ERT 452 - VIBRATION Munira Bt Mohamed Nazari School of Bioprocess, UniMAP 2012 1ERT 452.

ERT 452 - VIBRATION

Munira Bt Mohamed NazariSchool of Bioprocess, UniMAP 2012

1ERT 452

LECTURE 1:FUNDAMENTALS OF VIBRATION

2ERT 452

Topic Outline Introduction Basic Concepts of Vibration Classification of Vibration Vibration Analysis Procedure Spring Elements Mass or Inertia Elements Damping Elements Harmonic Motion Harmonic Analysis

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INTRODUCTION

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Brief History of VibrationBrief History of Vibration

Phenomenon of Vibration

Musical instrument (string)

Musical instrument (string)

Observed that if 2 string of different length are subject to the same tension, the shorter

one emits a higher note.

Observed that if 2 string of different length are subject to the same tension, the shorter

one emits a higher note.

Use monochordUse monochord

Frequency of vibration

Frequency of vibration

Pythagoras (582 - 507 BC)

Brief History of VibrationBrief History of Vibration(1564 – 1642) Galileo Galilei - Founder of modern experimental science.- Started experimenting on simple pendulum.- Study the behavior of a simple pendulum (observe pendulum movement of a lamp).- Describing resonance, frequency, length, tension and density of a vibrating stretched string.

(1642 – 1727) Sir Isaac Newton - Derive the equation of motion of a vibrating body.

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Brief History of VibrationBrief History of Vibration

(1902 – 1909) Frahm- Investigate the importance of torsional vibration study in the design of the propeller shafts of steamships.- Propose the dynamic vibration absorber, which involves the addition of a secondary spring-mass system to eliminate the vibration of main system.

7ERT 452

Importance of the Study of VibrationImportance of the Study of Vibration

Vibrations can lead to excessive deflections and failure on the machines and structures.

To reduce vibration through proper design of machines and their mountings.

To utilize profitably in several consumer and industrial applications.

To improve the efficiency of certain machining, casting, forging & welding processes.

To stimulate earthquakes for geological research and conduct studies in design of nuclear reactors.

8ERT 452

WHY???


Vibrational problems of prime movers due to inherent unbalance in the engine.

Wheel of some locomotive rise more than centimeter off the track – high speeds – due to imbalance.

Turbines – vibration cause spectacular mechanical failure.

9ERT 452

EXAMPLE OF PROBLEMS


Cause rapid wear.Create excessive noise. Leads to poor surface finish (eg: in metal cutting

process, vibration cause chatter).Resonance – natural frequency of vibration of a

machine/structure coincide with the frequency of the external excitation (eg: Tacoma Narrow Bridge – 1948)

10ERT 452

DISADVANTAGES

Applications

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BASIC CONCEPTS OF VIBRATION

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Basic Concepts of VibrationBasic Concepts of Vibration

Vibration = any motion that repeats itself after an interval of time. Vibratory System consists of:

1) spring or elasticity2) mass or inertia3) damper

Involves transfer of potential energy to kinetic energy and vice versa.

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Basic Concepts of VibrationBasic Concepts of Vibration Degree of Freedom (d.o.f.) = min. no. of independent coordinates required to determine completely the positions of all parts of a system

at any instant of time Examples of single degree-of-freedom systems:

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Examples of single degree-of-freedom systems:

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Examples of Two degree-of-freedom systems:

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Examples of Three degree of freedom systems:

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Basic Concepts of VibrationBasic Concepts of VibrationExample of Infinite number of degrees of freedom system:

Infinite number of degrees of freedom system are termed continuous or distributed systems.Finite number of degrees of freedom are termed discrete or lumped parameter systems.More accurate results obtained by increasing number of degrees of freedom.

