Post on 06-Jul-2020
Wave Motion: revision Professor Guy Wilkinson
guy.wilkinson@physics.ox.ax.uk Trinity Term 2014
Introduction Two lectures to remind ourselves of what we learned last term Will restrict discussion to the topics on the syllabus Will re-state main relations/facts/principles you should know. Most of these I will not re-derive (no time) – look back to last term’s notes. Essential information contained in these slides, but some points will be augmented by working through on board. Will illustrate with some typical problems from recent prelims papers. (No guarantee my answers are correct! If working through them we find bugs I shall fix the online version of the slides)
These slides & HT material on http://www.physics.ox.ac.uk/users/wilkinsong
2
Normal modes
4
Normal modes - syllabus
According to the course handbook you should know about the following
Let’s remind ourselves of the essentials, before looking at a few past problems
An illustrative example: coupled pendula
)(
)(
xykl
ymgym
xykl
xmgxm
xy
x y
F F
mg mg
ll
k
First thing to do is to write down the equations of motion:
5
Finding the normal modes
6
Recall there are two good ways to solve these problems:
1. The decoupling method
The equations are decoupled by finding the ‘mode (or normal) coordinates’. If the system is symmetric (i.e. equal masses, spring constants, length of pendula...) then a good bet will be
)(2
1
)(2
1
2
1
yxq
yxq
(the factor of 1/√2 is not necessary for solving the problem, and is often omitted. It’s there to make other expressions, i.e. those for the energy, come out right)
In practice then one adds and subtracts the equations of motion.
If the system is not symmetric then the normal modes will in general be more complicated. In that case I advise a second plan of attack...
2. The matrix method
I’ll remind you of this shortly.
Solving with decoupling method So for the coupled pendula we add and subtract the equations of motion:
yxq 1
1
2
11 qq
to obtain
with l
g2
1
where A1,2 & ϕ1,2 are constants set by inital conditions
)cos( 1111 tAq
7
)( xykl
xmgxm )( xyk
l
ymgym
2
2
22 qq m
k
l
g22
2 with
yxq 2
the solutions of which are clearly
)cos( 2222 tAq
Coupled pendula – the normal modes
)cos( 1111 tAql
g2
1yxq 1
yxq 2)cos( 2222 tAqm
k
l
g22
2
First normal mode: centre-of-mass motion
Second normal mode: relative motion
8
This correlated and anti-correlated motion is given by the form of the normal coordinates and is typical for the simple problems we will encounter.
Coupled pendula: the general solution General solution is a sum of the two normal modes
)cos()cos(
)cos()cos(
222111
222111
tAtAy
tAtAx
The constants A1 ϕ1, A2 and ϕ2 are set by the initial conditions
For example
0)0()0(;)0()0( yxayx
gives
0;0; 121 AaA
x
y
t
t
a
a
T=2π/ω1
9
In this case only the 1st normal mode is excited. (For other possibilities look back at HT lecture notes.)
Solving with matrix method
0
0
dt
ddt
d
2
2
2
2
y
x
m
k
l
g
m
km
k
m
k
l
g
)(
)(
xykl
ymgym
xykl
xmgxm
tieY
X
y
x
Re
0
0
2
2
Y
X
m
k
l
g
m
km
k
m
k
l
g
Expecting an oscillatory solution, so let’s try substituting one in, making use of complex notation
We obtain: eigenvector
equation
X & Y are complex constants
10
Solving with matrix method
0A 0]det[ A
02
2
m
k
l
g
m
km
k
m
k
l
g
So here:
11
We have equation of sort
Non-trivial solution requires matrix has no inverse
Eigenvalue equation m
k
m
k
l
g
2
l
g2
1
m
k
l
g22
2
Substitute these back into the eigenvector equn to find the amplitude ratios, i.e. the relationship between X and Y. Find X=Y and X=-Y, as before.
Illustrate these methods with a couple of past exam questions
12
• TT 2008, Q11 • TT 2011, Q8
We have reminded ourselves about:
Let’s apply these techniques to
Trinity Term 2009, Q11
13
14
Trinity Term 2009, Q11
)2(
)2(2
122
211
xxkxm
xxkxm
Equations of motion:
Different masses involved, so recommend matrix method.
tieX
X
x
x
2
1
2
1Re
0
0
22
2
1
2
2
X
X
m
k
m
km
k
m
k
Usual ansatz
yields eigenvector equation
which gives
2
3
2
32
m
k
2
3
2
32
m
kand
with the X1/X2 amplitude ratios obtained by substituting ω2 results back into eigenvector equation
31
1
2
1
X
X
13
1
2
1
X
X
15
Trinity Term 2009, Q11
So general solution is
)cos(13
)cos(31
1
t
At
Ax
)cos()cos(2 tAtAx
Use initial conditions to fix A+,- and Φ+,- . It is the requirement that the masses start at rest that determines Φ+,- = 0, while the initial displacements fix A+,-.
