Types of Waveforms
• Pulse: not periodic
– drops to zero
on both sides
displacement
travels in time
– f(x)
x
• Continuous (or periodic) wave:
displacement
travels in time
x
λ
June 27, 2011
6
Types of Waveforms
• Harmonic (aka sinusoidal)
– Special case of periodic
(Longitudinal and
transverse)
displacement
travels in time
A
x
λ
– Wave speed: v =
June 27, 2011
λ
T
= λf
7
Mathematical Description
• Pulse: f (x )
• Moving pulse:
– “+” direction f ( x − vt )
– “-”direction f ( x + vt )
travels in time
f(x) f(x-vt)
x
June 27, 2011
8
Simple Harmonic Wave
displacement
• Position at t=0:
–
 2πx 
y ( x,0 ) = A cos
 = A cos(kx )
 λ 
A
x
λ
• Travelling to the right with speed, v:
y ( x, t ) = y ( x − vt ) = A cos k ( x − vt ) =
–
A cos(kx − ωt )
ω = 2πf =
2π
T
• Angular frequency--->
2π
k
=
• Wave number---------> λ
ω
v
=
• Velocity-----------------> k = λf
June 27, 2011
8
displacement
A
t
T
Waves on a String under tension
• Small amplitude displacements
• transverse
• Given mass density, μ, tension, F
Speed: v =
F
–
µ
– Dimensions: [F]/[μ]=[v]
June 27, 2011
9
v
Average Power
(waves propagate energy)
v v
v v
• P=dE/dt= F ⋅ u (W = F ⋅ x )
∂y
– Speed of medium u =
= ωA sin (kx − ωt )
∂t
• Power fluctuates throughout the wave (book)
– P = FωkA2 sin 2 (kx − ωt )
• Interested in average power
1
– P = FωkA2
2
1
2 2
P
=
µω
Av
• Waves on a string
2
June 27, 2011
10
Wave Intensity
• Intensity= average energy/time/area (W/m^2)
– Area perpendicular to propagation direction
• Wavefront – surface of constant phase
– Plane wavefronts (plane wave) – constant
intensity
– Spherical wavefront (spherical wave) – intensity
decreases as distance from source increases
P
P
I= =
A 4πr 2
June 27, 2011
11
Superposition
• Wave interference = algebraic sum of
displacements
Wave1
Wave2
Constructive
(in phase)
∆φ = 2πn
June 27, 2011
12
Destructive
(180 degree out of phase)
∆φ = nπ
Beats
• Interference of 2 waves
w/slightly different
frequencies
• At a fixed position, x=0:
y (t ) = A cos ω1t + A cos ω2t
 ω1 − ω2   ω1 + ω2 
= 2 A cos
t  cos
t
 2
  2

“Amplitude”
June 27, 2011
13
Dispersion
• Wave frequency (speed) depends on
wavelength v = λf (λ )
• Waves on the surface
λg
v=
2π
June 27, 2011
14
Wave Equation
• All equations of the form f ( x ± vt ) satisfy
∂2 f
1 ∂2 f
= 2 2
2
∂x
v ∂t
• i.e. displacement of waves on a string
y ( x, t ) = A cos(kx ± ωt )
∂2 y µ ∂2 y
=
2
∂x
F ∂t 2
June 27, 2011
15
Sound Waves In Gas
• Longitudinal pressure wave
pressure
γP
• Speed of wave v =
ρ
density
• Coefficient, γ, depends on nature of gas
– Ideal γ = 5 / 3
– Diatomic γ = 7 / 3
June 27, 2011
16
Sound waves in Liquid/Solid
• Speed of sound
B
γP
v=
v=
(in
gas,
)
ρ
ρ
• Bulk modulus (wave speed depends on
direction)
∆P
B=−
∆V / V
– Units of pressure
– Measure of “stiffness” (large B, hard to compress)
• longitudinal & transverse
June 27, 2011
17
Sound Intensity
P
I=
Area
• Recall for a string,
• In terms of pressure,
1
2 2
P
=
µω
Av
,
2
Power
= Force ⋅ velocity = (∆P )( Area )u
Area
• Pressure, displacement vary sinusoidally
∆P = (∆P0 ) cos(kx − ωt )
s = − s0 sin (kx − ωt )
2
(
∆P0 )
• See pg 419-420 I =
1
2
= ρω 2 s0 v
2 ρv
2
June 27, 2011
18
Decibels (dB)
• Human ear can detect a wide range of
frequencies log scale!
