s1105 - a = 0 and x ( / 2) = b 2 = 1 b = 2 so that this...

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ADVANCED DYNAMICS — PHY 4241/5227 HOME AND CLASS WORK – SET 1 Solution for assignment 5. 1. The Lagrangian is L = 1 2 ˙ x 2 - 1 2 x 2 . Thus, the Euler-Lagrange equation becomes 0 = ∂L ∂x - d dt ∂L ˙ x = - x - ¨ x and the general solution is given by x ( t ) = A sin( t ) + B cos( t ). The integration constants are determined by the initial conditions x (0) = B = 0 and x ( π/ 2) = A = 1 so that the exact path is x ( t ) = sin( t ) . For this path the action is given by S [ x ( t )] = 1 2 Z π/ 2 0 dt h cos 2 ( t ) - sin 2 ( t ) i = 1 2 Z π/ 2 0 dt cos(2 t ) = 0 . 2. For a linear path x ( t ) = a + b t subject to the boundary conditions: x (0) =
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Unformatted text preview: a = 0 and x ( / 2) = b 2 = 1 b = 2 so that this path is x ( t ) = 2 t . For this path the action is given by S [ x ( t )] = 1 2 Z / 2 dt " 2 2- 2 2 t 2 # = 2 2 Z / 2 dt 1-t 2 = 2 2 " 2-1 3 2 3 # = 1 1- 2 12 ! = 1 - 12 = 0 . 0565 . . . . 3. Assuming that the previous result is in J s , we nd for the large number S/ h 5 . 37 10 32 ....
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This note was uploaded on 11/10/2011 for the course PHY 4241 taught by Professor Berg during the Spring '11 term at University of Florida.

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