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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 ....
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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.
 Winter '08
 Hunter
 Differential Equations, Equations

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