Phase Relations in the Synchronized Motion of Two-Pendulum System

Boris R. Andrievsky Institute for Problems of Mechanical Engineering of Russian Academy of Sciences (IPME RAS) 61 Bolshoy Av., St.Petersburg,199178 Russia E-mail: [email protected]; b [email protected]

Abstract actly the same frequency and 180 degrees out of phase. After external disturbance was made, the an- The synchronization phenomenon in nonlinear oscil- tiphase state was restored within a half of an hour lating system is studied by means of examination and remained indefinitely. Huygens’ synchronization of the coupled pendulums. Dependence of the phase observations have served to inspire study of sympa- shift between pendulum states on system parameters thetic rhythms of interacting nonlinear oscillators in and initial conditions is studied both analytically and many areas of science and technology. The onset of numerically. The harmonic linearization (HL) tech- synchronization and the selection of particular phase nique is applied for analytical examinations. relations have been studied in the numerous papers and monographs, see e.g. [5, 7, 4, 18, 21, 23]. Explo- ration of the synchronization phenomena is usually 1 Introduction. Resonance and focused on the study of the phase relations between synchronization motions of the coherent units. Different studies of the pair of coupled oscillators confirmed an obser- Nonlinear oscillations problem falls within the fields vation that the asynchronous mode of motion is a of nonlinear mechanics, nonlinear physics, as well as predominant one. At the same time some experi- nonlinear control theory. During the latest years the ments show that the synchronous mode can also be subject of nonlinear oscillations control has attracted observed. Therefore the problem of phase relations growing researchers’ attention among the other fields between coupled oscillators is still opened. of research where control theory methods were ap- This paper is focused on the study of phase rela- plied to physical phenomena exploration [13]. tions between coupled oscillators, excited by means The resonant property of nonlinear systems was in- of the feedback control torque. The model of 1-DOF tensively studied in the areas of charged-particle ac- pendulums coupled with a weak spring, “the Cou- celeration physics [17], plasma physics [10], celestial pled Pendulums of the Kumamoto University”, [22] mechanics [3], and automatic control. Various defi- is considered. nitions of the resonance phenomenon have been in- The Coupled Pendulums dynamics model is given in troduced to describe such properties of the nonlin- Sec. 2, where some design features of the system ear oscillations as stochastic resonance [23], chaotic are also briefly described. In Sec. 3, the phase re- resonance [9], autoresonance [17, 19], and feedback lations are studied analytically on the basis of the resonance [1, 11, 12, 14]. harmonic linearization (HL) method. The case of a The resonance property is allied to the synchroniza- single excitation is considered. It is shown that both tion one. Synchronization is usually treated as corre- synchronous and asynchronous modes can appear. sponding in time behavior of two or more processes. For the particular system the computer simulation General definitions of synchronization were proposed has been provided for checking up the analytical re- in [6, 8]. Starting with the work of Christian Huygens sults and receiving additional information on system [20], the synchronization phenomena attracted atten- behavior in Sec. 4. Some results of the computer si- tion of many researchers, see e.g. monographs [5, 24]. mulations for double-input excitation are presented In his study, C. Huygens described synchronization in Sec. 5. The summarized information is given in of the pair of pendulum clocks weakly coupled one Conclusion. with another owing to the common basement. Huy- gens had found the pendulum clocks swung in ex- 2 Coupled pendulums model 3.1 Case of the identical pendulums At first, consider the simplified case of the identical Let us consider the “Coupled Pendulums of the pendulums. Assume that J1 = J2 = J, m1 = m2 = Kumamoto University” [22]. This system consists m, l1 = l2 = l, ρ1 = ρ2 = ρ. For that case the trans- of a pair of 1-DOF pendulums, coupled with a weak fer function of the “linear part” of the closed-loop spring. A motor is attached to the first pendulum. system (2), (1) has a simple form and the harmonic The joint of the second pendulum is passive. balance equation can be easily solved. The transfer function W (s) from the input signal u to the output The system dynamics can be written as [22]: ϕ˙ 1 is as follows:

