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Lecture 14
1
Matrix mechanics
In this lecture we will use matrix algebra to the full extent. The matrix
version of equations of motion for the system shown in ﬁgure
??
is
±
m
0
0
2
m
²±
¨
x
1
¨
x
2
²
=
±

2
k
k
k

2
k
²±
x
1
x
2
²
Thus, using our deﬁnition of the mass matrix and the stiﬀness matrix,
we get
M
¨
x
=

Kx,
or
M
¨
x
+
Kx
= 0
.
multiplying the above equation by the matrix inverse of
M
, denoted
M

1
, and remembering that
M

1
M
=
I
, where
I
is the 2
×
2 identity
matrix, we obtain
¨
x
=

M

1
Kx,
(1)
where
x
= (
x
1
,x
2
)
T
is the vertical coordinate vector, and
M

1
is the
matrix inverse of the mass matrix
M
. Recall, that the matrix inverse
of a 2
×
2 matrix
A
=
±
a b
c d
²
is obtained by
A

1
=
1
det
(
A
)
±
d

b

c
a
²
=
1
ad

bc
±
d

b

c
a
²
Thus, the inverse of the mass matrix reads
M

1
=
1
det
(
M
)
±
2
m
0
0
m
²
=
1
2
m
2
±
2
m
0
0
m
²
=
±
1
/m
0
0
1
/
(2
m
)
²
Note that the equation (1) looks very much like our canonical equation
of motion for a single degree of freedom system,
¨
x
=

ω
2
n
x
1
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placed by the matrix product
M

1
K
. In fact, if there was no coupling
spring,
k
12
= 0, then we would have two
decoupled
equations:
±
m
0
0
2
m
²±
¨
x
1
¨
x
2
²
=
±

k
0
0

k
²±
x
1
x
2
²
Wouldn’t it be great if we had a way of taking the coupled equations
and ﬁnding new coordinates in which equations of motion decouple like
that? In fact, it is possible, due to same matrix magic we will perform
next
1.1
Normal mode coordinates
We’d like to ﬁnd coordinates
y
= (
y
1
,y
2
) such that
¨
y
=

Dy,
where
y
= (
y
1
,y
2
)
T
is the vertical ”new coordinate” vector,
D
is a
diagonal, positive matrix (i.e. all the entries of
D
that are not on the
main diagonal are zero, and entries on the main diagonal are either
zero or positive):
D
=
³
(
ω
1
)
2
0
0
(
ω
2
)
2
´
,
and
ω
1
,ω
2
are the frequencies for this 2 DOF system. This can be
achieved by setting
y
=
V

1
x
where
V
=
±
(
φ
1
)
1
(
φ
2
)
1
(
φ
1
)
2
(
φ
2
)
2
²
,
and
φ
1
= ((
φ
1
)
1
,
(
φ
1
)
2
)
T
,φ
2
= ((
φ
2
)
1
,
(
φ
2
)
2
)
T
are the columnvector
normal modes that we discussed in the last lecture! To show this, note
that
¨
y
=
V

1
¨
x
=
V

1
(

M

1
K
)
x
=

V

1
(
M

1
K
)
V y
From linear algebra we know that, if we set the columns of
V
to be the
eigenvectors of
M

1
K
, then
V

1
(

M

1
K
)
V
=
D
a diagonal matrix,
with eigenvalues of
M

1
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This note was uploaded on 03/23/2009 for the course ME 163 taught by Professor Mezic,i during the Winter '08 term at UCSB.
 Winter '08
 Mezic,I

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