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Oscillations of a pendulum with a periodically varying length and a model of swing
International Journal of Non-Linear Mechanics 34 (1999) 105—109
Oscillations of a pendulum with a periodically varying length and a model of swing M.A. Pinsky!,*, A.A. Zevin" !Mathematics Department, ºniversity of Nevada Reno, Reno, N» 89557, ºSA "¹ransmag Research Institute, ºkraine Academy of Sciences, 320005 Dniepropetrovsk, Piesarzhevskogo 5, ºkraine Received 18 August 1997
Abstract Qualitative analysis of a pendulum with a periodically varying length is conducted. It is proved that there are two periodic solutions having a prescribed amplitude A(n and a period ¹ which is an even multiple k of the excitation period. Stability analysis is carried out for the principal parametric oscillations (k"2). In this connection it is shown that such a pendulum cannot serve as a mathematical model of swing as it is generally considered. ( 1998 Elsevier Science Ltd. All rights reserved. Keywords: Global stability analysis; Parametric oscillations
1. Introduction Oscillations of a pendulum with a varying length are described by the equation (l2(ut)x@)@#gl(ut) sin x"0
(1)
where x is an angle coordinate, l is the length of the pendulum, g is the acceleration due to gravity and u is an excitation frequency. In this paper we study the existence and stability of periodic solutions of system (1). The known
Contributed by W. F. Ames. *Corresponding author. Tel.: 702-7846725; e-mail:pinsky@ unr.edu
analytical results on the problems (see, e.g. [1, 2]) are obtained by approximate methods for the case l(ut)"l #e cos ut where e is a small 0 parameter (so that the system is quasi-conservative). The distinctive features of our analysis are as follows: the variation of the length of the pendulum is not bounded, a wide class of the excitation (including, in particular, the harmonic one) is considered and, finally, a qualitative approach is employed with the aim of obtaining rigorous results. Suppose that l(ut)"l(!ut)"!l(ut#n), l(ut)'0, l@(ut))0 for t3(0, n/u).
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(2)
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M.A. Pinsky and A.A. Zevin / International Journal of Non-Linear Mechanics 34 (1999) 105–109
Thus, the function l(ut) is symmetric, periodic in t with a period ¹ "2n/u and changes monotoni0 cally between the maximum and minimal values l "l(0) and l "l(n). ` ~ We consider symmetric ¹-periodic solutions of the kind
2. Existence of periodic oscillations
x (t)"x (!t)"!x (t#¹/2), 1 1 1 x (t)'0 on (!¹/4, ¹/4) 1 and
¹heorem 1 For any A(n and even integer k, there exists u (A, k) such that Eq. (1) with u"u possesses i i a solution x (t) with the amplitude A and the period i ¹ "2kn/u . i i
(3)
x (t)"!x (!t)"!x (t#¹/2), 2 2 2 x (t)'0 on (0, ¹/2) (4) 2 where ¹"k¹ , k being the order of subharmonic 0 oscillations. It can be easily shown that the symmetry condition x(t)"!x(t#¹/2) may be satisfied only for even integer k. Note that, on a shift in t by ¹/4, solution (3) takes form (4). Physically, distinctions between these solutions are as follows: under oscillations x (t), the 1 length l(ut) takes the values l and l at the instan~ ` ces the pendulum passes, respectively, the equilibrium or extreme positions, while for the solution x (t), the situation is directly opposite (Fig. 1). 2
The following theorem establishes the existence of solutions (3) and (4) having a prescribed amplitude A"max x(t) and order k.
Proof Let x(t) be the solution of Eq. (1) with x(0)"A and x@(0)"0. If for some u, x(kn/2u)"0,
(5)
then, by Eqs. (1) and (2), x(t)"x(!t)"!x(kn/ u!t), so the corresponding solution x(t) is of the form (3) with the period ¹"2kn/u. As seen from Eq. (1), x(t) decreases for x(t)*0. Let ¸"¸(x, u)"l(ut(x)) where t(x) is the respective inverse function. The phase trajectory v(x, u) of the solution x(t) satisfies the equation v
d(¸2v) #g¸ sin x"0. dx
(6)
Integrating Eq. (6) premultiplied by ¸2, we obtain ¸4v2"2g
A
Px ¸3 sin x dx,
(7)
whence v2 (x)(v2(x, u)(v2 (x) for x3[0, A), ~ ` v2 (x)"2gl3 (cos x!cos A)/l4 , ~ ~ ` v2 (x)"2gl3 (cos x!cos A)/l4 . ~ ` `
(8)
Let t "t (A, u) be the first zero of x(t), then Eq. 0 0 (5) is equivalent to t "kn/2u. Expressing t by 0 0 v(x, u), we obtain the following equation for u: Fig. 1. The trajectories of a pendulum. ( —— ) x (t), ( — — — ) 1 x (t). 2
A
P0 D v(x, u) D~1 dx"kn/2u.
