in_class_prob_1_solution

# in_class_prob_1_solution - b = 0 The roots of this equation...

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ME 130: Applied Engineering Analysis (02) Fall 2008 In-class Problem # 1: Solution #1. A damped spring-mass system involves a mass of 30 kg, a spring with K = 2000 N/m, and a dashpot with C = 300 kg/s. (a) Write down the governing differential equation for the motion, y(t), of this system. (a) Determine a general solution for the motion of this system. (b) Determine the time between consecutive maximum amplitudes. Solution (a) As per the problem, it’s the case of damped free vibration. The governing differential equation is 67 . 66 = 30 2000 = b , 10 = 30 300 = a , Where 0 = y b + t d y d a + t d y d 0 = y m K + t d y d m C + t d y d 0 = y K + t d y d C + t d y d m 2 2 2 2 2 2 (b) For the above equation, the characteristic equation is: m 2 + a.m
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Unformatted text preview: + b = 0. The roots of this equation are: m 1 = - a/2 + 1/2x(a 2- 4b) 1/2 = - 10/2 + 1/2 x(10 2- 4 x 66.67) 1/2 = -5 + 6.46 i, and m 2 = -a/2 - 1/2x(a 2 – 4b) 1/2 = - 10/2 - 1/2x(10 2- 4x66.67) 1/2 = - 5 – 6.46 i. So, the general solution can be expressed as y(t) = e-5t x [ A Sin {6.46t} + B Cos (6.46t) ], where natural frequency of oscillation of the system is ω o = 6.46 rad/sec. Would need two initial conditions to solve for the constants A and B. (c) The time between two consecutive peaks is simply, ∆ t = 1/f = 1/( ω o /2 π ) = 2 π / ω o = 2 π /6.46 = 0.973 sec . C m K y(t)...
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## This note was uploaded on 09/08/2010 for the course ME 130 at San Jose State.

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