nonlinearly coupled transverse and in-plane vibrations of a spinning disk

Applied Mathematical Modelling 31 (2007) 54–77

www.elsevier.com/locate/apm

Nonlinearly coupled in-plane and transversevibrations of a spinning disk

Natalie Baddour *, Jean W. Zu

Department of Mechanical and Industrial Engineering, University of Toronto, 5 King’s College Road, Toronto, Ont., Canada M5S 3G8

Received 1 January 2003; received in revised form 1 March 2005; accepted 2 August 2005Available online 21 September 2005

Abstract

Previous nonlinear spinning disk models neglected the in-plane inertia of the disk since this permits the use of a stressfunction. This paper aims to consider the effect of including the in-plane inertia of the disk on the resulting nonlineardynamics and to construct approximate solutions that capture the new dynamics. The inclusion of the in-plane inertiaresults in a nonlinear coupling between the in-plane and transverse vibrations of the spinning disk. The full nonlinear par-tial differential equations are simplified to a simpler nonlinear two degrees of freedom model via the method of Galerkin. Acanonical perturbation approach is used to derive an approximate solution to this simpler nonlinear problem. Numericalsimulations are used to evaluate the effectiveness of the approximate solution. Through the use of these analytical andnumerical tools, it becomes apparent that the inclusion of in-plane inertia gives rise to new phenomena such as internalresonance and the possibility of instability in the system that are not predicted if the in-plane inertia is ignored. It is alsodemonstrated that the canonical perturbation approach can be used to produce an effective approximate solution.� 2005 Elsevier Inc. All rights reserved.

Keyword: Nonlinear spinning disk coupled vibrations

1. Introduction

Spinning disks can be found in many engineering applications. Common industrial applications include cir-cular sawblades, turbine rotors, brake systems, fans, flywheels, gears, grinding wheels, precision gyroscopesand computer storage devices. Spinning disks may experience severe vibrations which could lead to fatiguefailure of the system. Thus, the dynamics of spinning disks has attracted much research interest over the years.

Many authors have investigated the vibrations of spinning disks using linear theory. The original papers areby Lamb and Southwell [1] and by Southwell [2] where the disk was modelled as a spinning membrane withadded bending stiffness. Another popular approach in the literature is to model the spinning disks as a puremembrane with no bending stiffness [3–6]. The incorporation of both the bending stiffness of the disk and theeffect of rotation leads to a fourth order PDE that is difficult to solve. As a result, various researchers have

0307-904X/$ - see front matter � 2005 Elsevier Inc. All rights reserved.

doi:10.1016/j.apm.2005.08.004

* Corresponding author. Tel./fax: +1 709 753 3628.E-mail address: [email protected] (N. Baddour).

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N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77 55

applied different solution techniques to the linear analysis of the free transverse vibrations of spinning disks[7–10]. In addition to considering the transverse vibrations of a spinning disk, the problem of free planarvibrations has also been investigated using linear theory [11–15].

Although less extensive, nonlinear models have also been employed since it is known from the theory ofstationary disks that the linear theory breaks down when the transverse displacement is on the order of thethickness of the disk [16–19]. However, even in the nonlinear case the in-plane inertia of the disk has alwaysbeen neglected and the authors could not find any papers that considered its inclusion in the nonlinear dynam-ics of the rotating disk. This paper thus aims to investigate the effect of the inclusion of the in-plane inertia ofthe disk on the resulting dynamics.

In this paper, an approximate solution to the nonlinear coupled vibrations of a spinning disk is consideredand analyzed. The main focus is to investigate the effect of including the in-plane inertia. This is done by rep-resenting the solution as one mode for each of the transverse and in-plane vibrations and assuming a form forthe space-dependent part of the solution. The time-dependent parts of the in-plane and transverse problemsare then solved for. Through this method, three nonlinear partial differential equations become condensed intotwo nonlinear ordinary differential equations, for the time dependence of the transverse and in-plane vibra-tions. If the in-plane inertia of the plate is neglected, the same approach would yield a one degree of freedommodel. The two degree of freedom model can be used to determine the effect of the inclusion of in-plane inertiaon the dynamics of the problem. The thrust of this paper is to derive the simplified two degree of freedommodel, to consider analytical and numerical solutions and to compare these solutions to the one degree of free-dom model that results if in-plane inertia is neglected.

2. Equations of motion

Since all previous models in the literature ignore the in-plane inertia of the disk in order that a stress func-tion may be used, it is necessary to rederive the equations of motion so that the in-plane inertia of the disk maybe kept. This results in three equations of motion for the dynamics instead of the usual two—a considerablecomplication. This derivation was performed in detail in [20]. The resulting equations and accompanyingboundary conditions are given below as

