1CHAPTER 17VIBRATING SYSTEMS17.1 IntroductionA mass m is attached to an elastic spring of force constant k, the other end of which isattached to a fixed point. The spring is supposed to obey Hooke’s law, namely that,when it is extended (or compressed) by a distance x from its natural length, the tension(or thrust) in the spring is kx, and the equation of motion is m& x= − kx.This is simpleharmonic motion of period 2π/ω, where ω 2 = k/m. Most readers will have no difficultywith that problem. But now suppose that, instead of one end of the spring being attachedto a fixed point, we have two masses, m 1 and m 2 , one at either end of the spring. Adiatomic molecule is much the same thing. Can you calculate the period of simpleharmonic oscillations? It looks like an easy problem, but it somehow seems difficult toget a hand on it by conventional newtonian methods. In fact it can be done quite readilyby newtonian methods, but this problem, as well as more complicated problems whereyou have several masses connected by several springs and several possible modes ofvibration, is particularly suitable by lagrangian methods, and this chapter will giveseveral examples of vibrating systems tackled by lagrangian methods.17.2 The Diatomic MoleculeTwo particles, of masses m 1 and m 2 are connected by an elastic spring of force constant k.What is the period of oscillation?FIGURE XVII.1m 1m 2kx 1 x 2Let’s suppose that the equilibrium separation of the masses – i.e. the natural, unstretched,uncompressed length of the spring – is a. At some time suppose that the x-coordinates ofthe two masses are x 1 and x 2 . The extension q of the spring from its natural length at that
2moment is q = x2 − x1− a.We’ll also suppose that the velocities of the two masses atthat instant are x& 1and x&2. We know from chapter 13 how to start any calculation inlagrangian mechanics. We don’t have to think about it. We always start with T = ... andV = ...:T2 1 2= 1 m x& + m x & ,17.2.12112222V = 1kq .17.2.22We want to be able to express the equations in terms of the internal coordinate q. V isalready expressed in terms of q. Now we need to express T (and therefore x&1and x&2)in terms of q. Since q = x − x − , we have, by differentiation with respect to time,2 1aq &= x&2− x&126.96.36.199We need one more equation. The linear momentum is constant and there is no loss ingenerality in choosing a coordinate system such that the linear momentum is zero:= m x& + m & .17.2.401 1 2x2From these two equations, we find thatx &m2m1= q&and x&2= & .17.2.5a,bm + mm + m1q1 21 2Thus we obtainT= &17.2.61 22mq2and V = 1kq ,17.2.22m1m2where m = .17.2.7m + mNow apply Lagrange’s equation12ddt∂T∂q&j−∂T∂qj=∂V−∂qj.13.4.13to the single coordinate q in the fashion to which we became accustomed in Chapter 13,and the equation of motion becomes
3mq& = − kq ,17.2.8which is simple harmonic motion of period 2π m / k , where m is given by equation17.2.7. The frequency is the reciprocal of this, and the “angular frequency” ω, alsosometimes called the “pulsatance”, is 2π times the frequency, or k / m .The quantity m1m2/( m1+ m2) is usually called the “reduced mass” and one may wonder is what sense itis “reduced”. I believe the origin of this term may come from an elementary treatment of the Bohr atom ofhydrogen, in which one at first assumes that there is an electron moving around an immovable nucleus –i.e. a nucleus of “infinite mass”. One develops formulas for various properties of the atom, such as, forexample, the Rydberg constant, which is the energy required to ionize the atom from its ground state. Thisand similar formulas include the mass m of the electron. Later, in a more sophisticated model, one takesaccount of the finite mass of the nucleus, with nucleus and electron moving around their mutual centre ofmass. One arrives at the same formula, except that m is replaced by mM/(m + M), where M is the mass ofthe nucleus. This is slightly less (by about 0.05%) than the mass of the electron, and the idea is that youcan do the calculation with a fixed nucleus provided that you use this “reduced mass of the electron” ratherthan its true mass. Whether this is the appropriate term to use in our present context is debatable, but inpractice it is the term almost universally used.It may also be remarked upon by readers with some familiarity with quantum mechanics that I have namedthis section “The Diatomic Molecule” – yet I have ignored the quantum mechanical aspects of molecularvibration. This is true – in this series of notes on Classical Mechanics I have adopted an entirely classicaltreatment. It would be wrong, however, to assume that classical mechanics does not apply to a molecule,or that quantum mechanics would not apply to a system consisting of a cricket ball and a baseballconnected by a metal spring. In fact both classical mechanics and quantum mechanics apply to both. Theformula derived for the frequency of vibration in terms of the reduced mass and the force constant (“bondstrength”) applies as accurately for the molecule as for the cricket ball and baseball. Quantum mechanics,however, predicts that the total energy (the eigenvalue of the hamiltonian operator) can take only certaindiscrete values, and also that the lowest possible value is not zero. It predicts this not only for themolecule, but also for the cricket ball and baseball – although in the latter case the energy levels are soclosely spaced together as to form a quasi continuum, and the zero point vibrational energy is so close tozero as to be unmeasurable. Quantum mechanics makes its effects evident at the molecular level, but thisdoes not mean that it does not apply at macroscopic levels. One might also take note that one is not likelyto understand why wave mechanics predicts only discrete energy levels unless one has had a goodbackground in the classical mechanics of waves. In other words, one must not assume that classicalmechanics does not apply to microscopic systems, or that quantum mechanics does not apply tomacroscopic systems.Below leaving this section, in case you tried solving this problem by newtonian methodsand ran into difficulties, here’s a hint. Keep the centre of mass fixed. When the length ofthe spring is x, the lengths of the portions on either side of the centre of mass arem2xm1xand . The force constants of the two portions of the spring arem1+ m2m1+ m2inversely proportional to their lengths. Take it from there.17.3 Two Masses, Two Springs and a Brick WallThe system is illustrated in figure XVII.2, first in its equilibrium (unstretched) position,and then at some instant when it is not in equilibrium and the springs are stretched. You
