Mee5206 Chapter 2

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MEE 5206:

VIBRATION

Chapter 2Free Vibration of Single-Degree-of-Freedom Systems

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Mechanical Vibrations

Fourth EditionWilliam T. Thomson

REFERENCES

Third Edition

Daniel J. INMAN

Fifth EditionSingiresu S. Rao

TEXT BOOK

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Chapter 2Free Vibration of Single-Degree-of-FreedomSystems

Chapter Outline

Introduction

Free Vibration of an Undamped Translational System

Free Vibration of an Undamped Torsional System

(READING ASSIGNMENT)

Free Vibration with Viscous Damping

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4

Learning Objectives

After completing this chapter, the reader should be able

to do the following:

Derive the equation of motion of a single-degree-of-freedom system

using a suitable technique such as Newtons second law of motion,

D Alemberts principle, the principle of virtual displacements, andthe principle of conservation of energy.

Solve a spring-mass-damper system for different types of free-

vibration response depending on the amount of damping.

Compute the natural frequency, damped frequency, logarithmic

decrement, and time constant.

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Introduction

FreeVibration occurs when a system oscillates only under an initial

disturbance with no external forces acting after the initialdisturbance

Undampedvibrations result when amplitude of motion remains

constant with time (e.g. in a vacuum)

Dampedvibrations occur when the amplitude of free vibrationdiminishes gradually overtime, due to resistance offered by the

surrounding medium (e.g. air)

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Introduction

Several mechanical and structural systems can be idealized as

single degree of freedom systems, for example, the mass and

stiffness of a system

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2011 Mechanical Vibrations Fifth Edition in SI Units7

If mass mis displaced a distance when acted upon by a

resultant force in the same direction,

If mass mis constant, this equation reduces to

where is the acceleration of the mass

)(tx

)(tF

dt

txdmdt

dtF

)()(

(2.1))(

)( 2

2

xmdt

txdmtF

2

2 )(

dt

txdx

Equation of Motion Using Newtons Second Law of Motion:

Newtons Second law is the first basis for examining themotion of the system

Free Vibration of an Undamped TranslationalSystem

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For a rigid body undergoing rotational motion, Newtons Law gives

where is the resultant moment acting on the body and and

are the resulting angular displacement and angular

acceleration, respectively.

For undamped single degree of freedom system, the application of Eq.

(2.1) to mass myields the equation of motion:

)2.2()( JtM

M

22 /)( dttd

)3.2(0or)( kxxmxmkxtF


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)3.2(0or)( kxxmxmkxtF


Spring-Mass System in Horizontal Position:

Consider the undamped single degree of freedom systemshown in Fig. below.

The application of Eq. (2.1) to mass myields the equation ofmotion:

Figure 2.1

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Principle of Conservation of Energy

A system is said to be conservative if no energy is lost due to frictionor energy-dissipating nonelastic members.

If no work is done on the conservative system by external forces,

the total energy of the system remains constant. Thus the principle

of conservation of energy can be expressed as:

)6.2(0)(orconstant UTdt

dUT

Equation of Motion Using Energy Methods:

The differential equation of motion (Eq. (2.3)) can also bederived by using the principle of conservation of energy

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The kinetic and potential energies are given by:

Substitution of Eqs. (2.7) & (2.8) into Eq. (2.6) yields the desired

equation

)8.2(2

1

)7.2(21

2

2

kxU

xmT

)3.2(0 kxxm

Equation of Motion Using Energy Methods:


For the free vibration of undamped systme, the energy ispartly kinetic and partly potential

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The equations of motion, Eqs. (2.1) & (2.2) can be rewritten as

The application of DAlembertsprinciple to the system shown in Fig.

2.1(c) yields the equation of motion:

(2.4b)0)(

)2.4a(0)(

JtM

xmtF

)3.2(0or0 kxxmxmkx

Equation of Motion Using DAlemberts Principle :


DAlemberts principle states that if the resultant force actingon a body along with the inertia force is zero, then the bodywill be in static equilibrium.

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Consider the configuration of the spring-mass system shown in thefigure.

Equation of Motion of a Spring-Mass System in Vertical

Position:

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For static equilibrium,

The application of Newtons second law of motion to mass m gives

and since , we obtain

)9.2(stkmgW

Wxkxm st )(

)10.2(0 kxxm

Wk st

where w = weight of mass m,

= static deflection

g = acceleration due to gravityst

Equation of Motion of a Spring-Mass System in Vertical Position:

Notice that Eqs. (2.3) and (2.10) are identical. This indicates that when amass moves in a vertical direction, we can ignore its weight, provided wemeasurexfrom its static equilibrium position.

