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for vibrational motion, what term denotes the maximum displacement from the equilibrium position?

Engineering Vibration

Fourth Edition

DaniEl J. inman University of Michigan

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© 2014, 2008, 2001 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. The author and publisher have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these theories and programs.

Printed in the United States of America

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Library of Congress Cataloging-in-Publication Data on File

ISBN-13: 978-0-13-287169-3 ISBN-10: 0-13-287169-6

iii

Contents

Preface viii

1 IntroductIon to VIbratIon and the free resPonse 1

1.1 Introduction to Free Vibration 2

1.2 Harmonic Motion 13

1.3 Viscous Damping 21

1.4 Modeling and Energy Methods 31

1.5 Stiffness 46

1.6 Measurement 58

1.7 Design Considerations 63

1.8 Stability 68

1.9 Numerical Simulation of the Time Response 72

1.10 Coulomb Friction and the Pendulum 81

Problems 95

MATLAB Engineering Vibration Toolbox 115

Toolbox Problems 116

2 resPonse to harmonIc excItatIon 117

2.1 Harmonic Excitation of Undamped Systems 118

2.2 Harmonic Excitation of Damped Systems 130

2.3 Alternative Representations 144

2.4 Base Excitation 151

2.5 Rotating Unbalance 160

2.6 Measurement Devices 166

iv Contents

2.7 Other Forms of Damping 170

2.8 Numerical Simulation and Design 180

2.9 Nonlinear Response Properties 188

Problems 197

MATLAB Engineering Vibration Toolbox 214

Toolbox Problems 214

3 General forced resPonse 216

3.1 Impulse Response Function 217

3.2 Response to an Arbitrary Input 226

3.3 Response to an Arbitrary Periodic Input 235

3.4 Transform Methods 242

3.5 Response to Random Inputs 247

3.6 Shock Spectrum 255

3.7 Measurement via Transfer Functions 260

3.8 Stability 262

3.9 Numerical Simulation of the Response 267

3.10 Nonlinear Response Properties 279

Problems 287

MATLAB Engineering Vibration Toolbox 301

Toolbox Problems 301

4 multIPle-deGree-of-freedom systems 303

4.1 Two-Degree-of-Freedom Model (Undamped) 304

4.2 Eigenvalues and Natural Frequencies 318

4.3 Modal Analysis 332

4.4 More Than Two Degrees of Freedom 340

4.5 Systems with Viscous Damping 356

4.6 Modal Analysis of the Forced Response 362

Contents v

4.7 Lagrange’s Equations 369

4.8 Examples 377

4.9 Computational Eigenvalue Problems for Vibration 389

4.10 Numerical Simulation of the Time Response 407

Problems 415

MATLAB Engineering Vibration Toolbox 433

Toolbox Problems 433

5 desIGn for VIbratIon suPPressIon 435

5.1 Acceptable Levels of Vibration 436

5.2 Vibration Isolation 442

5.3 Vibration Absorbers 455

5.4 Damping in Vibration Absorption 463

5.5 Optimization 471

5.6 Viscoelastic Damping Treatments 479

5.7 Critical Speeds of Rotating Disks 485

Problems 491

MATLAB Engineering Vibration Toolbox 501

Toolbox Problems 501

6 dIstrIbuted-Parameter systems 502

6.1 Vibration of a String or Cable 504

6.2 Modes and Natural Frequencies 508

6.3 Vibration of Rods and Bars 519

6.4 Torsional Vibration 525

6.5 Bending Vibration of a Beam 532

6.6 Vibration of Membranes and Plates 544

6.7 Models of Damping 550

6.8 Modal Analysis of the Forced Response 556

vi Contents

Problems 566

MATLAB Engineering Vibration Toolbox 572

Toolbox Problems 572

7 VIbratIon testInG and exPerImental modal analysIs 573

7.1 Measurement Hardware 575

7.2 Digital Signal Processing 579

7.3 Random Signal Analysis in Testing 584

7.4 Modal Data Extraction 588

7.5 Modal Parameters by Circle Fitting 591

7.6 Mode Shape Measurement 596

7.7 Vibration Testing for Endurance and Diagnostics 606

7.8 Operational Deflection Shape Measurement 609

Problems 611

MATLAB Engineering Vibration Toolbox 615

Toolbox Problems 616

8 fInIte element method 617

8.1 Example: The Bar 619

8.2 Three-Element Bar 625

8.3 Beam Elements 630

8.4 Lumped-Mass Matrices 638

8.5 Trusses 641

8.6 Model Reduction 646

Problems 649

MATLAB Engineering Vibration Toolbox 656

Toolbox Problems 656

aPPendIx a comPlex numbers and functIons 657

aPPendIx b laPlace transforms 663

Contents vii

aPPendIx c matrIx basIcs 668

aPPendIx d the VIbratIon lIterature 680

aPPendIx e lIst of symbols 682

aPPendIx f codes and Web sItes 687

aPPendIx G enGIneerInG VIbratIon toolbox and Web suPPort 688

references 690

ansWers to selected Problems 692

Index 699

viii

Preface

This book is intended for use in a first course in vibrations or structural dynamics for undergraduates in mechanical, civil, and aerospace engineering or engineer- ing mechanics. The text contains the topics normally found in such courses in accredited engineering departments as set out initially by Den Hartog and refined by Thompson. In addition, topics on design, measurement, and computa- tion are addressed.

Pedagogy

Originally, a major difference between the pedagogy of this text and competing texts is the use of high level computing codes. Since then, the other authors of vibrations texts have started to embrace use of these codes. While the book is written so that the codes do not have to be used, I strongly encourage their use. These codes (Mathcad®, MATLAB®, and Mathematica®) are very easy to use, at the level of a programmable calculator, and hence do not require any prereq- uisite courses or training. Of course, it is easier if the students have used one or the other of the codes before, but it is not necessary. In fact, the MATLAB® codes can be copied directly and will run as listed. The use of these codes greatly enhances the student’s understanding of the fundamentals of vibration. Just as a picture is worth a thousand words, a numerical simulation or plot can enable a completely dynamic understanding of vibration phenomena. Computer calcula- tions and simulations are presented at the end of each of the first four chapters. After that, many of the problems assume that codes are second nature in solving vibration problems.

Another unique feature of this text is the use of “windows,” which are distributed throughout the book and provide reminders of essential informa- tion pertinent to the text material at hand. The windows are placed in the text at points where such prior information is required. The windows are also used to summarize essential information. The book attempts to make strong connections to previous course work in a typical engineering curriculum. In particular, refer- ence is made to calculus, differential equations, statics, dynamics, and strength of materials course work.

Preface ix

WHAT’S NEW IN THIS EDITION

Most of the changes made in this edition are the result of comments sent to me by students and faculty who have used the 3rd edition. These changes consist of improved clarity in explanations, the addition of some new examples that clarify concepts, and enhanced problem statements. In addition, some text material deemed outdated and not useful has been removed. The computer codes have also been updated. However, software companies update their codes much faster than the publishers can update their texts, so users should consult the web for updates in syntax, commands, etc. One consistent request from students has been not to reference data appearing previously in other examples or problems. This has been addressed by providing all of the relevant data in the problem statements. Three undergraduate engineering students (one in Engineering Mechanics, one in Biological Systems Engineering, and one in Mechanical Engineering) who had the prerequisite courses, but had not yet had courses in vibra- tions, read the manuscript for clarity. Their suggestions prompted us to make the fol- lowing changes in order to improve readability from the student’s perspective:

Improved clarity in explanations added in 47 different passages in the text. In addition, two new windows have been added.

Twelve new examples that clarify concepts and enhanced problem statements have been added, and ten examples have been modified to improve clarity.

Text material deemed outdated and not useful has been removed. Two sections have been dropped and two sections have been completely rewritten.

All computer codes have been updated to agree with the latest syntax changes made in MATLAB, Mathematica, and Mathcad.

Fifty-four new problems have been added and 94 problems have been modi- fied for clarity and numerical changes.

Eight new figures have been added and three previous figures have been modified. Four new equations have been added.

Chapter 1: Changes include new examples, equations, and problems. New textual explanations have been added and/or modified to improve clarity based on student sug- gestions. Modifications have been made to problems to make the problem statement clear by not referring to data from previous problems or examples. All of the codes have been updated to current syntax, and older, obsolete commands have been replaced.

Chapter 2: New examples and figures have been added, while previous exam- ples and figures have been modified for clarity. New textual explanations have also been added and/or modified. New problems have been added and older problems modified to make the problem statement clear by not referring to data from previ- ous problems or examples. All of the codes have been updated to current syntax, and older, obsolete commands have been replaced.

x Preface

Chapter 3: New examples and equations have been added, as well as new problems. In particular, the explanation of impulse has been expanded. In addition, previous problems have been rewritten for clarity and precision. All examples and problems that referred to prior information in the text have been modified to pres- ent a more self-contained statement. All of the codes have been updated to current syntax, and older, obsolete commands have been replaced.

Chapter 4: Along with the addition of an entirely new example, many of the examples have been changed and modified for clarity and to include improved information. A new window has been added to clarify matrix information. A fig- ure has been removed and a new figure added. New problems have been added and older problems have been modified with the goal of making all problems and examples more self-contained. All of the codes have been updated to current syntax, and older, obsolete commands have been replaced. Several new plots intermixed in the codes have been redone to reflect issues with Mathematica and MATLAB’s automated time step which proves to be inaccurate when using singu- larity functions. Several explanations have been modified according to students’ suggestions.

Chapter 5: Section 5.1 has been changed, the figure replaced, and the example changed for clarity. The problems are largely the same but many have been changed or modified with different details and to make the problems more self-contained. Section 5.8 (Active Vibration Suppression) and Section 5.9 (Practical Isolation Design) have been removed, along with the associated problems, to make room for added material in the earlier chapters without lengthening the book. According to user surveys, these sections are not usually covered.

Chapter 6: Section 6.8 has been rewritten for clarity and a window has been added to summarize modal analysis of the forced response. New problems have been added and many older problems restated for clarity. Further details have been added to several examples. A number of small additions have been made to the to the text for clarity.

