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ME429_Fall_2009_HW6_soln[1]

ME429_Fall_2009_HW6_soln[1] - 1 m 20 103 kg k 500 103 N/m n...

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Solution to ME429 Fall 2009 HW6 1) 3 20 10 kg m , 3 500 10 N/m k 3 5 10 rad/s n k m 250 Hz 1571 rad/s 2 2 2 1 where 1 2 n H r r r 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 F t 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 a F t e dt T   0 1 80000 T o a F t dt T 2 5 5 3 3 0 5 1 400 10 800 400 10 o o T T ip t ip t p T a te dt t e dt 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
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ME429_Fall_2009_HW6_soln[1] - 1 m 20 103 kg k 500 103 N/m n...

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