• Decibel unit defined as
β = 10 log(I I 0 )
where I 0 = 10−12W / m 2 is threshold of hearing
at 1kHz
• >40dB, perceived loudness doubles for every
10dB increase
June 27, 2011
19
Standing waves
Fundamental, m=1
• String (L) clamped
tightly at both ends
• No propagation
(superposition of
waves propagating in
opposite directions
• Allowed modes
mode number
mλ1
L=
June 27, 2011
First harmonic
Second harmonic
node
2
20
antinode
Standing waves
• String clamped at one • String open at both
end
ends
June 27, 2011
21
Standing waves are everywhere!
June 27, 2011
22
*GRADES – will be
curved, but only in
your favor
Review - waves
ω
• v = λf = k (different from velocity of medium)
∂ f
1 ∂ f
f ( x m vt )
• Wave equation – ∂x = v ∂t
A cos(kx − ωt )
– Wave on a string v = F µ
2
2
2
2
2
– Sound in gas v = γP ρ
– Sound in solid/liquid v = B ρ
(bulk modulus B = − ∆P (∆V /V ) )
k=
• Longitudinal and/or transverse
• Superposition (i.e. beats)
• Dispersion (wave packets spread out)
June 28, 2011
1
2π
λ
ω=
2π
T
ω1 + ω2
2
incident wave
reflected wave
• Propagation through
bounded medium
• Superposition of 2
waves travelling in
opposite directions
• 3 bounded scenarios
mλ
L=
2
mλ
L=
4
(m + 1 2)λ
L=
2
– Closed on both
– Open on both
– open on one
node
antinode
f 0 = v λmax
harmonics
Review – Standing
waves
amplitude vs instantaneous amplitude
June 28, 2011
2
Review - Intensity
1
• Average power (string)– P = µω 2 A2v
2
• “wavefront” – surfaces of constant phase
• Plane/spherical wave – plane/spherical
wavefronts
• Intensity – characterizes energy flow in >1D
– 3D
I=
P
P
=
A 4πr 2
(2D
I2D
• Intensity of sound
decibels β = 10 log(I I 0 )
2
(
∆P0 )
I=
1
2
= ρω 2 s0 v
2 ρv
2
June 28, 2011
P
P
= =
)
l 2πr
3
Doppler effect
• Wave propagates
symmetrically in every
direction from a point
source at rest
• When source/observer
moves, the observed
frequency and/or
wavelength is shifted
• Let u be speed of source or
observer
June 28, 2011
4
Doppler: source
moving
u
f ′ = f (1 + u / v )
• Source reference frame::
travel λ in one period, T
• During T, source covers
distance of uT
• Observed wavelength:
λ ′ = λ m uT
λ ′ = λ (1 m u / v )
• frequency f ′ = f (1 m u / v )
June 28, 2011
5
f ′ = f (1 − u / v )
Doppler Shift: Moving Observer
•
•
•
•
Shift in frequency only,
wavelength does not change
Speed observed = v+u
Observed period T ′ = λ
v
u
v+u
• Observed frequency shift
f ′ = f (1 ± u / v )
(negative sign means observer moving AWAY)
June 28, 2011
6
EM waves
June 28, 2011
7
4 laws of EM (up to now)
Gauss
no magnetic
v v
∇ ⋅ E = ρ ε 0 monopoles
v v
v v
v v
∇ ⋅ B = 0 Faraday ∫∫∫ ∇ ⋅ Fdτ = ∫∫ F ⋅ dA
v v
v
v v v
v v
∇ × E = − ∂B ∂t
∇ × F ⋅ dS = ∫ F ⋅ d l
∫∫
v v
v
∇ × B = µ0 J
Ampere
dB/dt
Changing E-fields
induce B-fields?
E
June 28, 2011
v v
∫ E ⋅ dA = qencl ε 0
v v
∫ B ⋅ dA = 0 Lenz’s law
v v
∫ E ⋅ d l = − ∂Φ B ∂t
v v
∫ B ⋅ d l = µ0 I encl
8
Modify Ampere’s law
• Ambiguity in ampere’s law
v v
∫ B ⋅ d l = µ0 I encl
v v
∫l1B ⋅ d l = µ0 I
v v
∫ B ⋅ d l = 0!
l1
l2
l2
• Displacement current
I D = ε 0 ∂Φ E ∂t
Amperean loop
v v
∫ B ⋅ d l = µ0 I encl + µ0ε 0 ∂Φ E ∂t
June 28, 2011
8
Maxwell’s Equations Simplest scenario: in
a vacuum
- complete
v v
Gauss
∫ E ⋅ dA = qencl ε 0
v v
∫ B ⋅ dA = 0
v v
∫ E ⋅ d l = − ∂Φ B ∂t Faraday
v v
∫ B ⋅ d l = µ0 I encl + µ0ε 0 ∂Φ E ∂t
Ampere (complete)
Electric and Magnetic fields on
equal footing
June 28, 2011
v v
∫ E ⋅ dA = 0
v v
∫ B ⋅ dA = 0
v v
∫ E ⋅ d l = − ∂Φ B ∂t
v v
∫ B ⋅ d l = µ0ε 0 ∂Φ E ∂t
Electromagnetic waves!
