oscillations, waves and optics - part ib physics a 2013 ...jsr/_downloads/owo_handout.pdfharmonic...
TRANSCRIPT
-
Oscillations, Waves and OpticsPart IB Physics A 2013/2014
John Richer [email protected]...
revision 247
Part IB Physics A: Oscillations, Waves and Optics H2(r247)
Welcome!This course will build on the foundation of the 1A course, but take the materialconsiderably further and in more mathematical detail.We will cover
Oscillations: driven and free
Waves: the new key ideas are polarisation, group velocity, dispersion, reflectionand transmission
Applications of Fourier ideas to waves and oscillations
Optics: mathemetical theory of diffraction — in the near and far fields
Interference of light
-
Damped Harmonic Oscillator H3(r247)
mx
k
v
Figure 1 : Simple oscillator
Consider a simple system with a restoring forceproportional to displacement (F ∝ −x). Thisharmonic oscillator has a quadratic potential(V ∝ x2).Assume that there is a resistive drag force with amagnitude proportional to the speed of theoscillation.Assume we apply some driving force F (t). Theequation of motion for a driven, damped oscillatoris then
F (t) = mẍ+ bẋ+ kx (1)
b is a constant giving the strength of thedamping. k specifies the strength of the restoringforce (or the curvature of the quadratic potentialfunction near equilibrium: V = 1
2kx2).
When does this equation apply? H4(r247)
Simple harmonic motion with negligible drag force is common near equilibriumbecause the potential will most likely be quadratic locally.
A mass on a spring undergoing small oscillations with a viscous damping force mayapproximate this ideal situation well.
But a pendulum is not a simple harmonic oscillator in general:the restoring force is proportional to (− sin θ) for displacement θ which is linear onlyfor small displacements when sin θ ≈ θ.damping may not be proportional to speed (e.g. wind resistance).
In general, numerical integration will be needed to solve such problems. (See theMATLAB example on the example sheet).
-
Canonical Form of Damped SHM H5(r247)
We define some quantities to put the equation of motion into a standard (“canonical”)form:
ω0 =
√k
m. (2)
This is the frequency of oscillation in the absence of damping and forcing.
γ =b
m(3)
this expresses the damping; these yield the canonical form:
ẍ+ γẋ+ ω20x =F
m(4)
N.B. This is consistent with Main’s book, but not Pain’s or perhaps the IA notes (where2γẋ appears in the equation).Note also that γ has dimensions of inverse time. It is also useful to define a newdimensionless parameter
Q =ω0γ
(5)
This is the Quality Factor.
Complex Notation: Phasor Diagrams H6(r247)
Re
Im
Figure 2 : phasor diagram
It is often convenient (but not necessary) torepresent an oscillating quantity as a vectorrotating anti-clockwise in the complex plane. Theobservable quantity (in this case the displacementx) is by convention equal to the real part of thecomplex representation.
x = <(x0 e
iωt)
where the complex amplitude is x0 = A eiφ:
x = <[A eiφeiωt
]= <
[A ei(ωt+φ)
]= A cos(ωt+ φ)
A = |x0| is the amplitude of the oscillation (areal value), and arg(x0) is its phase.If instead we use exp(−iωt), vector rotatesclockwise.
-
Speed and Acceleration H7(r247)
Using complex representation, the speed of the particle can be derived using
v =d
dt<(x0e
iωt)
= <(
iωx0eiωt)
= <(v0e
iωt)
where we introduce the complex velocity v0 = iωx0.
Note we have interchanged the order of differentiation and taking the real part:these operations commute.
Similarly, the complex acceleration is
a0 = iω0v0 = −ω2x0
Speed and Acceleration: Phasor Diagram H8(r247)
position
velocity
acceleration
ωt + φ
Figure 3 : Phasor diagram for velocity and acceleration
-
Free Oscillations: Review H9(r247)
With no driving force, the oscillator obeys the homogeneous equation
ẍ+ γẋ+ ω20x = 0 (6)
We look for solutions of the form x ∝ ept, finding
p2 + γp+ ω20 = 0 (7)
∴ p =−γ ±
√γ2 − 4ω202
= −γ2
(1±
√1− 4Q2
)where we used Q = ω0/γ. There are three regimes to consider, depending on the sign ofthe square root term:
Light Damping: γ < 2ω0 (Q > 0.5)
Critical Damping: has γ = 2ω0 (Q = 0.5)
Heavy Damping: has γ > 2ω0 (Q < 0.5)
Q = 0.5 separates the regimes. Q is the quality factor of the oscillator. High-Q meanslow damping, and vice versa.
Free Oscillations: Light Damping Q > 0.5 H10(r247)
We find p = − γ2± iωf , where the free
oscillation frequency ωf is
ωf =
√ω20 −
γ2
4= ω0
√1− 1
4Q2(8)
Note that ωf is very close to ω0 for Q valuesgreater than a few. (For Q = 5, the differenceis 0.5%). The solution involves 2 arbitraryconstants, set by the boundary conditions.We can write equivalently:
x(t) = e−γt/2(Aeiωf t +A∗e−iωf t
)x(t) = e−γt/2B cos(ωf t+ φ)
x(t) = e−γt/2 (C1 cosωf t+ C2 sinωf t)
x(t) = <(De−γt/2eiωf t
)where A and D are complex; and B, C1 andC2 are real. Simple algebra allows us to relatethe A, B, C1,2, and D.
0 5 10 15 20 25 30 35 40 45−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
t
disp
lace
men
tx
Damped oscillations, ω0 = 1
Q = 10
Figure 4 : Lightly damped free oscillations, starting fromẋ = 0, x = 1 at t = 0
-
Light Damping: estimating Q H11(r247)
0 5 10 15 20 25 30 35 40 45−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
t
disp
lace
men
tx
Damped oscillations, ω0 = 1
Q = 1Q = 3Q = 9
Q = 27
Figure 5 : Lightly damped free oscillations
Amplitude ∝ e−γt/2 ∝ e−ω0t/(2Q) i.e. decayswith a time constant τ = 2/γ.
When ω0t = Q, amplitude is e−1/2 ≈ 61% of
initial amplitude.
The energy U decays twice as fast,U ∝ e−ω0t/Q.So Q is the number of radians of phaseelapsed as the amplitude falls toe−0.5 ≈ 61% of its original valueand the energy is then e−1 ≈ 37% of theinitial energy.
e.g. The light blue curve has amplitude ≈ 0.6 after 4 complete oscillations, soQ ≈ 4× 2π ≈ 25 (close to true value of 27)
Free Oscillations: Heavy Damping Q < 0.5 H12(r247)
In this case, γ > 2ω0, and p has two real values.
No oscillations, only exponential decay occurs.
x(t) = C1e−µ1t + C2e
−µ2t
µ1,2 =1
2
(γ ±
√γ2 − 4ω20
)with C1 and C2 being real.
0 1 2 3 4 5 6 7 8 9
Time/s
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
Dis
pla
cem
ent
(m)
orS
pee
d(m
/s)
Heavy Damping: Q = 0.25, x(0) = 1, v(0) = 0
displacement
speed
0 1 2 3 4 5 6 7 8 9
Time/s
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
Dis
pla
cem
ent
(m)
orS
pee
d(m
/s)
Heavy Damping: Q = 0.25, x(0) = 0, v(0) = 1
displacement
speed
Figure 6 : Transient response of a heavily damped oscillator with m = 1 kg, k = 1 N m−1 , b = 4 kg s−1 .
-
Free Oscillations: Critical Damping Q = 0.5 H13(r247)
Only one solution for p, which is p = −γ/2 = −ω0. It is impossible to fit twoboundary conditions with this single exponential solution. The general solution canbe shown to be
x(t) = (C1 + C2ω0t) e−ω0t
This solutions yields the most rapid approach to equilibrium with noovershooting.
Note that the system returns to equilibrium with a time constant of 1/ω0 in criticaldamping conditions.
0 1 2 3 4 5 6 7 8 9
Time/s
−0.4
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
Dis
pla
cem
ent
(m)
orS
pee
d(m
/s)
Critical Damping: Q = 0.50, x(0) = 1, v(0) = 0
displacement
speed
0 1 2 3 4 5 6 7 8 9
Time/s
−0.4
−0.2
0.0
0.2
0.4
0.6
0.8
1.0
Dis
pla
cem
ent
(m)
orS
pee
d(m
/s)
Critical Damping: Q = 0.50, x(0) = 0, v(0) = 1
displacement
speed
Figure 7 : Transient response of a critically damped oscillator with m = 1 kg, k = 1 N m−1 , b = 2 kg s−1 .
Finding A Specific Solution from Initial Conditions: Example H14(r247)
Consider a lightly-damped oscillator, with initial conditions x(0) = 1, ẋ(0) = 0, i.e. it isdisplaced by unit amplitude and released. Write general solution (with real A, φ):
x(t) = A e−γt/2 cos(ωf t+ φ)
The initial, or boundary conditions imply the following.
Since x(0) = 1, we haveA cosφ = 1
Since ẋ(0) = 0, we have
−Aωf sinφ−γA
2cosφ = 0
These have solution
tanφ = − γ2ωf
; A =
√1 +
γ2
4ω2f=ω0ωf
Note that the initial phase is not zero, but close to zero if Q is large; and theamplitude is greater than 1.
-
The Driven Harmonic Oscillator H15(r247)
What happens when we apply a force F (t) to an oscillator?
F (t) = mẍ+ bẋ+ kx (9)
Let us initially restrict the driving force to be sinusoidal in form.We will see that this is not a serious restriction once we have the tools of Fourieranalysis to handWrite the real driving force as F = <
[F0e
iωt]
F0 is in general complex (driving sinusoid may not be at phase 0 at t = 0)Substitute into the equation of motion:
<{
[−ω2 + iγω + ω20 ]x0eiωt}
= <{F0m
eiωt}
ω20x0γωx0
−ω2x0
F0/m
φ
Figure 8 : Phasor diagram for driven oscillator
This implies (we can drop the
-
Driven Response: Amplitude H17(r247)
0 0.5 1 1.5 2 2.5 30
2
4
6
8
10
12
circle 3ptcircle 3ptcircle 3ptcircle 3pt
ω
|R|
γ = 0.1, Q = 10γ = 0.2, Q = 5γ = 0.5, Q = 2γ = 1, Q = 1
Max Response
Figure 9 : Driven Response –Amplitude. m = 1 kg, k = 1 N m−1 , ω0 = 1 rad s−1
|R| = 1m√
(ω20 − ω2)2 + γ2ω2(13)
Driven Response: Phase H18(r247)
0 0.5 1 1.5 2 2.5 3−1
−0.8
−0.6
−0.4
−0.2
0
ω
Phas
e/π
γ = 0.01, Q = 100γ = 0.1, Q = 10γ = 0.2, Q = 5γ = 0.5, Q = 2γ = 1, Q = 1
Figure 10 : Driven Response – Phase
arg(R) = arctan
[−γω
(ω20 − ω2)
](14)
-
Three Frequency Regimes of Driven Oscillator H19(r247)
ω � ω0: response in phase with the driving force. Motion controlled by springconstant/stiffness.
|R| ≈ 1/(mω20) = 1/k, argR ≈ 0
ω � ω0: response in anti-phase with the driving. Motion controlled by the inertiaof the system.
|R| ≈ 1/(mω2), argR ≈ −π
ω ∼ ω0: resonance, i.e. max value of |R|. Differentiate |R| to show this occurs atangular frequency ωa:
ωa = ω0
√1− γ
2
2ω20= ω0
√1− 1
2Q2(15)
i.e. ωa is smaller than ω0 if γ 6= 0. If damping is light,
|R| ≈ Q/(mω20), argR ≈ −π/2
i.e. the response lags π/2 behind the driving force, and the response is Q timeslarger than ω → 0 limit. So we can also regard Q as a measure of the amplification.