Chapters 7 and 8: These chapters were not changed, except to make minor corrections and additions as suggested by users.

Units

This book uses SI units. The 1st edition used a mixture of US Customary and SI, but at the insistence of the editor all units were changed to SI. I have stayed with SI in this edition because of the increasing international arena that our engineering graduates compete in. The engineering community is now completely global. For instance, GE Corporate Research has more engineers in its research center in India than it does in the US. Engineering in the US is in danger of becoming the ‘gar- ment’ workers of the next decade if we do not recognize the global work place. Our engineers need to work in SI to be competitive in this increasingly international work place.

Preface xi

Instructor Support

This text comes with a bit of support. In particular, MS PowerPoint presentations are available for each chapter along with some instructive movies. The solutions manual is available in both MS Word and PDF format (sorry, instructors only). Sample tests are available. The MS Word solutions manual can be cut and pasted into presentation slides, tests, or other class enhancements. These resources can be found at www.pearsonhighered.com and will be updated often. Please also email me at daninman@umich.edu with corrections, typos, questions, and suggestions. The book is reprinted often, and at each reprint I have the option to fix typos, so please report any you find to me, as others as well as I will appreciate it.

Student Support

The best place to get help in studying this material is from your instructor, as there is nothing more educational than a verbal exchange. However, the book was writ- ten as much as possible from a student’s perspective. Many students critiqued the original manuscript, and many of the changes in text have been the result of sug- gestions from students trying to learn from the material, so please feel free to email me (daninman@umich.edu) should you have questions about explanations. Also I would appreciate knowing about any corrections or typos and, in particular, if you find an explanation hard to follow. My goal in writing this was to provide a useful resource for students learning vibration for the first time.

AckNOWlEDgEmENTS

The cover photo of the unmanned air vehicle is provided courtesy of General Atomics Aeronautical Systems, Inc., all rights reserved. Each chapter starts with two photos of different systems that vibrate to remind the reader that the material in this text has broad application across numerous sectors of human activity. These photographs were taken by friends, students, colleagues, relatives, and some by me. I am greatly appreciative of Robert Hargreaves (guitar), P. Timothy Wade (wind mill, Presidential helicopter), General Atomics (Predator), Roy Trifilio (bridge), Catherine Little (damper), Alex Pankonien (FEM graphic), and Jochen Faber of Liebherr Aerospace (landing gear). Alan Giles of General Atomics gave me an informative tour of their facilities which resulted in the photos of their products.

Many colleagues and students have contributed to the revision of this text through suggestions and questions. In particular, Daniel J. Inman, II; Kaitlyn DeLisi; Kevin Crowely; and Emily Armentrout provided many useful comments from the perspective of students reading the material for the first time. Kaitlyn and Kevin checked all the computer codes by copying them out of the book towww.pearsonhighered.com

xii Preface

make sure they ran. My former PhD students Ya Wang, Mana Afshari, and Amin Karami checked many of the new problems and examples. Dr. Scott Larwood and the students in his vibrations class at the University of the Pacific sent many sug- gestions and corrections that helped give the book the perspective of a nonresearch insitution. I have implemented many of their suggestions, and I believe the book’s explanations are much clearer due to their input. Other professors using the book, Cetin Cetinkaya of Clarkson University, Mike Anderson of the University of Idaho, Joe Slater of Wright State University, Ronnie Pendersen of Aalborg University Esbjerg, Sondi Adhikari of the Universty of Wales, David Che of Geneva College, Tim Crippen of the University of Texas at Tyler, and Nejat Olgac of the University of Conneticut, have provided discussions via email that have led to improvements in the text, all of which are greatly appreciated. I would like to thank the review- ers: Cetin Cetinkaya, Clarkson University; Dr. Nesrin Sarigul-Klijn, University of California–Davis; and David Che, Geneva College.

Many of my former PhD students who are now academics cotaught this course with me and also offered many suggestions. Alper Erturk (Georgia Tech), Henry Sodano (University of Florida), Pablo Tarazaga (Virginia Tech), Onur Bilgen (Old Dominian University), Mike Seigler (University of Kentucky), and Armaghan Salehian (University of Waterloo) all contributed to clarity in this text for which I am grateful. I have been lucky to have wonderful PhD students to work with. I learned much from them.

I would also like to thank Prof. Joseph Slater of Wright State for reviewing some of the new materials, for writing and managing the associated toolbox, and constantly sending suggestions. Several colleagues from government labs and com- panies have also written with suggestions which have been very helpful from that perspective of practice.

I have also had the good fortune of being sponsored by numerous companies and federal agencies over the last 32 years to study, design, test, and analyze a large variety of vibrating structures and machines. Without these projects, I would not have been able to write this book nor revise it with the appreciation for the practice of vibration, which I hope permeates the text.

Last, I wish to thank my family for moral support, a sense of purpose, and for putting up with my absence while writing.

Daniel J. inman Ann Arbor, Michigan

1

Introduction to Vibration and the Free Response

1 Vibration is the subdiscipline of dynamics that deals with repetitive motion. Most of the examples in this text are mechanical or structural elements. However, vibration is prevalent in biological systems and is in fact at the source of communication (the ear vibrates to hear and the tongue and vocal cords vibrate to speak). In the case of music, vibrations, say of a stringed instrument such as a guitar, are desired. On the other hand, in most mechanical systems and structures, vibration is unwanted and even destructive. For example, vibration in an aircraft frame causes fatigue and can eventually lead to failure. An example of fatigue crack is illustrated in the circle in the photo on the bottom left. Everyday experiences are full of vibration and usually ways of mitigating vibration. Automobiles, trains, and even some bicycles have devices to reduce the vibration induced by motion and transmitted to the driver.

The task of this text is to teach the reader how to analyze vibration using principles of dynamics. This requires the use of mathematics. In fact, the sine function provides the fundamental means of analyzing vibration phenomena.

The basic concepts of understanding vibration, analyzing vibration, and predicting the behavior of vibrating systems form the topics of this text. The concepts and formulations presented in the following chapters are intended to provide the skills needed for designing vibrating systems with desired properties that enhance vibration when it is wanted and reduce vibration when it is not.

This first chapter examines vibration in its simplest form in which no external force is present (free vibration). This chapter introduces both the important concept of natural frequency and how to model vibration mathematically.

The Internet is a great source for examples of vibration, and the reader is encouraged to search for movies of vibrating systems and other examples that can be found there.

2 Introduction to Vibration and the Free Response Chap. 1

1.1 IntroductIon to Free VIbratIon

Vibration is the study of the repetitive motion of objects relative to a stationary frame of reference or nominal position (usually equilibrium). Vibration is evident everywhere and in many cases greatly affects the nature of engineering designs. The vibrational properties of engineering devices are often limiting factors in their per- formance. When harmful, vibration should be avoided, but it can also be extremely useful. In either case, knowledge about vibration—how to analyze, measure, and control it—is beneficial and forms the topic of this book.

Typical examples of vibration familiar to most include the motion of a guitar string, the ride quality of an automobile or motorcycle, the motion of an airplane’s wings, and the swaying of a large building due to wind or an earth- quake. In the chapters that follow, vibration is modeled mathematically based on fundamental principles, such as Newton’s laws, and analyzed using results from calculus and differential equations. Techniques used to measure the vibra- tion of a system are then developed. In addition, information and methods are given that are useful for designing particular systems to have specific vibrational responses.

The physical explanation of the phenomena of vibration concerns the inter- play between potential energy and kinetic energy. A vibrating system must have a component that stores potential energy and releases it as kinetic energy in the form of motion (vibration) of a mass. The motion of the mass then gives up kinetic en- ergy to the potential-energy storing device.

Engineering is built on a foundation of previous knowledge and the subject of vibration is no exception. In particular, the topic of vibration builds on pre- vious courses in dynamics, system dynamics, strength of materials, differential equations, and some matrix analysis. In most accredited engineering programs, these courses are prerequisites for a course in vibration. Thus, the material that follows draws information and methods from these courses. Vibration analysis is based on a coalescence of mathematics and physical observation. For example, consider a simple pendulum. You may have seen one in a science museum, in a grandfather clock, or you might make a simple one with a string and a marble. As the pendulum swings back and forth, observe that its motion as a function of time can be described very nicely by the sine function from trigonometry. Even more interesting, if you make a free-body diagram of the pendulum and ap- ply Newtonian mechanics to get the equation of motion (summing moments in this case), the resulting equation of motion has the sine function as its solution. Further, the equation of motion predicts the time it takes for the pendulum to repeat its motion. In this example, dynamics, observation, and mathematics all come into agreement to produce a predictive model of the motion of a pendulum, which is easily verified by experiment (physical observation).

Sec. 1.1 Introduction to Free Vibration 3

This pendulum example tells the story of this text. We propose a series of steps to build on the modeling skills developed in your first courses in statics, dy- namics, and strength of materials combined with system dynamics to find equations of motion of successively more complicated systems. Then we will use the tech- niques of differential equations and numerical integration to solve these equations of motion to predict how various mechanical systems and structures vibrate. The following example illustrates the importance of recalling the methods learned in the first course in dynamics.

Example 1.1.1

Derive the equation of motion of the pendulum in Figure 1.1.

m

l

O

g

mg

l

O

Fy

Fx

m

(b)(a)

Figure 1.1 (a) A schematic of a pendulum. (b) The free-body diagram of (a).

Solution Consider the schematic of a pendulum in Figure 1.1(a). In this case, the mass of the rod will be ignored as well as any friction in the hinge. Typically, one starts with a photograph or sketch of the part or structure of interest and is immediately faced with having to make assumptions. This is the “art” or experience side of vibration analysis and modeling. The general philosophy is to start with the simplest model possible (hence, here we ignore friction and the mass of the rod and assume the motion remains in a plane) and try to answer the relevant engineering questions. If the simple model doesn’t agree with the experiment, then make it more complex by relaxing the assump- tions until the model successfully predicts physical observation. With the assumptions in mind, the next step is to create a free-body diagram of the system, as indicated in Figure 1.1(b), in order to identify all of the relevant forces. With all the modeled forces identified, Newton’s second law and Euler’s second law are used to derive the equa- tions of motion.