(9)
M.A. Pinsky and A.A. Zevin / International Journal of Non-Linear Mechanics 34 (1999) 105–109
By inequality (8), the left side of this equation admits bilateral bounds independent of u, whereas the right side changes within (0,J) as u changes within (J, 0). Therefore, the required root u (A) 1 certainly exists which proves the existence of the solution x (t). 1 As mentioned above, solution (4) takes form (3) after a shift in time by ¹/4. So, the existence of the solution x (t) is established analogously. The the2 orem is proved. K From Eqs. (9) and (8) it follows that the period ¹"2kn/u of parametric oscillations of the kind (3) or (4) with an amplitude A satisfies the inequality (10) l3@2l~2¹ (A)(¹(A)(l3@2l~2¹ (A) ` ~ 0 ~ ` 0 where ¹ (A) is the corresponding period of free 0 oscillations of the pendulum having the unit length. Note that these bounds do not depend on the order k of the oscillations.
Consider now stability of the solutions x (t), i i"1, 2, with the period ¹"4n/u (i.e. k"2; oscillations of a parametrically excited system with such a period are called principal). As is known, stability is determined by the corresponding variational equation (l2y@)@#a y"0, i a (t)"gl(ut) cos x (t). (11) i i If Eq. (11) is unstable or stable, then the solution x (t) is unstable or stable to a first approximation. i The latter, generally, does not guarantee the Lyapunov stability of the solution x (t); however, i an introduction of arbitrary small dissipative forces makes this solution asymptotically stable [3]. So, the stability of Eq. (11) implies practically the asymptotic stability of the corresponding Mechanical system. By making the change u"yl, we reduce Eq. (11) to a Hill equation uA#r u"0, i r (t)"a (t)/l2(ut)!l@(ut)/l(ut). i i
Since x (t)"!x (t#¹/2) and l(ut)" i i l(ut#2n), the minimal period of a (t) and, therefore, i r (t) equals ¹/2"2n/u. So, the multipliers o i 1 and o of Eq. (12) are the roots of the character2 istic equation o2#2Co#1"0, C"!0.5[u (2n/u)#u@ (2n/u)] 2 1
(12)
(13)
where u (t) and u (t) are the solutions of Eq. (12) 1 2 with u (0)"1, u@ (0)"0 and u (0)"0, u@ (0)"1. 2 1 2 1 Eq. (12) is stable or unstable according as D C D(1 or D C D'1. Taking into account that the functions cos x and l(ut) are even, we find that, for solution (3) or (4), the function r (t) is also even. Then, as shown i in [4], C"!1!2u (n/u)u@ (n/u) 2 1 "1!2u (n/u)u@ (n/u). 2 1
3. Stability of periodic oscillations
107
(14)
Thus, to check the stability condition, it is sufficient to find the signs of u (n/u) and u@ (n/u). It appears i i that this can be done without calculating the solutions u (t) and u@ (t). i i Suppose that for the amplitudes of the solutions considered A )n/2, i"1, 2. i ¹heorem 2 The solution x (t) is stable, the solution x (t) is 1 2 unstable. Proof Let us multiply Eqs. (1) and (11) by y@(t) and x@(t) and combine them. The result can be written as follows: (l2x@y@)@#( gly sin x)@#(l2)@x@y@!gl@y sin x"0, whence l2x@y@ D t "!gly sin x D t t0 t0 t
Pt [ gl@y sin x!(l2)@x@y@] dt.