Velocity Response H20(r247)
We can derive the particle velocity’s complex amplitude via
v0 = iωx0 = iωF0R(ω)
=iωF0
m[(ω20 − ω2) + iγω]
=F0
m[(ω20 − ω2)/(iω) + γ](16)
The velocity resonance occurs where |v0| is at a maximum.
|v0| =|F0|
m√
(ω20 − ω2)2/ω2 + γ2, (17)
has maximum value at ωv = ω0 independent of the damping. From Eq. 16 we can seethat, at this frequency, the velocity is exactly in phase with the driving force, and hasmagnitude F0/γm.
-
Velocity Response: Amplitude H21(r247)
0 0.5 1 1.5 2 2.5 30
2
4
6
8
10
ω
|v0|
/|F
0|
Velocity Response: ω0 = 1
γ = 0.1, Q = 10γ = 0.2, Q = 5γ = 0.5, Q = 2γ = 1, Q = 1
Figure 11 : Velocity Response - Amplitude∣∣∣∣ v0F0∣∣∣∣ = 1
m
√(ω20−ω2)2
ω2+ γ2
(18)
This has a peak value at ω = ω0
Velocity Response: Phase H22(r247)
0 0.5 1 1.5 2 2.5 3
−0.4
−0.2
0
0.2
0.4
ω
arg(
v 0)/
π
γ = 0.1γ = 0.2γ = 0.3γ = 0.7
Figure 12 : Velocity Response - Phase
arg
(v0F0
)= arctan
[(ω20 − ω2)
γω
](19)
Maximum response at ω = ω0; velocity in-phase with the driving force at this velocityresonance.
-
Acceleration Response H23(r247)
We can derive the particle acceleration via
a0 = −ω2x0 = −ω2F0R(ω)
=−ω2F0
m[(ω20 − ω2) + iγω]
=−F0
m[(ω20 − ω2)/(ω2) +iγω
](20)
The acceleration resonance occurs where |a0| is at a maximum. Differentiate to findthat the maximum acceleration occurs at ωacc = where
ωacc = ω0
(1− 1
2Q2
)−1/2Note this is above the frequency of undamped, unforced oscillations ω0 and thatωaωacc = ω
20 .
Power: Multiplying Quantities using Complex Notation H24(r247)
The rate at which the driving force does work is “force × velocity”: P = Fv. We haveF (t) = <
[F0e
iωt]
and v(t) = <[v0e
iωt]
How should we evaluate the product? “Take the real part” does not commute with“multiply”. In general, to multiply two quantities represented in complex form, wemust. . .
Take the real parts first!
-
Phase differences and power factor H25(r247)
Mean power input depends on phase difference between the force and the velocity:if F0 = |F0|eiφF and v0 = |v0|eiφv then
〈P 〉 = 12
-
Power Resonance and Bandwidth H27(r247)
The mean power input is 12b|v0|2, so power resonance occurs at a frequency ωP which is
equal to the velocity resonance frequency, i.e. ωP = ωv = ω0.
0.5 1 1.5 2 2.5 30
1
2
3
4
5
ω/ω0
Pow
erin
put
Power resonance bandwidth
γ = 0.1, Q = 10γ = 0.2, Q = 5γ = 0.5, Q = 2
Figure 13 : Bandwidth of a driven damped oscillator
Power Resonance and Bandwidth H28(r247)
We can characterise the width of the resonance via its half power points, wherepower absorbed is half the value at power/velocity resonance.
This corresponds to the two frequencies ω+ and ω− at which |v0| = |vmax|/√
2.Using
|v0|2 =|F0|2
m2[(ω20 − ω2)2/ω2 + γ2].
we see thatγω± = ∓(ω20 − ω2±) (24)
Simple algebra shows the half-power bandwidth of the resonance is
∆ω = ω+ − ω− = γ (25)
As expected, the peak gets wider as the damping goes up.
The dimensionless ratio Q gives the ratio of the resonant frequency to the bandwidth
Q =ω0γ
=ω0∆ω
.
This provides another way of measuring Q.
-
General Response: driven and transient behaviour H29(r247)
Driven SHO equation has two kinds of solutions:
F (t) = mẍ+ bẋ+ kx (26)
free oscillations, F = 0: solution characterised by 2 free parameterssteady-state response for F 6= 0 — no free parameters
Because the equation is linear in x we can add any free solution to the drivensolution and still satisfy the equation.
The general solution is thus the sum of the complementary function, xc which is asolution to the homogeneous equation
mẍc + bẋc + kxc = 0 (27)
and a function that solves equ. (26), which is termed the particular integral
This must be the general solution as it will include two free parameters.
These are fixed by the initial conditions.
Driven near the resonance H30(r247)
0 10 20 30 40 50 60−10
−5
0
5
10
t
disp
lace
men
tx
Driven oscillation with transient; ω0 = 1, x(0) = 0, ẋ(0) = 0
Q = 3, ω = 1ω0Q = 1, ω = 1ω0
Q = 10, ω = 1ω0
Figure 14 : Oscillator driven from rest at x = 0
-
Driven H31(r247)
total
transient
steady-state
-2 0 2 4 6 8 10 12 14 16 18 20-1.0
-0.5
0.0
0.5
1.0
x(a
rbitr
ary
units
)
t (s)
Figure 15 : Oscillator driven from rest at x = 0 with Q = 1
Driven Oscillation: Light damping H32(r247)
0 5 10 15 20 25 30−2
−1
0
1
2
3
t
disp
lace
men
tx
Driven oscillation with transient; ω0 = 1, x(0) = 0, ẋ(0) = 0
Q = 40, ω = 0.7ω0Q = 40, ω = 1.5ω0
Figure 16 : Oscillator driven from rest
-
Driven Oscillation: Light damping H33(r247)
0 20 40 60 80 100 120 140 160 180 200−2
−1
0
1
2
3
t
disp
lace
men
tx
Driven oscillation with transient; ω0 = 1, x(0) = 0, ẋ(0) = 0
Q = 40, ω = 1.5ω0
Figure 17 : Oscillator driven from rest
Driven Electrical Circuits H34(r247)
R
L
C
V
Figure 18 : L-C-R circuit
VL + VR + VC = V (t)
Lq̈ +Rq̇ +1
Cq = V (t)
Our analysis of a driven mechanicaloscillator can be used in other areas ofphysics – electrical circuits, in particular.We simply need to make some translations:
displacement, x charge, qvelocity, v current, Iforce, F voltage, Vmass, m inductance, Ldamping, b resistance, Rspring constant, k (capacitance)−1, 1/C
-
Canonical Form of LCR circuit equation H35(r247)
We obtain the canonical form equ. (4)
ω0 =1√LC
γ = R/L
Q =ω0γ
=1
R
√L
C
so that we have
q̈ + γq̇ + ω20q = V/L
50 60 70 80 90 100 110 120 130 140 1500
2
4
6
8
10
ω
curr
entm
agni
tude
|I|/|V
|
LCR circuit: L = 0.1 C = 0.001
R = 0.1, Q = 100R = 0.2, Q = 50R = 0.5, Q = 20R = 1, Q = 10
Figure 19 : LCR resonance
Current peaks at the resonant frequency ω0 = (LC)−0.5, as does VR (’velocity’)
VC : ’displacement’ (peaks just below ω0)
VL: ’acceleration’ (peaks above ω0)
Power Bandwidth ∆ω given by Q = ω0/∆ω (a route to measuring Q)
-
Impedance H37(r247)
Impedance relates current to voltage: if V =
-
Atoms as Oscillators H39(r247)
How does light interact with atomic matter?
Consider a semi-classical model of a hydrogen atom: a uniform spherical electroncloud of radius a. It is driven by an electric field E0e
iωt (e.g. from a light wave).How does it respond?
If the proton moves a distance x from the centre of the electron cloud, it feels arestoring force
F =
(x3
a3e
)× e× 1
4π�0x2= kx
where the effective spring constant is thus k = e2/4π�0a3
We can thus find the natural undamped oscillation frequency ω20 = k/m:
ω20 =k
m=
e2
4π�0mea3
For a = 50 pm (Bohr radius) we get f0 ≈ 1× 1016 Hz.Optical light has fblue ≈ 8× 1014 Hz; suggests atoms should behave as oscillatorsdriven below resonance when illuminated by light — stiffness controlled; amplitudeindependent of ω. (This is a simple model for scattering).
Damping in Atomic Oscillators H40(r247)
What is the source of the damping? If a charge moves as x0eiωt, it will radiate
energy at an average rate
Prad =e2x20ω
4
12π�0c3
[Larmor’s Radiation formula — you don’t need to know this (Part II).]
The mean energy of the electron oscillating is W = meω20x
20/2 (time average of KE
+ PE).
HencedW
dt= −Prad = −
e2ω2
6π�0c3meW
Thus the energy decays exponentially.
But a lightly damped oscillator’s energy decays with time constant 1/γ, hence
γ =e2ω2
6π�0c3me≈ 1.3× 1010 s−1
Hence we estimate Q = ω0/γ for the atom as ∼ 106.
-
Anharmonic Oscillations (non-examinable) H41(r247)
In real world, perfect harmonic oscillationsare rare: need exactly quadratic potential.
The equation becomes non-linear: cannotsuperpose solutions
For example, a pendulum has a restoringforce ∝ sin θ ≈ θ − θ3/3! + . . .. If weassume the force is proportional to θ, thefractional error in the force (θ2/6) is oforder 0.5% for a swing of 10◦, and ∼ 5%for a swing of 30◦
Consider a symmetric, anharmonicpotential
V (x) =kx2
2
(1 +
α
2x2)
So the restoring force is kx(1 + αx2) andthe equation of motion is then
ẍ+ (1 + αx2)ω20x = 0 (31)
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
1.2
displacement x
rest
orin
gfo
rce
Symmetric, anharmonic potentials
linear restoring forcesin(x) force: ‘soft’ restoring force
‘hard’ restoring force (x + ex3)
Figure 20 : Soft and Hard Restoring Forces
If α > 1, the potential is hard —the spring stiffens with extension.α < 1 implies a soft potential.Apendulum is soft, for example.
