Golden ratio in a coupled-oscillator problem

Transcription

1 IOP PUBLISHING Eur. J. Phys. 28 (2007) EUROPEAN JOURNAL OF PHYSICS doi: / /28/5/013 Golden ratio in a coupled-oscillator proble Crystal M Mooran and John Eric Goff School of Sciences, Lynchburg College, Lynchburg, VA 24501, USA E-ail: goff@lynchburg.edu Received 31 May 2007, in final for 14 June 2007 Published 19 July 2007 Online at stacs.iop.org/ejp/28/897 Abstract The golden ratio appears in a classical echanics coupled-oscillator proble that any undergraduates ay not solve. Once the syetry is broen in a ore standard proble, the golden ratio appears. Several student exercises arise fro the proble considered in this paper. 1. Introduction The golden ratio eerges in a variety of contexts, fro art and architecture to nature and atheatics. Exaples where the golden ratio appears in physics include the study of the onset of chaos [1] and certain resistor networs [2]. Of the scores of boos devoted to the golden ratio, we particularly enjoy Livio s boo [3]. The irrational nuber ϕ = ( 5+1)/2 (ϕ = ) is the golden ratio. Aong its any interesting properties is that its inverse is related to itself by ϕ 1 = ϕ 1 (i.e. ϕ 1 = ). We will show in the next section that the golden ratio and its inverse appear in the solution to the coupled oscillator proble illustrated in figure 1. Two identical asses are allowed to oscillate while connected to two identical assless springs. The oscillation taes place along a horizontal frictionless table. Finding noral odes and corresponding frequencies is our goal. We have seen the aforeentioned proble as a hoewor proble in Taylor s text [4], though he does not ention the golden ratio 1. The proble also appears in other boos [5, 6], except that it is turned 90 so that the asses and springs hang in a unifor gravitational field. While we obtain the sae eigenfrequencies whether we solve the horizontal proble or the vertical proble, we find no ention of the golden ratio in the stateents of the vertical proble. While the proble we consider here is probably not solved that often by undergraduate and graduate classical echanics students, they ust surely solve one or both of the probles 1 Taylor provides a bac-of-the-boo answer to his proble 11.5 on page 766. However, he writes the eigenfrequencies in the for of equation (7) instead of equation (9), and there is no ention of the golden ratio /07/ $30.00 c 2007 IOP Publishing Ltd Printed in the UK 897

2 898 C M Mooran and J E Goff Figure 1. Our proble of interest. Both asses are identical as are both assless springs. The syste oscillates on a horizontal frictionless table. The box on the left represents an iovable wall Figure 2. A paradig of coupled oscillation seen in any textboos. The boxes represent iovable walls. The proble shown here reduces to our proble of interest for 1 = 2 =, 3 = 0, and 1 = 2 = Figure 3. This textboo paradig represents a odel of a triatoic olecule (perhaps CO 2, carbon dioxide, if 1 = 3 and 1 = 2 ; perhaps HCN, hydrogen cyanide, if one is interested in odeling an asyetric triatoic olecule). shown in figures 2 and 3. Several classical echanics textboos [4 11] solve one or both of the aforeentioned probles. If all asses in figures 2 and 3 are identical, as well as all the assless springs, both systes enjoy reflection syetry. While the syetry in the proble we solve is broen, we find it interesting that a nuber associated with aesthetic syetry appears in the solution. Section 2 will solve the proble given in figure 1 and section 3 will provide discussion for why the golden ratio appears in our solution. Section 4 will offer physics instructors soe proble extensions that can be used in the classroo or for hoewor assignents. 2. Proble solution The proble at hand is easily solved with either a Newtonian approach or a Lagrangian approach. We eploy Newton s second law here. If the position of the left ass is x 1 and the position of the right ass is x 2, Newton s second law gives ẍ 1 = x 1 (x 1 x 2 ) (1) for the left ass and ẍ 2 = (x 2 x 1 ) (2) for the right ass. A dot represents a total tie derivative. To solve the above coupled differential equations, we guess solutions of the for x 1 (t) = A 1 e iωt (3)

