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Analytical Mech Homework Solutions 182

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

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19 ( ) ( ) 2 2 2 1 1 2 2 2 T MX m X r Xr θ θ + + + ± ± ± ± ± ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 1 1 1 1 2 2 2 2 2 2 2 T M m X m r Xr MX m Xr m r θ θ θ + + + + + ± ± ± ± ± ± ± θ ± and the M -matrix is M m m m M The potential energy is ( ) 2 2 2 0 0 1 1 1 1 2 2 2 2 g V kX m r M m r θ ω + = + ± 2 The K -matrix is thus 2 0 2 0 0 0 M m ω K To solve for 2 ω , we set 2 0 ω = K M ( ) ( ) 2 2 2 0 2 2 2 0 0 M m m m ω ω ω ω ω = which yields … 2 2 4 2 2 2 2 4 2 0 0 0 0 0 mM m ω ω ω ω ω ω + = ( ) ( ) 4 2 2 2 2 2 0 0 0 0 0 M m M M ω ω ω ω + Ω + = Neglecting m with respect to M and simplifying yields … ( ) 4 2 2 2 2 2 0 0 0 0 0 ω ω ω ω + Ω + = Solving for 2 ω ( ) ( ) 2 2 2 2 2 2 0 0 0 0 0 0 2 4 2 2 ω ω ω ω + Ω + Ω = ± 2 ( ) ( ) 2 2 2 0 0 0 0 2 2 2 ω ω ω + Ω − Ω = ± 2 2 0 Thus, we get … 2 2 2 1 0 2 and ω ω ω = = Now, we solve for the amplitudes a of the normal mode vectors …
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