qð1� m2ÞE

o2u

ot2� v

oXot� X2ðr þ uÞ � 2X

ovot

� �

¼ ð1� mÞ2r2

o2u

oh2þ ð1þ mÞ

2ro2voroh

þ o2uor2þ 1

rð1� mÞ

2

owor

� �2

� ð3� mÞ2r

ovoh� ð1þ mÞ

2r2

owoh

� �2

� urþ ou

or

" #

þ ð1� mÞ2r2

owor

o2w

oh2þ ow

oro2wor2þ ð1þ mÞ

2r2

owoh

o2woroh

; ð1Þ

qð1� m2ÞE

o2vot2þ ðr þ uÞ oX

ot� X2vþ 2X

ouot

� �

¼ ð1þ mÞ2r

owor

o2woroh

þ ð1� mÞ2

o2vor2þ 1

rð1� mÞ

2rowoh

oworþ ð1� mÞ

2

ovorþ ð3� mÞ

2rouoh� ð1� mÞ

2

vr

� �

þ ð1þ mÞ2r

o2uoroh

þ 1

r2

o2v

oh2þ 1

r3

owoh

o2w

oh2þ ð1� mÞ

2rowoh

o2wor2

; ð2Þ

qð1� m2ÞE

o2wot2þ h2X2

3r2w� h2

3

o2

ot2r2w

� �

¼ � h2

3r4wþ 1

r4

owoh

3

2

owoh

o2w

oh2þ r

o2v

oh2

�� r

2

owoh

oworþ ð1þ mÞ

2rouohþ ð1þ mÞ

2r2 o2u

oroh

þ ð1� mÞ2

rvþ r2

2

owoh

o2wor2þ r2 o2w

orohoworþð1� mÞ

2r3 o2v

or2� ð1� mÞ

2r2 ov

or

�

56 N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77

þ 1

r4

owor

r2 o2w

orohowohþ ð1þ mÞr3 ou

orþ ð1þ mÞ

2r3 o

2voroh

�þð1� mÞ

2r2 o

2u

oh2þ r3

2

owor

� �2

þ r2

2

owor

o2w

oh2þ r4 o2u

or2þ 3r4

2

owor

o2wor2� ð1� mÞ

2r2 ov

oh

�þ 1

r4

o2w

oh2mr2 ou

orþ r uþ ov

oh

� ��

þ o2wor2

ouorþ m

ruþ ov

oh

� �� þ ð1� mÞ

r2

o2woroh

ouoh� vþ r

ovor

� �. ð3Þ

Note that the full nonlinear representation of the spinning disk problem requires the solution of three nonlin-early coupled partial differential equation, Eqs. (1)–(3). By way of contrast, the nonlinear model used by otherresearchers consists of two coupled partial differential equations [16]. The difference arises as a result of the useof Lagrangian coordinates as well as the inclusion of in-plane inertia and Coriolis terms.

2.1. Nonlinear boundary conditions

The boundary conditions at a free boundary obtained from applying Hamilton�s principle and integrationby parts are given below. Recall that here r must be evaluated on the free boundary. Thus, for a solid disk, r

below is the outer radius of the disk.

murþ ou

orþ m

rovohþ 1

2

owor

� �2

þ m2r2

owoh

� �2

¼ 0; ð4Þ

1

rouoh� v

� �þ ov

orþ 1

rowoh

owor¼ 0; ð5Þ

o2wor2þ m

r2

o2w

oh2þ m

rowor¼ 0; ð6Þ

o

orr2wþ ð1� mÞ

r2

o2

oh2

owor� w

r

� �þ ð1� m2Þq

EX2 ow

or� o3w

orot2

� �¼ 0. ð7Þ

For an annulus, the outer radius is free and the boundary conditions would be as given above. The innerboundary is clamped and the boundary conditions there would be w ¼ ow

or ¼ 0.

3. Simplification of solution to a 2 DOF system via Galerkin’s method

3.1. Strategy of solution

To consider the effect of the inclusion of in-plane inertia, we choose to proceed with Galerkin�s procedure toproduce a simplified problem. The result will be a two degree of freedom nonlinear model—one for each of thetime dependences of the in-plane and transverse vibrations. To this purpose, let us choose the displacements as

uðr; h; tÞ ¼ ueqðrÞ þ UðrÞ cosðmhÞcðtÞ; ð8Þvðr; h; tÞ ¼ �V ðrÞ sinðmhÞcðtÞ; ð9Þwðr; h; tÞ ¼ W ðrÞ cosðnhÞsðtÞ; ð10Þ

where ueq, U, V and W are known and will be discussed momentarily. The only unknowns here are the timedependences, c(t) and s(t). The ueq portion is the in-plane equilibrium displacement due to the spin of the disk.The implication here is that there is a symmetrical time-independent in-plane displacement that occurs in theplane of the disk as a result of its rotation (that is, due to the centrifugal force). The form for u and v followsfrom the solution of the in-plane linear problem [21,22,13] giving

umk ¼ AmkUmk

iV mk

� �eiðmh�pmk tÞ; ð11Þ

where Umk and Vmk are the in-plane mode shapes of the linear spinning disk and pmk is the in-plane frequencyof vibration. Taking the real part of Eq. (11) leads to the forms for u and v indicated in Eqs. (8) and (9). Now,


U and V will be taken to be Umk and Vmk, respectively, where Umk and Vmk are the (m, k)th in-plane modeshapes found by solving the free linear in-plane vibration problem. Other choices for U and V can also bemade and we are not limited to choosing them the be the in-plane mode shapes. However, the advantagesof taking U and V to be the in-plane mode shapes are that (i) there are proven orthogonality properties forthe in-plane mode shapes [21] and more importantly (ii) the in-plane mode shapes satisfy the boundary con-ditions rrr = rrh = 0 at the boundary of the disk. Similarly, W will be taken to be Wnk, where Wnk is the(n, k)th mode shape of the linear strain model of the spinning disk, obtained by dropping ALL nonlinear termsin Eq. (3), see [23]. That is,

W nkðrÞ ¼ J n P nka

ra

� �þ BnIn P nk

b

ra

� �; ð12Þ

where P nka and P nk

b are the (n, k)th solutions of the frequency equation for the linear strain model of the spin-ning disk [20]. Recall that the appropriate relationship between P nk

a and P nkb must be specified. Other choices

for the space-dependent portion of the solution can easily be made and substituted into the above expressions.Once a choice for the space-dependent part of the solution has been made, it still remains to use this choice

to derive the equations of motion for the temporal part of the solution. One option is to substitute the assumedforms of the solution into the equations of motion and then apply the method of Galerkin. A completelyequivalent calculation is to substitute the assumed form of the solution into the expressions for kinetic andpotential energies and using Lagrange�s equations. The advantage of actually calculating the potential and ki-netic energies of the system is that conserved quantities are more readily apparent. Such conservation laws canbe used to simplify the equations of motion and could aid in the finding of an analytic solution. Furthermore,the machinery of Hamilton�s equations and canonical perturbation theory is also available in this formulation.For these reasons, the simplified 2 DOF ordinary differential equations of motion are derived in this alterna-tive manner.

3.2. Galerkin method

Using the assumption of plane stress, along with linear stress–strain relationships and nonlinear von Kar-man strains, the potential and kinetic energies of the system can then be found in terms of the two degrees offreedom, c and s to given by

PE ¼ Q1c2 þ Q2s4 þ Q3s

2 þ Q4 þ Q10cs2; ð13ÞKE ¼ Q5c2 þ Q6s

2 þ Q7 _s2 þ Q8 _c2 þ Q9. ð14Þ

The coefficients in Eqs. (13) and (14) can be expressed as

Q1 ¼hEpð1� m2Þ

Z a

0

dUdr

� �2

þ Ur� m

Vr

� �2

þ ð1� mÞ2

oVor� V

rþ m

Ur

� �2"

þ 2mdUdr

Ur� m

Vr

� �#r dr; ð15Þ

Q2 ¼hEp

16ð1� m2Þ

Z a

0

3n4 Wr

� �4

þ 2n2 Wr

� �2dWdr

� �2

þ 3dWdr

� �4( )

r dr; ð16Þ

Q3 ¼ Qh3 þ Qh3

3 ; ð17Þ

where

Qh3

3 ¼h3Ep

3ð1� m2Þ

Z a

0

d2Wdr2

� �2

þ 2md2Wdr2

1

rdWdr� n2 W

r2

� �þ 1

rdWdr� n2 W

r2

� �2(

þ2ð1� mÞ nr

dWdr� n

Wr2

� �2)

r dr; ð18Þ


Qh3 ¼

hEpð1� m2Þ

Z a

0

dWdr

� �2dueq

drþ m

ueq

r

� �þ n2W 2

r2

ueq

rþ m

dueq

dr

� �( )r dr

¼ phZ a

0

reqrr

dWdr

� �2

þ reqhh

n2W 2

r2

( )r dr. ð19Þ

Here reqrr ; r

eqhh are the equilibrium stresses in the rotating disk resulting from the equilibrium displacement ueq.

