Application of the energy balance method to nonlinear vibrating equations

Current Applied Physics 10 (2010) 104–112

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Current Applied Physics journal homepage: www.elsevier.com/locate/cap

Application of the energy balance method to nonlinear vibrating equations I. Mehdipour a, D.D. Ganji b,*, M. Mozaffari b a b

Department of Mechanical Engineering, Semnan Azad University, Semnan, Iran Department of Mechanical Engineering, Babol University of Technology, Babol, Iran

a r t i c l e

i n f o

Article history: Received 9 November 2008 Received in revised form 27 April 2009 Accepted 11 May 2009 Available online 23 May 2009 PACS: 05.45.-a 82.40.Bj 05.45.+b

a b s t r a c t In this paper, He’s energy balance method is applied to nonlinear vibrations and oscillations. The method is applied to four nonlinear differential equations. It has indicated that by utilizing He’s energy balance method (HEBM), just one iteration leads us to high accuracy of solutions. It has illustrated that the energy balance methodology is very effective and convenient and does not require linearization or small perturbation. Contrary to the conventional methods, in energy balance method, only one iteration leads to high accuracy of the solutions. The results reveal that the energy balance method is very effective and simple. It is predicted that the energy balance method can be found wide application in engineering problems, as indicated in following examples. Ó 2009 Elsevier B.V. All rights reserved.

Keywords: Energy balance method Nonlinear oscillation Periodic solutions

1. Introduction This paper considers the following general nonlinear oscillators [2,3,6]:

u00 þ x20 u þ ef ðuÞ ¼ 0

ð1Þ

with initial conditions

uð0Þ ¼ A;

0

u ð0Þ ¼ 0:

u00 þ f ðuðtÞÞ ¼ 0: ð2Þ

Here f is a nonlinear function of u00 ; u0 and u; we limit ourselves to the simplest case, i.e., f depends upon only the function of u. If there is no small parameter in the equation, the traditional perturbation methods cannot be applied directly. Recently, considerable attention has been directed towards to analytical solutions for nonlinear equations without possible small parameters. The traditional perturbation methods have many shortcomings, and they are not useful to strongly nonlinear equations. To overcome the shortcomings, many new techniques have appeared in open literature such as Variational Iteration Method (VIM) [11–16], homotopy perturbation method [17–22] and bookkeeping parameter. Perturbation method [23], just to name a few, a review on some recently developed nonlinear analytical methods can be found in detail in [24–26]. In He’s energy balance method [1], a variational principle for the nonlinear oscillation is established, and then a Hamiltonian is constructed, from which the angular frequency can be readily ob* Corresponding author. Tel./fax: +98 111 3234205. E-mail address: [email protected] (D.D. Ganji). 1567-1739/$ - see front matter Ó 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.cap.2009.05.016

tained by collocation method. The results are valid not only for weakly nonlinear systems, but also for strongly nonlinear ones. Some examples reveal that even the lowest order approximations trigger high accuracy [4,6–10]. In this paper, we consider a general nonlinear oscillator in the form [5,6]:

ð3Þ

Since u and t are generalized dimensionless displacement and time variables, respectively. Its variational principle can be easily obtained:

 Z t 1  u02 þ FðuÞ dt; 2 Z0 FðuÞ ¼ f ðuÞdu:

JðuÞ ¼

ð4Þ

It is Hamiltonian, therefore, can be written in the form:

H¼

1 02 u þ FðuÞ ¼ FðAÞ 2

ð5Þ

or

RðtÞ ¼

1 02 u þ FðuÞ  FðAÞ ¼ 0: 2

ð6Þ

Oscillation systems contain two important physical parameters, i.e. the frequency x and the amplitude of oscillation, A. So let us consider such initial conditions:

uð0Þ ¼ A;

u0 ð0Þ ¼ 0:

ð7Þ

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Assume that its initial approximate guess can be expressed as:

uðtÞ ¼ A cos xt:

ð8Þ

u_ 1 ¼ u2 ;

Substituting Eq. (8) into u which term of Eq. (6), yields:

