AC 2010-162: PREDICTION COMPARISONS BETWEEN NON-LINEAR ANDLINEAR MODELS FOR DYNAMICS ENHANCED EDUCATION

Arnaldo Mazzei, Kettering UniversityARNALDO MAZZEI is an Associate Professor of Mechanical Engineering at KetteringUniversity. He received his Ph.D. in Mechanical Engineering from the University of Michigan in1998. He specializes in dynamics and vibrations of mechanical systems and stability ofdrivetrains with universal joints. His current work relates to modal analysis, stability ofdrivetrains, finite element analysis and CAE. He is a member of ASME, ASEE and SEM.

Richard Scott, University of MichiganRICHARD A. SCOTT received his Ph.D. in Engineering Science from The California Institute ofTechnology. He is a Professor of Mechanical Engineering at the University of Michigan, AnnArbor. He has obtained a teaching award from the College of Engineering and was selected asprofessor of the semester four times by the local chapter of Pi-Tau-Sigma.

© American Society for Engineering Education, 2010

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Prediction comparisons between non-linear and linear models for dynamics

enhanced education

Introduction

In previous works 1, 2, 3, 4 examples were given illustrating benefits of introducing modern software, such as MAPLE®, into undergraduate and beginning graduate mechanics courses. There are many articles on the use of simulation in engineering education. For example, Fraser et al. 5 give a very informative and useful discussion on the use of simulations in fluid mechanics. Student difficulties were assessed using questions from the Fluid Mechanics Concepts Inventory (FMCI). The impact of the simulation was assessed using a second administration of the FMCI. Similar work could probably be done using the Dynamics Concept Inventory of Gray et al. 6 However that is not the goal of the present work, which is to enlarge and augment presentations given in most standard texts. For example consider the amplitude of resonant motions as given by a linear damped model. Are the traditional predictions accurate? Since near resonance the underlying equations are non-linear, substantial differences from the linear model may (and usually are) found. With modern numerical simulation software (for example, MAPLE®) the non-linear ordinary differential equations of motion can be readily solved and differences assessed. In the following, three illustrative examples are presented. First a damped one-degree of freedom system with a non-linear spring is investigated. Many texts give expressions for power flow at resonance based on a linear model, however near resonance large amplitudes would be encountered and a non-linear model should be employed. Here numerical solutions to the differential equations are used to obtain the velocity, from which the power flow per cycle can be calculated. It is shown that for reasonable values of damping differences can be substantial. Other interesting problems involve the effects of non-linearities on frequencies and mode shapes in vibrations. A simple single degree of freedom model is developed here which shows the effect of the non-linearity on the period of vibrations. This is also done for a two degree of freedom example. Another interesting problem involves the linear normal modes of a two-degree of freedom mechanical system. It is well known that if the system is initialized in a normal mode, it remains in that mode. What happens if the underlying physical model is non-linear? Here, for purposes of classroom demonstration, simple models, showing some interesting effects, are developed and analyzed using MAPLE®.

Physical Examples

Power Flow in Linear and Non-linear oscillators The equation of motion for a non-linear (hardening spring) viscously damped system subjected to a harmonic forcing function can be written as:

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3

1 2 0sin( )mx cx k x k x f tψ− − − ?&& &

(1)

or

2 3 022 sin( )n n

fkx x x t

m mx|ψ ψ ψ− − − ?&&&

(2)

where n

ψ is the undamped natural frequency of the linear system | , is the damping ratio, m is

the system mass, 0

f is the amplitude of the forcing function (frequency ψ ) and 1

k and 2

k are

stiffnesses for the linear and non-linear terms respectively.

A non-dimensional version of equation (2) can be obtained by letting ntϖ ψ? , /

nξ ψ ψ? ,

0

1

dxx

f

k

? and 3

122

0

kk

fχ

?

