soln_hw7

# soln_hw7 - Solutions to Homework 7 Section 3.9 6 The...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to Homework 7 Section 3.9 6. The coefficients are calculated to be, m = 5 kg , γ = 50 Ns/m , and k = 490 N/m . The initial value problem is therefore 5 u 00 + 50 u + 490 u = 10sin( t/ 2) , u (0) = 0 m, u (0) = . 03 m/s. 8. a. The homogenous solution u c ( t ) = Ae- 5 t cos √ 72 t + Be- 5 t sin √ 72 t . Based on the method of undetermined coefficients, the particular solution is U ( t ) = 1 153281 [- 160cos( t/ 2)- 3128sin( t/ 2)] . The initial conditions give A = 160 / 163281 and B = 383443 √ 73 / 1118961300. The general solution is u = u c + U , therefore u ( t ) = 1 153281 "- 160 e- 5 t cos √ 73 t- 383443 √ 73 7300 e- 5 t sin √ 73 t # + U ( t ) . b. u c ( t ) is the transient part and U ( t ) is the steady state part of the response. c.See figure 1 d.Based on Eqs.(9) and (10) from the book, the amplitude of the forced responce is given by R = 2 / Δ, in which Δ = p 25(98- ω 2 ) 2 + 2500 ω 2 ....
View Full Document

## This note was uploaded on 10/23/2009 for the course MATH 22B taught by Professor Hunter during the Winter '08 term at UC Davis.

### Page1 / 5

soln_hw7 - Solutions to Homework 7 Section 3.9 6 The...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online