Anharmonic Oscillations (non-examinable) H42(r247)
Look for solutions when α� 1; andassume that the mass is at rest at t = 0for simplicity. We need a solution whichhas the correct periodicity and symmetry:a suitable form is
x(t) = A(cosωf t+ � cos 3ωf t+ . . .)
Two unknowns: for small α, expectωf ≈ ω0, and � to be small compared tounity. 0 1 2 3 4 5 6 7 8 9 10
−1
−0.5
0
0.5
1
time t
disp
lace
men
tx
cos(ωt)−0.1 cos(3ωt)
1.1 cos(ωt)− 0.1 cos(3ωt)
Figure 21 : Approximate solution for symmetric anharmonic potential
-
Anharmonic Oscillations (non-examinable) H43(r247)
The amplitude is of order A, so the anharmonic correction is small if |αA2| � 1Now substitute this expression into the equation of motion, working to first order inαA2 and �. We find that this obeys the equation if
ωf ≈ ω0(1 +3
8αA2)
� ≈ 132αA2
(See Main’s book for all the algebra)
So that the period is slightly longer than in harmonic potential if α < 0, as weexpect.
Moreover, the period is now amplitude-dependent.
Application to a Pendulum H44(r247)
The equation of motion is, denoting the angle of swing by x:
0 = ẍ+g
lsinx ≈ ẍ+ g
lx
(1− x
2
6
)
So we have, from equ. (31), α = −1/6 i.e. a soft potential. This yields
ωf ≈ ω0(
1− A2
16
), � ≈ − A
2
192
So if we put A = 10◦ ≈ 0.175 rad, we get a fractional change in period of 0.19%compared to the harmonic solution. (This is about 164 s per day).
We could also examine asymmetric potentials, in which case our expansion would includeeven harmonics. In the end, we will need to use numerical integration of the differentialequation for large amplitude anharmonic oscillations.
-
Anharmonic Oscillations (non-examinable) H45(r247)
What happens if we drive a non-linear oscillator with a sinusoidal forcing term? A wholerich set of dynamical phenomena emerge. Here’s an on-line demo.http://www.myphysicslab.com/pendulum2.html
New behaviours:
’Deterministic Chaos’: driven response can look very complex
Furthermore, even if initial conditions extremely close, solutions will diverge — TheButterfly Effect
Period doubling, quadrupling etc: the final driven solution can contain sub-harmonicsof the driving period.
Increasing the driving force H46(r247)
http://www.myphysicslab.com/pendulum2.html
-
Increasing the driving force H47(r247)
Dependence on initial conditions H48(r247)
Figure 22 : The Butterfly Effect: Initial amplitudes differ by only 1 part in 105
-
Linear System Driven at Multiple Frequencies H49(r247)
Consider driving a system with sum of two harmonic forces:
F (t) = F1 cos(ω1t+ α1) + F2 cos(ω2t+ α2).
Because the equation of motion is linear, the solution is found by adding the responsesas if the individual driving forces acted separately.Initially, suppose the driving forces have the same angular frequency ω1 = ω2 = ω. Then
x = x1 + x2 = A1 cos(ωt+ α1 + φ) +A2 cos(ωt+ α2 + φ)
with Ai = Fi|R| and φ = arg(R). We can draw a phasor diagram because thefrequencies are the same.
A1
A2A
α1α2
Figure 23 : Summing responses from two driving forces
Coherence H50(r247)
The resulting amplitude depends on the relative phase α2 − α1.
A2 = A21 +A22 + 2A1A2 cos(α2 − α1)
If we choose equal driving forces, F1 = F2, then
A2 = 2A21 [1 + cos(α2 − α1)]
In-phase driving: double the amplitude response, quadruple the power
Anti-phase driving: no response (net force is zero)
However, in practice, unless the driving terms F1 and F2 are highly stable, the termcos(α2 − α1) is likely to drift and average to zero over time. For example, twoindependent pendulums of equal length will eventually drift in phase due to tinydifferences in e..g their lengths. In this case, the driven response is the quadrature sum ofthe individual responses:
A =√A21 +A
22
We say the forces are incoherent, and in this case we add the energies not theamplitudes.To achieve coherence, we probably need to derive the driving forces from acommon source.Two oscillations may be coherent over short periods (so thatcos(α2 − α1) does not average to zero), but incoherent over longer times. Next year wewill quantify this.Note also that the changing phase α(t) implies the existence of extra frequencycomponents in the oscillation.
-
Different Driving Frequencies H51(r247)
The response is, as before, simply the sum of the individual responses:
x = x1 + x2 = A1 cos(ω1t+ α1 + φ) +A2 cos(ω2t+ α2 + φ)
In general, we get beating between the different frequencies, with a fast oscillation atangular frequency 1
2(ω1 + ω2) modulated by a slower envelope of angular frequency
12(ω1 − ω2)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
t
ampl
itud
e
Beats: oscillator driven at f1 = 9.5 f2 = 10.5
Figure 24 : Beats for equal amplitude responses A1 = A2
Waves H52(r247)
Wave are central to our understanding of the Universe.
A wave moves energy, and information, from one point to another without bulktranslation of the medium which it disturbs.The medium maybe a solid, a liquid, or agas; or the vacuum in the case of electromagnetic or quantum waves.
-
Transverse String Waves H53(r247)
T
T
x x+ xd
xq ~
¶Y¶
x
q ~¶Y¶
x+ xdx
Y
Figure 25 :
Transverse waves on a string are a good model: we can derive some general results.The transverse displacement is Ψ(x, t); the tension is T and the mass per unit length is ρ.If string’s gradient ψ′ � 1, then the tension is approximately uniform — because theextension caused by the wave is(∫ L
0
√1 + ψ′2dx
)− L = O(ψ′2).
A segment of length δx feels a restoring force towards the axis of
T
(∂Ψ
∂x
∣∣∣∣x
− ∂Ψ∂x
∣∣∣∣x+δx
)= T
(∂Ψ
∂x
∣∣∣∣x
−{∂Ψ
∂x
∣∣∣∣x
+∂2Ψ
∂x2
∣∣∣∣x
δx
})= −T ∂
2Ψ
∂x2
∣∣∣∣x
δx
Newton’s second law yields:
∂2Ψ
∂t2= v2
∂2Ψ
∂x2(32)
where v =√T/ρ.
Solutions To The 1-D Wave Equation H54(r247)
The (non-dispersive) wave equation occurs in many areas of physics.
∂2Ψ
∂t2= v2
∂2Ψ
∂x2(33)
We will see that all frequencies of wave travel at the same speed v; and so does ageneral disturbance. This is what we mean by non-dispersive. Suppose that adisturbance of shape f(x) propagates unchanged at speed v on a string:
Y( ,0)x
x
Y( , )x t
v t
Figure 26 : Wave motion.
We can writeΨ(x, t) = Ψ(x− vt, 0) = f(x± vt)
The sign indicates direction of motion, with the negative sign implying motion to +x. Wenow show that this waveform obeys the wave equation.
-
Propagation of a General Disturbance H55(r247)
Figure 27 :
Y( , )x t0
x
Y( , )x t0
t
x0
t0
snapshot at t0
waveform at x0
Figure 28 : Waveforms and snapshots.
Let u = x− vt, so that Ψ(x, t) = f(x− vt) = f(u). Then
∂Ψ
∂x=
df
du
∂u
∂x=df
du;
∂2Ψ
∂x2=d2f
du2∂u
∂x=d2f
du2.
and∂Ψ
∂t=df
du
∂u
∂t= −v df
du;
∂2Ψ
∂t2= −v d
2f
du2∂u
∂t= v2
d2f
du2
v is the wave speed or the phase speed, and is determined by the physical properties ofthe wave medium. The same equation will result if the wave is travelling to the left. Sowe have proved the result that f(x± vt) is a solution to the wave equation.
The Wave Equation H56(r247)
∂2Ψ
∂t2= v2
∂2Ψ
∂x2
Important features of this equation.
It is general — applies to any waveform.
Wave form will propagate without change of shape if v is constant
It is linear in Ψ, i.e. all the terms involving Ψ are raised to the first power only. Justas we have seen previously for simple oscillating systems, this leads to the principleof superposition. If Ψ1 and Ψ2 each satisfy the wave equation, then anycombination Ψ = αΨ1 + βΨ2 is also a solution. The linearity of the wave equationthus means we can use Fourier analysis to examine the properties of waves.
-
Harmonic Waves H57(r247)
A special case is a continuous sinusoidal wave – a harmonic wave.In a harmonic wave, the displacement Ψ(x, t) varies sinusoidally with time at any point x.The displacement at x = 0 will be given by
Ψ(0, t) = <(Aeiωt
)where A is a complex number. The wave (travelling in the positive x direction) will havethe general form f(x− vt), hence elsewhere it must be given by
Ψ(x, t) = <(Aeiω(t−x/v)
)= <
(Aei(ωt−kx)
)with k = v/ω. v is the wave or phase velocity, and k = 2π/λ is the wavenumber orwavevector.The phase of the wave is (ωt− kx). We can use various representations as we did foroscillators; in each case we have two free parameters to specify the harmonic wave, and athird piece of information is the direction of travel.
ψ(x, t) = A cos(ωt− kx+ φ)ψ(x, t) = BP cos(ωt− kx) +Bq sin(ωt− kx)
ψ(x, t) = Ceiωt−ikx + C∗e−iωt−ikx
ψ(x, t) = <(Deiωt−ikx
)
Waves in 2 and 3 dimensions H58(r247)
In two dimensions, the string becomes a stretched drumskin (assume horizontal). If themass per unit area is σ and the tension per unit length in the drumskin is γ, then thevertical displacement Ψ of a small rectangle (dx, dy) obeys
σdxdy∂2Ψ
∂t2= γdy
∂2Ψ
∂x2dx+ γdx
∂2Ψ
∂y2dy
Hence the wave speed is√γ/σ and
∂2Ψ
∂t2=γ
σ
(∂2Ψ
∂x2+∂2Ψ
∂y2
)= v2
(∂2Ψ
∂x2+∂2Ψ
∂y2
)In general, the wave equation has the form
c2∇2Ψ = ∂2Ψ
∂t2
-
Plane Waves H59(r247)
A plane is defined by the geometric constraint n̂ · r = d for some constant d, the distancefrom the origin to the plane.n̂ is the unit normal to the plane.Hence if we define a wavevector k = kn̂, where k = 2π/λ, then it is clear that k · r isconstant on this plane.
x
y
λ
k
dr n̂
Figure 29 : Plane wave front and wavevector
Plane Waves H60(r247)
A plane wave propagating in the direction k̂ can thus be written
ψ(r, t) = Aeiωt−ik·r (34)
and we can get wave travelling in the opposite direction −k̂ by
ψ(r, t) = Aeiωt+ik·r (35)
Note that to differentiate the first expression wrt time, multiply by iω.And to take its gradient, ∇ψ, multiply by −ik.So for example,
∇2ψ = (−ik) · (−ik)ψ = −(k2x + k2y + k2z)ψ = −k2ψ;
ψ̈ = (iω)(iω)ψ = −ω2ψ
Figure 30 : Wavefronts for a plane wave
-
Spherical Waves H61(r247)
Figure 31 :
Suppose a wave propagates from a point with spherical symmetry. The wave equation isnow (ignoring the θ, φ parts which do not vary):
c21
r2∂
∂r
(r2∂ψ
∂r
)=∂2ψ
∂t2
Consider a function
ψ(x, t) =f(r ± ct)
r(36)
where the function f describes a disturbance propagating outwards/inwards.
