vibraciones de dos grados de libertad

8/9/2019 vibraciones de dos grados de libertad

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3he vibrations o syst ms having more than one

degree o freedom

Many real systems can e represented y a single degree of freedom model. However,most actual systems have several bodies and several restraints and therefore severaldegrees of freedom. The number of degrees of freedom that a system possesses isequal to the number of independent coordina es necessary to describe the motion ofthe system. Since no body is completely rigid, and no spring is without mass, everyreal system has more than one degree of freedom, and sometimes it is not sufficientlyrealistic to approximate a system by a single degree of freedom model. Thus, it isnecessary to study the vibration o systems with more than one degree of freedom.

Each flexibly connected body in a multi-degree of freedom system can moveindependently of the other bodies, and only under certain conditions will all bodiesundergo an harmonic motion at the same frequency. Since all bodies move with thesame frequency, they all attain their amplitudes at the same time even if they do notall move in the same direction. When such motion occurs the frequency is called a

n tur lfrequency of the system, and the motion is a principal mode of vibration: thenumber o natural frequencies and principal modes that a system possesses is equalto the number of degrees of freedom of that system. The deployment of the systemat its lowest or first natural frequency is called its first mode, at the next highest orsecond natural frequency it is called the second mode and so on.

A two degree of freedom system will e considered initially. This is because theaddition of more degrees of freedom increases the labour of the sohion procedurebut does not introduce any new analytical principies.

Initially, we will obtain the equations ofmotion for a two degree offreedom model,

and from these find the natural frequencies and corresponding mode shapes.Sorne examples of two degree of freedom models of vibrating systems are shownin Figs 3.l aHh).


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The vibration of systems with two egrees of freedom 9

MFig. 3.l a) Linear undamped system, horizontal motion. Coordinates x 1 and x 2 .

m,

Fig. 3.l b) Linear undamped system, vertical motion. Coordinates y and y2

Fig. 3.1 c) Torsional undamped system. Coordinates e, and 8 2 .


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9 he vibratons of syst ms having more than one degree of freedom Ch . 3

Fig. 3.l d) Coupled pendula . Coordinates 1 and 2

,--....em /

G

X

Figt. 3 . l e) System with combined translation and rotation . Coordinates x and }

Fig. 3.1(f) Shear frame. Coordinates x 1 and x 2


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Sec 3 1] The vibration o systems with two degrees of freedom 9

Fig. 3.2 g) Two degree of freedom model, rotation plus translation. Coordinates y nd 8

antilever spring

Fig. 3.1 h) Two degree of freedom model, trans lation vibration. Coordinates x 1 nd x 2


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9 he vibrations o systems having more than one degree of freedom [Ch. 3

3 1 THE VIBR TION O SYSTEMS WITH TWO DEGREES O FREEDOM

3 1 1 Free vibration of an undamped systemOf the examples of two degree of freedom models shown in Figs. 3.1 a)- h),consider the system shown in Fig. 3.l a). I f x

1> x 2 the FBDs are as shown in

Figure 3.2.

k 1x 1 Q k x 1 2 )

k x 1 2 D k 2 x 2

a) b)

Fig. 3.2. a) Applied forces; b) effective forces.

The equations of motion are therefore,

for body 1, 3.1)

and

for body 2. 3.2)

The same equations are obtained if x 1 < x 2 is assumed because the direction ofthe central spring force is then reversed.

Equations 3.1) and 3.2) can be solved for the natural frequencies and correspondingmode shapes by assuming a solution of the form

x1

= A1sin wt

+t/1 and x

2= A

2sin wt

+t/J .

This assumes that x 1 and x 2 oscillate with the same frequency w and are either inphase or n out of phase. This is a sufficient condition to make w a natural frequency.

Substituting these solutions into the equations of motion gives

m1 A 1 w

2 sin wt + t/J = 1 A 1 sin wt + t/J)- k A 1 A 2 )sin wt + t/1and

m2 A 2 w

2 sin wt + t/1 = k A 1 - A 2 )sin wt + t/1)- k 2 A 2 sin wt + t/1 .

