Normal modes - Harvard University

1 Chapter 2 Normal modesDavid Morin, Chapter 1 we dealt with the oscillations of one mass. We saw that there were variouspossible motions, depending on what was influencing the mass (spring, damping, drivingforces). In this chapter we ll look at oscillations (generally without damping or driving)involving more than one object. Roughly speaking, our counting of the number of masseswill proceed as: two, then three, then infinity. The infinite case is relevant to a continuoussystem, because such a system contains (ignoring the atomic nature of matter) an infinitenumber of infinitesimally small pieces. This is therefore the chapter in which we will makethe transition from the oscillations of one particle to the oscillations of a continuous object,that is, to outline of this chapter is as follows. In Section we solve the problem of twomasses connected by springs to each other and to two walls. We will solve this in two ways a quick way and then a longer but more fail-safe way.

2 We encounter the important conceptsofnormal modesandnormal coordinates. We then add on driving and damping forces andapply some results from Chapter 1. In Section we move up a step and solve the analogousproblem involving three masses. In Section we solve the general problem involvingNmasses and show that the results reduce properly to the ones we already obtained in theN= 2 andN= 3 cases. In Section we take theN limit (which correspondsto a continuous stretchable material) and derive the all-importantwave equation. We thendiscuss what the possible waves can look Two massesFor a single mass on a spring, there is one natural frequency, namely k/m. (We ll considerundamped and undriven motion for now.) Let s see what happens if we have two equalmasses and three spring arranged as shown in Fig. 1. The two outside spring constantsm mk kFigure 1are the same, but we ll allow the middle one to be different.

3 In general, all three springconstants could be different, but the math gets messy in that the displacements of the left and right masses from their respectiveequilibrium positions. We can assume that all of the springs are unstretched at equilibrium,but we don t actually have to, because the spring force is linear (see Problem [to be added]).The middle spring is stretched (or compressed) byx2 x1, so theF=maequations on the12 CHAPTER 2. Normal MODEStwo masses arem x1= kx1 (x1 x2),m x2= kx2 (x2 x1).(1)Concerning the signs of the terms here, they are equal and opposite, as dictated byNewton s third law, so they are either both right or both wrong. They are indeed bothright, as can be seen by taking the limit of, say, largex2. The force on the left mass is thenin the positive direction, which is twoF=maequations are coupled, in the sense that bothx1andx2appear inboth equations. How do we go about solving forx1(t) andx2(t)?

4 There are (at least) twoways we can do First methodThis first method is quick, but it works only for simple systems with a sufficient amount ofsymmetry. The main goal in this method is to combine theF=maequations in well-chosenways so thatx1andx2appear only in certain unique combinations. It sometimes involves abit of guesswork to determine what these well-chosen ways are. But in the present problem,the simplest thing to do is add theF=maequations in Eq. (1), and it turns out that thisis in fact one of the two useful combinations to form. The sum yieldsm( x1+ x2) = k(x1+x2) = d2dt2(x1+x2) = km(x1+x2).(2)The variablesx1andx2appear here only in the unique combination,x1+x2. And further-more, this equation is simply a harmonic-motion equation for the quantityx1+x2. Thesolution is thereforex1(t) +x2(t) = 2 Ascos( st+ s),where s km(3)The s here stands for slow, to be distinguished from the fast frequency we ll findbelow.

5 And we ve defined the coefficient to be 2 Asso that we won t have a bunch of factorsof 1/2 in our final answer in Eq. (6) matter what complicated motion the masses are doing, the quantityx1+x2alwaysundergoes simple harmonic motion with frequency s. This is by no means obvious if youlook at two masses bouncing back and forth in an arbitrary other useful combination of theF=maequations is their difference, which conve-niently is probably the next thing you might try. This yieldsm( x1 x2) = (k+ 2 )(x1 x2) = d2dt2(x1 x2) = k+ 2 m(x1 x2).(4)The variablesx1andx2now appear only in the unique combination,x1 x2. And again,we have a harmonic-motion equation for the quantityx1 x2. So the solution is (the f stands for fast )x1(t) x2(t) = 2 Afcos( ft+ f),where f k+ 2 m(5)As above, no matter what complicated motion the masses are doing, the quantityx1 x2always undergoes simple harmonic motion with frequency TWO MASSES3We can now solve forx1(t) andx2(t) by adding and subtracting Eqs.