Spherical Waves H62(r247)
Then,∂ψ
∂r= − f
r2+f ′
r
The LHS of the wave equation is then equal to
c2
r2(−f ′ + f ′ + rf ′′
)=c2f ′′
r
The RHS of the wave equation is∂2ψ
∂t2=c2f ′′
r
Hence equ. (36) represents a valid general solution to the wave equation in 3-dimensions.In particular, we have a harmonic solution for outward propagating waves:
ψ(x, t) =Ae(iωt−ikr)
r
The 1/r amplitude dependence can be seen to conserve energy (i.e. this is the inversesquare law for power).
-
Cylindrical Waves H63(r247)
If we launch a wave from a line source instead of a point source, we can look forsolutions in cylindrical symmetry. The wave equation in cylindrical polar coordinates, withno angular dependence, reads
c21
r
∂
∂r
(r∂ψ
∂r
)=∂2ψ
∂t2
Let’s guess a possible solution
ψ(x, t) =f(r ± ct)√
r(37)
where the function f describes a disturbance propogating outwards/inwards. The√r will
conserve energy flow. The wave equation yields
c2f ′′√r
= c2f ′′√r
+c2f
4r5/2
So the function is not a solution. However, if the second term on the RHS is smallcompared to the first, the solution is approximately correct. We require∣∣∣∣ f4f ′′r2
∣∣∣∣� 1 → (kr)� 1where we have used f ′′ = −k2f (The full solution involves Bessel functions for small r.)Hence equ. (37) represents a valid (approximate) solution to the wave equation for r � λ.
Figure 32 : A cylindrical wave created by diffraction at a long slit.
ψ(x, t) ≈ f(r ± ct)√r
-
Polarisation H65(r247)
Figure 33 : Linearly polarised wave (from Hecht, Optics).
We can displace the string in a transverse direction either vertically or horizontally.Transverse means orthogonal to the direction of wave propagation.The waves in these directions are independent but have the same speed.:
v2ψ′′x = ψ̈x; v2ψ′′y = ψ̈y
The amplitude and relative phase of the displacement along these two directions definethe polarisation of the wave. The two oscillations must be coherent wrt one another tocreate a polarised wave.
Linear Polarisation H66(r247)
Achieved by oscillating the end of the string along a straight line — each point on willoscillate along a parallel line. Representing the y and z components of the displacementas
Ψy = Ay cos (ωt− kx)Ψz = Az cos (ωt− kx+ φ)
we find that linear polarisation arises where φ = 0 (or integer multiples of π). Clearly awave polarised along the y (z) axis can be described by Az = 0 (Ay = 0). Where bothAy and Az are non-zero, the wave will be polarised along an intermediate direction, withan amplitude A =
√A2y +A2z and an angle of polarisation θ = tan
−1(Az/Ay) to the yaxis. Thus any linearly polarised wave can be resolved into two orthogonal linearlypolarised components with the same phase.
Ay
Az
y
z
q
A
Figure 34 : Displacement vector at fixed position for a linearly polarised wave.
-
Circular Polarisation H67(r247)
If the two components are equal in magnitude, but in phase quadrature, (φ = (m+ 12)π)
we get circular polarisation.The displacement vector traces a corkscrew.
Figure 35 : Left-handed circular polarisation
Looking towards the wave source, right-circular polarisation corresponds to clockwisemotion of the displacement (electric field) vector. And vice versa for left-handed. (Radioconvention/IEE is opposite!)
y
z
A
Figure 36 : Displacement vector at fixed position for a circularly polarised wave.
Left and Right Circular Polarisation States H68(r247)
Figure 37 : Left-handed circularly polarised wave.
Figure 38 : Right-handed circularly polarised wave.
-
Elliptical Polarisation H69(r247)
General case: amplitudes and relative phase of the two components are arbitrary, and thedisplacement at any position follows an ellipse.
Ψy = Ay cos (ωt− kx)
Ψz = Az cos (ωt− kx+ φ) = Az (cos(ωt− kx) cosφ− sin(ωt− kx) sinφ)Eliminating ωt− kx gives
Ψz = Az
(ΨyAy
cosφ−
√1−
Ψ2yA2y
sinφ
).
We can rearrange this into the standard equation for an ellipse
Ψ2yA2y
+Ψ2zA2z− 2ΨyΨz
AyAzcosφ = sin2 φ; tan 2α =
2AyAz cosφ
A2y −A2z
a
Ay
Az
Yz
Yy
Y
Figure 39 : Elliptical Polarisation
Polarisation: some comments H70(r247)
Circular and linear polarisations are special cases of elliptical polarisation.
We can regard circular polarisation as the sum of two equal-magnitude, coherent,orthogonal linear polarisations with π/2 relative phase.
Equally, we can regard linear polarisation as the sum of equal amplitude, coherent,Left- and Right-handed circularly polarised waves.
Figure 40 : Linear as sum of two circular beams
You need three numbers to specify the elliptically polarised state: two amplitudes,and an angle.
Waves are often partially polarised. A single extra value defines the power in thisunpolarised component. You can think of unpolarised light arising from twoorthogonal, incoherent oscillations with equal mean square magnitudes.
-
Representation of Polarisation States H71(r247)
It is common to represent a polarised wave by expressing the magnitudes and relativephase of the y and z oscillations (for waves moving in the +x direction) as a2-component vector (Ψy,Ψz).
Linearly polarised along y :
(A0
)eiωt
Linearly polarised along z :
(0A
)eiωt
Usually we drop the eiωt term.
left circular:
(AiA
)right circular:
(−AiA
)
elliptical aligned with x/y axes:
(AiB
)(α, β) represents a wave linearly polarised at an angle arctan(β/α) to the y axis.For example we note that the sum of equal magnitude RCP and LCP waves yields alinearly polarised one.
Wave Impedance H72(r247)
Wave impedance is used to define the relationship between the force and the waveresponse. It is important in calculating power transfer in a wave, and governs whathappens when waves cross boundaries. For our string,
Z = Impedance =driving force
velocity response
T
q
Tsinq
Figure 41 : Transverse force in a wave.
Note this velocity is the transverse response ψ̇, not the wave speed ω/k. Consider afree end in the string. The transverse driving force F is F = −T sin θ ≈ −T ∂Ψ
∂x. Hence
Z =transverse driving force
transverse velocity= −T (∂Ψ/∂x)
(∂Ψ/∂t)
But for a waveform Ψ = f(u− vt) moving in direction +x, we have∂Ψ
∂x=df
duand
∂Ψ
∂t= −v df
du
Hence
Z =T
v=√Tρ = ρv. (38)
. . . and Z is negative for wave travelling to −x.
-
Power in String Wave H73(r247)
Energy is present in a string wave in the form of kinetic and potential energy. The powerinput into a wave is given by
power input = transverse force× transverse velocity = F u.
where u is the transverse velocity. The mean power input is 〈P 〉 = 12
-
Wave Reflection at Boundaries 2 H75(r247)
Continuity of Ψ at x = 0:A1e
iωt +B1eiωt = A2e
iωt
A1 +B1 = A2. (40)
Continuity of transverse force −TΨ′:
T (−ik1)A1 + T (ik1)B1 = T (−ik2)A2.
k1 = ω/v1 and k2 = ω/v2
∴ − Tv1ωA1 +
T
v1ωB1 = −
T
v2ωA2.
UseT/v1 = Z1 and T/v2 = Z2
HenceZ1(A1 −B1) = Z2A2. (41)
We can now solve equ. (41) and equ. (40) for the amplitude reflection andtransmission coefficients r and τ (the latter not to be confused with time t).
Wave Reflection at Boundaries 3 H76(r247)
We obtain:
r =B1A1
=Z1 − Z2Z1 + Z2
(42)
τ =A2A1
= 1 + r =2Z1
Z1 + Z2(43)
If Z2 =∞ (clamped string): r = −1 and τ = 0. Complete anti-phase reflection —phase change of π on reflection.
Figure 43 :
If Z2 = 0, i.e. the second string is massless (same as a free end if tension could bemaintained): r = 1, τ = 2 — complete in-phase reflection. No energy transmitted.
Figure 44 :
Z2 = Z1: r = 0 and τ = 1. Strings are matched.
-
Notes on Reflection and Transmission H77(r247)
Results are general and apply to a wide variety of wave systems.
The General Recipe: at interfaces need to consider (i) continuity of “displacement”,Ψ and (ii) continuity of “force” ∝ ZΨ for harmonic waves. Remember to usenegative Z for negative-going waves.
Depending on the precise problem, we may end up with Z1 and Z2interchanged. . . e.g. in an electromagnetic wave, E and H = E/Z are continuous atthe boundary. And on a transmission line, we have continuity of voltage V andcurrent I = V/Z (see the electromagnetism course).
Remember that the impedance is in general complex, so we can get arbitrary phasedifferences between the incident, reflected and transmitted waves.
We have only found the steady-state response.
Reflection and transmission of energy H78(r247)
The power transmitted by a harmonic wave is P = Zω2A2/2. So the incident energy flowis 1
2Z1ω
2A21, with12Z1ω
2B21 reflected and12Z2ω
2A22 transmitted. The Power ReflectionCoefficient R is thus
R =Z1B
21
Z1A21=
(Z1 − Z2Z1 + Z2
)2= R
and the Power Transmission Coefficient T is
T =Z2A
22
Z1A21=
4Z1Z2(Z1 + Z2)2
= T.
