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Unformatted text preview: ME163 Mechanical Vibrations (Winter 2009) HW9 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 directforced 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) = . 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 e a 2 τ sin( a 3 τ )sin( a 1 ( t τ )) dτ = 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!)....
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This note was uploaded on 03/23/2009 for the course ME 163 taught by Professor Mezic,i during the Winter '08 term at UCSB.
 Winter '08
 Mezic,I

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