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Transcript of Dalgalar II
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1
University of Cape Town
Department of PhysicsPHY2014F
Vibrations and Waves
Andy BufflerDepartment of Physics
University of Cape [email protected]
Part 2
Coupled oscillators
Normal modes of continuous systemsThe wave equation
Fourier analysis
covering (more or less)
French
Chapters 2, 5 & 6
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Problem-solving and homework
Each week you will be given a take-home problem set to
complete and hand in for marks ...
In addition to this, you need to work through the followingproblems inFrench, in you own time, at home. You will not
be
asked to hand these in for marks. Get help from you friends, the
course tutor, lecturer, ... Do not take shortcuts.Mastering these problems is a fundamental aspect
of this course.
The problems associated with Part 2
are:
2-2, 2-3, 2-4, 2-5, 2-6, 5-2, 5-8, 5-9, 6-1, 6-2, 6-6, 6-7, 6-10,
6-11, 6-14
You might find these tougher: 5-4, 5-5, 5-6, 5-7
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The superposition of periodic motions
Two superimposed vibrations of equal frequency
1 1 0 1cos( )x A t = +
2 2 0 2cos( )x A t = +
combination can be written as
0cos( )x A t = +
Using complex numbers:0 1( )
1 1
j tz A e
+=0 2( )
2 2
j tz A e
+=1 2z z z= +
{ }0 1 2 1( ) ( )1 2j t jz e A A e + = +
A A2
A1
0 1t +
2 1
2 1 =
Phase differenceThen
( )2 2 1sin sinA A = and
2 2 2
1 2 1 2 2 12 cos( )A A A A A = + +
French
page 20
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If we add two sinusoids of slightly different frequency and we observe beats
x1x2
x1
+x
2
t
t
Superposed vibrations of slightly different frequency: Beats
2cos t
1 2
cos cost t +1 2cos
2
t
1 2
2beatT
=
1cos t
1 2 1 21 2cos cos 2cos cos
2 2t t t t
+ + =
1
2
French
page 22
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Combination of two vibrations at right angles
1 1 1cos( )x A t = +2 2 2cos( )y A t = +
???
Write and1 0cos( )x A t=
Consider case where frequencies are equal and let initialphase difference be
2 0cos( )y A t = +
Case 1 : 0 = 1 0cos( )x A t=
2 0cos( )y A t=
2
1
Ay x
A= Rectilinear motion
2 = 1 0cos( )x A t=
2 0 2 0cos( 2) sin( )y A t A t = + =
Case 2 :
2 2
2 2
1 2
1x yA A
+ = Elliptical path inclockwise direction
French
page 29
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Case 3 : = 1 0cos( )x A t= 2
1
Ay xA
=
Combination of two vibrations at right angles 2
2 0 2 0cos( ) cos( )y A t A t = + =
3 2 = 1 0cos( )x A t=2 0 2 0cos( 3 2) sin( )y A t A t = + = +
Case 4 :
2 2
2 2
1 21
x y
A A + = Elliptical path inanticlockwise direction
4 = 1 0cos( )x A t=
2 0cos( 4)y A t = +
Case 5 :
Harder to see use a graphical approach
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Superposition of simple
harmonic vibrations at
right angles with an initialphase difference of 4
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Superposition of two perpendicular simple
harmonic motions of the same frequency forvarious initial phase differences.
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Abbreviated construction for the superposition of
vibrations at right angles
see French page 34.
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Perpendicular motions with different frequencies: Lissajous figures
See French page 35.
Lissajous figures for
with various
initial phase differences.
2 12 =
= 0 4 3 42
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2 1:
1:1
1:2
2:3
1:3
3:4
5:6
4:5
3:5
= 0 4 3 42
Lissajous figures
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Coupled oscillators
When we observe two weakly coupled identical oscillators
A and B, we see:
t
t
xA
xB
these functions arise mathematically from the addition of twoSHMs
of similar frequencies
so what are
these two SHMs?
These two modes are known as normal modes which are states
of the system in which all parts of the system oscillate with SHM
either in phase or in antiphase.
