Vibrations Fundamentals and Practice ch02

Vibrations Fundamentals and Practice ch02 Maintaining the outstanding features and practical approach that led the bestselling first edition to become a standard textbook in engineering classrooms worldwide, Clarence de Silva''s Vibration: Fundamentals and Practice, Second Edition remains a solid instructional tool for modeling, analyzing, simulating, measuring, monitoring, testing, controlling, and designing for vibration in engineering systems. It condenses the author''s distinguished and extensive experience into an easy-to-use, highly practical text that prepares students for real problems in a variety of engineering fields.

de Silva, Clarence W “Time Response”

Vibration: Fundamentals and Practice

Clarence W de Silva

Boca Raton: CRC Press LLC, 2000

2 Time Response

Vibrations are oscillatory responses of dynamic systems Natural vibrations occur in these systemsdue to the presence of two modes of energy storage Specifically, when the stored energy is convertedfrom one form to the other, repeatedly back and forth, the resulting time response of the system

is oscillatory in nature In a mechanical system, natural vibrations can occur because kinetic energy,which is manifested as velocities of mass (inertia) elements, can be converted into potential energy(which has two basic types: elastic potential energy due to the deformation in spring-like elements,and gravitational potential energy due to the elevation of mass elements against the Earth’s grav-itational pull) and back to kinetic energy, repetitively, during motion Similarly, natural oscillations

of electrical signals occur in circuits due to the presence of electrostatic energy (of the electriccharge storage in capacitor-like elements) and electromagnetic energy (due to the magnetic fields

in inductor-like elements) Fluid systems can also exhibit natural oscillatory responses as theypossess two forms of energy But purely thermal systems do not produce natural oscillations becausethey, as far as anyone knows, have only one type of energy These ideas are summarized in Appendix

A Note, however, that an oscillatory forcing function is able to make a dynamic system respondwith an oscillatory motion (usually at the same frequency as the forcing excitation) even in theabsence of two forms of energy storage Such motions are forced responses rather than natural orfree responses This book concerns vibrations in mechanical systems Nevertheless, clear analogiesexist with electrical and fluid systems as well as mixed systems such as electromechanical systems.Mechanical vibrations can occur as both free (natural) responses and forced responses innumerous practical situations Some of these vibrations are desirable and useful, and others areundesirable and should be avoided or suppressed The sound that is generated after a string of aguitar is plucked is a free vibration, while the sound of a violin is a mixture of both free and forcedvibrations These sounds are generally pleasant and desirable The response of an automobile after

it hits a road bump is an undesirable free vibration The vibrations felt while operating a concretedrill are desirable for the drilling process itself, but are undesirable forced vibrations for the humanwho operates the drill In the design and development of a mechanical system, regardless of whether

it is intended for generating desirable vibrations or for operating without vibrations, an analyticalmodel of the system can serve a very useful function The model will represent the dynamic system,and can be analyzed and modified more quickly and cost effectively than one could build and test

a physical prototype Similarly, in the control or suppression of vibrations, it is possible to design,develop, and evaluate vibration isolators and control schemes through analytical means before theyare physically implemented It follows that analytical models (see Appendix A) are useful in theanalysis, control, and evaluation of vibrations in dynamic systems, and also in the design anddevelopment of dynamic systems for desired performance in vibration environments

An analytical model of a mechanical system is a set of equations, and can be developed either

by the Newtonian approach where Newton’s second law is explicitly applied to each inertia element,

or by the Lagrangian or Hamiltonian approach, which is based on the concepts of energy (kineticand potential energies) These approaches are summarized in Appendix B A time-domain analyticalmodel is a set of differential equations, with respect to the independent variable time (t) Afrequency-domain model is a set of input-output transfer functions with respect to the independentvariable frequency (ω) The time response will describe how the system moves (responds) as afunction of time Both free and forced responses are useful The frequency response will describethe way the system moves when excited by a harmonic (sinusoidal) forcing input, and is a function

of the frequency of excitation This chapter introduces some basic concepts of vibration analysis

using time-domain methods The frequency-domain analysis will be studied in subsequent chapters(Chapters 3 and 4, in particular).