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Solutions for equation of motion (Eqs. 2.3 and 2.10)

Equation of Motion of a Spring-Mass System in Vertical Position:

The solution of Eqs. (2.3) can be found by assuming

(2.11))( stCetx

Where Cand sare constants to be determined. Substitution of Eq.

(2.11) into Eq. (2.3) gives(2.12)0zerobecannotsince,0)( 22 kmsCkmsC

nim

ks

2/1

And hence, 2/11i

2/1

mkn

where

Therefore,

Note: Eq. (2.12 is called characteristic equationcorresponding to eq. (2.3).

The two values of s is the roots of characteristicequation. It is also called eigenvalues or characteristicvalues of the problem

n, is the natural frequency of the system

The natural period & frequency

m

kT 2

k

mfn

2

1,

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2011 Mechanical Vibrations Fifth Edition in SI Units


From Eq (2.16), we have

Hence,

Solution of Eq. (2.3) is subjected to the initial conditions of Eq. (2.17)which is given by

)17.2()0(

)0(

02

01

xAtx

xAtx

n

)18.2(sincos)( 00 txtxtx nn

n

nxAxA /and 0201

Hence, Eq. (2.3) can be expressed as

By using the identities

)15.2()( 21 titinn eCeCtx

where C1and C2are constants

)16.2(sincos)( 21 tAtAtx nn whereA1andA2are new constants

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Another representation of eq. (2.16) by introducing the notions:

Eqs.(2.15), (2.16) & (2.18) are harmonic functions of time. Eq.

(2.16) can also be expressed as:

where and are new constants, amplitude and phase angle

respectively:

)21.2()cos()( tAtx n

0

)22.2(

2/12

02

0

2

2

2

1

n

xxAAA

nx

x

0

01tan

)sin(

)cos(

2

1

AA

AA

Harmonic Motion

A

anglePhaseAmplitudeA

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The nature of harmonic oscillation can be represented graphically asshown in the figure.

Similarly by using the relations: )cos(

)sin(

002

001

AA

AA

Harmonic Motion

)23.2()sin()( 00 tAtx n

)24.2(

2/12

02

0

2

2

2

1

n

xxAAA

0

01

0 tanx

x n

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Note the following aspects of spring-mass systems:

1. When the spring-mass system is in a vertical position

Circular natural frequency:

Spring constant, k:

Hence,

)26.2(2/1

m

kn

)27.2(stst

mgWk

)28.2(

2/1

st

n

g

Harmonic Motion

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1. When the spring-mass system is in a vertical position (Cont)

Natural frequency in cycles per second:

Natural period:

)29.2(2

12/1

st

n

gf

)30.2(21

2/1

gf

st

n

n

Harmonic Motion

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2. Displacement , Velocity and the acceleration of themass mat time tcan be obtained as:

)(tx )(tx

)31.2()sin()cos()()(

)2

cos()sin()()(

22

2

2

tAtAtdtxdtx

tAtAtdt

dxtx

nnn

nnnn

n

Harmonic Motion

)21.2(Eq.)cos()( tAtx n

)(tx

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3. If initial displacement is zero,

If initial velocity is zero,

0x

)32.2(sin2

cos)( 00 tx

txtx n

n

n

n

0x

)33.2(cos)( 0 txtx n

Harmonic Motion

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4. The response of a single degree of freedom system can berepresented by:

By squaring and adding Eqs. (2.34) & (2.35)

)35.2()sin(

)34.2()sin()(

A

y

A

xt

tAtx

n

n

nn

)36.2(1

1)(sin)(cos

2

2

2

2

22

A

y

A

x

tt nn

nxy /where

Harmonic Motion

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4. Phase plane representation of an undamped system

Harmonic Motion

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From the theory of torsion of circular shafts, we have the relation:

)37.2(0

l

GIMt

where

Mt= torque that produces the twist ,

G = shear modulus,

l= is the length of shaft,

I0= polar moment of inertia of cross section of shaft

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Polar Moment of Inertia:

Torsional Spring Constant:

)38.2(32

4

0

dI

)39.2(32

4

0

l

Gd

l

GIMk tt


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Equation of Motion:

Applying Newtons Second Law of Motion,

The natural circular frequency is

The period and frequency of vibration in cycles per second are:

)40.2(00 tkJ

)41.2(

2/1

0

J

ktn

)43.2(2

1,)42.2(2

2/1

0

2/1

0

J

kf

k

J tn

t

n

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Note the following aspects of this system:1) If the cross section of the shaft supporting the disc is not circular,

an appropriate torsional spring constant is to be used.