French
page 121
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t
t
A B
xA
xB
Coupled oscillators
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The double mass-spring oscillator
m
Axk
m
Bx k
Individually we know that andA Amx kx= B Bmx kx=
For both oscillators:0
k
m
=
Now add a weak coupling force:
m
Axk
m
Bx kck
For mass A: ( )A A c B Amx kx k x x= +
or where ,2 20 ( )A A B Ax x x x= + 2
0k
m = 2 ck
m =
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2 2
0 ( )A A B Ax x x x= +
The double mass-spring oscillator 2
For mass A:
For mass B:2 2
0( )
B B B Ax x x x=
two coupled differential equations how to solve ?Adding them:
22
02( ) ( )
A B A B
dx x x x
dt+ = +
Subtract B from A:2
2 2
02( ) ( ) 2 ( )
A B A B A Bd x x x x x xdt
=
Define two new variables:1 A Bq x x= +
2 A Bq x x=
Then and2
21 0 12
d qqdt =
2
2 22 0 22 ( 2 )d q
qdt = +
called normalcoordinates
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The double mass-spring oscillator 3
The two equations are now decoupled
221
12 s
d qq
dt
=
Write 222
22 f
d qq
dt=
2 2
0s =
2 2 2
02
f = +
s
= slow
f
= fast
which have the solutions:
1 1cos( )
sq C t = +
2 2cos( )fq D t = +
Since and 2 A Bq x x=
We can write and
1 A Bq x x= +
( )
1
1 22A
x q q= +( )
1
1 22B
x q q=
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The double mass-spring oscillator 4
1 2
cos( ) cos( )2 2A s f
C Dx t t = + + +
Then1 2cos( ) cos( )
2 2B s f
C Dx t t = + +
SoxA
andxB
have been expressed as the sum and difference of
two SHMs
as expected from observation.
C,D, 1
and 2
may be determined from the initial conditions.
whenxA
=xB
,then q2
= 0
there is no contribution from the
fast mode and the two masses move in phase
the coupling spring
does not change length and has no effect on the motion 0s =
whenxA
= xB
,then q1
= 0
there is no contribution from the
slow mode
the coupling spring gives an extra force
each mass
experiences a force giving( )2c
k k x +
2 2
02= +
2 2 cf
k k
m
+
=
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The double mass-spring oscillator 5
symmetric mode antisymmetric
mode mixed mode
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The double mass-spring oscillator 6
We now have a system with two natural frequencies, and
experimentally find two resonances.
Frequency
Amplitude
F h
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Pitch and bounce oscillator
d
Axk
m
Bx
L
k
Two normal modes (by inspection):
Bouncing
A Bx x=
Restoring force = 2kx2
2 2
d x
m kxdt = Pitching
A B
x x= Ax
Bx Centre of mass stationary
2
bounce
2k
m =
I =
21 12 12
( )kd d mL = 2
26kdmL
=2
2
pitch 26k dm L
=
French
page 127
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N= 2
( )1 0 01
2 sin2 2 1
= =
+
( )2 0 022 sin 3
2 2 1 = =+
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N
= 4
N
= 3
French
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N-coupled oscillators
fixed fixedTension T
l
1 2 3 p1 p p+1 N
consider transverse displacements that are small.
Each bead has mass m
1 2 3 p1 p p+1 N
1p
p
Transverse force on pth particle:1sin sinp p pF T T = +
1 1p p p py y y yT T
l l
+ = +
for small
y
French
page 136
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N-coupled oscillators 2
2
1 12p p p p p
p d y y y y yF m T Tdt l l
+
= = +
( )
2
2 2
0 0 1 12 2 0p
p p p
d y
y y ydt + + =
where ,20
T
ml = p
= 1, 2
N
a set ofN
coupled differential equations.
Normal mode solutions: sinp py A t=
Substitute to obtainN
simultaneous equations
( ) ( )2 2 20 0 1 12 0p p pA A A + + =
or2 21 1 0
2
0
2p p
p
A AA
+ +=
i 3
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N-coupled oscillators 3
From observation of physical systems we expect sinusoidalshape functions
of the form sinpA C p=
Substitute into
2 2
1 1 02
0
2p p
p
A A
A
+
+=
And apply boundary conditions and0 0A = 1 0NA + =
and
n
= 1, 2, 3,
N
(modes)
( )02 sin 2 1nn
N
= +
find that1
n
N
=
+
There areN modes: sin sin sin1
pn pn n n n
pny A t C t
N
= =
+
N l d ill t 4
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02
n
0 1 2 3 N+1 n
( )02 sin 2 1nn
N
= +
N-coupled oscillators 4
For smallN:
N l d ill t 5
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Forn
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N
coupled oscillators haveN
normal modes and henceN
resonances
response
N-coupled oscillators 6
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Continuous systems
Continuous systems
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Continuous systems
Consider a string stretched between two rigid supports
x = 0 x =L
tension T
String has mass m
and mass per unit length m L=
Suppose that the string is disturbed in some way:
y
x
The displacementy
is a function ofx
and t : ( , )y x t
TNormal modes of a stretched string
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T
T
+
y
x xx x+
Consider the forces on a
small length of string
Restrict to small amplitude disturbances then is small andcos 1 = sin tan yx
= = =
The tension T
is uniform throughout the string.