9
EM Waves
• Coupled differential equations
v v
Take the curl of both sides
∇⋅E = 0
v v
∇⋅B = 0
v v
v
∇ × E = − ∂B ∂t
v v
v
∇ × B = µ0ε 0 ∂E ∂t
v v v
v v v0 2v
∇ × (∇ × E ) = ∇(∇ ⋅ E ) − ∇ E
v
2
v
v
v
∂
∂
∂ E
− (∇ × B ) = − (µ0ε 0 ∂E ∂t ) = − µ0ε 0 2
∂t
∂t
∂t
∂2 f
1 ∂2 f
= 2 2
2
∂x
c ∂t
• In one dimension
v
v
2
2
∂ E
∂ E
= µ0ε 0 2
2
∂x
∂t
June 28, 2011
c=
10
1
µ0ε 0
= 3.0 × 10 m / s
8
EM plane Wave
• E,B harmonic waves, oscillate in perpendicular
directions
• We use x for the propagation direction
v
E ( x, t ) = E p cos(kx − ωt ) ˆj Amplitudes related
speed of wave
v
B ( x, t ) = B p cos(kx − ωt )kˆ
y
E
c =ω k
E
In phase
h
Δx
Δx
h
z
B
June 28, 2011
E p = cB p
11
x
Electromagnetic Spectrum
• Continuum of allowed wavelengths
(frequencies) as long as c = λf
June 28, 2011
12
Polarization
• Specifies direction of E field
v
E ( x, t ) = E p cos(kx − ωt )iˆ
– Polarization is perpendicular to B, v
• Unpolarized: superposition of waves with a
random orientation of polarizations
– i.e. visible light from sun, light bulb
• Unpolarized becomes polarized…
– Reflection from surfaces (partial)
– POLARIZER (crystal with preferred direction of
transmission, called transmission axis)
June 29, 2011
1
Polarizers
nothing
E0 cos θ
E0
• Polarizers reorient a component of the
incident wave
– The output light is completely polarized along the
polarizer axis
– Polarizers change the light
– Polarizers are not filters
• Filters can only decrease (or leave unchanged) the light
• Filters cannot add light
no filter can add red light
Filters:
white
light
June 29, 2011
2
blue
light
?
nothing
Polarizers – Law of
Malus
• Only component of E along
transmission axis will pass through
unattenuated
E f = E0 cos θ0
• Intensity (S) varies as amplitude squared
S = S0 cos θ0
2
• For unpolarized light, average over ever direction to
get emerging intensity
Sunpolarized
June 29, 2011
3
1
= S0
2
Producing EM Waves
• Changing B E; changing EB
– Accelerated electrical charges
• If motion of charges is periodic, EM with that
frequency
• Systems are “good” transmitter/receivers if
size of system d~λ
• Directional antenna (TV bunny)
• Approximate plane wave
June 29, 2011
4
Energy in EM Waves
• Energy density (energy/volume)
(
v2 v2
1
u = ε 0 E + B µ0
2
)
A
• Want intensity (energy/time/area)
dU = udVol = uAdx
•
dU
dx
= uA
dx c
dt
v2 v2
1 dU c
S=
= ε 0 E + B µ0
A dt 2
v2
rewrite
S = ε 0c E
v v
)
(
• Can
S=
June 29, 2011
EB
µ0
5
S=
v2
cB
µ0
dx
c
(E
p
= cB p )
v
E ( x, t ) = E p cos(kx − ωt ) ˆj
v
B ( x, t ) = B p cos(kx − ωt )kˆ
Poynting Vector
• Energy/area/time
v v
v E×B
S=
y
µ0
• Points in direction of propagation (RHR)
v
z
• Wavevector k
v
S
v
E
v
B
June 29, 2011
6
x
Average intensity (magnitude)
• Poynting vector magnitude in terms of E
v
S =
v v
EB
µ0
=
E p Bp
µ0
cos (kx − ωt )
• Average value cos^2=1/2 2
S=
June 29, 2011
S=
E p Bp
2 µ0
S=
7
