Chem Differential Eq HW Solutions Fall 2011 23

# Chem Differential Eq HW Solutions Fall 2011 23 - 00028 cos5...

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

Section 2.7 Forced Oscillations 23 Clear a, mu, p, k, alph, bet, capa, capb, b, y mu 1; c 5 100; k 1001 100; p Pi; a0 0; a n_ 0; b n_ 2 Pi n 1 Cos nPi ; alph0 a0 k; capa n_ k mu nPi p ^2 capb n_ cnPi p alph n_ capa n a n capb n b n capa n ^2 capb n ^2 bet n_ capa n b n capb n a n capa n ^2 capb n ^2 1001 100 n 2 n 20 1 Cos n 10 n 2 400 1001 100 n 2 2 2 1001 100 n 2 1 Cos n n n 2 400 1001 100 n 2 2 It appears that α n = - (1 - cos( )) 10 ± n 2 400 + ( 1001 100 - n 2 ) 2 ² π and β n = 2 ( 1001 100 - n 2 ) (1 - cos( )) n ± n 2 400 + ( 1001 100 - n 2 ) 2 ² π Note how these formulas yield 0 when n is even. The Frst two nonzero modes of the modiFed solution are y 1 ( t )= α 1 cos t + β 1 sin t = - . 0007842 cos t + . 14131 sin t and y 5 ( t )= α 5 cos 5 t + β 5 sin5 t -
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 00028 cos5 t-. 01698 sin5 t. (c) In what follows, we use 10 nonzero terms of the original steady-state solution and compare it with 10 nonzero terms of the modiFed steady-state solution. The graph of the original steady-state solution looks like this: steadystate t_ Sum alph n Cos n t bet n Sin n t , n, 1, 20 ; Plot Evaluate steadystate t , t, 0, 4 Pi 2 4 6 8 10 12-0.4-0.2 0.2 0.4...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online