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Solution Chap 1,2,4 - 2 1 Problems and Solutions Section...

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2 - 1 Problems and Solutions Section 2.1 (2.1 through 2.15) 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 (2.11) 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: ! = 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 ! n for each of the five cases.
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2 - 2 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 ! n = 1000 10 + 1 3 = 9.837 rad/s, ! = 2 ! n = 19.675 rad/s X = f 0 ! n 2 " ! 2 = 2.338 9.837 2 " 19.675 2 = " 8.053 # 10 " 3 m, v 0 ! n = 0.02033 m Thus the response as given by equation (2.11) is x ( t ) = 0.02033sin9.837 t + 8.053 ! 10 " 3 (cos9.837 t " cos19.675 t ) 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 ! 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 " ! 2 t # $ % & ( sin ! n + ! 2 t " # $ % & 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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