#
0
(15)
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M.A. Pinsky and A.A. Zevin / International Journal of Non-Linear Mechanics 34 (1999) 105–109
Put in Eq. (15) i"1, t "0, y"y (t) (y (0)"0, 0 2 2 y@ (0)"1). The inequality y@ (t)'0 cannot break 2 2 for some t 3(0, n/u], otherwise the left side of * Eq. (15) equals zero (x@ (0)"0, y@ (t )"0) while 1 2 * the right side is negative (y (t)'0, sin x (t)'0, 2 1 l@(t)(0, y@ (t)'0 on (0, t )). Thus, y@ (¹/4)'0; 2 * 2 observing that u(t)"y(t)l(t) and l@(0)"l@(n)"0, we find u@ (n/u)'0. 2
(16)
Setting a* (t)"gl(t)x~1 (t) sin x (t) we find that x (t) i i i i satisfies the equation (l2y@)@#a*y"0. i
(17)
Since cos x("(sin x)/x for D x D(n, then a* (t)'a (t). According to the Sturm theorem, the i i distances between neighbouring zeros of a solution y(t) of Eq. (11) decrease with an increase in a (t). i Taking into account that the solution x (t) of Eq. 1 (17) is even, x (t)'0 on (0, n/u) and x (n/u)"0, 1 1 we find that y (t)'0 on (0, n/u] where y (t) is the 1 1 solution of Eq. (11) with y (0)"1, y@ (0)"0. So, 1 1 y (n/u)'0 and, therefore, 1 u (n/u)'0. 1
(18)
From Eqs. (16), (18) and (14) it follows that C(1. Since a (t)'0 (by condition, D A D)n/2), 1 i then y@ (t)(0 for y (t)'0, so y @ (n/u)(0 and, 1 1 1 hence, u@ (n/u)(0. As shown above, y@ (t)'0 on 2 1 [0, n/u], so y (n/u)'0 and u (n/u)'0. With 2 2 these inequalities, from Eq. (14) one has C'!1. Thus, for the solution x (t), the stability condition 1 D C D(1 is satisfied. Consider now the solution x (t). Recalling that 2 x (t) satisfies Eq. (17) with i"2, x (t)'0 on 2 2 (0, 2n/u), x (0)"x (2n/u)"0 and a (t)(a* (t) 2 2 2 2 on (0, 2n/u), one can analogously show that y (n/u, 2n/u)'0 where y (t, n/u) is the solution 1 1 of Eq. (11) with y (n/u, n/u)"1, y@ (n/u, n/u)"0; 1 1 therefore, for the corresponding solution u (t, n/u) 1 of Eq. (12), one has u (2n/u, n/u)'0. 1
(19)
Put in Eq. (15) i"2, t "n/u, y"y (t, n/u) 0 2 where y (n/u, n/u)"0, y@ (n/u, n/u)"1. Here 2 2 the inequality y@ (t, n/u)'0 cannot hold for 2
[n/u, 2n/u], otherwise for t"2n/u, the left side of Eq. (15) is negative (x@ (n/u)"0, x@ (2n/u)(0 2 2 while the right side is positive (x (2n/u)" 2 0, l@(ut)'0, x@ (t)(0 on (n/u, 2n/u)). Therefore, 2 y@ (t*, n/u)"0 for some t*(2n/u. Since 2 y (t, n/u)'0 for (n/u, 2n/u] (otherwise, two zeros 2 of the solution y (t, n/u) lie between the zeros 2 of the solution y (t, n/u) which is impossible), 1 y@ (t, n/u)(0 on (t*, 2n/u] and, therefore, 2 u@ (2n/u, n/u)(0. (20) 2 Shifting in t by n/u, we find that, for the solution x (t), in Eq. (14) u (n/u)" u (2n/u, n/u)'0 and 2 1 1 u@ (n/u)"u@ (2n/u, n/u)(0, so C'1 that testifies 2 2 to the instability of the solution x (t). The theorem 2 is proved. K Thus, only the oscillations under which the pendulum passes the equilibrium position with the minimal length appear to be stable.
