Free Vibrations

Mechanical Vibrations(ME 421)

Section – 3 Single Degree of Freedom Systems: Free Vibrations

Instructor: Muhammad Haider

Book:Mechanical Vibrations, by S.S. Rao, Fifth Edition, Chapter 2

SDOF:Free Vibrations

Course ContentsS.No. Description

1. Basic Concepts

2. Harmonic Motion, Complex Algebra and Fourier Series

3. Single Degree of Freedom Systems: Free Vibrations

4.Single Degree of Freedom Systems: Harmonically Excited Vibrations

5.Two Degree of Freedom Systems: Natural Frequencies and Mode

Shapes

6.Two Degree of Freedom Systems: Coupling, Orthogonality and

Forced Response

7. Multi Degree of Freedom Systems

8. Lagrangian Method


• Introduction

• Free Vibration of an Undamped Translational System

• Free Vibration of an Undamped Torsional System

• First Order System Response

• Rayleigh’s Energy Method

• Free Vibrations with Viscous Damping

• Graphical Representation of Characteristic Roots

• Parameter Variations and Root Locus Representation

• Stability of Systems

Lecture Outline


Free Vibrations of UndampedTranslational System

a) Equation of Motion Using Newton’s Second Law of Motion

b) Equation of Motion Using Other Methods

i. D’Alembert’s Principle

ii. Principle of Virtual Displacements

iii. Principle of Conservation of Energy

c) Equation of Motion of a Spring-Mass System in Vertical Position

d) Solution

e) Harmonic Motion



The column of the water tank shown in Figure is 300 ft high and is made of

reinforced concrete with a tubular cross section of inner diameter 8 ft and outer

diameter 10 ft. The tank weighs 6x105 lb when filled with water. By neglecting the

mass of the column and assuming the Young s modulus of reinforced concrete as

4x106 psi, determine the following:

a) The natural frequency and the natural time period of transverse vibration of the water tank.

b) The vibration response of the water tank due to an initial transverse displacement of 10 in.

c) The maximum values of the velocity and acceleration experienced by the water tank.

Example 2.1

Free Vibrations of Undamped Translational System


Three springs and a mass are attached to a rigid, weightless bar PQ as shown in

Figure.

a) Find the natural frequency of vibration of the system.

Problem 2.7




Find the natural frequency of the pulley system shown in

Figure by neglecting the friction and the masses of the

pulleys.

Problem 2.13



Derive the expression for the natural frequency of the system shown in Figure.

Note that the load W is applied at the tip of beam 1 and midpoint of beam 2.

Problem 2.17

Review Examples: 2.2-2.5; Practice Problems: 2.2-2.6, 2.9,2.11, 2.12, 2.18, 2.22, 2.28, 2.33, 2.36, 2.38-2.40, 2.43, 2.44, 2.49-2.52, 2.54, 2.56, 2.58, 2.60

• If a rigid body oscillates about a specific reference axis, the resulting motion is called torsional vibration.• the displacement of the body is measured in terms of an angular

coordinate


Free Vibrations of Undamped Torsional System

• Single DOF Torsional System• A disc mounted at one end of a solid circular

shaft, the other end of which is fixed • Disc has a polar mass moment of 𝐽𝑜 and its

angular displacement (shaft twist) is given by 𝜃

Equation of Motion• Using Newton’s second law, following equation can be

derived,

𝐽𝑜 ሷ𝜃 + 𝑘𝑡𝜃 = 0• The equation is analogous to SDOF translational

system where 𝐽𝑜, 𝜃 and 𝑘𝑡 corresponds to 𝑚, 𝑥 and 𝑘respectively.