Q4 ¼2hEpð1� m2Þ

Z a

0

2mueq

rdueq

drþ ueq

r

� �2

þ dueq

dr

� �2( )

r dr; ð20Þ

Q5 ¼ qhpX2

Z a

0

U 2 þ V 2�

r dr; ð21Þ

Q6 ¼qh3pX2

3

Z a

0

dWdr

� �2

þ n2 W 2

r2

" #r dr; ð22Þ

Q7 ¼qh3p

3

Z a

0

dWdr

� �2

þ n2 W 2

r2

" #r dr þ qhp

Z a

0

W 2r dr; ð23Þ

Q8 ¼ qhpZ a

0

U 2 þ V 2�

r dr; ð24Þ

Q9 ¼ 2qhpX2

Z a

0

ueq þ r� 2

r dr; ð25Þ

Q10 ¼ 0 for m 6¼ 2n; ð26Þ

Q10 ¼hEp

2ð1� m2Þ

Z a

0

(Ur

n2 Wr

2dWdr� W

r

� �þ m

dWdr

dWdr� 2n2 W

r

� �� : ð27Þ

þ Vr

nWr

2n2 Wr� dW

dr

� �þ nm

dWdr

Wr� 2

dWd

� ��

þ dUd

dWd

� �2

� mn2 Wr

� �2" #

þ nð1� mÞ dVdr

Wr

dWdr

)r dr otherwise. ð28Þ

The above calculations indicate that Q10 = 0 for m 5 2n. This point is worth some consideration since it con-siderably affects the resulting dynamics if Q10 is indeed zero. In this case, Q10 = 0 follows as a result of ourchoosing trigonometric functions for the h component of the trial solution to the problem and the fact thatthe sine and cosine functions possess orthogonality properties. A slightly different choice for the h componentof the solution would not necessarily lead to Q10 = 0.

Note that for a stationary disk, where X = 0, it would follow that Q4 = 0 unless the disk is pre-stressed andthat Q5 = Q6 = Q9 = 0. All other expressions are the same for a stationary disk, although the actual modeshapes used in these expressions will be different.

Once expressions for the kinetic and potential energies are found, the Lagrangian can be constructed andthe equations of motion for the simplified system can be found.

If new variables are introduced such that

c ¼ cIffiffiffiffiffiffiffiffi2Q8

p ; ð29Þ

s ¼ sIffiffiffiffiffiffiffiffi2Q7

p . ð30Þ

Then the Lagrangian from equation can be written as

L ¼_sI� 2

2þ

_cI� 2

2� Q4 � Q9ð Þ � x2

c

2cI� 2 � k3

4sI� 4 � x2

s

2sI� 2 � k1cI sI

� 2; ð31Þ


where

x2c ¼

Q1 � Q5

Q8

; ð32Þ

x2s ¼

Q3 � Q6

Q7

; ð33Þ

k1 ¼Q10

2Q7

ffiffiffiffiffiffiffiffi2Q8

p ; ð34Þ

k3 ¼Q2

Q27

. ð35Þ

Thus the equations of motion for the system become

€cI þ x2ccI þ k1ðsIÞ2 ¼ 0; ð36Þ

€sI þ x2ss

I þ 2k1cIsI þ k3ðsIÞ3 ¼ 0. ð37Þ

In future discussions, cq and sq will be replaced with c and s for ease of notation.It is important to note that if Q10 = 0 then k1 = 0 and the two equations of motion (36) and (37) are uncou-

pled. That is, for Q10 = 0, the system behaves like two independent oscillators, where the in-plane oscillator islinear and the transverse one is a nonlinear duffing oscillator. What are the physical implications of this?Recall that to arrive at Eqs. (36) and (37), two particular modes were selected at the outset. If k1 = 0, the impli-cation is that the vibration of these particular modes are not coupled. Since there is an infinite number ofin-plane and transverse modes of vibration, this would not necessarily imply that none of the in-plane andtransverse modes of vibrations are coupled. However, it does not follow that all the transverse modes arecoupled to all the in-plane modes and this is the physical implication of k1 = 0.

3.3. Simplifying the coefficients

After some calculation and simplification, it can be rigorously shown that xc and xs are exactly what wewould expect them to be—namely the frequencies of vibration of the linear in-plane and transverse vibrations.

3.3.1. Calculating xc

It can be shown that Q1 reduces to the sum of a boundary term and an integral term

Q1 ¼ Qboundary1 þ Qintegral

1 ; ð38Þ
where Qboundary
1 ¼ hprUrrrja0 þ hprV rrhja0 ¼ 0 by virtue of the boundary conditions. If U = Umk, V = Vmk issubstituted into Qintegral

1 , denoting the (m,k)th modeshape, then it can be shown that Qintegral1 becomes

Qintegral1 ¼ �hqpAkk; ð39Þ

where Akk ¼ ðp2mk � X2Þakk � pmk for a spinning disk and Akk ¼ �p2

mkakk for a stationary disk. Here pmk is thelinear in-plane frequency of vibration and

ajk ¼Z a

0

U mjUmk þ V mjV mk

� r dr. ð40Þ

Thus for a spinning disk, x2c can be written as

x2c ¼

Q1 � Q5

Q8

¼ Q1

Q8

� X2 ð41Þ

¼� p2

k � X2�

akk � pmk

� �hpq

hpqakk� X2 ð42Þ

¼ pmk

akk� p2

mk ¼ pmk1

akk� pmk

� �; ð43Þ


which is precisely the linear in-plane frequency of vibration as shown in [21]. For a stationary disk, it can beshown that x2

c ¼ p2mk. In other words, xc is precisely the frequency of vibration of the in-plane mode that was

chosen. This is a naturally intuitive result, which the calculations justify.