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CLASSIFICATION OF VIBRATION

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20

Classification of VibrationClassification of Vibration

Free Vibration:A system is left to vibrate on its own after an initial disturbance and no external force acts on the system. E.g. simple pendulum

Forced Vibration:A system that is subjected to a repeating external force. E.g. oscillation arises from diesel enginesResonance occurs when the frequency of the external force coincides with one of the natural frequencies of the system

21

Undamped Vibration:When no energy is lost or dissipated in friction or other resistance during oscillations

Damped Vibration: When any energy is lost or dissipated in friction or other resistance during oscillations

Linear Vibration:When all basic components of a vibratory system, i.e. the spring, the mass and the damper behave linearly


22

Nonlinear Vibration:If any of the components behave nonlinearly

Deterministic Vibration:If the value or magnitude of the excitation (force or motion) acting on a vibratory system is known at any given time

Nondeterministic or random Vibration: When the value of the excitation at a given time cannot be predicted


23

Examples of deterministic and random excitation:


VIBRATION ANALYSIS PROCEDURE

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25

Vibration Analysis ProcedureVibration Analysis Procedure

Step 1: Mathematical Modeling

Step 2: Derivation of Governing Equations

Step 3: Solution of the Governing Equations

Step 4: Interpretation of the Results

Derive system/component

Free body diagram(FBD)Find the response

(solve problem method)Response (result):

Displacement, velocities & acceleration

26

Example of the modeling of a forging hammer:

Vibration Analysis ProcedureVibration Analysis Procedure

27

Example 1.1Example 1.1Mathematical Model of a MotorcycleMathematical Model of a Motorcycle

Figure below shows a motorcycle with a rider. Develop a sequence of three mathematical models of the system for investigating vibration in the vertical direction. Consider the elasticity of the tires, elasticity and damping of the struts (in the vertical direction), masses of the wheels, and elasticity, damping, and mass of the rider.

28

Example 1.1 Example 1.1 SolutionSolution

We start with the simplest model and refine it gradually. When the equivalent values of the mass, stiffness, and damping of the system are used, we obtain a single-degree of freedom model of the motorcycle with a rider as indicated in Fig.(b). In this model, the equivalent stiffness (keq) includes the stiffness of the tires, struts, and rider. The equivalent damping constant (ceq) includes the damping of the struts and the rider. The equivalent mass includes the mass of the wheels, vehicle body and the rider.

29


30

This model can be refined by representing the masses of wheels, elasticity of tires, and elasticity and damping of the struts separately, as shown in Fig.(c). In this model, the mass of the vehicle body (mv) and the mass of the rider (mr) are shown as a single mass, mv + mr. When the elasticity (as spring constant kr) and damping (as damping constant cr) of the rider are considered, the refined model shown in Fig.(d) can be obtained.


31


32

Note that the models shown in Figs.(b) to (d) are not unique. For example, by combining the spring constants of both tires, the masses of both wheels, and the spring and damping constants of both struts as single quantities, the model shown in Fig.(e) can be obtained instead of Fig.(c).


SPRING ELEMENTS

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Linear spring is a type of mechanical link that is generally assumed to have negligible mass and damping. Spring force is given by:

34

1.1kxF

F = spring force,k = spring stiffness or spring constant,

andx = deformation (displacement of one end

with respect to the other)

Spring ElementsSpring Elements

35

Work done (U) in deforming a spring or the strain (potential) energy is given by:

When an incremental force ΔF is added to F:

2.121 2kxU

3.1...)(!2

1

)()(

)(

22

2

*

*

*

*

xdx

Fd

xdxdF

xF

xxFFF

x

x


36

37

Static deflection of a beam at the free end is given by:

Spring Constant is given by:

6.13

3

EIWl

st

W = mg is the weight of the mass m,E = Young’s Modulus, andI = moment of inertia of cross-section of beam

7.13

3l

EIWk

st


38

Combination of Springs:

1) Springs in parallel – if we have n spring constants k1, k2, …, kn in parallel, then the equivalent spring constant keq is:

11.121 ... neq kkkk


39

Combination of Springs:

2) Springs in series – if we have n spring constants k1, k2, …, kn in series, then the equivalent spring constant keq is:

17.11

...111

21 neqkkkk


40

Example 1.3Example 1.3Torsional Spring Constant of a Propeller ShaftTorsional Spring Constant of a Propeller Shaft

Determine the torsional spring constant of the speed propeller shaft shown in Fig. 1.25.