2 2 2  s Js +ρs+β +kl0 J1ϕ¨1 + ρ1ϕ˙ 1 + m1gl1 sin ϕ1 W (s)= , (4)  2 2 2 2 2  = f(ϕ , ϕ ) cos ϕ + u(t), (Js +ρs+β ) Js +ρs+β +2kl0 1 2 1 (1) J2ϕ¨2 + ρ2ϕ˙ 2 + m2gl2 sin ϕ2 where parameter β2 = mgl is introduced. The value  √  = −f(ϕ1, ϕ2) cos ϕ2, of β/ J can be treated as eigen angular frequency 2 of the small oscillations for (each) single pendulum. f(ϕ1, ϕ2) = −kl0(sin ϕ1 − sin ϕ2), The feedback (3) is positive and the harmonic bal- where u is an external torque (developed by the ance equation can be written as W (jω) = 1/q(a), drive motor); J , g, ρ , k represent inertia, gravita- i i where j2 = −1; q(a) is a harmonic linearization tional acceleration, frictional coeﬃcient and coupling gain (also known as a describing function). For strength, respectively; m denotes mass of the bob; i the considered relay control law (3) this gain has a i = 1, 2 is a number of the pendulum. The motor 4γ form: q(a) = , where a denotes the oscillations torque u(t) is considered as a control action, applied πa to the system (1). The following linearized model is magnitude of the system’s linear part output (i.e. also used: a = lim ϕ˙ 1(t)). Solutions ω =ω ˜, a =a ˜ of the t→∞  harmonic balance equation are the sought quantities J1ϕ¨1 + ρ1ϕ˙ 1 + m1gl1ϕ1  of the frequency and amplitude of oscillations. The  +kl2(ϕ − ϕ ) = u(t), 0 1 2 (2) simple calculations lead to the following equations: J ϕ¨ + ρ ϕ˙ + m gl ϕ  2 2 2 2 2 2 2   2 2 4 2 2 2 2  +kl (ϕ − ϕ ) = 0.  πaJ ω + πa − Jβ − ρ − J β + 2kl0 0 2 1    2 2 2 2 Equations (2) are obtained via linearization of (1) at  +4ργ ω + πaβ β + 2kl0 = 0, the lower equilibrium point ϕ = ϕ = 0. 1 2  2 − πaρ+2γJω3 + πaβ2ρ + (β2 + 2kl2)ρ  0  Let the system be excited by means of the relay feed-  2 2  −4γ(β + kl ) ω = 0. back control action 0 (5) u = γ signϕ ˙ , (3) 1 Omitting the trivial (and unstable) solutionω ˜ = 0, a˜ = 0 one may rewrite the second equation in (5) as where parameter γ gives the control torque magni- follows: tude. The law (3) is a particular case of the speed- gradient control algorithms [11, 12, 13]. It had been 2 2 2 (−ω J + β + kl0)(πaρ − 2γ) = 0. (6) presented and studied in the earlier published papers [1, 2, 11, 15, 16]. Let us consider two cases:

2 2 2 1. −ω J + β + kl0 = 0 and 3 Examination of synchronous oscillations 2. πaρ − 2γ = 0. mode via harmonic linearization method

Let us examine the motion of the system (1), (3), Both of the above equalities satisfy Eq. (6). zeroing in phase shifts between the pendulum de- ﬂections in the steady-state mode. The harmonic 2 2 2 1. Assume that ω = (β + kl0)/J. Substituting linearization method is applied to provide the an- this expression for ω2 in (5) one gets: alytical expressions for amplitude and frequency of oscillations and the phase shift between the pendu- 4γρ(β2 + kl2) a˜ = 0 (7) lum states. 2 2 2 2 4 π ρ (β + kl0) + k l0J Reverting to the expression for ω one obtains: It is valid that k0 > k∗, but for the small ρ it can be taken k0 ≈ k∗. For a numerical example let us pick 1 q ω˜ = J β2 + kl2 . (8) up the system parameters given in [22]: J = J1 = 0 J 0 0.0071 kgm, m = m1 = 0.3 kg, l = l1 = 0.150 m, l0 = Eqs. (7), (8) give amplitude and frequency of 0.02 m, ρ = ρ1 = 0.013 Nms. Computing from (12), one of two possible oscillating motions of the (13) gives k∗ = 286 N/m, k0 = 287 N/m. Note that system. such the values of k correspond to a “stiﬀ” spring; in the paper [22] the value k = 80 ÷ 120 N/m is 2. Assume now that πaρ−2γ = 0. Then it is valid: considered. 2γ a˜ = . (9) ρπ Let us make some resumptive remarks.

