Analytical Mech Homework Solutions 28

# Analytical Mech Homework Solutions 28 - = tan 1 ( c 2m ) m...

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

() 1 22 2 tan cm mc ωα φ αω α = −−+ k Using sin , cos 1 φφ += () () 2 2 2 2 2 2 F k c A m ω  =− + +  D {} 1 2 2 2 2 2 F A k c m = −−++ − D and xt () ( ) cos t A e t + the transient term. 3.19 (a) 1 2 2 21 8 lA T g π  ≈−   for 4 A = , 2 1.041 l T g ≈× (b) 2 2 4 1.084 l g T Using 2 l g = D T gives 2 2 4 l g T = D , approximately 8% too small. (c) 3 2 32 A B λ D and 2 6 = D 2 192 B A A = for 4 A = , 0.0032 B A = 3.20 in t n n f tc e = . . . 0, 1, 2, n =±± cos sin nn ft c nt c i =+ ∑∑ , n 0, 2, = ±± . . . and 2 2 1 T in t T n t e T = cf , d t 0, 2, n . . . 1 cos sin TT n i t n t d t f t n ωω −− ∫∫ t d t The first term on is the same for and n c n n ; the second term changes sign for
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