)cos()cos(3
02 tt
xx
)cos(
13
1)cos(
31
1
3
01 tt
xx
16
Trinity Term 2011, Q8
17
Trinity Term 2011, Q8
)(22
)(2
122
211
xxkxm
xxkxm
Different masses involved, so recommend matrix method.
Equations of motion:
tieX
X
x
x
2
1
2
1Re
Usual ansatz
yields eigenvector equation
0
022
2
1
2
2
X
X
m
k
m
km
k
m
k
Solving this gives in the case mk /3 21 2XX
The other solution, with is . This is non-oscillatory ! 021 XX
tmkA
x
tmkAx
)/3(cos2
)/3(cos
2
1
vtBx
vtBx
2
1So have and
with A,B,Φ and v constants to be determined from initial conditions
18
Trinity Term 2011, Q8
tm
krtmk
rx
tm
krtmk
rx
3
20)/3(cos
20
3
20)/3(cos
10
2
1
i.e. anticorrelated oscillations superimposed upon spinning motion
Velocity of mass 2m is zero when 2
3
3
k
mt
Other topics
19
• Reminder about energy • Driving term – TT2012, Q10
Energy of system with normal modes
kxyyxm
k
l
gmyxmU
)(
2
1)(
2
1 2222
)(2
1
)(2
1
21
21
qqy
qqx
This a bit opaque. How does it look in terms of normal coordinates?
Recall (here with normalisation factors included):
and l
g2
1
m
k
l
g22
2
This gives a much more transparent expression, and without cross-terms:
So total energy of system = sum of energies in each excited mode
2
2
2
2
2
2
2
1
2
1
2
12
1
2
1
2
1
2
1qmqmqmqmU
energy in mode 1 energy in mode 2
20
Energy will include kinetic and potential contributions, with each coordinate featuring. So returning to the case of the two coupled pendula:
21
Trinity Term 2012, Q10
22
Equations of motion:
Trinity Term 2012, Q10
2
2
01
2
01 3 yyy
)(23 2
02
2
0
2
012 thyyy
These are inhomogeneous, due to presence of h(t) term. To find the normal modes we must find the complementary functions, i.e. solve homogenous case:
mLT 2/2
0 with
2
2
01
2
01 3 yyy
2
2
0
2
012 3 yyy
Symmetric system, so this is easily done by the decoupling method
with and with pp 2
02 qq 2
0421 yyp 21 yyq
02 p 02 q
23
Trinity Term 2012, Q10
Finding steady state amplitude means finding the particular integral
)Re()]4)(2/[(2 22
0
22
00
4
01
tiehy
)Re()]4)(2/[()3(2 22
0
22
0
22
00
4
02
tiehy
when and 4/01 hy 02 )4/3( hy 0
when 03
)Re(2 01
tiehy 02 y
2h0
h0
L
2L L
and
Waves
24
25
Waves - syllabus
According to the course handbook you should know about the following
Let’s remind ourselves of the essentials, before looking at a few past problems
26
Waves - syllabus
Focus first on the below topics
and illustrate by looking at Long Vacation 2011, Q9
Waves on a stretched string Consider a segment of string of linear density ρ stretched under tension T
x
T
y
xx
T
,
small
27
2
2
2
2
t
y
Tx
y
Key point – means that no net horizontal force at first order, only vertical force, and we can make approximations such as sinδθ ≈ δθ
For full derivation see HT lecture notes or text book, e.g. French
2
2
22
2 1
t
y
cx
y
Tc
This is the wave equation, which (anticipating solution) we can write
with
d’Alembert solution of wave equation
28
2
2
22
2 1
t
y
cx
y
ctxv
ctxu
y is a function of x and t. Define new variables so that y is now a function of u and v
02
vu
y)()(),( vgufvuy With chain rule
we can show
)()(),( ctxgctxftxy
So general solution of wave equation is
Here f and g are any functions of (x-ct) and (x+ct), determined by initial conditions. A common scenario is that they are sinusoidal.