:

32 sin( )

d d d dx xx x| χ ξϖ− − − ?&& &

(3)

where, now, the dots refer to differentiation with respect to the non-dimensional time ϖ .

Consider the linear model ( 0χ ? ). The steady state response can readily be shown to be:

2 2 2

sin( ) sin( )

(1 ) 4X

D

ξϖ λ ξϖ λ

ξ |

/ /? ?

/ −

(4)

1

2

2

1tan

|λ

ξ

/ ?

/

(5)

The instantaneous power (force times velocity) is given, in dimensionless form, by:

∗ +sin cos( )sin( )D

Xξ

ξϖ ξϖ λ ξϖ? ? /&P

(6)

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Then the average power over a cycle is given by:

2

0 0

π cos( )sin( ) sin( )D D

T

avgdt dt

ρξ ξ

ξϖ λ ξϖ λ? ? / ?∫ ∫P P

(7)

This is maximum when 2

ρλ ? , which occurs, as equation (5) shows, when 1ξ ? . That is right

at resonance (n

ψ ψ? , at which the linear model may be unsuitable!) and gives maximum

average power: 2avg

ρ|

?P .

As an example, take 0.1| ? . Then, for the linear case, 5 15.71avg

ρ? ΒP .

For the non-linear case, a numerical solution of the differential equation (3) can be obtained,

once χ is specified. In the sequel 0.1χ ? . Here MAPLE® 7 is used. There are several tasks, namely:

1. Determination of the non-linear resonant frequency (frequency at which the peak

occurs).

For a non-linear hardening oscillator the amplitude of the system varies with frequency as shown below in Figure 1. Note that as the frequency increases a drop occurs as it passes the peak value.

Figure 1 – Typical forced response for a non-linear system8

Frequency

Res

po

nse

Am

pli

tud

e

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This can be made as the basis of determining the resonant frequency for the nonlinear system.

The frequency can be obtained by varying the forcing frequency and noting when an amplitude drop occurs.

Figure 2 – Frequency response amplitude9 – Non-linear model versus linear model

For comparison purposes, the frequency responses for both the linear and non-linear models are shown in Figure 2. For the non-linear response, note the “bending” of the peak to the right (due to the non-linear stiffening effect) and the drop on the amplitude

around a frequency value 1.325ξ ? . Numerical obtained responses for the non-linear model at two distinct frequencies, below and above this estimated value, are shown in Figure 3. Again, a clear drop on the amplitude is seen confirming the resonance value.

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1.32ξ ?

1.34ξ ?

Figure 3 – Numerical obtained responses for non-linear model

2. Determination of the steady state response.

Figure 4 shows the numerical solution for equation (3)

( 0.1, 0.1, 1.325)χ | ξ? ? ? . Inspection of the figure, when the transient dies out,

shows that the steady state response can be modeled by (note that a phase shift is required and it can be obtained from the plot):

3.45 sin(1.325 1.1925)X ϖ? /

(8)

This is verified by plotting this equation and the numerical solution together. Differentiation of the approximate solution above, with respect to non-dimensional time, gives the approximate velocity:

4.57 cos(1.325 1.1925)X ϖ? /&

(9)

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3.45 sin(1.325τ)

Figure 4 – Non-linear response at resonance frequency

3. Determination of the average power at resonance. Using equation (9) the average power can be calculated by (recall that the period

is 2 2 1.325

ρ ρξ? ):

∗ +∗ +2

1.325

0

4.57cos 1.325 1.1925 sin(1.325 )avg

dt

ρ

ϖ ϖ? /∫P

(10)

Evaluation of the integral gives 10.07avgΒP . This is considerably lower (36%)

than the value obtained using the linear model. This significant difference demonstrates that the linear model may be unsuitable for the power flow problem.

Non-linear effects on vibrations

The next examples are meant to illustrate some effects of non-linearities on vibrations.

