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nagle_differential_equations_ISM_Part34

nagle_differential_equations_ISM_Part34 - Exercises 4.9 It...

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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 = C 2 1 + C 2 2 = 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 0 . 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
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Chapter 4 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. y (0) = ( C 1 cos 8 t + C 2 sin 8 t ) t =0 = C 1 = 1 y (0) = ( - 8 C 1 sin 8 t + 8 C 2 cos 8 t ) t =0 = 8 C 2 = 0 C 1 = 1 C 2 = 0 and so y ( t ) = cos 8 t .
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