except that the natural frequency (which is a scalar quantity) is re
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
¨
x
1
¨
x
2
=

k
0
0

k
x
1
x
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
=
(
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
,
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
K
on the diagonal. Thus,
¨
y
=

Dy,
2