2

PE=1

2TA

2k2

cos2kx '!"t( )dx '

x

x+ l#

$ PE / l! =1

4kTA

2k2

!2y

!x2="

T

!2y

!t2#

1

c2

!2y

!t2

Note

kinetic energy per unit length :dK

dx=

1

2ρ

(

∂y

∂t

)

2

potential energy per unit length :dU

dx=

1

2T

(

∂y

∂x

)

2

dE

dx=

dK

dx+

dU

dx

• For y(x, t) = f(x− ct), c =√

T/ρ, one has

∂y

∂t= −cf ′ ,

∂y

∂x= f ′

So1

2ρ

(

∂y

∂t

)

2

=1

2ρ c2 (f ′)2 =

1

2T (f ′)2 =

1

2T

(

∂y

∂x

)

2

⇒dK

dx=

dU

dx

Total energy per wavelength, E/!= KE+PE=1

2"A2# 2

Energy flow

=1

2!" 2

A2v

Energy flow/unit time =1

2!A

2" 2#$%

&'(

vt / t

Distance travelled = vt

=1

2T!kA

2

=1

2!" 3

A2/ k v =

!

k

=1

2Tk

2A2v Tk

2= !" 2