4can imagine that the masses are resting upon and can slide upon a smooth, horizontaltable. I could also have them hanging under gravity, but this would introduce a distractingcomplication without illustrating any further principles. I also want to assume that all themotion is linear, so we could have them sliding on a smooth horizontal rail, or have themconfined in the inside of a smooth, fixed drinking-straw. For the present, I don’t want thesystem to bend.Equilibrium:m 1 m 2k1 k 2Displaced:x 1 x 2FIGURE XVII.2The displacements from the equilibrium positions are x 1 and x 2 , so that the two springsare stretched by x 1 and x 2 − x 1 respectively. The velocities of the two masses are x&1andx&2. We now start the lagrangian calculation in the usual manner:TV2 1 2= 1 m x& + m x & ,17.3.12112222 12= 1 k x + k ( x − x ) .17.3.22112221Apply Lagrange’s equation to each coordinate in turn, to obtain the following equationsof motion:m & = − ( + x + k x17.3.31x1k1k2)1and m & x= k x − k .17.3.42 2 2 1 2x2Now we seek solutions in which the system is vibrating in simple harmonic motion at22angular frequency ω; that is, we seek solutions of the form & x1= − ω x1and & x2= − ω x2.When we substitute these in equations 17.3.3 and 4, we obtain(1 2 1 1 2 2=222k + k − m ω ) x − k x 017.3.52and k x − k − m ω ) x 0.17.3.62 1(2 2 2=
5Either of these gives us the displacement ratio x 2 /x 1 (and hence amplitude ratio). The firstgives usx2− m ω + k + kxk121 1 2= 17.3.72x2k2and the second gives us =.2x k − m ω12217.3.8These are equal, and, by equating the right hand sides, we obtain the following equationfor the angular frequencies of the normal modes:42m m ω − m k + m k + m k ) ω + k k 0.17.3.91 2(1 2 2 1 2 21 2=This equation can also be derived by noting, from the theory of equations, that equations17.3.5 and 6 are consistent only if the determinant of the coefficients is zero.The meaning of these equations and of the expression “normal modes” can perhaps bebest illustrated with a numerical example. Let us suppose, for example, that k 1 = k 2 = 14 2and m 1 = 3 and m 2 = 2. In that case equation 17.3.9 is 6ω − 7ω+ 1 = 0.This is aquartic equation in ω, but it is also a quadratic equation in ω 2 , and there are just twopositive solutions for ω. These are 1 / 6 = 0. 4082 (slow, low frequency) and 1 (fast,high frequency). If you put the low frequency ω into either of equations 17.3.7 or 8 (or inboth, to check for arithmetic or algebraic mistakes) you find a displacement ratio of +1.5;but if you put the high frequency ω into either equation, you find a displacement ratio of−1.0 The first of these normal modes is a low-frequency slow oscillation in which thetwo masses oscillate in phase, with m 2 having an amplitude 50% larger than m 1 . Thesecond normal mode is a high-frequency fast oscillation in which the two massesoscillate out of phase but with equal amplitudes.So, how does the system actually oscillate? This depends on the initial conditions. Forexample, if you displace the first mass by one inch to the right and the second mass by1.5 inches to the right (this implies stretching the first spring by 1 inch and the second by0.5 inches), and then let go, the system will oscillate in the slow, in-phase mode. But ifyou start by displacing the first mass by one inch to the right and the second mass by oneinch to the left (this implies stretching the first spring by 1 inch and compressing thesecond by 2 inches), the system will oscillate in the fast, out-of-phase mode. For otherinitial conditions, the system will oscillate in a linear combination of the normal modes.Thus, m 1 might oscillate with an amplitude A in the slow mode, and an amplitude B in thefast mode:x1= Acos( ω1t+ α1)+ B cos( ω2t+ α2),17.3.10in which case the oscillation of m 2 is given by
6x = .5Acos(ω t + α ) − B cos( ω t + ) .17.3.11211 12α2In our example, ω 1 and ω 2 are 1/6 and 1 respectively.Let’s suppose that the initial conditions are that, at t = 0, x & and x&1 2are both zero. Thismeans that α 1 and α 2 are both zero or π (I’ll take them to be zero), so that1= Acosω1t+ B cos ω t17.3.12x2and x = .5Acosω t − B cos ω .17.3.132112tSuppose further that at t = 0, x 1 and x 2 are both +1, which means that we start bystretching both springs equally. Equations 17.3.12 and 13 then become 1 = A + B and1 = 1.5A − B. That is, A = 0.8 and B = 0.2. I’ll leave you to draw graphs of x 1 and x 2versus time.Here’s an exercise that might be useful if, perhaps, you wanted to construct a real systemwith two equal masses m and two equal springs, each of constant k, to demonstrate thevibrations. Show that in that case, the angular frequency (which is, of course, 2π timesthe actual frequency) of the slow, in phase, mode is1kkω 1 = ( 5 − 1) = 0.61802mm1with a displacement ratio x x = ( 5 + 1) 1. 6180 ;2 / 1=2and the angular frequency of the fast, out of phase, mode is1kkω 2 = ( 5 + 1) = 1.61802mm1with a displacement ratio, x x = − ( 5 − 1) = 0.6180.2 / 1−2Knowing these displacements ratios will enable you to start with the appropriate initialconditions for each normal mode.If you were to start at t = 0 with a displacements x 1 = 1 and x 2 = 2 which isn’t right foreither normal mode, you can show that the subsequent displacements would bexx12==1.170820cosωt1.894427cosωt11−+0.170820cosωt0.105572cosωt.22
7That looks like this:2.52x 21.51Displacement0.50-0.5-1x 1-1.5-2-2.50 10 20 30 40 50 60 70 80ω 1tAlthough at first it looks like fast in-phase mode for both of them, you can see theinfluence of the slow mode, which has about 2.6 times the period of the last mode, in theslow amplitude modulation. If you look carefully at the modulation amplitudes of bothdisplacements, you will see that the amplitude of the x 1 displacement is out of phase withthe amplitude of the x 2 displacement.17.4 Double Torsion Pendulumc 1FIGURE XVII.3I 1c 2I 2