Substitutinga ˜ given in (9) for a in the ﬁrst equa- 1. There exists a single limit cycle for k ∈ [0, k ]. It tion of (5) and omitting the negative solution ∗ has the frequency ω =ω ˜ = J −1pJ(β2 + kl2) one obtains the following expression forω ˜: 0 0 (see Eq. (8)) and the amplitude given by Eq. 1 q (7). It can be easily checked that this limit cycle ω˜ = ±κ − 2ρ2 + 4Jkl2 + 4β2J, (10) 1,2 2J 0 is stable.

p 4 2 2 2 2 2 2 4 2. If k > k then the limit cycles with the frequen- where κ=2 ρ −4ρ Jkl0 −4ρ β J +4J k l0. ∗ ciesω ˜1,ω ˜2 ( see (10)) born. These cycles have the same oscillation amplitudes, see (9). The cy- Note that (7), (8) make sense for all (positive) pa- cle of item 1 with the frequencyω ˜0 becomes un- rameter values, while the expression (10) forω ˜1,2 stable. Accordingly with the HL method, both and, therefore, the formula (9) fora ˜ make sense if the limit cycles with the frequenciesω ˜1,ω ˜2 are all radicands in (10) are nonnegative. Let us find stable. To divide the system state space on the the explicit conditions that mark out the region of regions of attraction of each limit cycle some dif- existence the solutions ω1,2 at the system parame- ferent methods must be used, because the har- ters space. Regarding κ in (10) as a function of the monic linearization technique does not give a coupling gain k and supposing that κ ≥ 0, one gets definite answer in that case. the following inequalities with respect to k: √ √ ρρ − 2β J ρρ + 2β J Note that the formulae (7), (9) give oscillation am- k ≤ , k ≥ . (11) 2 2 plitudes for the output of the “linear part” of the 2l0J 2l0J system. In the considered case this output isϕ ˙ 1. To find the amplitude of ϕ1(t) the amplitudes given by Examination the linearized model equation (2) gives the formulae (7), (9) should be divided by the corre- that√ for each uncoupled oscillator it is valid that ρ < sponding frequencies obtained from (8), (10). There- 2β J. It is naturally to assume that the friction fore, the amplitude of oscillations with the frequency parameter ρ of the pendulums is small (so that the ω˜1 ( see (10)) is less that one with the frequencyω ˜2. pendulums free motion is an oscillatory one) and to pick up the second inequality from (11). It can be Let us find the phase shift between the processes easily checked that for all positive k, satisfying (11), ϕ1(t) and ϕ2(t) in the steady-state oscillation mode. the radicands in (10) are also positive. Summarizing, For the sake of it let us find the transfer functions of note that there exists a “critical” (or a bifurcation) the system (2) from input u to the outputs ϕ1 and ϕ2: ∆ value of k = k∗, defined as Js2 + ρs + β2 + kl2 √ W (s) = 0 , (14) ρρ + 2β J ϕ1 A(s) k = , (12) ∗ 2l2J kl2 0 W (s) = 0 , (15) ϕ2 A(s) so that for k > k∗ there exist solutions with the frequenciesω ˜ ,ω ˜ (see (10)) and the amplitude (9). 1 1 where common denominator A(s) is Evidently, for k = k∗ it is validω ˜1 =ω ˜2, but these values differ fromω ˜, obtained from (8) for k = k . ∗ A(s) = Js2 + ρs + β2)Js2 + ρs + β2 + 2kl2 It is easy to find that the solutions of (8) and (10) 0 have a common point if (see also (4)). The phase shifts ψ1, ψ2 between the input u(t) and the output processes ϕ (t), ϕ (t) are ρ p 2 2 1 2 k = k0 = ρ + ρ + 4β J . (13) 2 2 founded, correspondingly, as ψ1(ω) = arg − Jω + 2Jl0 2 2 jρω + β + kl0 − arg A(jω), ψ2(ω) = − arg A(jω). Therefore, the phase shift ∆ψ = ψ1 −ψ2 between the pendulum turning angles ϕ1(t) and ϕ2(t) depends on oscillations frequency ω and can be found as

2 2 2 ∆ψ(ω) = arg − Jω + jρω + β + kl0 = ρω  arctan , if ω2 < kl2 + β2,  U(ω) 0  π (16) = , if ω2 = kl2 + β2, 2 0  ρω 2 2 2  π + arctan , if ω > kl + β , Figure 1: Oscillations frequenciesω ˜ versus k.  U(ω) 0 i

2 2 2 where U(ω) = kl0 + β − Jω .

