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T
T
2
2
22
21
t
y
vx
y
x
x!
x
xmaF!
Standing Waves Progressive Waves
tL
xnn[
Tcossin
vtxf
Stretched string
Newtons 2ndWave Equation
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nnn
nn tx
Atxy H[
P
T
! cos
2sin,
g
!
!
1
cos2
sin,
n
nn
n
n tx
txy H[
P
T
Anywave motiononthe stringcanbe described bya sum of these modes!
Fourier Analysis
Normal Modes:
n = 1 n = 2 n = 3
JosephFourier, 1768-1830
Oneequation,infinityunknowns
thanksfornothin Joe!
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!
!1
sin
nn L
xnBxy
T
Focus on x first!
How to find a specific Bn?
!
!
L
n
n
L
dxL
xn
L
xndx
L
xnxy
0 1
*
0
*
sinsinsinTTT
? AJUJUJU !coscossinsin
2
1
dx
L
x
L
xL
g
!
!
1
**
coscos2
TT
This reduces series to 1 term!
Mathematically
Multiplyby nth harmonic and integrate.
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L
n
n
L
xnn
nn
L
L
xnn
nn
LB
01
*
*
*
*sinsin
2
g
!
!T
T
T
T
g
!
!1
*
*
*
*00sinsin
2n
n nnnn
Lnn
nn
LBT
TT
T
*nn {
*nn !
0! 0! all terms zero!
uh oh.0! 0!0
xgxf
xg
xf
cxcx d
d
pp
limlim
LHopitals Rule:
Iff(c)=0 andg(c)=0 then
x
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1
cos
2
* T
T
LBn!
2
*Ln!
2
sin*
0
* LBdx
L
xnxy n
L
!
T
dxL
xnxy
LB
L
n
!0
sinT
!
!
1
sinn
nL
xnBxy
T
Any shape between 0 andL
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n=1 n=2 n=3 n=4 n=5 n=6
Positivecontribution
zerocontribution
zerocontribution
zerocontribution
zerocontribution
zerocontribution
zerocontribution
Positivecontribution
zerocontribution
zerocontribution
zerocontribution
zerocontribution
Graphically.
0sinsin0
21 !
L
dxL
xn
L
xn TTorthogonal
n*=1
n*=2
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0 L
dxL
xncx
LB
L
n
!0
sin2 T
Integrate by parts:
? A d!db
a
b
a
b
a
dxxgxfxgxfdxxgxf
xxx
c
00
ccT
T
T
T
c = slope
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LL
nL
xn
n
L
L
xn
n
Lx
L
c
00
ccT
T
T
T
L
nL
xn
n
L
L
xnx
n
c
0
sicos2
T
T
T
T
? A000cos2 TT
nLn
Bn
TT
n
n
LBn cos!
g
!
! sin
nL
xnBxy
T
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-1.2
-0.8
-0.4
0
0.4
0.8
1.2
-3 -2 -1 0 1 2 3
x
y
1st
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-1.2
-0.8
-0.4
0
0.4
0.8
1.2
-3 -2 -1 0 1 2 3
x
y
1st
+ 2nd
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-1.2
-0.8
-0.4
0
0.4
0.8
1.2
-3 -2 -1 0 1 2 3
x
y
1st
+ 2nd
+ 3rd
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-1.2
-0.8
-0.4
0
0.4
0.8
1.2
-3 -2 -1 0 1 2 3
x
y
1st
+ 2nd
+ 3rd
+ 4th
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-1.2
-0.8
-0.4
0
0.4
0.8
1.2
-3 -2 -1 0 1 2 3
x
y
1st
+ 2nd
+ 3rd
+ 4th
+ 5th
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Barrys profile
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0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 3 0 0 3 5 0 0
5 0
1 0 0
1 5 0
2 0 0
2 5 0
3 0 0
3 5 0
Barrys profile
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0 50 100 150 20 0 250 300 350 400 0
20
40
60
80
10 0
12 0
Barrys 1st harmonic
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0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 3 0 0 3 5 0 4 0 0 0
2 0
4 0
6 0
8 0
1 0 0
1 2 0
Barrys 1st and 2nd harmonic
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0 50 1 00 150 20 0 250 300 350 400 -100
-5 0
0
50
10 0
15 0
20 0
25 0
Barrys 1st, 2nd, and 3rd harmonic
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clearload f;f=f-172;plot(f)
B(200)=0;y(355)=0;for n = 1:200
for x = 1:355B(n)=B(n)+(2/355)*f(x)*sin(n*pi*x/355);
end
end
figureplot(B,'.')for x = 1:355
for n = 1:200
y(x)=y(x)+B(n)*sin(n*pi*x/355);endendfigureplot(y)
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Frenchs Fourier Foible:
g
!
!
1
si
n
nLxnBxy T dxL
xnxyL
L
n !0
sin T
Good choice if boundaries are at 0: Not so good for other shapes..
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More General:function periodic on intervalx = L to x = L.
g
!
g
!
!
11
0sincos
2 nn
n
n
L
xnB
L
xnA
Axy
TT
dxL
xnxy
LB
L
L
n
!
Tsi
1
dxL
xnxy
LA
L
L
n
!T
cos1
dxxyL
A
L
L
!1
0
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Something hard:
0 L/8 L
H
x
!L
L
L
dxL
HdxL
A8
8
0
0011
80
HA !
!L
L
L
n dxLxn
Ldx
LxnH
LA
8
8
0
cos01cos1 TT
8
0
sin
L
n
L
xn
n
L
L
HA
!T
T
!8
sinT
T
n
n
HAn
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0si
8
0
!
L
ndx
L
xnH
LB
T
8
0
cos
L
nL
xn
n
L
LB
!T
T
!8
c s1 TT
nn
Bn
g
!
g
!
!11
sin8
cos1cos8
sin8 nn L
xnn
n
H
L
xnn
n
HHxy
TT
T
TT
T
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0 L
0
H
MATLAB summation of 1 to 100 terms.
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Time dependence?
Make y(x) an initial condition: y(x,0) = y(x)
The normal modes will oscillate at [n:
g
!
g
!
!11
0 cossincoscos2
,n
nnnn
n
nn txkBtxkAA
txy [[
P
T[
vvk
2||
Given the wavelengths, the frequencies are known.
This is NOT a consequence of Fourier Analysis
2
2
22
2
t
y
vx
y
x
x!
x
xThis IS a consequence of the wave equation:
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0 L
0
H
18,000,000 sinusoids in MATLAB:
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Any well behaved repetitive function can bedescribed as an infinite sum of sinusoids withvariable amplitudes (a Fourier Series). On astretched string these correspond to the
normal modes. Fourier analysis can describearbitrary string shapes as well as progressivewaves and pulses.
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