4. On the model of swing Swinging is usually realized by periodic squatting and raising; as a result, the distance of the centre of gravity of the system to the suspension point changes periodically. This has led to widely adopted modelling of the swing phenomenon by a pendulum with a periodically varying length. When swinging, one squats and raises during the motion down and up, respectively. So the period of the excitation is half of the oscillations one, and the distance takes its maximum value when the swing passes the equilibrium position (Fig. 2). Thus, such oscillations correspond to the solution x (t) con2 sidered above (Fig. 1). However, according to Theorem 2, this solution is unstable and, therefore, is not physically realizable. The contradiction obtained stems from the fact that, actually, a pendulum with a periodically varying length cannot serve as a model of a swing. To clarify this conclusion, let us first consider acceleration of a swing from the equilibrium position to high energy. It is quite evident that the onset and duration of every squatting and raising are governed by the current position of the swing, so the corresponding length is of the form l"l(x, x@). As is
M.A. Pinsky and A.A. Zevin / International Journal of Non-Linear Mechanics 34 (1999) 105–109
109
Fig. 2. To the model of swing.
known, the oscillation period of a conservative pendulum increases with the amplitude, so the distances between neighbouring zeros of the solution x(t) also increases in the course of acceleration, i.e., the function l"l(x(t), x@(t)) is not even periodic. Consider now ¹-periodic oscillations of the swing. Here the function l"l(x(t), x@(t)) is ¹/2periodic, so the oscillations are identical with that of the corresponding pendulum. However, if, for example, because of some disturbances the duration of a motion to the equilibrium position increases by a small value e, then the successive raising begins by e later. Thus, in the presence of perturbations, the periodicity of the length l"l(x(t), x@(t)) is violated, whereas in a parametrically excited pendulum the length l"l(ut) does not depend on the motion. Therefore, a swing and a pendulum with a periodically varying length are
quite different systems from the stability standpoint. Thus, the considerations presented show that, in fact, swing is a self-oscillatory system.
References [1] H. Kauderer, Nichtlineare Mechanik, Berlin, 1958. [2] V.M. Volosov, B.I. Morgunov, Method of Averaging in the Theory of Non-linear Oscillatory Systems (in Russian), Moscow, Izd. MGU, 1971. [3] A.A. Zevin, Qualitative investigation of stability of periodic oscillations and rotations in parametrically excited nonlinear second-order systems, Mech. Solids (Mech. Tverdogo Tela) 18 (2) (1983) 34—40. [4] A.A. Zevin, Existence and stability of forced oscillations in non-linear systems with one degree of freedom, Int. J. NonLinear Mech. 30 (3) (1995) 205—221.
Oscillations of a pendulum with a periodically varying length and a model of swing M.A. Pinsky!,*, A.A. Zevin" !Mathematics Department, ºniversity of Nevada Reno, Reno, N» 89557, ºSA "¹ransmag Research Institute, ºkraine Academy of Sciences, 320005 Dniepropetrovsk, Piesarzhevskogo 5, ºkraine Received 18 August 1997
Abstract Qualitative analysis of a pendulum with a periodically varying length is conducted. It is proved that there are two periodic solutions having a prescribed amplitude A(n and a period ¹ which is an even multiple k of the excitation period. Stability analysis is carried out for the principal parametric oscillations (k"2). In this connection it is shown that such a pendulum cannot serve as a mathematical model of swing as it is generally considered. ( 1998 Elsevier Science Ltd. All rights reserved. Keywords: Global stability analysis; Parametric oscillations
1. Introduction Oscillations of a pendulum with a varying length are described by the equation (l2(ut)x@)@#gl(ut) sin x"0
(1)
where x is an angle coordinate, l is the length of the pendulum, g is the acceleration due to gravity and u is an excitation frequency. In this paper we study the existence and stability of periodic solutions of system (1). The known
Contributed by W. F. Ames. *Corresponding author. Tel.: 702-7846725; e-mail:pinsky@ unr.edu
analytical results on the problems (see, e.g. [1, 2]) are obtained by approximate methods for the case l(ut)"l #e cos ut where e is a small 0 parameter (so that the system is quasi-conservative). The distinctive features of our analysis are as follows: the variation of the length of the pendulum is not bounded, a wide class of the excitation (including, in particular, the harmonic one) is considered and, finally, a qualitative approach is employed with the aim of obtaining rigorous results. Suppose that l(ut)"l(!ut)"!l(ut#n), l(ut)'0, l@(ut))0 for t3(0, n/u).