Substituting ωi (i = 0, 1, 2) for ω in (16) one ﬁnds the phase shifts ∆ψi for the self-excited oscillations.

In the case of k < k∗, the oscillations frequency is 2 2 2 defined by (8). Hence it is valid thatω ˜ = kl0 + β and, accordingly with (16), ∆ψ(˜ω) = π/2. Thus, if the spring elasticity is “small” (i.e. k < k∗) the phase shift between pendulums turning angles is a quarter- Figure 2: Oscillations amplitudesa ˜ϕ,i versus k; period. If k is greater than k∗ the oscillation frequen- γ = 0.1 Nm. cies areω ˜1 orω ˜2. Substitutingω ˜1,ω ˜2 from (10) for ω in (16) one can find the phase shift as a function of ∆ system parameters. Define ∆ψk,1 = ∆ψ(ω) , ω=˜ω1(k) ∆ ∆ψk,2 = ∆ψ(ω) . Evidently, if k → ∞ it is ω=˜ω2(k) valid that ∆ψk,1 → π, and ∆ψk,2 → 0. Therefore, for “large” coupling parameter k two stable oscillating modes with different frequencies appear. One of these modes trends toward to the sinphase motion

(∆ψk ≈ 0), another one trends to the antiphase mo- Figure 3: Phase shifts ∆ψi versus k. tion (∆ψk ≈ π).

Let us make some graphic illustrations. For the given above system parameters the values ofω ˜i,a ˜ϕ,i ≡ a˜i/ω˜i and ∆ψi are calculated and plotted as the functions of the coupling parameter k in Figs. 1 – 3. The oscillations amplitudes are found for γ = 0.1 Nm in (3). The bifurcation value of k = k∗ is also marked.

3.2 Case of the non-identical pendulums Consider now the case of different pendulums param- Figure 4: Oscillations frequenciesω ˜i versus k. eters. Such the case is more complex that the fore- Non-identical pendulums. going one and for the sake of brevity let us avoid any analytical expressions and be confined study- ing the numerical example. Assume that in (1) the moments of inertia and the frictional coefficients of the pendulums are given as: J1 = 0.0071 kgm, J2 = 0.0068 kgm, ρ1 = 0.013 Nms, ρ2 = 0.009 Nms, l1 = l2 = 0.150 m, m1 = m2 = 0.3 kg.

The plots of the solutionsω ˜ =ω ˜(k),a ˜ =a ˜(k) of the harmonic balance equation for that case and the corresponding phase shifts ∆ψ are shown in Figs. 4 Figure 5: Oscillations amplitudesa ˜ϕ,i versus k; – 6. It is seen that the bifurcation is “hard” for that γ = 0.1 Nm. Non-identical pendulums. case: there exists a mode hopping in the bifurcation Figure 6: Phase shifts ∆ψi versus k. Non-identical pendulums.

Figure 11: Phase shift v.s. k.

Figure 12: Amplitude of oscillations v.s. k.

Figure 7: Nyquist plots for k = 50, 300, 600 N/m.

Figure 8: Oscillations frequency Ω v.s. k and ϕ2(0).

Figure 9: Phase shift ∆ψ v.s. k and ϕ2(0).

Figure 13: “Topographical maps” for ∆ψ.

