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Solutions_HW12

# Solutions_HW12 - Homework 12 Solutions Problem 1 a The...

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Homework 12 - Solutions Problem 1 a) The Lagrangian is given by L = 1 2 I ˙ θ 2 + V 0 λ cos ( λ 1 / 2 θ ). Since θ is small, we can expand the cosine, L = 1 2 I ˙ θ 2 + V 0 λ 1 - ( λ 1 / 2 θ ) 2 2! + ( λ 1 / 2 θ ) 4 4! - ( λ 1 / 2 θ ) 6 6! + · · · = V 0 λ + 1 2 I ˙ θ 2 - V 0 θ 2 2! + λV 0 θ 4 4! - λ 2 V 0 θ 6 6! + · · · (1) Apart from the first term on the RHS which has no effect to the equation of motion, we can see that the Lagrangian now is the summation of leading order, perturbation and smaller perturbation term. b) The equation of motion can be obtained from d dt ∂L ˙ θ - ∂L ∂θ = 0 (2) and it gives I ¨ θ + V 0 λ 1 / 2 sin ( λ 1 / 2 θ ) = 0 (3) Expand the sine function because θ is small, sin ( λ 1 / 2 θ ) = λ 1 / 2 θ - ( λ 1 / 2 θ ) 3 3! + ( λ 1 / 2 θ ) 5 5! + · · · (4) We can write θ as θ = θ 0 + λθ 1 + λ 2 θ 2 + · · · , hence ¨ θ = ¨ θ 0 + λ ¨ θ 1 + λ 2 ¨ θ 2 + · · · , such that we will have I ( ¨ θ 0 + λ ¨ θ 1 + λ 2 ¨ θ 2 + · · · ) = - V 0 n θ 0 + λ θ 1 - θ 0 3! + λ 2 θ 2 - θ 1 2! + θ 0 5! + · · · o (5) Equating the coefficient of λ , we will have ¨ θ 0 = - V 0 I θ 0 (6) ¨ θ 1 = - V 0 I θ 1 - θ 3 0 3! (7) ¨ θ 2 = - V 0 I θ 2 - θ 2 0 θ 1 2! + θ 5 0 5! (8) 1

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And the last computation can be done using Mathematica to get θ 0 ( t ) = θ 0 (0) cos (Ω t ) (9) θ 1 ( t ) = 1 192 n θ 3 0 (0) + 192 θ 1 (0)
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Solutions_HW12 - Homework 12 Solutions Problem 1 a The...

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