nagle_differential_equations_ISM_Part35

nagle_differential_equations_ISM_Part35 - Chapter 4 12. The...

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Unformatted text preview: Chapter 4 12. The equation of the motion of this mass-spring system is (1 / 4) y 00 + 2 y + 8 y = 0 , y (0) =- 1 / 2 , y (0) =- 2 . Clearly, this is an underdamped motion because b 2- 4 mk = (2) 2- 4(1 / 4)(8) =- 4 < . So, we use equation (16) in the text for a general solution. With =- b 2 m =- 2 ( 1 / 2) =- 4 and = 1 2 m 4 mk- b 2 = 2 4 = 4 , equation (16) becomes y ( t ) = ( C 1 cos 4 t + C 2 sin 4 t ) e- 4 t . From the initial condiions, y (0) = ( C 1 cos 4 t + C 2 sin 4 t ) e- 4 t t =0 = C 1 =- 1 / 2 y (0) = [(4 ( C 2- C 1 ) cos 4 t- 4( C 1 + C 2 ) sin 4 t ] e- 4 t t =0 = 4 ( C 2- C 1 ) =- 2 C 1 =- 1 / 2 , C 2 =- 1 , and so y ( t ) =- 1 2 cos 4 t + sin 4 t e- 4 t . The maximum displacement to the left occurs at the first point t * of local minimum of y ( t ). The critical points of y ( t ) are solutions to y ( t ) =- 2 e- 4 t (cos 4 t- 3 sin 4 t ) = 0 . Solving for t , we conclude that the first point of local minimum is at t * = arctan(1 / 3) 4 . 08 (sec) . 14. For the damping factor, Ae- ( b/ 2 m ) t , lim b Ae- ( b/ 2 m ) t = Ae- (0 / 2 m ) t = Ae = A since the exponential function is continuous on (- , ). For the quasifrequency, we have lim b 4 mk- b 2 4 m = 4 mk 4 m = p (4 mk ) / (2 m ) 2 2 = p k/m 2 . 166 Exercises 4.10 16. Since the period P = 2 / = 2 p m/k , we have a system of two equations to determine m (and k ). 2 r m k = 3 2 r m + 2 k = 4 ....
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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.

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nagle_differential_equations_ISM_Part35 - Chapter 4 12. The...

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