Net horizontal force is zero: cos( ) cos 0T T + =
Vertical force: sin( ) sinF T T = +
Thenx x x
y yF T Tx x
+ =
g
French
page 162
Normal modes of a stretched string 2
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x x x
y yF T T
x x+
=
Use( ) ( )dg g x x g x
dx x
+ =
Then2
2
yF T x
x=
or ( )
2 2
2 2
y y
x T xt x
=
giving
2 2
2 2
y y
x T t
=
Check: has the
dimensions
T21 v
Then is the speed at which awave propagates along
the string see later
v T =
Write2 2
2 2 2
1y y
x v t
=
One dimensional wave equation
Normal modes of a stretched string 2
: mass per unit length
Normal modes of a stretched string 3
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Look the standing wave (normal mode) solutions
Normal mode: all parts of the system move in SHM at the
same frequency
Write: ( , ) ( )cosy x t f x t=( )f x is the shape
function
substitute into wave equation
2 2
2 2( , ) ( ) cosy x t d f x tx dx
=
( )
22
2
( , )
( ) cos
y x t
f x tt
=
22
2 2
( ) 1cos ( )cos
d f xt f x t
dx v =
which must be true for all t
then
2 2
2 2 ( )d f
f xdx v
=
No a odes o a st etc ed st g 3
Normal modes of a stretched string 4
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2 2
2 2 ( )
d f
f xdx v
=
which has the same form as the eq. of SHM:2
2
02
d xx
dt=
has general solution:0sin( )x A t = +
Thus we must have: ( ) sinf x A x
v
= +
Apply boundary conditions: y
= 0 at x = 0 and x
=L
(0) 0f = and ( ) 0f L =
x
= 0,f
=0 : 0 sin 0A
v
= +
i.e.
0 sin Lv
=
0 =
x = L,f =0 : L nv
=i.e. n = 1,2,3,
g
Normal modes of a stretched string 5
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n
n v
L
=Write n = 1,2,3,( ) sin sinn n
x n v n xf x A A
v L L
= =
Therefore
shape function, or eigenfunction
x
= 0
x
=L
n = 1
n = 3n
= 2
n = 5n
= 4
( )1( ) sinf x A x L=( )2( ) sin 2f x A x L=
( )3( ) sin 3f x A x L=
( )4( ) sin 4f x A x L=
( )5( ) sin 5f x A x L=
1
v L =
3 3 v L =
2 2 v L =
5 5 v L =
4 4 v L =
g
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Normal modes of a stretched string
n
= 1
n
= 2
n
= 3
n = 4
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Normal modes of a stretched string 6
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Full solution for our standing waves:
( , ) sin cosn nn x
y x t A tL
=
n
n v
L
=
The mode frequencies are evenly spaced:1n n =
n
0 1 2 3 n
3
2
1
(recall the beaded string)
This continuum approach breaks
down as the wavelength approaches
atomic dimensions
also if there is
any stiffness in the spring whichadds an additional restoring forcewhich is more pronounced in the
high frequency modes.
Normal modes of a stretched string 7
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All motions of the system can be made up from thesuperposition of normal modes
1
( , ) sin cos( )n n nn
n xy x t A t
L
=
= +
with n
n v
L
=
Note that the phase angle is back since the modes may
not be in phase with each other.
Whispering galleries
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Whispering galleries
best example is the inside
dome of St. Pauls cathedral.
If you whisper just inside the
dome, then an observer close to
you can hear the whisper
coming from the oppositedirection
it has travelled
right round the inside of the
dome.