Note that describes forward-going wave and )( ctxf
)( ctxg describes a backwards-going one !
with c the (phase) velocity of the wave
Sinusoidal waves
)()(),( ctxgctxftxy
A very common functional dependence for f and g...
...is sinusoidal. In this case it is usual to write:
)cos()cos(),( tkxBtkxAtxy
or Asin(kx-ωt) ... etc (choice doesn’t matter, unless we are comparing one wave with another and then relative phases become important)
with k and ω (and A and B) constants
• speed of wave
• frequency
where ω is angular frequency
• wavelength where k is the wave-number (or wave-vector if also used to indicate direction of wave)
kc /
2//1 Tf
k/2
29
Notation choices
)cos(),( tkxAtxy Sinusoidal solution (writing here, for compactness, only the forward-going solution)
Using the relationships between k,ω, λ & c this can be expressed in many forms
)](cos[),( ctxkAtxy
Also note that sometimes it is convenient to write
)cos(),( kxtAtxy
A very frequent approach is to use complex notation (we already made use of this when analysing normal modes, and you will have seen it in circuit analysis)
)](exp[Re),( tkxiAtxy
or if it’s important to pick out sine function. Note that often the ‘Re’ or ‘Im’ is implicit, and it gets omitted in discussion.
)](exp[Im),( tkxiAtxy
30
Energy and impedance for travelling wave on string
31
Energy stored in a mechanical wave
Integrate kinetic energy and potential energy densities over an integer number of wavelengths to show they contribute equally and give
2
2
1
d
ddensity KE
t
y
x
K
2
2
1
d
ddensity PE
x
yT
x
U
nA 22
2
1Total energy in n wavelengths =
t
yx
yT
v
FZ
y
y
so with
)sin(),( tkxAtxy )(
T
c
TTkZ
The characteristic impedance Z is defined as the applied driving force acting in the y-direction divided by the velocity of the string in the y-direction
Characteristic impedance
(Note sign on driving force which ensures Z positive for forward wave!)
Power flow = kTA 2
2
1
Long Vacation 2011, Q9
32
Waves at boundaries
33
This question a good opportunity to remind ourselves what happens to waves at boundaries, so let’s answer it in a more general way than is being asked by first allowing strings to be different (e.g. different densities)
x=0
ρ1
ρ2
x→
M
So we must allow for reflected and transmitted waves. To satisfy boundary conditions all waves must have same frequencies, but their velocity and wave-vector will depend on which string they are on
)](exp[Re 1xktiA
)](exp['Re 1xktiA )](exp[''Re 2xktiA
Incident
Reflected Transmitted
Complex notation convenient for these problems. Pay attention to signs!
34
Long Vacation 2011, Q9 Write down boundary conditions
M
x=0 θ1 θ2
T T
1. Equation of motion for mass
021
0
2
2or1
2
)sinsin(
x
x
TTt
yM
x
y
11
11
tan
tansin
x
y
22
22
tan
tansin
0
12
0
2
2or1
2
xxx
y
x
yT
t
yM
2. Strings stay stuck to mass
),0(),0( 21 tyty
Apply to expressions for incident, reflected and transmitted waves
')('' 11
2
2 TAikTAikmTikA
''' AAA
(1)→
(2)→
35
Long Vacation 2011, Q9
Evaluate r ≡ A’/A and t ≡ A’’/A
mikkT
Tk
A
At
2
21
1
)(
2''
mikkT
imkkT
A
Ar
2
21
2
21
)(
)('
Specialising to case where strings are identical and so k1=k2 =k gives
ip
ip
mikT
imr
12 2
2
ipmikT
kTt
1
1
2
22 kT
mp
2
2with and
(Aside: one is often asked about transmitted and reflected energy.
So make sure you remember wave power ) 2)Amplitude(2
1kT
Phase of r = )(tan2
1 p • m low, phase is –π/2
• m high, phase is -π
but this case is slightly artificial, as no reflection in case m=0!
36
Waves - syllabus Now let’s consider
with reference to questions: TT 2009, Q6; TT 2010, Q6; TT 2012, Q8
Wave packets and group velocity
37
A single wave cannot transmit information. To do that we need a wave packet. Any wave packet can be formed from a sum of single waves. Simplest example:
Sum together two waves which differ by 2δω and 2δk in angular frequency and wave-number, respectively:
txkkAy
txkkAy
)()(sin
)()(sin
2
1
)sin()cos(221 tkxtxkAyyy
y1
y2
y
Describes envelope – so envelope moves with velocity and indeed
)sin()cos(2 tkxtxkA
Group velocity while phase velocity k
vgd
d
kvp
k
Dispersion
38
Dispersion is when there is not a linear relationship between ω and k.
