Analytical Mech Homework Solutions 176

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

Ask a homework question - tutors are online