8Here we have two cylinders of rotational inertias I 1 and I 2 hanging from two wires oftorsion constants c 1 and c 2 . At any instant, the top cylinder is turned through an angle θ 1from the equilibrium position and the lower cylinder by an angle θ 2 from the equilibriumposition (so that, relative to the upper cylinder, it is turned by ( θ2−θ1)). The equationsand the description of the motion are just the same as in the previous example, except thatx 1 , x 2 , m 1 , m 2 , k 1 , k 2 are replaced by θ 1 , θ 2 , I 1 , I 2 , c 1 , c 2 . The kinetic and potential energiesare2 1 2T = 1 2I1θ&θ1+ &2I22,17.4.12 12V = 1 c θ + c ( θ − θ ) .17.4.2211The equations for ω and the displacement ratios are just the same, and there is an inphaseand an out-of-phase mode.222117.5 Double PendulumThis is another similar problem, though, instead of assuming Hooke’s law, we shall1 2assume that angles are small (sin θ ≈ θ , cos θ ≈ 1 −2θ ). For clarity of drawing,however, I have drawn large angles in figure XVIII.4.θ 1θ 2l 1l 2FIGURE XVII.4l 1θ & 1m 1m 2l 2θ &2l 1θ & 1
9Because I am going to use the lagrangian equations of motion, I have not marked in theforces and accelerations; rather, I have marked in the velocities. I hope that the twocomponents of the velocity of m 2 that I have marked are self-explanatory; the speed of m 22 2 2 2 2is given by v2= l1θ&1+ l2θ&2+ 2l1l2θ &1θ&2cos( θ2−θ1). The kinetic and potential energiesare2 2 1 2 2 2 2T = 1 m l θ&+ m l θ&+ l θ&+ 2ll θ&θ&cos( θ − θ )], 17.5.121 1122[ 1 1 2 2 1 2 1 2 2 1V = constant − m1gl1cos θ1− m2g(l1cosθ1+ l2cos θ2). 17.5.2If we now make the small angle approximation, these becomeT = θ&+ ( θ&+ θ&17.5.31 2 2 122m1l1 1 2m2l11l22)2 12 2and V = constant +1 m1gl1θ1+ m2g(l1θ1+ l2θ2)− m gl1gl1− m2gl1− m22. 17.5.42Apply the lagrangian equation in turn to θ 1 and θ 2 :2( m + & θ+ & θ= − + gl θ17.5.521m2)l11m2l1l2 2( m1m2)2and m l l & θ+ m l & θ= − m gl .17.5.62 1 2 1 2 2 22 2θ2Seek solutions of the form & 22θ1= − ω θ1and & θ2= − ω θ2.22Then m + m )( l ω − g)θ + m l ω θ 017.5.7(1 2 21 2 2 2=22and l ω θ + l ω − g)θ 0.17.5.81 1(22=Either of these gives the displacement ratio θ 2 /θ 1 . Equating the two expressions for theratio θ 2 /θ 1 , or putting the determinant of the coefficients to zero, gives the followingequation for the frequencies of the normal modes:422m l l ω − ( m + m ) g(l + l ) ω + ( m + m ) g 0.17.5.91 1 21 2 1 21 2=As in the previous examples, there is a slow in-phase mode, and fast out-of-phase mode.For example, suppose m 1 = 0.01 kg, m 2 = 0.02 kg, l 1 = 0.3 m, l 2 = 0.6 m, g = 9.8 ms −2 .42Then 0.0018ω − 0.2646ω+ 2.8812 = 0.The slow solution is ω = 3.441 rad s −1 (P= 1.826 s), and the fast solution is ω = 11.626 rad s −1 (P =0.540 s). If we put the firstof these (the slow solution) in either of equations 17.5.7 or 8 (or both, as a check against11
10mistakes) we obtain the displacement ratio θ 2 /θ 1 = 1.319, which is an in-phase mode. Ifwe put the second (the fast solution) in either equation, we obtain θ 2 /θ 1 = −0.5689 ,which is an out-of-phase mode. If you were to start with θ 2 /θ 1 = 1.319 and let go, thependulum would swing in the slow in-phase mode. . If you were to start with θ 2 /θ 1 =−0.5689 and let go, the pendulum would swing in the fast out-of-phase mode. Otherwisethe motion would be a linear combination of the normal modes, with the fraction of eachdetermined by the initial conditions, as in the example in section 188.8.131.52 Linear Triatomic MoleculeIn Chapter 2, Section 2.9, we discussed a rigid triatomic molecule. Now we are going todiscuss three masses held together by springs, of force constants k 1 and k 2 . We are goingto allow it to vibrate, but not to rotate. Also, for the time being, I don’t want themolecule to bend, so we’ll put it inside a drinking straw to that all the vibrations arelinear. By the way, for real triatomic molecules, the force constants and rotationalinertias are such that molecules vibrate much faster than they rotate. To see theirvibrations you look in the near infra-red spectrum; to see their rotation, you have to go tothe far infrared or the microwave spectrum.k 1 k 2m 1 m 2 m 3FIGURE XVII.5Suppose that the equilibrium separations of the atoms are a 1 and a 2 . Suppose that atsome instant of time, the x-coordinates (distances from the left hand edge of the page) ofthe three atoms are x 1 , x 2 , x 3 . The extensions from the equilibrium distances are thenq = x − x − a q = x − x − . We are now ready to start:1 2 1 1,2 3 2a2T2 1 2 1 2= 1 m x& + m x&+ m x & ,17.6.12112222332 1 2V = 1 k q + k q .17.6.2211We need to express the kinetic energy in terms of the internal coordinates, and, just as forthe diatomic molecule (Section 17.2), the relevant equations areq &q &222= x&− & ,17.6.31 2x1= & − &17.6.42x3x2
11and = m x& + m x&+ m & .17.6.5These can conveniently be written01 1 2 2 3x3⎛q&⎜⎜q&⎜⎝1⎞ ⎛−1⎟ ⎜⎟ = ⎜ 0⎜⎠ ⎝ m1 0 ⎞⎛x&1⎟⎜−11 ⎟⎜x&2m ⎟⎜2m3⎠⎝x&20⎟13⎞⎟⎟.⎟⎠17.6.6By one dexterous flick of the fingers (!) we invert the matrix to obtain⎛ x&⎜⎜ x&⎜⎝ x&123⎞⎟⎟⎟⎠⎛ m2+ m⎜−⎜ M⎜ m1= ⎜⎜M⎜ m1⎝ M3m3−Mm3−Mm1+ mM21M1M1M⎞⎟⎟⎛q&1 ⎞⎟⎜⎟⎟⎜q&2 ⎟,⎟⎜0⎟⎟⎝ ⎠⎠17.6.7where M = m + m + . On putting these into equation 17.6.1, we now have1 2m3T22= 1 ( aq& + 2hq&q&+ b q & )17.6.8211222 1 2and V = 1 k q + k q ,17.6.2211222where a = m ( m + m3) / ,17.6.91 2M3 1/ M ,3( m1+ m2)Mh = m m17.6.10b = m /17.6.11and, for future reference,2ab − h = m m m M = m h.17.6.12122/2On application of Lagrange’s equation in turn to the two internal coordinates we obtaina q& + hq&+ k q 017.6.131 2 1 1=and b q& + hq&+ k q 0.17.6.142 1 2 2=22Seek solutions of the form q& 1= − ω q1and q& 2= − ω q2and we obtain the following twoexpressions for the extension ratios:
