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Problem Set 3

# Problem Set 3 - October 9 2003 Physics 16 Problem Set 3 3-1...

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October 9, 2003 Physics 16 Problem Set 3 3-1. a. L = T - V 2 2 2 1 1 2 2 2 1 1 1 1 ( ) 2 2 2 L m x m x k x x = + - - & & 2 2 2 2 1 1 2 2 2 1 1 2 1 1 1 1 1 2 2 2 2 2 L m x m x kx kx kx x = + - - + & & Euler-Langrange Equations: 1 1 2 1 m x kx kx = - && 2 2 1 2 m x kx kx = - && This is indeed what we would expect from f=ma , using Hooke's Law. b. Substituting x 2 out: 2 2 2 1 1 2 1 1 1 1 ( ) 2 2 2 L ky m x m x y = - + + - & & & 2 2 2 2 1 1 2 1 2 2 1 1 1 1 1 2 2 2 2 L ky m x m x m y m x y = - + + + - & & & & & Since x 1 does not appear in the Langragian, we can find the generalized momentum conjugate: 1 1 2 1 1 ( ) ( ) L m x m x y x t = + - & & & & Substituting x 1 out: 2 2 2 2 2 1 2 1 1 1 ( ) 2 2 2 L ky m x m x y = - + + + & & & 2 2 2 2 2 2 1 2 1 1 2 1 1 1 1 2 2 2 2 L ky m x m x m y m x y = - + + + + & & & & & 2 2 1 2 2 ( ) ( ) L m x m x y x t = + + & & & & But 1 x y - & & is the velocity of m 2 , and 2 x y + & & is the velocity of m 1 , so these are both equal to each other and to the total momentum of the system. c. We know that 1 1 2 2 0 m x m x + = & & , so 2 2 1 1 2 1 ( [0]) [0] m x x x x m - = - + . Integrating this expression from time = 0 to t, we get that 2 2 1 1 2 1 ( [0]) [0] m x x x x m - = - + . Going back to

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our differential equation from part a, we can solve these two equations simultaneously by substituting the preceding expression for x 2 . L @ t D =- 1 2 k H - x 1 @ t D + x 2 @ t D L 2 + 1 2 m 1 x 1 t @ t D 2 + 1 2 m 2 x 2 t @ t D 2 L @ t D = - 1 2 k i k x 2 @ 0 D - x 1 @ t D - m 1 H - x 1 @ 0 D + x 1 @ t D L m 2 y { 2 + 1 2 m 1 x 1 t @ t D 2 + 1 2 m 2 x 2 t @ t D 2 Now we can use the Euler-Lagrange equation to find that: m 1 x 1 1 @ t D ==- k i k - 1 - m 1 m 2 y { i k - x 1 @ t D - m 1 H - x 1 @ 0 D + x 1 @ t D L m 2 + x 2 @ 0 D y { The solution to this differential equation (using Mathematica) is: x 1 @ t D = k m 1 x 1 @ 0 D + k m 2 x 2 @ 0 D k m 1 + k m 2 + C @ 1 D Cos A D k m 1 + k m 2 t t m 1 t m 2 E + C @ 2 D Sin A D k m 2 + k m 1 t t m 1 t m 2 E We know from the form of this equation that the frequency of this oscillating mode is just: k m 1 + k m 2 t m 1 t m 2 This has the correct units of inverse seconds.
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Problem Set 3 - October 9 2003 Physics 16 Problem Set 3 3-1...

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