Steady state (numerical response)

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A feature of non-linear oscillations is that natural frequencies and mode shapes depend on vibration amplitudes. Showing this analytically is usually beyond the scope of most undergraduate, and beginning graduate, courses. A goal here, which is modest in scope, is to illustrate some important effects numerically, thereby enriching the students understanding and knowledge. The firs example is to show that the period depends on amplitude. This has been shown previously by the authors 1, but for sake of completeness, another case is treated here. Consider a simple pendulum restrained by a spring (non-linear) as shown in Figure 5.

Figure 5 – Simple pendulum restrained by spring

Taking the spring force to be given by:

3

1 2springF k kφ φ? −

(11)

where φ is the spring deflection, the equation of motion can be shown to be, using, for example, energy methods:

∗ + ∗ +2 2 4 3

1 2sin cos( ) 2 (sin( )) cos( ) sin 0ML k L k L MgLσ σ σ σ σ σ− − − ?&&

(12)

- L

- M

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Introducing the dimensionless variables ∗ +1 /k M tϖ ? , 2

2 1(2 ) /k L kδ ? and

1( ) / ( )Mg k Lι ? ,

equation (12) becomes:

∗ + ∗ +3sin cos( ) (sin( )) cos( ) sin 0σ σ σ δ σ σ ι σ− − − ?&&

(13)

Equation (13) is now solved using MAPLE® for the following initial conditions:

a. Small: 0.17 10 radσ ″? Β , 0σ ?& (spring force in linear range).

b. Large: 0.79 45radσ ″? Β , 0σ ?& (spring force in non-linear range).

Figure 6 – Restrained pendulum amplitude response for different initial conditions

Figure 6 show the results for the pendulum amplitude of vibrations for both the small and large initial conditions. The periods are 4.46 and 4.86 for the small and large conditions respectively. The difference is about 9%. Clearly the period depends on the amplitude. The goal of the second example is to show some effects of non-linearity on normal modes of vibration. Consider two pendulums coupled by a non-linear spring (see Figure 7). The spring force is given by equation (11).

Large initial conditions

Small initial conditions

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Figure 7 – Coupled pendulum

The equations of motion can be derived by Newtonian methods (graduate students may want to use a Lagrangian approach). They are given by:

∗ + ∗ +∗ + ∗ + ∗ + ∗ + ∗ +32 2 4

1 1 2 1 1 2 2 1 1 1sin sin cos 2 sin( ) sin( ) cos 0ML k a k a MgLsinη η η η η η η η/ / / / − ?&&

∗ + ∗ + ∗ + ∗ +32 2 4

2 1 2 1 2 2 2 1 2 2sin( ) sin( ) cos 2 sin( ) sin( ) cos( ) 0ML k a k a MgLsinη η η η η η η η− / − / − ?&&

(14)

Using a non-dimensional time Τg

tL

?

, non-dimensional versions of equations (14) can be

written as:

∗ + ∗ +3

1 2 1 1 2 1 1 1Γ 0sin sin cos sin sin cos sinη χ η η η η η η η/ / / / − ?&&

∗ + ∗ +3

2 2 1 2 2 1 2 2Γ 0sin sin cos sin sin cos sinη χ η η η η η η η− / − / − ?&&

(15)

where, now, the dots represent differentiation with respect to the non-dimensional time Τand

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2

1k a

gMLχ ? ,

2

2

1

2Γ

k ak

χ

?

.

Linearized versions of equations (15) are as follows:

∗ + ∗ +1 1 21 0η χ η χ η− − / ?&&

∗ + ∗ +2 1 21 0η χ η χ η/ − − ?&&

(16)

The linear natural frequencies and mode shapes can be found by assuming solution forms:

∗ + ∗ +1 2sin Τ , sin ΤA s B sλ λ? ?

(17)

Substituting into equations (16) gives:

] _21 0A s Bχ χ− / − / ?

] _ 21 0A B sχ χ/ − − / ?