In this example Euler’s second law takes the form of summing moments about point O. This yields

ΣMO = J𝛂

4 Introduction to Vibration and the Free Response Chap. 1

where MO denotes moments about the point O, J = ml2 is the mass moment of inertia of the mass m about the point O, l is the length of the massless rod, and 𝛂 is the angu- lar acceleration vector. Since the problem is really in one dimension, the vector sum of moments equation becomes the single scalar equation

Jα(t) = -mgl sin θ(t) or ml2θ $ (t) + mgl sin θ(t) = 0

Here the moment arm for the force mg is the horizontal distance l sin θ, and the two overdots indicate two differentiations with respect to the time, t. This is a second-order ordinary differential equation, which governs the time response of the pendulum. This is exactly the procedure used in the first course in dynamics to obtain equations of motion.

The equation of motion is nonlinear because of the appearance of the sin(θ) and hence difficult to solve. The nonlinear term can be made linear by approximating the sine for small values of θ(t) as sin θ ≈ θ. Then the equation of motion becomes

θ $ (t) +

g

l θ(t) = 0

This is a linear, second-order ordinary differential equation with constant coefficients and is commonly solved in the first course of differential equations (usually the third course in the calculus sequence). As we will see later in this chapter, this linear equa- tion of motion and its solution predict the period of oscillation for a simple pendulum quite accurately. The last section of this chapter revisits the nonlinear version of the pendulum equation.

n

Since Newton’s second law for a constant mass system is stated in terms of force, which is equated to the mass multiplied by acceleration, an equation of motion with two time derivatives will always result. Such equations require two constants of integration to solve. Euler’s second law for constant mass systems also yields two time derivatives. Hence the initial position for θ(0) and velocity of θ

# (0) must be

specified in order to solve for θ(t) in Example 1.1.1. The term mgl sin θ is called the restoring force. In Example 1.1.1, the restoring force is gravity, which provides a potential-energy storing mechanism. However, in most structures and machine parts the restoring force is elastic. This establishes the need for background in strength of materials when studying vibrations of structures and machines.

As mentioned in the example, when modeling a structure or machine it is best to start with the simplest possible model. In this chapter, we model only sys- tems that can be described by a single degree of freedom, that is, systems for which Newtonian mechanics result in a single scalar equation with one displacement coor- dinate. The degree of freedom of a system is the minimum number of displacement coordinates needed to represent the position of the system’s mass at any instant of time. For instance, if the mass of the pendulum in Example 1.1.1 were a rigid body, free to rotate about the end of the pendulum as the pendulum swings, the angle of rotation of the mass would define an additional degree of freedom. The problem would then require two coordinates to determine the position of the mass in space, hence two degrees of freedom. On the other hand, if the rod in Figure 1.1 is flexible,

Sec. 1.1 Introduction to Free Vibration 5

its distributed mass must be considered, effectively resulting in an infinite number of degrees of freedom. Systems with more than one degree of freedom are dis- cussed in Chapter 4, and systems with distributed mass and flexibility are discussed in Chapter 6.

The next important classification of vibration problems after degree of freedom is the nature of the input or stimulus to the system. In this chapter, only the free response of the system is considered. Free response refers to analyzing the vibration of a system resulting from a nonzero initial displacement and/or velocity of the system with no external force or moment applied. In Chapter 2, the response of a single-degree-of-freedom system to a harmonic input (i.e., a sinusoidal applied force) is discussed. Chapter 3 examines the response of a sys- tem to a general forcing function (impulse or shock loads, step functions, random inputs, etc.), building on information learned in a course in system dynamics. In the remaining chapters, the models of vibration and methods of analysis become more complex.

The following sections analyze equations similar to the linear version of the pen- dulum equation given in Example 1.1.1. In addition, energy dissipation is introduced, and details of elastic restoring forces are presented. Introductions to design, measure- ment, and simulation are also presented. The chapter ends with the introduction of high-level computer codes (MATlAb®, Mathematica, and Mathcad) as a means to visualize the response of a vibrating system and for making the calculations required to solve vibration problems more efficiently. In addition, numerical simulation is intro- duced in order to solve nonlinear vibration problems.

1.1.1 the Spring–Mass Model

From introductory physics and dynamics, the fundamental kinematical quantities used to describe the motion of a particle are displacement, velocity, and accelera- tion vectors. In addition, the laws of physics state that the motion of a mass with changing velocity is determined by the net force acting on the mass. An easy de- vice to use in thinking about vibration is a spring (such as the one used to pull a storm door shut, or an automobile spring) with one end attached to a fixed object and a mass attached to the other end. A schematic of this arrangement is given in Figure 1.2.

fk

mg

m

m

0

(a) (b)

Figure 1.2 A schematic of (a) a single-degree-of-freedom spring–mass oscillator and (b) its free-body diagram.

6 Introduction to Vibration and the Free Response Chap. 1

Ignoring the mass of the spring itself, the forces acting on the mass consist of the force of gravity pulling down (mg) and the elastic-restoring force of the spring pulling back up (fk). Note that in this case the force vectors are collinear, reducing the static equilibrium equation to one dimension easily treated as a scalar. The nature of the spring force can be deduced by performing a simple static experiment. With no mass attached, the spring stretches to the position labeled x0 = 0 in Figure 1.3. As successively more mass is attached to the spring, the force of gravity causes the spring to stretch further. If the value of the mass is recorded, along with the value of the displacement of the end of the spring each time more mass is added, the plot of the force (mass, denoted by m, times the acceleration due to gravity, denoted by g) versus this displacement, denoted by x, yields a curve similar to that illustrated in Figure 1.4. Note that in the region of values for x between 0 and about 20 mm (millimeters), the curve is a straight line. This indicates that for deflections less than 20 mm and forces less than 1000 N (newtons), the force that is applied by the spring to the mass is pro- portional to the stretch of the spring. The constant of proportionality is the slope of the straight line between 0 and 20 mm. For the particular spring of Figure 1.4, the constant is 50 N>mm, or 5 * 104 N>m. Thus, the equation that describes the force applied by the spring, denoted by fk, to the mass is the linear relationship

fk = kx (1.1)

The value of the slope, denoted by k, is called the stiffness of the spring and is a property that characterizes the spring for all situations for which the displacement is less than 20 mm. From strength-of-materials considerations, a linear spring of stiffness k stores potential energy of the amount 12 kx

2.

x0

g

x1 x2 x3

Figure 1.3 A schematic of a massless spring with no mass attached showing its static equilibrium position, followed by increments of increasing added mass illustrating the corresponding deflections.

x

fk

20 mm0

103 N

Figure 1.4 The static deflection curve for the spring of Figure 1.3.

Sec. 1.1 Introduction to Free Vibration 7

Note that the relationship between fk and x of equation (1.1) is linear (i.e., the curve is linear and fk depends linearly on x). If the displacement of the spring is larger than 20 mm, the relationship between fk and x becomes nonlinear, as indi- cated in Figure 1.4. Nonlinear systems are much more difficult to analyze and form the topic of Section 1.10. In this and all other chapters, it is assumed that displace- ments (and forces) are limited to be in the linear range unless specified otherwise.

Next, consider a free-body diagram of the mass in Figure 1.5, with the mass- less spring elongated from its rest (equilibrium or unstretched) position. As in the earlier figures, the mass of the object is taken to be m and the stiffness of the spring is taken to be k. Assuming that the mass moves on a frictionless surface along the x direction, the only force acting on the mass in the x direction is the spring force. As long as the motion of the spring does not exceed its linear range, the sum of the forces in the x direction must equal the product of mass and acceleration.

Summing the forces on the free-body diagram in Figure 1.5 along the x direc- tion yields

mx $ (t) = -kx(t) or mx$(t) + kx(t) = 0 (1.2)

where x $ (t) denotes the second time derivative of the displacement (i.e., the accel-

eration). Note that the direction of the spring force is opposite that of the deflection (+ is marked to the right in the figure). As in Example 1.1.1, the displacement vec- tor and acceleration vector are reduced to scalars, since the net force in the y direc- tion is zero (N = mg) and the force in the x direction is collinear with the inertial force. both the displacement and acceleration are functions of the elapsed time t, as denoted in equation (1.2). Window 1.1 illustrates three types of mechanical sys- tems, which for small oscillations can be described by equation (1.2): a spring–mass system, a rotating shaft, and a swinging pendulum (Example 1.1.1). Other examples are given in Section 1.4 and throughout the book.

One of the goals of vibration analysis is to be able to predict the response, or motion, of a vibrating system. Thus it is desirable to calculate the solution to equation (1.2). Fortunately, the differential equation of (1.2) is well known and is covered extensively in introductory calculus and physics texts, as well as in texts on differential equations. In fact, there are a variety of ways to calculate this solution. These are all discussed in some detail in the next section. For now, it is sufficient to present a solution based on physical observation. From experience

y

x

�kx mg

N

m

x0

k

0

�� Friction-free surface

Rest position

(a) (b)

Figure 1.5 (a) A single spring–mass system given an initial displacement of x0 from its rest, or equilibrium, position and zero initial velocity. (b) The system’s free- body diagram.

8 Introduction to Vibration and the Free Response Chap. 1

watching a spring, such as the one in Figure 1.5 (or a pendulum), it is guessed that the motion is periodic, of the form

x(t) = A sin(ωnt + ϕ) (1.3)

This choice is made because the sine function describes oscillation. Equation (1.3) is the sine function in its most general form, where the constant A is the amplitude, or maximum value, of the displacement; ωn, the angular natural fre- quency, determines the interval in time during which the function repeats itself; and ϕ, called the phase, determines the initial value of the sine function. As will be discussed in the following sections, the phase and amplitude are determined by the initial state of the system (see Figure 1.7). It is standard to measure the time t in seconds (s). The phase is measured in radians (rad), and the frequency is measured in radians per second (rad>s). As derived in the following equation, the frequency ωn is determined by the physical properties of mass and stiffness (m and k), and the constants A and ϕ are determined by the initial position and velocity as well as the frequency.