• Natural frequency of the system is, 𝜔𝑛 =𝑘𝑡

𝐽𝑜

• Period of vibration, 𝜏𝑛 = 2𝜋𝐽𝑜

𝑘𝑡

• Frequency of vibration, 𝑓𝑛 =1

2𝜋

𝑘𝑡

𝐽𝑜

• 𝑀𝑡 = (𝐺𝐼𝑜

𝑙)𝜃 and 𝐼𝑜 for a circular shaft 𝐼𝑜 =

𝜋𝑑4

32

• ⇒ 𝑘𝑡 =𝑀𝑡

𝜃=

𝐺𝐼𝑜

𝑙=

𝐺𝜋𝑑4

32𝑙

• Also, 𝐽𝑜 for the disc is given by 𝐽𝑜 =𝜌ℎ𝜋𝐷4

32=

𝑊𝐷2

8𝑔




Free Vibrations of Undamped Torsional SystemSolution• General solution can be obtained using

𝜃 𝑡 = 𝐴1 cos𝜔𝑛𝑡 + 𝐴2 sin𝜔𝑛𝑡• 𝜔𝑛 is calculated using geometrical and material

properties• Constants 𝐴1 and 𝐴2 are calculated using initial

conditions

• 𝜃 𝑡 = 0 = 𝜃𝑜 and ሶ𝜃 𝑡 = 0 =𝑑𝜃

𝑑𝑡𝑡 = 0 = ሶ𝜃𝑜

• And constants 𝐴1 and 𝐴2becomes𝐴1 = 𝜃𝑜

𝐴2 =ሶ𝜃𝑜

𝜔𝑛

Practice Problems: 2.64, 2.65, 2.67, 2.68, 2.69, 2.72, 2.73, 2.74, 2.76, 2.79, 2.81

Compound Pendulum• Any rigid body pivoted at a point other than its center of mass will oscillate

about the pivot point under its own gravitational force. Such a system is known as a compound pendulum



• With center of oscillation O, center of mass G, coordinate 𝜃, d, 𝐽𝑜, system equation is

𝐽𝑜 ሷ𝜃 + 𝑊𝑑 sin 𝜃 = 0• The equation is nonlinear, it can be

approximated by using small angle approximation i.e sin 𝜃 = 𝜃

𝐽𝑜 ሷ𝜃 + 𝑊𝑑𝜃 = 0• Natural frequency becomes

𝜔𝑛 =𝑊𝑑

𝐽𝑜

12

=𝑚𝑔𝑑

𝐽𝑜

12

• Comparing with the natural frequency of a

simple pendulum, 𝜔𝑛 =𝑔

𝑙

1

2, gives

𝑙 =𝐽𝑜𝑚𝑑

Compound Pendulum



• If 𝐽𝑜is replaced by 𝑚𝑘𝑜2, where 𝑘𝑜 is the radius

of gyration of the body about O, we get

𝜔𝑛 =𝑚𝑔𝑑

𝐽𝑜

12

=𝑔𝑑

𝑘𝑜2

12

𝑙 =𝑘𝑜2

𝑑• If 𝑘𝐺 denotes the radius of gyration of the body

about G, then we have

𝑘𝑜2 = 𝑘𝐺

2 + 𝑑2 ⇒ 𝑙 =𝑘𝐺2

𝑑+ 𝑑

• If the line OG is extended to point A such that

𝐺𝐴 =𝑘𝐺2

𝑑⇒ 𝑙 = 𝐺𝐴 + 𝑑 = 𝑂𝐴

• Natural frequency then can be written as

𝜔𝑛 =𝑔

𝑘𝑜2/𝑑

12

=𝑔

𝑙

12=

𝑔

𝑂𝐴

12

• No matter whether the body is pivoted from O or A, its natural frequency is the same. The point is called center of percussion.

Center of Percussion• The center of percussion is the point on an object where a perpendicular impact

will produce translational and rotational forces which perfectly cancel each other out at some given pivot point, so that the pivot will not be moving momentarily after the impulse.



Center of Percussion• Compound pendulum and center of percussion find applications in many

real life scenarios• Hammer• Cricket/Baseball Bat• Izod (Impact) testing of materials• Automobile



• First Order System?• A system having only one

energy storage element• Consider a rotor shaft with bearings

𝐽 ሶ𝜔 + 𝑐𝑡𝜔 = 0• where 𝜔 is the angular velocity of

the rotor and ሶ𝜔 is the angular acceleration

• Assume that we give an initial angular speed 𝜔 𝑡 = 0 = 𝜔𝑜 to the system and then see how it behaves.

