9. Electromagnetic Waves
Waves along a rope
Stadium waves (medium is people)
• It appears like a ripple is moving:
– but particles in the medium only do a local cyclic movement.
– only the disturbance (wave) is moving.
Wave motion
A transverse wave
A longitudinal wave
Suppose a displacement of a wave propagating in z is given by 𝑓 𝑧, 𝑡
the initial value of the wave (t=0) is 𝑓 𝑧, 0 = 𝑔 𝑧
If the velocity of the wave is 𝑣 , at time t it has traveled a distance 𝑣𝑡
So 𝑓 𝑧, 𝑡 = 𝑓 𝑧 − 𝑣𝑡, 0 = 𝑔(𝑧 − 𝑣𝑡)
represents a wave of fixed shape travelling in the z direction at speed v
e.g. 𝑓1 𝑧, 𝑡 = 𝐴𝑠𝑖𝑛 𝑎 𝑧 − 𝑣𝑡 ; . 𝑓2 𝑧, 𝑡 = 1 𝑏 𝑧−𝑣𝑡 2 are waves
Waves on a stretched string
Net transverse force on the string segment of Δz =ΔF
2
2
2 2 2
2 2 2
sin sin tan tan
If mass per unit length is , ( )
z z z
F F fF T T T T T T z
z z z
f f fF z
t z T t
:Wave equation it admits a solution of the foam
𝑓 𝑧, 𝑡 = 𝑔(𝑧 − 𝑣𝑡)
2 2
2 2
f f
z T t
2 2 2
2 2 2
2 2 2 2 22
2 2 2 2 2 2
-
1 and =
f dg u dg f dg u dgv u z vt
z du z du t du t du
f dg d g u d g
z z du du z du
f dg d g u d g f f Tv v v v
t t du du t du z v t
also has a solution of the form 𝑓 𝑧, 𝑡 = ℎ 𝑧 + 𝑣𝑡which represents a wave travelling in –z direction
2 2
2 2 2
1
f f
z v t
So the general solution to wave equation
𝑓 𝑧, 𝑡 = 𝑔 𝑧 − 𝑣𝑡 + ℎ 𝑧 + 𝑣𝑡
22
2 2
1
ff
v t
In 3 dimensions wave is
𝑓 𝑟, 𝑡 = 𝑔 𝑟 − 𝑣𝑡 + ℎ 𝑟 + 𝑣𝑡
9.1.2 Sinusoidal Waves
pattern repeats itself after a distance equal to the wavelength λ
k𝜆 = 2𝜋 ⇒ 𝜆 =2𝜋
𝑘
• Period T =2𝜋
𝜔=2𝜋
𝑘𝑣frequency = ν =
1
𝑇=
𝑘𝑣
2𝜋=
𝑣
𝜆
• angular frequency ω = 𝑘𝑣 =2𝜋
𝜆𝑣 = 2𝜋𝜈
• For the wave in –z direction : 𝑓 𝑧, 𝑡 = 𝐴𝑐𝑜𝑠 𝑘𝑧 + 𝜔𝑡 + 𝛿
=𝐴𝑐𝑜𝑠 −𝑘𝑧 − 𝜔𝑡 + 𝛿wave with –k propagates in -z
𝑓 𝑧, 𝑡 = 𝐴𝑐𝑜𝑠 𝑘 𝑧 − 𝑣𝑡 + 𝛿= 𝐴𝑐𝑜𝑠 𝑘𝑧 − 𝜔𝑡 + 𝛿
amplitude
phase
wave number angular frequency
k =2p
l
v =w
k
w = 2p f
These are purely
kinematic, they
work for any wave.
v = flHow far the wave
moves per cycle
How many cycles per second
θ
z = x + iy
z* = x - iy
z2
= zz* = (x + iy)(x - iy) = x2 + y2real
Imaginary
z = reiq
r
2 3 4 5
2 3 4 5
2 4 3 5
( ) ( ) ( ) ( )1 ( ) ...
2! 3! 4! 5!
1 ...2! 3! 4! 5!
1 ...2! 4! 3! 5!
cos sin
i i i i ie i
i i i
i
i
Complex number notation
Since 𝑒𝑖𝜃= 𝑐𝑜𝑠𝜃 + 𝑖𝑠𝑖𝑛𝜃
( )
( )
( , ) cos( ) Re[ ]
introduce complex wave function ( , )
with complex amplitude
then ( , ) Re( ( , ))
i kz t
i kz t
i
f z t A kz t Ae
f z t Ae
Ae
f z t f z t
3 1 2 1 2 3
( ) ( ) ( )3 1 2 3 3 1 2
Example:
combining two waves f f +f Re(f ) Re(f ) Re(f )
( , ) + = where
i kz t i kz t i kz tf z t Ae A e A e A A A
i.e. just add the complex amplitudes. Combined wave still have the same frequency and wavelength
Boundary conditions:
• Consider an incident wave 𝑓𝐼 𝑧, 𝑡 = 𝐴𝐼𝑒𝑖(𝑘1𝑧−𝜔𝑡) 𝑧 < 0
• Gives rise to a reflected wave 𝑓𝑅 𝑧, 𝑡 = 𝐴𝑅𝑒𝑖(−𝑘1𝑧−𝜔𝑡) 𝑧 < 0
• And a transmitted wave 𝑓𝑇 𝑧, 𝑡 = 𝐴𝑇𝑒𝑖(𝑘2𝑧−𝜔𝑡)
𝑓𝐼 𝑧, 𝑡
𝑓𝑇 𝑧, 𝑡
𝑓𝑅 𝑧, 𝑡
1 2 1
2 1 2
k v
k v
All waves have to have the same frequency, but on two sides wave lengths and velocities are different
1( ) ( )
( )
net disturbance is:
+ for 0
( , )
for 0
R
T
i k z t i k z tI R
i k z tT
A e A e z
f z t
A e z
• At the boundary (0 , ) (0 , )f t f t
0 0
and f f
z z
+ = I R TA A A1 1 2 = I R Tk A k A k A
1 2 1 2 1 2
1 2 1 2 2 1 1 2
2 2 = , = or = , = R I T I R I T I
k k k v v vA A A A A A A A
k k k k v v v v
2 1 2
2 1 1 2
2= , = R I T Ii i i i
R I T I
v v vA e A e A e A e
v v v v
In terms of real amplitudes and phases
If the 2nd string is lighter than the 1st 𝜇2 <𝜇1 ⇒ 𝑣2 > 𝑣1 all
have the same phase angle 𝛿𝐼 = 𝛿𝑇 = 𝛿𝑅2 1 2
2 1 1 2
2= , = R I T I
v v vA A A A
v v v v
If the 2nd string is heavier than the 1st 𝑣2< 𝑣1 so the reflected wave is out
of phase by 𝜋 to incident 𝛿𝐼 = 𝛿𝑇 = 𝛿𝑅 + 𝜋
)))
since cos( ) cos( )I Ikz t kz t
The reflected wave is upside down12
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