3.3.2. Calculating xs

Let us now consider how to calculate xs. Recall that xs is defined by

x2s ¼

Q3 � Q6

Q7

ð44Þ

¼ Qh3

3 � Q6

Q7

þ Qh3

Q7

. ð45Þ

That is, x2s can be considered to have two components; a �plate� part where all terms are proportional to h3 and

a �membrane� part which is proportional to h. Clearly the plate part is given byQh3

3�Q6

Q7and the membrane part is

given byQh

3

Q7. If we take W to be the eigenfunctions of the linear strain model of the spinning disk [23], then

W ¼ W nk ¼ AnkJ n P ara

� �þ CnkIn P b

ra

� �. ð46Þ

Let knk denote the (n,k)th frequency of vibration of the linear strain model of the spinning disk. Then follow-ing some algebra it can be shown that the plate portion of x2

s is given by

Qh3

3 � Q6

Q7

¼ k2nk. ð47Þ

That is, the plate portion of x2s is the square of the frequency of the linear strain model of the spinning plate.

Considering the fact that the Wnk were chosen to be the mode shapes of this model, this is an intuitive result.Now consider the membrane portion x2

s .

Qh3

Q7

¼phR a

0req

rrdWdr

� 2 þ reqhh

n2W 2

r2

n or dr

qh3p3

R a0

dWdr

� 2 þ n2 W 2

r2

h ir dr þ qhp

R a0

W 2r dr. ð48Þ

Now, for W = Wnk, where the Wnk are the eigenfunctions of the linear strain model of the spinning plate, theabove expression does not permit as nice an interpretation as the plate component of x2

s did. However, for thesake of illustration, suppose that in Eq. (48) we take W ¼ W m

nk, the eigenfunctions of the linear spinning mem-

brane. First, the term proportional to h3 in the denominator of Eq. (48) would not be present for a membrane.Thus, the denominator becomes qhp

R a0

W 2r dr. Furthermore, if the eigenfunctions of the spinning membraneare normalized, then Eq. (48) can be shown to reduce to

Qh3

Q7

¼qph km

nk

� 2

qph¼ km

nk

� 2; ð49Þ

that is, the square of the natural frequency of the (n,k)th mode of the spinning membrane. Again, this is anatural result. However, in this development, it is the eigenfunctions of the linear strain model of the spinningdisk that are being used and thus the membrane component of x2

s will not give exactly ðkmnkÞ

2 but rather someapproximation to it.

As another note, suppose that we had taken W ¼ W NLSMnk , the eigenfunctions of the nonlinear strain model,

obtained by dropping only the terms nonlinear in w and its derivatives in Eq. (3). Note that these eigenfunc-tions are not actually available to us, but it is instructive to see what would happen is we could use them in thiscalculation. A quick review of the above developments would indicate that the result to be expected isx2

s ¼ ðkNLSMnk Þ2.


4. Canonical perturbation approach of the simplified 2 DOF system

The coupled 2 DOF equations of motion derived in the above sections are given by

€cþ x2ccþ k1s

2 ¼ 0; ð50Þ€sþ x2

ssþ 2k1csþ k3s3 ¼ 0. ð51Þ

Clearly, xc and xs are the natural frequencies of the linear in-plane and transverse modes that were chosen.The expressions for k1 and k3 are complicated, but it suffices to note that they will depend on the amplitudes ofthe chosen modes.

Eqs. (50) and (51) can also be arrived at from the point of view of Hamilton�s equations. That is, Eqs. (50)and (51) are equivalent to

_c ¼ oHopc

;

_s ¼ oHops

;

_pc ¼ �oHoc

;

_ps ¼ �oHos

;

ð52Þ

where pc and ps are the conjugate momenta to c and s, and the Hamiltonian is given by

H ¼ p2c

2þ p2

s

2þ x2

c

2c2 þ x2

s

2s2 þ k1cs2 þ k3

4s4. ð53Þ

Eq. (52) represent a general Hamiltonian system. Since the equations of motion of the 2 DOF model can bederived from a Hamiltonian, they are a Hamiltonian system. The general objective here is to transform thegiven Hamiltonian system into a simpler Hamiltonian system and to use the result to construct an approxi-mate solution.

4.1. Transformation of the Hamiltonian

Generally, a transformation of variables from generalized coordinates q, p to qq, pq can be of the form

qI ¼ qIðq; p; tÞ;pI ¼ pIðq; p; tÞ

ð54Þ

with corresponding transformed differential equations

_qI ¼ _qIðq; p; tÞ;_pI ¼ _pIðq; p; tÞ.

ð55Þ

Suppose that Eq. (55) has the same symmetry as the original equations. That is, suppose that there exists afunction Hq(qq,pq, t) such that

_qI ¼ oHI

opI;

_pI ¼ � oHI

oqI.

ð56Þ

Then the transformation (54) is known as a canonical transformation [24], transforming Hamiltonian differ-ential equations into another system of differential equations of the same type.

One possible approach to solving the equations of motion is to find a canonical transformation such thatthe transformed Hamiltonian Hq is a constant function, independent of time and any of the generalized


coordinates or momenta. Then Eq. (56) reduce to the simple form of _qI ¼ _pI ¼ 0 and can be easily integrated.In reality, finding such a transformation can be just as difficult as solving the original equations of motion.However, it is possible to use a canonical transformation that effectively transforms a simpler Hamiltonianto aid in the construction of an approximate solution for a problem with a more complicated Hamiltonian.Even though such a transformation will not solve the more complicated problem, it will help in the construc-tion of an approximate solution. This is the approach that will be undertaken here.

Consider the Hamiltonian of the 2 DOF model:

H ¼ p2c

2þ p2

s

2þ x2

c

2c2 þ x2

s

2s2 þ k1cs2 þ k3

4s4. ð57Þ

This Hamiltonian can be expressed as

H ¼ H ð0Þ þ H ð1Þ; ð58Þ

where

H ð0Þ ¼ p2c

2þ p2

s

2þ x2

c

2c2 þ x2

s

2s2; ð59Þ

H ð1Þ ¼ k1cs2 þ k3

4s4. ð60Þ

The motivation for doing this is that H(0) can be recognized as the Hamiltonian of two uncoupled linear har-monic oscillators. H(1) brings in coupling by way of the third order k1 term. The fourth order term in H(1) doesnot couple the two oscillators. Rather, it has the effect of turning the linear s oscillator into a nonlinear duffingoscillator. The k1 term is the only term that couples the two oscillators.

