
264 Classical Mechanics
© 2010 Taylor & Francis Group, LLC
Substituting Equation 8.60 into Equation 8.59, we obtain
()
()
(
−+
−+−+
−−−
mkAkA
kA MkAkA
kA m
ω
ω
ω
2
12
1
2
2
0
22
3
0
kA
(8.61)
The secular equation is
−+ −
−−+−
−−+
=
mk k
kMkk
kmk
ω
ω
ω
2
2
2
0
2
0
0
(8.62)
or
ω
2
(−mω
2
+ k)(−mMω
2
+ kM + 2km) = 0 (8.62a)
from which we find the normal frequencies of the system
ωω ω
12 3
==
,,km km
/
. (8.63)
1. Setting ω = 0 in Equation 8.61, we find that A
1
= A
2
= A
3
. Thus, this mode is no oscillation
at all but is a pure translation of the system as a whole (Figure 8.6a).
2. Setting
ω=
/
in Equation 8.61, we find A
2
= 0 and A
1
= −A
3
. Thus, the center mass M
is at rest ...