Two Coupled Oscillators / Normal Modes

Transcription

1 Lecure 3 Phys 3750 Two Coupled Oscillaors / Normal Modes Overview and Moivaion: Today we ake a small, bu significan, sep owards wave moion. We will no ye observe waves, bu his sep is imporan in is own righ. The sep is he coupling ogeher of wo oscillaors via a spring ha is aached o boh oscillaing objecs. Key Mahemaics: We gain some experience wih coupled, linear ordinary differenial euaions. In paricular we find special soluions o hese euaions, known as normal modes, by solving an eigenvalue problem. I. Two Coupled Oscillaors Le's consider he diagram shown below, which is nohing more han copies of an harmonic oscillaor, he sysem ha we discussed las ime. We assume ha boh oscillaors have he same mass m and spring consan k s. Noice, however, ha because here are wo oscillaors each has i own displacemen, eiher or. k s m m k s 0 0 Based on he discussion las ime you should be able o immediaely wrie down he euaions of moion (one for each oscillaing objec) as & + ω 0, and (a) & + ω 0, (b) where ω k s m. As we saw las ime, he soluion o each of heses euaions is harmonic moion a he (angular) freuency ω. As should be obvious from he D M Riffe -- /4/03

2 Lecure 3 Phys 3750 picure, he moion of each oscillaor is independen of he oher oscillaor. This is also refleced in he euaion of moion for each oscillaor, which has nohing o do wih he oher oscillaor. Le's now make hings a bi more ineresing by adding in anoher spring ha connecs he wo oscillaing objecs ogeher, as illusraed in he following picure. To make hings even more ineresing we assume ha his new spring has a differen consan k s. However, o keep hings simple we assume ha he middle spring provides no force if 0. Tha is, his spring is neiher sreched or compressed if is lengh is eual o he is lengh when wo objecs are a euilibrium. k s m k m s k s 0 0 Thinking abou his picure we should realize ha he wo euaions of moion will no longer be independen. Tha is, he euaion of moion for he firs objec will depend (somehow) upon wha he second objec is doing, and vice versa. Le's use Newon's second law o wrie down he euaion moion for each objec. Recall ha Newon's second law for eiher objec ( i, ) can be wrien as Fi & i, () m where F i is he ne force on objec i. The ricky par, if here is a ricky par, is o deermine he sum F i on each objec. The ne force on he firs objec comes from he spring on he lef and he spring in he middle. Wih a lile hough you should realize ha his ne force is F D M Riffe -- /4/03

3 Lecure 3 Phys 3750 F s s ( k k ). (3a) Make sure ha you undersand he signs of all he erm on he rhs of his euaion. Noice ha he force provided by he middle spring depends no only on he firs objec's displacemen bu also on he second objec's displacemen. Similarly, he ne force on he second objec is F k s k s ( ). (3b) Subsiuing hese wo forces ino E. (), once for each objec, we obain he wo euaions of moion, & + ω + ω ( ) 0 (4a) for he firs objec and & + ω + ω ( ) 0 (4b) for he second. Here ω k m. Given he symmery of he problem, i migh no surprise you ha you can obain one euaion of moion from he oher wih he ransformaion in he subscrips ha label he objecs. So now we have a considerably more complicaed problem: as expeced from looking a he drawing above, he euaion of moion for each objec depends upon wha he oher objec is doing. Specifically, each euaion of moion depends upon he displacemen of he oher objec. II. Normal Modes A. Harmonic Ansaz So wha are he soluions o hese differenial euaions? Well, we will evenually wrie down he general soluion (nex lecure). Bu righ now we are going o look a a special class of soluions known as normal-mode soluions, or simply, normal modes. A normal mode is a soluion in which boh masses harmonically oscillae a he same freuency. We sae why hese special soluions are exremely useful a he end of he lecure. For now le's see if we can find hem. We use he complex form of harmonic moion and wrie e i Ω 0 and (5a) D M Riffe -3- /4/03