(18)

For non-zero solutions, the determinant of the coefficients must be zero. This gives a polynomial in s , from which the natural frequencies can be obtained. Equations (18) give the associated

mode shapes, namely the ratio BA

.

For 0.1χ ? , one obtains 1.0,1 .095s ? .

The associated mode shapes are given by:

1. 1.0s ? : 1.0BA? , first mode.

This is a mode in which both masses are in phase.

2. 1.095s ? : 1.0BA? / , second mode.

This is a mode in which the masses are out of phase.

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If the initial conditions are:

1 2(0) , (0)

4 4ρ ρλ λ? ?

(19)

∗ + ∗ +1 20 , 0

4 4ρ ρλ λ? ? /

(20)

The linear model predicts that the motions will be purely first and second ones, respectively.

Figure 8 – Coupled pendulum linear mode shapes – first and second modes

Figure 8 shows that this is indeed true. Now consider the non-linear model, which is solved subjected to the same initial conditions (19)

and (20) with Γ 0.55? . As seen in Figure 9 the non-linear mode shapes still show motions in the first and second mode forms, respectively. This is not surprising in view of the symmetrical nature of the problem. However the period is different. For instance, the period of vibrations for the non-linear response

1λ in the first mode is 5% larger than its linear counterpart. In the second mode the period is

26% smaller.

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Figure 9 – Coupled pendulum non-linear mode shapes – first and second modes

Figure 10 – Coupled pendulum mode shapes comparison – linear versus non-linear

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Consider now a non-symmetrical situation in which the length of one bar is twice that of the other. The non-linear (non-dimensional) equations of motion are:

∗ + ∗ +3

1 2 1 1 2 1 1 14 Γ 2 0sin sin cos sin sin cos sinη χ η η η η η η η/ / / / − ?&&

∗ + ∗ +3

2 2 1 2 2 1 2 2Γ 0sin sin cos sin sin cos sinη χ η η η η η η η− / − / − ?&&

(21)

The linearized equations are:

∗ + ∗ +1 1 24 2 0η χ η χ η− − / ?&&

∗ + ∗ +2 1 21 0η χ η χ η/ − − ?&&

(22)

In this case the natural frequencies and associated mode shapes, for 0.1χ ? , are given by:

1. 0.7215847572s ? : 1.05.7932

BA? , first mode.

2. 1.050864139s ? : 1.00.0432

BA? / , second mode.

Taking Γ 0.11? and initializing in the first linear mode, Figure 11 shows the response for both the linear and non-linear cases. Significant differences are seen. The response does not remain in a first mode configuration. Initializing in the second linear mode, Figure 12 gives the response for the linear and non-linear cases. Again, significant differences are seen. For the non-symmetrical problem, initializing in a linear normal mode does not lead to that configuration in the non-linear case.

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Figure 11 – Coupled non-symmetrical pendulum: linear and non-linear mode shapes – first mode

Figure 12 – Coupled non-symmetrical pendulum: linear and non-linear mode shapes – second mode

Page 15.970.15

Conclusions

In the preceding some examples were developed to demonstrate how numerical simulation software can be used to investigate some interesting non-linear effects in mechanical systems. Three illustrative examples were discussed. A damped one-degree of freedom system with a non-linear spring was investigated in order to address the issue of power flow at resonance. Numerical solutions to the differential equations were used to obtain the velocity, from which the power flow per cycle was calculated. It was shown that, for reasonable values of damping, differences between the linear and non-linear model results can be substantial. Some effects of non-linearities on frequencies and mode shapes in vibrations were also treated. By using two simple systems, i.e., a simple pendulum and a symmetric coupled pendulum, interesting results were obtained, such as demonstrating the dependency of the period on the amplitudes of vibrations. A non-symmetrical model was then treated and it was shown that the linear normal mode response is not maintained when the model becomes non-linear. References

1. A. Mazzei and R. A. Scott, "Enhancing student understanding of mechanics using simulation software," Proceedings of the 2006 American Society for Engineering