Longitudinal vibrations of a rodFrench
page 170
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x x x+
x x + + + x +
1F 2F
section of massive rod
section is displaced and stretched
by an unbalanced force
Average stress =x
Average strain = Y
x
Y: Youngs modulus
stress at = (stress atx) +x x+ (stress)
xx
page 170
Longitudinal vibrations of a rod 2
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If the cross sectional area of the rod is
1F Yx
=
2
2 2F Y Y x
x x
= +
and
2
1 2 2F F Y x ma
x
= =
2 2
2 2Y x xx t
=
or2 2
2 2
x Y t
=
2 2
2 2 2
1
x v t
=
Yv =
Longitudinal vibrations of a rod 3
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Apply boundary conditions: one end fixed and the other free
x =L :x
= 0 : i.e. 0 =
n
= 1,2,3,
( , ) ( )cosx t f x t =Look for solutions of the type:
(0, ) 0t =
0F Yx
= =
( ) sinf x A xv
= +
where
then cos 0L
v
=
or ( )12L nv
=
( ) ( )1 12 2n
n v n Y
L L
= =The natural angular frequencies
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x = 0 x =Ln
= 1
n = 3
n
= 2
n
= 5
n
= 4
12
Y
L
=
232
YL
=
3
5
2
Y
L
=
Normal modes for different boundary conditions
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45n
= 1 n
= 2 n
= 3
Simplysupported
Clampedone end
Free bothends
Clampedboth ends
The elasticity of a gasl
French
page 176
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A , p Bulk modulus:dp
K VdV=
Kinetic theory of gases: Pressure
2 21
rms rms3 3
m
p v vAl= =If then
21rms2k
E mv=2
3
kEpA l
=
Now move piston so as to compress the gas work done on
gas:
kW pA l E = =
Then ( )2
2 2 2 5( )
3 3 3 3
kk
E l l lp E pA l p p
A l A l A l l
= = =
adiabatic
5
3
pK V p
V
= =
giving
and1.667K p
v = =
Sound waves in pipesFrenchpage 174
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A sound wave consists of a series of compressions and
rarefactions of the supporting medium (gas, liquid, solid)
In this wave individual molecules move longitudinally with
SHM. Thus a pressure maximum represents regions in which themolecules have approached from both sides, receding from the
pressure minima.
wave propagation
p g
Longitudinal wave on a spring
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Standing sound waves in pipes
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Pressurep
p0
Flow
velocity u
0
x
x
x
x
0 :t=
2 :t T=p
u
Standing sound waves in pipes 2
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Consider a sound wave in a pipe. At the closed end the flow
velocity is zero (velocity node, pressure antinode).At the open end the gas is in contact with the atmosphere,
i.e.p
=p0
(pressure node and velocity antinode).
p
u
p0
0
pressure node
velocity antinode
pressure antinode
velocity node
Open end Closed end
Standing sound waves in pipes 3
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Pipe closed at
both ends
Pipe open at
both ends
Pipe open at
one end
n n vL =
2 2
n nvL
f
= =
2
nvf
L=
( ) ( )2 1 2 1
4 4
n n vL
f
= =
( )2 1
4
n vf
L
=
( )2 12
nn v
L =
Sound
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Audible sound is usually a longitudinal compression wave in air towhich the eardrum responds.
Velocity of sound (at NTP) ~ 330 m s-1
By considering the transport of energy by a compression wave,
can show that 2 2 22 mP f Avs =
whereA
is cross sectional area of air column andsm
is
maximum displacement of air particle in longitudinal wave
Then intensity = 2 2 22 mP
f vsA
= unit: W m-2
Sound 2
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The human ear detects sound from ~10-12
W m-2
to ~1 W m-2
use a logarithmic scale forI :
10
0
10logI
I
=
decibels
whereI0
= reference intensity
= 10-12
W m-2
Intensity level or loudness:
Musical sounds
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Waveforms from real musical instruments are complex, and maycontain multiple harmonics, different phases, vibrato, ...
Pitch is the characteristic of a sound which allows sounds to be
ordered on a scale from high to low (!?). For a pure tone, pitch
is
determined mainly by the frequency, although sound level may
also change the pitch. Pitch is a subjective sensation and is a
subject in psychoacoustics.
The basic unit in most musical scales is the octave. Notes judged
an octave apart have frequencies nearly (not exactly) in the ratio2:1. Western music normally divides the octave into 12 intervals
called semitones ... which are given note names (A through G withsharps and flats) and designated on musical scales.
Musical sounds ...2
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Timbre is used to denote tone quality or tone colour of asound and may be understood as that attribute of auditorysensation whereby a listener can judge that two sounds are
dissimilar using any criteria other than pitch, loudness or
duration. Timbre depends primarily on the spectrum of the
stimulus, but also on the waveform, sound pressure and
temporal characteristics of the stimulus.
