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Unformatted text preview: Homework 12  Solutions Problem 1 a) The Lagrangian is given by L = 1 2 I 2 + V cos ( 1 / 2 ). Since is small, we can expand the cosine, L = 1 2 I 2 + V 1 ( 1 / 2 ) 2 2! + ( 1 / 2 ) 4 4! ( 1 / 2 ) 6 6! + = V + 1 2 I 2 V 2 2! + V 4 4! 2 V 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  L = 0 (2) and it gives I + V 1 / 2 sin ( 1 / 2 ) = 0 (3) Expand the sine function because is small, sin ( 1 / 2 ) = 1 / 2  ( 1 / 2 ) 3 3! + ( 1 / 2 ) 5 5! + (4) We can write as = + 1 + 2 2 + , hence = + 1 + 2 2 + , such that we will have I ( + 1 + 2 2 + ) = V n + 1 3! + 2 2 1 2! + 5! + o (5) Equating the coefficient of , we will have = V I (6) 1 = V I 1 3 3!...
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This note was uploaded on 12/29/2011 for the course PHYSICS 601 taught by Professor Hassam during the Fall '11 term at Maryland.
 Fall '11
 Hassam
 mechanics, Work

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