Education Annual Conference & Exposition, Chicago - IL, 2006. 2. A. Mazzei and R. A. Scott, "Broadening student knowledge of dynamics by means of

simulation software," Proceedings of the 2007 American Society for Engineering

Education Annual Conference & Exposition, Honolulu - HI, 2007. 3. A. Mazzei and R. A. Scott, "Introduction of modern problems into beginning mechanics

curricula," Proceedings of the 2008 American Society for Engineering Education Annual

Conference & Exposition, Pittsburgh - PA, 2008. 4. A. Mazzei and R. A. Scott, "Introduction of some optimization and design problems into

undergraduate solid mechanics," Proceedings of the 2009 American Society for

Engineering Education Annual Conference & Exposition, Austin - TX, 2009. 5. D. M. Fraser, R. Pillay, L. Tjatindi, and J. M. Case, "Enhancing the Learning of Fluid

Mechanics using Computer Simulations," Journal of Engineering Education, vol. Oct, 2007.

6. www.esm.psu.edu/dci/ 7. www.maplesoft.com 8. J. M. Thompson and H. B. Stewart, Nonlinear Dynamics and Chaos. New York: John

Wiley, 1986. 9. J. J. Thomsen, Vibrations and Stability, 1st ed: McGraw-Hill, 1997. Appendix A

Example 01 – Worksheet: #power non-linear restart: with(plots): with(DEtools):with(plottools):

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#nonlinear eq eqn:=diff(x(t),t,t)+.2*diff(x(t),t)+x(t)+.1*x(t)^3=sin(s*t); eqn1:=subs(s=1.325,eqn); ic:=x(0)=0,D(x)(0)=0; sol1:=dsolve({eqn1,ic},numeric,output=listprocedure); odeplot(sol1,[t,x(t)],0..400,numpoints=1000,labels=["non-dimensional frequency","non-dimensional steady state amplitude"],labeldirections=[horizontal,vertical],color=black,axes=boxed); #STEADY STATE AMPLITUDE =3.45 eqn2:=subs(s=1.325,eqn); sol2:=dsolve({eqn2,ic},numeric); odeplot(sol2,[t,x(t)],0..200,numpoints=1000); #BIG DROP OFF IN STEADY STATE AMPLITUDE;GONE THRU RESONANCE #SO LETS ASSUME NONLINEAR RES FREQ IS S=1.325 #BACK TO SOL1 odeplot(sol1,[t,x(t)],55..70,numpoints=1000,axes=boxed,labels=["Non-dimensional time","Amplitude"]); #visual inspection gives #HARMONIC STEADY STATE RESPONSE WITH PERIOD 65.1-60.41; #can be refined #the forcing freq is 2*3.14159/1.325; #close enough steady state res has same freq as input # so xsn:=3.45*sin(1.325*t); #check p1:=odeplot(sol1,[t,x(t)],100 ..120,numpoints=1000,labels=["Non-dimensional time","Amplitude"]):display(%); p2:=plot(xsn,t=100..120,numpoints=1000,color=blue): display({p1,p2},axes=boxed,labels=["Non-dimensional time","Amplitude"]); #phase prob #try xsn1:=3.45*sin(1.325*(t-.9)); p3:=plot(xsn1,t=100..120,color =blue): plots[display]({p1,p3},axes=boxed,labels=["Non-dimensional time","Amplitude"]); #very close lets take xsn1 as the steady state nonlinear response #power #instantaneous power vsn1:=diff(xsn1,t); pp:=vsn1*sin(1.325*t); #ave power at NONLINEAR RESONANCE #period=2*Pi/1.325; ppave:=evalf(int(pp,t=00..2*Pi/1.325)); #big drop(linear was 15 BUT LINEAR AMPLITUDE WAS 5) # duffing equation restart: with(linalg):with(plots):with(DEtools):with(plottools): eq01:=(diff(x(t), `$`(t, 2)))+2*beta*(diff(x(t), t))+x(t)+delta*x(t)^3-sin(nu*t); delta:=0.1;beta:=0.1; eq01; nu:=0;n:=400;number:=100; for i from 1 to number do eq01; sol001:=dsolve({eq01,x(0)=0,D(x)(0)=0},{x(t)}, type=numeric, method=gear,output=listprocedure):xt:=eval(x(t),sol001); xxt:=[seq(abs(xt(j)),j=300..n)]:f1:=max((xxt));p[i]:=[nu,f1]; odeplot(sol001,[t,x(t)],0..n,numpoints=2500,color=black,labels=["non-dimensional time","amplitude"],axes=boxed);

