hw8soln

# hw8soln - ME 375 HOMEWORK #8 SOLUTION PROBLEM 1: (30%) (a)...

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ME 375 HOMEWORK #8 SOLUTION Fall 2009 PROBLEM 1 : (30%) (a) Problem 8.10 in Palm textbook, page 472. 8.10 Given 10 20 ( ) x cx x f t ++ = ±± ± Find the resonant frequency and peak magnitude. Solution: Write the diff. eq. as: 22 2( ) nn s n x xx K f t ζω ω ++ = ± Thus, ( ) 10 s c x K f t = ± where 1 2 1.414 rad/sec, , and 20 20 2 ns c K ωζ == = = . The resonant frequency and peak magnitude are given by 2 12 rn =− and 2 21 s r K M ζ = Thus, a) For = 0 . 1, r = 1 . 4 and M r = 0 . 25. b) For = 0 . 3, r = 1 . 28 and M r = 0.087.

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(b) Problem 8.11 in Palm textbook, page 472. 8.11 Given 10 20 ( ) x cx x f t ++ = ±± ± , where () 11s in f tt ω = Find the damping constant c so that the maximum steady-state amplitude of x is no greater than 3. Solution: Write the diff. eq. as: 22 ( ) 10 s c x xx K f t ++ = ± where 1 2 1.414 rad/sec, , and 20 20 2 ns c K ωζ == = = . The peak magnitude is given by 1 21 4 01 s r K M ζ ζζ −− . The peak amplitude of x is 11 M r . Thus, 2 11 3 40 1 < Rewriting this inequality gives: 42 119 119 1 0 +< Solving for ζ gives two values: 0.092 and 0.996. Since a resonant frequency does not exist for ζ > 0.707, the answer must be: 0.092 > Solving for c
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## This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue.

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hw8soln - ME 375 HOMEWORK #8 SOLUTION PROBLEM 1: (30%) (a)...

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