Since these solutions are true for all values of t,A 1 k + k 1 - m 1w 2 ) + A 2 - k = O

and

3.3)

3.4)


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Sec 3.1] The vibration of systems with two degrees. of freedom 93

A 1 and 2 can be eliminated by writing

\

k k 1 m 1w2

- k \ = O- k k k 2 - m 2 w

23.5)

This is the characteristic or frequency equation. Alternatively, we may write

A 1/A 2 = k/ k k 1 - m1 w2

) from 3.3)

and

3.6)

A/A 2 = k 2 k - m2 w2 )/k from 3.4)

Thus

k/ k k 1 - m1w2

) = k 2 k - m2 w2 )/k

and

k+ k 1 - m 1w2 ) k 2 k - m2 w

2 k 2 =O 3.7)

This result is the frequency equation and could also be obtained by multiplying outthe above determinant, equation 3.5).

The solutions to equation 3.7) give the natural frequencies of free vibration for

the system in Fig. 3.1 a). The corresponding mode shapes are found by substitutingthese frequencies, in turn, into either of equations 3.6).

Consider the case when k 1 = k 2 =k , and m 1 = m2 =m. The frequency equation is2 k - mw 2 k2 =O; that is,

m2 w 4 - 4mkw 2 3k 2 = O or mw 2 - k) mw 2 - 3k) = O.

Therefore, either mw 2 - k = O or mw 2 - 3k = O.Thus

w 1 = .j k/m)rad/s and w 2 = .j 3k/m) rad/s.

f

w = .j k/m)rad/s,

and if

w = .j 3k/m)rad/s, from 3.6)

This gives the mode shapes corresponding to the frequencies w1

and w2

Thus, thefirst mode of free vibration occurs a t a frequency 1/2n).j k/m)Hz and A 1 /A 2 )

1 = 1,that is, the bodies move in phase with each other and with the same amplitude as ifconnected by a rigid link, Fig. 3.3. The second mode of free vibration occurs at afrequency 1/2n).j 3k/m)Hz and A/A 2 )

11 = - 1 that is, the bodies move exactly outof phase with each other, but with the same amplitude, see Fig. 3.3.


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94 The vibrations o systems having more than one degree o freedom [Ch. 3

l v ~x

1st mode

1/ .= Hz1 27T

.

t t

Fig . 3.3 Natural frequencies and mode shapes for two degree of freedom translation vibration system .Bodies of equal mass and springs of equal stiffne ss.

3 1 2 Free motion

The two modes of vibration can be written

and

where the ratio A 1 /A 2 is specified for each mode.Since each solution satisfies the equation of motion the general solution is

where A 1 A 2 1 2 are found from the initial conditions.

A2 y -A2

A1 y +A1


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9 he vibrations of syst ms having more than one degree of freedom Ch. 3

3 1 3 Coordinate coupliogIn sorne systems the motion is such that the coordinates are coupled in the equationsof motion. Consider the system shown in Fig. 3.l e); only motion in the plane of thefigure is considered, horizontal motion being neglected because the lateral stiffnessof the springs is assumed to be negligible. The coordinates of rotation, O, andtranslation, x are coupled as shown in Fig. 3.4. G is the centre of mass of the rigidbeam of mass m and moment of inertia 1 about G.

Fig. 3.4. Two degree of freedom model, rotation plus translation.

The FBDs are shown in Fig. 3.5; since the weight of the beam is supported by thesprings, both the initial spring forces and the beam weight may be omitted.

la l lb

m i

Fig. 3.5. (a) Applied forces. (b) Effective force and moment.

For small amplitudes of oscillation (so that sin ~ O the equations of motion are

x = k 1 x L 10 k 2 x L 2 0

and

that is,

F en x

M en z


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Sec. 3.1) The vibration of systems with two degrees of freedom 9

andX 2 2l u - k 1L 1 - k 2 L 2 )x + k 1 L 1 + k 2 L 2 )0 =O.

l t will be noticed that these equations can be uncoupled by making k 1L 1 = k 2 L 2 ;if this is arranged, translation x motion) and rotation 8 motion) can take placeindependently. Otherwise translation and rotation occur simultaneously.