To see if equation (1.3) is in fact a solution of the equation of motion, it is substituted into equation (1.2). Successive differentiation of the displacement, x(t) in the form of equation (1.3), yields the velocity, x

# (t), given by

x # (t) = ωnA cos(ωnt + ϕ) (1.4)

and the acceleration, x $ (t), given by

x $ (t) = -ωn2A sin(ωnt + ϕ) (1.5)

Window 1.1 Examples of Single-Degree-of-Freedom Systems (for small displacements)

(a) (b) (c)

Spring–mass mx � kx � 0

m

k

x(t)

J

k

�(t)

Shaft and disk J� � k� � 0

Torsional stiffness

g m

Simple pendulum � � (g/l)� � 0

Gravity l � length

Sec. 1.1 Introduction to Free Vibration 9

Substitution of equations (1.5) and (1.3) into (1.2) yields

-mωn2A sin(ωnt + ϕ) = -kA sin(ωnt + ϕ)

Dividing by A and m yields the fact that this last equation is satisfied if

ωn2 = k m

, or ωn = A km (1.6) Hence, equation (1.3) is a solution of the equation of motion. The constant ωn

characterizes the spring–mass system, as well as the frequency at which the motion repeats itself, and hence is called the system’s natural frequency. A plot of the solu- tion x(t) versus time t is given in Figure 1.6. It remains to interpret the constants A and ϕ.

The units associated with the notation ωn are rad>s and in older texts natural frequency in these units is often referred to as the circular natural frequency or cir- cular frequency to emphasize that the units are consistent with trigonometric func- tions and to distinguish this from frequency stated in units of hertz (Hz) or cycles per second, denoted by fn, and commonly used in discussing frequency. The two are related by fn = ωn>2π as discussed in Section 1.2. In practice, the phrase natu- ral frequency is used to refer to either fn or ωn, and the units are stated explicitly to avoid confusion. For example, a common statement is: the natural frequency is 10 Hz, or the natural frequency is 20π rad>s.

Recall from differential equations that because the equation of motion is of second order, solving equation (1.2) involves integrating twice. Thus there are two constants of integration to evaluate. These are the constants A and ϕ. The physical significance, or interpretation, of these constants is that they are determined by the initial state of motion of the spring–mass system. Again, recall Newton’s laws, if no force is imparted to the mass, it will stay at rest. If, however, the mass is displaced to a position of x0 at time t = 0, the force kx0 in the spring will result in motion. Also, if the mass is given an initial velocity of v0 at time t = 0, motion will result because

2 4 6 8 10

A 1.5

x(t) (mm)

Time (s)

1

0.5

0

�0.5

�1

�A �1.5

12

T � 2��n

Figure 1.6 The response of a simple spring–mass system to an initial displacement of x0 = 0.5 mm and an initial velocity of v0 = 212 mm>s. The natural frequency is 2 rad>s and the amplitude is 1.5 mm. The period is T = 2π>ωn = 2π>2 = πs.

10 Introduction to Vibration and the Free Response Chap. 1

of the induced change in momentum. These are called initial conditions and when substituted into the solution (1.3) yield

x0 = x(0) = A sin(ωn0 + ϕ) = A sin ϕ (1.7)

and

v0 = x # (0) = ωnA cos(ωn0 + ϕ) = ωnA cos ϕ (1.8)

Solving these two simultaneous equations for the two unknowns A and ϕ yields

A = 2ωn2 x02 + v02

ωn and ϕ = tan-1

ωnx0 v0

(1.9)

as illustrated in Figure 1.7. Here the phase ϕ must lie in the proper quadrant, so care must be taken in evaluating the arc tangent. Thus, the solution of the equation of motion for the spring–mass system is given by

x(t) = 2ωn2 x02 + v02

ωn sin aωnt + tan-1

ωnx0 v0

b (1.10)

and is plotted in Figure 1.6. This solution is called the free response of the system, be- cause no force external to the system is applied after t = 0. The motion of the spring– mass system is called simple harmonic motion or oscillatory motion and is discussed in detail in the following section. The spring–mass system is also referred to as a simple harmonic oscillator, as well as an undamped single-degree-of-freedom system.

v0

v0

x0

x0

� 90º

A � x0 2 �

v0 n

2

n�

n�

(a)

x0

x0

90º

A � x0 2 �

v0 n

2

v0 n�

(b)

Figure 1.7 The trigonometric relationships between the phase, natural frequency, and initial conditions. Note that the initial conditions determine the proper quadrant for the phase: (a) for a positive initial position and velocity, (b) for a negative initial position and a positive initial velocity.

Sec. 1.1 Introduction to Free Vibration 11

Example 1.1.2

The phase angle ϕ describes the relative shift in the sinusoidal vibration of the spring– mass system resulting from the initial displacement, x0. Verify that equation (1.10) satisfies the initial condition x(0) = x0.

Solution Substitution of t = 0 in equation (1.10) yields

x(0) = A sin ϕ = 2ωn2 x02 + v02

ωn sin atan-1 ωn x0

v0 b

Figure 1.7 illustrates the phase angle ϕ defined by equation (1.9). This right triangle is used to define the sine and tangent of the angle ϕ. From the geometry of a right triangle, and the definitions of the sine and tangent functions, the value of x(0) is computed to be

x(0) = 2ωn2 x02 + v02

ωn

ωn x02ωn2 x02 + v02 = x0 which verifies that the solution given by equation (1.10) is consistent with the initial displacement condition.

n

Example 1.1.3

A vehicle wheel, tire, and suspension assembly can be modeled crudely as a single- degree-of-freedom spring–mass system. The (unsprung) mass of the assembly is measured to be about 30 kilograms (kg). Its frequency of oscillation is observed to be 10 Hz. What is the approximate stiffness of the suspension assembly?

Solution The relationship between frequency, mass, and stiffness is ωn = 2k>m, so that

k = mωn2 = (30 kg) a10 cycle

s # 2π rad

cycle b

2

= 1.184 * 105 N>m

This provides one simple way to estimate the stiffness of a complicated device. This stiffness could also be estimated by using a static deflection experiment similar to that suggested by Figures 1.3 and 1.4.

n

Example 1.1.4

Compute the amplitude and phase of the response of a system with a mass of 2 kg and a stiffness of 200 N>m, to the following initial conditions:

a) x0 = 2 mm and v0 = 1 mm>s b) x0 = -2 mm and v0 = 1 mm>s c) x0 = 2 mm and v0 = -1 mm>s

Compare the results of these calculations.

12 Introduction to Vibration and the Free Response Chap. 1

Solution First, compute the natural frequency, as this does not depend on the initial conditions and will be the same in each case. From equation (1.6):

ωn = B km = B 200 N>m2 kg = 10 rad>s Next, compute the amplitude, as it depends on the squares of the initial conditions and will be the same in each case. From equation (1.9):

A = 2ωn2 x02 + v02

ωn =

2102 # 22 + 12 10

= 2.0025 mm

Thus the difference between the three responses in this example is determined only by the phase. Using equation (1.9) and referring to Figure 1.7 to determine the proper quadrant, the following yields the phase information for each case:

a) ϕ = tan-1 aωn x0 v0

b = tan-1 a(10 rad>s) (2 mm) 1 mm>s b = 1.521 rad (or 87.147°)

which is in the first quadrant.

b) ϕ = tan-1 aωn x0 v0

b = tan-1 a(10 rad>s) (-2 mm) 1 mm>s b = -1.521 rad (or -87.147°)

which is in the fourth quadrant.

c) ϕ = tan-1 aωn x0 v0

b = tan-1 a(10 rad>s)(2 mm) -1 mm>s b = (-1.521 + π) rad (or 92.85°)

which is in the second quadrant (position positive, velocity negative places the angle in the second quadrant in Figure 1.7 requiring that the raw calculation be shifted 180°).

Note that if equation (1.9) is used without regard to Figure 1.7, parts b and c would result in the same answer (which makes no sense physically as the responses each have different starting points). Thus in computing the phase it is important to consider which quadrant the angle should lie in. Fortunately, some calculators and some codes use an arc tangent function, which corrects for the quadrant (for instance, MATlAb uses the atan2(w0*x0, v0) command).

The tan(ϕ) can be positive or negative. If the tangent is positive, the phase angle is in the first or third quadrant. If the sign of the initial displacement is positive, the phase angle is in the first quadrant. If the sign is negative or the initial displacement is negative, the phase angle is in the third quadrant. If on the other hand the tangent is negative, the phase angle is in the second or fourth quadrant. As in the previous case, by examining the sign of the initial displacement, the proper quadrant can be determined. That is, if the sign is positive, the phase angle is in the second quadrant, and if the sign is negative, the phase angle is in the fourth quadrant. The remaining possibility is that the tangent is equal to zero. In this case, the phase angle is either zero or 180°. The initial velocity determines which quadrant is correct. If the initial displacement is zero and if the initial velocity is zero, then the phase angle is zero. If on the other hand the initial velocity is negative, the phase angle is 180°.

n

Sec. 1.2 Harmonic Motion 13

The main point of this section is summarized in Window 1.2. This illustrates harmonic motion and how the initial conditions determine the response of such a system.

Window 1.2 Summary of the Description of Simple Harmonic Motion

sin(�nt � �)

v0 � initial velocity� � tan�1 v0 �nx0

Maximum velocity

Period

Time, t

x0 Initial

displace- ment

� �n

2� �nSlope here

is v0

Displacement, x(t)

AmplitudeT �

x(t) � �n �n

2 x0 2 + v0

21

A� �n 2 x0

2 � v0 2

�n 1

1.2 HarMonIc MotIon

The fundamental kinematic properties of a particle moving in one dimension are displacement, velocity, and acceleration. For the harmonic motion of a simple spring–mass system, these are given by equations (1.3), (1.4), and (1.5), respectively. These equations reveal the different relative amplitudes of each quantity. For systems with natural frequency larger than 1 rad>s, the relative amplitude of the velocity response is larger than that of the displacement response by a multiple of ωn, and the acceleration response is larger by a multiple of ωn2. For systems with frequency less than 1, the velocity and acceleration have smaller relative ampli- tudes than the displacement. Also note that the velocity is 90° (or π>2 radians) out of phase with the position [i.e., sin(ωnt + π>2 + ϕ) = cos(ωnt + ϕ)], while the acceleration is 180° out of phase with the position and 90° out of phase with the velocity. This is summarized and illustrated in Window 1.3.

