20
12
01
a
and
a
==
Thus, the normal modes are approximately …
00
10
it
i t
e
and
e
ω
Ω
QQ
Note that we could have guessed this almost immediately.
The above assumption is
tantamount to omitting the cross elements
.
This completely eliminates the small
coupling between the two oscillators, which reduces the matrix
to the purely
diagonal terms …
2
m
−
2
−
KM
()
22
0
0
0
0
M
m
ωω
−
Ω−
, which leads directly to the above solution.
22.
m
m
2m
k
K
k
θ
3
θ
2
θ
1
We “scale” the force constants and masses to 1 unit,
namely,
11
m
and
k
=
=
.
Let
k
1
K and
l
′ =
=
such that
3l
= circumference
So:
222
123
1
2
2
T
θ
=+
+
±±±
2222
2
21
13
31
23
32
11111
1
22222
2
VK
2
K
θθ
=−+−+−+−+
−+
−
Collecting terms …
() ()
1
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.
 Fall '11
 JohnAnderson
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