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nagle_differential_equations_ISM_Part35

# nagle_differential_equations_ISM_Part35 - Chapter 4 12 The...

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Chapter 4 12. The equation of the motion of this mass-spring system is (1 / 4) y + 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 < 0 . So, we use equation (16) in the text for a general solution. With α = - b 2 m = - 2 ( 1 / 2) = - 4 and β = 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 0 . 08 (sec) . 14. For the damping factor, Ae - ( b/ 2 m ) t , lim b 0 Ae - ( b/ 2 m ) t = Ae - (0 / 2 m ) t = Ae 0 = A since the exponential function is continuous on ( -∞ , ). For the quasifrequency, we have lim b 0 4 mk - b 2 4 = 4 mk 4 = (4 mk ) / (2 m ) 2 2 π = k/m 2 π . 166

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Exercises 4.10 16. Since the period P = 2 π/ω = 2 π m/k , we have a system of two equations to determine m (and k ). 2 π m k = 3 2 π m + 2 k = 4 . Dividing the second equation by the first one, we eliminate k and get m + 2 m = 4 3 m + 2 m = 16 9 9 m + 18 = 16 m m = 18 7 (kg) . 18. As it was noticed in the discussion concerning an overdamped motion, a general solution to the equation my + by + ky = 0 has the form y ( t ) = C 1 e r 1 t +
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nagle_differential_equations_ISM_Part35 - Chapter 4 12 The...

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