Forced Mechanical Vibrations

The model and solution

We consider the damped, mass-spring system with a sinusoidal external force:

where $F_0,\gamma >0$$F_0,\gamma >0$.

By using the method of undetermined coefficients, one can shown that

where the complementary solution, $x_c(t)$$x_c(t)$, is the solution of the corresponding homogeneous equation.

In real-world damped systems, we must have $x_c(t)\to 0$$x_c(t) \to 0$ as $t\to \mathrm{\infty }$$t \to \infty$, and therefore, $x_c(t)$$x_c(t)$ is called the transient part of the solution. The transient part depends only on the parameters of the system ($m$$m$, $b$$b$, and $k$$k$) and the initial conditions. The other term of the solution is called the steady-state part or the steady-periodic part of the solution.

The expression

is called the gain factor. Notice that the gain factor depends on the system parameters and the frequency of the external force.

Practical resonance

Let's find the maximum value of $G(\gamma )$$G(\gamma)$.

Upon differentiating, we find that

and $G^{\prime }(\gamma )=0$$G'(\gamma)=0$ if and only if

We use ${\gamma }_r$$\gamma_r$ to denote the frequency on the right.

Case 1

If the system is over-damped ($b^2>4mk$$b^2>4mk$) or critically damped ($b^2=4mk$$b^2=4mk$), then ${\gamma }_r$$\gamma_r$ is imaginary, and we must have the maximum $G(\gamma )$$G(\gamma)$ when $\gamma =0$$\gamma=0$. In this case, the maximum gain is $1/k$$1/k$.

Case 2

If the system is under-damped ($b^2<4mk$$b^2<4mk$), we have the maximum $G(\gamma )$$G(\gamma)$ when $\gamma ={\gamma }_r$$\gamma=\gamma_r$. In this case, the maximum gain is

and ${\gamma }_r$$\gamma_r$ is called the resonance frequency. For undamped systems, maximum gain occurs when the frequency of the external force is equal to the resonance frequency. Notice that as $b\to 0$$b \to 0$, the gain increases without bound and ${\gamma }_r\to \sqrt{k/m}$$\gamma_r \to \sqrt{k/m}$ ($\sqrt{k/m}=\omega$$\sqrt{k/m}=\omega$ is the natural frequency of the system).

Example

Solve: $10x^{″}+3x^{\prime }+49x=20\cos 4t,x(0)=5,x^{\prime }(0)=0$$10x''+3x'+49x=20 \cos 4t, \quad x(0)=5, \, x'(0)=0$

Here is the Sage code.

t=var("t")
x=function("x")(t)
de=10*diff(x,t,2)+3*diff(x,t)+49*x==20*cos(4*t)
desolve(de,x,[0,5,0])

The solution is

which can be written in the approximate form

The first term is the transient part. Notice that it decays with time. The second term is the steady-state part. It dominates the solution for large $t$$t$.

The graph of $x(t)$$x(t)$ is shown here.

In this example, the resonance frequency is $\sqrt{971/200}\approx 2.203$$\sqrt{971/200} \approx 2.203$, which is not particularly close to the frequency of the external force ($\gamma =4$$\gamma=4$). The maximum gain is rather small.