Thus the first step in our approximate solution will be to find a canonical transformation that transformsH(0) into a constant. This can easily be done by solving the Hamilton–Jacobi equation associated with H(0)

[24]. Now this canonical transformation is equally valid no matter which Hamiltonian is employed. It justhas the particular advantage of simplifying this particular Hamiltonian H(0), but it can be used with the entireoriginal Hamiltonian H. After thus transforming the Hamiltonian, approximate solutions for the ensuingHamiltonian equations will be sought.

Solving the Hamilton–Jacobi equation associated with H(0) yields

c ¼ffiffiffiffiffiffiffi2a1

p

xcsin xcðt � b1Þ; ð61Þ

s ¼ffiffiffiffiffiffiffi2a2

p

xssin xsðt � b2Þ. ð62Þ

Furthermore, the generalized momenta can also be transformed

pc ¼oSoc¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2a1 � x2

cc2

q¼

ffiffiffiffiffiffiffi2a1

pcos xcðt � b1Þ; ð63Þ

ps ¼oSos¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2a2 � x2

ss2

q¼


pcos xsðt � b2Þ. ð64Þ

Eqs. (61)–(64) are essentially a canonical transformation from the old variables c, s, pc, ps to the new variablesa1, a2, b1, b2, where these are constants of integration. It can be verified that if (61)–(64) are substituted intoH(0), the result is the constant a1 + a2. This is not coincidence: the transformation was specifically constructedto do so. In fact, Eqs. (61)–(64) are precisely the solution to the uncoupled two oscillator problem. The trans-formed Hamiltonian for the uncoupled two oscillator problem is given by H ð0Þ þ oS

ot , which is precisely zero (byconstruction).

Now, while this canonical transformation transforms H(0) into zero, it does not do so for the entireHamiltonian H = H(0) + H(1). The transformed Hamiltonian becomes HI ¼ H ð0Þ þ oS

ot þ H ð1Þ ¼ H ð1Þða1;a2; b1; b2; tÞ. Using these results, the �perturbing Hamiltonian� H(1) can be expressed as


H ð1Þ ¼ � a2

8xcx4s

�8k1


px2

s sinðxct � xcb1Þ þ 4k1


px2

s sinððxc þ 2xsÞt � xcb1 � 2xsb2Þ�

þ 4k1


px2

s sinððxc � 2xsÞt � xcb1 þ 2xsb2Þ � 3k3a2xc � k3a2xc cosð4xst � 4xsb2Þ

þ 4k3a2xc cosð2xst � 2xsb2Þ�

.

Note from this definition of H(1) that there is a difference in its form depending on whether or not xc = 2xs.Although it may not be obvious yet, if xc = 2xs, then the system will display internal resonance and H(1) willtake on a different form.

From the above, using the new variables a1,a2, b1 and b2, Hamilton�s equations can be written

_a1 ¼oH ð1Þ

ob1

¼ 1

2a2k1


p

x2s

ð�2 cos �xct þ xcb1ð Þ

þ cos �xct þ xcb1 � 2xst þ 2xsb2ð Þ þ cos xct � xcb1 � 2xst þ 2xsb2ð ÞÞ; ð65Þ

_a2 ¼oH ð1Þ

ob2

¼ 1

2

a2

x1x32

ð2k1


px2

s cos �xct þ xcb1 � 2xst þ 2xsb2ð Þ

� 2k1


px2

s cos xct � xcb1 � 2xst þ 2xsb2ð Þ � k3a2xc sin �4xst þ 4xsb2ð Þ

þ 2k3a2xc sin �2xst þ 2xsb2ð ÞÞ; ð66Þ

_b1 ¼ �oH ð1Þ

oa1

¼ 1

4a2k1

ffiffiffi2p

x2sxc

ffiffiffiffiffia1p ð2 sin �xct þ xcb1ð Þ

� sin �xct þ xcb1 � 2xst þ 2xsb2ð Þ þ sin xct � xcb1 � 2xst þ 2xsb2ð ÞÞ; ð67Þ

_b2 ¼ �oH ð1Þ

oa2

¼ 1

4xcx4s

ð�2k1


px2

s sin �xct þ xcb1 � 2xst þ 2xsb2ð Þ

þ 4k1

ffiffiffi2p ffiffiffiffiffi

a1

px2

s sin �xct þ xcb1ð Þ þ 2k1


px2

s sin xct � xcb1 � 2xst þ 2xsb2ð Þ

� 3k3a2xc � k3a2xc cos �4xst þ 4xsb2ð Þ þ 4k3a2xc cos �2xst þ 2xsb2ð ÞÞ. ð68Þ

Eqs. (65)–(68) are exact—no approximations have been made at this point. We have used a valid transforma-tion to change the variables and obtained the equations of motion for the transformed variables. If these equa-tions could be solved, they would provide the solution to the original problem although practically this is noteasier than the original problem. The utility of the transformation is that Eqs. (65)–(68) offer a good startingpoint for approximate solutions and make certain features of the solutions easy to spot.

To construct an approximate solution, a1, a2, b1, b2 on the right hand side of Eqs. (65)–(68) are replaced bytheir initial constant values. Eqs. (65)–(68) are then integrated to yield the first order canonical perturbationsolution. For a second order canonical perturbation solution, a1, a2, b1, b2 on the right hand side of Eqs. (65)–(68) are replaced with the first order perturbation solution found in the previous step. This procedure can becontinued indefinitely, although in practice this gets unwieldy after a few iterations.

4.2. Canonical perturbation solution

If the initial conditions for (50) and (51) are given by

sð0Þ ¼ s0; ð69Þdsð0Þ

dt¼ 0; ð70Þ

cð0Þ ¼ c0; ð71Þdcð0Þ

dt¼ 0; ð72Þ


then the corresponding initial values for a1, a2, b1, b2 can be worked out from the transformation equations(61)–(64) to be

a01 ¼

1

2x2

cc20; ð73Þ

b01 ¼ �

p2xc

; ð74Þ

a02 ¼

1

2x2

ss20; ð75Þ

b02 ¼ �

p2xs

. ð76Þ

Once the a1, a2, b1, b2 are replaced in the right hand side of Eqs. (65)–(68) with their initial values a01; a