41


We need to consider the segments 12 and 23 of the shaft as springs in combination. From Fig. 1.25, the torque induced at any cross section of the shaft (such as AA or BB) can be seen to be equal to the torque applied at the propeller, T. Hence, the elasticities (springs) corresponding to the two segments 12 and 23 are to be considered as series springs. The spring constants of segments 12 and 23 of the shaft (kt12 and kt23) are given by:

42


m/rad-N109012.8

)3(32

)15.025.0()1080(

32

)(

6

449

23

4

23

4

23

23

23

23

l

dDG

l

GJk

t

m/rad-N105255.25

)2(32

)2.03.0()1080(

32

)(

6

449

12

4

12

4

12

12

12

12

l

dDG

l

GJk

t

43


Since the springs are in series, Eq. (1.16) gives

m/rad-N105997.6

)109012.8105255.25(

)109012.8)(105255.25(

6

66

66

2312

2312

tt

tt

t kk

kkk

eq

44

Example 1.5Example 1.5Equivalent Equivalent kk of a Crane of a Crane

The boom AB of crane is a uniform steel bar of length 10 m and x-section area of 2,500 mm2. A weight W is suspended while the crane is stationary. Steel cable CDEBF has x-sectional area of 100 mm2. Neglect effect of cable CDEB, find equivalent spring constant of system in the vertical direction.

A vertical displacement x of pt B will cause the spring k2 (boom) to deform by x2 = x cos 45º and the spring k1 (cable) to deform by an amount x1 = x cos (90º – θ). Length of cable FB, l1 is as shown.

45

m3055.12

426.151135cos)10)(3(2103

1

2221

l

l


46


The angle θ satisfies the relation:

The total potential energy (U):

0736.35,8184.0cos

10cos)3)((23 21

221

ll

1.E)]90cos([21

)45cos(21 2

22

1 xkxkU

N/m106822.10355.12

)10207)(10100( 696

1

111

lEA

k

N/m101750.510

)10207)(102500( 796

2

222

lEA

k

47


Potential Energy of the equivalent spring is:

By setting U = Ueq, hence:

2.E21 2xkU eqeq

N/m104304.26 6eq k

MASS OR INERTIA ELEMENTS

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49

Using mathematical model to represent the actual vibrating system

E.g. In figure below, the mass and damping of the beam can be disregarded; the system can thus be modeled as a spring-mass system as shown.

Mass or Inertia ElementsMass or Inertia Elements

50

Combination of MassesE.g. Assume that the mass of

the frame is negligible compared to the masses of the floors. The masses of various floor levels represent the mass elements, and the elasticities of the vertical members denote the spring elements.


51

Case 1: Translational Masses Connected by a Rigid Bar

Velocities of masses can be expressed as:

18.111

331

1

22 x

ll

xxll

x


52

and,

By equating the kinetic energy of the system: 19.11xxeq

20.121

21

21

21 2

eqeq233

222

211 xmxmxmxm

21.13

2

1

32

2

1

21eq m

ll

mll

mm


Case 2: Translational and Rotational Masses Coupled Together

53

meq = single equivalent translational mass = translational velocity = rotational velocityJ0 = mass moment of inertiaJeq = single equivalent rotational mass

x


54

Case 2: Translational and Rotational Masses Coupled Together1. Equivalent translational mass:Kinetic energy of the two masses is given by:

Kinetic energy of the equivalent mass is given by:

22.121

21 2

02 JxmT

23.121 2

eqeqeq xmT


55

Case 2: Translational and Rotational Masses Coupled Together

2. Equivalent rotational mass:Here, and , equating Teq and T gives eq Rx

25.1

21

21

21

20eq

20

22eq

mRJJor

JRmJ

24.120

eqR

Jmm

Since and , equating Teq & T gives R

x xx eq


56

Example 1.7Example 1.7Cam-Follower MechanismCam-Follower Mechanism

A cam-follower mechanism is used to convert the rotary motion of a shaft into the oscillating or reciprocating motion of a valve.