Energy is conserved:R+ T = 1
If Z1 and/or Z2 are complex, recall the average power input is12
-
Impedance Matching 1 H79(r247)
Often useful to match two media to minimise reflections – many applications in optics,acoustics, electrical transmission. . . Consider a wave incident on a layer of thickness l,impedance Z2, filling 0 ≤ x ≤ l, on top of an infinite substrate Z3:
z x
y
e−ik1x
re+ik1x
ae−ik2x
be+ik2x
τe−ik3(x−l)
impedance Z1 impedance Z2 impedance Z3
l
Figure 45 : Reflection from an intermediate layer
To simplify algebra: drop the implicit exp(iωt); set the incident amplitude to unity (sothat the reflected amplitude is r); also added a phase shift exp(ik3l) to the transmissioncoefficient. Note the three wavevectors k1, k2 and k3. It is useful to define
γ = e−ik2l
There are four unknowns: r, τ, a, b in the steady-state solution. We apply four boundaryconditions to find them.
Impedance Matching 2 H80(r247)
Ψ continuity at x = 0:
1 + r = a+ b
Ψ continuity at x = l with γ = e−ik2l:
γa+b
γ= τ
−TΨ′ continuity at x = 0:
Z1(1− r) = Z2(a− b)
−TΨ′ continuity at x = l:
Z2
(γa− b
γ
)= Z3τ
Solve these 4 equations for (a, b, r, τ) — generally messy. But if we choose a quarterwavelength of material in the layer, k2l = (m+
12)π for integer m, then
γ = e−iπ/2 = −i and they simplify to:
1 + r = a+ b; Z1(1− r) = Z2(a− b)
a− b = τ i; Z2(a+ b) = Z3τ i
∴ r =Z1 − Z
22Z3
Z1 +Z22Z3
=Z1 − ZeffZ1 + Zeff
where Zeff = Z22/Z3. So the layer plus quarter-wave of substrate have an effective
impedance of Z22/Z3. Hence by choosing l and Z2 we can match two media and stopreflections. We use a quarter wave λ2/4 section of impedance Z2 =
√Z1Z3.
-
Impedance Matching 3 H81(r247)
Quarter-wave matching works because the waves reflected at the first and secondboundary are out of phase if l = λ2/4.
We have again solved the steady-state problem — a transient will occur as wavehits boundary.
Common application: anti-reflection lens coatings
Figure 46 : Quarter-Wavelength coated lens
Need to use result that Z = Z0/n. If we have a quarter wave coating of MagnesiumFlouride (n = 1.38) on crown glass (n = 1.52), it has effective refractive indexne = 1.38
2/1.52 = 1.25, yielding a power reflection coefficient of 1% (compared to4% without).
Can also match impedances with a gradual impedance change.
» l
Figure 47 : Gradual impedance change.
Longitudinal Waves H82(r247)
Longitudinal Waves have displacement of the medium parallel to the wavevector:sound waves are the most common example
Corresponds to a wave of rarefaction and compression
No polarisation possible
In a sound wave, the pressure wave and displacement wave are π/2 out of phase:maximum pressure occurs at minimum displacement.
x
displacementin -directionx
x
x
compression rarefaction
Figure 48 : Displacement of gas in a longitudinal wave.
-
Compression of an Ideal Gas H83(r247)
How does a gas react to compression/rarefaction? Two limits:
Isothermal process: heat Q flows rapidly out of/into gas to keep temperatureconstant. The gas law applies: pV = nRT = constant.
Adiabatic process: the compression/rarefaction is so fast that heat cannot flowinto/out of the gas before the next phase of the pressure wave passes through. Thework done on the gas is stored as extra internal energy U :
−p dV = dU
A perfect gas has internal energy is U = CvT , where Cv is its heat capacity atconstant volume.
dU = Cv dT = −p dV .Using pV = nRT , we obtain
Cv dT +nRT
VdV = 0; ∴
dT
T+nR
Cv
dV
V= 0
∴ T VnRCv = constant
∴ p VCv+nRCv = pV γ = constant
where the heat capacity at constant pressure Cp = Cv + nR, and γ = Cp/Cv. Fordiatomic molecules at room temperature (air) γ ≈ 7/5. For atomic gas, γ ≈ 5/3.
We will see that sound waves are usually adiabatic.
Wave Equation for Sound in a Gas: the Strain H84(r247)
Consider dynamics of a parcel of gas lying (when undisturbed) between x and x+ ∆x. Ithas cross section ∆S and moves in the x direction only.
x x+ Dx
( + + )p SY DY DP P
x a x+ ( ) x x+a a+ +D D
( + )p SY DP
Figure 49 :
The wave displaces gas originally at x to x+ a(x, t), and creates an extra pressureΨp(x, t), so the total gas pressure locally is p+ Ψp. When the wave is present, the parcelextends from x+ a(x) to x+ ∆x+ a+ ∆a = x+ ∆x+ a(x) + ∂a
∂x∆x. The volume
change of the element is given by
∆S∆a = ∆S∂a
∂x∆x.
The fractional change in volume (the volume strain) is
∆V
V=
∆S ∂a∂x
∆x
∆S∆x=∂a
∂x(44)
-
Wave Equation for Sound in a Gas: the Stress H85(r247)
x x+ Dx
( + + )p SY DY DP P
x a x+ ( ) x x+a a+ +D D
( + )p SY DP
Figure 50 :
In a gas, the stress is equal to the pressure, and is isotropic. The force imbalance on theends of the parcel is
Fnet = (p+ Ψp)∆S −(p+ Ψp +
∂Ψp∂x
∆x
)∆S = −∂Ψp
∂x∆x∆S
i.e. the pressure gradient times the volume.This force accelerates the parcel of gas:
−∂Ψp∂x
= ρ∂2a
∂t2(45)
Wave Equation for Sound in a Gas H86(r247)
We now need to relate the stress to the strain. Assume the waves are adiabatic, so thatheat cannot flow quickly in/out of the compression regions: pV γ is constant. (We cancheck this by experiment.) Differentiating:
dp V γ + pγV γ−1dV = 0
∴ dp = Ψp = −γp∆V
V= −γp∂a
∂x. (46)
using equ. (44). Note that dp is the pressure change associated with the wave: dp ≡ Ψp.We need the spatial derivative of Ψp to form the equation of motion:
∂Ψp∂x
= −γp∂2a
∂x2− γ ∂p
∂x
∂a
∂x.
The ratio of the 2nd term to the first term on the RHS is ∼ Ψp/p ∼ γka ∼ a/λ and isnegligible in most practical situations i.e. we usually have a� λ. Using equ. (45), weobtain
γp∂2a
∂x2∆x∆S = ρ∆x∆S
∂2a
∂t2.
-
Wave Equation for Sound in a Gas H87(r247)
Finally, we have the wave equation for (small amplitude) sound waves:
∂2a
∂x2=
ρ
γp
∂2a
∂t2(47)
Note the speed is independent of wavelength/frequency i.e. there is no dispersion.
This is the wave equation, with wave speed v given by
v =
√γp
ρ=
√γRT
M(48)
where M is the molar mass. Note the dependences on temperature, mass and γ.
The mean squared speed of molecules of mass m in an ideal gas is given by theequipartition theorem:
1
2m〈c2〉 = 3
2kBT ; ∴ crms =
√〈c2〉 > =
√3RT
M.
Hence the r.m.s. molecular speeds will be close to, but slightly larger than the wavespeed. Remember, though, that the passage of a sound wave does not require thenet movement of gas molecules from one point to another.
Sound speed: examples H88(r247)
Speed of sound in dry air at 20 ◦C and atmospheric pressure (101.325 kPa) is
v =
√1.4× 101.325× 103
1.204= 343 m s−1
Accurately confirmed by experiment — so adiabatic assumption is good. (γ iseffectively 1 for isothermal change, so speed difference would be about 20%).
Speed rises by about 0.6 m s−1 per ◦C of temperature rise.
In fact, Newton was the first to estimate the speed of sound; he assumed isothermalchanges. His estimate was ∼ 15% too low because of this.Speed of sound in Helium at same pressure and temperature is
v =
√1.667× 8.314× 293
0.004= 1004 m s−1
Solar corona: γ = 5/3, ionised hydrogen (m = 1amu), T = 1× 106 K, andv = 1.18× 105 m s−1
In cold interstellar gas clouds: T ≈ 8 K, and v = 75 m s−1
-
The Displacement Wave and Impedance H89(r247)
Our wave equation was expressed in terms of the molecular displacements a but this isvery hard to measure. We normally measure the pressure fluctuation with e.g. amicrophone. We relate the two via
Ψp = −γp∂a
∂x.
If we assume a harmonic wave traveling to +x of the form
a = a0e(iωt−ikx)
thenΨp = iγpka
i.e. the pressure leads the displacement by π/2 and has amplitude γpka0. Thecharacteristic impedance Z can be defined as the driving force divided by responsespeed, in the usual way. For this harmonic wave we have
Z =force
velocity=
∆SΨpȧ
=∆Siγpka
iωa= ∆S
γp
v= ∆S
√γpρ
Often, we use the impedance per unit area L (often just called the acoustic impedance):
L = √γpρ = γpv
In dry air at 20 ◦C, 101.325 kPa: L ≈ 413 kg m−2 s−1.
Power in an acoustic wave H90(r247)
The power transferred by the wave is force time velocity in the usual way. For a harmonicsound wave, the particle speed and pressure are
a(x, t) = a0 exp(iωt− ikx)
ȧ(x, t) = iωa0 exp(iωt− ikx)
Ψp = A0 exp(iωt− ikx) = iγpka0 exp(iωt− ikx)
where A0 = iγpka0 is the pressure amplitude of the wave. Note that the speed and forceare in phase.The mean power per unit area, or intensity I of the wave, is thus
I =1
2
-
The Decibel Scale for Sound H91(r247)
We can specify the sound pressure level, SPL of a sound in decibels with respect tosome reference level; remember the decibel scale is a relative scale only.
SPL in decibels = 10 log10
(p2rmsp2ref
)= 20 log10
(prmspref
)(49)
We usually use a reference level of pref = 20 µPa: note this is a root-mean-square (rms)value. pref is approximately the threshold of human hearing at 1 kHz In a young adult.This is only 2× 10−10 atmospheres!.Equivalently, we can express the sound as a power level in decibels:
dBA level = 10 log10
(I
Iref
)(50)
with the reference intensity
Iref = p2ref/L = 1× 10−12 W m−2
where we used for air L = 400 kg m−2 s−1.
Human Hearing H92(r247)
The human ear can respond to pressurewave amplitudes from 1 pW m−2 to1 W m−2, and frequencies from 30 Hz to20 kHz.
Figure 51 : Human Hearing Response with Age
-
Typical Sound Characteristics H93(r247)
Middle C has a frequency (in some music systems) of 256 Hz, so has a wavelength of1.34 m in air at 20 ◦C.
