This preview shows page 1. Sign up to view the full content.
8
Equilibrium @
0
φ
=
2
3
0
cos2
e
GM a
V
r
′′≈
and
2
0
3
0
e
GM a
r
′
≈
V
2
1
3
M
Im
a
==
0
3
0
3
e
VG
Mr
ω
′′
M
3
0
0
2
2
3
e
r
T
GM
π
11.13
The amplitude of the symmetric component is
A
1
and the amplitude of the antisymmetric
component is
A
2
(See Equations 11.3.19a through 11.3.20b).
[]
2
2
112
1
(0)
(0)
4
Axx
=+
0
11 2
1
(0)
(0)
22
A
=
2
2
221
1
(0)
(0)
4
=−
0
22 1
1
(0)
(0)
A
=
12
AA
∴
=
From Equation 11.3.18 the solution for
x
1
is …
()
0
11
cos
2
A
2
x
tt
t
(The phase
2
δ
is 180
o
, which insures that
10
(0)
x
A
=
)
0
1
2cos
22 2
A
xt
ωω
+−
=
Letting …
1
and
2
=∆
=
x
tA
t
t
This is the end of the preview. Sign up
to
access the rest of the document.
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
 Work

Click to edit the document details