0020-7462/98/$19.00 ( 1998 Elsevier Science Ltd. All rights reserved. PII: S 0 0 2 0 - 7 4 6 2 ( 9 8 ) 0 0 0 0 5 - 5
(2)
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M.A. Pinsky and A.A. Zevin / International Journal of Non-Linear Mechanics 34 (1999) 105–109
Thus, the function l(ut) is symmetric, periodic in t with a period ¹ "2n/u and changes monotoni0 cally between the maximum and minimal values l "l(0) and l "l(n). ` ~ We consider symmetric ¹-periodic solutions of the kind
2. Existence of periodic oscillations
x (t)"x (!t)"!x (t#¹/2), 1 1 1 x (t)'0 on (!¹/4, ¹/4) 1 and
¹heorem 1 For any A(n and even integer k, there exists u (A, k) such that Eq. (1) with u"u possesses i i a solution x (t) with the amplitude A and the period i ¹ "2kn/u . i i
(3)
x (t)"!x (!t)"!x (t#¹/2), 2 2 2 x (t)'0 on (0, ¹/2) (4) 2 where ¹"k¹ , k being the order of subharmonic 0 oscillations. It can be easily shown that the symmetry condition x(t)"!x(t#¹/2) may be satisfied only for even integer k. Note that, on a shift in t by ¹/4, solution (3) takes form (4). Physically, distinctions between these solutions are as follows: under oscillations x (t), the 1 length l(ut) takes the values l and l at the instan~ ` ces the pendulum passes, respectively, the equilibrium or extreme positions, while for the solution x (t), the situation is directly opposite (Fig. 1). 2
The following theorem establishes the existence of solutions (3) and (4) having a prescribed amplitude A"max x(t) and order k.
Proof Let x(t) be the solution of Eq. (1) with x(0)"A and x@(0)"0. If for some u, x(kn/2u)"0,
(5)
then, by Eqs. (1) and (2), x(t)"x(!t)"!x(kn/ u!t), so the corresponding solution x(t) is of the form (3) with the period ¹"2kn/u. As seen from Eq. (1), x(t) decreases for x(t)*0. Let ¸"¸(x, u)"l(ut(x)) where t(x) is the respective inverse function. The phase trajectory v(x, u) of the solution x(t) satisfies the equation v
d(¸2v) #g¸ sin x"0. dx
(6)
Integrating Eq. (6) premultiplied by ¸2, we obtain ¸4v2"2g
A
Px ¸3 sin x dx,
(7)
whence v2 (x)(v2(x, u)(v2 (x) for x3[0, A), ~ ` v2 (x)"2gl3 (cos x!cos A)/l4 , ~ ~ ` v2 (x)"2gl3 (cos x!cos A)/l4 . ~ ` `
(8)
Let t "t (A, u) be the first zero of x(t), then Eq. 0 0 (5) is equivalent to t "kn/2u. Expressing t by 0 0 v(x, u), we obtain the following equation for u: Fig. 1. The trajectories of a pendulum. ( —— ) x (t), ( — — — ) 1 x (t). 2
A
P0 D v(x, u) D~1 dx"kn/2u.
(9)
M.A. Pinsky and A.A. Zevin / International Journal of Non-Linear Mechanics 34 (1999) 105–109
By inequality (8), the left side of this equation admits bilateral bounds independent of u, whereas the right side changes within (0,J) as u changes within (J, 0). Therefore, the required root u (A) 1 certainly exists which proves the existence of the solution x (t). 1 As mentioned above, solution (4) takes form (3) after a shift in time by ¹/4. So, the existence of the solution x (t) is established analogously. The the2 orem is proved. K From Eqs. (9) and (8) it follows that the period ¹"2kn/u of parametric oscillations of the kind (3) or (4) with an amplitude A satisfies the inequality (10) l3@2l~2¹ (A)(¹(A)(l3@2l~2¹ (A) ` ~ 0 ~ ` 0 where ¹ (A) is the corresponding period of free 0 oscillations of the pendulum having the unit length. Note that these bounds do not depend on the order k of the oscillations.
