Vibrations in Mechanical Systems

Section 11.3 Vibrations in Mechanical Systems

Consider a spring that resists compression as well as extension, that is suspended vertically from a fixed support, and a body of mass \(m\) that is attached at the lower end of the spring. We make the assumption that the mass \(m\) is much larger than the mass of the spring so that we can neglect the mass of the spring. If there is no motion then the system is in static equilibrium, as illustrated in Figure 11.3.1. If the mass is pulled down further and released, then it will undergo an oscillatory motion.

Suppose there is no friction to slow down the motion of the mass, then we say that the system is undamped. We will determine the motion of this mechanical system by considering the forces acting on the mass during the motion. This will lead to a differential equation relating the displacement as a function of time. The most obvious force is that of gravitational attraction acting on the mass \(m\) and is given by

\begin{equation*} F_1=mg\text{,} \end{equation*}

where \(g\) is the acceleration of gravity. The next force to be considered is the spring force acting on the mass and is directed upward if the spring is stretched and downward if it is compressed. It obeys Hooke’s law

\begin{equation*} F_2=ks\text{,} \end{equation*}

where \(s\) is the amount the spring is stretched when \(s>0\) and is the amount it is compressed when \(s\lt 0\text{.}\)

When the system is in static equilibrium and the spring is stretched by the amount \(s_0\text{,}\) the resultant of the spring force and the gravitational force is zero, which is expressed by the equation

\begin{equation*} mg-ks_0=0\text{.} \end{equation*}

Let \(s=U(t)\) denote the displacement from static equilibrium with the positive \(s\) direction pointed downward as indicated in Figure 10.17(b).The spring force can be written as

\begin{equation*} F_2=-k[s_0+U(t) ] =-ks_0-kU(t)\text{,} \end{equation*}

and the resultant force \(F_{R}\) is

\begin{equation} F_{R}=F_1+F_2=mg-ks_0-kU(t) =-kU(t)\text{.}\tag{11.3.1} \end{equation}

The differential equation for motion is obtained by using Newton’s second law, which states that the resultant of the forces acting on the mass at any instant satisfies

\begin{equation} F_{R}=ma\text{.}\tag{11.3.2} \end{equation}

The distance from equilibrium at time \(t\) is measured by \(U(t)\text{,}\) so the acceleration \(a\) is given by \(a=U\,''(t)\text{.}\) Equations (11.3.1) and (11.3.2) then give

\begin{equation*} F_R = -kU(t) = mU\,''(t) \end{equation*}

Therefore, the undamped mechanical system is governed by the linear differential equation

\begin{equation*} mU\,''(t) + kU(t) = 0\text{.} \end{equation*}

The general solution to undamped system is known to be

\begin{equation*} U(t) = A\cos \omega t+B\sin \omega t, \text{ where } \omega = \sqrt{\frac{k}{m}} \end{equation*}

Subsection 11.3.1 Damped System

If we consider frictional forces that slow down the motion of the mass, then we say that the system is damped. This is visualized by connecting a dashpot to the mass, as indicated in Figure 11.3.2. For small velocities it is assumed that the frictional force \(F_3\) is proportional to the velocity, that is,

\begin{equation*} F_3=-cU\,'(t) \end{equation*}

Figure 11.3.2. The spring mass dashpot system

The damping constant \(c\) must be positive, for if \(U\,'(t)>0\text{,}\) then the mass is moving downward and hence \(F_3\) must point upward, which requires that \(F_3\) is negative. The result of the three forces acting on the mass is given by

\begin{equation*} F_1+F_2+F_3 = -kU(t)-cU\,'(t) = mU\,''(t) = F_R\text{,} \end{equation*}

so the damped mechanical system is governed by the differential equation

\begin{equation*} mU\,''(t) + cU\,'(t) + kU(t) = 0\text{.} \end{equation*}

Subsection 11.3.2 Forced Vibrations

The vibration discussed earlier are called free vibrations because all the forces that affect the motion of the system are internal to the system. We extend our analysis to cover the case in which an external force \(F_4=F(t)\) acts on the mass (see Figure 11.3.3). Such a force might occur from vibrations of the support to which the top of the spring is attached, or from the effect of a magnetic field on a mass made of iron. As before, we sum the forces \(F_1\text{,}\) \(F_2\text{,}\) \(F_3\text{,}\) and \(F_4\) and set this equal to the resultant force \(F_{R}\) and obtain

\begin{equation*} F_1+F_2+F_3+F_4=F_{R} = -KU(t) -cU\,'(t) + F(t) = mU\,''(t) \end{equation*}

Figure 11.3.3. The dashpot system with an external force

Thus, the forced motion of the mechanical system satisfies the nonhomogeneous linear differential equation

\begin{equation} mU\,''(t) + cU\,'(t) + kU(t) = F(t)\text{.}\tag{11.3.3} \end{equation}

The function \(F(t)\) is called the input or driving force and the solution \(U(t)\) is called the output or response. Of particular interest are periodic inputs \(F(t)\) that can be represented by Fourier series.

