245_pdfsam_math 54 differential equation solutions odd

245_pdfsam_math 54 differential equation solutions odd - A...

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Exercises 4.9 15. Since the exponential function is never zero, from the equation of motion (16) on page 213 in the text we conclude that the mass passes the equilibrium position, that is, y ( t ) = 0, if and only if sin( ωt + φ )=0 . Therefore, the time between two successive crossings of the equilibrium position is π/ω ,which is a half of the quasiperiod P . So, we can Fnd the quasiperiod P by multiplying the time between two successive crossings of the equilibrium position by two. Whenever P is computed, we can measure the displacement y ( t )atanymoment t (with y ( t ) 6 = 0) and then at the moment t + P . Taking the quotient y ( t + P ) y ( t ) = Ae ( b/ 2 m )( t + P ) sin[ ω ( t + P )+ φ ] Ae ( b/ 2 m ) t sin( ωt + φ ) = e ( b/ 2 m ) P , we can calculate the damping coefficient b as b = 2 m ln[ y ( t + P ) /y ( t )] P . EXERCISES 4.9:
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Unformatted text preview: A Closer Look at Forced Mechanical Vibrations, page 227 1. The frequency response curve (13) on page 223, with m = 4, k = 1, and b = 2, becomes M ( ) = 1 p ( k m 2 ) 2 + b 2 2 = 1 p (1 4 2 ) 2 + 4 2 . The graph of this function is shown in igure B.26 in the answers in the text. 3. The auxiliary equation in this problem is r 2 +9 = 0, which has roots r = 3 i . Thus, a general solution to the corresponding homogeneous equation has the form y h ( t ) = C 1 cos 3 t + C 2 sin 3 t. We look for a particular solution to the original nonhomogeneous equation of the form y p ( t ) = t s ( A cos 3 t + B sin 3 t ) , 241...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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