Question

The motion of a mass-spring system, in a certain system of units, is governed by the differentia equation

2 y''+4 y'+10 y=40 sin 2 t
with initial condition y(0)=-0.375, y'(0)=1.5. Calculate the steady-state solution.

Answer 1

The steady-state solution for the given mass-spring system is y(t) = -0.15 sin(2t) - 0.375 cos(2t) + 0.3 sin(2t) - 0.75 cos(2t).

Step-by-step explanation:

To find the steady-state solution, we consider the particular solution of the differential equation. The given non-homogeneous differential equation is 2y'' + 4y' + 10y = 40 sin(2t). To solve for the particular solution, we assume a sinusoidal form for y_p(t), given by y_p(t) = A sin(2t) + B cos(2t). Substituting this into the differential equation, we find the values of A and B. After solving, we get A = 0.3 and B = -0.75.

Now, the steady-state solution is the sum of the homogeneous and particular solutions. The homogeneous solution is obtained by setting the right-hand side of the differential equation to zero. The characteristic equation is given by 2r^2 + 4r + 10 = 0, and its roots are complex conjugates with real part -1. Thus, the homogeneous solution is of the form y_h(t) = C_1 e^(-t) cos(3t) + C_2 e^(-t) sin(3t). Applying the initial conditions y(0) = -0.375 and y'(0) = 1.5, we find C_1 = -0.15 and C_2 = 0.

Therefore, the complete steady-state solution is y(t) = y_h(t) + y_p(t) = -0.15 e^(-t) cos(3t) + 0.3 sin(2t) - 0.375 e^(-t) sin(3t) - 0.75 cos(2t).