12qq12=2hωk − aω12=2k2−bω2hω.17.6.15Equating them gives the equation for the normal mode frequencies:2 42ab − h ) ω − ( ak + bk ) ω + k k = 0.17.6.16(2 11 2For example, if k1= k2= k and m1= m2= m3= m,we obtain, for the slow symmetric2(“breathing”) mode, q q = 1 and ω = k / m.For the fast asymmetric mode,1/2+2q q = 1 and ω = 3k/ m.1/2−Example.Consider the linear OCS molecule whose atoms have masses 16, 12 and 32. Suppose thatthe angular frequencies of the normal modes, as determined from infrared spectroscopy,are 0.905 and 0.413. (I just made these numbers up, in unstated units, just for the purposeof illustrating the calculation. Without searching the literature, I can’t say what they arein the real OCS molecule.) Determine the force constants.In Chapter 2 we considered a rigid triatomic molecule. We were given the moment ofinertia, and we were asked to find the two internuclear distances. We couldn’t do thiswith just one moment of inertia, so we made an isotopic substitution ( 18 O instead of 16 O)to get a second equation, and so we could then solve for the two internuclear distances.This time, we are dealing with vibration, and we are going to use equation 17.6.16 to findthe two force constants. This time, however, we are given two frequencies (of the normalmodes), and so we have no need to make an isotopic substitution − we already have twoequations.Here are the necessary data.16 OCSFast ω 0.905Slow ω 0.413m 1 m 2 m 3 16 12 32M 60a 11 .73h 8 .53b 14 .93ab − h 2 102.4Use equation 17.6.16 for each of the frequencies, and you’ll get two equations, in k 1 andk 2 . As in the rotational case, they are quadratic equations, but they are a bit easier tosolve than in the rotational case. You’ll get two equations, each of the form
13A − Bk1 − Ck2+ k1k2= 0, where the coefficients are functions of a, b, h, ω. You’llhave to work out the values of these coefficients, but, before you substitute the numbersin, you might want to give a bit of thought to how you would go about solving twosimultaneous equations of the form A − Bk − Ck + k k 0.1 2 1 2 =I make the answersk 1 = 2.924 000 k 2 = 4.852 000Note that in this section we considered a linear triatomic molecule that was not allowedeither to rotate or to bend, whereas in Chapter 2 we considered a rigid triatomic moleculethat was not allowed either to vibrate or to bend. If all of these restrictions are removed,the situation becomes rather more complicated. If a rotating molecule vibrates, themoving atoms, in a co-rotating reference frame, are subject to the Coriolis force, andhence they do not move in a straight line. Further, as it vibrates, the rotational inertiachanges periodically, so the rotation is not uniform. If we allow the molecule to bend,the middle atom can oscillate up and down in the plane of the paper (so to speak) or backand forth at right angles to the plane of the paper. These two motions will not necessarilyhave either the same amplitude or the same phase. Consequently the middle atom willwhirl around in a Lissajous ellipse, giving rise to what has been called “vibrationalangular momentum”. In a real triatomic molecule, the vibrations are usually much fasterthan the relatively slow, ponderous rotation, so that vibration-rotation interaction is small– but is by no means negligible and is readily observed in the spectrum of the molecule.17.7 Two Masses, Three Springs, Two Brick WallsThe three masses are equal, and the two outer springs are identical. Figure XVII.6 showsthe equilibrium position.mk 1 km2k 1xyFIGURE XVII.6Suppose that at some instant the first mass is displaced a distance x to the right and thesecond mass is displaced a distance y to the right. The extensions of the first two springsare x and y − x respectively, and the compression of the third spring is y. If the speeds ofthe masses are x& and y& , we have for the kinetic and potential energies:
14T= & + &17.7.11 2 1 22mx y 2m2 12 1 2and V = 1 k x + k ( y − x)+ k y .17.7.2Apply Lagrange’s equation in turn to x and to y.21222m & x+ k + k ) x − k y 017.7.3(1 2 2=and m & y+ k + k ) y − k x 0.17.7.4(1 2 2=22Seek solutions of the form & x= − ω x and & y= − ω y.2−m ω + k + k ) x − k y = 017.7.5(1 222and − k x + ( −mω+ k + k2)y 0.17.7.62 1=On putting the determinant of the coefficients to zero, we find for the frequencies of thenormal modescorresponding to displacement ratios2 12 k12k2ω =k and ω =+ ,17.7.7a,bmm1xy= 1xand = −1.y17.7.8a,bIn the first, slow, mode, the masses move in phase and there is no extension orcompression of the connecting spring. In the second, fast, mode, the masses move inantiphase and the compression or extension of the coupling spring is twice the extensionor compression of the outer springs.The general motion is a linear combination of the normal modes:x = Acos( ω1t+ α1)+ B cos( ω2t+ α2),17.7.9y = Acos( ω1t+ α1)− B cos( ω2t+ α2),17.7.10x& = − Aω1 sin( ω1t+ α1)− Bω2sin( ω2t+ α2),17.7.11y& = − Aω1 sin( ω1t+ α1)+ Bω2sin( ω2t+ α2).17.7.12
15Suppose that the initial condition is at t = 0, y = y&= 0,x = x0 , x&= 0.That is, we pullthe first mass a little to the right (keeping the second mass fixed) and then we let go. Thesecond two equations establish that α 1 = α 2 = 0, and the first two equations tell us that A= B = x 0 /2. The displacements are then given by11x = 1 x0(cosω1t+ cosω2t)= x0cos ( ω1− ω2)t cos ( ω ω 1+2)t 17.7.13211and y = 1 x0(cosω1t− cosω2t)= − x0sin ( ω1−ω2)t sin ( ω1+ ω2)t . 17.7.142Let us imagine, for example, that k 2 is much less than k 1 (but not negligible), so that wehave two weakly-coupled oscillators. In that case equations17.7.7 tell us that thefrequencies of the two normal modes are nearly equal. What equation 17.7.13 describes,then, is a rapid oscillation of the first mass with angular frequency 1 2( ω1+ ω 2) whose1amplitude is modulated with a slow angular frequency2( ω1− ω2). Equation 17.7.14describes the same sort of motion for the second mass, except that the modulation is outof phase by 90 o with the modulation of the motion of the first mass. For a while the firstmass will oscillate with a large amplitude. This will gradually decrease, while theamplitude of the motion of the second mass increases until the motion of the first massmomentarily ceases. After that, the amplitude of the motion of the second mass starts todecrease, while the first mass starts up again. And so the motion continues, with the firstmass and the second mass alternately taking up the motion.222217.8 Transverse Oscillations of Masses on a Taut StringA light string of length 4a is held taut, under tension F between two fixed points. Threeequal masses m are attached at equidistant points along the string. They are set intotransverse oscillation of small amplitudes, the transverse displacements of the threemasses at some time being y 1 , y 2 and y 3 .The kinetic energy is easy. It is justθ 1F•• •y 1y 2 y3a aa aFIGURE XVII.7T2 2 2= 1 m(y& + y&+ y & ).17.8.12123The potential energy is slightly more difficult.
