Vibration in Systems with Two Degrees of Freedom

From ControlTheoryPro.com

Jump to: navigation, search
Symbol.gif
Vibration in Systems with Two Degrees of Freedom
Green carrot left.gif
Green carrot.jpg
In order to prevent spam, users must register before they can edit or create articles.


See Also



Contents

1 Introduction to Vibration in Systems with Two Degrees of Freedom

Figure 1: Linear Undamped System with Two Degrees of Freedom
Figure 2: Linear Undamped System with Two Degrees of Freedom Free Body Diagram, (a) Applied Forces; (b) Effective Forces

Many real world systems can be modeled with single degree of freedom systems. Either 1 system or multiple systems under the principle of superposition. In particular MIMO systems can often be decoupled into multiple SISO systems.

However, not all systems can be adequately modeled with a single degree of freedom so in this article we work through an example system with two degrees of freedom. The Beard text book works this example through the differential equations. A much simpler approach for the controls engineer is to skip the full blown solution to the differential equation. We skip the differential equation solution and all of that math by creating State Space equations directly from the equations of motion.

2 Equations of Motion for Vibration in Systems with Two Degrees of Freedom[1]

A system model with two degrees of freedom is depicted in Figure 1. The two masses are connected by three springs to two walls and each other. There is no damping in the system.

If we consider the case where x1 > x2 then the free body diagrams become those seen in Figure 2. The equations of motion are

LaTeX: m_{1}\ddot{x}_{1}=-k_{1}x_{1}-k\left(x_{1}-x_{2} \right) for body 1



LaTeX: m_{2}\ddot{x}_{2}=k\left(x_{1}-x_{2} \right)-k_{2}x_{2} for body 2


The same equations are obtained for x2 > x1. In this case the direction of the central spring force is reversed.

As with all differential equations, we first assume a solution form. The above equations can be solved for the natural frequency and corresponding mode shapes by assuming a solution of the form

LaTeX: x_{1}=A_{1}\mbox{ sin}\left( \omega t + \psi \right) and LaTeX: x_{2}=A_{2}\mbox{ sin}\left( \omega t + \psi \right)


Assumptions:

  1. x1 and x2 oscillate with the same frequency ω and are either in phase or π out of phase

This is a sufficient condition to make ω a natural frequency.

Substituting these solutions into the equations of motion we get

LaTeX: -m_{1}A_{1}\omega^2\mbox{ sin} \left( \omega t + \psi \right)=-k_{1}A_{1}\mbox{ sin} \left( \omega t + \psi \right)-k\left( A_{1}-A_{2} \right)\mbox{ sin} \left( \omega t + \psi \right)


and

LaTeX: -m_{2}A_{2}\omega^2\mbox{ sin} \left( \omega t + \psi \right)=k\left( A_{1}-A_{2} \right)\mbox{ sin} \left( \omega t + \psi \right)-k_{2}A_{2}\mbox{ sin} \left( \omega t + \psi \right)-


Since these solutions are true for all values on t,

LaTeX: A_{1}\left(k+k_{1}-m_{1}\omega^2 \right)+A_{2}\left( -k \right)=0


and

LaTeX: A_{1}\left(-k \right)+A_{2}\left( k_{2}+k-m_{2}\omega^2 \right)=0


A1 and A2 can be eliminated by writing

LaTeX: \begin{vmatrix} k+k_{1}-m_{1}\omega^2 & -k \\ -k & k+k_{2}-m_{2}\omega^2 \end{vmatrix} = 0 Eqn. 1


This is the characteristic or frequency equation. Alternatively, we may write

LaTeX: \frac{A_{1}}{A_{2}}=\frac{k}{k+k_{1}-m_{1}\omega^2} Eqn. 2a



LaTeX: \frac{A_{1}}{A_{2}}=\frac{k_{2}+k-m_{2}\omega^2}{k} Eqn. 2b


Thus

LaTeX: \frac{k}{k+k_{1}-m_{1}\omega^2}=\frac{k_{2}+k-m_{2}\omega^2}{k}


and

LaTeX: \left( k+k_{1}-m_{1}\omega^2 \right) \left( k_{2}+k-m_{2}\omega^2 \right)-k^2=0 Eqn. 3


This result is the frequency equation and could also be obtained by multiplying out the above determinant (Eqn. (1)).

