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HW4soln - ME 375 HOMEWORK 4 SOLUTIONS Spring 2009 Out...

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ME 375 HOMEWORK # 4 Spring 2009 SOLUTIONS Out: February 4, 2009 Due: February 11, 2009 (at the beginning of class) PROBLEM 1: (40%) Consider the following set of differential equations that represent an engine supported by elastomeric mounts and excited by a reciprocating force, f ( t ) (see Figure 1): = + + 0 ) ( 5 . 0 5 . 0 5 . 0 5 . 1 1 . 0 0 0 0 0 0 0 1 t f y x y x y x ± ± ± ± ± ± These equations describe an engine, which is isolated from its chassis by a mount that possesses frequency dependent damping and stiffness. Answer the following questions. Figure 1 (a) Solve for the Laplace transforms, X ( s ) and Y ( s ), of x ( t ) and y ( t ), respectively, in terms of the Laplace transform, F ( s ), of f ( t ) and the initial conditions of the system. (b) For f ( t ) = 0 N with initial conditions y (0) = 0 m, x (0) = 0 m, and 1 . 0 ) 0 ( = x ± m/s, solve for x ( t ). (c) For f ( t ) equal to 0.1 δ ( t ), where δ ( t ) is the unit impulse, and zero initial conditions, solve for x ( t ). How does this result compare to the answer for part (b)? K 1 = 1 N/m K 2 = 0.5 N/m C 2 = 0.1 Ns/m M = 1 kg x(t) y(t) f(t)
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2 (a) The Laplace transform is applied to the differential equations to solve part (a). Then the like terms are grouped on the left hand side of the equation so that X(s) and Y(s) can be found in terms of the Laplace transform of the input force F(s) and initial conditions. ( ) ( ) + + + + + + + = + + + + + + + = + + + + + + = + + + + = + + = + + + + = + + ) 0 ( 1 . 0 ) 0 ( ) 0 ( ) ( 5 . 1 5 . 0 5 . 0 5 . 0 1 . 0 5 5 . 1 5 10 ) ( ) ( ) 0 ( 1 . 0 ) 0 ( ) 0 ( ) ( 5 . 1 5 . 0 5 . 0 5 . 0 1 . 0 5 . 0 15 . 0 5 . 0 1 . 0 1 ) ( ) ( ) 0 ( 1 . 0 ) 0 ( ) 0 ( ) ( 5 . 1 5 . 0 5 . 0 5 . 0 1 . 0 25 . 0 5 . 0 1 . 0 5 . 1 1 ) ( ) ( ) 0 ( 1 . 0 ) 0 ( ) 0 ( ) ( 5 . 0 1 . 0 5 . 0 5 . 0 5 . 1 ) ( ) ( ) 0 ( 1 . 0 ) 0 ( ) 0 ( ) ( ) ( ) ( 5 . 0 1 . 0 5 . 0 5 . 0 5 . 1 ) 0 ( 1 . 0 ) 0 ( ) 0 ( ) ( ) ( ) ( 5 . 0 5 . 0 5 . 0 5 . 1 ) ( ) ( 1 . 0 0 0 0 ) ( ) ( 0 0 0 2 2 3 2 2 3 2 2 1 2 2 2 y x sx s F s s s s s s Y s X y x sx s F s s s s s s Y s X y x sx s F s s s s s Y s X y x sx s F s s s Y s X y x sx s F s Y s X s s y x sx s F s Y s X s Y s X s s Y s X s ± ± ± ± ± ± The denominator in this final equation is a third order polynomial. Its roots can be easily calculated using the Matlab roots command. The roots are given by s1=-4.9021, s2=-0.0490+1.0088j, and s3=- 0.0490-1.0088j. Given these roots of the Laplace transform denominator polynomial, we can immediately determine that the final solutions for x ( t ) and y ( t ) must contain a rapidly decaying exponential (s1) and an exponentially decaying sinusoid (s2 and s3). As you solve these Laplace transform problems, be sure to think about the preliminary steps like solving for the roots of the denominator polynomial because these steps provide insight into the ultimate form of the solution.
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