RðtÞ ¼

The numerical solution (with Runge–Kutta method of order 4) for nonlinear equation is:

1 2 2 x A sin2 xt þ FðA cos xtÞ  FðAÞ ¼ 0: 2

ð9Þ

If, by any chance, the exact solution had been chosen as the trial function, then it would be possible to make R zero for all values of t by appropriate choice of x. Since (7) is only an approximation to the exact solution, R cannot be made zero everywhere. Collocation at xt ¼ p=4 gives:

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðFðAÞ  FðA cos xtÞÞ x¼ ; 2 A2 sin xt

u1 ð0Þ ¼ A; ð13Þ   1 ku1 þ e1 u1 u22 þ 2e2 u31 u22 þ e3 u31 þ e4 u51 ; u_ 2 ¼  1 þ e1 u21 þ e2 u41 u2 ð0Þ ¼ 0:

ð14Þ

The solution of nonlinear equation with the energy balance method is:

€ þ ku þ e1 u2 u € þ e1 uu_ 2 þ e2 u4 u € þ 2e2 u3 u_ 2 þ e3 u3 þ e4 u5 ¼ 0; u _ uð0Þ ¼ A; uð0Þ ¼0 ð15Þ

ð10Þ

in which u and t are generalized dimensionless displacement and time variables, respectively. Its variational principle can be easily obtained:

where T ¼ 2p=x is period of the nonlinear oscillator. Its period can be written in the form of:

2p T ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :

ð11Þ

2ðFðAÞFðA cos xtÞÞ A2 sin2 xt

JðuÞ ¼

 Z t 1 k e3 e4  u_ 2 ð1 þ e1 u2 þ e2 u4 Þ þ u2 þ u4 þ u6 dt: 2 2 4 6 0

ð16Þ

Its Hamiltonian, therefore, can be written in the form: 2. Applications

1 2 k e3 e4 u_ ð1 þ e1 u2 þ e2 u4 Þ þ u2 þ u4 þ u6 2 2 4 6 k 2 e3 4 e4 6 ¼ A þ A þ A 2 4 6

H¼

In order to assess the advantages and the accuracy of the energy balance method, we will consider the following four examples:

ð17Þ

or

2.1. Example 1 From Hamden [26,27], it is known that the free vibrations of an autonomous conservative oscillator with inertia and static type fifth-order non-linearities is expressed by:

RðtÞ ¼

1 2 k e3 e4 k u_ ð1 þ e1 u2 þ e2 u4 Þ þ u2 þ u4 þ u6  A2 2 2 2 4 6 

e3 4

A4 

e4 6

A6 ¼ 0:

ð18Þ

Oscillatory systems contain two important physical parameters, i.e. the frequency x and the amplitude of oscillation, A. So let us consider such initial conditions:

€ þ ku þ e1 u2 u € þ e1 uu_ 2 þ e2 u4 u € þ 2e2 u3 u_ 2 þ e3 u3 þ e4 u5 ¼ 0; u _ uð0Þ ¼ A; uð0Þ ¼ 0: ð12Þ Motion is assumed to start from the position of maximum displacement with zero initial velocity. k is an integer which may take values of k ¼ 1, 0 or 1, and e1 ; e2 ; e3 and e4 are positive parameters.

_ uð0Þ ¼ 0:

uð0Þ ¼ A;

ð19Þ

Assume that its initial approximate guess can be expressed as:

uðtÞ ¼ A cos xt:

ð20Þ

Substituting Eq. (20) into Eq. (18), yields: Table 1 Values of dimensionless parameters

ei in Eq. (12) for four modes.

