Analytical Mech Homework Solutions 178

Analytical Mech Homework Solutions 178 - 15 mx1 + kx1 k (...

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15 () 11 2 1 0 mx kx k x x +− −= ±± 2 20 mx kx kx +−= 22 1 3 2 0 mx k x x k x x −− = 12 23 kx mx kx kx −+ + − = 33 23 0 mx k x x kx + = 23 3 kx mx kx = The secular equation (Equation 11.4.12) is … 2 2 2 02 mk k km kk k ω −+ + + = 0 2 0 3 2 2 mkk ωω −+−−+= 2 , 2 2 mk o rmkk = −+ −= 0 2 2 k m == 2 = ± k 0 k m From Equation 11.5.17 ( N = 1 and n = 3) 10 2s i n 8 π = Because 1c o s sin θ = 0 o s 4 2 1 2 2 =− 0 i n 2 2 82 0 = 30 0 31 c o s 3 i n 2 4 0 212 2 2 =+ = + 2
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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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