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Unformatted text preview: Exercises 4.9 It follows that period = 2 π ω = 2 π 5 natural frequency = ω 2 π = 5 2 π . A general solution, given in (4) in the text, becomes y ( t ) = C 1 cos ωt + C 2 sin ωt = C 1 cos 5 t + C 2 sin 5 t. We find C 1 and C 2 from the initial conditions. y (0) = ( C 1 cos 5 t + C 2 sin 5 t ) t =0 = C 1 = 1 / 4 y (0) = ( 5 C 1 sin 5 t + 5 C 2 cos 5 t ) t =0 = 5 C 2 = 1 ⇒ C 1 = 1 / 4 C 2 = 1 / 5 . Thus, the solution to the initial value problem is y ( t ) = 1 4 cos 5 t 1 5 sin 5 t. The amplitude of the motion therefore is A = q C 2 1 + C 2 2 = r 1 16 + 1 25 = √ 41 20 . Setting y = 0 in the above solution, we find values of t when the mass passes through the point of equilibrium. 1 4 cos 5 t 1 5 sin 5 t = 0 ⇒ tan 5 t = 5 4 ⇒ t = πk arctan(5 / 4) 5 , k = 1 , 2 ,.... (Time t is nonnegative.) The first moment when this happens, i.e., the smallest value of t , corresponds to k = 1. So, t = π arctan(5 / 4) 5 ≈ . 45 (sec) . 4. The characteristic equation in this problem, r 2 + br + 64 = 0, has the roots r = b ± √ b 2 256 2 . Substituting given particular values of b , we find the roots of the characteristic equation and solutions to the initial value problems in each case. 161 Chapter 4 b = 0 b = 0 b = 0. r = ± √ 256 2 = ± 8 i. A general solution has the form y = C 1 cos 8 t + C 2 sin 8 t . Constants C 1 and C 2 can be found from the initial conditions....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

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