14 Introduction to Vibration and the Free Response Chap. 1

Window 1.3 The Relationship between Displacement, Velocity, and Acceleration

for Simple Harmonic Motion

�A

0

A

t Displacement

x(t) � A sin (�nt � �)

��A

0

�A

t

��2A

0

�2A

t Acceleration

x(t) � ��2nA sin (�nt � �)

Velocity x(t) � �nA cos (�nt � �)

..

.

Sec. 1.2 Harmonic Motion 15

The angular natural frequency, ωn, used in equations (1.3) and (1.10), is mea- sured in radians per second and describes the repetitiveness of the oscillation. As indicated in Window 1.2, the time the cycle takes to repeat itself is the period, T, which is related to the natural frequency by

T = 2π rad

ωn rad>s =

2π ωn

s (1.11)

This results from the elementary definition of the period of a sine function. The frequency in hertz (Hz), denoted by fn, is related to the frequency in radians per second, denoted by ωn:

fn = ωn 2π

= ωn rad>s

2π rad>cycle = ωn cycles

2π s =

ωn 2π

(Hz) (1.12)

Equation (1.2) is exactly the same form of differential equation as the linear pendulum equation of Example 1.1.1 and of the shaft and disk of Window 1.1(b). As such, the pendulum will have exactly the same form of solution as equation (1.3), with frequency

ωn = Agl rad>s The solution of the pendulum equation thus predicts that the period of oscillation of the pendulum is

T = 2π ωn

= 2πA lg s where the non-italic s denotes seconds. This analytical value of the period can be checked by measuring the period of oscillation of a pendulum with a simple stop- watch. The period of the disk and shaft system of Window 1.1 will have a frequency and period of

ωn = AkJ rad>s and T = 2πA Jk s respectively. The concept of frequency of vibration of a mechanical system is the single most important physical concept (and number) in vibration analysis. Measurement of either the period or the frequency allows validation of the analyti- cal model. (If you made a 1-meter pendulum, the period would be about 2 s. This is something you could try at home.)

As long as the only disturbance to these systems is a set of nonzero initial con- ditions, the system will respond by oscillating with frequency ωn and period T. For

16 Introduction to Vibration and the Free Response Chap. 1

the case of the pendulum, the longer the pendulum, the smaller the frequency and the longer the period. That’s why in museum demonstrations of a pendulum, the length is usually very large so that T is large and one can easily see the period (also a pendulum is usually used to illustrate the earth’s precession; Google the phrase Foucault Pendulum).

Example 1.2.1

Consider a small spring about 30 mm (or 1.18 in) long, welded to a stationary table (ground) so that it is fixed at the point of contact, with a 12-mm (or 0.47-in) bolt welded to the other end, which is free to move. The mass of this system is about 49.2 * 10−3 kg (equivalent to about 1.73 ounces). The spring stiffness can be measured using the method suggested in Figure 1.4 and yields a spring constant of k = 857.8 N>m. Calculate the natural frequency and period. Also determine the maximum amplitude of the response if the spring is initially deflected 10 mm. Assume that the spring is oriented along the direc- tion of gravity as in Window 1.1. (Ignore the effect of gravity; see below.)

Solution From equation (1.6) the natural frequency is

ωn = B km = B 857.8 N>m49.2 * 10-3 kg = 132 rad>s In hertz, this becomes

fn = ωn 2π

= 21 Hz

The period is

T = 2π ωn

= 1 fn

= 0.0476 s

To determine the maximum value of the displacement response, note from Figure 1.6 that this corresponds to the value of the constant A. Assuming that no initial velocity is given to the spring (v0 = 0), equation (1.9) yields

x(t)max = A = 2ωn2 x02 + v02

ωn = x0 = 10 mm

Note that the maximum value of the velocity response is ωnA or ωnx0 = 1320 mm>s and the acceleration response has maximum value

ωn2A = ωn2x0 = 174.24 * 103 mm>s2

Since v0 = 0, the phase is ϕ = tan−1(ωnx0>0) = π>2, or 90°. Hence, in this case, the response is x(t) = 10 sin(132t + π>2) = 10 cos(132t) mm.

n

Does gravity matter in spring problems? The answer is no, if the system oscillates in the linear region. Consider the spring of Figure 1.3 and let a mass of value m extend

Sec. 1.2 Harmonic Motion 17

the spring. let Δ denote the distance deflected in this static experiment (Δis called the static deflection); then the force acting upon the mass is kΔ. From static equilibrium the forces acting on the mass must be zero so that (taking positive down in the figure)

mg – kΔ = 0

Next, sum the forces along the vertical for the mass at some point x and apply Newton’s law to get

mx $ (t) = -k(x + Δ) + mg = -kx + mg – Δk

Note the sign on the spring term is negative because the spring force opposes the motion, which is taken here as positive down. The last two terms add to zero (mg – kΔ = 0) because of the static equilibrium condition, and the equation of motion becomes

mx $ (t) + kx(t) = 0

Thus gravity does not affect the dynamic response. Note x(t) is measured from the elongated (or compressed if upside down) position of the spring–mass system, that is, from its rest position. This is discussed again using energy methods in Figure 1.14.

Example 1.2.2

(a) A pendulum in brussels swings with a period of 3 seconds. Compute the length of the pendulum. (b) At another location, assume the length of the pendulum is known to be 2 meters and suppose the period is measured to be 2.839 seconds. What is the acceleration due to gravity at that location?

Solution The relationship between period and natural frequency is given in equation (1.11). (a) Substitution of the value of natural frequency for a pendulum and solving for the length of the pendulum yields

T = 2π ωn

1 ωn2 = g

l =

4π2

T 2 1 l =

gT 2

4π2 =

(9.811 m>s2 )(3)2s2 4π2

= 2.237 m

Here the value of g = 9.811 m>s2 is used, as that is the value it has in brussels (at 51° latitude and an altitude of 102 m). (b) Next, manipulate the pendulum period equation to solve for g. This yields

g

l =

4π2

T 2 1 g =

4π2

T 2 l =

4π2

(2.839)2 s2 (2)m = 9.796 m>s2

This is the value of the acceleration due to gravity in Denver, Colorado, United States (at an altitude 1638 m and latitude 40°).

These sorts of calculations are usually done in high school science classes but are repeated here to underscore the usefulness of the concept of natural frequency and period in terms of providing information about the vibration system’s physical properties. In addition, this example serves to remind the reader of a familiar vibration phenomenon.

n

18 Introduction to Vibration and the Free Response Chap. 1

The solution given by equation (1.10) was developed assuming that the response should be harmonic based on physical observation. The form of the response can also be derived by a more analytical approach following the theory of elementary differ- ential equations (see, e.g., boyce and DiPrima, 2009). This approach is reviewed here and will be generalized in later sections and chapters to solve for the response of more complicated systems.

Assume that the solution x(t) is of the form

x(t) = aeλt (1.13)

where a and λ are nonzero constants to be determined. Upon successive differen- tiation, equation (1.13) becomes x

# (t) = λaeλt and x$(t) = λ2aeλt. Substitution of the

assumed exponential form into equation (1.2) yields

mλ2aeλt + kaeλt = 0 (1.14)

Since the term aeλt is never zero, expression (1.14) can be divided by aeλt to yield

mλ2 + k = 0 (1.15)

Solving this algebraically results in

λ = {A – km = {A km j = {ωn j (1.16) where j = 1-1 is the imaginary number and ωn = 1k>m is the natural frequency as before. Note that there are two values for λ, λ = +ωnj and λ = -ωnj, because the equation for λ is of second order. This implies that there must be two solutions of equation (1.2) as well. Substitution of equation (1.16) into equation (1.13) yields that the two solutions for x(t) are

x(t) = a1e+jωnt and x(t) = a2e-jωnt (1.17)

Since equation (1.2) is linear, the sum of two solutions is also a solution; hence, the response x(t) is of the form

x(t) = a1e+jωnt + a2e-jωnt (1.18)

where a1 and a2 are complex-valued constants of integration. The Euler relations for trigonometric functions state that 2j sin θ = (eθj – e–θj) and 2 cos θ = (eθj + e–θj), where j = 1-1. [See Appendix A, equations (A.18), (A.19), and (A.20), as well as Window 1.5.] Using the Euler relations, equation (1.18) can be written as

x(t) = A sin(ωnt + ϕ) (1.19)

where A and ϕ are real-valued constants of integration. Note that equation (1.19) is in agreement with the physically intuitive solution given by equation (1.3). The re- lationships among the various constants in equations (1.18) and (1.19) are given in Window 1.4. Window 1.5 illustrates the use of Euler relations for deriving harmonic functions from exponentials for the underdamped case.

Sec. 1.2 Harmonic Motion 19

Window 1.4 Three Equivalent Representations of Harmonic Motion

The solution of mx $ + kx = 0 subject to nonzero initial conditions can be

written in three equivalent ways. First, the solution can be written as

x(t) = a1ejωnt + a2e-jωnt, ωn = A km, j = 1-1 where a1 and a2 are complex-valued constants. Second, the solution can be written as

x(t) = A sin(ωnt + ϕ)

where A and ϕ are real-valued constants. last, the solution can be written as

x(t) = A1 sin ωnt + A2 cos ωnt

where A1 and A2 are real-valued constants. Each set of two constants is deter- mined by the initial conditions, x0 and v0. The various constants are related by the following:

A = 2A21 + A22 ϕ = tan-1 aA2A1 b A1 = (a1 – a2)j A2 = a1 + a2

a1 = A2 – A1j

2 a2 =

A2 + A1j 2

all of which follow from trigonometric identities and Euler’s formulas. Note that a1 and a2 are a complex conjugate pair, so that A1 and A2 are both real numbers provided that the initial conditions are real valued, as is normally the case.