The follower system consists of a pushrod of mass mp, a rocker arm of mass mr, and mass moment of inertia Jr about its C.G., a valve of mass mv, and a valve spring of negligible mass.

Find the equivalent mass (meq) of this cam-follower system by assuming the location of meq as (i) pt A and (ii) pt C.

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Example 1.7Example 1.7Cam-Follower MechanismCam-Follower Mechanism

58

The kinetic energy of the system (T) is:

If meq denotes equivalent mass placed at pt A, with the kinetic energy equivalent mass system Teq is:

1.E21

21

21

21 2222

rrrrvvpp xmJxmxmT

xx eq

2.E21 2

eqeqeq xmT

Example 1.7Solution

59

By equating T and Teq, and note that

Similarly, if equivalent mass is located at point C, hence,

11

3

1

2 and,,,lx

llx

xllx

xxx rrvp

3.E21

23

21

22

21

eql

lm

l

lm

l

Jmm rv

rp

,eq vxx

4.E21

21 2

eq2eqeqeq vxmxmT

Example 1.7Solution

60

Equating (E.4) and (E.1) gives

5.E

2

21

23

2

2

122

eq

l

lm

ll

ml

Jmm rp

rv

Example 1.7Solution

DAMPING ELEMENTS

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62

Damping ElementsDamping Elements

Viscous Damping: Damping force is proportional to the velocity of the vibrating body in a fluid medium such as air, water, gas, and oil.

Coulomb or Dry Friction Damping: Damping force is constant in magnitude but opposite in direction to that of the motion of the vibrating body between dry surfaces.

Material or Solid or Hysteretic Damping:Energy is absorbed or dissipated by material during deformation due to friction between internal planes.

63

Hysteresis loop for elastic materials:


Construction of Viscous Dampers

ERT 452 64


µ

Plate be moved with a velocity v in its own plane

Fixed planeVelocity of intermediate fluid layersare assumed to vary linearly

65

Shear Stress ( ) developed in the fluid layer at a distance y from the fixed plate is:

where du/dy = v/h is the velocity gradient.• Shear or Resisting Force (F) developed at the bottom

surface of the moving plate is:

where A is the surface area of the moving plate and is the damping constant.

26.1dydu

27.1cvhAv

AF


hA

c

66

If a damper is nonlinear, a linearization process is used about the operating velocity (v*) and the equivalent damping constant is:

29.1*vdv

dFc


67

Example 1.10Example 1.10Equivalent Spring and Damping Constants of a Machine Equivalent Spring and Damping Constants of a Machine

Tool SupportTool Support

A precision milling machine is supported on four shock mounts, as shown in Fig. 1.37(a). The elasticity and damping of each shock mount can be modeled as a spring and a viscous damper, as shown in Fig. 1.37(b). Find the equivalent spring constant, keq, and the equivalent damping constant, ceq, of the machine tool support in terms of the spring constants (ki) and damping constants (ci) of the mounts.

68

Example 1.10Example 1.10Equivalent Spring and Damping Constants of a Machine Equivalent Spring and Damping Constants of a Machine

Tool SupportTool Support

69

The free-body diagrams of the four springs and four dampers are shown in Fig. 1.37(c). Assuming that the center of mass, G, is located symmetrically with respect to the four springs and dampers, we notice that all the springs will be subjected to the same displacement, , and all the dampers will be subject to the same relative velocity , where and denote the displacement and velocity, respectively, of the center of mass, G. Hence the forces acting on the springs (Fsi) and the dampers (Fdi) can be expressed as

Example Example 1.10 Solution1.10 Solution

xxxx

70


71

Let the total forces acting on all the springs and all the dampers be Fs and Fd, respectively (see Fig. 1.37d). The force equilibrium equations can thus be expressed as


E.1)(4,3,2,1;

4,3,2,1;

ixcF

ixkF

idi

isi

E.2)(4321

4321

ddddd

sssss

FFFFF

FFFFF

72


E.3)(xcF

xkF

eqd

eqs

where Fs + Fd = W, with W denoting the total vertical force (including the inertia force) acting on the milling machine. From Fig. 1.37(d), we have

Equation (E.2) along with Eqs. (E.1) and (E.3), yield

E.4)(4

4

4321

4321

cccccc

kkkkkk

eq

eq

Parallel

73

where ki = k and ci = c for i = 1, 2, 3, 4.