For damped mechanical systems that are driven by a periodic input \(F(t)\text{,}\) the general solution involves a transient part that vanishes as \(t \to +\infty\text{,}\) and a steady state part that is periodic. The transient part of the solution \(U_h(t)\) is found by solving the homogeneous differential equation

\begin{equation*} mU_h\,''(t) +cU_h\,'(t) + kU_h(t) = 0\text{.} \end{equation*}

This homogeneous equation has the characteristic equation \(m\lambda^2+c\lambda+k=0\text{,}\) and it’s roots are \(\lambda =\frac{-c\pm \sqrt{c^2-4mk}}{2m}\text{.}\) The coefficients \(m\text{,}\) \(c\text{,}\) and \(k\) are all positive, and there are three cases to consider.

Case.

If \(c^2-4mk>0\text{,}\) the roots are real and distinct, and since \(\sqrt{c^2-4mk}\lt c\text{,}\) it follows that the roots \(\lambda_1\) and \(\lambda_2\) are negative real numbers. Thus, for this case, we have

\begin{equation*} \lim_{t \to +\infty}U_h(t) = \lim_{t \to +\infty}(A_1e^{\lambda_1t}+A_2e^{\lambda_2t}) = 0\text{.} \end{equation*}

Case.

If \(c^2-4mk=0\text{,}\)the roots are real and equal, then \(\lambda_1=\lambda_2=\lambda\text{,}\) where \(\lambda\) is a negative real number. Again, for this case we find that

\begin{equation*} \lim_{t \to +\infty}U_h(t) =\lim_{t \to +\infty}(A_1e^{\lambda t}+A_2e^{\lambda t}) = 0\text{.} \end{equation*}

Case.

If \(c^2-4mk\lt 0\text{,}\) the roots are complex, then \(\lambda =-\alpha \pm \beta i\) , where \(\alpha\) and \(\beta\) are positive real numbers, and it follows that

\begin{equation*} \lim_{t \to +\infty}U_h(t) =\lim_{t \to +\infty}(A_1e^{-\alpha t}\cos \beta t+A_2e^{-\alpha t}\sin \beta t) = 0\text{.} \end{equation*}

In all three cases, we see that the homogeneous solution \(U_h(t)\) decays to \(0\) as \(t \to +\infty\text{.}\) The steady state solution \(U_p(t)\) can be obtained by representing \(U_p(t)\) by its Fourier series and substituting \(U_p\,''(t)\text{,}\) \(U_p\,'(t)\) , and \(U_p(t)\) into the nonhomogeneous differential equation and solving the resulting system for the Fourier coefficients of \(U_p(t)\text{.}\) The general solution to Equation (11.3.3) is then given by

\begin{equation*} U(t) = U_h(t) + U_p(t) \end{equation*}

Example 11.3.4.

Find the general solution to \(U\,''(t) + 2U\,'(t) + U(t) = F(t)\text{,}\) where \(F(t)\) is given by the Fourier series \(F(t)=\sum\limits_{n=1}^{\infty}\frac{1}{(2n-1)^2}\cos [(2n-1) t]\text{.}\)

Solution.

First we solve \(U_h\,''(t) + 2U_h\,'(t) + U_h(t) = 0\) for the transient solution. The characteristic equation is \(\lambda^2+2\lambda +1=0\text{,}\) which has a double root \(\lambda =-1\text{.}\) Hence

\begin{equation*} U_h(t) = A_1e^{-t} + A_2te^{-t}\text{.} \end{equation*}

The steady state solution is obtained by assuming that \(U_p(t)\) has the Fourier series representation

\begin{equation*} U_p(t) =\frac{a_0}{2}+\sum\limits_{n=1}^{\infty}a_n\cos nt+\sum\limits_{n=1}^{\infty}b_n\sin nt\text{,} \end{equation*}

and that \(U_h\,'(t)\) and \(U_h\,''(t)\) can be obtained by termwise differentiation:

\begin{align*} 2U_p\,'(t) \amp = 2\sum_{n=1}^{\infty}nb_n\cos nt-2\sum\limits_{n=1}^{\infty}na_n\sin nt, \text{ and }\\ U_p\,''(t) \amp = -\sum_{n=1}^{\infty}n^2a_n\cos nt-\sum\limits_{n=1}^{\infty}n^2b_n\sin nt\text{.} \end{align*}