6 (3) and (5). Theresult isx1(t) =Ascos( st+ s) +Afcos( ft+ f),x2(t) =Ascos( st+ s) Afcos( ft+ f).(6)The four constants,As,Af, s, fare determined by the four initial conditions,x1(0),x2(0), x1(0), x1(0).The above method will clearly work only if we re able to guess the proper combinations oftheF=maequations that yield equations involving unique combinations of the and subtracting the equations worked fine here, but for more complicated systemswith unequal masses or with all the spring constants different, the appropriate combinationof the equations might be far from obvious. And there is no guarantee that guessing aroundwill get you anywhere. So before discussing the features of the solution in Eq. (6), let s takea look at the other more systematic and fail-safe method of solving Second methodThis method is longer, but it works (in theory) for any setup. Our strategy will be to lookfor simple kinds of motions where both masses move with the same frequency.

7 We willthen build up the most general solution from these simple motions. For all we know, suchmotions might not even exist, but we have nothing to lose by trying to find them. We willfind that they do in fact exist. You might want to try to guess now what they are for ourtwo-mass system, but it isn t necessary to know what they look like before undertaking s guess solutions of the formx1(t) =A1ei tandx2(t) =A2ei t. For bookkeepingpurposes, it is convenient to write these solutions in vector form:(x1(t)x2(t))=(A1A2)ei t.(7)We ll end up taking the real part in the end. We can alternatively guess the solutione twithout thei, but then our will come out to be imaginary. Either choice will get the jobdone. Plugging these guesses into theF=maequations in Eq. (1), and canceling the factorofei t, yields m 2A1= kA1 (A1 A2), m 2A2= kA2 (A2 A1).(8)In matrix form, this can be written as( m 2+k+ m 2+k+ )(A1A2)=(00).

8 (9)At this point, it seems like we can multiply both sides of this equation by the inverse ofthe matrix. This leads to (A1, A2) = (0,0). This is obviously a solution (the masses justsit there), but we re looking for a nontrivial solution that actually contains some only way to escape the preceding conclusion thatA1andA2must both be zero is ifthe inverse of the matrix doesn t exist. Now, matrix inverses are somewhat messy things(involving cofactors and determinants), but for the present purposes, the only fact we need toknow about them is that they involve dividing by the determinant. So if the determinant is4 CHAPTER 2. Normal modes zero, then the inverse doesn t exist. This is therefore what we want. Setting the determinantequal to zero gives the quartic equation, m 2+k+ m 2+k+ = 0 = ( m 2+k+ )2 2= 0= m 2+k+ = = 2=kmork+ 2 m.(10)The four solutions to the quartic equation are therefore = k/mand = (k+ 2 ) the case where 2=k/m, we can plug this value of 2back into Eq.

9 (9) to obtain (1 1 1 1)(A1A2)=(00).(11)Both rows of this equation yield the same result (this was the point of setting the determinantequal to zero), namelyA1=A2. So (A1, A2) is proportional to the vector (1,1).For the case where 2= (k+ 2 )/m, Eq. (9) gives ( 1 1 1 1)(A1A2)=(00).(12)Both rows now yieldA1= A2. So (A1, A2) is proportional to the vector (1, 1).With s k/mand f (k+ 2 )/m, we can write the general solution as the sumof the four solutions we have found. In vector notation,x1(t) andx2(t) are given by(x1(t)x2(t))=C1(11)ei st+C2(11)e i st+C3(1 1)ei ft+C4(1 1)e i ft.(13)We now perform the usual step of invoking the fact that the positionsx1(t) andx2(t)must be real for allt. This yields that standard result thatC1=C 2 (As/2)ei sandC3=C 4 (Af/2)ei f. We have included the factors of 1/2 in these definitions so that wewon t have a bunch of factors of 1/2 in our final answer. The imaginary parts in Eq.

10 (13)cancel, and we obtain(x1(t)x2(t))=As(11)cos( st+ s) +Af(1 1)cos( ft+ f)(14)Therefore,x1(t) =Ascos( st+ s) +Afcos( ft+ f),x2(t) =Ascos( st+ s) Afcos( ft+ f).(15)This agrees with the result in Eq. (6).As we discussed in Section , we could have just taken the real part of theC1(1,1)ei standC3(1, 1)ei ftsolutions, instead of going through the positions must be real , you should continue using the latter reasoning until you re comfortable with theshort cut of taking the real :Note that Eq. (9) can be written in the form,(k+ k+ )(A1A2)=m 2(A1A2).(16) TWO MASSES5So what we did above was solve for theeigenvectorsandeigenvaluesof this matrix. The eigenvectorsof a matrix are the special vectors that get carried into a multiple of themselves what acted on bythe matrix. And the multiple (which ism 2here) is called the eigenvalue. Such vectors are indeedspecial, because in general a vector gets both stretched (or shrunk)androtated when acted on bya matrix.