Analytical Mech Homework Solutions 176

# Analytical Mech Homework Solutions 176 - 13 For the second...

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13 For the second eigenvector (the symmetric mode, j = 2) … Inserting 2 2 51 7 4 k m ω =  into the first of the two homogeneous equations yields 12 22 7 2 4 kk ak a  −+ =   22 12 31 7 4 aa + = Letting a 12 = 1, then a 22 = 1.781 (Thus, in the symmetric normal mode, the amplitude of the vibration of the second mass is 1.781 that of the first mass and in phase with it.) The two eigenvectors (Equation 11.4.13 and see accompanying Table) are … () ( 11 11 1 1 21 1 cos cos 0.281 a Qt a ) 1 t δω  =− =  δ ) 2 t 12 22 2 2 22 1 cos cos 1.781 a a = 11.18 () 2 Tm l m ll 2 θ φ =++ ±± ± 2 cos cos cos Vm g l m g l l θφ + For small angular displacements … 2 2 1 2 21 1 2 mm LTV l l l m g l m g l 2 θθ   =−≈ + + + +     ± 2 1 2 ,2 dL L ml ml l l mgl dt 1 ∂∂ =+ + = ± 0 12 g ++ = 2 2 , L ml l l mgl dt φφ = ± 0 ll g
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