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Unformatted text preview: 18 2 0
Thus K = mω0 2 The above matrix equation is thus …
0 1
a 0
2 2
2 2 1 1 mω0 − mω a = 0
0 1 1 1 2 2 (ω0 2 − ω 2 )
−ω 2 a1 = 0 ω0 2 − ω 2 a2 −ω 2 From the top equation in the matrix equation above, we get …
2 (ω0 2 − ω 2 ) a1 − ω 2 a2 = 0
2
2
a2 2 (ω0 − ω )
=
a1
ω2 The amplitudes for the two normal modes can be found by setting ω 2 = ω12 or ω2 2 .
In each case we set a1 = 1 , which we are free to do since only ratios of the amplitudes can
be determined.
For ω 2 = ω12 = 2 − 2 ω0 2 we obtain … ( a2 = ) 2 (ω0 2 − ω12 ) ω 2
1 ( ) 2 1 − 2 − 2 = 2
=
2− 2 ( ) Thus, for mode 1, the lower frequency, symmetric mode … 1 iω t
q1 = e 1 2 ( ) For ω 2 = ω2 2 = 2 + 2 ω0 2 we obtain …
a2 = 2 ( ω0 2 − ω 2 2 ) ω 2
1 ( ) 2 1 − 2 + 2 =− 2
=
2+ 2 ( ) Thus, for mode 2, the higher frequency, antisymmetric mode … 1 iω 2 t
q2 = e
− 2
(b) In this case, m
<< 1 or M >> m . We also assume that the spring is “slack”,
m+M k
g
<< , an assumption not stated in the problem (which needs to be rectified in
M +m
r
the next edition, I suppose)
k
k
g
and Ω0 2 =
Also, 2mg ≠ kr , so we let ω0 2 =
≈
M +m M
r i.e., The kinetic energy is ...
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 Fall '11
 JohnAnderson
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