11
Normal
Modes

Matrix
Methods
Fall
2003
Matrix methods provide a very powerful tool for analyzing normal modes of coupled
oscillator systems.
We'll introduce this general approach by means of an example, the
same system we discussed in detail in
Section 8 of these notes.
Our development will
parallel the previous treatment, but we'll use matrix language.
Example
For the system discussed in Section 8,
the equations of motion
(from
Σ
F
=
ma
)
are
mx
k
k
x
k x
k x
k
k
x
&&
(
' )
'
,
'
(
' )
.
1
1
2
2
1
2
= 
+
+
=

+
(1)
The equations of motion can be written as a single matrix equation:
m
m
x
x
k
k
k
k
k
k
x
x
0
0
1
2
1
2
F
H
G
I
K
J
F
H
G
I
K
J
= 
+


+
F
H
G
I
K
J
F
H
G
I
K
J
'
'
'
'
.
(2)
Each side of this equation is a column matrix.
You should verify that when the matrix
products are carried out, equating the first (top) elements on the two sides gives the first
of Eqs. (1), and equating the second (bottom) elements gives the second.
We define the matrices
M,
K,
and
x
as follows:
M
K
x
=
F
H
G
I
K
J
=
+


+
F
H
G
I
K
J
=
F
H
G
I
K
J
m
m
k
k
k
k
k
k
x
x
0
0
1
2
,
'
'
'
'
,
(3)
Then
Eq. (2)
can be written simply as
Mx
Kx
.
= 
(4)
For other systems with different arrangements of masses and springs, the
M
K
matrices will be different.
But if the spring forces are
linear
functions of the coordinates,
the equations of motion can always be written in the form of Eq. (4).
So the following
development is general, and is not restricted to the specific example cited above.
Note that the (1,1) element of
K
represents the negative of the force on mass
1
when
it
is displaced a distance
x
1
The
(1,2) element is the negative of the force on mass
when mass
2
is displaced
a distance
x
2
,
and so on.
This gives an alternative way to
determine the elements of
K
instead of working out Eqs. (1)
from
Σ
F
=
ma
.
Caution
:
Some books define
K
with the opposite sign, so there is no minus in Eq. (4).
(See, for example, Edwards and Penney, p. 321.)
We prefer to retain the minus sign, so
that Eq. (4) has the same form as
the equation of motion
kx
= 
for a single
particle.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document112
11
Normal Modes  Matrix Methods
General Formulation
For a normal mode, each
x
must vary sinusoidally, and all
x
's
must have the same
frequency.
Therefore we try to find a solution
of Eq. (4) in the form
x
x
C
a
a
t
1
2
1
2
F
H
G
I
K
J
=
F
H
G
I
K
J
+
cos
,
ϖ
ϕ
b
g
or
x
a
=
+
C
t
cos
.
ϖ
ϕ
b
g
(5)
In this expression,
C
is a scalar amplitude factor (determined by initial conditions), and
a
is a column matrix (or vector) whose elements
a
1
,
a
2
give the ratios of the amplitudes
of the
x
for each normal mode.
For example, if
a
2
=
2
a
1
,
the motion of the mass
with coordinate
x
2
is in phase with that of
x
1
but with amplitude twice as great.
Taking the second time
derivative of Eq. (5),
we get
&&
cos
cos
,
.
x
a
a
= 
+
= 
+
=
ϖ
ϖ
ϕ
λ
ϖ
ϕ
λ
ϖ
2
2
C
t
C
t
b
g
b
g
where
(6)
To test whether
Eq. (5)
really
is
a solution of Eq. (4), we substitute
Eqs. (5) and (6)
into
Eq. (4).
After dividing out the common factor
C
cos (
ϖ
t
+
ϕ
),
we get

= 
λ
Ma
Ka
,
or
K
M a

=
λ
b
g
0.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Clarke
 Eigenvectors, Matrices, Power, Heat, Singular value decomposition, Diagonal matrix, Orthogonal matrix

Click to edit the document details