19
()
( )
2
22
11
2
TM
X m
Xr
X
r
θ
≈+ +
+
±±
±
( )
1
1
1
2
2
2
m
Xm
r X
r M
X
r m
r
θθ
+ ≈ +
+
±
±
±
±
and the
M
matrix is
M
m
mm
≈
M
The potential energy is
2
00
1
1
2
2
g
Vk
r
M
m
r
θω
≈+
=
+
±
2
Ω
The
K
matrix is thus
2
0
2
0
0
0
M
m
ω
≈
Ω
K
To solve for
2
, we set
2
0
−=
KM
2
0
2
0
0
Mm
ωω
−−
=
−Ω
−
which yields …
4
4
2
0
0
0
mM
m
Ω+ − Ω−
−
=
42
2
2
2
2
0
0
0
M
M
+
Ω
+Ω =
Neglecting m with respect to M and simplifying yields …
422
2
2
2
00 0
0
0
−+
Ω
+
Ω
=
Solving for
2
…
2
2
0
0
2
4
+Ω
−
Ω
=±
2
2
2
2
2
0
Ω
Thus, we get …
2
10
2
and
==
Now, we solve for the amplitudes
a
of the normal mode vectors …
2
0
a
2
0
1
2
2
0
0
a
a
=
−
Using the first of the above matrix equations for
1
2
0
gives …
21
01
a
and
a
Using the second equation for
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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
 Energy, Potential Energy, Work

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