2.1 UNDAMPED OSCILLATOR

Consider the mechanical system that is schematically shown in Figure 2.1 The inputs (or excitation)applied to the system are represented by the force f(t) The outputs (or response) of the system arerepresented by the displacement y The system boundary demarcates the region of interest in thisanalysis This boundary could be an imaginary one What is outside the system boundary is theenvironment in which the system operates An analytical model of the system can be given by one

or more equations relating the outputs to the inputs If the rates of changes of the response (outputs)are not negligible, the system is a dynamic system In this case, the analytical model in the timedomain becomes one or more differential equations rather than algebraic equations System param-eters (e.g., mass, stiffness, damping constant) are represented in the model, and their values should

be known in order to determine the response of the system to a particular excitation State variablesare a minimum set of variables that completely represent the dynamic state of a system at anygiven time t These variables are not unique (more than one choice of a valid set of state variables

is possible) The concepts of state variables and state models are introduced in Appendix A andalso in this chapter For a simple oscillator (a single-degree-of-freedom mass-spring-damper system

as in Figure 2.1), an appropriate set of state variables would be the displacement y and the velocity An alternative set would be and the spring force

This chapter provides an introduction to the response analysis of mechanical vibrating systems inthe time domain In this introductory chapter, single-degree-of-freedom systems that require only onecoordinate (or one independent displacement variable) in their model, are considered almost exclusively.Higher-degree-of-freedom systems will be analyzed elsewhere in the book (e.g., Chapter 5) Mass(inertia) and spring are the two basic energy storage elements in a mechanical vibrating system Amass can store gravitational potential energy as well when located against a gravitational force Theseelements are analyzed first In a practical system, mass and stiffness properties can be distributed(continuous) throughout the system But in this present analysis, lumped-parameter models areemployed where inertia, flexibility, and damping effects are separately lumped into single parameters,with a single geometric coordinate used to represent the location of each lumped inertia

This chapter section first shows that many types of oscillatory systems can be represented bythe equation of an undamped simple oscillator In particular, mechanical, electrical, and fluid systemsare considered Please refer to Appendix A for some foundation material on this topic The conser-vation of energy is a straightforward approach for deriving the equations of motion for undampedoscillatory systems (or conservative systems) The equations of motion for mechanical systems can

FIGURE 2.1 A mechanical dynamic system.

be derived using the free-body diagram approach with the direct application of Newton’s secondlaw An alternative and rather convenient approach is the use of Lagrange equations, as described

in Appendix B The natural (free) response of an undamped simple oscillator is a simple harmonic motion This is a periodic, sinusoidal motion This simple time response is also discussed

2.1.1 E NERGY S TORAGE E LEMENTS

Mass (inertia) and spring are the two basic energy storage elements in mechanical systems Theconcept of state variables can be introduced as well through these elements (see Appendix A fordetails), and will be introduced along with associated energy and state variables

∫

Since the integral of a finite quantity over an almost zero time interval is zero, these results implythat a finite force will not cause an instantaneous change in velocity in an inertia element Inparticular, for a mass element subjected to finite force, since the integral on the RHS of equation(2.4) is zero, one obtains

(2.5)

Spring (k)

Consider a massless spring element of lumped stiffness k, as shown in Figure 2.3 One end of thespring is fixed and the other end is free A force f is applied at the free end, which results in adisplacement (extension) x in the spring

Hooke’s law gives

∫

From these results, it follows that at finite velocities, there cannot be an instantaneous change inthe force of a spring In particular, from equation (2.9) one sees that at finite velocities of a spring

(2.10)Also, it follows that

(2.11)

Gravitational Potential Energy

The work done in raising an object against the gravitational pull is stored as gravitational potentialenergy of the object Consider a lumped mass m, as shown in Figure 2.4, that is raised to a height

y from some reference level The work done gives

Hence,

(2.12)

2.1.2 C ONSERVATION OF E NERGY

There is no energy dissipation in undamped systems, which contain energy storage elements only

In other words, energy is conserved in these systems, which are known as conservative systems.For mechanical systems, conservation of energy gives

(2.13)These systems tend to be oscillatory in their natural motion, as noted before Also, as discussed inAppendix A, analogies exist with other types of systems (e.g., fluid and electrical systems) Considerthe six systems sketched in Figure 2.5

System 1 (Translatory)