RðtÞ ¼

Mode

e1

e2

e3

e4

1 2 3

0.326845 1.642033 4.051486

0.129579 0.913055 1.665232

0.232598 0.313561 0.281418

0.087584 0.204297 0.149677

1 ðA sin xtÞ2 ð1 þ e1 ðA cos xtÞ2 þ e2 ðA cos xtÞ4 Þ 2 k e3 e4 þ ðA cos xtÞ2 þ ðA cos xtÞ4 þ ðA cos xtÞ6 2 4 6 k e3 e4  A2  A4  A6 ¼ 0: 2 4 6

ð21Þ

1.5 1.0

u(t)

0.5 0.0 -0.5 -1.0 -1.5

Runge-Kutta Energy Balance

0

2

4

6

8

10

12

14

t(s) Fig. 1. The Comparison between energy balance method solution and the numerical solution which solved by Runge–Kutta method of order 4, solid line: numerical solution and dashed line: the energy balance solution. With k ¼ 1, A = 1 for mode-1.

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u(t)

0.5 0.0 -0.5 -1.0 Runge-Kutta Energy Balance

-1.5

0

2

4

6

8

10

12

14

16

t(s) Fig. 2. The comparison between energy balance method solution and the numerical solution which obtained by Runge–Kutta method of order 4, solid line: numerical solution and dashed line: the energy balance solution. With k ¼ 1, A = 1 for mode-2.

1.5 1.0

u(t)

0.5 0.0 -0.5 -1.0 Runge-Kutta Energy Balance

-1.5

0

5

10

15

20

25

t(s) Fig. 3. The comparison between energy balance method solution and numerical solution which solved by Runge–Kutta method of order 4, solid line: numerical solution and dashed line: the energy balance solution. With k ¼ 1, A = 1 for mode-3.

Which trigger the following results: vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi uk 2 2 4 6 e3 4 e4 6 2 u t2 ðA  ðAcos xtÞ Þ þ 4 ðA  ðAcos xtÞ Þ þ 6 ðA  ðAcos xtÞ Þ: x¼ 2 4 A sin xt 1 þ e1 ðA cos xtÞ þ e2 ðAcos xtÞ

If we collocate at xt ¼ p4, we obtain:

xEBM

ffi pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 12k þ 9e3 A2 þ 7e4 A4 ¼ : 3 4 þ 2e1 A2 þ e2 A4

ð23Þ

ð22Þ Substituting Eq. (23) into Eq. (20) yields:

0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 3 12k þ 9e3 A2 þ 7e4 A4 A uðtÞ ¼ A cos @ t : 3 4 þ 2e1 A2 þ e2 A4

ð24Þ

Motion is assumed to start from the position of maximum displacement with zero initial velocity. k is an integer which may take values of k ¼ 1, 0 or 1, and e1 ; e2 ; e3 and e4 are positive parameters. The values of parameters e1 ; e2 ; e3 and e4 associated with each of the four calculation modes are shown in Table 1. Additionally, the comparison between these methodologies can be found in Fig. 1–3). 2.2. Example 2 We consider the motion equation of the pendulum with harmonic stringer point in Fig. 4. The motion equation of pendulum with harmonic stringer point is:

Fig. 4. Pendulum with harmonic stringer point: yðtÞ ¼ Y cos x0 t[28].

  2 €h þ g  x0 Y cos x0 t sin h ¼ 0; l l

hð0Þ ¼ A;

_ hð0Þ ¼ 0:

ð25Þ

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0.2

θ(t)

0.1

0.0

-0.1

-0.2 Runge-Kutta Energy Balance

-0.3

0

1

2

3

4

5

3

4

5

t(s)

a) A=π/12 1.0 0.8 0.6 0.4

θ(t)

0.2 0.0 -0.2 -0.4 -0.6 -0.8

Runge-Kutta Energy Balance

-1.0 0

1

2

t(s)

b) A= π/4 1.5

1.0

θ(t)

0.5

0.0

-0.5

-1.0 Runge-Kutta Energy Balance

-1.5 0

1

2

3

4

5

t(s)

c) A=5 π/12 Fig. 5. The comparison between energy balance method and numerical solution performed by Runge–Kutta method of order 4, solid line denotes the numerical solution anddashed line denotes the energy balance solution.