The solutions to Eqn. (3) give the natural frequencies of free vibration for the system in Figure 1. The corresponding mode shapes are found by substituting these frequencies, in turn, into either Eqn. 2a or 2b.

Consider the case when k1=k2=k, and m1=m2=m. The frequency equation is

LaTeX: \left( 2k-m\omega^2 \right)^2-k^2=0


expanding leads to

LaTeX: m^2\omega^4 - 4mk\omega^2 + 3k^2=0 or LaTeX: \left( m\omega^2 - k \right)\left( m\omega^2 - 3k \right) = 0


Therefore, either

LaTeX: m\omega^2-k=0 or LaTeX: m\omega^2-3k=0


Thus

LaTeX: \omega_{1}=\sqrt{\frac{k}{m}} rad/s and LaTeX: \omega_{2}=\sqrt{\frac{3k}{m}} rad/s


If

LaTeX: \omega=\sqrt{\frac{k}{m}} then LaTeX: \left( \frac{A_{1}}{A_{2}} \right)=+1



LaTeX: \omega=\sqrt{\frac{3k}{m}} then LaTeX: \left( \frac{A_{1}}{A_{2}} \right)=-1


This gives the mode shapes corresponding to the frequencies ω1 and ω2. Thus, the first mode of free vibration occurs at a frequency

LaTeX: \left( \frac{1}{2\pi} \right)\sqrt{\frac{k}{m}} Hz and LaTeX: \left( \frac{A_{1}}{A_{2}} \right)^I = 1


this is, the bodies move in phase with each other and with the same amplitude as if connected by a rigid link. The second mode of free vibration occurs at a frequency

LaTeX: \left( \frac{1}{2\pi} \right)\sqrt{\frac{3k}{m}} Hz and LaTeX: \left( \frac{A_{1}}{A_{2}} \right)^{II} = -1


that is, the bodies move exactly out of phase with each other but with the same amplitude.

2.1 State Space Equations for Systems with Two Degrees of Freedom

If we revisit the original equations of motion for this system with two degrees of freedom

LaTeX: m_{1}\ddot{x}_{1}=-k_{1}x_{1}-k\left(x_{1}-x_{2} \right) for body 1



LaTeX: m_{2}\ddot{x}_{2}=k\left(x_{1}-x_{2} \right)-k_{2}x_{2} for body 2


then we can form State Space equations directly from these equations.

For body 1 the equation of motion for this system with two degrees of freedom

LaTeX: \begin{align} \ddot{x}_{1} & = \frac{-k_{1}}{m_{1}}x_{1} + \frac{-k}{m_{1}}x_{1} + \frac{k}{m_{1}}x_{2} \\ & = \left( \frac{-k_{1}}{m_{1}} + \frac{-k}{m_{1}} \right)x_{1} + \frac{k}{m_{1}}x_{2} \end{align}


For body 1 the equation of motion for this system with two degrees of freedom

LaTeX: \begin{align} \ddot{x}_{2} & = \frac{k}{m_{2}}x_{1} - \frac{k}{m_{2}}x_{2} + \frac{-k_{2}}{m_{2}}x_{2} \\ & = \frac{k}{m_{2}}x_{1} + \left( \frac{-k}{m_{2}} + \frac{-k_{2}}{m_{2}} \right)x_{2} \end{align}


The state space equations become

LaTeX: \begin{Bmatrix} \ddot{x_{1}} \\ \dot{x_{1}} \\ \ddot{x_{2}} \\ \dot{x_{2}} \end{Bmatrix} = \begin{bmatrix} 0 & \frac{-k_{1}-k}{m_{1}} & 0 & \frac{k}{m_{1}} \\ 1 & 0 & 0 & 0 \\ 0 & \frac{k}{m_{2}} & 0 & \frac{-k-k_{2}}{m_{2}} \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{Bmatrix} \dot{x_{1}} \\ x_{1} \\ \dot{x_{2}} \\ x_{2} \end{Bmatrix} + Bu



LaTeX: y=Cx + Du


where:

B, C, and D are unknown at this stage.

