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
2 - 3 2.4 An airplane wing modeled as a spring-mass system with natural frequency 40 Hz is driven harmonically by the rotation of its engines at 39.9 Hz. Calculate the period of the resulting beat. Solution: Given: n ! = 2 (40) = 80 rad/s, = 2 (39.9) = 79.8 rad/s Beat period: T b = " # n 4 = 8 . 79 80 4 " = 20 s. 2.5 Derive Equation 2.13 from Equation 2.12 using standard trigonometric identities. Solution: Equation (2.12): x ( t ) = 2 2 0 " n f [cos t – cos n t ] Let A = 2 2 0 " n f x ( t ) = A [cos t – cos n t ] = A [1 + cos t – (1 + cos n t )] = A [2cos 2 t 2 – 2cos 2 t n 2 ] = 2 A [(cos 2 t 2 - cos 2 2 n cos 2 t 2 ) - (cos 2 t n 2 - cos 2 t n 2 cos 2