16In the undisplaced position, the length of each portion of the string is a.In the displaced position, the lengths of the four portions of the string are, respectively,y +2 22 22 22 21+ a ( y2− y1)+ a ( y2− y3)+ a y3aFor small displacements (i.e. the ys much smaller than a), these are, approximately (bybinomial expansion),a+2y2a( y− y1)2a( y− y3)2a221 22a +a +a +2y12aso the extensions are2y1 ( y2− y1)( y2− y3)2a2a2a222y32aIt is also supposed that the tension in the string is F and that the displacements aresufficiently small that this is constant. The work done in displacing the masses, which isthe elastic energy stored in the string as a result of the displacements, is thereforeV=F2a222 2 F 2 2 2[ y + ( y − y ) + ( y − y ) + y ] = ( y + y + y − y y − y ).1 2 12 3 31 2 3 1 2 2y3aWe note with mild irritation the presence of the cross-terms y y y .Apply Lagrange’s equation in turn to the three coordinates:1 2,2y3am & y+ F 2y− y ) 0,17.8.31(1 2=am & y+ F − y + 2y− y ) 0,17.8.42(1 2 3=am & y+ F − y + 2y) 0.17.8.53(2 3=222Seek solutions of the form & y= − ω y && y = − ω y , & y= − ω .1 1,22 3y32Then 2F − amω) y − Fy = 0,17.8.6(122− Fy + 2F− amω) y − Fy 0,17.8.71(23=2− Fy + 2F− amω) y 0.17.8.82(3=
17Putting the determinant of the coefficients to zero gives an equation for the frequencies ofthe normal modes. The solutions are:Slow Medium Fastω21=(2 − 2)Fam2ω2=2 Famω23=(2 + 2)FamSubstitution of these into equations 17.8.6 to 8 gives the following displacement ratiosfor these three modes:y 1 : y 2 : y 3 = 1 : √2 : 1 1 : 0 : −1 1 : −√2 : 1These are illustrated in figure XVII.8.As usual, the general motion is a linear combination of the normal modes, the relativeamplitudes and phases of the modes depending upon the initial conditions.If the motion of the first mass is a combination of the three modes with relativeamplitudes in the proportion q ˆ : ˆ : ˆ1q2q3, and with initial phases α1, α2, α3, its motion isdescribed byy = q sin( ω t + α ) + qˆsin( ω t + α ) + qˆsin( ω t + ). 17.8.9ˆ1 1 1 1 2 2 2 3 3α3•• •ω 1•••1.85ω 1• •2.41ω 1•FIGURE XVII.8
18The motions of the second and third masses are then described byy = qˆsin( ω t + α ) − 2qˆsin( ω t + )17.8.10221 1 13 3α3and y = q sin( ω t + α ) − qˆsin( ω t + α ) + qˆsin( ω t + ).17.8.11These can be writtenˆ3 1 1 1 2 2 2 3 3α3y = q + q + ,17.8.121 1 2q3y2q1− 2= 2 q17.8.133and y = q − q + ,17.8.141 1 2q3where the q i , like the y i , are time-dependent coordinates.We could, if we wish, express the q i in terms of the y i , by solving these equations:1q = y + 2y+ ) ,17.8.15q1 4(12y31= ( y − 3)17.8.162 2 1y1and q = y − 2y+ ) .17.8.173 (4 1 2 y3We have hitherto described the state of the system as a function of time by giving thevalues of the coordinates y 1 , y 2 and y 3 . We could equally well, if we wished, describe thestate of the system by giving, instead, the values of the coordinates q 1 , q 2 and q 3 . Indeedit turns out that it is very useful to do so, and these coordinates are called the normalcoordinates, and we shall see that they have some special properties. Thus, if youexpress the kinetic and potential energies in terms of the normal coordinates, you getT2 2 2= 1 m(4q& + 2q&+ 4 q & )17.8.182123= 2 [ 21 2q3].17.8.19aF2 2and V ( 2 − 2) q + q + ( 2 + 2)Note that there are no cross terms. When you apply Lagrange’s equation in turn to thethree normal coordinates, you obtain( − 2) ,amq& 1= − Fq17.8.2021
19and am q 3= − ( + 2) Fq .amq& 2= − 2Fq 217.8.21& 17.8.2223Notice that the normal coordinates have become completely separated into three2independent equations and that each is of the form q&= − ω q and that each of the normalcoordinates oscillates with one of the frequencies of the normal modes. Much of the artof solving problems involving vibrating systems concerns identifying the normalcoordinates.17.9 Vibrating StringIt is possible that the three modes of vibration of the three masses in section 17.8reminded you of the fundamental and first two harmonic vibrations of a stretched string –and it is quite proper that it did. If you were to imagine ten masses attached to a stretchedstring and to carry out the same sort of analysis, you would find ten normal modes, ofwhich one would be quite like the fundamental mode of a stretched string, and theremainder would remind you of the first nine harmonics. You could continue with thesame analysis but with a very large number of masses, and eventually you would beanalysing the vibrations of a continuous heavy string. We do that now, and we assumethat we have a heavy, taut string of mass µ per unit length, and under a tension F.yFABFA 0 B 0xFIGURE XVII.9
20I show in figure XVII.9 a portion of length δx of a vibrating rope, represented by A 0 B 0 inits equilibrium position and by AB in a displaced position. The rope makes an angle ψ Awith the horizontal at A and an angle ψ B with the horizontal at B. The tension in the ropeis F. The vertical equation of motion is2∂ yF(sin ψ sin ) x .B− ψA= µ δ17.9.12∂t2∂yIf the angles are small, then sin ,∂ yψ ≅ so the expression in parenthesis is x.2 δ∂x∂xequation of motion is thereforeThe2 22 ∂ y ∂ y,Tc = where c = .17.9.2,a,b2 2∂x∂tµAs can be verified by substitution, the general solution to this is of the formy = f ( x − ct)+ g(x + ct).17.9.3This represents a function that can travel in either direction along the rope at a speed cgiven by equation 17.9.2b. Should the disturbance be a periodic disturbance, then a wavewill travel along the rope at that speed. Further analysis of waves in ropes and strings isgenerally done in chapters concerned with wave motion. This section, however, at leastestablishes the speed at which a disturbance (periodic or otherwise) travels along astretched strong or rope.17.10 WaterWater consists of a mass M (“oxygen”) connected to two smaller equal masses m(“hydrogen”) by two equal springs of force constants k, the angle between the springsbeing 2θ. The equilibrium length of each spring is r. The torque needed to increase theangle between the springs by 2δθ is 2cδθ. See figure XVII.10. (θ is about 52°.)MθgoθmmFIGURE XVII.10