Note: If the center of mass, G, is not located symmetrically with respect to the four springs and dampers, the ith spring experiences a displacement of and the ith damper experiences a velocity of where and can be related to the displacement and velocity of the center of mass of the milling machine, G. In such a case, Eqs. (E.1) and (E.4) need to be modified suitably.


xx

ix

ix

ix

ix

HARMONIC MOTION

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Periodic Motion: motion repeated after equal intervals of timeHarmonic Motion: simplest type of periodic

motionDisplacement (x):

Velocity:

Acceleration:

75

30.1sinsin tAAx

31.1cos tAdtdx

32.1sin 222

2xtA

dt

xd

Harmonic MotionHarmonic Motion

(On horizontal axis)

76

• Scotch yoke mechanism:The similarity between cyclic (harmonic) and sinusoidal motion.


77

Complex number representation of harmonic motion:

where i = √(–1) and a and b denote the real and imaginary x and y components of X, respectively.

35.1ibaX


78

Also, Eqn. (1.36) can be expressed as

Thus,

43.1sincos

36.1sincos

iAeiAX

iAAX

47.12,1;)( 22 jbaA jjj

48.12,1;tan 1

j

a

b

j

jj


79

Operations on Harmonic Functions:Rotating Vector, 51.1tiAeX

54.1cos]Re[ntDisplaceme tAAe ti

55.190cos

sin]Re[Velocity

tA

tAAei ti

56.1180cos

cos

]Re[onAccelerati

2

2

2

tA

tA

Ae ti


Where Re denotes the real part.

80

• Displacement, velocity, and accelerations as rotating vectors

• Vectorial addition of harmonic functions


81

Example 1.11Example 1.11Addition of Harmonic MotionsAddition of Harmonic Motions

E.1)()()()cos()(21

txtxtAtx

Find the sum of the two harmonic motions

).2cos(15)( and cos10)(21

ttxttx

Solution:

Method 1: By using trigonometric relations: Since the circular frequency is the same for both x1(t) and x2(t), we express the sum as

82

Example 1.11 SolutionExample 1.11 Solution

E.2)()2sinsin2cos(cos15cos10

)2cos(15cos10sinsincoscos

ttt

ttttA

E.3)()2sin15(sin

)2cos1510(cos)sin(sin)cos(cos

t

tAtAt

That is,

That is,

By equating the corresponding coefficients of cosωt and sinωt on both sides, we obtain

E.4)(1477.14

)2sin15(2cos1510

2sin15sin

2cos1510cos

22

A

A

A

83


E.5)(5963.74

2cos15102sin15

tan 1

and

Method 2: By using vectors: For an arbitrary value of ωt, the harmonic motions x1(t) and x2(t) can be denoted graphically as shown in Fig. 1.43. By adding them vectorially, the resultant vector x(t) can be found to be

E.6)()5963.74cos(1477.14)( ttx

84


E.7)(15ReRe)(

10ReRe)()2()2(

22

11

titi

titi

eeAtx

eeAtx

Method 3: By using complex number representation: the two harmonic motions can be denoted in terms of complex numbers:

The sum of x1(t) and x2(t) can be expressed as

E.8)(Re)( )( tiAetx

where A and α can be determined using Eqs. (1.47) and (1.48) as A = 14.1477 and α = 74.5963º

85

Definitions of Terminology:Amplitude (A) is the maximum displacement of a

vibrating body from its equilibrium position

Period of oscillation (T) is time taken to complete one cycle of motion

Frequency of oscillation (f) is the no. of cycles per unit time

59.12T

60.12

1

Tf


86

Definitions of Terminology:Natural frequency is the frequency which a system oscillates

without external forces

Phase angle () is the angular difference between two synchronous harmonic motions

62.1sin

61.1sin

22

11

tAx

tAx


87

Definitions of Terminology:Beats are formed when two harmonic motions, with

frequencies close to one another, are added


88

Definitions of Terminology:Decibel is originally defined as a ratio of electric

powers. It is now often used as a notation of various quantities such as displacement, velocity, acceleration, pressure, and power

1.69)(log20dB

1.68)(log10dB

0

0

XX

PP

where P0 is some reference value of power and X0 is specified reference voltage.