This equation is as known as Mathieu equation or the system with dependent coefficients to time. The numerical solution (with Runge–Kutta method of order 4) of pendulum with harmonic stringer point is:

h_ 1 ¼ h2 ; h1 ð0Þ ¼ A;   g x2 Y h_ 2 ¼   0 cos x0 t sin h1 ; l l

ð26Þ h2 ð0Þ ¼ 0:

ð27Þ

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The solution of Mathieu equation of pendulum with harmonic stringer point according to the energy balance method is:

The numerical solution (with Runge–Kutta method of order 4) of nonlinear equation is:

  2 €h þ g  x0 Y cos x0 t sin h ¼ 0; l l

h_ 1 ¼ h2 ;

hð0Þ ¼ A;

_ hð0Þ ¼ 0:

ð28Þ

In which h and t are generalized dimensionless displacements and time variables, respectively. Its variational principle can be easily obtained as:

JðhÞ ¼

   Z t 1 g x20 Y  h_ 2   cos x0 t cos h dt: 2 l l 0

ð29Þ

Its Hamiltonian, therefore, can be written in the form:

    1 g x20 Y g x2 Y  cos x0 t cos h ¼   0 cos x0 t cos A H ¼ h_ 2  2 l l l l ð30Þ or

RðtÞ ¼

  1 _2 g x20 Y h   cos x0 t cos h 2 l l   g x20 Y þ  cos x0 t cos A ¼ 0: l l

ð31Þ

_ hð0Þ ¼ 0:

ð32Þ

Assume that its initial approximate guess can be expressed as:

hðtÞ ¼ A cos xt:

ð33Þ

Substituting Eq. (33) into Eq. (31), yield:

RðtÞ ¼

  1 2 2 g x20 Y 2 A x sin xt   cos x0 t cosðA cos xt Þ 2 l l   g x20 Y þ  cos x0 t cos A ¼ 0: l l

xEBM

4

ð40Þ

€h þ 4k sin h  3F 0 sin x0 t ¼ 0; 3m ml

hð0Þ ¼ A;

_ hð0Þ ¼ 0:

ð34Þ

JðhÞ ¼

 Z t 1 4k 3F 0  h_ 2  cos h  ðsin x0 tÞh dt: 2 3m ml 0

1 _ 2 4k 3F 0 h  cos h  ðsin x0 tÞh 2 3m ml 4k 3F 0 ¼ cos A  ðsin x0 tÞA 3m ml

or

1 _ 2 4k 3F 0 4k h  cos h  cos A ðsin x0 tÞh þ 2 3m 3m ml 3F 0 ðsin x0 tÞA ¼ 0: þ ml

RðtÞ ¼

_ hð0Þ ¼ 0:

ð45Þ

Assume that its initial approximate guess can be expressed as:

ð36Þ

1 _ 2 4k 3F 0 h  cosðA cos xtÞ  ðsin x0 tÞðA cos xtÞ 2 3m ml 4k 3F 0 cos A þ ðsin x0 tÞA ¼ 0: þ 3m ml

ð46Þ

For the following value parameters, we compared the numerical solution with the energy balance method’s, see Fig. 5a–c:

Y ¼ 0:25 m; g ¼ 9:81 m=s2 :

2.3. Example 3 We consider the physical model of nonlinear equation in the following figure with FðtÞ ¼ F 0 sin x0 t; indicated in Fig. 6. The motion equation is:

ð38Þ

ð47Þ

We obtain the following results: pffiffiffi 2 x¼ A sin xt rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4k 3F 0 ðcosðA cos xtÞ  cos AÞ þ ðA sin x0 tðcos xtÞ  Aðsin x0 tÞÞ:  3m ml ð48Þ

ð37Þ

_ hð0Þ ¼ 0:

ð44Þ

Oscillatory systems contain two important physical parameters, i.e. the frequency x and the amplitude of oscillation, A. So let us consider such initial conditions:

RðtÞ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 v !ffi 1 pffiffiffi u  u g x2 Y 2 A 2 hðtÞ ¼ A cos @ t  0 cos x0 t cos  cos A t A: A l 2 l

hð0Þ ¼ A;

ð43Þ

Substituting Eq. (46) into Eq. (44), yields:

Substituting Eq. (36) into Eq. (33), yields:

€h þ 4k sin h  3F 0 sin x0 t ¼ 0; 3m ml

ð42Þ

H¼

ð35Þ

we obtain:

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi pffiffiffi u  2 2u g x Y A 2 ¼ t  0 cos x0 t cos  cos A : A l l 2

L ¼ 1 m; x0 ¼ 1 rad=s;

ð41Þ

In which h and t are generalized dimensionless displacements and time variables, respectively. Its variational principle can be easily obtained:

hðtÞ ¼ A cos xt:

ffi pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   2 g x20 Y  cos x0 t ðcosðA cos xt Þ  cos AÞ: x¼ A sin xt l l If we collected at xt ¼

h2 ð0Þ ¼ 0:

The solution of nonlinear equation with the energy balance method is:

hð0Þ ¼ A;

We obtain the following result:

p,

ð39Þ

Its Hamiltonian, therefore, can be written in the form:

Oscillatory systems contain two important physical parameters, i.e. the frequency x and the amplitude of oscillation, A. So let us consider such initial conditions:

hð0Þ ¼ A;

; h1 ð0Þ ¼ A;

4k 3F 0 h_ 2 ¼  sin h1 þ sin x0 t; 3m ml

Fig. 6. the physical model of nonlinear equation [28].

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0.2

θ(t)

0.1

0.0

-0.1

-0.2 Runge-Kutta Energy Balance

-0.3 0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.8

1.0

1.2

t(s)

a) A=π/12 0.6

0.4

θ(t)

0.2

0.0

-0.2

-0.4 Runge-Kutta Energy Balance

-0.6

0.0

0.2

0.4

0.6

t(s)

b) A= 5π/36 1.5

1.0

θ(t)

0.5

0.0

-0.5

-1.0 Runge-Kutta Energy Balance

-1.5 0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

t(s)

c) A= π/3 Fig. 7. The Comparison between energy balance method solution and numerical solution performed with Runge–Kutta method of order 4, solid line: numerical solution and dashed line: the energy balance solution.

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If we collocate at xt ¼ p4, we obtain:

xEBM

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! !ffi pffiffiffi pffiffiffi u 2u 4k 3F 2 2 0 ¼ t cos A sin x0 t A  cos A þ 1 : A 3m ml 2 2 ð49Þ

Substituting Eq. (49) into Eq. (46) yields: ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 v ! !ffi 1 pffiffiffi pffiffiffi u 2 2 2u 4k 3F 0 t @ cos hðtÞ ¼ A cos A sin x0 t A  cos A þ  1 t A: A 3m ml 2 2

ð50Þ For this following value of parameters we compared the numerical solution and energy balance method, see Fig. 7a–c:

L ¼ 1 m; m ¼ 10 kg; k ¼ 1000 N=m; F0 ¼ 1 N; x0 ¼ 1 rad=s: Fig. 8. the physical model of Duffing equation with constant coefficient [28].

2.4. Example 4 In this example we have Duffing equation with constant coefficient, see Fig. 8:

€x þ

k1 k2 3 F 0 x ¼ xþ sin x0 t; 2 m m 2mh

xð0Þ ¼ A;

_ xð0Þ ¼ 0:

ð51Þ

Oscillatory systems contain two important physical parameters, i.e. the frequency x and the amplitude of oscillation, A. So let us consider such initial conditions:

xð0Þ ¼ A;

_ xð0Þ ¼ 0:

ð58Þ

The numerical solution (with Rung–Kuta method of order 4) of Duffing equation with constant coefficient is:

Assume that its initial approximate guess can be expressed as:

x_ 1 ¼ x2 ;

ð52Þ

Substituting Eq. (59) into Eq. (57), yields:

ð53Þ

RðtÞ ¼

x1 ð0Þ ¼ A;

k1 k2 3 F 0 x_ 2 ¼  x1  x þ sin x0 t; 2 1 m m 2mh

x2 ð0Þ ¼ 0:

xðtÞ ¼ A cos xt:

The solution of nonlinear equation with energy balance method is:

€x þ

k1 k2 3 F 0 x ¼ xþ sin x0 t; 2 m m 2mh

xð0Þ ¼ A;

