210
6 Free Vibration of Multiple Degrees of Freedom System
The frequency equation becomes
−
mp
2
2k
+
−
mp
2
k
+ 1
2
= 0
or
p
4
−
5k
m
p
2
+
k
2
m 2 = 0
Solution of Eq. (a) yields
p
2
1, 2 =
k
2m
,
2k
m
The normal modes are calculated as follows:
{Z }
L
1 =
1 1/2k
0 − 1
0
N
R
0
Let N 0 = 1, then {Z }
L
1 =
1/2k
1
{Z }
R
1 =
1
0
− 2mp
2 1
1/2k
1
=
1/2k
−
mp
2
k
+ 1
{Z }
L
2 =
1 1/k
0 1
1/2k
−
mp
2
k
+ 1
=
1
2k
+
1
k
−
mp
2
k
+ 1
−
mp
2
k
+ 1
{Z }
R
2 =
1 0
− mp
2 1
1
2k
+
1
k
−
mp
2
k
+ 1
−
mp
2
k
+ 1
=
⎧
⎨
⎩
1
2k
+
1
k
−
mp
2
k
+ 1
− mp
2
1
2k
+
1
k
−
mp
2
k
+ 1
+
−
mp
2
k
+ 1
⎫
⎬
⎭
Substituting the value of the first frequency, p
2
1 =
k
2m
in the above relations, we
get
{Z }
L
1 =
1
2k
1
, {Z }
R
1 =
⎧
⎨
⎩
1
2k
1
2
⎫
⎬
⎭
, {Z }
L
2 =
⎧
⎨
⎩
1
k
1
2
⎫
⎬
⎭