# SolSec3.7 - 3 42 Problems and Solutions for Section...

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3- 42 Problems and Solutions for Section 3.7 (3.39 through 3.45) 3.39 Using complex algebra, derive Eq. (3.89) from (3.86) with s = j ω . Solution: From Eq. (3.86): () k cs ms s H + + = 2 1 Substituting sj = yields Hj mj cj k km c j ωω = + + = −− 11 2 2 The magnitude is given by m j cj k c j dr = + + = 2 2 12 / c = + 1 2 2 2 which is Eq. (3.89)

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3- 43 3.40 Using the values indicated in Fig. 3.18, plot the inertance transfer function's magnitude and phase. Solution: From Fig. 3.18 1 20 5 25 08 1 46 0087 k k k m m c c n =⇒= === = ≈⇒ = . .. ωω ω The inertance transfer function is given by Eq. (3.88): () k cs ms s s H s + + = 2 2 2 Substitute sj = to get the frequency response function. The magnitude is given by: jH j km c ()() = + = + 2 2 2 2 2 2 2 2 2 05 008 . The phase is given by φ = tan Imaginary part of frequency response function Real part of frequency response function -1 Multiply the numerator and denominator of j k m c j 2 2 by to get j c j c = −− + + 2 23 2 2 2 So, =
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## This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec3.7 - 3 42 Problems and Solutions for Section...

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