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nu:=nu+0.02; end do; mat01 := array(1..number): for j from 1 to number do mat01[j] := p[j] end do: print(mat01); fig01:=pointplot(mat01,color=black,labels=["Non-dimensional Frequency","Amplitude"],labeldirections=[horizontal,vertical],symbol='box',symbolsize=15):display(fig01); eq02:=(diff(x(tau), `$`(tau, 2)))+2*beta*(diff(x(tau), tau))+x(tau)+delta*x(tau)^3-sin(omega/omega0*tau); amp:=1/(sqrt((1-(omega/omega0)^2)^2+(2*beta*omega/omega0)^2)); omega0:=1;delta:=0;beta:=0.10; eq02; amp; fig02:=plot(amp,omega=0..2): plots[display]([fig01,fig02],labels=["Non-dimensional frequency","Non-dimensional steady state amplitude"],labeldirections=[horizontal,vertical]);

Example 02 – Worksheet: restart: #angle:=evalf(convert(.175,degrees)); angle1:=evalf(convert(10*degrees,radians)); angle2:=evalf(convert(45*degrees,radians)); with(DEtools):with(plots): alpha:=1;beta:=.1;gamma1:=1; eq1:=diff(x(t),t,t)+alpha*sin(x(t))*cos(x(t))+beta*sin(x(t))^3*cos(x(t))+gamma1*sin(x(t))=0; ic1:=x(0)=angle1,D(x)(0)=0; ic2:=x(0)=angle2,D(x)(0)=0; sol1:=dsolve({eq1,ic1},{x(t)},type=numeric,method=rkf45,output=listprocedure); odeplot(sol1,[t,x(t)],0..15); plot1:=odeplot(sol1,[t,x(t)],0..30,color=blue,numpoints=100,view=[0..10,-1...1]):display(%); yproc1:=rhs(sol1[2]); t1:=fsolve(yproc1(x)=0, x=0..2); t2:=fsolve(yproc1(x)=0, x=3..4); t3:=fsolve(yproc1(x)=0, x=5..6); period1:=(t3-t1); sol2:=dsolve({eq1,ic2},{x(t)},type=numeric,method=rkf45,output=listprocedure); odeplot(sol2,[t,x(t)],0..15); plot2:=odeplot(sol2,[t,x(t)],0..30,color=orange,numpoints=100,view=[0..10,-1...1]):display(%); yproc2:=rhs(sol2[2]); t1:=fsolve(yproc2(x)=0, x=0..2); t2:=fsolve(yproc2(x)=0, x=3..4); t3:=fsolve(yproc2(x)=0, x=5..7); period2:=(t3-t1); display([plot1,plot2],labels=["Non-dimensional time","Angle - rad"],labeldirections=[horizontal,vertical]); difference:=((period2-period1)/period1)*100;

Example 03 – Worksheet: restart: with(linalg):with(DEtools):with(plots): #linear modes eq1:=diff(s1(t),t,t)+(1+alpha)*s1(t)-alpha*s2(t)=0; eq2:=diff(s2(t),t,t)+(1+alpha)*s2(t)-alpha*s1(t)=0; alpha:=0.1; ic1:=s1(0)=evalf(Pi/4),D(s1)(0)=0,s2(0)=evalf(Pi/4),D(s2)(0)=0;

