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Analytical Mech Homework Solutions 178

# Analytical Mech Homework Solutions 178 - 15 mx1 kx1 k x2 x1...

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15 ( ) 1 1 2 1 0 mx kx k x x + = ±± 1 1 2 2 0 mx kx kx + = ±± ( ) ( ) 2 2 1 3 2 0 mx k x x k x x + = ±± 1 2 2 3 2 0 kx mx kx kx + + = ±± ( ) 3 3 2 3 0 mx k x x kx + + = ±± 2 3 3 2 0 kx mx kx + + = ±± The secular equation (Equation 11.4.12) is … 2 2 2 2 0 2 0 0 2 m k k k m k k k m k ω ω ω + + + = 0 2 0 ( ) ( ) 3 2 2 2 2 2 2 m k k m k ω ω + + = ( ) 2 2 2 2 0, 2 2 m k or m k k ω ω + = + = 2 2 0 2 2 k m ω ω = = 2 2 2 m k ω + = ± k ( ) ( ) 2 2 0 2 2 2 2 k m ω ω = ± = ± From Equation 11.5.17 ( N = 1 and n = 3) 1 0 2 sin 8 π ω ω = Because 1 cos 2 2 sin θ θ = 1 0 0 1 cos 4 2 2
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