Two consequences:
1. Phase velocity, ω/k, depends on ω and k. e.g. Light in medium m has refractive index n and velocity cm, where
That’s why a prism splits light.
2. Group velocity ≠ phase velocity
nccm /
kvg
d
dIf and it follows that
kvp
dk
dvkvv
p
pg
d
dvvv
p
pg
d
dn
nn
cvg 1and and
(more important that you can derive these, rather than learn them!)
Dispersion question: TT 2009, Q6
39
(1)
4222222 4 DBAc 2
2
2 )2(
d
dv
c
DA
and (1) → so
kk d
d
d
d
d
d
Now and 2
2
d
d
kk
hence result
Standing waves: TT 2010, Q6
40
0x Lx
tkxA
tkxAtkxAtxy
cossin2
)sin()sin(),(
Sum right- and left-going travelling waves of same amplitude and frequency:
Standing waves: every point on the string moves with a certain time dependence (cosωt), but the amplitude depends on its position along the string (sinkx)
factorised spatial & temporal dependence
...
41
Standing waves: TT 2010, Q6 ...
0x Lx
Ends of string being fixed determine boundary conditions: 0),(),0( tLyty
angular frequency of normal modes
tkxAtxy cossin2),( nLkn L
Tnn
and
and putting in numbers, with n=1, gives T=71 N
Wave equation revisited – solving by separation of variables
42
2
2
22
2 1
t
y
cx
y
We have already solved the wave equation using the d’Alembert approach
Can also be solved by looking for solutions which have the ‘separated’ form
)()(),( tTxXtxy
i.e. that factorise into functions that are separate functions of x and t. This is just the situation that applies to standing waves !
T
T
cX
X 2
1
set each side equal to some separation constant –k2
kxBkxAxX sincos)( cktEcktDtT sincos)(
with A,B,D and E constants defined by initial conditions
This yields and
Standing waves: TT 2012 Q8
43
44
Standing waves: TT 2012 Q8 Let’s rearrange question so we can discuss the relevant topics more clearly
Going from general solution to specific solution through applying initial conditions and monitoring subsequent evolution with time
Energy of system
45
Standing waves: TT 2012 Q8 Know from separation of variables that a solution to wave equation is
)sincos)(sincos(),( kctDkctCkxBkxAtxy
and we also have four boundary conditions:
1. String initially at rest, i.e. for all x
2. y(0,t)=0
3. y(L,t)=0 where n any integer. This is the eigenvalue eqn. and discretises k. Each value of n corresponds to a normal mode.
4. Form of initial displacement involves normal modes 1 and 3. From these we fix coefficient of mode 1 to be 1 and 3 to be -2/3, and all others 0.
0/ ty 0D
0 A
nkL
L
ct
L
x
L
ct
L
xtxy
3cos
3sin
3
2cossin),( hence
This first returns to initial displacement when cLt /2
Energy of standing waves
46
Normal mode n for our string, with given boundary conditions:
L
ctn
L
xnFtxy nn
cossin),(
Calculate kinetic energy, Kn, and potential energy, Un, for each mode
xt
yK
L
nn d
2
1
0
2
x
x
yTU
L
nn d
2
1
0
2
L
Tnn
4
22
nnnnn
LFUKE
Evaluate and sum with
What about the case when several normal modes are excited (as in question)?
1n
nEENote that all cross-terms have vanished due to the orthogonality of sines
mnxL
xm
L
xnL
withdsinsin0
i.e. all these terms are zero
So total energy is weighted sum of all the excited normal modes
47
Standing waves: TT 2012 Q8
xx
tyTtU
L
d)0(
2
1)0(
0
2
Initial energy of string is all in PE
L
TtU
4
5)0(
2
1n
nEEThis result makes sense as it equals total energy of system as calculated from
4
22
nnn
LFE
and
Kinetic energy of each standing wave, i.e. kinetic energy of each mode
xt
yK
L
nn d
2
1
0
2
L
ctncn
LAK nn
222sin)(
4
where An is the amplitude coefficient for mode n (here A1=1 and A3=-2/3 and others =0)