How powerful is the human voice? 60dB at 1 m seems a reasonable average. Now60dB (relative to 1 pW m−2) corresponds to a flux of 1 µW m−2, so a total power of4πR2 times the flux, yielding ∼ 10 µWWhat is the pressure level you are nearing now in the lecture theatre? What is themolecular displacement?
The Sound Pressure Level for 60dB is 1000pref = 20 mPa i.e. only 2× 10−7 ofatmsospheric pressure.
The displacement comes from
Ψp = −γp∂a
∂x
so |a| = |Ψp|/(γpk) ∼ 30 nm i.e. 300 atomic diameters.Barely audible sounds, 10dB say, have molecular amplitudes of order 1× 10−10 m!
A Dangerous Experiment H94(r247)
(For Ball organisers) you may get about a few watts of audio power out of largespeakers, so standing 1m away, the intensity is of order 1 W m−2 which is 120dB(and a pressure amplitude of 20 Pa).
A dangerous experiment. . .
(Interestingly, 20 Pa is the atmospheric pressure difference over a height of only∼ 2 m).
-
Sound waves in solids and liquids 1 H95(r247)
Longitudinal waves also exist in liquids and solids. The properties of the medium aredescribed by a generalisation of equ. (46), the relationship between the pressure due tothe wave and the strain in the medium
Ψp = −K∂a
∂x
where K is the relevant elastic modulus. This gives the general result
v =
√K
ρ
where ρ is the density.For gases and liquids, the pressure is isotropic, but the expansion and rarefaction takesplace only in the direction of passage of the wave. Hence the fractional volume change isdirectly proportional to the fractional change in length of the element(∆V/V = ∆a/∆x = ∂a/∂x). The relevant modulus is the bulk modulus B:
dp = −B dVV
where B = γp for a gas undergoing adiabatic changes (and B = p for isothermal ones).Solids are more complicated as they can support shear stresses. Shear waves travel moreslowly than compression waves and are important in earthquakes.
Sound waves in solids and liquids 2 H96(r247)
When a thin bar is compressed longitudinally it can expand in the transverse direction(the ratio of the two strains is known as Poisson’s ratio). However, in a bulk solid, themedium cannot expand sideways, hence more longitudinal pressure is required to producea longitudinal strain, and the modulus in therefore larger. (See Physics B: Dynamics).We can ignore these effects if the solid is a thin bar and we look at compression wavestravelling along the bar. The longitudinal stress τ (force/area) in the solid is then relatedto the strain ∂a
∂xby
τ = Y∂a
∂x
where Y is Young’s modulus for the material.The equation of motion then becomes
Y∂2a
∂x2= ρ
∂2a
∂t2
with a wave speed
v =
√Y
ρ
where ρ is the density.
-
Longitudinal Waves Usually Travel Faster than Transverse Waves H97(r247)
Figure 52 :
On a stretched string, transverse waves have
vT =
√T
ρ=
√220 N
0.6 g m−1≈ 600 m s−1
whereas longitudinal waves have
vT =
√Y
ρ=
√200 GPa
7700 kg m−3≈ 5000 m s−1
The fundamental mode oscillations both have awavelength of about 1m for a violin (twice theviolin length), so the frequencies are about600 Hz and 5 kHz.
Note also the acoustic impedance L = Y/v =√Y ρ for waves in a solid will be much
larger than in a gas typically. For steel, ρ = 7700 kg m−3, so L ≈ 4× 107 kg m−2 s−1,which is 105 time larger than for air. So an open air-steel interface is essentially a freeend with a pressure node, and displacement antinode occurring there.
Standing Waves in One Dimension H98(r247)
x = 0 x b=
n = 1
n 3=
n = 2
Figure 53 : Standing waves on a string.
A standing wave is a wave confinedwithin a region of space by applyingboundary conditions, e.g. a wave on aviolin string
The standing wave can be found as asuperposition of forward and backwardtraveling waves. Simplest case to analyseis where perfect reflection occurs at theboundary.
On a string of length b, with boundaryconditions are Ψ(0, t) = Ψ(b, t) = 0:
Ψ = A cos(ωt− kx)−A cos(ωt+ kx)= 2A sinωt sin kx.
The boundary conditions are satisfied ifkb = nπ, where n = 1, 2, 3 . . ., i.e.k = nπ
b. For a wave speed v, the allowed
frequencies are given by
ω =nvπ
b.
-
Standing Waves in 2 and 3 Dimensions H99(r247)
Easiest to look for solutions to the wave equation
c2∇2Ψ = ∂2Ψ
∂t2
which satisfy the boundary conditions. For waves confined in a 3-d box, for example, thewavefunctions are given by
Ψ = A sin kxx sin kyy sin kzz cosωt.
For a box with a corner at the origin, of dimensions a, b, c, the boundary conditions aresatisfied by a wavevector
k =
(lπ
a,mπ
b,nπ
c
)where l, m, n = 1, 2, 3 . . .. In two dimensions, this gives the patterns of oscillation of arectangular drum.
Figure 54 : Standing Waves on a rectangular and a round drum
Damped Waves H100(r247)
Like oscillators, real waves usually lose energy through damping. A string wave will loseenergy via air resistance: there will be an extra transverse force on an element dx ofstring. Assume it is proportional to the transverse speed, so has form −βψ̇dx. Therevised wave equation is now
∂2Ψ
∂t2= v2
∂2Ψ
∂x2− βρ
∂Ψ
∂t(51)
which we can rewrite as
∂2Ψ
∂t2+ Γ
∂Ψ
∂t= v2
∂2Ψ
∂x2(52)
where Γ = β/ρ expresses the damping. To find solutions, try a harmonic waveψ = D exp(iωt− ikx): we find
ω2 − iΓω = v2k2 (53)
so that k must be complex in general (ω is real). Thus we must have exponential andoscillating parts in the solution for Ψ. Splitting k into real and imaginary parts:k = kr − iki, we find that
k2r − k2i =ω2
v2; 2krki =
Γω
v2(54)
-
Lightly Damped Waves H101(r247)
0 5 10 15 20 25 30 35 40
−1
−0.5
0
0.5
1
x
ψ(x)
A Damped Travelling Wave: kR = 1, kI = 0.1
t = 0t = T/4t = T/2
Figure 55 : A lightly-damped wave
If we assume light damping (Γ� ω), we can see from equ. (53) that kr ≈ ω/v, and thuski ≈ Γ/(2v): so the solution is
Ψ = exp(−kix)
-
Heavily Damped Waves H103(r247)
0 1 2 3 4 5 6 7
−1
−0.5
0
0.5
1
x
ψ(x)
A Heavily Damped Wave: kR = 1, kI = 1
t = 0t = T/4t = T/2
Figure 56 : A heavily-damped wave
Ψ = exp(−k′x)
-
Reflection H105(r247)
Consider now an undamped wave system (perfect string, impedance Z0) meeting adamped one (string with lateral resistance, impedance Z).We will get reflections at the boundary, and can use the usual relation for the reflectioncoefficient:
r =Z1 − Z2Z1 + Z2
Light Damping, Γ� ω: we obtain
r(ω) =iΓ
4ω
so |r| is small, and has phase π/2: most power is transmitted, little is reflected.Heavy Damping (Γ� ω): we obtain
r(ω) =1− (1− i)
√Γ2ω
1 + (1− i)√
Γ2ω
≈ −1
so the wave is almost all reflected in anti-phase. (You will see in Electromagnetismthat this is a good model for reflection of light from a good conductor.)
Phase Speed of Damped String Waves H106(r247)
In the lightly damped case, we saw that the wave propagates with an exponentiallydecaying amplitude:
Ψ = exp(−kix)
-
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
kr
phas
esp
eed
v φ=
v( 1+
Γ24v
2 k2 r
) −1/2
Dispersion of Waves on a Damped String, v = 1
Γ = 0.0Γ = 0.2Γ = 0.8
Figure 57 : Speed of waves on a damped string
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
kr
ω=
vkr
( 1+Γ2
4v2 k
2 r
) −1/2
Dispersion of Waves on a Damped String, v = 1
Γ = 0.0Γ = 0.2Γ = 0.8
Figure 58 : Dispersion Relation for Waves on a damped string
-
Second Dispersion Example: A Stiff String H109(r247)
As a second example of dispersion, consider a string which resists bending but has noexternal resistance (no damping). There exists a torque on any element of string when itis curved. This introduces a second term into the wave equation (details of derivation notimportant here — see dynamics course):
∂2Ψ
∂t2=T
ρ
(∂2Ψ
∂x2− α∂
4Ψ
∂x4
)= v2
(∂2Ψ
∂x2− α∂
4Ψ
∂x4
)(55)
Trying a harmonic wave solution Ψ ∝
-
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
1.2
k
v φ=
v√1+
αk2
Dispersion of Waves on a Stiff String, v = 1
α = 0.5 anomalous dispersionα = 0.0 non-dispersive
α = −0.5 normal dispersion
Figure 60 : Phase Speed on a stiff string
Implications for Tuning a Piano H112(r247)
A piano string of length l has fixed ends. Its fundamental mode (middle C, 256 Hzsay) is a standing wave of wavelength λ = 2l; its first harmonic has wavelengthλ = l; etc.
If the phase speed as a function of wavelength is v(λ), the frequencies of these areffund = v(2l)/2l = and fharm = v(l)/l.
We want to tune the next C string one octave higher, i.e. at twice the frequency.But do we tune to twice ffund or to fharm? If v is constant, these are the same andpiano tuning is easy.
But piano strings are stiff (α > 0) so that vφ increases with frequency, and sofharm/ffund > 2. Suppose fharm = 542 Hz So we can either
tune the second string’s fundamental have twice the frequency of the first’sfundamental (i.e. to 512 Hz); ortune the second string to match the frequency of the first harmonic of the first string.(i.e. to 542 Hz)
In fact we choose the second so you won’t hear any beats (in this case, at 30 Hz)between these two modes when the strings are played simultaneously, so things willsound nice.