Figure 2.5 (a) shows a translatory mechanical system (an undamped oscillator) that has just onedegree of freedom x This can represent a simplified model of a rail car that is impacting against

a snubber The conservation of energy (equation (2.13)) gives

FIGURE 2.4 A mass element subjected to gravity.

f( )0+ = f( )0−

x( )0+ =x( )0−

Energy : E=∫ fdy=∫mgdy

Gravitational potential energy : PE=mgy

KE+PE=const

Here, m is the mass and k is the spring stiffness Differentiate equation (2.14) with respect to time t

to obtain

Since ≠ 0 at all t, in general, one can cancel it out Hence, by the method of conservation

of energy, one obtains the equation of motion

(2.15)

System 2 (Rotatory)

Figure 2.5(b) shows a rotational system with the single degree of freedom θ It may represent a

simplified model of a motor drive system As before, the conservation energy gives

(2.16)

In this equation, J is the moment of inertia of the rotational element and K is the torsional stiffness

of the shaft Then, by differentiating equation (2.16) with respect to t and canceling , one obtains

the equation of motion

(2.17)

System 3 (Flexural)

Figure 2.5(c) is a lateral bending (flexural) system, which is a simplified model of a building

structure Again, a single degree of freedom x is assumed Conservation of energy gives

(2.18)

Here, m is the lumped mass at the free end of the support and k is the lateral bending stiffness of

the support structure Then, as before, the equation of motion becomes

(2.19)

System 4 (Swinging)

Figure 2.5(d) shows a simple pendulum It may represent a swinging-type building demolisher or

a skilift and has a single-degree-of-freedom θ Thus,

12

12

12

12

2

˙cos

θ

θ

Gravitational

Here, m is the pendulum mass, l is the pendulum length, g is the acceleration due to gravity, and

Eref is the PE at the reference point, which is a constant Hence, conservation of energy gives

(2.20)

Differentiate with respect to t after canceling the common ml:

Since ≠ 0 at all t, the equation of motion becomes

(2.21)

This system is nonlinear, in view of the term sin θ For small θ, sin θ is approximately equal to θ.Hence, the linearized equation of motion is

(2.22)

FIGURE 2.5 Six examples of single D.O.F oscillatory systems: (a) translatory, (b) rotatory, (c) flexural,

(d) swinging, (e) liquid slosh, and (f) electrical.

12

System 5 (Liquid Slosh)

Consider a liquid column system shown in Figure 2.5(e) It may represent two liquid tanks linked

by a pipeline The system parameters are

Area of cross section of each column = A

Mass density of liquid = ρ

Length of liquid mass = l

Then,

Note that the center of gravity of each column is used in expressing the gravitational PE Hence,

conservation of energy gives

(2.23)Differentiate:

But, one has

12

12

The constitutive equation for the inductor is

(2.26)The constitutive equation for the capacitor is

Now consider the energy conservation approach for this electrical circuit, which will give the

same result Note that power is given by the product vi.

Capacitor

(2.29)

Here, v denotes vC Also,

(2.30)

Since the current i is finite for a practical circuit, then

Hence, in general, the voltage across a capacitor cannot change instantaneously In particular,

dv dt

dv dt

i C

Here, v denotes v L Also,

(2.32)

Since v is finite in a practical circuit, then

Hence, in general, the current through an inductor cannot change instantaneously In particular,

(2.33)

Since the circuit in Figure 2.5(f) does not have a resistor, there is no energy dissipation As a result,conservation energy gives

(2.34)

Differentiate equation (2.34) with respect to t.

Note that v = v c in this equation

Substitute the capacitor constitutive equation (2.27)

Since i ≠ 0 in general, one can cancel it Now, by differentiating equation (2.27), one has Substitute this in the above equation to obtain

(2.35)Similarly, one obtains

iv Li di dt

di

dt C

d v dt

This is the equation of an undamped, simple oscillator For a mechanical system of mass m and stiffness k,