One subjective rating scale for timbre (von Bismarck, 1974)
dull
compact
full
colourful
sharp
scattered
empty
colourless
Two dimensional systems Frenchpage 181
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the membrane has mass per unit area ,
and a surface tension S
which gives a force Sl
perpendicular to a length l in the surface
y
x
y
x
The forces on the shaded
portion are
Sx
Sx
SySy
Consider an elastic
membrane clamped
at its edges
Two dimensional systems 2
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If the membrane is displaced
from thez
= 0 plane then a
cross section through the
shaded area shows:
+
z
x xx x+
Sy
Sy
looks exactly like the case of the stretched string.2
2zS y xx The transverse force on the element will be
And if we looked at a cross section perpendicular to this
the transverse force will be2
2
zS x y
y
Two dimensional systems 3
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The mass of the element is .x y
Thus2 2 2
2 2 2
z z zS y x S x y x y
x y t
+ =
or2 2 2
2 2 2
z z z
x y S t
+ =
a two dimensional wave equation
with the wave velocity Sv
=
L k f l d l i f h f
Two dimensional systems 4
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( , , ) ( ) ( )cosz x y t f x g y t=2 2
2 2( )cos
z d fg y t
x dx
=
( )2
2
2( ) ( ) cos
zf x g y t
t =
Look for normal mode solutions of the form:
2 2
2 2( )cos
z d gf x t
y dy
=
2 2
2 2( )cos ( )cosd f d g
g y t f x tdx dy + =2
2( ) ( )cosf x g y t
v
2 2 2
2 2 2
1 1d f d g
f dx g dy v
+ = i.e.
In a similar fashion to the 1D case, find
1
1( ) sinnx
n xf x A
L
=
and2
2( ) sinny
n yg y B
L
=
Two dimensional systems 5
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then1 2 1 2
1 2,( , , ) sin sin cosn n n n
x y
n x n yz x y t C tL L
=
1 2
22
1 2,n n
x y
n v n v
L L
= +
where the normal mode frequencies are
e.g. for a membrane having sides 1.05L and 0.95L
then1 2
2 21 2
,1.05 0.95
n n
n nv
L
= +
Normal modes of a rectangular membrane
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up
down1,1
2,1 2,2
3,1 3,2
Normal modes of a circular membrane
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1,0 2,0 3,0
1,1 2,1 2,2
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Modes of vibration of a 38 cm cymbal. The first 6 modes
resemble those of a flat plate ... but after that the resonances
tend to be combinations of two or more modes.
Normal modes of a circular drum
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Chladni plates
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Soap films
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Holographic interferograms
of the top and bottom plates
of a violin at several resonances.
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Holographic interferograms
of a classical guitar top
plate at several resonances.
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Holographic interferograms
showing the vibrations of a 0.3 mm
thick trombone driven acoustically at 240 and 630 Hz.
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Time-average hologram interferograms
of inextensional
modes in a C5
handbell
Normal modes of a square membrane
v2n area per point =
2v
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one point per
normal mode
1 2
2 2
1 2,
2 2n n
n v n vf
L L
= +
12
v
L
22
v
L
32
v
L
4 2
v
L
0
12
v
L 2 2
v
L 3 2
v
L 4 2
v
L 5 2
v
L00 1 2 3 4 5
0
1
2
3
4
4,3f
1n
Normal modes having the same frequency are said to be degenerate
2L
Normal modes of a square membrane for large n1 and n2
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1
n
2ndf
f
area =
2
2
v
L
area per mode =
14(2 )f df
Number of modes with
frequencies betweenf
and (f+ df) =
2
14
2(2 )
Lf df
v
2
2
2 L f df
v
=
Three dimensional systems
Consider some quantity which depends on x y z and t
French
page 188
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Consider some quantity
which depends onx
, y , z
and t
,
e.g. the density of air in a room.
In three dimensions:2 2 2 2
2 2 2 2 2
1
x y z v t
+ + =
which can be written:2
2
2 2
1
v t
=
The solutions for a rectangular enclosure:
1 2 3
22 2
31 2, ,n n n
x y z
n vn v n vL L L
= + +
and for a cube: 1 2 32 2 2
, , 1 2 3n n n
v
n n nL
= + +
How many modes are there with frequencies in the range
Three dimensional systems 2
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f and (f+ df) ?Set up an imaginary cubic lattice with spacing 2v L2n
1n
3n
df
f
and consider positive frequencies only.