-
Group Velocity (Take One) H113(r247)
With no dispersion, a wavepacket or wave group propagates without change of shape atthe speed given by the wave equation: Ψ̈ = v2Ψ′′. We have proved this earlier.But if v depends on wavelength, the wave group may (a) travel at a different speed, and(b) may change its shape. This is the phenomenon of dispersion.Water Wave Dispersion videoConsider two equal amplitude waves propagating together, with slightly differentfrequencies, ω1 ≈ ω2:
Ψ = cos(ω1t− k1x) + cos(ω2t− k2x)
Define the sum and difference frequencies and wavevectors:
ω+ =1
2(ω1 + ω2), ω− =
1
2(ω1 − ω2), k+ =
1
2(k1 + k2), k− =
1
2(k1 − k2).
then simple algebra shows the usual beat phenomenon:
Ψ = 2 cos(ω+t− k+x) cos(ω−t− k−x)
There is a high-frequency wave with speed vφ = ω+/k+, which is very close either of thephase velocities ω1/k1 or ω2/k2. This is modulated by a lower-frequency travellingenvelope with speed vg = ω−/k−.
vφ =ω
k(56)
vg =ω−k−
=ω1 − ω2k1 − k2
; vg =dω
dk(57)
Group Velocity: Take Two H114(r247)
A more general argument is as follows. The group velocity vg is the speed at which a’bump’ in the wave travels. We can regard the wave being composed of many individualharmonic waves of different frequencies (Fourier components); each has its own phase(ωt− kx+ φ).If these phases are equal for the constituent waves, they will add up in phase to make a’bump’.Thus we can find the bump where ωt− kx+ φ is equal for all the constituent waves: thisoccurs when [
d
dω(ωt− kx+ φ)
]ω0
= 0
where we evalue at ω0, the angular frequency of the typical wave in the group. Hence
t− x(
dk
dω
)ω0
= 0
which shows the speed of the ’bump’ is the group velocity:
vg =dω
dk
http://bit.ly/vXfBFE
-
Group Velocity: Take Three H115(r247)
Figure 61 : A wave group.
Consider a wave packet consisting of a carrier wave with wavenumber k0 multiplied byan envelope f(x) (in this example, f is a Gaussian). At t = 0:
Ψ(x, 0) = <[f(x)eik0x
].
Using the Fourier transform, write envelope in terms of its components:
Ψ(x, 0) = <[(∫ ∞
−∞F (k1)e
ik1xdk1
)eik0x
]= <
[∫ ∞−∞
F (k1)ei(k0+k1)xdk1
]F (k1) is the Fourier transform of f(x) (ignore factors of e.g.
√2π)
Each sinusoid with wavenumber k0 + k1 will propagate at its own phase velocity, so wecan write the waveform at some later time t as
Ψ(x, t) = <[∫ ∞−∞
F (k1)ei[(k0+k1)x−(ω0+ω1)t]dk1
]= <
[∫ ∞−∞
F (k1)ei(k1x−ω1t)dk1e
i(k0x−ω0t)]
(58)where ω0 = ω(k0) and ω0 + ω1 = ω(k0 + k1) for a dispersion relation ω = ω(k).
Now assume that F (k1) is non-zero over a small wavenumber range ±∆k (∆k � k0), soonly need consider waves in the band k0 ±∆k. (This is same as assuming f(x) is manycarrier wavelengths across). Now expand the dispersion relation ω(k) about k0:
ω = ω0 +∂ω
∂k
∣∣∣∣k0
k1 +1
2
∂2ω
∂k2
∣∣∣∣k0
k21 + · · · (59)
Working to first order in k1:ω1 ≈ vgk1 (60)
where the group velocity is vg =∂ω∂k
∣∣k0
. Substituting equ. (60) into equ. (58) we have
Ψ(x, t) = <[∫ ∞−∞
F (k1)eik1(x−vgt)dk1e
i(k0x−ω0t)]
= <[∫ ∞−∞
F (k1)e−ik1vgteik1xdk1e
i(k0x−ω0t)]
Recognising the k1 integral as an inverse Fourier transform, we can make use of theconvolution theorem and the fact that the Fourier transform of a delta-function is acomplex exponential to yield
Ψ(x, t) = <[f(x) ∗ δ(x− vgt)ei(k0x−ω0t)
]= <
[f(x− vgt)eik0(x−vpt)
](61)
where vp = ω0/k0.
-
Notes on Group Velocity H117(r247)
equ. (61) says that after a time t the carrier wave eik0x has propagated a distancevpt while the modulating envelope f(x) has propagated a distance vgt. For anon-dispersive wave ω/k = ∂ω
∂kfor all frequencies, and so the envelope will propagate
at the same speed as the carrier wave, but for a dispersive wave the two velocitiescan be different, and so the wave crests will move relative to the envelope.
Typically we use the envelope as a tracer of the information carried by the wavepacket, so the group velocity will be the speed of interest when looking at thepropagation of information.
The range of wavevectors ∆k in a wave group is inversely related to the spatialextent of the group ∆x0 at t = 0:
∆k∆x ≈ 1
Note that the above result is restricted to wave groups consisting of sinusoidalcomponents with a very narrow spread of frequencies. A broad-band signal can bethought of as consisting of a number of narrow-band wave groups, and if the groupvelocity is different between these groups (i.e. the second-order term in theexpansion in equ. (59) is non-negligible) then the different wave groups will begin tospread apart as time progresses and the envelope will become distorted.
Estimate of rate of spreading of wave packet H118(r247)
By approximating the dispersion relation as a linear function near the carrier wavevectork0 (i.e ω = ω0 + vgk1), we have shown that a wave group propagates at vg =
∂ω∂k
.If we include a higher order (quadratic) term in the Taylor expansion, we also find thatthe envelope elongates and changes shape — it disperses.A simple way to estimate this is to note that if the group contain wavevectors in a bandof size 2∆k about k0, then there exists a range of group velocities in the group, withtypical extreme values being
vg1 =∂ω
∂k
∣∣∣∣k0+∆k
, vg2 =∂ω
∂k
∣∣∣∣k0−∆k
After a time t, the packet will thus spread by an amount equal to t times this differencebetween these two speeds; expanding the group velocity close to k0 we predict that thewavepacket will have a length given by
∆x ≈ ∆x0 + (vg1 − vg2)t ≈ ∆x0 + 2|β|∆k t ≈ ∆x0 + 2|β|t
∆x0
where ∆x0 is the initial length of the group, and
β =∂2ω
∂k2
∣∣∣∣k0
and we used ∆x0∆k ≈ 1.
-
Water Waves 1 H119(r247)
Water waves are surprisingly complex to analyse (see Main: chapter 13), but it’s worthknowing their basic properties (but not how to derive them). In general, the water movesin both longitudinal and transverse directions, with these oscillations in quadrature –making ellipses/circles.In deep water (that is, many wavelengths deep) water has a dispersion relation:
ω2 = gk +σk3
ρ
where we recognise the first is due to gravitational forces, and the second is caused by thesurface tension σ of the water. The two terms are equal when k2 = ρg/σ, whichcorresponds for water (σ = 0.073 N m−1) to a wavelength of λ0 = 17 mm.
Water Waves: Ripples on Deep Water H120(r247)
Figure 62 : Ripples
Ripples have shorter wavelengths than λ0 and are surface tension driven: the dispersionrelation is
ω ≈
√σk3
ρ, vφ ≈
√σk
ρ, vg ≈
3
2
√σk
ρ
so that their phase speed and group speed decrease as wavelength increases. This isa case of ’anomalous’ dispersion. Here, vg > vφ, and the ripples appear to movebackwards through the wavepacket. At 1 mm wavelength, the phase speed is 0.68 m s−1.ripples from a drop video
http://bit.ly/sLKVZN
-
Water Waves: Gravity Waves in Deep Water H121(r247)
Figure 63 : Deep ocean waves
Gravity waves have longer wavelengths than λ0: propagation is controlled by thegravitational force. The dispersion relation is
ω ≈√gk, vφ ≈
√g
k, vg ≈
1
2
√g
k
so that their phase speed and group speed increase as the wavelength increases. Thisis a case of ’normal’ dispersion. In this case, vg = vφ/2, and the individual wavesappear to move forwards through the wavepacket. At 100 m wavelength, the phasespeed is 12.5 m s−1 (about 28mph).Ripples on a lake?Water Wave Dispersion video (speedboat)
Shallow Water Gravity Waves H122(r247)
If the wavelength λ exceeds the water depth h, gravity waves have a dispersionrelationship
ω2 = ghk2(
1− h2k2
3
)and the wave motion is mainly longitudinal.
For very shallow water, we can neglect the second term and ω ≈ k√gh, and we have
approximately non-dispersive waves with phase/group speed√gh.
Water waves approaching the shore slow down as h decreases, and so must increasein amplitude (height) to conserve water. (They often then break).
A tsunami is caused by an earthquake which creates a wave of huge wavelength(∼ 100 km); so the ’shallow water’ approximation holds even in the deep ocean. Fora typical Pacific Ocean depth of h = 4 km, the wavespeed is about 200 m s−1 (450mph). The amplitude in deep ocean is ∼ 1 m, and the period is ∼ 10 minutes, soships hardly notice.
There is very little dispersion, so the wavegroup arrives almost simultaneously atthe shore line even after a long (say 12 hour) journey, where it slows and grows inamplitude — hence causing much damage.
http://bit.ly/tJrU2Vhttp://bit.ly/vXfBFE
-
Guided Waves H123(r247)
It is useful to confine waves to transmit energy or information:
Electromagnetic Waveguides
Optical fibres
Stethoscopes
Note this list list does not include the national grid, or ethernet cables, both of which aretransmission lines and have different characteristics.The boundaries of the guide set conditions which affect the wave’s propagation.
Figure 64 : Waveguide Examples
Simple Model: Guided waves on a membrane H124(r247)
Consider waves on a large two-dimensional elastic membrane in the xy plane), with massρ per unit area, and under tension T per unit length. Its motion obeys
T
ρ∇2Ψ = v2∇2Ψ = ∂
2Ψ
∂t2
So the membrane waves are non-dispersive, have speed v, and obey ω = vk.
k kx y+ k - kx y
x
y
y = 0
y b=
Figure 65 : Two-dimensional waveguide.
Now let us clamp it along two parallel edges at y = 0 and y = b. Consider two travellingwaves with wavevector k on the membrane, moving at equal angles ±θ to x:
ΨA = +A exp i(ωt− kxx− kyy), ΨB = −A exp i(ωt− kxx+ kyy).
These waves have k = (kx,±ky) = (k cos θ,±k sin θ), and the total displacement is
Ψ = A exp i(ωt− kxx) [exp (−ikyy)− exp(ikyy)] = −2iA sin (kyy) exp i(ωt− kxx).
This is a wave travelling in the x direction of wavelength 2π/kx, with amplitudemodulated by a standing wave in the y direction.
-
Dispersion Relation for Guided Waves H125(r247)
You can think of the constituent waves being multiply reflected at the boundaries with aphase change of π at each as they move down the guide. The boundary conditionsrequire that Ψ = 0 at y = 0 and y = b, i.e. that sin kyb = 0. This implies
ky =nπ
b
for integer n. Only discrete values of ky are allowed. To determine the dispersionrelation we substitute Ψ into the wave equation:
v2(k2x + k
2y
)= v2k2 = ω2
But ky is quantised, so we obtain the dispersion relation
ω2 = v2(k2x +
n2π2
b2
). (62)
And the phase velocity of the waves travelling along the guide is
vp =ω
kx=
ω√(ω2
v2− n2π2
b2
) (63)
Dispersion H126(r247)
The guided waves are dispersive. Differentiating equ. (62), we find
2ωdω
dkx= 2v2kx
∴
(ω
kx
)dω
dkx= v2
The product of the phase and group velocities is v2 for this dispersion relation:vpvg = c
2, and
vg =v2
ω
√(ω2
v2− n
2π2
b2
).
which has several branches fixed by the integer n.
