Key_to_ME130_F08_Exam1

Key_to_ME130_F08_Exa - Dr Saxena ME130_F08 ME130 APPLIED ENGINEERING ANALYSIS Key to Exam#1 Fall 2008 Section 02#1 A 5-kg mass is hanging from a

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Dr. Saxena ME130_F’08 1 ME130 – APPLIED ENGINEERING ANALYSIS Key to Exam #1: Fall 2008 – Section 02: #1 A 5-kg mass is hanging from a spring with a spring constant K = 30 N/m, and there is a dash pot attached to the mass with a damping coefficient C = 25 Kg/sec. The initial conditions are: y(0) = 1 m, and y (0) = 0. (a) Derive the governing differential equation for the displacement y(t) of this mass (b) Solve the equation to determine the displacement y(t). (c) Calculate the percent reduction in the amplitude in 1 sec. Solution: (a) From the free body diagram F = F I R S R D = 0, where R D = Spring force = K y(t), K = spring constant R D = Viscous (damping) force = C dy/dt, C = damping coefficient F I = Inertia force = m d 2 y/dt 2 So, the governing differential equation becomes 2 2 dy M+ C + K y = 0 dt dt (Note: the weight is offset by initial stretch of the spring) (b) This is a second order homogeneous differential equation, with constant coefficients, with the initial conditions given as: y(0) = 1 m, and y’(0) = 0 The corresponding characteristic equation is: m 2 + am + b = 0, where a = C/M = 25/5 = 5, and b = K/M = 30/5 = 6. The two roots of this equation are: () ( ) 3 6 4 5 2 1 2 5 b 4 a 2 1 2 5 m 2 6 4 5 2 1 2 5 b 4 a 2 1 2 a m 2 2 2 2 2 1 = × = = = × + = + = Therefore, the general solution of this equation is expressed as t 3 t 2 t m t m e B e A e B e A ) t ( y 2 1 + = + = Using the given initial conditions, the two constants are A = 3, and B = 2. Therefore, the specific solution of the differential equation is t 3 t 2 e 2 e 3 ) t ( y =
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This note was uploaded on 09/08/2010 for the course ME 130 at San Jose State University .

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Key_to_ME130_F08_Exa - Dr Saxena ME130_F08 ME130 APPLIED ENGINEERING ANALYSIS Key to Exam#1 Fall 2008 Section 02#1 A 5-kg mass is hanging from a

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