This equation is identically satisfied for all t Hence, the general solution of equation (2.37) is

indeed equation (2.39), which is periodic and sinusoidal

This response is sketched in Figure 2.6 Note that this sinusoidal oscillatory motion has a

frequency of oscillation of ω rad/s) Hence, a system that provides this type of natural motion is called a simple oscillator In other words, the response exactly repeats itself in time periods of T,

corresponding to a cyclic frequency (Hz) The frequency ω is in fact the angular frequency

given by ω = 2πf Also, the response has an amplitude A, which is the peak value of the sinusoidal

response Now, suppose that the response curve is shifted to the right through φ/ω Consider the

resulting curve to be the reference signal (with signal value = 0 at t = 0, and increasing) It should

be clear that the response shown in Figure 2.6 leads the reference signal by a time period of φ/ω

This can be verified from the fact that the value of the reference signal at time t is the same as that

of the signal in Figure 2.6 at time t – φ/ω Hence, φ is termed the phase angle of the response, and

it represents a phase lead

The left-hand portion of Figure 2.6 is the phasor representation of a sinusoidal response In

this representation, an arm of length A rotates in the counterclockwise direction at angular speed

ω This is the phasor The arm starts at an angular position φ from the horizontal axis, at time t = 0 The projection of the arm onto the vertical (x) axis is the time response In this manner, the phasor

representation can conveniently indicate the amplitude, frequency, phase angle, and the actual time

response (at any time t) of a sinusoidal motion.

As noted previously, a repetitive (periodic) motion of the type (2.39) is called simple harmonic

motion, meaning it is a pure sinusoidal oscillation at a single frequency.

Next, it is shown that the amplitude A and the phase angle φ both depend on the initial conditions.Substitute the ICs (2.40) into equation (2.39) and its time derivative to get

(2.41)(2.42)

ωn k m

f T

= 1

x o=Asinφ

v o= ωA ncosφ

Now divide equation (2.41) by (2.42), and also use the fact that sin2φ + cos2φ = 1 and obtain

Hence,

(2.43)

(2.44)

E XAMPLE 2.1

A simple model for a tracked gantry conveyor system in a factory is shown in Figure 2.7

The carriage of mass (m) moves on a frictionless track The pulley is supported on frictionless bearings, and its axis of rotation is fixed Its moment of inertia about this axis is J The motion of the carriage is restrained by a spring of stiffness k1, as shown The belt segment that drives thecarriage runs over the pulley without slip, and is attached at the other end to a fixed spring of

stiffness k2 The displacement of the mass is denoted by x and the corresponding rotation of the

pulley is denoted by θ When x = 0 (and θ = 0), the springs k1 and k2 have an extension of x10 and

x20, respectively, from their unstretched (free) configurations Assume that the springs will remain

in tension throughout the motion of the system

a Using Newton’s second law, first principles, and free-body diagrams, develop an

equiv-alent equation of motion for this system in terms of the response variable x What is the

equivalent mass, and what is the equivalent stiffness of the system?

b Verify the result in part (a) using the energy method

FIGURE 2.6 Free response of an undamped simple oscillator.

o

n

x v x

A

v A

ω

Phase : φ=tan− 1ωn o

o x v

c What is the natural frequency of vibration of the system?

d Express the equation of the system in terms of the rotational response variable θ

What is the natural frequency of vibration corresponding to this rotational form of the systemequation?

What is the equivalent moment of inertia and the equivalent torsional stiffness of the rotationalform of the system?

S OLUTION

A free-body diagram for the system is shown in Figure 2.8

a Hooke’s law for the spring elements:

(i)

(ii)

Newton’s second law for the inertia elements:

(iii) (iv)

Compatibility:

FIGURE 2.7 A tracked conveyor system.

FIGURE 2.8 A free-body diagram for the conveyor system.

F1=k x1( 10+x)

F2=k x2( 20−x)

mx˙˙= −F F1

J˙˙θ =rF2−rF

(v) Straightforward elimination of F1, F2, F, and θ in (i) to (v), using algebra, gives

(vi)

It follows that

b Total energy in the system:

Differentiate w.r.t time:

Substitute the compatibility relation, = r , to get

Eliminate the common velocity variable (which cannot be zero for all t) Obtain

which is the same result as before

c The natural frequency (undamped) of the system is

(vii)

d Substitute for x and its derivatives into (vi) using the compatibility condition (v) to obtain

The natural frequency

12

12

is identical to the previous answer (vii) This is to be expected, as the system has not

changed (only the response variable was changed)