02; b

01; b

02,

then Eqs. (65)–(68) can be easily integrated to give the first order canonical perturbation solutions for a1, a2,b1, b2:

a11 ¼ �

k1xcc0s20 cosððxc þ 2xsÞtÞ

4ðxc þ 2xsÞ� k1xcc0s2

0 cosððxc � 2xsÞtÞ4ðxc � 2xsÞ

� k1c0s20

2cosðxctÞ

þ k1c0s20ðx2

c � 2x2sÞ

x2c � 4x2

s

þ x2cc2

0

2; ð77Þ

a12 ¼ �

k3s40

32cosð4xstÞ �

k3s40

8cosð2xstÞ �

k1xsc0s20 cosððxc þ 2xsÞtÞ

2ðxc þ 2xsÞ

þ k1xsc0s20 cosððxc � 2xsÞtÞ

2ðxc � 2xsÞ� 2x2

sk1c0s20

x2c � 4x2

s

þ 5k3s40

32þ x2

ss20

2; ð78Þ

b11 ¼ �

k1s20 sinðxctÞ2c0x3

c

� k1s20 sinððxc þ 2xsÞtÞ

4c0x2cðxc þ 2xsÞ

� k1s20 sinððxc � 2xsÞtÞ

4c0x2cðxc � 2xsÞ

� p2xc

; ð79Þ

b12 ¼ �

k1c0 sinðxctÞxcx2

s

� k1c0 sinððxc þ 2xsÞtÞ2x2

sðxc þ 2xsÞ� k1c0 sinððxc � 2xsÞtÞ

2x2sðxc � 2xsÞ

� k3s20 sinð2xstÞ

4x3s

� k3s20 sinð4xstÞ32x3

s

� 3k3s20t

8x2s

� p2xs

. ð80Þ

Clearly, a quick glance at some of the values in the denominators reveals that the above expressions fora1

1; a12; b

11; b

12 are not valid if the resonance condition of xc = 2xs holds. In that case, the limit of the above

expressions as xc ! 2xs is taken. This yields the first order canonical perturbation solution for the internalresonance case:

a11;IR ¼ k1c0s

20ð1� cos4ðxstÞÞ þ 2x2

sc20; ð81Þ

a12;IR ¼ k1c0s

20ðcos2ðxstÞ � cos4ðxstÞÞ þ

k3s40

41� cos4ðxstÞ�

þ x2ss

20

2; ð82Þ

b11;IR ¼ �

s20

16c0x3s

2k1 sinðxstÞcos3ðxstÞ þ k1 sinðxstÞ cosðxstÞ þ4c0px2

s

s20

þ k1xst� �

; ð83Þ

b12;IR ¼ �

ð4k1c0 þ k3s20Þ

4x3s

sinðxstÞcos3ðxstÞ �ð4k1c0 þ 3k3s2

0Þ8x3

s

sinðxstÞ cosðxstÞ

� 4pxs þ 4k1c0t þ 3k3s20t

8x2s

. ð84Þ

This procedure thus gives a first order solution for both the internal resonance case and the non-internal res-onance case. To find the second order solution, a1, a2, b1, b2 on the right hand side of Eqs. (65)–(68) are re-placed with their first order values a1

1; a12; b

11; b

12. Once again, the right hand sides of Eqs. (65)–(68) can be

integrated to yield the second order perturbation solution a21; a

22; b

21; b

22, although this can be difficult to achieve

analytically. In general, the canonical perturbation approach easily provides a first order approximate solutionwhile higher order approximations are possible but more difficult with this approach.


Recall that a1,a2, b1,b2 do not provide the solution to the original problem directly. These are obtainedfrom the transformation equations, namely

c ¼ffiffiffiffiffiffiffi2a1

p

xcsin xcðt � b1Þ; ð85Þ

s ¼ffiffiffiffiffiffiffi2a2

p

xssin xsðt � b2Þ. ð86Þ

Recall that a01; a

02; b

01; b

02 are constants, thus the resulting solution for c and s for the two uncoupled oscillators

are simply periodic trigonometric functions, namely the linear solution for simple harmonic oscillators that areso familiar. For the coupled problem, a1

1; a12; b

11; b

12 are actually periodic functions themselves. Thus the result-

ing solutions for c and s are now periodic trigonometric functions with periodically changing amplitudes.

5. Exact solution of the 1 DOF model

Recall that the nonlinear model of a spinning disk used in the literature neglects the presence of in-planeinertia. The model derived here, Eqs. (50) and (51), can be reduced to the model used in the literature by ignor-ing the in-plane inertia in Eq. (50). Eq. (50) would then become

x2ccþ k1s

2 ¼ 0. ð87Þ

This equation can be solved for c and the result substituted into Eq. (51) to yield the one degree of freedommodel (1 DOF):

€sþ x2ssþ k3 �

2k21

x2c

� �s3 ¼ 0. ð88Þ

Note that this is the equation of motion of a nonlinear one degree of freedom system and will be referred to asthe one degree of freedom (1 DOF) model, describing the dynamics of the system that ignores the effect ofin-plane inertia.

The Hamiltonian associated with Eq. (88) can be written as

H 1 DOF ¼p2

s

2þ x2

s

2s2 þ k3

4� k2

1

2x2c

� �s4. ð89Þ

Recall that ps ¼ dsdt and that the Hamiltonian must be conserved since it does not explicitly depend on time,

that is H1 DOF is a constant. Using those two facts, Eq. (88) can be integrated to give s as a function of time.If the initial conditions are given by

sð0Þ ¼ s0; ð90Þdsð0Þ

dt¼ 0; ð91Þ

then the solution is given by

sðtÞ ¼ s0cnðxð1Þt; kð1ÞÞ; ð92Þ

where

xð1Þ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2

s þ s20 k3 �

2k21

x2c

� �s; ð93Þ

kð1Þ ¼ s0

xð1Þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

2k3 �

2k21

x2c

� �s. ð94Þ


In Eq. (92), cn is the Jacobian elliptic function that is a generalization of the cosine function. The functioncn(x(1)t,k(1)) is a periodic function with period

T ð1Þ ¼ 4Kð1Þ

xð1Þ; ð95Þ

where K(1)(k(1)) = F(k(1),p/2) is the complete elliptic integral of the first kind. Note that the above solution isonly valid for real k(1). That is, for k3x2

c � 2k21 > 0. The solution to the 1 DOF model is thus a periodic func-

tion which reduces to the cosine function when k(1) = 0. Furthermore, the frequency of this function is givenby x(1) which depends on xs, as we would expect, as well as s0, k1, k3 and xc. Here, k1 and k3 may depend onthe amplitudes of vibration and thus we have the familiar result of the frequency of vibration being dependenton the amplitude.

6. Numerical simulations

In this section, numerical simulations of the 1 DOF and 2 DOF models are considered, along with theapproximate solution. All numerical results were obtained by using MATLAB. For numerical solutions tothe nonlinearly coupled equations, fourth and fifth order Runge–Kutta formulas were employed via built-in MATLAB functions.