Assuming x 1 = A 1 sin wt + t/1 and } = A 2 sin wt + t/1 , substituting into theequations of motion gives

-m w 2 A 1 + k 1 + k 2)A 1 - k 1L 1 - k 2 L 2 A 2 =O,and

- l w 2 A 2 - k 1 L 1 - k2 L 2 )A 1 + k 1 Li + k 2 L; A 2 =O.That is,

A 1 k 1 + k 2 - mw 2 ) + A 2 k 1L 1 - k 2 L 2 =O,and

A 1 k 1 L 1 - k 2 L 2 + A 2 k 1 Li + k 2 L ; - 2 ) =O.Hence the frequency equation is

\

k 1 k 2 - m w2

k 1L 1 2 L 2 ) 12 2 2 =O.

k 14 - k 2 L 2 ) k 1L 1 + k 2 L 2 - IwFor each natural frequency, there is a corresponding mode shape, given by A 1 /A 2

Example 8A system is modelled as a straight link AB 1m long of mass 10 kg, supportedhorizontally by springs A and B of stiffness k 1 and k 2 respectively. The moment ofinertia of AB about its centre of mass Gis kgm 2 , and Gis located t distance afrom A and b from B

Find the relationship between k 1 , k2 , a, and b so that one mode of free vibration

shall be translational motion only and the other mode rotation only.l f a = 0.3 m and k 1 = 13 kN/m, find k 2 to give these two modes of vibration andcalculate the two natural frequencies.

The frequency equation is section 3.1.3)

l

k 1 + k 2 - mw 2 - k 1a - k 2 b) 1 = 0- k 1 a - k 2 b) k 1 a

2 + k 2 b 2 - Jw 2 For the modesto be uncoupled k 1 a = k 2 b and then

w 1 = .j k 1 + k2 )/m), bouncing mode)and

rocking mode).

Since a = 0.3 m, b = 0.7 m and k 1 = 13 kN/m,


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98 he vibrations of systems having more than one degree o freedom [Ch. 3

k 2 = 13 x 0.3)/0. 7 = 5 57 kN/m and then

w 1 = J 18570/10) = 43 1 radjs; so f 1 = 6 86 Hz

and w 2 = J 13 x 0 3 2 5.57 x OY l0 3 ) = 62.45 radjs; so f 2 = 9.94 Hz

xample 9

When transported, a space vehicle is supported in a horizontal position by twosprings, as shown. The vehicle can be considered to be a rigid body of mass m andradius of gyration h about an axis normal to the plane of the figure through the masscentre G. The rear support has a stiffness k 1 and is a ta distance from G while thefront support has a stiffness k 2 and is at a distance b from G. The only motionspossible for the vehicle are vertical translation and rotation in the vertical plane.

Write the equations of small amplitude motion of the vehicle and obtain thefrequency equation in terms of the given parameters.

Given that k 1a = k 2 b determine the natural frequencies of the free vibrations ofthe vehicle and sketch the corresponding modes of vibration. Also state or sketchthe modes of vibration if k 1 k 2 b

8

The FBDs are as below:

G

n 2

rm y

:

t 1


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Sec. 3.11 The vibration of syst ms with two degrees of freedom

The equations of motion are

k 1 y afJ) k i y - b }) = -my,and

k 1 y afJ a k i y - bfJ)b = - m h2

{J.Assuming

y Y sin vt

these give

and }=e sin vt,

Y k 1 k 2 mv 2 ) E> k 1 k 2 b) =O

and

Y k1 a - k 2 b) E> k 1 a 2 k 2 b 2 - mh 2 v2 ) = O.The frequency equation is, therefore,

k 1 k 2 - mv2 ) k 1a

2 k 2 b2

- mh 2 v2 k 1 k 2 b f =O.

f k 1a = k 2 b, motion is uncoupled so

v1 = k 1 : k) rad/s and J k a 2 k b 2 )Vz = 1 mhz 2 rad/s.v1 is the frequency of a bouncing or translation mode no rotation):

v2 is the frequency of a rotation mode no bounce)

F en y

M en z


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100 The vibrations of systems having more than one degree of freedom [Ch 3

f k a - k 2 b the modes are coupled:

1 1 1Example 20In a study of earthquakes a building is idealized as a rigid body of mass M supportedon two springs one giving translational stiffness k and the other rotational stiffnesskT, as shown.