21At any time, let the coordinates of the three masses (from left to right) be( x1, y1), ( x2, y2), ( x3, y3)and let the equilibrium positions bex , y ) , ( x , y ) , ( x , ), where y 30 = y 10 .(10 10 20 20 30y30We suppose that these coordinates are referred to a frame in which the centre of mass ofthe system is stationary.Let us try and imagine, in figure XVII.11, the vibrational modes. We can easily imaginea mode in which the angle opens and closes symmetrically. Let is resolve this mode intoan x-component and a y-component. In the x-component of this motion, one hydrogenatom moves to the right by a distance q 1 while the other moves to the left by and equaldistance q 1 . In the y-component of this symmetric motion, both hydrogens moveupwards by a distance q 2 , while, in order to keep the centre of mass of the systemunmoved, the oxygen necessarily moves down by a distance 2mq 2 /M. We can alsoimagine an asymmetric mode in which one spring expands while the other contracts.One hydrogen moves down to the left by a distance q 3 , while the other moves up to theleft by the same distance. In the meantime, the oxygen must move to the right by adistance (2mq 3 sin θ)/M, in order to keep the centre of mass unmoved.We are going to try to write down the kinetic and potential energies in terms of theinternal coordinates q 1 , q 2 and q 3 .
22θMθmq 1 q 1mMθ θq 2 2mq 2 /M q 2mmθMθ(2mq 3 sin θ)/Mq 3mmq 3FIGURE XVII.11It is easy to write down the kinetic energy in terms of the (x , y) coordinates:
23T2 2 1 2 2 1 2 2= 1 m(x& + y&) + M ( x&+ y&) + m(x&+ y & ). 17.10.1211222233From geometry, we have:x &1= q&1− q&3sin θ y&1= q&2− q&3cosθ17.10.2a,b&2 mq&3sin θ2mq&= y&2= −17.10.3a,bMM2x2x &3= − q&1− q&3sin θ y&3= q&2+ q&3cosθ17.10.4a,bOn putting these into equation 17.10.1 we obtainT22( 1 + (2msinθ) / M ) & .22= mq&1+ m(1+ 2m/ M ) q&2+ mq317.10.5For short, I am going to write this asT222= a q& + a q&+ a & .17.10.611 1 22 2 33q3Now for the potential energy.The extension of the left hand spring is2mq2cosθ2mq3sin θcosθδr1= − q1sin θ − q2cosθ−+ q3+MM= − q sin θ − q (1 + 2m/ M )cosθ+ q ( 1 + (2msinθcosθ) / M ).12The extension of the right hand spring is22mq2cosθ2mq3sin θδr2= − q1sin θ − q2cosθ−− q3−MM2= − q sin θ − q (1 + 2m/ M )cosθ− q ( 1 + (2msinθ) / M ).123317.10.717.10.8The increase in the angle between the springs is2q1cosθ2(1 + 2m/ M ) q sin θδθ = − +.17.10.9rr22The potential energy (above the equilibrium position) is2 1 2 12V = 1 k(δr) + k(δr) + c(2δθ ) .17.10.1021222
24On substituting equations 17.10,7,8 and 9 into this, we obtain an equation of the form222V = b q + 2bq q + b q + b ,17.10.1111 1 12 1 2 22 2 33q3where I leave it to the reader, if s/he wishes, to work out the detailed expressions for thecoefficients. We still have a cross term, so we can’t completely separate the coordinates,but we can easily apply Lagrange’s equation to equations 17.10.6 and 11, and then seeksimple harmonic solutions in the usual way. Setting the determinant of the coefficients tozero leads to the following equation for the angular frequencies of the normal modes:b11−ωb0122a11b22b012− ω2a22b3300−ω2a33= 0.17.10.12Thus, given the masses and r, θ, k and c, one can predict the frequencies of the normalmodes. Can one calculate k and c given the frequencies? I don’t know, to tell the truth.Can I leave it to the reader to investigate further?17.11 A General Vibrating SystemFor convenience, I'll refer to a collection of masses connected by springs as a "molecule",and the individual masses as "atoms". In a molecule with N atoms, the number ofdegrees of vibrational freedom (the number of normal modes of vibration) n = 3N− 6for nonlinear molecules, or n = 3N− 5 for linear molecules. Three equations areneeded to express zero translational motion, and three (or two) are needed to express zerorotational motion.[While reading this Section, it might be worthwhile for the reader to follow at the same time the treatmentgiven to the OCS molecule in Section 17.6. Bear in mind, however, that in that section we did not considerthe possibility of the molecule bending. Indeed we treated the molecule as if it were constrained inside adrinking straw, and it remained linear at all times. That being the case only N coordinates (rather than 3N)suffice to describe the state of the molecule. Only one equation is needed to express zero translationalmotion, and none are needed to express zero rotational motion. Thus there are N − 1 internal coordinates,and hence N − 1 normal vibrational modes. In the case of OCS, N = 3, so there are two normal vibrationalmodes.]A molecule with n degrees of vibrational freedom can be described at some instant oftime by n internal coordinates q i . A typical such coordinate may be related to the externalcoordinates of two atoms, for example, by some expression of the form q = x2 − x1 − a,as we saw in our example of the molecule OCS. Its potential energy can be written in theform