HARMONIC ANALYSIS

ERT 452 89

90

• A periodic function:

Harmonic AnalysisHarmonic Analysis

91


• Fourier Series Expansion: If x(t) is a periodic function with period , its Fourier Series representation is given by

1

0

21

21

0

)70.1()sincos(2

...2sinsin

...2coscos2

)(

n nntnbtna

a

tbtb

tataa

tx

92

•Gibbs Phenomenon: An anomalous behavior observed from a periodic function that is being represented by Fourier series.

As n increases, the approximation can be seen to improve everywhere except in the vicinity of the discontinuity, P. The error in amplitude remains at approximately 9 percent, even when . k


93

•Complex Fourier Series:The Fourier series can also be represented in terms of complex numbers.

)79.1(sincos

)78.1(sincos

tite

titeti

ti

Also,

)81.1(2

sin

)80.1(2

cos

i

eet

eet

titi

titi

and


94

•Frequency Spectrum:

Harmonics plotted as vertical lines on a diagram of amplitude (an and bn or dn and Φn) versus frequency (nω).


95

• Representation of a function in time and frequency domain:


96

• Even and odd functions:

1

0 )88.1(cos2

)(

)87.1()()(

n ntna

atx

txtx

Even function & its Fourier series expansion

Odd function & its Fourier series expansion

1)90.1(sin)(

)89.1()()(

n ntnbtx

txtx


97

• Half-Range Expansions:The function is extended to include the interval as shown in the figure. The Fourier series expansions of x1(t) and x2(t) are known as half-range expansions.

0 to


98

• Numerical Computation of Coefficients.If x(t) is not in a simple form, experimental determination of the amplitude of vibration and numerical integration procedure like the trapezoidal or Simpson’s rule is used to determine the coefficients an and bn. )99.1(

2sin

2

)98.1(2

cos2

)97.1(2

1

1

10

iN

i in

iN

i in

N

i i

tnx

Nb

tnx

Na

xN

a


99

Example 1.12Example 1.12Fourier Series ExpansionFourier Series Expansion

Determine the Fourier series expansion of the motion of the valve in the cam-follower system shown in the Figure.

100

Example 1.12Example 1.12SolutionSolution

E.1)()()(

)()(tan

1

2

21

tyl

ltx

ltx

lty

E.2)(0;)(

tt

Yty

If y(t) denotes the vertical motion of the pushrod, the motion of the valve, x(t), can be determined from the relation:

or

where

and the period is given by . 2

101

1

2

l

YlA

E.3)(0;)(

tt

Atx

By defining

x(t) can be expressed as

Equation (E.3) is shown in the Figure. To compute the Fourier coefficients an and bn, we use Eqs. (1.71) to (1.73):

E.4)(2

)(/2

0

2

/2

0

/2

00A

tAdt

tAdttxa


102

E.5)(.. 2, ,1,0

sincos

2cos

coscos)(

/2

0

22

/2

0

/2

0

/2

0

n

n

tnt

n

tnAdttnt

A

dttnt

Adttntxan

E.6)(.. 2, ,1,

cossin

2sin

sinsin)(

/2

0

22

/2

0

/2

0

/2

0

nn

A

n

tnt

n

tnAdttnt

A

dttnt

Adttntxbn


103

Therefore the Fourier series expansion of x(t) is

E.7)(...3sin3

12sin

2

1sin

2

...2sin2

2sin2

)(

tttA

tA

tAA

tx

The first three terms of the series are shown plotted in the figure. It can be seen that the approximation reaches the sawtooth shape even with a small number of terms.


ERT 452 104

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ERT 452 - VIBRATION Munira Bt Mohamed Nazari School of Bioprocess, UniMAP 2012 1ERT 452.

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