3 Golden ratio in a coupled-oscillator proble 899 and x 2 (t) = A 2 e iωt, (4) where A 1 and A 2 are coplex aplitudes and ω is a frequency to be deterined. Inserting equations (3) and (4) into equations (1) and (2), we find a atrix equation given by ( ) ω 2 +2ω0 2 ω0 2 (A1 ) ( ) 0 =, (5) ω0 2 ω 2 + ω0 2 A 2 0 where ω0 2 = /. Note the asyetry in the 2 2 atrix; naely, a factor of 2 appears in the top left entry while no such factor appears in the lower right entry. One can easily trace the origin of that factor of 2 bac to the right-hand side of equation (1) where there are two x 1 ters. Physically, the factor of 2 arises because the left ass feels forces fro two springs while the right ass only feels a force fro one spring. To avoid the trivial solution (i.e. A 1 = A 2 = 0), the deterinant of the 2 2atrixin equation (5) ust vanish. The quadratic equation in ω 2 one gets is (ω 2 ) 2 3ω 2 0 (ω2 ) + ( ω 2 0) 2 = 0. (6) The two positive solutions are 3 ± 5 ω ± = ω 0, (7) 2 where we label the two eigenfrequencies ω + and ω. The eigenfrequencies are siplified by noting that eaning 3 ± 5 = 1 2 ( 5 ± 1) 2, (8) ω ± = ω 0 2 ( 5 ± 1). (9) We now use the golden ratio (ϕ = ( 5+1)/2) and its inverse (ϕ 1 = ϕ 1) to write the eigenfrequencies in the copact way ω ± = ϕ ±1 ω 0. (10) Thus, the eigenfrequencies of the proble shown in figure 1 are related to the golden ratio in a very siple anner. Going bac to equation (5) and finding the relationship between A 1 and A 2 for each eigenfrequency allows us to find the noral odes. The (unnoralized) noral odes we find are η ± = ϕ ±1 x 1 x 2. (11) By the definition of noral odes, the decoupled differential equations satisfied by η ± are η ± + ω 2 ± η ± = 0, (12) where ω ± are given by equation (10). Thus, the golden ratio appears in the noral odes, too. If the noral ode associated with ω ± is excited, the aplitude of the left ass is ϕ ±1 ties the aplitude of the right ass.

4 900 C M Mooran and J E Goff y z Figure 4. The golden section requires (y + z)/y = y/z = ϕ. 3. Discussion To understand how the golden ratio enters into our proble, factor equation (6)as ( ω 2 ω 0 ω ω0 2 )( ω 2 + ω 0 ω ω0 2 ) = 0. (13) One of the two factors in the above equation ust vanish. Setting the left factor to zero and solving for the positive root gives ω + fro equation (9). If one instead loos for the positive root by equating the right factor to zero, one gets ω fro equation (9). Factoring equation (6) is where we ae the connection with the golden ratio. Figure 4 shows the so-called golden section. What aes the section golden is that the ratio of y + z to y is the sae as the ratio of y to z. That ratio is defined as the golden ratio. Matheatically, y + z = y = ϕ. (14) y z Setting up the quadratic equation for y gives y 2 zy z 2 = 0. (15) The above equation s positive root is ϕz. Note that the for of the above equation is exactly the sae as the left factor in equation (13). Had we wished to solve equation (14) as a quadratic equation for z, we would have obtained z 2 + yz y 2 = 0. (16) In other words, we obtain the exact sae for as that of the right factor in equation (13). The above equation s positive root is ϕ 1 y. Thus, the golden ratio enters our physical proble because the quadratic equation that one ust solve to find the eigenfrequencies is exactly the sae as the quadratic equation one ust solve to obtain the golden ratio fro the golden section. We thus see the golden ratio in our echanical syste of interest. 4. Classroo extensions The first exercise an instructor ight give a class is to derive equation (10). This is a standard undergraduate exercise in which a student aes use of ideas fro linear algebra. A second exercise is to derive equation (11), or at least verify that after solving equation (11)forx 1 and x 2 in ters of η + and η and plugging bac into equations (1) and (2) that equation (12)isthe result. This last exercise is very straightforward, but it requires students to use an interesting property of the golden ratio, naely ϕ 1 = ϕ 1. If an instructor is looing for ways to incorporate coputational techniques into a classical echanics course, he or she could have students use sybolic software pacages such as Matheatica, MATLAB and JMathLib and their linear algebra tools to derive the noralized noral odes, i.e. the noralized version of equation (11). 2 A sipler coputational exercise is to ipose initial conditions on the syste and as students to plot x 1 (t) and x 2 (t) versus 2 Follow the procedure given in, for exaple, the text [8] by Thornton and Marion. See pp