Given that l is the mass moment of inertia of the building about its mass centreG write down the equations of motion using coordinates x for the translation fromthe equilibrium position and J for the rotation of the building .

Hence determine the frequency equation of the motion.


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Sec 3.1] he vibration of systems with two degrees of freedom 1 1

The FBDs are as follows.

Assume small e (earthquakes), hencem(.i + h ) = -kx ,

and

The equations of motion are therefore

mhi + x + kx = Oand

f

and

e= A sin wt and X= Az sin wt,

-mhw 2 A 1 - mw2 A 2 + kA 2 =O

+mhw 2 A2 + mh 2 + / 0 )w 2 A 1 + mgh- kT A 1 =O.The frequency equation is

1

-mhw 2 m w 2 l Omh 2 + / 0 w 2 + mgh- kT mhw 2

That is, mhw 2 + k - mw 2 )[ mh 2 + 10 w 2 + mgh - kT ] = O or,m 0 w

4- w 2 [mkh 2 + 10 m2 gh + m k T ] - mghk + kkT =O.

F en x

M en z


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102 The vibrations of systems having more than one degree of freedom [Ch. 3

3.1.4 orced vibratiooHarmonic excitation of vibration in a system may be generated in a number of ways,for example by unbalanced rotating or reciprocating machinery, or it may arise fromperiodic excitation containing a troublesome harmonic component.

A two degree of freedom model of a dynamic system excited by an harmonic forceF sin vt is shown in Fig. 3 6 Damping is assumed to be negligible. The force has aconstant amplitude F and a frequency v/2n Hz.

+vef-x,

Fig. 3 6 Two degree of freedom model with forced excitation.

The equations of motion are

m 1x1 = k 1 x 1 - k x 1 - x 2 sin vt,

andm2 x2 = k x 1 - x 2 k 2 x 2

Since there is zero damping, the motions are either in phase or n out of phase withthe driving force, so that the following solutions may be assumed:

x 1 = A 1 sin vt and x 2 = 2 sin vt.

Substituting these solutions into the equations of motion gives

A 1 k 1k

1 v2

A 2 - k ) = F,and

Thus

and

where

Regla de Kramer

solucin 1 g.d.l.


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4 The vibrations of syst ms having more than one degree of freedom [Ch. 3

The frequency equation is

k 1 k 2 - m2 v2 ) k 1 - m 1 v

2 k =O.

The critica speeds are those which correspond to the natural frequencies and henceexcite resonances. The frequency equation simplifies to

m 1m2 v4

- m 1k 1 m 1k 2 m2 k 1)v2 k k2 =O.

Hence substituting the given data,

500 X 400 X V4 - (500 X 60 500 X 170 400 X 60) 103 v2 60 X 170 X 10 6 = 0.

That is 0.2v 4 - 139v 2 1O200 = O which can be sol ved by the formula. Thusv = 16.3 radjs or 20.78 radjs, and f = 2.59 Hz or 3.3 Hz.

Now if the trailer is drawn at v kmjh, or vj3.6 mjs the frequency is v/ 3.6 x 5)Hz.

Therefore the critica speeds areV = 18 X 2.59 = 46.6 kmjh,

and

v2 = 18 x 3.3 = 59.4 km/h.

Towing the trailer at either of these speeds will excite a resonance in the system.From the equations of motion,

{ k k 2 }A 2 2 2 A3k 1 k 2 m 2 v ) k 1 m 1 v k 1

{10200 }

= 0.2v 4 - 139v 2 10200 A 3

At 50 kmjh, v = 17.49 radjs.Thus A 1 = 0.749 A 3 . Since A 3 = 0.1 m, the amplitude of the trailer vibration is

0.075 m. This motion is n out of phase with the road undulations.

3 1 5 The undamped dynamic vibration absorberf a single degree of freedom system or mode of a multi-degree of freedom system is

excited into resonance, large amplitudes of vibration result with accompanying highdynamic stresses and noise and fatigue problems. In most mechanical systems this isnot acceptable.

