Analytical Mech Homework Solutions 182

Analytical Mech Homework Solutions 182 - 19 2 1 1 MX 2 m X...

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19 () ( ) 2 22 11 2 TM X m Xr X r θ ≈+ + + ±± ± ( ) 1 1 1 2 2 2 m Xm r X r M X r m r θθ + ≈ + + ± ± ± ± and the M -matrix is M m mm    M The potential energy is 2 00 1 1 2 2 g Vk r M m r θω ≈+ = + ± 2 The K -matrix is thus 2 0 2 0 0 0 M m ω K To solve for 2 , we set 2 0 −= KM 2 0 2 0 0 Mm ωω −− = −Ω which yields … 4 4 2 0 0 0 mM m  Ω+ − Ω− =  42 2 2 2 2 0 0 0 M M + +Ω = Neglecting m with respect to M and simplifying yields … 422 2 2 2 00 0 0 0 −+ + = Solving for 2 2 2 0 0 2 4 +Ω 2 2 2 2 2 0 Thus, we get … 2 10 2 and == Now, we solve for the amplitudes a of the normal mode vectors … 2 0 a 2 0 1 2 2 0 0 a a = Using the first of the above matrix equations for 1 2 0 gives … 21 01 a and a Using the second equation for
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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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