Often when computing frequencies from equation (1.16) such as λ2 = -4, there is a temptation to write that the frequency is ωn = {2. This is incorrect be- cause the {sign is used up when the Euler relation is used to obtain the function sin ωnt from the exponential form. The concept of frequency is not defined until it appears in the argument of the sine function and, as such, is always positive.

Precise terminology is useful in discussing an engineering problem, and the sub- ject of vibration is no exception. Since the position, velocity, and acceleration change continually with time, several other quantities are used to discuss vibration. The peak value, defined as the maximum displacement, or magnitude A of equation (1.9), is

20 Introduction to Vibration and the Free Response Chap. 1

often used to indicate the region in space in which the object vibrates. Another quan- tity useful in describing vibration is the average value, denoted by x‾, and defined by

x‾ = lim T S ∞

1 T

L T

0 x(t) dt (1.20)

Note that the average value of x(t) = A sin ωnt over one period of oscillation is zero. Since the square of displacement is associated with a system’s potential

energy, the average of the displacement squared is sometimes a useful vibration property to discuss. The mean-square value (or variance) of the displacement x(t), denoted by x‾

2, is defined by

x‾ 2 = lim

T S ∞

1 T

L T

0 x2(t) dt (1.21)

The square root of this value, called the root mean-square (rms) value, is commonly used in specifying vibration. because the peak value of the velocity and accelera- tion are multiples of the natural frequency times the displacement amplitude [i.e., equations (1.3)–(1.5)], these three peak values often differ in value by an order of magnitude or more. Hence, logarithmic scales are often used. A common unit of measurement for vibration amplitudes and rms values is the decibel (db). The deci- bel was originally defined in terms of the base 10 logarithm of the power ratio of two electrical signals, or as the ratio of the square of the amplitudes of two signals. Following this idea, the decibel is defined as

db K 10 log10 a x1 x0 b

2

= 20 log10 x1 x0

(1.22)

Here the signal x0 is a reference signal. The decibel is used to quantify how far the measured signal x1 is above the reference signal x0. Note that if the measured signal is equal to the reference signal, then this corresponds to 0 db. The decibel is used extensively in acoustics to compare sound levels. Using a db scale expands or com- presses vibration response information for convenience in graphical representation.

Example 1.2.3

Consider a 2-meter long pendulum placed on the moon and given an initial angular displacement of 0.2 rad and zero initial velocity. Calculate the maximum angular veloc- ity and the maximum angular acceleration of the swinging pendulum (note that gravity on the earth’s moon is gm = g>6, where g is the acceleration due to gravity on earth). Solution From Example 1.1.1 the equation of motion of a pendulum is

$ θ(t) +

gm l

θ(t) = 0

Sec. 1.3 Viscous Damping 21

This equation is of the same form as equation (1.2) and hence has a solution of the form

θ(t) = A sin(ωnt + ϕ), ωn = Agml From equation (1.9) the amplitude is given by

A = Cωn2x02 + v02ωn2 = x0 = 0.2 rad From Window 1.3 the maximum velocity is just ωnA or

vmax = ωnA = B gml (0.2) = (0.2)B 9.8>62 = 0.18 rad>s The maximum acceleration is

amax = ωn2A = gm l

A = 9.8>6

2 (0.2) = 0.163 rad>s2

n

Frequencies of concern in mechanical vibration range from fractions of a hertz to several thousand hertz. Amplitudes range from micrometers up to meters (for systems such as tall buildings). According to Mansfield (2005), human beings are more sensitive to acceleration than displacement and easily perceive vibration around 5 Hz at about 0.01 m>s2 (about 0.01 mm). Horizontal vibration is easy to experience near 2 Hz. Work attempting to characterize comfort levels for human vibrations is still ongoing.

1.3 VIScouS daMpIng

The response of the spring–mass model (Section 1.1) predicts that the system will oscillate indefinitely. However, everyday observation indicates that freely oscillat- ing systems eventually die out and reduce to zero motion. This observation suggests that the model sketched in Figure 1.5 and the corresponding mathematical model given by equation (1.2) need to be modified to account for this decaying motion. The choice of a representative model for the observed decay in an oscillating system is based partially on physical observation and partially on mathematical convenience. The theory of differential equations suggests that adding a term to equation (1.2) of the form cx

# (t), where c is a constant, will result in a solution x(t) that dies out.

Physical observation agrees fairly well with this model and is used successfully to model the damping, or decay, in a variety of mechanical systems. This type of damp- ing, called viscous damping, is described in detail in this section.

22 Introduction to Vibration and the Free Response Chap. 1

While the spring forms a physical model for storing potential energy and hence causing vibration, the dashpot, or damper, forms the physical model for dissipating energy and thus damping the response of a mechanical system. An example dashpot consists of a piston fit into a cylinder filled with oil as indicated in Figure 1.8. This piston is perforated with holes so that motion of the piston in the oil is possible. The laminar flow of the oil through the perforations as the piston moves causes a damp- ing force on this piston. The force is proportional to the velocity of the piston in a direction opposite that of the piston motion. This damping force, denoted by fc, has the form

fc = cx # (t) (1.23)

where c is a constant of proportionality related to the oil viscosity. The constant c, called the damping coefficient, has units of force per velocity, or N s>m, as it is customarily written. However, following the strict rules of SI units, the units on damping can be reduced to kg>s, which states the units on damping in terms of the fundamental (also called basic) SI units (mass, time, and length).

In the case of the oil-filled dashpot, the constant c can be determined by fluid principles. However, in most cases, fc is provided by equivalent effects occurring in the material forming the device. A good example is a block of rubber (which also provides stiffness fk) such as an automobile motor mount, or the effects of air flowing around an oscillating mass. In all cases in which the damping force fc is proportional to velocity, the schematic of a dashpot is used to indicate the presence of this force. The schematic is illustrated in Figure 1.9. Unfortunately, the damping coefficient of a system cannot be measured as simply as the mass or stiffness of a system can be. This is pointed out in Section 1.6.

Using a simple force balance on the mass of Figure 1.9 in the x direction, the equation of motion for x(t) becomes

mx $ = -fc – fk (1.24)

or

mx $ (t) + cx# (t) + kx(t) = 0 (1.25)

Mounting point

Seal Case

Orifice

Piston

Mounting point

Oil

x(t)

Figure 1.8 A schematic of a dashpot that produces a damping force fc(t) = cx # (t),

where x(t) is the motion of the case relative to the piston.

Sec. 1.3 Viscous Damping 23

subject to the initial conditions x(0) = x0 and x # (0) = v0. The forces fc and fk are

negative in equation (1.24) because they oppose the motion (positive to the right). Equation (1.25) and Figure 1.9, referred to as a damped single-degree-of-freedom system, form the topic of Chapters 1 through 3.

To solve the damped system of equation (1.25), the same method used for solving equation (1.2) is used. In fact, this provides an additional reason to choose fc to be of the form cx

# . let x(t) have the form given in equation (1.13), x(t) = aeλt.

Substitution of this form into equation (1.25) yields

(mλ2 + cλ + k) aeλt = 0 (1.26)

Again, aeλt ≠ 0, so that this reduces to a quadratic equation in λ of the form

mλ2 + cλ + k = 0 (1.27)

called the characteristic equation. This is solved using the quadratic formula to yield the two solutions

λ1, 2 = – c

2m { 1

2m 2c2 – 4km (1.28)

Examination of this expression indicates that the roots λ will be real or complex, de- pending on the value of the discriminant, c2 – 4km. As long as m, c, and k are positive real numbers, λ1 and λ2 will be distinct negative real numbers if c2 – 4km 7 0. On the other hand, if this discriminant is negative, the roots will be a complex conjugate pair with a negative real part. If the discriminant is zero, the two roots λ1 and λ2 are equal negative real numbers. Note that equation (1.15) represents the characteristic equa- tion for the special undamped case (i.e., c = 0).

In examining these three cases, it is both convenient and useful to define the critical damping coefficient, ccr, by

ccr = 2mωn = 21km (1.29)

x(t)

k

c

m

y

x

fk

fc

mg

N

Friction-free surface

(a) (b)

Figure 1.9 (a) The schematic of a single-degree-of-freedom system with viscous damping indicated by a dashpot and (b) the corresponding free-body diagram.

24 Introduction to Vibration and the Free Response Chap. 1

where ωn is the undamped natural frequency in rad>s. Furthermore, the nondimen- sional number ζ, called the damping ratio, defined by

ζ = c

ccr =

c 2mωn

= c

21km (1.30) can be used to characterize the three types of solutions to the characteristic equa- tion. Rewriting the roots given by equation (1.28) yields

λ1, 2 = -ζωn { ωn2ζ2 – 1 (1.31) where it is now clear that the damping ratio ζ determines whether the roots are complex or real. This in turn determines the nature of the response of the damped single-degree-of-freedom system. For positive mass, damping, and stiffness coef- ficients, there are three cases, which are delineated next.

1.3.1 underdamped Motion

In this case, the damping ratio ζ is less than 1 (0 6 ζ 6 1) and the discriminant of equation (1.31) is negative, resulting in a complex conjugate pair of roots. Factoring out (-1) from the discriminant in order to clearly distinguish that the second term is imaginary yields

2ζ2 – 1 = 2(1 – ζ2)(-1) = 21 – ζ2 j (1.32) where j = 1-1. Thus the two roots become λ1 = -ζωn – ωn21 – ζ2 j (1.33) and

λ2 = -ζωn + ωn21 – ζ2 j (1.34) Following the same argument as that made for the undamped response of equation (1.18), the solution of (1.25) is then of the form

x(t) = e-ζωnt(a1ej21 – ζ2ωnt + a2e-j21 – ζ2ωnt) (1.35) where a1 and a2 are arbitrary complex-valued constants of integration to be deter- mined by the initial conditions. Using the Euler relations (see Window 1.5), this can be written as

x(t) = Ae-ζωnt sin(ωdt + ϕ) (1.36)

where A and ϕ are constants of integration and ωd, called the damped natural fre- quency, is given by

ωd = ωn21 – ζ2 (1.37) in units of rad>s.