4 Lecure 3 Phys 3750 () e i Ω. (5b) 0 Noice ha he (unknown) freuency of oscillaion Ω of boh oscillaors is he same, a key feaure of a normal mode. Also, because we are using he complex form of harmonic moion, he ampliudes 0 and 0 may be complex, bu hey oo are unknown a his poin. Keep in mind ha E. (5) is only he form of a normal-mode soluion: i is only a soluion if i saisfies he euaions of moion. In oher words, we now need o find values of he freuency Ω and ampliudes 0 and 0 ha saisfy Es. (4a) and (4b), he euaions of moion. So le's subsiue E. (5) ino E. (4) and see wha ha ells us abou Ω, 0, and 0. Carrying ou he subsiuion and calculaing he derivaives yields Ω Ω iω iω ( e e ) 0 iω + ω e + ω, and (6a) iω 0e iω iω iω + ω e + ω ( e e ) 0. (6b) iω 0e Dividing by i e Ω (Is his legal?) and rearranging some erms gives ( ) + ω Ω ω 0 ω, and (7a) 0 0 ( ω + ω Ω ) 0 ω +. (7b) 0 0 So wha do we have here? We have wo algebraic euaions and hree unknowns Ω, 0, and 0. The problem seems a bi underspecified, and i is: as we shall see below, we will only be able o solve for Ω and he raio 0 0. B. Eigenvalue Problem If you have previously sudied differenial euaions and linear algebra you may be inclined o wrie E. (7) in marix noaion as ω + ω ω ω Ω 0 0, (8) which you would recognize as an eigenvalue problem. Generally, an eigenvalue problem is one where some linear operaor (in his case a marix) operaing on some objec (in his case a D column vecor) produces a consan (in his case Ω ) imes he original objec. Generally, an N -dimensional linear-algebra eigenvalue problem D M Riffe -4- /4/03

5 Lecure 3 Phys 3750 has N soluions [which consis of special values of he consan (or eigenvalue) Ω 0 and ampliudes K 0 0n (or eigenvecor M )]. 0N C. Eigenvalues Well, ha was a lo of erminology, bu wha abou he soluion? Well, le's rewrie E. (8) as ω + ω Ω ω ω + ω Ω (9) Now his is ineresing. Expressed in his way, we have he produc of wo uaniies eual o zero. There are wo ways ha E. (9) can be rue. The firs, which is he rivial (i.e., unineresing) soluion, is Physically, his corresponds o no moion of he sysem prey unineresing! The nonrivial way ha E. (9) can be saisfied is if he deerminan of he marix is zero. Tha is ω + ω Ω ω ω ω + ω Ω 0. (0) For a marix he deerminan is easily calculaed, E. (0) can be expressed as A C B D AD BC, so in his case ( ω + ω Ω ) ω 4 0. () E. () [or E. (0)] is known as he characerisic euaion for he eigenvalue problem. This is grea! We now have an euaion for he eigenvalue Ω and hus he normal-mode freuency Ω, ω + ω Ω ± ω. () Solving E. () for Ω produces he wo eigenvalues ω Ω ω, ω +, (3) which gives us four soluions for he normal-mode freuency D M Riffe -5- /4/03

6 Lecure 3 Phys 3750 Ω ± ω, ± ω + ω. (4) D. Eigenvecors and Normal Modes So now ha we have he eigenvalues ω and ω +, we need o find he eigenvecor associaed wih each eigenvalue. To do his we subsiue each eigenvalue ino eiher E. (7a) or (7b) (I doesn' maer which, you ge he same euaion in eiher case).. firs normal mode For he firs eigenvalue, Ω ω, his subsiuion produces ω ω 0, (5) 0 0 which gives us he resul for he ampliudes 0 0, (6) and so he eigenvecor associaed wih he firs normal mode is. This resul ells us ha boh oscillaors oscillae idenically [check ou E. (5) wih he resul of E. (6)] if his normal mode is excied. Tha is, he objecs oscillae wih exacly he same ampliude and he same phase. Now because he eigenvalue Ω ω corresponds o wo normal-mode freuencies Ω ±ω, his firs eigensoluion of he linear algebra problem gives us o wo linearlyindependen soluions o he euaions of moion. A e i and A e i ω (7a),(7b) ω is he firs soluion, and i iω B e and B e (8a),(8b) ω is he second, where he ampliudes A and B are arbirary. Euaions (7) and (8) can be wrien in linear-algebra inspired noaion as A e i ω + (9) and D M Riffe -6- /4/03

7 Lecure 3 Phys 3750 iω B e, (0) respecively. The + and denoe he + ω and ω soluions. However, because oscillaions a freuency ω are really jus oscillaions a ω, any linear combinaion of hese wo soluions really jus oscillaes a he freuency ω, and so any linear combinaion of E. (9) and (0) can be hough of as he firs normal-mode soluion a freuency ω. Tha is, he mos general way we can wrie he firs normal-mode soluion is + + i ω i ω ( A e + B e ), () where A and B are unspecified consans. Noe ha arbirariness of A and B is consisen wih our knowledge ha for an harmonic oscillaor he freuency is independen of he ampliude. Noe also ha (if we wish o a his poin) we can * specify he soluion o be real, in which case we would se B. A. second normal mode Le's now look a he second normal-mode soluion, which corresponds o he second eigenvalue ω +. As before, we subsiue his eigenvalue ino E. (7a), which gives us or ω ω 0 () , (3) and so he eigenvecor associaed wih he second normal mode is. So if his normal mode is excied he wo oscillaors oscillae wih he same ampliude, bu wih opposie phase (or a phase difference of π ). Tha is, when he firs oscillaor is moving o he lef he second is moving o he righ wih he same magniude in is displacemen, and vice versa. As in he case of he firs eigenvalue, here are wo linearly-independen soluions corresponding o he wo freuencies ± ω +. The firs soluion is D M Riffe -7- /4/03