Common approaches of developing equations of motion for mechanical systems are summarized

in Box 2.1

BOX 2.1 Approaches for Developing Equations of Motion

1 Conservative Systems (No Nonconservative Forces/No Energy Dissipation):

Kinetic energy = T

Potential energy = V

Conservation of energy: T + V = const

Differentiate with respect to time t

2 Lagrange’s Equations:

Lagrangian L = T – V

n = number of degrees of freedom

Qi = generalized force corresponding to generalized coordinate q i

Find Q i using: δW = Σ Q iδqi

where δW = work done by nonconservative forces in a general incremental motion

Torques Angular momentum (About centriod or a fixed point)

=

=

d dt d dt

2.2 HEAVY SPRINGS

A heavy spring has its mass and flexibility properties continuously distributed throughout its body

In that sense, it has an infinite number of degrees of freedom, and a single coordinate cannotrepresent its motion However, for many practical purposes, a lumped-parameter approximationwith just one lumped mass to represent the inertial characteristics of the spring may be sufficient.Such an approximation can be obtained using the energy approach Here, the spring is represented

by a lumped-parameter “model” such that the original spring and the model have the same net

kinetic energy and potential energy This energy equivalence is used in deriving a lumped mass

parameter for the model Although damping (energy dissipation) is neglected in the present analysis,

it is not difficult to incorporate that as well in the model

2.2.1 K INETIC E NERGY E QUIVALENCE

Consider the uniform, heavy spring shown in Figure 2.9, with one end fixed and the other end

moving at velocity v Note that:

Equivalent lumped mass concentrated at the free end spring mass

Note: This derivation assumes that one end of the spring is fixed and, furthermore, that the conditions

are uniform along the spring.

An example of utilizing this result is shown in Figure 2.10 Here, a system with a heavy springand a lumped mass is approximated by a light spring (having the same stiffness) and a lumped mass

FIGURE 2.9 A uniform heavy spring.

x

l v m

1

3 2

2

= ×1

3

Another example is shown in Figure 2.11 In this case, it is not immediately clear which ofthe approximations shown on the right-hand side is most appropriate.

E XAMPLE 2.2

A uniform heavy spring of mass m s and stiffness k is attached at one end to a mass m that is free

to roll on a frictionless horizontal plane The other end is anchored to a vertical post A schematicdiagram of this arrangement is shown in Figure 2.12

The unstretched length of the spring is l Assume that when the velocity of the connected mass

is v, the velocity distribution along the spring is given by

FIGURE 2.10 Lumped-parameter approximation for an oscillator with heavy spring.

FIGURE 2.11 An example where the lumped-parameter approximation for a spring is ambiguous.

FIGURE 2.12 A heavy spring connected to a rolling stock.

where x is the distance of a point along the spring, as measured from the fixed end Determine an equivalent lumped mass located at the moving end of the spring (i.e., at the moving mass m) to

represent the inertia effects of the spring What are the limitations of your result?

S OLUTION

Consider an element of length δx at location x of the spring Since the spring is uniform, the element

mass = Also, according to the given assumption, the element velocity = Hence,the kinetic energy of the spring is

It follows that the equivalent lumped mass to be located at the moving end of the spring is This result is valid only for the assumed velocity distribution, and corresponds to the first mode

of motion only In fact, a linear velocity distribution would be more realistic in this low-frequency(quasi-static motion) region, which will give an equivalent lumped mass of , as seen before.Such approximations will not be valid for high frequencies (say, higher than )



2.3 OSCILLATIONS IN FLUID SYSTEMS

As discussed in Appendix A, fluid systems can undergo oscillations (vibrations) quite analogous

to mechanical and electrical systems Again, the reason for their natural oscillation is the ability

to store and repeatedly interchange two types of energy — kinetic energy and potential energy.The kinetic energy comes from the velocity of fluid particles during motion The potential energyarises primarily from the following three main sources

1 Gravitational potential energy

2 Compressibility of the fluid volume

3 Flexibility of the fluid container

A detailed analysis of these three effects is not undertaken here However, one sees from theexample in Figure 2.5(e) how a liquid column can oscillate due to repeated interchange betweenkinetic energy and gravitational potential energy Now consider another example