I f neither the excitation frequency nor the natural frequency can conveniently bealtered, this resonance condition can often be successfully controlled by adding afurther single degree of freedom system. Consider the model of the system shown in

Fig. 3.7, whereK

andM

are the effective stiffness and mass of the primary systemwhen vibrating in the troublesome mode.The absorber is represented by the system with parameters k and m. From section3.1.4 it can be seen that the equations of motion are

MX = X k X x) Fsin vt for the primary system

Regla de Kramer


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Sec. 3 .1] The vibration of systems with two degrees of freedom 1 05

Primary system

Fig. 3 7 System with undamped vibration absorber .

and

m i k X x),

for the vibration absorber. Substituting

XX 0 sin

vtand

x= x 0 sin

vt

gves

and

Thus

F k mv 2 o =

A

and

kXo=A

where

and A = O is the frequency equation .t can e seen that not only does the system now possess two natural frequencies,

0 1 and 0 2 instead of one, but by arranging for k - mv2 = O 0 can e made zero.

Regla de Kramer

solucin 1 g.d.l.


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Thus if .J k/m) = .J K/M), the response of the primary system at its originalresonance frequency can be made zero. This is the usual tuning arrangement for anundamped absorber because the resonance problem in the primary system is onlysevere when v .J K/ M) radjs. This is shown in Fig. 3 8

o

\ Prim ry s y s mL response

\

/ \\\

\\

wW Ml= Oc ml

Primary syS1etTI responsewhen tuned absorberattached

--V

Fig. 3 8 Amplitude-frequency response for system with and without tuned absorber.

When 0 =O, 0 = -F /k so that the force in the absorber spring, kx 0 is F ;thus the absorber applies a force to the primary system which is equal and oppositeto the exciting force. Hence the body in the primary system has a net zero excitingforce acting on it and therefore zero vibration amplitude.

fan absorber is correctly tuned w 2 = K/M = k/m, and ifthe mass ratio Jl =m/ M,the frequency equation = O is

This is a quadratic equation in v/w) 2 Hence


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Sec. 3.1] The vibration o syst ms with two degrees o freedom 1 7

and the natural frequencies 1 and 2 are found to be

~ 2= [ 1 J ~ 2 J / 2or a small ~ 1 and 2 are very close to each other and near to w increasinggives better separation between 1 and 2 as shown in Fig. 3 9

n n 1

o 0.2 0.5

Fig. 3 9 Effect of absorber mass ratio on natural frequencies.

This effect is of great importance in those systems where the excitation frequencymay vary; if is small resonances at 1 or 2 may be excited. l t should be notedthat since

;y= 1 ) - J ~ 2and

Then


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That is

Al so

These relationships are very useful when designing absorbers. f the proximity ofQ 1 and Q 2 to is likely to be a hazard damping can be added in parallel with the

absorber spring to limit the response at these frequencies. Unfortunately if dampingis added the response at frequency will no longer be zero. A design criterion thathas to be carefully considered is the possible fatigue and failure ofthe absorber spring:this could ha ve severe consequences. In view of this sorne damped absorber systemsdispense with the absorber spring and sacrifice sorne of the absorber effectiveness.This is particularly important in torsional systems where the device is known as aLanchester damper.

ExampleThe figure represents a pump of mass m 1 which rests on springs having a stiffnessk 1 so that only vertical motion can occur. Given that the damping is negligible andthe mass m 2 is ignored derive an expression for the frequency of the harmonicdisturbing force at which the pump will execute vertical oscillations of verylarge theoretically infinite-amplitude.

Given that an undamped dynamic absorber of mass m2 is then connected to thepump by a spring of stiffness k 2 as shown pro ve that the amplitude of the oscillations

of the pumpis

reduced to zero when

where v is the natural frequency of the free vibrations of the pump in the absence ofthe dynamic vibration absorber.

The pump has a mass of 130 kg and ro tates at a constant speed of 2400 revolutionsper minute but due to a rotating un balance very large amplitudes of pump vibrationon the spring supports result. An undamped vibration absorber is to be fitted so thatthe nearest natural frequency of the system is at least 20 per cent removed from therunning speed of 2400 revolutions per minute. Find the smallest absorber massnecessary and the corresponding spring stiffness.