-
0 1 2 3 4 5 6 7 8 9 100
2
4
6
8
10
kx
ω=
v√ (k2 x
+n2
π2 /
b2)
Dispersion of waveguide modes: v = 1, b = π
n = 1n = 2n = 3n = 4
Figure 66 : Dispersion Relation in Waveguides
Note the disallowed range of frequencies below ωc = π/b (which is unity in this example).
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
kx
v φ
Waveguide phase and group speeds: v = 1, b = π
vφn = 1vgn = 1
vφ, n = 2vg, n = 2vφ, n = 3vg, n = 3
Figure 67 : Phase and group Speed in Waveguides
Note that vφvg = 1 for this dispersion relation.
-
n = 1 n = 2
x
y
t = 0
t = / 2p w
t = /p w
x = 0y = 0
x = 0y = 0
2 /p k1 2 /p k2
n = 3
k v b12 2 2 2 2= / - /w p k v b2
2 2 2 2 2= / - 4 /w p
k v b -ve22 2 2 2 2= / - 9 / =w p
b
t = /p w no displacement
no displacementt = 0
t = / 2p w
t = 3 / 2p w
1 / a
2 sin(2 / ) sin exp -A y b t xp w a
2 sin( / ) sin( - )A y b t k xp w 1 2 sin(2 / ) sin( - )A y b t k xp w 2
Figure 68 : Modes in a waveguide.
Guided Wave Properties 1 H130(r247)
There is a series of waveguide modes specified by n, with different dispersionrelations and displacement patterns.
The wavelength of the guided wave is λx = 2π/kx, which exceeds λ.
l
lx
wave crest
Figure 69 : Wavelength in a guided wave
Hence the phase velocity, vp is greater than v, and the group velocity vg is less thanv.
The group velocity can be expressed v cos θ where tan θ = ky/kx. This makes sense— the constituent plane waves travel at angle θ to x.
-
Guided Wave Properties 2 H131(r247)
As kx → 0, vp →∞ and vg → 0 (i.e. dω/dkx = 0 at kx = 0). This does not violatespecial relativity, since energy and information are transmitted at vg.
As kx →∞, the behaviour approaches that of an unguided wave, with vp andvg → v.Below a cutoff angular frequency ωc =
nπvb
, k2x becomes negative and propagationis not possible. The lowest cut-off frequency occurs for n = 1; for ω < (πv/b),propagation is not possible.
If a spread of frequencies exists, can excite multiple modes. These have differentspeeds, and the signal will distort (disperse) as it travels. To avoid this problem,ensure only one mode n = 1 is possible by choosing a frequency below the cutofffrequency of mode n = 2:
πv
ω0< b <
2πv
ω0
The guide is then single-moded for the given frequency ω0.
Optical Fibres H132(r247)
Optical fibres are an extremely important example of a waveguide: a cylindrical silicacore (typical diameter 9 µm), is clad by a lower refractive index glass. Usually,infrared waves are guided down the core of the fibre.
The data is sent as a series of pulses, and the data rate which can be achieved over agiven length of fibre is determined by the dispersion of these pulses. The aim is tochoose a wavelength where the dispersion is a minimum (but also need low loss).
core
cladding
Figure 70 : Schematic representation of an optical fibre.
Different propagation modes exist, with different group velocities. Most fibres aredesigned so only one propagating mode exists. There is dispersion (a) from thewaveguide dispersion relation, and (b) due to the refractive index of glass dependingon wavelength. Try to design a fibre so these two effects have opposite signs andtherefore cancel. Current optical fibre systems can carry data at rates of 100 Gbits/sby modulating a single-wavelength laser source operating at 1550 nm. WithWavelength Division Multiplexing (WDM), using e.g. 16 wavelengths, can get 1.6Terabits/s.
-
Evanescent Waves: some notes H133(r247)
We showed that the dispersion relation for guided waves is
k2x = k2 − k2y =
ω2
v2− n
2π2
b2. (64)
Below the lowest cut-off frequency, ωc = nvπ/b with n = 1, we see that k2x becomes
negative, so that kx = ±iα for some real α. Hence
Ψ = A sin kyy exp(iωt) exp(±|kx|x)
This is not a propagating wave — there no phase change with x; all parts of themembrane move up and down in phase together, with the amplitude decayingexponentially with distance along the wave. This is an evanescent wave.
The positive exponential solution is usually excluded if the membrane extends overmany times 1/|kx| — otherwise energy would not be conserved.Since the phase is now independent of position (as in standing waves), there is nonet power flow along the guide. As a consequence, a travelling wave with ω < ωcwhich is incident on the guide will undergo perfect reflection (with a phase shift) anda standing wave will be set up outside the guide.
x
Y
outside guide inside guide
Figure 71 : Travelling wave with ω < ωc incident on a waveguide.
transmitted wave
reflected wave
incident wave 1
2
A B C
x
Figure 72 : Evanescent wave between two discontinuities.
If a system has a second discontinuity beyond which it is again possible to have atravelling wave, some of the incident wave would be transmitted into this region, (C), andthere will only be partial reflection at the boundary AB. There are two evanescent wavesin region B; wave 1 decays as exp(−αx), and wave 2 increases as exp(+αx). Thecombination of these two gives a position dependence of the phase angle for the totaldisturbance in region B and, as a result, a net flow of energy from region A to region C isnow possible.This can be applied to quantum tunnelling.
-
Superposition H135(r247)
The equation of motion for the damped driven oscillator
F (t) = mẍ(t) + bẋ(t) + kx(t)
is linear.
if F1(t) gives displacement x1(t)and F2(t) gives displacement x2(t)then c1F1(t) + c2F2(t) gives displacement c1x1(t) + c2x2(t).
Consider a force which is the sum of two sinusoidal driving forces at different frequencies.We can calculate the individual responses to the two forces (using the response functionR(ω)), and add the results.
F1eiω1t → R(ω1)F1eiω1t
+ +F2e
iω2t → R(ω2)F2eiω2t‖ ‖
F1eiω1t + F2e
iω2t → R(ω1)F1eiω1t +R(ω2)F2eiω2t
This leads to the idea that if we can decompose a force into sinusoids, we can determinethe response to that force in terms of the sum of the responses to each of the sinusoids.
Fourier Series H136(r247)
What is the response of an oscillator to e.g. a square wave driving force? A periodicforce can be represented as a sum of discrete sinuosoidal components. The Fourier seriesof a periodic function f(t), period T , angular frequency ω0 = 2π/T is
f(t) =1
2A0 +
∞∑n=1
[An cos
(2πnt
T
)+Bn sin
(2πnt
T
)](65)
i.e. a superposition of sinusoidal waves with frequencies ω0, 2ω0, 3ω0 . . ..To find the coefficients An, Bn, we note these orthogonality properties:∫ T/2
−T/2 cos(
2πmtT
)cos(
2πntT
)dt = 0 for m 6= n
= T/2 for m = n∫ T/2−T/2 cos
(2πmtT
)sin(
2πntT
)dt = 0 for all m, n
∫ T/2−T/2 sin
(2πmtT
)sin(
2πntT
)dt = 0 for m 6= n
= T/2 for m = n.
-
Fourier Series 2 H137(r247)
Hence, to find the value of a particular coefficient Am or Bm, we multiply the definingequation equ. (65) by cos
(2πmtT
)or sin
(2πmtT
)and integrate from −T/2 to T/2 to give
An =2
T
∫ T/2−T/2
f(t) cos
(2πnt
T
)dt (66)
Bn =2
T
∫ T/2−T/2
f(t) sin
(2πnt
T
)dt (67)
These integrals ’project out’ the coefficient we need — all the other sinusoidalcomponents in f(t) are orthogonal to the component we are interested in, so make nocontribution to the integral.
Fourier Series for a Square Wave H138(r247)
-2
-1
0
1
2
= + + + ...TT/2
Figure 73 : Fourier analysis of a square wave.
Symmetry considerations are often helpful: since the square wave drawn is an oddfunction (f(x) = −f(−x)), An = 0 for all n. And
Bn =4
πnfor n odd
Bn = 0 for n even,
∴ f(t) =4
π
(sin (ω0t) +
1
3sin (3ω0t) +
1
5sin (5ω0t) + . . .
)
-
Square wave driving of a damped oscillator H139(r247)
-2
-1
0
1
2
= + +
= + +
t
R R( )w0 R(3 )w0 R(5 )w0x
t
F
0 1 2 3 4 5 6
|R|
Frequency / w0
Figure 74 : Response function for a damped oscillator with Q = 1.7, and the response of this oscillator to a square wave at frequency ω0
Since |R| tends to zero at high frequencies, first few Fourier components often give agood approximation to the response.
Complex coefficients H140(r247)
Rather than write the series as a sum of sine and cosine terms, it often convenient todescribe the Fourier components with complex coefficients which represent the amplitudeand phase of each component.
f(t) =
∞∑n=−∞
Cnei2πnt/T =
∞∑n=−∞
Cneinω0t. (68)
The coefficients Cn are given by multiplying by e−imω0t and integrating from −T/2 to
T/2,
Cn =1
T
∫ T/2−T/2
f(t)e−inω0tdt.
This results relies on the fact that∫ T/2−T/2 e
inω0te−imω0tdt = 0 for m 6= n= T for m = n.
(69)
We are interested in functions f(t) which are real, and in this case we find thatC−m = C
∗m. This ensures that the imaginary parts of the terms in equ. (68) cancel to
zero when summing over positive and negative n.
-
Example Fourier Series H141(r247)
T-T T/8- /8T t
f t( )
0
A
0
A/4
Cn
0 8-8 n
Figure 75 : Fourier coefficients for a periodic function.
This function repeats with period T = 2π/ω0, and is non-zero for −T/8 < t < T/8. ItsFourier coefficients are
Cn =1
T
∫ T/8−T/8
Ae−inω0t dt =A
T
[e−inω0t
−inω0
]T/8−T/8
=A
πnsin(nπ/4)
The coefficient is zero whenever n is a multiple of 4.
Fourier Transforms H142(r247)
If a function is not-periodic, it is still useful to be able to decompose it into sinusoidalwaves.Taking the limit of a Fourier series as T →∞, the fundamental frequency tends to zero,and the frequency components become infinitesimally close together. The sum overdiscrete frequency components then becomes an integral over a continuous spectrum offrequency components.We write the function as
f(t) =1√2π
∫ ∞−∞
g(ω)eiωtdω (70)
where
g(ω) =1√2π
∫ ∞−∞
f(t)e−i