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hw_9_solu - ME163 Mechanical Vibrations(Winter 2009 HW-9...

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ME163 Mechanical Vibrations (Winter 2009) HW-9 Due 3.12.09 1. Time Solution, three approaches. In this problem we will collect and compare three different approaches to obtaining the time solution to the equation: ¨ x + 2 ζω n ˙ x + ω 2 n x = 1 ω f sin( ω f t ) where ζ = 0 . 1, ω n = 10, and ω f = 1 / 5. The initial conditions are x (0) = ˙ x (0) = 0. (a) Calculate x(12), x(24), x(36) by assuming a solution x ( t ) = x sin ( ω f t - φ ) and using this assumed solution (and x and φ that you solve for) to obtain the three values. This is a standard direct-forced problem. Solving for the two unknowns we have x = F/k radicalbig (1 - Ω 2 ) 2 + (2 ζ Ω) 2 φ = arctan 2 ζ Ω (1 - Ω 2 ) Here, Ω = ω f ω n = 0 . 02, and k = mwn 2 = 100. Using this we can evaluate the answer for the three specified times. x (12) = 0 . 0339 x (24) = - 0 . 0498 x (36) = 0 . 0396 We do not repeat these numbers...they are all equal for each approach, see Figure 1 . (b) Calculate x(12), x(24), x(36) using the convolution approach (use m = 1.0). The following integral may be useful. I = integraldisplay t 0 e - a 2 τ sin( a 3 τ ) sin( a 1 ( t - τ )) = sin( a 1 t ) a 3 3 - 2 a 1 a 2 cos( a 1 t ) a 3 - a 2 1 sin( a 1 t ) a 3 + a 2 2 sin( a 1 t ) a 3 a 4 1 + 2 a 2 2 a 2 1 - 2 a 2 3 a 2 1 + a 4 2 + a 4 3 + 2 a 2 2 a 2 3 + O ( ε ) where O ( ε ) is a messy term which is small in size and therefore you can neglect it. To obtain the answer, we use x ( t ) = 1 ω n ω d I ( a 1 = ω = 1 / 5 , a 2 = ζω n = 1 , a 3 = ω d = 9 . 9499 , t ). (c) Calculate x(12), x(24), x(36) using the Laplace approach. It may be useful to perform a partial fraction expansion to obtain the solution. A good assumed expansion is X ( s ) = C 1 ( s + a ) + C 2 ( s + b ) + C 3 ( s + c ) + C 4 ( s + d ) where a,b,c,d and the C’s may be complex numbers (the answer in the end should be real!). Taking the Laplace transform of each side we have X ( s ) = 1 ( s 2 + ω 2 f )( s 2 + 2 ζω n s + ω 2 n ) ME163 Mechanical Vibrations 1 HW9
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Equating this to the recommended form for the partial fraction expansion we solve for the unknowns: a = 0 + 0 . 2000 i b = 0 - 0 . 2000 i c = 1 . 0000 - 9 . 9499 i d = 1 . 0000 + 9 . 9499 i C 1 = - 0 . 0001 + 0 . 0250 i C 2 = - 0 . 0001 - 0 . 0250 i C 3 = 1 . 0008 e - 004 + 4 . 9265 e - 004
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