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soll:=dsolve({eq1,eq2,ic1},numeric,output=procedurelist); pp1:=odeplot(soll,[t,s1(t)],0..25,numpoints=5000,color=black,legend=["phi1-linear-mode1"],labels=["Non-dimensional time","Amplitude"]):display(%); pp2:=odeplot(soll,[t,s2(t)],0..25,numpoints=5000,color=orange,legend=["phi2-linear-mode1"],labels=["Non-dimensional time","Amplitude"]):display(%); display(pp1,pp2,labels=["Non-dimensional time", "Amplitude"]); ic2:=s1(0)=evalf(Pi/4),D(s1)(0)=0,s2(0)=-evalf(Pi/4),D(s2)(0)=0; sol2:=dsolve({eq1,eq2,ic2},numeric,output=procedurelist); pp3:=odeplot(sol2,[t,s1(t)],0..25,numpoints=5000,color=black,legend=["phi1-linear-mode2"],labels=["Non-dimensional time","Amplitude"]):display(%); pp4:=odeplot(sol2,[t,s2(t)],0..25,numpoints=5000,color=cyan,legend=["phi2-linear-mode2"],labels=["Non-dimensional time","Amplitude"]):display(%);#display(%); display(pp3,pp4,labels=["Non-dimensional time", "Amplitude"]); gamma1:=(1+alpha)*0.5; eq1n1:=diff(s1(t),t,t)-alpha*(sin(s2(t))-sin(s1(t)))*cos(s1(t))-gamma1*(sin(s2(t))-sin(s1(t)))^3*cos(s1(t))+sin(s1(t))=0; eq2n2:=diff(s2(t),t,t)+alpha*(sin(s2(t))-sin(s1(t)))*cos(s2(t))+gamma1*(sin(s2(t))-sin(s1(t)))^3*cos(s1(t))+sin(s2(t))=0; sol3:=dsolve({eq1n1,eq2n2,ic1},numeric,output=procedurelist); pp5:=odeplot(sol3,[t,s1(t)],0..25,numpoints=5000,color=red,legend=["phi1-non-linear-mode1"],labels=["Non-dimensional time","Amplitude"]):display(%); pp6:=odeplot(sol3,[t,s2(t)],0..25,numpoints=5000,color=green,legend=["phi2-non-linear-mode1"],labels=["Non-dimensional time","Amplitude"]):display(%); display(pp5,pp6,labels=["Non-dimensional time", "Amplitude"]); sol4:=dsolve({eq1n1,eq2n2,ic2},numeric,output=procedurelist); pp7:=odeplot(sol4,[t,s1(t)],0..25,numpoints=5000,color=red,legend=["phi1-non-linear-mode2"],labels=["Non-dimensional time","Amplitude"]):display(%); pp8:=odeplot(sol4,[t,s2(t)],0..25,numpoints=5000,color=blue,legend=["phi2-non-linear-mode2"],labels=["Non-dimensional time","Amplitude"]):display(%); display(pp7,pp8,labels=["Non-dimensional time", "Amplitude"]); # display(pp1,pp5,labels=["Non-dimensional time", "Amplitude"],title=["Mode Comparison - Initializing: First Mode"]); display(pp2,pp6,labels=["Non-dimensional time", "Amplitude"],title=["Mode Comparison - Initializing: First Mode"]); display(pp3,pp7,labels=["Non-dimensional time", "Amplitude"],title=["Mode Comparison - Initializing: Second Mode"]); display(pp4,pp8,labels=["Non-dimensional time", "Amplitude"],title=["Mode Comparison - Initializing: Second Mode"]); # restart: with(linalg):with(DEtools):with(plots): #linear modes eq1:=4*diff(s1(t),t,t)+(2+alpha)*s1(t)-alpha*s2(t)=0; eq2:=diff(s2(t),t,t)+(1+alpha)*s2(t)-alpha*s1(t)=0; alpha:=0.1; ic1:=s1(0)=evalf(5.793154380),D(s1)(0)=0,s2(0)=evalf(1),D(s2)(0)=0; soll:=dsolve({eq1,eq2,ic1},numeric,output=procedurelist); pp1:=odeplot(soll,[t,s1(t)],0..50,numpoints=5000,color=black,legend=["phi1-linear-mode1"],labels=["Non-dimensional time","Amplitude"]):display(%); pp2:=odeplot(soll,[t,s2(t)],0..50,numpoints=5000,color=orange,legend=["phi2-linear-mode1"],labels=["Non-dimensional time","Amplitude"]):display(%); display(pp1,pp2,labels=["Non-dimensional time", "Amplitude"]); ic2:=s1(0)=evalf(-0.4315438243e-1),D(s1)(0)=0,s2(0)=evalf(1),D(s2)(0)=0; sol2:=dsolve({eq1,eq2,ic2},numeric,output=procedurelist);