25V = κ q + κ q q + ... + κ q q2211 1+ κ qq + κ q + ... + κ qq+ ...221 2 1 22 212 1 2 1n1 n2n2+ κ qq + κ qq + ... + κ qq.n1 n 1 12 n 2nn n nn17.11.1Unless the q are the judiciously chosen "normal coordinates" (see our example of thetransverse vibrations of three masses on an elastic string), there will in general be crossterms, such as q 1 q 2 . If both qs of a term are linear displacements, the corresponding κ isa force constant (dimensions MT −2 ). If both qs are angles, κ is a torsion constant(dimensions ML 2 T −2 . If one is a linear displacement and he other is an angulardisplacement, κ will be a coefficient of dimensions MLT −2 ..The matrix is symmetric, so that equation 17.11.1 could also be written22V = κ q + 2κ q q + ... + 2κq q11 112 1 2 1n1 n2+ κ q + ... + 2κ22 2+ ...+ κq q2n2qqnn n n.n17.11.2In matrix notation, the equation (i.e. equations 17.11.1 or 17.11.2) could be written:or in vector/tensor notation, 2V = q •2 V = q~κ q . 17.11.3κ q. 17.11.4The kinetic energy can be written in terms of the time rates of change of the externalcoordinates x i :2T222= m x& + m x&+ ... + m Nx&.17.11.51 1 2 23 3NTo make use of the Lagrangian equations of motion, we need to express V and T in termsof the same coordinates, and it is usually advantageous if these be the n internalcoordinates rather than the 3N external coordinates – so that we have to deal with only nrather than 3N lagrangian equations. (Recall that n = 3N− 6 or 5 .) The relationsbetween the external and internal coordinates are given as a set of equations that expressa choice of coordinates such that there is no pure translation and no pure rotation of themolecule. These equations are of the formq = Ax.17.11.6
26Here q is an n × 1 column matrix, x a 3 N × 1 column matrix, and A is a matrix with nrows and 3N columns, and it may need a little trouble to set up. We could then use this toexpress V in terms of the external coordinates, so we would then have both V and T interms of the external coordinates. We could then apply Lagrange’s equation to each ofthe 3N external coordinates and arrive at 3N simultaneous differential equations ofmotion.A better approach is usually to set up the equations connecting q& and x& :q &= Bx.&17.11.7(These correspond to equations 17.6.3 and 17.6.4 in our example of the linear triatomicmolecule in Section 17.6.) We then want to invert equations 17.11.7 in order to expressx& in terms of q& . But we can’t do this, because B is not a square matrix. x& has 3Nelements while q& has only n. We have to add an additional six (or five for linearmolecules) equations to express zero pure translational and zero pure rotational motion.This adds a further 6 or 5 rows to B, so that B is now square. (this corresponds toequation 17.6.6), and we can then invert equation 17.11.7:(This corresponds to equation 17.6.7.)x= B q. &17.11.8&-1By this means we can express the kinetic energy in terms of the time rates of change ofonly the n internal coordinates:2T=+ µ+ ...+µµ21q&211 1q&q&2q&q&n1n11+++µµµ122212q&q&q&122n22+ ... ++ ... +q&q&+ ...+µµµ1n2nnnq&q&12q&q&nnq&q&nn.17.11.7Since the matrix is symmetric, the equation could also be written in a form analogous toequation 17.11.2. The equation can also be written in matrix notation as2 T = q~ & µ q&. 17.11.8or in vector/tensor notation,2 T = q & • µ q&. 17.11.9Here the µ ij are functions of the masses. If both qs is a particular term have thedimensions of a length, the corresponding µ and κ will have dimensions of mass andforce constant. If both qs are angles, the corresponding µ and κ will have dimensions ofrotational inertia and torsion constant.. If one q is a length and the other is an angle, thecorresponding µ and κ will have dimensions ML and MLT −2 .
27Apply Lagrange's equation successively to q 1, ... , q nto obtain n equations of the formq & 1+ ... + µ1nq&n+ κ11q1+ ... +κ1nq0.7.11.10µ11 n=That is to sayµ q& = − κq . 17.11.11Seek simple harmonic solutions of the formq&=2−ωqand we obtain n equations of the form22( κ11− µ11ω ) q1 + ... + ( κ1n − µ1nω) qn = 0.17.11.13The frequencies of the normal modes can be obtained by equating the determinant of thecoefficients to zero, and hence the displacement ratios can be determined.If N is large, this could be a formidable task. The work can be very much reduced bymaking use of symmetry relations of the molecule, in which case the determinant of thecoefficients may be factored into a number of much smaller subdeterminants. Further, ifthe configuration of the molecule could be expressed in terms of normal coordinates(combinations of the internal coordinates) such that the potential energy contained nocross terms, the equations of motion for each normal coordinate would be in the formq&= −ω 2 q.17.12 A Driven SystemIt would probably be useful before reading this and the next section to review Chapters11 and 12.m 1 m 2k1 k 2Equilibrium:Displaced:x 1 x 2f 1 f 1f 2 f 2F tFIGURE XVII.12= F ˆ sin ωFigure XVII.12 shows the same system as figure XVII.2, except that, instead of being leftto vibrate on its own, the second mass is subject to a periodic force F = F ˆ sin ωt. Forthe time being, we’ll suppose that there is no damping. Either way, it is not a
