Physics 325 Spring 2011 Problem Session 10 Solutions

Physics 325 Spring 2011 Problem Session 10 Solutions - E is...

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Physics 325 Spring 2011 Discussion 10 April 11, 2011 A one-dimensional system has the Lagrangian: L = 1 2 m ˙ x 2 + k 2 x 2 , where k is a positive constant. (a) Write down the equation of motion. (b) Let G = ± m ˙ x 2 - k x 2 ² t - mx ˙ x. Take the time derivative of G , and show that it is zero by using the equation of motion. Note that the quantity inside the parantheses is proportional to the energy. (c) Since the Lagrangian does not have explicit time-dependence, the energy E is a constant of motion. Suppose that we have the initial conditions x (0) = R , and ˙ x (0) = 0 . Solve for x ( t ) . Hint: If you use the fact that E and G are constants of motion, you can solve for x ( t ) without having to solve any differential equation. (d) When will the particle reach the origin? Solutions : (a) The Euler-Lagrange equation of motion is d dt ± ∂L ˙ x ² - ∂L ∂x = 0 d dt ( m ˙ x ) - ± - k x 3 ² = 0 m ¨ x + k x 3 = 0 (b) The energy is E = 1 2 m ˙ x 2 - k 2 x 2 . 1
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Unformatted text preview: E is a constant of motion, G = m x 2-k x 2 t-mx x = 2 Et-mx x G = 2 E-m x 2-mx x Now substitute x from the equation of motion. G = 2 E-m x 2-x -k x 3 G = 2 E- m x 2-k x 2 G = 2 E-2 E = 0 (c) Calculate E and G at t = 0 . E = 0-k 2 R 2 =-k 2 R 2 G = 2 E (0)-mx (0) x (0) = 0 Thus, for any time t , we have E =-k 2 R 2 = 1 2 m x 2-k 2 x 2 G = 0 =-2 k 2 R 2 t-mx x From the second equation, x =-kt mxR 2 Substitute this into the energy equation.-k 2 R 2 = 1 2 m -kt mxR 2 2-k 2 x 2-k 2 R 2 = k 2 t 2 2 mR 4 x 2-k 2 x 2-1 R 2 = kt 2 mR 4 x 2-1 x 2-1 R 2 =-1 x 2 1-kt 2 mR 4 x ( t ) = R r 1-kt 2 mR 4 (d) From previous part, x = 0 when t = R 2 r m k . 2...
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This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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Physics 325 Spring 2011 Problem Session 10 Solutions - E is...

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