2) The polar mass moment of inertia of a disc is given by

3) An important application of a torsional pendulum is in a mechanicalclock

g

WDDhJ

832

44

0

where = mass density

h = thickness

D = diameter

W = weight of the disc

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A machine oscillates in simple harmonic motion and appears to be

well modeled by an undamped single-degree-of-freedom oscillation.Its acceleration is measured to have an amplitude of 10,000 mm/s2

at 8 Hz. What is the machine's maximum displacement?

Example

The equations of motion for position and acceleration are [From Eq. (2.21) and

(2.31)]:

Solution:

)21.2()cos()( tAtx n

)31.2()cos()cos()()(22

2

2

tAtAtdt

xdtx nnn n

The amplitude of acceleration is:22 m/s000,10An

The natural frequency of the system is: rad/s16Hz)8(22 fn

Hence, the machine displacement is:

mm96.3rad/s16

m/s000,10m/s000,102

2

2

2

n

A

F Vib ti ith Vi D i

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Equation of Motion:

where c = damping

From the figure, Newtons law yieldsthat the equation of motion is

)58.2(xcF

)59.2(0

kxxcxm

kxxcxm

F Vib ti ith Vi D i

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We assume a solution in the form

The characteristicequation is

The roots and solutions are

)60.2()( stCetx

)61.2(02 kcsms

)62.2(

222

422

2,1

m

k

m

c

m

c

m

mkccs

)63.2()(and)( 21 2211tsts

eCtxeCtx

where C and s are undetermined constants

F Vib ti ith Vi D i

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Thus the general solution is:

where C1and C2are arbitrary constants to be determined from the initial conditions of the system

)64.2(

)(

22

21

22

2

22

1

21

tm

k

m

c

m

ctm

k

m

c

m

c

tsts

eCeC

eCeCtx


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The critical damping ccis defined as the value of the dampingconstant cfor which the radical in Eq.(2.62) becomes zero:

The damping ratio is defined as:

)65.2(22202

2

ncc mkm

m

kmc

m

k

m

c

)66.2(/ ccc

Critical Damping Constant and Damping Ratio:


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Thus the general solution for Eq.(2.64) is

Assuming that 0, consider the following 3 cases:

Case 1: Underdamped system

For this condition, is negative and the roots are

)/2oror1( mkmc/cc c

)69.2()(1

2

1

1

22tt nn

eCeCtx

n

n

is

is

2

2

2

1

1

1


12


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The solution can be written in different forms:

)/2oror1( mkmc/cc c

)70.2(1cos1sin

1sin1cos

)(

0

2

0

2

2

2

2

1

1

2

1

1

1

2

1

1

22

22

teX

tXe

tCtCe

eCeCe

eCeCtx

n

t

nt

nn

t

titit

titi

n

n

n

nnn

nn

where (C1,C2), (X,),

and (X0, 0)are arbitrary constants



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For the initial conditions at t = 0,

and hence the solution becomes

)/2oror1( mkmc/cc c

)71.2(1

and2

00201

n

nxxCxC

)72.2(1sin1

1cos)(2

2

002

0

t

xx

txetx nn

nn

tn



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Eq.(2.72) describes a damped harmonic motion. Its amplitudedecreases exponentially with time, as shown in the figure below.

The frequency of damped vibrationis:

)/2oror1( mkmc/cc c

)76.2(1 2 nd


n

d

2 Damped period


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Case 2: Critically damped system

In this case, the two roots are:

Due to repeated roots, the solution of Eq.(2.59) is given by

)77.2(2

21 nc

mcss

)/2oror1( mkmc/ccc

)78.2()()(21

tnetCCtx



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Application of initial conditions gives:

Thus the solution becomes:

)/2oror1( mkmc/ccc

)79.2(and 00201 xxCxC n

)80.2()( 000t

nnetxxxtx



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It can be seen that the motion represented by Eq.(2.80) is aperiodic (i.e., non-periodic).

Since , the motion will eventually diminish to zero,as indicated in the figure below.

