Forced Vibrations

The last case we consider is when an external force acts on the system. In the case of the motorcycle suspension system, for example, the bumps in the road act as an external force acting on the system. Another example is a spring hanging from a support; if the support is set in motion, that motion would be considered an external force on the system.

We model these forced systems with the nonhomogeneous differential equation:

[latex]mx^{\prime\prime}+bx^\prime+kx=f(t)[/latex],

where the external force is represented by the [latex]f(t)[/latex] term. As we saw earlier, differential equations such as this have solutions of the form:

[latex]x(t)=c_1x_1(t)+c_2x_2(t)+x_p(t)[/latex],

where [latex]c_1x_1(t)+c_2x_2(t)[/latex] is the general solution to the complementary equation and [latex]x_p(t)[/latex] is a particular solution to the nonhomogeneous equation.

If the system is damped, [latex]\underset{t\to\infty}{\text{lim}}(c_1x_1(t)+c_2x_2(t))=0[/latex]. Since these terms do not affect the long-term behavior of the system, we call this part of the solution the transient solution. The long-term behavior of the system is determined by [latex]x_p(t)[/latex], so we call this part of the solution the steady-state solution.

transient and steady-state solutions

For a damped forced system with solution [latex]x(t)=c_1x_1(t)+c_2x_2(t)+x_p(t)[/latex]:

• The transient solution is [latex]c_1x_1(t)+c_2x_2(t)[/latex], which approaches zero as [latex]t\to\infty[/latex]
• The steady-state solution is [latex]x_p(t)[/latex], which determines the long-term behavior

A mass of [latex]1[/latex] slug stretches a spring [latex]2[/latex] ft and comes to rest at equilibrium. The system is attached to a dashpot that imparts a damping force equal to eight times the instantaneous velocity of the mass. Find the equation of motion if an external force equal to [latex]f(t)=8\sin(4t)[/latex] is applied to the system beginning at time [latex]t=0[/latex]. What is the transient solution? What is the steady-state solution?

Show Solution

We have [latex]mg=1(32)=2k[/latex], so [latex]k=16[/latex] and the differential equation is

[latex]x^{\prime\prime}+8^\prime+16x=8\sin(4t)[/latex].

The general solution to the complementary equation is

[latex]c_1e^{-4t}+c_2te^{-4t}[/latex].

Assuming a particular solution of the form [latex]x_p(t)=A\cos(4t)+B\sin(4t)[/latex] and using the method of undetermined coefficients, we find [latex]x_p(t)=-\frac14\cos(4t)[/latex], so

[latex]x(t)=c_1e^{-4t}+c_2te^{-4t}-\frac14\cos(4t)[/latex].

At [latex]t=0[/latex], the mass is at rest in the equilibrium position, so [latex]x(0)=x^\prime(0)=0[/latex]. Applying these initial conditions to solve for [latex]c_1[/latex] and [latex]c_2[/latex], we get

[latex]x(t)=\frac14e^{-4t}+te^{-4t}-\frac14\cos(4t)[/latex].

The transient solution is [latex]\frac14e^{-4t}+te^{-4t}[/latex]. The steady-state solution is [latex]-\frac14\cos(4t)[/latex].