Analytical Mech Homework Solutions 174

Analytical Mech - 11 Chapter 11(continued 11.16 1 1 T = m1 x12 m2 x2 2 2 2 1 1 1 2 V = k1 x12 k x2 x1 k2 x2 2 2 2 2 L = T V d L L = k1 x1 k x2 x1 =

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11 Chapter 11 (continued) 11.16 22 11 2 2 Tm x m x =+ ±± () 2 2 1 1 2 Vk x k xx k x =+ − + LTV =− 1 dL mx dt x = ± 2 1 1 L kx k x x x + 2 1 0 x +− −= 2 dt x = ± 2 1 2 L x x 2 1 x ++ − 2 2 0 mk k k km k k ω ′′ −+ + = −− + + ( )( ) 42 12 2 1 1 2 1 2 0 mm m k k m k k k k k k k ′′ ′ +++ ++   2 = ( ) 12 1 2 0 m k k k kk k k k ωω ++++ = () () 2 21 1 2 1 2 2 4 0 2 mk k mmk k k kk + = = 2 For 2 ,2 , ,2 , mmm m k k k k == === () () ( ) ( ) 2 222 2 2 23 4 6 4 4 2 2 6 mk mk m k m k m k k m +±+− + = 2 10 6 44 00 0 2 k and where m ==
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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