SolSec3_1 - 3Chapter Three Solutions Problem and Solutions...

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3- 1 Chapter Three Solutions Problem and Solutions for Section 3.1 (3.1 through 3.14) 3.1 Calculate the solution to !! x + 2 ! x + 2 x = ! t " # ( ) x 0 ( ) = 1 ! x 0 ( ) = 0 and plot the response. Solution: Given: !! x + 2 ! x + 2 x = ! t " # ( ) x 0 ( ) = 1, ! x 0 ( ) = 0 ! n = k m = 1.414 rad/s, " = c 2 km = 0.7071, ! d = ! n 1 # " 2 = 1 rad/s Total Solution: x t ( ) = x h t ( ) + x p t ( ) Homogeneous: From Equation (1.36) x h t ( ) = Ae ! "# n t sin # d t + \$ ( ) A = v 0 + "# n x 0 ( ) 2 + x 0 # d ( ) 2 # d 2 , \$ = tan ! 1 x 0 # d v 0 + "# n x 0 % & ( ) * = .785 rad + x h t ( ) = 1.414 e ! t sin t + .785 ( ) Particular: From Equation. (3.9) x p t ( ) = 1 m ! d e " #! n t " \$ ( ) sin ! d t " \$ ( ) = 1 1 ( ) 1 ( ) e " t " % ( ) sin t " % ( ) But, sin " t ( ) = " sin t So, x p t ( ) = " e " t " % ( ) sin t & x t ( ) = 1.414 e ! t sin t + 0.785 ( ) 0 < t < " x t ( ) = 1.414 e ! t sin t + 0.785 ( ) ! e ! ( t ! " ) sin t t > " This is plotted below using the Heaviside function.

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3- 3 3.2 Calculate the solution to !! x + 2 ! x + 3 x = sin t + ! t " # ( ) x 0 ( ) = 0 ! x 0 ( ) = 1 and plot the response. Solution: Given: !! x + 2 ! x + 3 x = sin t + ! t " # ( ) , x 0 ( ) = 0, ! x 0 ( ) = 0 ! n = k m = 1.732 rad/s, " = c 2 km = 0.5774, ! d = ! n 1 # " 2 = 1.414 rad/s Total Solution: x t ( ) = x h + x p 1 0 < t < ! x t ( ) = x h + x p 1 + x p 2 t > ! Homogeneous: Eq. (1.36) x h t ( ) = Ae ! "# n t sin # d t + \$ ( ) = Ae ! t sin 1.414 t + \$ ( ) Particular: #1 (Chapter 2) x p 1 ( t ) = X sin ! t " # ( ) , where ! = 1 rad/s . Note that f 0 = F 0 m = 1 \$ X = f 0 ! n 2 " ! 2 ( ) 2 + 2 %! n ! ( ) 2 = 0.3536, and # = tan " 1 2 %! n ! ! n 2 " ! 2 & ( ( ) * + + = 0.785 rad \$ x p 1 t ( ) = 0.3536sin t " 0.7854 ( ) Particular: #2 Equation 3.9 x p 2 t ( ) = 1 m ! d e " #! n t " \$ ( ) sin ! d t " % ( ) = 1 1 ( ) 1.414 ( ) e " t " \$ ( ) sin1.414 t " \$ ( ) & x p 2 t ( ) = 0.7071 e " t " \$ ( ) sin1.414 t " \$ ( ) The total solution for 0< t < π becomes: x t ( ) = Ae ! t sin 1.414 t + " ( ) + 0.3536sin t ! 0.7854 ( ) ! x t ( ) = ! Ae ! t sin(1.414 t + " ) + 1.414 Ae ! t cos 1.414 t + " ( ) + 0.3536cos t ! 0.7854 ( ) x 0 ( ) = 0 = A sin " ! 0.25 # A = 0.25 sin " ! x 0 ( ) = 1 = ! A sin " + 1.414 A cos " + 0.25 # 0.75 = 0.25 ! 1.414 0.25 ( ) 1 tan " # " = 0.34 and A = 0.75 Thus for the first time interval, the response is x t ( ) = 0.75 e ! t sin 1.414 t + 0.34 ( ) + 0.3536sin t ! 0.7854 ( ) 0 < t < " Next consider the application of the impulse at t = π :

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3- 4 x t ( ) = x h + x p 1 + x p 2 x t ( ) = ! 0.433 e ! t sin 1.414 t + 0.6155 ( ) + 0.3536sin t ! 0.7854 ( ) ! 0.7071 e ! t ! " ( ) sin 1.414 t ! " ( ) t > " The response is plotted in the following (from Mathcad): 3.3 Calculate the impulse response function for a critically damped system. Solution: The change in the velocity from an impulse is v 0 = ˆ F m , while x 0 = 0. So for a critically damped system, we have from Eqs. 1.45 and 1.46 with x 0 = 0: x ( t ) = v 0 te ! " n t # x ( t ) = ˆ F m te ! " n t