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pp3:=odeplot(sol2,[t,s1(t)],0..50,numpoints=5000,color=black,legend=["phi1-linear-mode2"],labels=["Non-dimensional time","Amplitude"]):display(%); pp4:=odeplot(sol2,[t,s2(t)],0..50,numpoints=5000,color=cyan,legend=["phi2-linear-mode2"],labels=["Non-dimensional time","Amplitude"]):display(%);#display(%); display(pp3,pp4,labels=["Non-dimensional time", "Amplitude"]); gamma1:=(1+alpha)*0.1; eq1n1:=4*diff(s1(t),t,t)-alpha*(sin(s2(t))-sin(s1(t)))*cos(s1(t))-gamma1*(sin(s2(t))-sin(s1(t)))^3*cos(s1(t))+2*sin(s1(t))=0; eq2n2:=diff(s2(t),t,t)+alpha*(sin(s2(t))-sin(s1(t)))*cos(s2(t))+gamma1*(sin(s2(t))-sin(s1(t)))^3*cos(s1(t))+sin(s2(t))=0; sol3:=dsolve({eq1n1,eq2n2,ic1},numeric,output=procedurelist); pp5:=odeplot(sol3,[t,s1(t)],0..50,numpoints=5000,color=red,legend=["phi1-non-linear-mode1"],labels=["Non-dimensional time","Amplitude"]):display(%); pp6:=odeplot(sol3,[t,s2(t)],0..50,numpoints=5000,color=green,legend=["phi2-non-linear-mode1"],labels=["Non-dimensional time","Amplitude"]):display(%); display(pp5,pp6,labels=["Non-dimensional time", "Amplitude"]); sol4:=dsolve({eq1n1,eq2n2,ic2},numeric,output=procedurelist); pp7:=odeplot(sol4,[t,s1(t)],0..50,numpoints=5000,color=red,legend=["phi1-non-linear-mode2"],labels=["Non-dimensional time","Amplitude"]):display(%); pp8:=odeplot(sol4,[t,s2(t)],0..50,numpoints=5000,color=blue,legend=["phi2-non-linear-mode2"],labels=["Non-dimensional time","Amplitude"]):display(%); display(pp7,pp8,labels=["Non-dimensional time", "Amplitude"]); # display(pp1,pp5,labels=["Non-dimensional time", "Amplitude"],title=["Mode Comparison - Initializing: First Mode"]); display(pp2,pp6,labels=["Non-dimensional time", "Amplitude"],title=["Mode Comparison - Initializing: First Mode"]); display(pp3,pp7,labels=["Non-dimensional time", "Amplitude"],title=["Mode Comparison - Initializing: Second Mode"]); display(pp4,pp8,labels=["Non-dimensional time", "Amplitude"],title=["Mode Comparison - Initializing: Second Mode"]); #

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