Consider now stability of the solutions x (t), i i"1, 2, with the period ¹"4n/u (i.e. k"2; oscillations of a parametrically excited system with such a period are called principal). As is known, stability is determined by the corresponding variational equation (l2y@)@#a y"0, i a (t)"gl(ut) cos x (t). (11) i i If Eq. (11) is unstable or stable, then the solution x (t) is unstable or stable to a first approximation. i The latter, generally, does not guarantee the Lyapunov stability of the solution x (t); however, i an introduction of arbitrary small dissipative forces makes this solution asymptotically stable [3]. So, the stability of Eq. (11) implies practically the asymptotic stability of the corresponding Mechanical system. By making the change u"yl, we reduce Eq. (11) to a Hill equation uA#r u"0, i r (t)"a (t)/l2(ut)!l@(ut)/l(ut). i i
Since x (t)"!x (t#¹/2) and l(ut)" i i l(ut#2n), the minimal period of a (t) and, therefore, i r (t) equals ¹/2"2n/u. So, the multipliers o i 1 and o of Eq. (12) are the roots of the character2 istic equation o2#2Co#1"0, C"!0.5[u (2n/u)#u@ (2n/u)] 2 1
(12)
(13)
where u (t) and u (t) are the solutions of Eq. (12) 1 2 with u (0)"1, u@ (0)"0 and u (0)"0, u@ (0)"1. 2 1 2 1 Eq. (12) is stable or unstable according as D C D(1 or D C D'1. Taking into account that the functions cos x and l(ut) are even, we find that, for solution (3) or (4), the function r (t) is also even. Then, as shown i in [4], C"!1!2u (n/u)u@ (n/u) 2 1 "1!2u (n/u)u@ (n/u). 2 1
3. Stability of periodic oscillations
107
(14)
Thus, to check the stability condition, it is sufficient to find the signs of u (n/u) and u@ (n/u). It appears i i that this can be done without calculating the solutions u (t) and u@ (t). i i Suppose that for the amplitudes of the solutions considered A )n/2, i"1, 2. i ¹heorem 2 The solution x (t) is stable, the solution x (t) is 1 2 unstable. Proof Let us multiply Eqs. (1) and (11) by y@(t) and x@(t) and combine them. The result can be written as follows: (l2x@y@)@#( gly sin x)@#(l2)@x@y@!gl@y sin x"0, whence l2x@y@ D t "!gly sin x D t t0 t0 t
Pt [ gl@y sin x!(l2)@x@y@] dt.
#
0
(15)
108
M.A. Pinsky and A.A. Zevin / International Journal of Non-Linear Mechanics 34 (1999) 105–109
Put in Eq. (15) i"1, t "0, y"y (t) (y (0)"0, 0 2 2 y@ (0)"1). The inequality y@ (t)'0 cannot break 2 2 for some t 3(0, n/u], otherwise the left side of * Eq. (15) equals zero (x@ (0)"0, y@ (t )"0) while 1 2 * the right side is negative (y (t)'0, sin x (t)'0, 2 1 l@(t)(0, y@ (t)'0 on (0, t )). Thus, y@ (¹/4)'0; 2 * 2 observing that u(t)"y(t)l(t) and l@(0)"l@(n)"0, we find u@ (n/u)'0. 2
(16)
Setting a* (t)"gl(t)x~1 (t) sin x (t) we find that x (t) i i i i satisfies the equation (l2y@)@#a*y"0. i
(17)
Since cos x("(sin x)/x for D x D(n, then a* (t)'a (t). According to the Sturm theorem, the i i distances between neighbouring zeros of a solution y(t) of Eq. (11) decrease with an increase in a (t). i Taking into account that the solution x (t) of Eq. 1 (17) is even, x (t)'0 on (0, n/u) and x (n/u)"0, 1 1 we find that y (t)'0 on (0, n/u] where y (t) is the 1 1 solution of Eq. (11) with y (0)"1, y@ (0)"0. So, 1 1 y (n/u)'0 and, therefore, 1 u (n/u)'0. 1
(18)
From Eqs. (16), (18) and (14) it follows that C(1. Since a (t)'0 (by condition, D A D)n/2), 1 i then y@ (t)(0 for y (t)'0, so y @ (n/u)(0 and, 1 1 1 hence, u@ (n/u)(0. As shown above, y@ (t)'0 on 2 1 [0, n/u], so y (n/u)'0 and u (n/u)'0. With 2 2 these inequalities, from Eq. (14) one has C'!1. Thus, for the solution x (t), the stability condition 1 D C D(1 is satisfied. Consider now the solution x (t). Recalling that 2 x (t) satisfies Eq. (17) with i"2, x (t)'0 on 2 2 (0, 2n/u), x (0)"x (2n/u)"0 and a (t)(a* (t) 2 2 2 2 on (0, 2n/u), one can analogously show that y (n/u, 2n/u)'0 where y (t, n/u) is the solution 1 1 of Eq. (11) with y (n/u, n/u)"1, y@ (n/u, n/u)"0; 1 1 therefore, for the corresponding solution u (t, n/u) 1 of Eq. (12), one has u (2n/u, n/u)'0. 1