14

s l

s l

s l

E XAMPLE 2.3

A university laboratory has developed a procedure for optimal cutting (portion control) of fish forcan filling, with the objective of minimizing the wastage (overfill) and regulatory violations (under-fill) The procedure depends on the knowledge of the volumetric distribution of a dressed (cleaned)fish In fact, a group of volumeteric models is developed through off-line experimentation so thatextensive measurements need not be made on-line, during processing One set of such off-lineexperiments consists of dipping a fish into a tank of water in fixed increments and measuring thevolume of water that is displaced An illustration of the experimental setup is given in Figure 2.13(a)

One day, an adventurous student decided to try a different test with the experimental system

Instead of a fish, he used a cylindrical wooden peg of uniform cross section and height h Realizing

that the object could not be completely immersed in water, he pushed it down by hand, in theupright orientation (see Figure 2.13(b)) The object oscillated up and down while floating in thetank Let ρb and ρl be the mass densities of the body (peg) and the liquid (water), respectively

a Clearly stating the assumptions that are made, obtain an expression for the naturalfrequency of oscillations

b If the object is slightly tilted to one side would it return to its upright configuration?Explain

(a)

(b)

FIGURE 2.13 (a) An experimental system for determining the volumetric distribution of a fish body.

(b) Buoyancy experiment.

S OLUTION

a Suppose that, under equilibrium in the upright position of the body, the submersed length

is l The mass of the body is

(i)

where A is the area of cross section (uniform).

By Archimedes principle, the buoyancy force R is equal to the weight of the liquid

displaced by the body Hence,

Substitute equations (ii) and (iii) to obtain

Substitute equation (i):

=ρρ

b

Note that this result is independent of the area of cross section of the body.

Assumptions:

1 The tank is very large compared to the body The change in liquid level is negligible

as the body is depressed into the water

2 Fluid resistance (viscous effects, drag, etc.) is negligible

3 Dynamics of the liquid itself are negligible Hence, “added inertia” due to liquidmotion is neglected

The buoyancy force R acts through the centroid of the volume of displaced water [Figure2.14(b)]. Its line of action passes through the central axis of the body at point M The point is known as the metacenter Let C be the centroid of the body.

FIGURE 2.14 (a) Upright oscillations of the body (b) Restoring buoyancy couple due to a stable metacenter.

FIGURE 2.15 A damped simple oscillator and its free-body diagram.

ρ

n l

b

g h

=

If M is above C, then, when tilted, there will be a restoring couple that will tend to restore the

body to its upright position Otherwise, the body will be in an unstable situation, and the buoyancycouple will tend to tilt it further toward a horizontal configuration



2.4 DAMPED SIMPLE OSCILLATOR

Now consider the free (natural) response of a simple oscillator in the presence of energy dissipation(damping)

Assume viscous damping, and consider the oscillator shown in Figure 2.15 The free-bodydiagram of the mass is shown separately

The following notation is used in this book

ωn = undamped natural frequency

ωd = damped natural frequency

ωr = resonant frequency

ω = frequency of excitation

The concept of resonant frequency will be addressed in Chapter 3

The viscous damping constant is denoted by b (but sometimes c will be used instead of b, as

done in some literature)

Apply Newton’s second law From the free-body diagram in Figure 2.15, one has the equation

of motion

or

(2.46)or

Also note that ζ is called the damping ratio The formal definition and the rationale for this

terminology will be discussed later

Assume an exponential solution:

(2.50)

This is justified by the fact that linear systems have exponential or oscillatory (i.e., complexexponential) free responses A more detailed justification will be provided later

Substitute equation (2.50) into (2.47) to obtain

Note that Ceλt is not zero in general It follows that when λ satisfies the equation

(2.51)

then equation (2.50) will represent a solution of equation (2.47)

Equation (2.51) is called the characteristic equation of the system This equation depends on

the natural dynamics of the system, not forcing excitation or initial conditions

Solution of equation (2.51) gives the two roots:

(2.52)

These are called eigenvalues or poles of the system.