ε 0cE p
2
v 2
cB p
2 µ0
2
2 lasers emit light of the
same color, but the E field
in laser 1 is 2x that of laser
2. compare their:
B-fields
intensities
wavelengths
Waves from point source
• Spherical waves
Power
S=
4πr 2
• Energy spreads in space over time, amplitude
does not change
• Field strength falls off as 1/r
– EM waves dominate in all but the immediate
vicinity of accelerating charges
June 29, 2011
8
Wave Momentum
• EM wave energy and momentum related by
p=
U
c
– If average energy/t/area= S
• Average momentum/t/area=>
v
v
• Newton’s second law F = dp dt
• Force/Area = Pressure,
S
c
– RADIATION PRESSURE
• EM wave delivers 2x pressure to object on
reflection
June 29, 2011
9
Ch 17 Review (Sound & more Waves)
• Speed of (longitudinal) wave is
v=
– Gases
– Liq/Sol
v=
γP
ρ
(in air, 340 m/s)
B
B=−
ρ
∆P
∆V / V
• Intensity
–
2
(
∆P0 )
I=
2 ρv
=
1
ρω 2 s0 2v
2
• Decibels
– β = 10 log(I I 0 )
• Standing waves
• Doppler effect
– Moving source f ′ = f (1 ± u / v )
– Moving Observer f ′ = f (1 ± u / v )
July 5, 2011
1
Examples
• Ch 17-22
A 1-dB increase in sound level is about
the minimum change the human ear
can perceive. By what factor is the
sound intensity increased for this dB
• Ch 17 46
Estimate the fundamental frequency of
the human vocal tract by assuming it
to be a cylinder 15 cm long that is
closed on one end.
change?
L = λmax 4
∆β = β 2 − β1 = 10 log(I 2 I1 )
I 2 I1 = 10
I 2 I1 = 10
1
10
∆β
L = 3λ 4
10
= 1.26dB
L = 5λ 4
v = λmax f 0 = 340m / s
f0 = v
July 5, 2011
2
4L
=
(340m / s )
1m 

415cm ×

100cm 

= 570 Hz
Ch 34 Review (EM Waves)
• Maxwell’s Complete Equations
v v
∫ E ⋅ dA = qencl ε 0
v v
B
∫ ⋅ dA = 0
v v
Displacement
E
⋅
d
l
=
−
∂
Φ
∂
t
B
∫
current
v v
∫ B ⋅ d l = µ0 I encl + µ0 (ε 0 ∂Φ E ∂t )
v v
∫ E ⋅ dA = 0
v v
B
∫ ⋅ dA = 0
v v
E
∫ ⋅ d l = − ∂Φ B ∂t
v v
∫ B ⋅ d l = µ0ε 0 ∂Φ E ∂t
v
v
ˆ
E ( x, t ) = E0 cos(kx − ωt ) j B ( x, t ) = B0 cos(kx − ωt )kˆ
polarization
• Intensity:
– Poynting vector
v v
v E×B
S=
• S → S cos θ (S → S 2)
U
p
=
• EM momentum
c
2
0
0
– Rad Press
July 5, 2011
0
S
c 3
0
µ0
v
v
2
∂ E 1 ∂ E
= 2 2
2
c ∂t
∂x
2
c=
E0 = cB0
1
µ0ε 0
=
ω
k
Examples
• Ch 34-64
• Ch 34-6
Find the angle between 2 polarizers if
An E-field points into the page and
upolarized light incident on the pair
occupies a circular region of radius
emerges with 10% of its incident
1m. There is B-field forming CCW
intensity.
loops. The B-field strength 50cm from
After 1st polarizer…
center is 2.0-uT. (a) What is the rate of
1
change of E-field? (b) Is the E-field
S1 = S0
2
increasing or decreasing?
B
v v
dE/dt=? ∫ B ⋅ d l = µ0ε 0 ∂Φ E ∂t
xxx
xxxxx
x x x xr x x x
Rx x x x x
xxx
B (2πr ) = µ0ε 0 (∂E ∂t )πr 2
B (@ r ) = µ0ε 0 r 2 (∂E ∂t )
(∂E ∂t ) = 2 Bc 2
r
CCW B dE/dt out of page by RHR!
So strength of E - decreasing
July 5, 2011
4
After 2nd polarizer
S2 = S1 cos2 θ0
= (S0 2 ) cos2 θ0
S2
cos2 θ0
= 0.10 =
2
S0