(19)
Put in Eq. (15) i"2, t "n/u, y"y (t, n/u) 0 2 where y (n/u, n/u)"0, y@ (n/u, n/u)"1. Here 2 2 the inequality y@ (t, n/u)'0 cannot hold for 2
[n/u, 2n/u], otherwise for t"2n/u, the left side of Eq. (15) is negative (x@ (n/u)"0, x@ (2n/u)(0 2 2 while the right side is positive (x (2n/u)" 2 0, l@(ut)'0, x@ (t)(0 on (n/u, 2n/u)). Therefore, 2 y@ (t*, n/u)"0 for some t*(2n/u. Since 2 y (t, n/u)'0 for (n/u, 2n/u] (otherwise, two zeros 2 of the solution y (t, n/u) lie between the zeros 2 of the solution y (t, n/u) which is impossible), 1 y@ (t, n/u)(0 on (t*, 2n/u] and, therefore, 2 u@ (2n/u, n/u)(0. (20) 2 Shifting in t by n/u, we find that, for the solution x (t), in Eq. (14) u (n/u)" u (2n/u, n/u)'0 and 2 1 1 u@ (n/u)"u@ (2n/u, n/u)(0, so C'1 that testifies 2 2 to the instability of the solution x (t). The theorem 2 is proved. K Thus, only the oscillations under which the pendulum passes the equilibrium position with the minimal length appear to be stable.
4. On the model of swing Swinging is usually realized by periodic squatting and raising; as a result, the distance of the centre of gravity of the system to the suspension point changes periodically. This has led to widely adopted modelling of the swing phenomenon by a pendulum with a periodically varying length. When swinging, one squats and raises during the motion down and up, respectively. So the period of the excitation is half of the oscillations one, and the distance takes its maximum value when the swing passes the equilibrium position (Fig. 2). Thus, such oscillations correspond to the solution x (t) con2 sidered above (Fig. 1). However, according to Theorem 2, this solution is unstable and, therefore, is not physically realizable. The contradiction obtained stems from the fact that, actually, a pendulum with a periodically varying length cannot serve as a model of a swing. To clarify this conclusion, let us first consider acceleration of a swing from the equilibrium position to high energy. It is quite evident that the onset and duration of every squatting and raising are governed by the current position of the swing, so the corresponding length is of the form l"l(x, x@). As is
M.A. Pinsky and A.A. Zevin / International Journal of Non-Linear Mechanics 34 (1999) 105–109
109
Fig. 2. To the model of swing.
known, the oscillation period of a conservative pendulum increases with the amplitude, so the distances between neighbouring zeros of the solution x(t) also increases in the course of acceleration, i.e., the function l"l(x(t), x@(t)) is not even periodic. Consider now ¹-periodic oscillations of the swing. Here the function l"l(x(t), x@(t)) is ¹/2periodic, so the oscillations are identical with that of the corresponding pendulum. However, if, for example, because of some disturbances the duration of a motion to the equilibrium position increases by a small value e, then the successive raising begins by e later. Thus, in the presence of perturbations, the periodicity of the length l"l(x(t), x@(t)) is violated, whereas in a parametrically excited pendulum the length l"l(ut) does not depend on the motion. Therefore, a swing and a pendulum with a periodically varying length are
quite different systems from the stability standpoint. Thus, the considerations presented show that, in fact, swing is a self-oscillatory system.
References [1] H. Kauderer, Nichtlineare Mechanik, Berlin, 1958. [2] V.M. Volosov, B.I. Morgunov, Method of Averaging in the Theory of Non-linear Oscillatory Systems (in Russian), Moscow, Izd. MGU, 1971. [3] A.A. Zevin, Qualitative investigation of stability of periodic oscillations and rotations in parametrically excited nonlinear second-order systems, Mech. Solids (Mech. Tverdogo Tela) 18 (2) (1983) 34—40. [4] A.A. Zevin, Existence and stability of forced oscillations in non-linear systems with one degree of freedom, Int. J. NonLinear Mech. 30 (3) (1995) 205—221.