When λ1≠λ2, the general solution is

(2.53)

The two unknown constants C1 and C2 are related to the integration constants, and can be determined

by two initial conditions, which should be known

If λ1 = λ2 = λ; one has the case of repeated roots In this case, the general solution (2.53) does

not hold because C1 and C2 would no longer be independent constants, to be determined by twoinitial conditions The repetition of the roots suggests that one term of the homogenous solution

should have the multiplier t (a result of the double-integration of zero) Then, the general solution is

(2.54)One can identify three categories of damping level, as discussed below, and the nature of theresponse will depend on the particular category of damping

ζ =1

2

b km

2.4.1 C ASE 1: U NDERDAMPED M OTION (ζ < 1)

In this case, it follows from equation (2.52) that the roots of the characteristic equation are

(2.55)

where, the damped natural frequency is given by

(2.56)Note that λ1 and λ2 are complex conjugates The response (2.53) in this case can be expressed as

(2.57)

The term within the square brackets of equation (2.57) has to be real because it represents the time

response of a real physical system It follows that C1 and C2, as well, have to be complex conjugates

˙x

x o =A1

Note that the response x → 0 as t →∞ This means the system is asymptotically stable.

2.4.2 L OGARITHMIC D ECREMENT M ETHOD

The damping ratio ζ can be experimentally determined from the free response by the logarithmicdecrement method To illustrate this approach, note from equation (2.63) that the period of dampedoscillations is

1 2 2 2

2 1 2 2 2

1 2 and

T d

= 2πω

t d

t nT

d

( )+

sinsin

sin[ωd(t+nT)+φ]=sin(ωd t+ +φ 2nπ)=sin(ωd+φ)

Take the natural logarithm of equation (2.67), the logarithmic decrement:

This is the basis of the logarithmic decrement method of measuring damping Start by measuring

a point x(t) and another point x(t + nT) at n cycles later For high accuracy, pick the peak points

of the response curve for the measurement of x(t) and x(t + nT) From equation (2.68), it is clear

that for small damping, ζ = α = per-radian logarithmic decrement

2.4.3 C ASE 2: O VERDAMPED M OTION (ζ > 1)

In this case, roots λ1 and λ2 of the characteristic equation (2.51) are real Specifically,

(2.70)(2.71)and the response (2.53) is nonoscillatory Also, it should be clear from equations (2.70) and (2.71)that both λ1 and λ2 are negative Hence, x → 0 as t →∞ This means the system is asymptotically stable.From the initial conditions

2 2 x t x t( ) ( +nT)=r

2

π ζζ

− = nlnr=

α

=+

2 21

(ii)

Then subtract (iii) from (ii) to obtain

Similarly, multiply the first IC(i) by λ2 and subtract from (ii) One obtains

Hence,

(2.73)

2.4.4 C ASE 3: C RITICALLY D AMPED M OTION (ζ = 1)

Here, we have repeated roots, given by

(2.74)The response for this case is given by (see equation (2.54))

(2.75)Since the term goes to zero faster than t goes to infinity, one has

Hence, the system is asymptotically stable

Now use the initial conditions x(0) = x o, (0) = vo One obtains

Hence,

(2.76)(2.77)

Note: When ζ = 1, one has the critically damped response because below this value, the response

is oscillatory (underdamped), and above this value, the response is nonoscillatory (overdamped)

It follows that one can define the damping ratio as

(2.78)

2.4.5 J USTIFICATION FOR THE T RIAL S OLUTION

In the present analysis, the trial solution (2.50) has been used for the response of a linear systemhaving constant parameter values A justification for this is provided now

First-Order System

Consider a first-order linear system given by (homogeneous, no forcing input)

(2.79)This equation can by written as

Integrate:

Here, ln C is the constant of integration Hence,

(2.80)This is then the general form of the free response of a first-order system It incorporates one constant

of integration and, hence, will need one initial condition

Second Order System

One can write the equation of a general second-order (homogenous, unforced) system in theoperational form

(2.81)

By reasoning as before, the general solution would be of the form Here, C1 and

C2 are the constants of integration, which are determined using two initial conditions

Repeated Roots

The case of repeated roots deserves a separate treatment First consider

(2.82)

Damping constant for critically damped condition

2

2 =0

Integrate twice: (2.83)

Note the term with t in this case Hence, a suitable trial solution for the system

(2.84)

The main results for free (natural) response of a damped oscillator are given in Box 2.2