Volume of shell =21
8(4 )f df
3
2vL
Volume per mode =
Number of modes with
frequencies betweenf
and (f+ df) =
3
218
2(4 )
Lf df
v
3 2
34 L f dfv=
Number of modes with 24 V f df =
Three dimensional systems 3
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frequencies betweenf and (f+ df) 3v=
holds for any volume V
provided its dimensions are much
greater than the wavelengths involved.
need to multiply by a factor of 2 when dealing withelectromagnetic radiation (2 polarization states)
Ultraviolet catastrophe for blackbody radiation
Equipartition
theorem: in thermal equilibrium each mode has an
average energy in each of its two energy stores
Hence, energy density of radiation in a cavity:
12 Bk T
( )2
123
42 2
V f df df kT
c
=
or
2
3
8 fkTc
= f
experiment
!?
Planck was able to show, effectively by
assuming that energy was emitted an
absorbed in quanta of energy hf that the
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absorbed in quanta of energy hf
, that the
average energy of a cavity mode was not kT
but1hf kT
hf
e
where Plancks constant h
= 6.67
10-34
J K-1
Then
2
3
8
1hf kTf df hf
df c e
=
which agrees extremely well with experiment.
energy
density
no. of
modes inrangef
tof
+df
average
energy permode
Plancks law
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Introduction to Fourier methods
We return to our claim that any physically observed shape
French
page 189
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y p y y p
function of a stretched string can be made up from normal
mode shape functions.
x
( )f x
i.e.1
( ) sinnn
n xf x B
L
=
=
a surprising claim
?
first find
n
multiply both sides by
and integrate over the range x
= 0 to x = L
1
sin
n x
L
1 1
10 0( )sin sin sin
L L
n
n
n x n x n xf x dx B dxL L L
=
=
1 1( )sin sin sin
L L
n
n x n x n xf x dx B dx
=
Fourier methods 2
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10 0 nL L L=
If the functions are well behaved, then we can re-order things:
1 1
10 0
( )sin sin sin
L L
n
n
n x n x n xf x dx B dxL L L
= =
[n1
is a particular integer and n
can have any value between 1 and .]
Integral on rhs:
( ) ( )1 110 0
1sin sin cos cos2
L L
n n x n n xn x n x dx dxL L L L
+ =
Fourier methods 3
Both (n1 + n) and (n1 n) are
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integers, so the functions
on the intervalx
= 0 toL
must look like from which it is evident that
( )1cos
n n x
L
( )1
0
cos 0
L n n xdx
L
=
Except for the special case when n1
and n are equal then ( )1
cos 1n n x
L
=
and
( )1
0
cos
L n n xdx L
L
=
Fourier methods 4
Thus all the terms in the summation are zero, except for the
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single case when n1
= n i.e.
1 2n
LB=
( ) ( )1
1 11
0 0
1( )sin cos cos
2
L L
n
n n x n n xn xf x dx B dx
L L L
+ =
1
1
0
2 ( )sin
L
n n xB f x dxL L
=
i.e.
We have found the value of the coefficient for some
particular value ofn1
the same recipe must work for any
value, so we can write:
0
2( )sin
L
n
n xB f x dx
L L
=
Fourier methods 5
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The important property we have used is that the functions
and
are orthogonal over the intervalx
= 0 tox
=L.
i.e.
1sinn x
L
sinn x
L
1
0
sin sin
Ln x n x
dxL L
=
10 if n n
1if2
L n n=
Read French pages 195-6
Fourier methods 6
The most general case (where there can be nodal or
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g (
antinodal
boundary conditions atx
= 0 andx
=L) is
0
1( ) cos sin2
n n
n
A n x n xf x A BL L
=
= + +
where
0
2
( )sin
L
n
n x
B f x dxL L
=
0
2
( )cos
L
n
n x
A f x dxL L
=
Fourier methods 7
One of the most commonly encountered uses of Fourier methods is
the representation of periodic functions of time in terms of sine
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and cosine functions
Put2
T
=
This is the lowest frequency in
clearly there are
higher frequencies
by the same method as before, write
( )f t
0
1
2 2( ) cos sin
2n n
n
A nt nt f t A B
T T
=
= + +
0
1cos sin2
n n
n
AA n t B n t
== + +
where and ( )0
2( )sin
T
n f t n t dt
T
= ( )0
2( )cos
T
nA f t n t dt
T
=
T( )f t
t
Waveforms of ...
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a flute
a clarinet
an oboe
a saxophone
Fast Fourier transform experiments, 10 March 2008
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Fast Fourier transform experiments, 10 March 2008
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Fast Fourier transform experiments, 10 March 2008
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Fast Fourier transform experiments, 10 March 2008
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Fast Fourier transform experiments, 10 March 2008
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Odd functions
An odd periodic function ( ) ( )f t f t =
h
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where 2 2T t T