SolSec 2.1 - Chapter 2 Problems Problems and Solutions...

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Chapter 2 Problems Problems and Solutions Section 2.1 (2.1 through 2.10) 2.1 To familiarize yourself with the nature of the forced response, plot the solution of a forced response of equation (2.2) with ω = 2 rad/s, given by equation (2.1) for a variety of values of the initial conditions and ω n as given in the following chart: Case x 0 v 0 f 0 ω n 1 0.1 0.1 0.1 1 2 -0.1 0.1 0.1 1 3 0.1 0.1 1.0 1 4 0.1 0.1 0.1 2.1 5 1 0.1 0.1 1 Solution: Given: ω = 2 rad/sec. From equation (2.11): x ( t ) = n v 0 sin n t + ( x 0 - 2 2 0 n f ) cos n t + 2 2 0 n f cos t Insert the values of x 0 , v 0 , f 0 , and for each of the five cases.
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Chapter 2 Problems 2.2 Repeat the calculation made in Example 2.1.1 for the mass of a simple spring-mass system where the mass of the spring is considered and known to be 1 kg. Solution: Given: m sp = 1 kg, Example 1.4.4 yields that the effective mass is m e = m + 3 sp m = 10 + 3 1 = 10.333 kg. Thus the natural frequency, X and the coefficients in equation (2.11) for the system now become ωω ω nn X fv = + == = = =− × = 1000 10 1 3 9 837 2 19 675 7 667 10 0 02 0 22 3 0 .. rad/s, rad/s m, m Thus the response as given by equation (2.11) is xt t t t ( ) . sin . . (cos . cos . ) =+ × 0 02 9 837 7 667 10 9 837 19 675 3 m 2.3 A spring-mass system is driven from rest harmonically such that the displacement response exhibits a beat of period of 0.2 π s. The period of oscillation is measured to be 0.02 s. Calculate the natural frequency and the driving frequency of the system. Solution: Given: Beat period: T b = 0.2 s, Oscillation period: T 0 = 0.02 s Equation (2.13): x ( t ) = 2 2 0 2 n f sin n t 2 sin n t + 2 So, T b = 0.2 = n 4 n = 2 . 0 4 = 20 rad/s T 0 = 0.02 = + n 4 + n = 02 . 0 4 = 200 rad/s Solving for n and gives: Natural frequency : n = 110 rad/s Driving frequency : = 90 rad/s
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Chapter 2 Problems 2.4
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SolSec 2.1 - Chapter 2 Problems Problems and Solutions...

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