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Sec 3 1] he vibration of systems with two degrees of freedom 1 09

The pu p can be modelled as below:

sin l t

The equation of motion is

so that if

x 1 X 1 sin vt


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11 O The vibrations of systems having more than one degree of freedom [Ch. 3

When

v = J ~

X OO.

That is resonance occurs when = = J k1/m

1). With a vibration absorber added

the system is

Fsinllf

The FBDs are therefore if x 2 > x 1 is assumed

ok 2 x 2 x1)

Fsinvt k 2 x 2 x1)

The equations of motion are thusm

2i

2 = --k 2 x 2 - x 1 ),


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Sec. 3.1 1 The vibration of systems with two degrees of freedom 111

or

and

or

m 1x1 + 1 + k 2 x 1 - k 2 x 2 = F sin vt.Assuming x 1 = X 1 sin vt and x 2 = X 2 sin vt these equations give

and

That is

Thus if

1 =0.

Now the frequency equation is

[ k 1 + k 2 m 1 v2 ] k 2 - m 2 v2 k = Of we put

mz kzf l = = -

m k 1

this becomes

or

so that

and

~ Y =2; J f.1.2

: 4f.l.).The limiting condition for the smallest absorber mass is v 1 /il = 0.8 because thenv 2 /il = 1.25, which is acceptable. Thus


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o.64_P Y= P2:pand

= 0.2.Hence

m = 0.2 x 130 = 26 kg,

and

k = 80n:)2 m = 1642 kN/m.

Example 23A system has a violent resonance at 79 Hz. As a trial remedy a vibration absorberis attached which results in a resonance frequency of 65Hz. How many such absorbersare required if no resonance is to occur between 60 and 120 Hz?

Sin ce

and

nln2 =w

in the case of one absorber, with w = 79 Hz and 0 1 = 65 Hz,

79 20 2 =6 5 =96Hz.

l so

~ G:r= 2 p.so = 0.154.

In the case of n absorbers, if

=60Hz,79 2

2 = 6o = 104 Hz too low).

So require 2 = 120Hz and then = 792 /120) =52 Hz. Hence

52)2

(120)279 79 = 2 p .

Thus

and0.74

n = 0.154 = 4.82.

Thus five absorbers are required.


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Sec. 3 .11 The vibration of systems with two degrees of freedom 113

Example 24A machine too of mass 3000 kg has a large resonance vibration in the vertical directionat 120Hz . To o ~ t r o lthis resonance, an undamped vibration absorber of mass 600 kgis fitted tuned to 120Hz. Find the frequency range in which the amplitude of themachine vibration is less with the absorber fitted than without.

If X 0 ) with absorber = X 0 ) without absorber,

Multiplying out and putting J = m M gives

4 v)2- 4 + p + 2 = o.Sin ce

600J = 3000 = 0"2'

( \ )2 4 + J~ = : k / (p 2 + 8p) = 1.05 0.32.Thus

and

\

- = 1.17 or 0.855,w

f = 102 Hz or 140Hz , where v = 2rrfThus the required frequency range is 102-140 Hz.

3.1.6 System with viscous damping

(phase requires - ve sign).

I f a system possesses damping of a viscous nature , the damping can be modelledsimilarly to that in the system shown in Fig. 3.10.

Fig . 3.10. Two degre e of freedom viscous damped model with forced excitation.


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Sec 3 2] The vibration of systems with more than two degrees of freedom 115

For light damping the damped frequency for each mode is approximately equalto the undamped frequency, that is b 1 = w 1 and b = w

The right-hand side of the s-plane Re s) +ve represents a root with a positiveexponent, that is, a term which grows with time, so unstable motion may exist.

The left-hand side contains roots with a negative exponent so stable motion exists.See also Fig. 2.20.

AH passive systems have negative real parts and are therefore stable but sornesystems such as rolling wheels and rockets can be become unstable, and thus it isimportant that the stability of a system is considered. This can be conveniently doneby plotting the roots of the frequency equation on the s-plane, as abo ve. This techniquehas. particular application in the study of control system stability.

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vibraciones de dos grados de libertad