Lab #2 - Two Degrees-of-Freedom OscillatorLast Updated: February 13, 2009INTRODUCTIONThe system illustrated in Figure (2.1) has two degrees-of-freedom. This means that twois the minimum number of coordinates necessary to uniquely specify the position of thesystem. The purpose of this laboratory is to introduce you to some of the properties oflinear vibrating systems with two or more degrees-of-freedom. You have already seen a onedegree-of-freedom vibrating system (the mass-spring-dashpot system) and should have somefamiliarity with the ideas of natural frequency and resonance. These ideas still apply to anundamped linear system with two or more degrees-of-freedom.The new idea for many degrees-of-freedom systems is the concept of modes (also called normalmodes). Each normal mode consists of a mode shape and corresponding natural frequency.The system will exhibit resonance if forced at one of its natural frequencies. The number ofmodes a system has is equal to the number of degrees-of-freedom. Thus the system belowhas two modes and two natural frequencies.k1 k2 k3M1 M2x1 x2Figure 2.1: A simple two-degree-of-freedom system.The primary goals of this laboratory are for you to learn the concept of normal modes in atwo degrees-of-freedom system – the simplest system which exhibits such modes. You willlearn this by experimentation and calculation.PRE-LAB QUESTIONSRead through the laboratory instructions and then answer the following questions:1. Are the number of degrees of freedom of a system and the number of its normal modesrelated? Explain.2. How can a normal mode be recognized physically?3. What do you expect to happen when you drive a system at one of its natural frequen-cies?3334 Lab #2 - Two Degrees-of-Freedom Oscillator4. Draw a free body diagram and derive the equations of motion for a three degrees-of-freedom system, with three different masses, four equal springs, and no forcing. Putthem in matrix form. Your result should resemble equation (2.4) except your matrixwill be 3x3 and you will have no f(t) term. Substitute in the normal mode solution(see (2.7)) to get an eigenvalue problem (see (2.5)). How would the eigenvalues andeigenvectors of your matrix relate to the mode shapes and natural frequencies?5. Using MATLAB, find the eigenvalues and eigenvectors of the following matrix andprint the results (HINT: Type help eig for assistance).[A] =1 22 1(2.1)NORMAL MODESA normal mode is a special type of vibration what occurs when all of the points in the systemare moving in simple harmonic motion. In addition, In a normal mode vibration all pointsmove with the same angular frequency ω and are exactly in-phase or exactly out of phase.An example on the following page (See 2.2) illustrates a normal mode vibration for a twodegrees-of-freedom-system. Note:• Both masses are moving in simple harmonic motion.• The system has a period of T = 4π (sec), and thus an angular frequency of ω =2πT=12.• When one mass is at its maximum displacement, the other is at its minimum displace-ment - thus the masses are totally out of phase.If we wanted to write out the equation of motion for this system, we would need a statevector x(t) with two elements x1(t) and x2(t) - one to represent the position of each mass asa function of time. That equation might look something like this:x(t) =x1(t)x2(t)=−22sin (12t) (2.2)Here, ω is the natural frequency of the normal mode (the same for all masses), and the vectorc=−22is its mode shape. The mode shape c tells you phase and relative amplitude ofmotion of each mass - here both masses have the same relative amplitude (|c1| = |c2|) thoughin general that is not the case. Since in t his example c1has the opposite sign of c2the twomasses are completely out of phase.TAM 203 Lab Manual 35M1 M2x2 =t = 0t = πt = 2πt = 3πt = 4πx1 = 0 0M1 M2M1 M2M1 M2M1 M2x1 = -2x2 = 2x2 =x1 = 0 0x2 =x1 = 0 0x2 = -2x1 = 20 −202 x1(t)x2(t)2ππ3π 4πxtFigure 2.2: A normal mode vibration of a two-degree-of-freedom system.36 Lab #2 - Two Degrees-of-Freedom OscillatorDERIVING THE EQUATIONS OF MOTIONWe will now derive the equations of motion for a driven two degrees-of-freedom system. Thediagram and physical setup are shown in Figures 2.3 and 2.5.k1 k2 k3M1 M2x1 x2x3Figure 2.3: Illustration of a coupled mass-spring system.Here, rather than having the rightmost spring attached to a fixed support, we have it attachedto a sinusoidally driven support whose position is x3(t). Do not be fooled into thinking thatx3counts as a degree of freedom - here we know how we are driving the system and so x3isa given. Look back over Lab 1 if you are confused about this point - we use the same trickthere to drive a one degree-of-freedom system. Now, we will draw the free-body diagram foreach mass and work out its equation of motion. To help get the signs right, assume thatM1M2k1 x1 k2 ( x2 - x1 )k2 ( x2 - x1 ) k3 ( x3 - x2 )Figure 2.4: The free-body diagrams for masses m1and m2.the displacements are all positive (i.e. to the right) with x1< x2< x3. This puts all of thesprings into tension relative to their equilibrium condition. The equations of motion for eachmass respectively (assuming equal spring constants) arek (x2− x1) − kx1= m1¨x1(2.3a)k (x3− x2) − k (x2− x1) = m2¨x2(2.3b)We can rewrite this in matrix form as¨x1¨x2=−2km1km1km2−2km2x1x2+0kx3(t)m2(2.4)or as¨x= [A] x + f(t) (2.5)TAM 203 Lab Manual 37Where the matrix [A] contains information about the system response to forcing and thevector f(t) contains information about the external forcing.SOLVING THE EQUATIONS OF MOTION USING NORMAL MODESTo make matters easier, let’s consider the case where there is no external forcing, thusf(t) = 0 and our equation of motion (2.5) reduces to:¨x= [A] x (2.6)Now we’ll look for the normal mode solutions of the system. Remember - a normal modevibration is when both masses are moving in simple harmonic motion with the same angularfrequency ω, but potentially different amplitudes of motion ci. Before we gave an exampleof a normal mode solution. Here is the general form of a normal mode solution for a twodegrees-of-freedom system:x(t) =x1(t)x2(t)=c1c2(A cos (ωt) + B sin (ωt)) (2.7)Once again, ω is the natural frequency of the mode which tells you the angular frequencywith which every mass vibrates, and cis the mode shape which tells you the phase andrelative amplitude of motion of each mass. If we plug in our ansatz