2.4.6 S TABILITY AND S PEED OF R ESPONSE

The free response of a dynamic system (particularly a vibrating system) can provide valuableinformation concerning the natural characteristics of the system The free (unforced) excitation can

be obtained, for example, by giving an initial-condition excitation to the system and then allowing

it to respond freely Two important characteristics that can be determined in this manner are:

1 Stability

2 Speed of response

The stability of a system implies that the response will not grow without bounds when the excitationforce itself is finite This is known as bounded-input-bounded-output (BIBO) stability In particular,

if the free response eventually decays to zero, in the absence of a forcing input, the system is said

to be asymptotically stable It was shown that a damped simple oscillator is asymptotically stable,but an undamped oscillator, while being stable in a general (BIBO) sense, is not asymptoticallystable It is marginally stable

Speed of response of a system indicates how fast the system responds to an excitation force

It is also a measure of how fast the free response (1) rises or falls if the system is oscillatory; or(2) decays, if the system is non-oscillatory Hence, the two characteristics — stability and speed

of response — are not completely independent In particular, for non-oscillatory (overdamped)systems, these two properties are very closely related It is clear then, that stability and speed ofresponse are important considerations in the analysis, design, and control of vibrating systems.The level of stability of a linear dynamic system depends on the real parts of the eigenvalues(or poles, which are the roots of the characteristic equations) Specifically, if all the roots have realparts that are negative, then the system is stable Also, the more negative the real part of a pole,the faster the decay of the free response component corresponding to that pole The inverse of thenegative real part is the time constant Hence, the smaller the time constant, the faster the decay

of the corresponding free response and, hence, the higher the level of stability associated with thatpole One can summarize these observations as follows:

Level of stability: Depends on decay rate of free response (and hence on time constants

or real parts of poles)Speed of response: Depends on natural frequency and damping for oscillatory systems and

decay rate for non-oscillatory systemsTime constant: Determines stability and decay rate of free response (and speed of

response in non-oscillatory systems)Now consider the specific case of a damped simple oscillator given by equation (2.47)

dx

dt =C; x=Ct+D

d dt

BOX 2.2 Free (Natural) Response of a Damped Simple Oscillator

Poles are complex conjugates: –ζωn± jωd

Damped natural frequency

ICs give:

Logarithmic decrement per radian:

where = decay ratio over n complete cycles.

For small

Case 2: Overdamped (ζ > 1)

Poles are real and negative:

Two identical poles:

mx bx kx x x««+ «+ =0 or x + 2«« ζωn«+ωn2 =0

ωn k m

12

12

sin

and and and tantanφ = A

A

1 2

ζ

αα

Case 1(ζ < 1): The free response is given by

(2.85)

The system is asymptotically stable The larger the ζωn the more stable the system Also, the speed

of response increases with both ωd and ζωn

Case 2 (ζ > 1): The response is non-oscillatory, and is given by

2 2

Now, bring the square-root term to the LHS and square it.

Hence,

or

This completes the proof

To explain this result further, consider an undamped (ζ = 0) simple oscillator of naturalfrequency ωn Its poles are at ± jωn (on the imaginary axis of the pole plane) Now add dampingand increase ζ from 0 to 1 Then the complex conjugates poles –ζωn± jωd will move away fromthe imaginary axis as ζ increases (because ζωn increases) and, hence, the level of stability willincrease When ζ reaches the value 1 (critical damping), one obtains two identical and real poles

at –ωn When ζ is increased beyond 1, the poles will be real and unequal, with one pole having amagnitude smaller than ωn and the other having a magnitude larger than ωn The former (closer tothe “origin” of zero) is the dominant pole, and will determine both stability and the speed ofresponse of the overdamped system It follows that as ζ increases beyond 1, the two poles willbranch out from the location –ωn, one moving toward the origin (becoming less stable) and theother moving away from the origin It is now clear that as ζ is increased beyond the point of criticaldamping, the system becomes less stable Specifically, for a given value of ζu < 1.0, there is a value

of ζo > 1, governed by (2.87), above which the overdamped system is less stable and slower thanthe underdamped system

The system is underdamped in this case

Case 2:

Then,

The system is overdamped in this case

Case 1: The characteristic equation is

or

The roots (eigenvalues or poles) are

The free (no force) response is given by

The amplitude A and the phase angle φ can be determined using initial conditions

Case 2: The characteristic equation is