Sec. 1.3 Viscous Damping 25

The constants A and ϕ are evaluated using the initial conditions in exactly the same fashion as they were for the undamped system as indicated in equations (1.7) and (1.8). Set t = 0 in equation (1.36) to get x0 = A sin ϕ. Differentiating (1.36) yields

x # (t) = -ζωnAe-ζωnt sin(ωdt + ϕ) + ωdAe-ζωnt cos(ωdt + ϕ)

let t = 0 and A = x0>sin ϕ in this last expression to get x # (0) = v0 = -ζωnx0 + x0ωd cot ϕ

Window 1.5 Euler Relations and the Underdamped Solution

An underdamped solution of mx $ + cx# + kx = 0 to nonzero initial conditions

is of the form

x(t) = a1eλ1t + a2eλ2t

where λ1 and λ2 are complex numbers of the form

λ1 = -ζωn + ωdj and λ2 = -ζωn – ωdj

where ωn = 2k>m, ζ = c>(2mωn), ωd = ωn21 – ζ2, and j = 1-1. The two constants a1 and a2 are complex numbers and hence represent four unknown constants rather than the two constants of integration required to solve a second-order differential equation. This demands that the two com- plex numbers a1 and a2 be conjugate pairs so that x(t) depends only on two undetermined constants. Substitution of the foregoing values of λi into the solution x(t) yields

x(t) = e-ζωnt(a1eωdjt + a2e-ωd jt)

Using the Euler relations eϕj = cos ϕ + j sin ϕ and e–ϕj = cos ϕ – j sin ϕ, x(t) becomes

x(t) = e-ζωnt3(a1 + a2) cos ωdt + j(a1 – a2) sin ωdt4 Choosing the real numbers A2 = a1 + a2 and A1 = (a1 – a2)j, this becomes

x(t) = e-ζωnt(A1 sin ωdt + A2 cos ωdt)

which is real valued. Defining the constant A = 2A 12 + A 22 and the angle ϕ = tan−1(A2>A1) so that A1 = A cos ϕ and A2 = A sin ϕ, the form of x(t) becomes [recall that sin a cos b + cos a sin b = sin(a + b)]

x(t) = Ae-ζωnt sin (ωdt + ϕ)

where A and ϕ are the constants of integration to be determined from the ini- tial conditions. Complex numbers are reviewed in Appendix A.

26 Introduction to Vibration and the Free Response Chap. 1

Solving this last expression for ϕ yields

tan ϕ = x0ωd

v0 + ζωnx0

With this value of ϕ, the sine becomes

sin ϕ = x0ωd2(v0 + ζωnx0)2 + (x0ωd)2

Thus the value of A and ϕ are determined to be

A = B (v0 + ζωnx0)2 + (x0ωd)2ωd2 , ϕ = tan-1 x0ωdv0 + ζωnx0 (1.38) where x0 and v0 are the initial displacement and velocity. A plot of x(t) versus t for this underdamped case is given in Figure 1.10. Note that the motion is oscillatory with exponentially decaying amplitude. The damping ratio ζ determines the rate of decay. The response illustrated in Figure 1.10 is exhibited in many mechanical sys- tems and constitutes the most common case. As a check to see that equation (1.38) is reasonable, note that if ζ = 0 in the expressions for A and ϕ, the undamped rela- tions of equation (1.9) result.

Time (s)

Displacement (mm)

1.0

0.0

�1.0 Figure 1.10 The response of an underdamped system: 0 6 ζ 6 1.

1.3.2 overdamped Motion

In this case, the damping ratio is greater than 1 (ζ 7 1). The discriminant of equa- tion (1.31) is positive, resulting in a pair of distinct real roots. These are

λ1 = -ζωn – ωn2ζ2 – 1 (1.39) and

λ2 = -ζωn + ωn2ζ2 – 1 (1.40)

Sec. 1.3 Viscous Damping 27

The solution of equation (1.25) then becomes

x(t) = e-ζωnt(a1e-ωn2ζ2 – 1t + a2e+ωn2ζ2 – 1t) (1.41) which represents a nonoscillatory response. Again, the constants of integration a1 and a2 are determined by the initial conditions indicated in equations (1.7) and (1.8). In this nonoscillatory case, the constants of integration are real valued and are given by

a1 = -v0 + ( -ζ + 2ζ2 – 1)ωnx0

2ωn2ζ2-1 (1.42) and

a2 = v0 + (ζ + 2ζ2 – 1)ωnx0

2ωn2ζ2-1 (1.43) Typical responses are plotted in Figure 1.11, where it is clear that motion does not involve oscillation. An overdamped system does not oscillate but rather returns to its rest position exponentially.

Displacement (mm)

1

1. x0 = 0.3, v0 = 0

3. x0 = �0.3, v0 = 0 2. x0 = 0, v0 = 1

2

3

Time (s)�0.4

�0.2

0.0

0.2

0.4

0 1 2 3 4 5 6

Figure 1.11 The response of an overdamped system, ζ 7 1, for two different values of initial displacement (in mm) both with the initial velocity set to zero and one case with x0 = 0 and v0 = 1 mm>s.

1.3.3 critically damped Motion

In this last case, the damping ratio is exactly one (ζ = 1) and the discriminant of equation (1.31) is equal to zero. This corresponds to the value of ζ that separates oscillatory motion from nonoscillatory motion. Since the roots are repeated, they have the value

λ1 = λ2 = -ωn (1.44)

The solution takes the form

x(t) = (a1 + a2t)e-ωnt (1.45)

28 Introduction to Vibration and the Free Response Chap. 1

where, again, the constants a1 and a2 are determined by the initial conditions. Substituting the initial displacement into equation (1.45) and the initial velocity into the derivative of equation (1.45) yields

a1 = x0, a2 = v0 + ωnx0 (1.46)

Critically damped motion is plotted in Figure 1.12 for two different values of initial conditions. It should be noted that critically damped systems can be thought of in several ways. They represent systems with the smallest value of damping rate that yields nonoscillatory motion. Critical damping can also be thought of as the case that separates nonoscillation from oscillation, or the value of damping that provides the fastest return to zero without oscillation.

0 0.5 1 1.5 2 2.5 3 3.5

1. x0 � 0.4 mm, v0 � � 1mm/s

3. x0 � 0.4 mm, v0 � �1mm/s 2. x0 � 0.4 mm, v0 � 0 mm/s

Time (s)

�0.1

0.2

Displacement (mm)

1

2

3

0.4

0.6

Figure 1.12 The response of a critically damped system for three different initial velocities. The physical properties are m = 100 kg, k = 225 N>m, and ζ = 1.

Example 1.3.1

Recall the small spring of Example 1.2.1 (i.e., ωn = 132 rad>s). The damping rate of the spring is measured to be 0.11 kg>s. Calculate the damping ratio and determine if the free motion of the spring–bolt system is overdamped, underdamped, or critically damped.

Solution From Example 1.2.1, m = 49.2 * 10−3 kg and k = 857.8 N>m. Using the definition of the critical damping coefficient of equation (1.29) and these values for m and k yields

ccr = 21km = 22(857.8 N>m)(49.2 * 10-3 kg) = 12.993 kg>s

If c is measured to be 0.11 kg>s, the critical damping ratio becomes

ζ = c

ccr =

0.11(kg>s) 12.993(kg>s) = 0.0085

or 0.85% damping. Since ζ is less than 1, the system is underdamped. The motion resulting from giving the spring–bolt system a small displacement will be oscillatory.

n

Sec. 1.3 Viscous Damping 29

The single-degree-of-freedom damped system of equation (1.25) is often writ- ten in a standard form. This is obtained by dividing equation (1.25) by the mass, m. This yields

x $ +

c m

x # +

k m

x = 0 (1.47)

The coefficient of x(t) is ωn2, the undamped natural frequency squared. A little manipulation illustrates that the coefficient of the velocity x

# is 2ζωn. Thus equation

(1.47) can be written as

x $ (t) + 2ζωnx

# (t) + ωn2x(t) = 0 (1.48)

In this standard form, the values of the natural frequency and the damping ratio are clear. In differential equations, equation (1.48) is said to be in monic form, meaning that the leading coefficient (coefficient of the highest derivative) is one.

Example 1.3.2

The human leg has a measured natural frequency of around 20 Hz when in its rigid (knee-locked) position in the longitudinal direction (i.e., along the length of the bone) with a damping ratio of ζ = 0.224. Calculate the response of the tip of the leg bone to an initial velocity of v0 = 0.6 m>s and zero initial displacement (this would cor- respond to the vibration induced while landing on your feet, with your knees locked from a height of 18 mm) and plot the response. last, calculate the maximum accelera- tion experienced by the leg assuming no damping.

Solution The damping ratio is ζ = 0.224 6 1, so the system is clearly underdamped.

The natural frequency is ωn = 20 1

cycles

s 2π rad cycles

= 125.66 rad>s. The damped natural

frequency is ωd = 125.6621 – (0.224)2 = 122.467 rad>s. Using equation (1.38) with v0 = 0.6 m>s and x0 = 0 yields

A = 230.6 + (0.224)(125.66)(0)42 + 3(0)(122.467)42

122.467 = 0.005 m

ϕ = tan-1 a (0)(ωd) v0 + ζωn(0)

b = 0

The response as given by equation (1.36) is

x(t) = 0.005e-28.148t sin (122.467t)

This is plotted in Figure 1.13. To find the maximum acceleration rate that the leg expe- riences for zero damping, use the undamped case of equation (1.9):

A = Bx02 + a v0ωn b2, ωn = 125.66, v0 = 0.6, x0 = 0

30 Introduction to Vibration and the Free Response Chap. 1

A = v0

ωn m =

0.6 ωn

m

max (x $ ) = 0 -ωn2 A 0 = ` -ωn2 a0.6ωn b ` = (0.6)(125.66 m>s

2) = 75.396 m>s2

In terms of g = 9.81 m>s2, this becomes

maximum acceleration = 75.396 m>s2 9.81 m>s2 g = 7.69 g>s

n

Example 1.3.3

Compute the form of the response of an underdamped system using the Cartesian form of the solution given in Window 1.5.

