ME429_Fall_2009_HW6_soln[1] - SolutiontoME429Fall2009HW6 1)...

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Solution to ME429 Fall 2009 HW6 1)  3 20 10 kg m , 3 500 10 N/m k  3 51 0r a d / s n k m 250 Hz 1571 rad/s   2 2 2 1 where 12 n Hr rr Since maximum allowable error up to 250 Hz is specified as 1%, one should use the following relation;  0.99 1.01 H to get ζ max . Note that 1 H is required for no error. Once the inequality is studied, one can obtain 0.652 0.726 so that max 0.01 E 2) a)  1 2Re o ip t Fourier o p p Ft a a e where 2 , 0.15 s o T T As only five terms will be used in the analysis, summation will end with p = 5 for this problem.
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 0 1 o T ip t p aF t e d t T  0 1 80000 T o t d t T        2 55 33 0 5 1 400 10 800 400 10 oo TT ip t ip t p T a t ed t t t T Note that the calculation of this integral by hand requires integration by parts. Steps of that procedure will not be given here and are assigned as exercise. However; one may find useful to choose the variable in integration by parts as follows:    , at at te dt v t du e dt vdu uv udv In this solution, all procedure is carried by MATLAB and the relevant code is supplied in the end.
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ME429_Fall_2009_HW6_soln[1] - SolutiontoME429Fall2009HW6 1)...

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