5 Golden ratio in a coupled-oscillator proble x/x x 1 / x 0 x 2 / x ω 0 t Figure 5. Equations (18)and(19) are plotted against the diensionless tie variable t = ω 0 t. Figure 6. A challenge proble. Shown above are N identical asses on N identical assless springs. tie. For exaple, if the right ass in figure 1 is displaced to the right a distance x 0 fro its equilibriu position while the left ass is held in its equilibriu position, the initial conditions that ust be iposed are x 1 (0) = 0, x 2 (0) = x 0, ẋ 1 (0) = 0, and ẋ 2 (0) = 0. (17) Students should then be able to derive the positions as functions of tie to be x 1 ( t) = 1 2ϕ 1 [cos(ϕ 1 t) cos(ϕ t)] (18) and 1 x 2 ( t) = 2ϕ 1 [ϕ cos(ϕ 1 t) + ϕ 1 cos(ϕ t)]. (19) Here, we use diensionless variables such that x 1 = x 1 /x 0, x 2 = x 2 /x 0 and t = ω 0 t.notealso that (2ϕ 1) is just a fancy way of writing 5. Students can then plot the above equations and get what we show in figure 5. Note that there is never a coplete transfer of energy fro one ass to the other; i.e., one ass is never copletely otionless at its equilibriu position. Finally, a ore challenging student exercise is to consider the proble illustrated in figure 6. That figure shows N identical asses and N identical assless springs. That syste

6 902 C M Mooran and J E Goff has a set of N eigenfrequencies {ω j } N j=1. Once students deterine the N eigenfrequencies, they can show that N ω j = 1, (20) j=1 where ω j = ω j /ω 0. Students can chec their results for the N = 2 case we exained in this paper 3. Note fro equation (10) that ω + ω = ω 2 0. References [1] Tabor M 1989 Chaos and Integrability in Nonlinear Dynaics (New Yor: Wiley) pp [2] Srinivasan T P 1992 Fibonacci sequence, golden ratio, and a networ of resistors A.J.Phys [3] Livio M 2002 The Golden Ratio: The Story of Phi, the World s Most Astonishing Nuber (New Yor: Broadway Boos) [4] Taylor J R 2005 Classical Mechanics (California: University Science Boos) (see proble 11.5 on pp 448 9) [5] Arya A P 1998 Introduction to Classical Mechanics (Englewood Cliffs, NJ: Prentice-Hall) (see proble 14.9 on pp 608 9) [6] Chow T L 1995 Classical Mechanics (New Jersey: Wiley) (see proble 11.9 on p 431) [7] Syon K R 1971 Mechanics (Longan, MA: Addison-Wesley) p 191 [8] Thornton S T and Marion J B 2004 Classical Dynaics of Particles and Systes (California: Thoson Broos/Cole) pp 469 and 491 [9] Fowles G R and Cassiday G L 2005 Analytical Mechanics (California: Thoson Broos/Cole) pp 472 and 496 [10] Goldstein H, Poole C and Safo J 2002 Classical Mechanics (New Yor: Addison-Wesley) p 253 [11] Fetter A L and Waleca J D 1980 Theoretical Mechanics of Particles and Continua (New Yor: McGraw-Hill) (see proble 4.2 on p 130) 3 We offer students a couple of hints that ay help the obtain equation (20). In the liit of large N, figure 6 loos lie a loaded string with one end attached to a fixed wall. Follow the procedure in, for exaple, the text [8] by Thornton and Marion, beginning on p 498. (Thornton and Marion consider a loaded string fixed at both ends. While their proble is different than ours, the ethod of solution is not.) One can show that the eigenfrequencies can be written in ters of sine functions. To prove equation (20), try either an analytic proof or a geoetric proof using a unit circle.