)/2oror1( mkmc/ccc

te tn as0

Comparison of motions withdifferent types of damping




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Case 3: Overdamped system

The roots are real and distinct and are given by:

In this case, the solution Eq.(2.69) is given by:

0101

2

2

2

1

n

n

s

s

)/2oror1( mkmc/cc c

)81.2()(1

2

1

1

22 tt nneCeCtx



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Case 3: Overdamped system

For the initial conditions at t = 0,

)/2oror1( mkmc/cc c

)82.2(12

1

12

1

2

0

2

0

2

2

0

2

01

n

n

n

n

xx

C

xxC



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For a damped system, m, c, and k are known to be m = 1 kg, c = 2 kg/s,

k = 10 N/m. Calculate the value of and n. Is the system overdamped,

underdamped, or critically damped?

Example

p g


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Using Eq.(2.70),

The logarithmic decrement can be obtained from Eq.(2.84):

)84.2(

)83.2(

)cos(

)cos(

1

1

2

1

020

010

2

1

dn

dn

n

n

n

ee

eteX

teX

x

x

t

td

t

d

t

)85.2(2

2

1

2

1

2ln

222

1

m

c

x

x

dn

ndn

It is used to find the damping ratio of an underdamped system in the time domain

It is the natural log of the ratio of the amplitudes of any two successive peaks

Logarithmic Decrement ( ):

p g


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p g

For small damping,

Hence,

or

Thus

)86.2(1if2

)92.2(ln1

1

1

mx

x

m

)87.2(

2 22

)88.2(2

where m is an integer

Logarithmic Decrement ( ):


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Energy dissipated in Viscous Damping:

In a viscously damped system, the rate of change of energy withtime is given by:

The energy dissipated in a complete cycle is:

)93.2(velocityforce

2

2

dt

dx

ccvFvdt

dW

)94.2()(cos 220

22

2

)/2(0

XctdtcXdtdtdxcW ddddt

d


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Consider the system shown in the figure.

The total force resisting the motion is

If we assume simple harmonic motion is

Eq.(2.95) becomes

)95.2(xckxcvkxF

)96.2(sin)( tXtx d

)97.2(cossin tXctkXF ddd


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The energy dissipated in a complete cycle will be

)98.2()(cos

)(cossin

2/2

0

22

/2

0

2

/2

0

XctdtXc

tdttkX

FvdtW

dt

ddd

t dddd

t

d

d

d


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Computing the fraction of the total energy of the vibrating systemthat is dissipated in each cycle of motion,

The loss coefficient is defined as

)99.2(constant422

2

2

2

1 22

2

m

c

Xm

Xc

W

W

d

d

d

)100.2(2

)2/(tcoefficienloss

W

W

W

W

where Wis either the max potential energy or the max kinetic energy


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Torsional systems with Viscous Damping:

Consider a single degree of freedom torsional system with a viscousdamper as shown in figure.


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The viscous damping torque is given by

The equation of motion can be derived as:

)101.2(tcT

)102.2(00 tt kcJ

where J0= mass moment of inertia of disckt= spring constant of system

= angular displacement of disc


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In the underdamped case, the frequency of damped vibration isgiven by

where

and

)103.2(1 2 nd

)104.2(0J

ktn

)105.2(22 00 Jk

cJc

cc

t

t

n

t

tc

t

ctc= critical torsional damping constant


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Example 2.11

Shock Absorber for a Motorcycle

An underdamped shock absorber is to be designed for a motorcycle ofmass 200kg (shown in Fig.(a)). When the shock absorber is subjectedto an initial vertical velocity due to a road bump, the resulting

displacement-time curve is to be as indicated in Fig.(b). Find thenecessary stiffness and damping constants of the shock absorber if thedamped period of vibration is to be 2 s and the amplitude x1 is to bereduced to one-fourth in one half cycle (i.e., x1.5 = x1/4). Also findthe minimum initial velocity that leads to a maximum displacement of250 mm.


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Example 2.11



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Example 2.11


Solution

Since , the logarithmic decrementbecomes

16/4/,4/ 15.1215.1 xxxxx

(E.1)1

27726.216lnln

22

1

x

x


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Example 2.11


Solution

From which can be found as 0.4037 and the damped period ofvibration is given by 2 s. Hence,