6.1. Resonance cases

For the first case, the following values for the constants are taken: k1 = 2, k3 = 2, xc = 4, xs = 2, c0 = 0.2,s0 = 0.2. This is clearly a case where xc = 2xs, thus internal resonance is expected.

Consider first a comparison of the numerical solution of the 1 DOF and 2 DOF models. This is shown for sand c in Figs. 1 and 2, respectively. Since the approximation c ¼ � k1

x2c

is used in order to obtain the 1 DOFmodel, it is to be expected that there is a large discrepancy between c(t) as obtained from the solutions ofthe 1 DOF model and the 2 DOF model. This is in fact shown explicitly in Fig. 2. What is interesting andnot predicted by the 1 DOF model is the internal resonance. From Fig. 1 it can be seen that while the ampli-tude of s(t) remains constant for the 1 DOF model, the 2 DOF model displays a periodically varying ampli-tude, in keeping with the phenomena of internal resonance. It can be seen from Figs. 1 and 2 that energy is

2 DOF 1 DOF

0 5 10 15 20 25 30 35 40 45 50–0.5

–0.4

–0.3

–0.2

–0.1

0

0.1

0.2

0.3

0.4

0.5Tau for 1 and 2 DOF models

Time

Tau

Fig. 1. The numerical solution of s for the 1 DOF and 2 DOF models.

2 DOF 1 DOF

0 5 10 15 20 25 30 35 40 45 50–0.25

–0.2

–0.15

–0.1

–0.05

0

0.05

0.1

0.15

0.2c for 1 and 2 DOF models

Time

c(t)

Fig. 2. The numerical solution of c for the 1 DOF and 2 DOF models.


transferred back and forth from the s oscillator to the c oscillator. Indeed, each achieves its maximum ampli-tude variation when the other achieves its minimum amplitude variation.

Now that we have a better understanding of how the exact solution of the 2 DOF model behaves, let usconsider how the approximate solution fares in capturing the periodic variation of amplitude. The exact solu-tion is compared to the first and second order canonical perturbation solutions in Figs. 3 and 4. Clearly, thefirst order canonical perturbation solution does not have the correct amplitude variation. However, it can beseen from the figures that the second order canonical perturbation solution is a much better approximation tothe actual solution since it begins to capture the variation of amplitude of the solution.

As a second example, consider the case k1 = 4, k3 = 6, xc = 2, xs = 1, c0 = 0.1, s0 = 0.1. This is also a casewhere xc = 2xs, thus internal resonance is expected.

Tau for Exact, First and Second Order Canonical Perturbation solutions

Exact First Order Canonical Second Order Canonical

0 5 10 15 20 25 30 35 40 45 50–0.5

–0.4

–0.3

–0.2

–0.1

0

0.1

0.2

0.3

0.4

0.5

Time

Tau

Fig. 3. Comparison of s with numerical solution, first and second order canonical perturbation solution.

c for Exact, First and Second Order Canonical Perturbation solutions

0 5 10 15 20 25 30 35 40 45 50–0.25

–0.2

–0.15

–0.1

–0.05

0

0.05

0.1

0.15

0.2

0.25

Time

c

Fig. 4. Comparison of c with numerical solution, first and second order canonical perturbation solution.


The graphs of s and c for the exact solutions of the 1 DOF and 2 DOF model are shown in Figs. 5 and 6respectively. Also, the graphs of s and c for the exact solution of the 2 DOF model as well as the first andsecond order canonical perturbation solutions are shown in Figs. 7 and 8, respectively.

Many of the comments made regarding the previous case can also be made here. Note how the 1 DOFmodel does not predict the variation of the amplitude. In addition, the first order canonical perturbation solu-tion does not correctly reflect this same change in the amplitude of the solution. Note that once again the sec-ond order canonical perturbation solution fares better at representing the solution of the 2 DOF model thanthe first order canonical perturbation solution.

2 DOF1 DOF

0 5 10 15 20 25 30 35 40 45 50–0.25

–0.2

–0.15

–0.1

–0.05

0

0.05

0.1

0.15

0.2


Time

Tau


2DOF1DOF

0 5 10 15 20 25 30 35 40 45 50–0.15

–0.1

–0.05

0

0.05

0.1


Time

c(t)




0 5 10 15 20 25 30 35 40 45 50–0.25

–0.2

–0.15

–0.1

–0.05

0

0.05

0.1

0.15

0.2

0.25

Time

Tau



6.2. Non-resonance cases

As another example, consider the case k1 = 2, k3 = 2, xc = 2, xs = 4, c0 = 0.1, s0 = 0.1. This is not a casewhere xc = 2xs, thus no resonance is expected.

The graphs of s and c for the exact solutions of the 1 DOF and 2 DOF model are shown in Figs. 9 and 10,respectively. Also, the graphs of s and c for the exact solution of the 2 DOF model, the first and second ordercanonical perturbation solutions are shown in Figs. 11 and 12, respectively.

As expected, the solutions for c for the exact 1 DOF and 2 DOF models differ significantly, but this is ex-pected as a result of the assumption required to arrive at the 1 DOF model. Other than that discrepancy, allother solutions match perfectly.


0 5 10 15 20 25 30 35 40 45 50–0.15

–0.1

–0.05

0

0.05

0.1

0.15

Time

c


Tau for 1 and 2 DOF models

2 DOF1 DOF

0 2 4 6 8 10 12 14 16 18 20–0.15

–0.1

–0.05

0

0.05

0.1

0.15

Time

Tau



Consider a similar example with k1 = 2, k3 = 2, xc = 2, xs = 4, c0 = 0.4, s0 = 0.4. This is the same as theprevious case, with a slight change in the initial conditions. The graphs of s and c for the exact solutionsof the 1 DOF and 2 DOF model are shown in Figs. 13 and 14, respectively. The graphs of s and c for theexact solution of the 2 DOF model, the first order canonical perturbation solution and the second order per-turbation solution are shown in Figs. 15 and 16, respectively. Note that the second order canonical perturba-tion solutions and the first order canonical perturbation solution produce results that are in good agreementwith the exact solution. Once again, there is good agreement between the approximate solutions and the exactsolution. Note also how the frequency of the solutions depends on the initial conditions. The only change inthis case from the previous case is in the initial conditions.