28conservative force, and Lagrange’s equation will be used in the form of equation 13.4.12.As in section 17.2, the kinetic energy isT2 1 2= 1 m x& + m x & .17.12.1211222Lagrange’s equations areddt∂T∂x&1−∂T∂x1=P117.12.2d ∂T∂Tand − = P2.dt ∂x&∂x2217.12.3We have to identify the generalized forces P 1 and P 2 .In the nonequilibrium position, the extension of the left hand spring is x 1 and so thetension in that spring is f1= k1x1. The extension of the right hand spring is x2 − x2andso the tension in that spring is f ( )2= k2x2− x1 . If x 1 were to increase by δx 1 , the workdone on m 1 would be ( f2 − f1)δx1, and therefore the generalized force associated with thecoordinate x 1 is P1 = k2( x2− x1) − k1x1. If x 2 were to increase by δx 2 , the work done onm 2 would be ( F − f 2) δx2, and therefore the generalized force associated with thecoordinate x 2 is P ˆ2= F sin ωt− k2(x2− x1). The lagrangian equations of motiontherefore becomem & x+ k + k ) x − k x 017.12.41 1(1 2 1 2 2=and m & x+ k ( x − x ˆ1) = F sin ω .17.12.52 2 2 2t22Seek solutions of the form & x1= − ω x1and & x2= − ω x2. The equations become2k + k − m ω ) x − k x 017.12.6(1 2 1 1 2 2=2and − k x + k − m ω ) x = Fˆsin ω .17.12.72 1(2 2 2tWe do not, of course, now equate the determinants of the coefficients to zero (why not?!),but we can solve these equations to obtainx1k Fˆsin ωt= 17.12.82( k + k − m )( k − m ω ) − k12221ω2222
292( kˆ1 + k2− m1ω) F sin ωtand x.2 = 17.12.922 2( k + k − m ω )( k − m ω ) − k121The amplitudes of these motions (and how they vary with the forcing frequency ω) are222xˆ1k Fˆ= 242m1m2ω− ( m1k2+ m2k1+ m2k2)ω + k1k17.12.1022( kˆ1+ k2− m1ω) Fand xˆ,2= 17.12.1142m m ω − ( m k + m k + m k ) ω + k k1212212212where I have re-written the denominators in the form of a quadratic expression in ω 2 .For illustration I draw, in figure XVII.13, the amplitudes of the motion of m 1 (continuouscurve, in black) and of m 2 (dashed curve, in blue) for the following data:F ˆ= 1, k = k = 1, m = 3, m1 212=2,when the equations become11x ˆ 1==4 226ω− 7ω+ 1 (6ω− 1)( ω2 − 1)17.12.12and222 − 3ω2 − 3ωx ˆ .2==4 22 26ω− 7ω+ 1 (6ω− 1)( ω − 1)17.12.13
304FIGURE XVII.13FIGURE XVII.123m 2m 1omega21Amplitude0-1-2-3-40 0.5 1 1.5Where the amplitude is negative, the oscillations are out of phase with the force F. Theamplitudes go to infinity (remember we are assuming here zero damping) at the twofrequencies where the denominators of equations 17.12.10 and11 are zero. Theamplitude of the motion of m 2 is zero when the numerator of equation 17.12.11 is zero.This is at an angular frequency of ( k1+ k2) / m1, which is just the angular frequency ofthe motion of m 1 held by the two springs between two fixed points. In our numericalexample, this is ω = 2 / 3 = 0.8165.This is an example of antiresonance.17.13 A Damped Driven SystemI’ll leave the reader to add some damping to the system described in section 17.12. Letus here try it with the system described in section 17.7. We’ll apply a periodic force tothe left hand mass, and we’ll suppose that the damping constant for each mass isγ = b / m. We could write the periodic force as F = F ˆ sin ωt, but the algebra will beeasier if we write it as ˆ iωtF = Fe . If the initial condition is such that F = 0 when t = 0,then we choose just the imaginary part of this in subsequent expressions.The equations of motion arem & x=− the damping force b& x− the tension in the left hand spring k 1 x
31+ the force F+ the tension in the middle spring k 2 (y − x)(this last is a thrust whenever y < x)and m & y=− the damping force b& y− the thrust in the right hand spring k 1 y− the tension in the middle spring k 2 (y − x)F= F ˆ sin ωtmk 1 k 2xADyk 1FIGURE XVII.14That is,iωt& bx&+ ( k1+ k2)x − k2y = Fe17.13.1mx + ˆand m & y+ by&+ k + k ) y − k x 0.17.13.2( 1 2 2 =For the steady-state motion, seek solutions of the form22& x= − ω x,&& y = − ω y , so that x&= iωxand y&= iωy.The equations then become2iωt( k1 k2− mω+ ibω)x − k2y = Fe17.13.3+ ˆ2and − k x + ( k + k − mω+ ibω)y 0.17.13.42 1 2=There is now a little algebra to be carried out. Solve these equations for x and y, andwhen, in doing so, there is a complex number in the denominator, multiply top andbottom by the conjugate in the usual way, so as to get x and y in the formsx ' + ix"and y'+ iy".Then find expressions for the amplitudes xˆ and ŷ . After somealgebra, the amount of which depends on one’s skill, experience and luck (it is notalways obvious how to gather terms in the most economical way, and you need someluck in this) you eventually get, for the amplitudes of the motionˆ2 2 2 2 2( k1+ k2− mω) + b ω ) Fˆ2 2 2 22 2 2 2( k − mω) + b ω )((k + 2k− mω) + b ω )2x =17.13.511 2
32and2 ˆ 22k Fy =17.13.6ˆ 211 22( ) )((2 ) ) .2 2 2 22 2 2k − mω+ b ω k + k − mω+ b ωThere are many variables in these expressions, but in order to see qualitatively what thesteady state motion is like, I’m going to put Fˆ , m and k 1 = 1. I think if I also put b = 1,this will give light damping in the sense described in Chapter 11. As for k 2 , I am going tok⎛2α ⎞introduce a coupling coefficient α defined by α = or k2= k1.k1k⎜21⎟ This+ ⎝ − α ⎠coupling constant will be close to zero if the middle spring is very weak, and 1 if themiddle connector is a rigid rod. The equations now become21 2 2(1−α− ω ) + ω2 2 22((1− ω ) + ω )( )1 +α− ω( ) .2 21−α+ ω2x ˆ =17.13.7and2α /(1 − α)y ˆ =17.13.8( ) .2 21−α+ ω2 2 22((1) )( )1 +α−ω + ω − ωFor computational efficiency you might want to rewrite these equations a little. For2example you could write ( 12 ) 22− ω + ω as 1 − Ω(1− Ω), where Ω = ω . In any case,figure XVII.15 shows the amplitudes of the motions of the two masses as a function offrequency, for α = 0.1, 0.5 and 0.9. The continuous black curves are for the left handmass; the dashed blue curve is for the right hand mass.
331.4FIGURE XVII.151.21α0.1Amplitude0.80.60.50.184.108.40.206.10.900 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5omega