8 Lecure 3 Phys 3750 A e i ω + ω and A e i ω + ω (4a),(4b) and he second is i B e ω + ω i ω and B e ω +, (5a),(5b) or in linear-algebra noaion A e i ω + ω (6) and i B ω + e ω. (7) As before, he general form of his normal-mode soluion is () () () i ω + i ω + ( Ae + Be + ) (8) + The graphs on he following page plo he ime-dependen ampliudes and ( ) for he wo normal modes for he following values of he arbirary consans: A B for he firs normal mode and A B for he second normal mode. (Wih hese choices he soluions are real.) For hese graphs we have also se m, k s, and k eual o, so ha s ω and ω + ω 3. (Admiedly, we have no specified unis here, bu if sandard SI unis are used for he mass, spring consans, and ime, hen he uni for displacemen is meers.) As he graphs show, in he firs normal mode he wo objecs oscillae idenically, while in he second normal mode hey oscillae exacly opposiely. As we will see in he nex lecure, a grea usefulness of he normal mode soluions is ha ANY soluion of Es. (4a) and (4b), he euaions of moion for his coupled oscillaor sysem, can be wrien as a linear combinaion of hese wo normal-mode soluions. Indeed, his propery of normal-mode soluions is so imporan ha i will be a heme hroughou he course. D M Riffe -8- /4/03

9 Lecure 3 Phys 3750 FIRST NORMAL MODE DISPLACEMENT 0 () () TIME SECOND NORMAL MODE DISPLACEMENT 0 () () TIME Exercises θ *3. Saring wih he Euler formula e i cos( θ ) + i sin( θ ) (and is complex conjugae), θ θ wrie cos ( θ ) and sin ( θ ) in erms of e i and e i. i ω i ω *3. Wrie he expression Ae + Be in he form C cos ( ω ) + Dsin( ω). Tha is, find * C and D in erms of A and B. From his resul show ha if B A hen C and D i are boh real (which means ha i Ae ω + Be ω is real). *3.3 In he graph of he firs normal-mode soluion, ( ) and ( ) boh look like cosine funcions. Show for A B, ha he soluion i ω i ω ( A e + Be ) can indeed be wrien as ( ) cos ω ( ). D M Riffe -9- /4/03

10 Lecure 3 Phys 3750 *3.4 If we ake he limi k s 0 in he coupled oscillaor problem, wha does his correspond o physically? Wha happens o he normal mode freuencies? Does his make sense? (Noe: if wo normal modes have he same freuency, hen hey are said o be degenerae, and any linear combinaion of hose wo normal modes is also a normal mode.) **3.5 Three coupled oscillaors. In his problem you will find he normal modes of hree coupled oscillaors, as illusraed below. Assume ha each objec has mass and each spring consan is. k s m 3 (The following seps lead you hrough he same procedure as is used in he noes o solve he wo-oscillaor problem in order o solve his problem. I will be mos helpful if you carefully sudy ha procedure before ackling his problem.) (a) Using Newon's second law, wrie down he euaion of moion for each objec [in he form of E. (4) in he noes]. (b) Assume a normal-mode ype soluion and find he hree algebraic euaions [euivalen o E. (7) in he noes] ha govern Ω and he ampliudes 0, 0, and 03. (c) Wrie your euaions in (b) in marix form [euivalen o E. (9) in he noes]. (d) Find he characerisic euaion [euivalen o E. () in he noes] ha deermines he hree eigenvalues. (Hin: you will need o calculae he deerminan of a 3 3 marix.) (e) Solve he characerisic euaion and show ha he hree eigenvalues for his problem are Ω ( ) ω, ω,( + ) ω. (f) For each eigenvalue, find he eigenvecor associaed wih ha eigenvalue. (g) Wrie down he 3 normal mode soluions in he form of Es. () and (8) in he noes. (h) As precisely as possible, describe he moion of he hree objecs for each of he normal modes. D M Riffe -0- /4/03