Solution From basic trigonometry sin(x + y) = sin x cos y + cos x sin y. Applying this to equation (1.36) with x = ωdt and y = ϕ yields

x(t) = Ae-ζωnt sin (ωdt + ϕ) = e-ζωnt(A1 sin ωdt + A2 cosωdt)

where A1 = A cos ϕ and A2 = A sin ϕ, as indicated in Window 1.5. Evaluating the initial conditions yields

x(0) = x0 = e0(A1 sin 0 + A2 cos 0)

Displacement (mm)

Time (s)�5

�4

�3

�2

�1

5

4

3

2

1

0

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14

Figure 1.13 A plot of displacement versus time for the leg bone of Example 1.3.2.

Sec. 1.4 Modeling and Energy Methods 31

Solving yields A2 = x0. Next, differentiate x(t) to get

x # = -ζωne-ζωnt (A1 sinωdt + A2 cos ωdt) + ωde-ζωnt (A1 cos ωdt – A2 sin ωdt)

Applying the initial velocity condition yields

v0 = x # (0) = -ζωn (A1 sin 0 + x0 cos 0) + ωd (A1 cos 0 – x0 sin 0)

Solving this last expression yields

A1 = v0 + ζωnx0

ωd

Thus the free response in Cartesian form becomes

x(t) = e-ζωnt av0 + ζωnx0 ωd

sin ωdt + x0 cos ωdtb

n

1.4 ModelIng and energy MetHodS

Modeling is the art or process of writing down an equation, or system of equa- tions, to describe the motion of a physical device. For example, equation (1.2) was obtained by modeling the spring–mass system of Figure 1.5. by summing the forces acting on the mass along the x direction and employing the experimental evidence of the mathematical model of the force in a spring given by Figure 1.4, equation (1.2) can be obtained. The success of this model is determined by how well the solution of equation (1.2) predicts the observed and measured behavior of the actual system. This comparison between the vibration response of a device and the response predicted by the analytical model is discussed in Section 1.6. The majority of this book is devoted to the analysis of vibration models. However, two methods of modeling—force balance and energy methods—are presented in this section. Newton’s three laws form the basis of dynamics. Fifty years after Newton, Euler published his laws of motion. Newton’s second law states: the sum of forces acting on a body is equal to the body’s mass times its acceleration, and Euler’s second law states: the rate of change of angular momentum is equal to the sum of external moments acting on the mass. Euler’s second law can be manipulated to reveal that the sum of moments acting on a mass is equal to its rotational in- ertia times its angular acceleration. These two laws require the use of free-body diagrams and the proper identification of forces and moments acting on a body, forming most of the activity in the study of dynamics.

An alternative approach, studied in dynamics, is to examine the energy in the system, giving rise to what is referred to as energy methods for determining the equations of motion. The energy methods do not require free-body diagrams

32 Introduction to Vibration and the Free Response Chap. 1

but rather require an understanding of the energy in a system, providing a useful alternative when forces are not easy to determine. More comprehensive treatments of modeling can be found in Doebelin (1980), Shames (1980, 1989), and Cannon (1967), for example. The best reference for modeling is the text you used to study dynamics. There are also many excellent descriptions on the Internet which can be found using a search engine such as Google.

The force summation method is used in the previous sections and should be familiar to the reader from introductory dynamics. For systems with constant mass (such as those considered here) moving in only one direction, the rate of change of momentum becomes the scalar relation

d dt

(mx # ) = mx$

which is often called the inertia force. The physical device of interest is examined by noting the forces acting on the device. The forces are then summed (as vectors) to produce a dynamic equation following Newton’s second law. For motion along the x direction only, this becomes the scalar equation

a i

fxi = mx $ (1.49)

where fxi denotes the ith force acting on the mass m along the x direction and the summation is over the number of such forces. In the first three chapters, only single- degree-of-freedom systems moving in one direction are considered; thus, Newton’s law takes on a scalar nature. In more practical problems with many degrees of freedom, energy considerations can be combined with the concepts of virtual work to produce lagrange’s equations, as discussed in Section 4.7. lagrange’s equations also provide an energy-based alternative to summing forces to derive equations of motion.

For rigid bodies in plane motion (i.e., rigid bodies for which all the forces act- ing on them are coplanar in a plane perpendicular to a principal axis) and free to rotate, Euler’s second law states that the sum of the applied torques is equal to the rate of change of angular momentum of the mass. This is expressed as

a i

M0i = Jθ $ (1.50)

where M0i are the torques acting on the object about the point 0, J is the moment of inertia (also denoted by I0) about the rotation axis, and θ is the angle of rotation. The sum of moments method was used in Example 1.1.1 to find the equation of motion of a pendulum and is discussed in more detail in Example 1.5.1.

If the forces or torques acting on an object or mechanical part are difficult to determine, an energy approach may be more efficient. In this method, the differen- tial equation of motion is established by using the principle of energy conservation. This principle is equivalent to Newton’s law for conservative systems and states that

Sec. 1.4 Modeling and Energy Methods 33

the sum of the potential energy and kinetic energy of a particle remains constant at each instant of time throughout the particle’s motion:

T + U = constant (1.51)

where T and U denote the total kinetic and potential energy, respectively. Conservation of energy also implies that the change in kinetic energy must equal the change in potential energy:

U1 – U2 = T2 – T1 (1.52)

where U1 and U2 represent the particle’s potential energy at the times t1 and t2, respectively, and T1 and T2 represent the particle’s kinetic energy at times t1 and t2, respectively. For periodic motion, energy conservation also implies that

Tmax = Umax (1.53)

Since energy is a scalar quantity, using the conservation of energy principle yields a possibility of obtaining the equation of motion of a system without using force or moment summations.

Equations (1.51), (1.52), and (1.53) are three statements of the conservation of energy. Each of these can be used to determine the equation of motion of a spring– mass system. As an illustration, consider the energy of the spring–mass system of Figure 1.14 hanging in a gravitational field of strength g. The effect of adding the mass m to the massless spring of stiffness k is to stretch the spring from its rest position at 0 to the static equilibrium position Δ. The total potential energy of the spring–mass system is the sum of the potential energy of the spring (or strain energy; see, e.g., Shames, 1989) and the gravitational potential energy. The potential energy of the spring is given by

Uspring = 1 2 k(Δ + x)2 (1.54)

The gravitational potential energy is

Ugray = -mgx (1.55)

where the negative sign indicates that x = 0 is the reference for zero potential energy. The kinetic energy of the system is

T = 12 mx # 2 (1.56)

g

mg

(b)(a)

k k�

x(t)

m m �

0

Figure 1.14 (a) A spring–mass system hanging in a gravitational field. Here Δ is the static equilibrium position and x is the displacement from equilibrium. (b) The free-body diagram for static equilibrium.

34 Introduction to Vibration and the Free Response Chap. 1

Substituting these energy expressions into equation (1.51) yields

12 mx # 2 – mgx + 12 k(Δ + x)2 = constant (1.57)

Differentiating this expression with respect to time yields

x # (mx

$ + kx) + x# (kΔ – mg) = 0 (1.58)

Since the static force balance on the mass from Figure 1.14(b) yields the fact that kΔ = mg, equation (1.58) becomes

x # (mx

$ + kx) = 0 (1.59)

The velocity x # cannot be zero for all time; otherwise, x(t) = constant and no vibra-

tion would be possible. Hence equation (1.59) yields the standard equation of motion

mx $ + kx = 0 (1.60)

This procedure is called the energy method of obtaining the equation of motion. The energy method can also be used to obtain the frequency of vibration

directly for conservative systems that are oscillatory. The maximum value of sine (and cosine) is one. Hence, from equations (1.3) and (1.4), the maximum displace- ment is A and the maximum velocity is ωnA (recall Window 1.3). Substitution of these maximum values into the expression for Umax and Tmax and using the energy equation (1.53) yields

12 m(ωnA) 2 = 12 kA

2 (1.61)

Solving equation (1.61) for ωn yields the standard natural frequency relation ωn = 1k>m. Example 1.4.1

Figure 1.15 is a simple single-degree-of-freedom model of a wheel mounted on a spring. The friction in the system is such that the wheel rolls without slipping. Calculate the natural frequency of oscillation using the energy method. Assume that no energy is lost during the contact.

x(t)

r k

m, J

Figure 1.15 The rotational displacement of the wheel of radius r is given by θ(t) and the linear displacement is denoted by x(t). The wheel has a mass m and a moment of inertia J. The spring has a stiffness k.

Solution From introductory dynamics, the rotational kinetic energy of the wheel is T rot =

1 2 Jθ

# 2, where J is the mass moment of inertia of the wheel and θ = θ(t) is the angle of rotation of the wheel. This assumes that the wheel moves relative to the surface without slipping (so that no energy is lost at contact). The translational kinetic energy of the wheel is TT =

1 2 mx

# 2.

Sec. 1.4 Modeling and Energy Methods 35

The rotation θ and the translation x are related by x = rθ. Thus x# = rθ # and

T rot = 1 2 Jx

# 2>r 2 . At maximum energy x = A and x# = ωn A, so that

Tmax = 1 2

mx # max 2 +

1 2

J

r 2 x # max 2 =

1 2

(m + J>r 2)ωn2A2

and

Umax = 1 2 kxmax

2 = 12 kA 2

Using conservation of energy in the form of equation (1.53) yields Tmax = Umax, or

1 2

am + J r 2 bωn2 = 12 k

Solving this last expression for ωn yields

ωn = A km + J>r 2 the desired frequency of oscillation of the suspension system.

The denominator in the frequency expression derived in this example is called the effective mass because the term (m + J>r2) has the same effect on the natural fre- quency as does a mass of value (m + J>r2).

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