# SolSec3_4 - 3 50 Problems and Solutions for Section...

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3- 50 Problems and Solutions for Section 3.4 (3.35 through 3.38) 3.35 Calculate the response of m !! x + c ! x + kx = F 0 ! ( t ) where Φ ( t ) is the unit step function for the case with x 0 = v 0 = 0. Use the Laplace transform method and assume that the system is underdamped. Solution: Given: m !! x + c ! x + kx = F 0 μ ( t ) !! x + 2 !" n ! x + " n 2 x = F 0 m μ ( t ) ( ! < 1) Take Laplace Transform: s 2 X ( s ) + 2 !" n sX ( s ) + " n 2 X ( s ) = F 0 m 1 s # \$ % & ( X ( s ) = F 0 / m s 2 + 2 !" n s + " n 2 ( ) s = F 0 m " n 2 # \$ % & ( " n 2 s s 2 + 2 !" n s + " n 2 ( ) Using inverse Laplace tables, x ( t ) = F 0 k ! F 0 k 1 ! " 2 e ! "# n t sin # n 1 ! " 2 t + cos ! 1 ( " ) ( )

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3- 51 3.36 Using the Laplace transform method, calculate the response of the system of Example 3.4.4 for the overdamped case ( ζ > 1). Plot the response for m = 1 kg, k = 100 N/m, and ζ = 1.5. Solution: From example 3.4.4, m !! x + c ! x + kx = ! ( t ) !! x + 2 "# n ! x + # n 2 x = 1 m ! ( t ) ( " > 1) Take Laplace Transform: s 2 X ( s ) + 2 !" n sX ( s ) + " n 2 X ( s ) = 1 m X ( s ) = 1/ m s 2 + 2 !" n s + " n 2 = 1/ m ( s + a )( s + b ) Using inverse Laplace tables, a = ! "# n + # n " 2 ! 1 , b = ! "# n ! # n " 2 ! 1 x ( t ) = e ! "# n t 2 m # n " 2 ! 1 e # n " 2 ! 1 t ! e ! # n " 2 ! 1 t \$ % & ( ) Inserting the given values yields: x ( t ) = e ! 15 t 22.36 e 11.18 t ! e ! 11.18 t " # \$ % m
3- 52 3.37 Calculate the response of the underdamped system given by m !! x + c ! x + kx = F 0 e ! at using the Laplace transform method. Assume a > 0 and that the initial conditions are all zero. Solution: Given: m !! x + c ! x + kx = F 0 e ! at a > 0, initial conditions = 0 Rewrite: !! x + 2 !" n ! x + " n 2 x = F 0 m e # at Take Laplace Transform:

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