2 DOF1 DOF

0 2 4 6 8 10 12 14 16 18 20–0.15

–0.1

–0.05

0

0.05

0.1


Time

c(t)




0 2 4 6 8 10 12 14 16 18 20–0.15

–0.1

–0.05

0

0.05

0.1

0.15

Time

Tau



Consider a similar example with k1 = 2, k3 = 4, xc = 1, xs = 3, c0 = 1, s0 = 1. There is no internal reso-nance predicted in this case. However, a quick calculation will reveal that potential energy function has a max-imum and that the initial energy in the system is greater than this maximum. Thus, we should expect instabilityin the system.

The graphs of s and c for the exact solutions of the 1 DOF and 2 DOF model are shown in Figs. 17 and 18,respectively. Also, the graphs of s and c for the exact solution of the 2 DOF model, the first and second ordercanonical perturbation solutions are shown in Figs. 19 and 20, respectively.

A quick glance at the figures reveals that the 2 DOF model does indeed display the expected instability.Indeed, the solutions for s and c both blow up in finite time for the 2 DOF model. However, it should be noted



0 2 4 6 8 10 12 14 16 18 20–0.15

–0.1

–0.05

0

0.05

0.1

0.15

Time

c


2 DOF1 DOF

0 2 4 6 8 10 12 14 16 18 20–0.5

–0.4

–0.3

–0.2

–0.1

0

0.1

0.2

0.3

0.4


Time

Tau



that neither the 1 DOF model, nor any of the approximate solutions display the same behaviour. The 1 DOFmodel solutions and the approximate solutions remain bounded.

On the other hand, if the same case is considered with only a change in initial conditions to c0 = 0.1,s0 = 0.1, the only thing this changes is the initial energy in the system so that the initial energy is now less thenthe maximum value of the potential energy. With the initial energy being less than the maximum value of thepotential energy, there is no instability in the system. Only periodic solutions are observed with the 1 DOF and2 DOF models producing the same solution for s. The graphs of the solutions resemble those of the previouscase and are thus not included. As for the previous case, the approximate solutions and the exact solutionsagree quite well.

0 2 4 6 8 10 12 14 16 18 20–0.5

–0.4

–0.3

–0.2

–0.1

0

0.1

0.2

0.3


Time

c(t)



0 2 4 6 8 10 12 14 16 18 200.5

0.4

0.3

0.2

0.1

0

0.1

0.2

0.3

0.4

0.5

Time

Tau



7. Summary of comparison between 1 DOF and 2 DOF models

By employing analytical and numerical tools, a better picture of the similarities and differences between the1 DOF and 2 DOF models begins to emerge. First, the 1 DOF model predicts a periodic solution for the timedependence of the transverse vibrations, s. Although the frequency of vibration depends on various parame-ters of the solution, the amplitude of the solution is constant. For this model, the time dependence of the in-plane vibrations, c is linearly related to s since the in-plane inertia is dropped. Although this will not accuratelyrepresent the dynamics of the in-plane vibrations, it simplifies the overall problem enough to permit a closed-form solution to the transverse problem. If k1 = 0, for the 1 DOF model we get the seemingly nonsensicalresult that c = 0. This is actually a reasonable statement which follows from the underlying assumption.

0 2 4 6 8 10 12 14 16 18 20–0.5

–0.4

–0.3

–0.2

–0.1

0

0.1

0.2

0.3

0.4c for Exact, First and Second Order Canonical Perturbation

Time

c


2 DOF1 DOF

0 5 10 15 20 25 30–50

0

50

100

150

200

250Tau for 1 and 2 DOF models

Time

Tau



Namely, by ignoring the in-plane inertia, we are essentially stating that the in-plane problem is a static prob-lem. Further assuming that k1 = 0 states that there is no coupling between the in-plane and transverse vibra-tions. Hence, no coupling to a static problem leads to the prediction that the static problem remainsunchanged. In other words, c = 0.

It becomes quickly apparent that solutions of the 2 DOF model that includes the effect of in-plane inertiawill exhibit a broader range of possible behaviour than for the 1 DOF model. This model is best viewed as anonlinear duffing oscillator for the transverse vibrations and a linear harmonic oscillator for the in-planevibrations. These are nonlinearly coupled. For k1 = 0, there is no coupling between the two oscillators. Thus,the two oscillators can oscillate independently. The first difference between the 1 DOF and 2 DOF oscillator

2 DOF1 DOF

0 5 10 15 20 25 30–7

–6

–5

–4

–3

–2

–1

0

1x 10

4 c for 1 and 2 DOF models

Time

c(t)




0 5 10 15 20 25 30–50

0

50

100

150

200

250

Time

Tau



becomes apparent: depending on the modes in question, it is possible that these modes are coupled. It is alsopossible that the vibrations of the chosen modes are not coupled. For the case where the modes are not cou-pled, the results for the vibrations of the transverse oscillator are very similar to the results of the 1 DOF trans-verse oscillator. In both cases, they are independent duffing oscillators.

For the case of coupled oscillators, the possibility of internal resonance between the oscillators arises. Thisrepresents a marked difference from the 1 DOF model where there is no possibility of internal resonance sincethere is only one degree of freedom. In the case of internal resonance, there is a sort of �energy sharing� be-tween the two oscillators. The amplitudes of vibrations of both are not constant but rather vary periodically.This is reminiscent of beats between two linear harmonic oscillators. For the linear harmonic oscillator, the


0 5 10 15 20 25 30–7

–6

–5

–4

–3

–2

–1

0

1x 10

4

c for Exact, First and Second Order Canonical Perturbation

Time

c



beat frequency is easily found, whereas for the nonlinearly coupled oscillators it is not. For the non-resonantcase numerical simulations showed that the predictions of the 1 DOF and 2 DOF models may vary in theirpredictions of the amplitude of the solution.

For the 2 DOF model, energy analysis and numerical simulations predict the possibility of instability in thesystem. It can also be seen that sometimes the 1 DOF model predicts a periodic solution while the 2 DOFmodel predicts instability, a marked difference.

8. Conclusion

In conclusion, by including the effect of in-plane inertia the dynamics of the system includes phenomena notpreviously predicted. New phenomena such as internal resonance or non-uniformity of the amplitude of thesolution arise. Depending on initial conditions, system instability is a possibility of the 2 DOF model that isnot predicted by the 1 DOF model. Such phenomena are not predicted by the simpler model where in-planeinertia is neglected.

Further, the efficacity of a canonical perturbation approach in deriving approximate solutions to the non-linear problems was investigated. It was generally observed that a second order canonical perturbation solu-tion can effectively capture the nonlinear system dynamics in the cases where the system is expected to remainstable.

References

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