SolSec 2.3 - 2- 24 Problems and Solutions Section 2.3 (2.32...

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2- 24 Problems and Solutions Section 2.3 (2.32 through 2.36) 2.32 Referring to Figure 2.10, 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, ! = n = k m = 10 rad/s = tan ! 1 cw k ! m " 2 # $ % & ( = tan ! 1 (40,000) (10,000 ! 10,000) # $ % & ( = 90 ° = ) 2 rad From the figure: X = F o k ! m 2 ) 2 + ( c ) 2 = 10 (10,000 ! 10,000) 2 + (40,000) 2 X = 0.00025 m c ω X F 0 ( k - m ω 2 ) X φ
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2- 25 2.33 Use the graphical method to compute the phase shift for the system of Problem 2.32 if ω = ω n /2 and again for the case ω = 2 ω n . Solution: From Problem 2.32 ! n = 10 rad/s (a) = n 2 = 5 rad/s X = 10 (10,000 ! 2500) 2 + (20,000) 2 = .000468 m kX = (10,000)(.000468) = 4.68 N c ω X = (4000)(5)(.000468) = 9.36 N m 2 X = (100) 2 ) 5 ( (.000468) = 1.17 N From the figure given in problem 2.32: = tan ! 1 9.36 4.68 ! 1.17 " # $ % = 69.4 ° = 1.21 rad (b) = 2 n = 20 rad/s X = 10 (10000 ! 40000) 2 + (80000) 2 = .000117 m kX = (10000)(.000117) = 1.17 N c X = (4000)(20)(.000117) = 9.36 N m 2 X = (100) 2 ) 20 ( (.000117) = 4.68 N From the figure: = tan ! 1 9.36 1.17 ! 4.68 " # $ % & = ! 69.4 ° = ! 1.21 rad
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2- 26 2.34 A body of mass 100 kg is suspended by a spring of stiffness of 30 kN/m and dashpot of damping constant 1000 N s/m. Vibration is excited by a harmonic force of amplitude 80 N and a frequency of 3 Hz. Calculate the amplitude of the displacement for the vibration
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SolSec 2.3 - 2- 24 Problems and Solutions Section 2.3 (2.32...

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