# solution

Two similar mechanical systems are shown in Figure P39. In both cases the input is the displacement y(t) of the base and the spring constant is nonlinear, so the differential equations are nonlinear. Their equations of motion are, for part (a), and for part (b)

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Order Paper Now  The only difference between these systems is that the system in Figure P39a has an equation of motion containing the derivative of the input function y(t). A step function is difficult to use with a numerical solution method, especially when the input derivative is present, due to the discontinuity at t = 0. So we will model the unit-step input with the function y(t) = 1 Ă˘Ë†â€™ eĂ˘Ë†â€™t/ĂŹâ€ž. The parameter ĂŹâ€ž should be chosen to be small compared to the oscillation period and the time constant, both of which we do not know. We can make an estimate by using the characteristic roots of the linearmodel obtained by setting k2 = 0. Use the values m = 100, c = 600, k1 = 8000, and k2 = 24,000. Choose the parameter ĂŹâ€ž to be small compared to the period and time constant of the linear model with k2 = 0. Plot the solution x(t) for both systems on the same graph. Use zero initial conditions

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