2.1.1 Most likely values for the rest of the State Space equations

If we assume a 1 dimensional force which is positive with positive x1 acting on m1 and an analogous force on x2 then

LaTeX: B=\begin{bmatrix} \frac{1}{m_{1}} & 0 \\ 0 & 0 \\ 0 & \frac{1}{m_{2}} \\ 0 & 0 \end{bmatrix}


and

LaTeX: u=\begin{Bmatrix} F_{1} \\ F_{2} \end{Bmatrix}



If we then also assume that the most likely desired outputs are x1 and x2 then

LaTeX: D=0



LaTeX: C=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}


Therefore

LaTeX: y=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{Bmatrix} \dot{x_{1}} \\ x_{1} \\ \dot{x_{2}} \\ x_{2} \end{Bmatrix} = \begin{Bmatrix} x_{1} \\ x_{2} \end{Bmatrix}



3 Free Motion[2]

The two modes of vibration can be written

LaTeX: \begin{Bmatrix} x_{1} \\ x_{2} \end{Bmatrix}^{I} = \begin{Bmatrix} A_{1} \\ A_{2} \end{Bmatrix}^{I}\mbox{ sin} \left( \omega_{1} t + \psi_{1} \right)


and

LaTeX: \begin{Bmatrix} x_{1} \\ x_{2} \end{Bmatrix}^{II} = \begin{Bmatrix} A_{1} \\ A_{2} \end{Bmatrix}^{II}\mbox{ sin} \left( \omega_{2} t + \psi_{2} \right)


where the ratio A1/A2 is specified for each mode.

Since each solution satisfies the equation of motion, the general solution is

LaTeX: \begin{Bmatrix} x_{1} \\ x_{2} \end{Bmatrix} = \begin{Bmatrix} A_{1} \\ A_{2} \end{Bmatrix}^{I}\mbox{ sin} \left( \omega_{1} t + \psi_{1} \right) + \begin{Bmatrix} A_{1} \\ A_{2} \end{Bmatrix}^{II}\mbox{ sin} \left( \omega_{2} t + \psi_{2} \right)


where A1, A2, ψ1, and ψ2 are found from the initial conditions.

For example, for the system considered above, if one body is displaced a distance X and released,

LaTeX: x_{1}\left( 0 \right)=X and LaTeX: x_{2} \left( 0 \right)=\dot{x}_{1} \left( 0 \right)=\dot{x}_{2} \left( 0 \right)=0


Remembering that in this system

LaTeX: \omega_{1}=\sqrt{\frac{k}{m}},
LaTeX: \omega_{2}=\sqrt{\frac{3k}{m}},
LaTeX: \left( \frac{A_{1}}{A_{2}} \right)_{\omega_{1}}=+1, and
LaTeX: \left( \frac{A_{1}}{A_{2}} \right)_{\omega_{2}}=-1

we can write

LaTeX: x_{1}=\mbox{sin}\left( \sqrt{\frac{k}{m}}t+\psi_{1} \right)+\mbox{sin}\left( \sqrt{\frac{3k}{m}}t+\psi_{2} \right)


and

LaTeX: x_{2}=\mbox{sin}\left( \sqrt{\frac{k}{m}}t+\psi_{1} \right)-\mbox{sin}\left( \sqrt{\frac{3k}{m}}t+\psi_{2} \right)


Substituting the initial conditions x1(0) = X and x2(0) = 0 gives

LaTeX: X=\mbox{sin}\psi_{1} + \mbox{sin}\psi_{2}


and

LaTeX: 0=\mbox{sin}\psi_{1} - \mbox{sin}\psi_{2}


that is

LaTeX: \mbox{sin}\psi_{1}=\mbox{sin}\psi_{2}=\frac{X}{2}


The remaining conditions give cos ψ1 = cos ψ2 = 0.

Hence

LaTeX: x_{1}=\left( \frac{X}{2} \right)\mbox{ cos}\sqrt{\frac{k}{m}}t + \left( \frac{X}{2} \right)\mbox{ cos}\sqrt{\frac{3k}{m}}t


and

LaTeX: x_{2}=\left( \frac{X}{2} \right)\mbox{ cos}\sqrt{\frac{k}{m}}t - \left( \frac{X}{2} \right)\mbox{ cos}\sqrt{\frac{3k}{m}}t


That is, both natural frequencies are excited and the motion of each body has two harmoinc components.

4 Notes

Beards, C. F. 1995 Engineering Vibration Analysis with Applications to Control Systems. ISBN 034063183X

4.1 References

  1. Beards, pp. 92-94
  2. Beards, pp. 94-95