_ xð0Þ ¼ 0:

ð54Þ

1 2 2 k1 k2 2 ðA cos xtÞ4 A x sin xt þ ðA cos xtÞ2 þ 2 2 2m 8mh F0 k1 2 k2  ðsin x0 tÞA cos xt  A4 A  2 m 2m 8mh F0 þ ðsin x0 tÞA ¼ 0: m

In which and t are generalized dimensionless displacements and time variables, respectively. Its variational principle can be easily obtained:

 Z t 1 k1 2 k2 4 F 0  x_ 2 þ x  x tÞx dt: x þ ðsin 0 2 2 2m m 0 8mh

ð56Þ

ð62Þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 00 v !ffi1 1 pffiffiffi u uk1 A2 3k2 A4 F 0 2 2 xðtÞ ¼ A cos @@ t þ A sin x0 t þ  1 At A: A 4m 32mh2 m 2 ð63Þ

or

1 2 k1 2 k2 4 F 0 k1 2 x  ðsin x0 tÞx  x_ þ x þ A 2 2 2m m 2m 8mh k2 F0  A4 þ ðsin x0 tÞA ¼ 0 2 m 8mh

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi pffiffiffi u 2 4 2 2u k A 3k A F 1 2 0 ¼ t þ A sin x0 t 1 : þ 2 A 4m 32mh2 m

Substituting Eq. (62) into Eq. (59) yields:

H¼

RðtÞ ¼

ð61Þ

If we collocate at xt ¼ p4, we obtain:

xEBM ð55Þ

Its Hamiltonian, therefore, can be written in the form:

1 2 k1 2 k2 4 F 0 x  ðsin x0 tÞx x_ þ x þ 2 2 2m m 8mh k1 2 k2 F 0 ¼ A4  ðsin x0 tÞA A þ 2 2m m 8mh

ð60Þ

We obtain the following result:

pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 k1 2 k2 F0 x¼ ðA4  ðA cos xtÞ4 Þ þ A sin x0 tðcos xt  1Þ: ðA  ðA cos xtÞ2 Þ þ 2 A sin xt 2m m 8mh

JðxÞ ¼

ð59Þ

For this following value of parameters we compared the numerical solution with energy balance method, Fig. 9a–c:

L ¼ 1 m; h ¼ 0:9 m; m ¼ 10 kg; k1 ¼ 1000 N=m; ð57Þ

k2 ¼ 1100 N=m; F0 ¼ 1N; x0 ¼ 1 rad=s:

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x(t)

0.02 0.00 -0.02 -0.04 -0.06 -0.08 -0.10

Runge-Kutta Energy Balance

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.0

1.2

1.4

1.0

1.2

1.4

t(s)

a) A=0.9tan (π/36) 0.20 0.15 0.10

x(t)

0.05 0.00 -0.05 -0.10 -0.15 -0.20

Runge-Kutta Energy Balance

0.0

0.2

0.4

0.6

0.8

t(s)

b) A=0.9tan (π/18) 0.3

0.2

x(t)

0.1

0.0

-0.1

-0.2 Runge-Kutta Energy Balance

-0.3

0.0

0.2

0.4

0.6

0.8

t(s)

c) A=0.9tan (π/12) Fig. 9. The comparison between energy balance method solution and numerical solution obtained by Runge–Kutta method of order 4, solid line: numerical solution and dashed line: the energy balance solution.

3. Conclusions In this paper, the energy balance method has been successfully used to study the nonlinear vibrating equations. The method, which is proved to be a powerful mathematical tool to study nonlinear vibrating equations, can be easily extended to any nonlinear equation. It has demonstrated the accuracy and efficiency of method by solving some examples. It has showed that the obtained solutions

are valid for the whole domain. The examples indicate that even lowest orders of approximations obtained by the present theory are trigger high accuracy. According to tables associated with above examples which are the comparison between energy balance method and numerical Runge–Kutta method of order 4, we conclude that both results obtained by both methods are accurate and close to each other. These examples illustrate the efficiency of the energy balance method in comparison with other methods.

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