rad/s4338.3

)4037.0(12

2

1

222

2

2

n

nd

d


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Example 2.11


Solution

The critical damping constant can be obtained as

Thus the damping constant is

The stiffness is

s/m-N54.373.1)4338.3)(200(22 nc mc

s/m-N4981.554)54.1373)(4037.0( ccc

N/m2652.2358)4338.3)(200( 22 nmk


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Example 2.11


Solution

The displacement of the mass will attain its max value at time t1is

sec3678.0)9149.0(sin

9149.0)4037.0(1sinsin

1sin

1

1

2

11

2

1

t

tt

t

d

d


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Example 2.11


Solution

The envelope passing through the max points is

Since x = 250mm,

The velocity of mass can be obtained by

(E.2)1 2 tnXex

m4550.0)4037.0(125.0 )3678.0)(4338.3)(4037.0(2 XXe

(E.3))cossin()(

sin)(

ttXetx

tXetx

dddn

t

dt

n

n


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Example 2.11


Solution

When t = 0,

m/s4294.1

)4037.0(1)4338.3)(4550.0(

1)0(2

2

0

nd XXxtx

Graphical Representation of Characteristic Rootsand Corresponding Solutions

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and Corresponding Solutions

Roots of the Characteristic Equation

The free vibration of a single-degree-of-freedom spring-mass-

viscous-damper system is governed by Eq. (2.59):

whose characteristic equation can be expressed as (Eq. (2.61)):

2.1060 kxxcxm

2.10802

0

22

2

nn wsws

kcsms


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Roots of the Characteristic Equation

The roots of Eq. (2.107) or (2.108) are given by (see Eqs. (2.62)

and (2.68)):

2.1101,

2

4

,

2

21

2

21

nn iwwss

m

mkcc

ss



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Graphical Representation of Roots and Corresponding

Solutions

The response of the system is given by

Following observations can be made by examining Eqs. (2.110) and

(2.111):

1. The roots lying farther to the left in the s-plane indicate that the

corresponding responses decay faster than those associated

with roots closer to the imaginary axis.

2. If the roots have positive real values of sthat is, the roots liein the right half of the s-planethe corresponding response

grows exponentially and hence will be unstable.

2.11121 21tsts eCeCtx



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Solutions

3. If the roots lie on the imaginary axis (with zero real value), the

corresponding response will be naturally stable.

4. If the roots have a zero imaginary part, the corresponding response

will not oscillate.

5. The response of the system will exhibit an oscillatory behavior only

when the roots have nonzero imaginary parts.

6. The farther the roots lie to the left of the s-plane, the faster the

corresponding response decreases.

7. The larger the imaginary part of the roots, the higher the frequency ofoscillation of the corresponding response of the system.

p g


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Solutions

p g

Parameter Variations and Root LocusRepresentations

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p

Interpretations of in thes-plane

The angle made by the line OA with the imaginary axis is given by

The radial lines pass through the origin correspond to different

damping ratios

The time constant of the system is defined as

and,, dn ww

2.113sin

sin

1

n

n

w

w

nw

1


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Interpretations of in thes-plane and,, dn ww


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Interpretations of in thes-plane and,, dn ww


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Interpretations of in thes-plane

Different lines parallel to the imaginary axis denote reciprocals of

different time constants

and,, dn ww


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Root Locus and Parameter Variations

A plot or graph that shows how changes in one of the parameters

of the system will modify the roots of the characteristic equation of

the system is known as the root locus plot.

Variation of the damping ratio:

We vary the damping constant from zero to infinity and study the

migration of the characteristic roots in the s-plane.

From Eq. (2.109) when c = 0,

2.1152

42,1 niw

m

k

m

mks


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Noting that the real and imaginary parts of the roots in Eq. (2.109)

can be expressed as

For , we have

2.116124

and2

22

dnn wwm

cmkwm

c

10 2.117222 nd ww


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The radius vector will make an angle with the positive imaginary

axis with

The two roots trace loci or paths in the form of circular arcs as thedamping ratio is increased from zero to unity as shown

21with

cos,sin

n

n

nn

d

ww

www


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Example 2.13

Study of Roots with Variation of c

Plot the root locus diagram of the system governed by the equation by

varying the value of c >0

0273 2 cs


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Example 2.13

Study of Roots with Variation of cSolution

The roots of equation are given by

We start with a value of C = 0 and the roots is as shown in the figure.

Eq. (E.2) gives the roots as indicated in the Table.

E.26

3242

2,1 ccs


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Example 2.13

Study of Roots with Variation of cSolution


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Variation of the spring constant:

Since the spring constant does not appear explicitly in Eq. (2.108),

we consider a specific form of the characteristic equation (2.107)

as:

The roots of Eq. (2.121) are given by

2.1210162 kss

2.12264824256162,1 kks


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Variation of the mass:

To find the migration of the roots with a variation of the mass m,

we consider a specific form of the characteristic equation, Eq.

(2.107), as

whose roots are given by

2.123020142 sms

2.1242

80196142,1

ms


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Some values of m and the corresponding roots given by Eq.

(2.124) are shown in Table.


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Mee5206 Chapter 2

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Transcript of Mee5206 Chapter 2