Substituting these expansions into the differential equation results in

\begin{align*} F(t) \amp = \frac{a_0}{2} + \sum_{n=1}^{\infty}\big[(1-n^2) a_n+2nb_n\big] \cos nt\\ \amp +\sum\limits_{n=1}^{\infty}\big[-2na_n+(1-n^2) b_n\big] \sin nt\text{.} \end{align*}

Equating the coefficients with the given series for \(F(t)\text{,}\) we find that \(\frac{a_0}{2}=0\text{,}\) and that

\begin{align*} (1-n^2) a_n+2nb_n \amp = \begin{cases} \frac{1}{n^2} \amp \text{ when } n \text{ is odd } ,\\ 0 \amp \text{ when } n \text{ is even, } \end{cases}\\ -2na_n+(1-n^2) b_n \amp =0 \text{ for all } n\text{.} \end{align*}

Solving this linear system for \(a_n\) and \(b_n\text{,}\) we find that

\begin{align*} a_n \amp = \begin{cases} \frac{1-n^2}{n^2(1+n^2)^2} \amp \text{ for } n \text{ odd } ,\\ \;\; 0 \amp \text{ for } n \text{ even } , \end{cases}\\ b_n \amp = \begin{cases} \frac{2n}{n^2(1+n^2)^2} \amp \text{ for } n \text{ odd } ,\\ \;\; 0 \amp \text{ for } n \text{ even } . \end{cases} \end{align*}

The general solution is given by

\begin{align*} U(t) = A_1e^{-t}+A_2te^{-t} \amp + \sum_{n=1}^{\infty}\frac{1-(2n-1)^2}{(2n-1)^2[1+(2n-1)^2]^2}\cos [(2n-1) t]\\ \amp + \sum\limits_{n=1}^{\infty}\frac{2(2n-1)}{(2n-1)^2[1+(2n-1)^2]^2}\sin [(2n-1) t]\text{.} \end{align*}

Exercises Exercises

1.

For the exercises in this section solve for \(U(t)\) given the following Fourier series for \(F(t)\text{.}\)

(a)

\(F(t) = \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin(nt)\text{,}\) where \(F(t)=\frac{t}{2}\) for \(-\pi \lt t \lt \pi\text{.}\)

(b)

\(F(t) = \sum\limits_{n=1}^\infty \frac{\sin(\frac{n\pi}{2}}{n}\cos(nt)\text{,}\) where \(F(t) = \begin{cases}-\frac{\pi}{4}, \amp \text{ for } \frac{\pi}{2} \lt t \lt \pi, \\ +\frac{\pi}{4}, \amp \text{ for } -\frac{\pi}{2} \lt t \lt \frac{\pi}{2},\\ -\frac{\pi}{4}, \amp \text{ for } -\pi \lt t \lt -\frac{\pi}{2}. \end{cases}\)

(c)

\(F(t) = \sum\limits_{n=1}^\infty \frac{4\sin(\frac{n\pi}{2}}{n^2\pi}\sin(nt)\text{,}\) shown in Figure 11.3.5.

(d)

\(F(t) = \sum\limits_{n=1}^\infty \frac{1-(-1)^n}{2n^2}\cos(nt)\text{,}\) shown in Figure 11.3.6.

Figure 11.3.5. Graph of \(U(t)\) for the exercises

Figure 11.3.6. Graph of \(F(t)\) for the exercises

\begin{task} \(U\,''(t) + 2U\,'(t) + 2U(t) = F(t)\text{.}\)

\begin{task}[(a)]

Solution.

\(U_h(t)=c_1e^{-t}\sin(t) + c_2e^{-t}\cos(t)\text{,}\) \(U_p(t) = \sum\limits_{n=1}^\infty \frac{2(-1)^n}{4+n^4}\cos(nt) + \sum\limits_{n=1}^\infty \frac{(-1)^{n}(n^2-1)}{n(4+n^2)}\sin(nt)\text{,}\) \(U(t)= c_1e^{-t}\sin(t) + c_2e^{-t}\cos(t) + \sum\limits_{n=1}^\infty \frac{2(-1)^n}{4+n^4}\cos(nt) + \sum\limits_{n=1}^\infty \frac{(-1)^n(n^2-2)}{n(4+n^2)}\sin(nt)\text{.}\)

