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s11f1027 - ADVANCED DYNAMICS PHY 4936 Solution 27: (A) Let...

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ADVANCED DYNAMICS — PHY 4936 Solution 27: (A) Let λ = ω 2 . We have to solve det ± ± ± ± 5 - λ 1 - λ 1 - λ 2(1 - λ ) ± ± ± ± = 0 , which leads to the quadratic equation 0 = (5 - λ ) 2 (1 - λ ) - (1 - λ ) (1 - λ ) = 10 - 2 λ - 10 λ + 2 λ 2 - 1 + 2 λ - λ 2 = λ 2 - 10 λ + 9 with the solutions ω 2 1 , 2 = λ 1 , 2 = 5 ± 5 2 - 9 = 5 ± 4 ω 2 1 = λ 1 = 9 , ω 2 2 = λ 2 = 1 . (B) Using normal coordinates Θ 1 = Re [ C 1 exp( 1 t )] and Θ 2 = Re [ C 2 exp( 2 t )], the general solutions are the superpositions x 1 = 4 11 Θ 1 + 4 12 Θ 2 x 2 = 4 21 Θ 1 + 4 22 Θ 2 It turns out that we have to take the minor of the second row, because for the first row the 4 12 and 4 22 minors are both zero. Solutions are then x 1 = 8 Θ 1 + 0 = 8 Θ 1 x 2 = - 4 Θ 1 + 4 Θ 2 . (C) Substituting these equations into the Lagrangian L = 1 2 ( ˙ x 1 ˙ x 1 + ˙ x 1 ˙ x 2 + ˙
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This note was uploaded on 12/11/2011 for the course PHY 4936 taught by Professor Berg during the Fall '11 term at FSU.

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s11f1027 - ADVANCED DYNAMICS PHY 4936 Solution 27: (A) Let...

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