SolSec 2.3

SolSec 2.3 - 2 24 Problems and Solutions Section 2.3(2.26...

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2- 24 Problems and Solutions Section 2.3(2.26 through 2.30) 2.26 Referring to Figure 2.9, draw the solution for the magnitude X for the case m = 100 kg, c = 4000 N s/m, and k = 10,000 N/m. Assume that the system is driven at resonance by a 10-N force. Solution: Given: m = 100 kg, c = 4000 N s/m, k = 10000 N/m, o F = 10 N, ωω == n k m = 10 rad/s φ = = = −− tan tan (, ) (, , ) 1 2 1 40 000 10 000 10 000 90 2 cw km ω π rad From the figure: X F c o = −+ = () ( ) ( ,, ) ( , ) 22 2 2 2 10 10 000 10 000 40 000 X = 0.00025 m c ω X F 0 ( k - m ω 2 ) X φ

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2- 25 2.27 Use the graphical method to compute the phase shift for the system of Problem 2.26 if ω = ω n /2 and again for the case ω = 2 ω n . Solution: From Problem 2.26, ω n = 10 rad/s (a) = n 2 = 5 rad/s X = −+ = 10 10 000 2500 20 000 000468 22 (, ) (, ) . m kX = (10,000)(.000468) = 4.68 N c X = (4000)(5)(.000468) = 9.36 N mX 2 = (100) 2 ) 5 ( (.000468) = 1.17 N From the figure given in problem 2.26: φ = = tan . .. 1 936 468 117 69 4 1 21 rad (b) ωω == 0 n rad/s X = = 10 10000 40000 80000 000117 () ( ) . m kX = (10000)(.000117) = 1.17 N c X = (4000)(20)(.000117) = 9.36 N 2 = (100) 2 ) 20 ( (.000117) = 4.68 N From the figure: = =− °=− tan .
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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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SolSec 2.3 - 2 24 Problems and Solutions Section 2.3(2.26...

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