Figure 10: Oscillations frequency v.s. k. point k = k∗ ≈ 300 N/m. Accordingly with the alytically obtained results, the solid lines correspond HL method it can be easily shown that the oscilla- to numerical results. The different initial conditions tions with “intermediate” frequencyω ˜0 (see Fig. 4) on ϕ2 are taken, therefore the solid curves Fig. 10 are unstable. The stable oscillations with the both are projections of the surface shown in Fig. 8 on the frequenciesω ˜1 andω ˜2 can be engendered in the sys- (k, Ω)-plane. For more detailed information about tem if k > k∗. Note that the greater frequencyω ˜1 phase shift dependence on the initial conditions, let corresponds to the approximately antiphase motion us consider Figs. 13, where the “topographical maps” (∆ψ1 plot in Fig. 6), while the frequencyω ˜2 corre- (contour plots) for the phase shift ∆ψ on the plane sponds to the inphase one (∆ψ2 plot). This effect ph1(0), ϕ2(0) are pictured. As it is seen from the has a lucid physical explanation: in the antiphase plots, for the small values of k the phase shift is ap- mode the spring torque is added to the torque of the proximately 50 degrees and less. For the values of gravitational force. k greater than the bifurcation one, the phase shift depends also on initial conditions. The bifurcation diagram can be elucidated by means of the Nyquist plots of the linear system W (jω). For the excitation law (3), the HL of the nonlinear ele- 5 Phase relations. Double-input system case ment is real and merely amplitude-dependent, therefore the limit cycles corresponds to the positive roots Consider now a model of the pendulum system with of V (ω), where it is defined W (jω) =∆ U(ω) + jV (ω). two external torques u1 and u2: Nyquist plots for the cases of k = 50, 300, 600 N/m  are shown in Fig 7. The asterisks mark solutions for J1ϕ¨1 + ρ1ϕ˙ 1 + m1gl1 sin ϕ1  stable oscillations, while points correspond to the un-  = f(ϕ1, ϕ2) cos ϕ1 + u1(t), stable ones. J2ϕ¨2 + ρ2ϕ˙ 2 + m2gl2 sin ϕ2   = −f(ϕ1, ϕ2) cos ϕ2 + u2(t). f(ϕ , ϕ ) = −kl2(sin ϕ − sin ϕ ), (17) 4 Numerical analyses of the phase relations. 1 2 0 1 2 The single-input system case Simirarly to (3) the excitation law is taken as u = γ signϕ ˙ , u = γ signϕ ˙ . (18) The HL method is the approximate one. Strictly 1 2 1 2 2 2 speaking, its application even for oscillation stability analyses based on heuristic reasons. Besides, we need Some numerical results of examination the closed- to get a definite answer about the posed problem of loop system (17), (18) are presented in Figs. 14, 15. phase relations: what kind of motion (sinphase or It is seen that antiphase synchronization occurs for antiphase) appears in the system? Let us use the “small” coupling k and for any initial conditions. For computer simulations for solving a problem for the greater k the steady-state motion can be sinphase (if particular set of the system parameters. The values the initial angles have the same sign), or antiphase of parameters are given above in subsection 3.2. (in the case of the opposite signs of the initial angles).

In our experiments, the closed-loop system (1), (3) was repeatedly simulated for the different initial Conclusions conditions ϕ1(0), ϕ2(0) and the coupling parameter k. The simulations was made for time interval In the present work a coupled pendulum system ex- t ∈ [0, tfin], tfin = 50 s. Time histories of ϕ1(t), cited via relay differential feedback is studied analyt- ϕ2(t) for the latest τ seconds (τ = 15 s) was pro- ically, via the HL method and numerically, by means cessed to find oscillation parameters (frequency Ω, of the computer simulations. Following conclusions angular amplitudes, phase shift). Some results are can be done: presented in Figs. 8 – 13.

In Figs. 8, 9 the 3D-plots of the oscillations fre- 1. There exists frequency synchronization in the steady-state mode. The pendulum deviation quency and phase shift versus initial value of ϕ2 and angles have the same oscillation frequency and coupling gain k are pictured. Initial condition on ϕ1 is taken equal to one radian. The control magnitude some constant phase shift. γ = 0.1 Nm. One can observe that both sinphase 2. If only one pendulum is excited, for the small and antiphase steady-state motions occur if the cou- gain k and small angular deflections the phase pling gain is sufficiently large. For comparison of the shift is about π/2 (a quarter-period). In the analytical results of Sec. 3 and the numerical ones let same case if the amplitudes are “large” the ap- us examine Figs. 10 – 12. The dashed lines show an- proximately sinphase oscillations occur. dation for Basic Research (grant 02-01-00765) and Presidium of Russian Academy of Sciences (scientific program 19, project 1.4).

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