\begin{task}[(c)] \(U_h(t)=c_1e^{-t}\sin(t) + c_2e^{-t}\cos(t)\text{,}\) \(U_p(t) = -\sum\limits_{n=1}^\infty \frac{8\sin(\frac{n\pi}{2})}{n(4+n^2)}\cos(nt) + \sum\limits_{n=1}^\infty \frac{4\sin(\frac{n\pi}{2})(2-n^2)}{n^2(4+n^4)\pi}\sin(nt)\text{,}\) \(U(t)= c_1e^{-t}\sin(t) + c_2e^{-t}\cos(t) + \sum\limits_{n=1}^\infty \frac{2(-1)^n}{4+n^4}\cos(nt) + \sum\limits_{n=1}^\infty \frac{(-1)^n(n^2-2)}{n(4+n^2)}\sin(nt)\text{.}\)

\begin{task}[(c)] Alternative Answer: Alternative Answer: \(U_p(t) = -\sum\limits_{n=1}^\infty \frac{8(-1)^n}{(2n-1)(16n^4-32n^3+24n^20-8n+5)\pi}\cos\big((2n-1)t\big)\) \(+\sum\limits_{n=1}^\infty \frac{4(-1)^n(4n^2-4n-1)}{(2n-1)^2(16n^4-32n^3+24n^20-8n+5)\pi}\sin\big((2n-1)t\big)\text{,}\) \(U(t)=c_1e^{-t}\sin(t)+c_2e^{-t}\cos(t)+\sum\limits_{n=1}^\infty\frac{8(-1)^n}{(2n-1)16n^4-32n^3+24n^20-8n+5)\pi}\cos\big((2n-1)t\big)\) \(+\sum\limits_{n=1}^\infty \frac{4(-1)^n(4n^2-4n-1)}{(2n-1)^2(16n^4-32n^3+24n^20-8n+5)\pi}\sin\big((2n-1)t\big)\text{.}\)

2.

\(U\,''(t) + 3U\,'(t) + 2U(t) =F (t)\text{.}\)

3.

\(U\,''(t) + 4U\,'(t) + 4U(t) =F (t)\text{.}\)

\begin{task}[(a)]

Solution 1.

\(U_h(t)=c_1e^{-2t} + c_2te^{-2t}\text{,}\) \(U_p(t) = -\sum\limits_{n=1}^\infty \frac{4(-1)^n}{(4+n^2)^2}\cos(nt) + \sum\limits_{n=1}^\infty \frac{(-1)^n(n^2-4)^2}{n(4+n^2)^2}\sin(nt)\text{,}\) \(U(t)= c_1e^{-2t} + c_2te^{-2t} - \sum\limits_{n=1}^\infty \frac{4(-1)^n}{(4+n^2)^2}\cos(nt) + \sum\limits_{n=1}^\infty \frac{(-1)^n(n^2-4)}{n(4+n^2)^2}\sin(nt)\text{.}\)

Solution 2.

\(U_h(t)=c_1e^{-2t} + c_2te^{-2t}\text{,}\) \(U_p(t) = -\sum\limits_{n=1}^\infty \frac{16\sin(\frac{n\pi}{2})}{n(4+n^2)^2\pi}\cos(nt) + \sum\limits_{n=1}^\infty \frac{4\sin(\frac{n\pi}{2}(4-n^2))}{n^2(4+n^2)^2\pi}\sin(nt)\text{,}\) \(U(t) = c_1e^{-2t} + c_2te^{-2t} - \sum\limits_{n=1}^\infty \frac{16\sin(\frac{n\pi}{2})}{n(4+n^2)^2\pi}\cos(nt) + \sum\limits_{n=1}^\infty \frac{4\sin(\frac{n\pi}{2})(4-n^2)}{n^2(4+n^2)^2\pi}\sin(nt)\text{.}\)

\begin{task}[(c)] Alternative Answer: Alternative Answer: \(U_p(t) = -\sum\limits_{n=1}^\infty \frac{16(-1)^n}{2n-1)(4n^2-4n+5)^2\pi}\cos\big((2n-1)t\big) + \sum\limits_{n=1}^\infty \frac{4(-1)^n(4n^2-4n-3)}{4n^2-4n+5)^2\pi}\sin\big((2n-1)t\big)\text{,}\) \(U(t) = c_1e^{-2t} + c_2te^{-2t} + \sum\limits_{n=1}^\infty \frac{16(-1)^n}{2n-1)(4n^2-4n+5)^2\pi}\cos\big((2n-1)t\big)\) \(+ \sum\limits_{n=1}^\infty \frac{4(-1)^n(4n^2-4n-3)}{4n^2-4n+5)^2\pi}\sin\big((2n-1)t\big)\text{.}\)

Prev Top Next