• Important to note here is that we are interested in only angular velocity

• The solution can be obtained by assuming the following trial solution

𝜔 𝑡 = 𝐴𝑒𝑠𝑡

• where A and s are unknown constants


Response of First-Order Systems

• By using initial condition, 𝜔 𝑡 = 0 = 𝜔𝑜

the equation can be written as𝜔 𝑡 = 𝜔𝑜𝑒

𝑠𝑡

• By substituting trial solution into the equation of motion, we get

𝜔𝑜𝑒𝑠𝑡(𝐽𝑠 + 𝑐𝑡) = 0

• Since 𝜔𝑜 = 0 means “not motion” so we assume that the second factor will be zero

(𝐽𝑠 + 𝑐𝑡) = 0

• From here, 𝑠 = −𝑐𝑡

𝐽

• Thus our solution becomes 𝜔 𝑡 = 𝜔𝑜𝑒−𝑐𝑡𝐽𝑡

• It is convenient to describe the term using time constant 𝜏, defined as the value of time for which the exponent power becomes -1, for 𝑡 = 𝜏

𝜔 𝑡 = 𝜔𝑜𝑒−𝑐𝑡𝐽 𝜏 = 𝜔𝑜𝑒

−1 = 0.368𝜔𝑜


Response of First-Order Systems

Practice Problems: 2.83, 2.84

• Energy method can be used to determine natural frequencies of single degree of freedom system

• Energy conservation for an undamped vibrating system can be written as𝑇1 + 𝑈1 = 𝑇2 + 𝑈2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

• where subscripts 1 and 2 denote two different instants of time• Now assume that 1 represent time when the mass is passing through its

static equilibrium position, we will have𝑈1 = 0 ; 𝑇1 = 𝑇𝑚𝑎𝑥

• Similarly, let 2 represents the time corresponding to the maximum displacement of mass

𝑇2 = 0 ;𝑈2 = 𝑈𝑚𝑎𝑥

• Putting values, we get𝑇1 + 0 = 0 + 𝑈2 ⇒ 𝑇𝑚𝑎𝑥 = 𝑈𝑚𝑎𝑥

• By using above relationship, we can directly calculate natural frequency of the system

• The method is also known as Rayleigh’s Energy Method


Rayleigh’s Energy Method


Find the natural frequency of transverse vibration of the

water tank considered in Example 2.1 by including mass

of the column

Example 2.9

Rayleigh’s Energy Method

Review Examples: 2.7, 2.8Practice Problems: 2.86-2.89, 2.91

• A system with Viscous damping can be represented as

• Also known as Mass-spring-damper system• Force due to viscous damping can be estimated by

𝐹 = −𝑐 ሶ𝑥• where 𝑐 is the damping constant • Equation of motion for the system using Newton’s

Law yield𝑚 ሷ𝑥 = −𝑐 ሶ𝑥 − 𝑘𝑥

• which can be rewritten as𝑚 ሷ𝑥 + 𝑐 ሶ𝑥 + 𝑘𝑥 = 0

Solution• We assume a solution in the form

𝑥 𝑡 = 𝐶𝑒𝑠𝑡

• where C and s are undetermined constants• Inserting this function into the equation leads to

characteristic equation𝑚𝑠2 + 𝑐𝑠 + 𝑘 = 0


Free Vibration with Viscous Damping

• The roots of which are

𝑠1,2 =−𝑐 ± 𝑐2 − 4𝑚𝑘

2𝑚= −

𝑐

2𝑚±

𝑐

2𝑚

2

−𝑘

𝑚

• These roots give two solutions of the EoM𝑥1 𝑡 = 𝐶1𝑒

𝑠1𝑡 and 𝑥2 𝑡 = 𝐶2𝑒𝑠2𝑡

• Thus the general solution of the equation is given by a combination of two solutions

𝑥 𝑡 = 𝐶1𝑒𝑠1𝑡 + 𝐶2𝑒

𝑠2𝑡

= 𝐶1𝑒−

𝑐2𝑚+

𝑐2𝑚

2−𝑘𝑚 𝑡

+ 𝐶2𝑒−

𝑐2𝑚−

𝑐2𝑚

2−𝑘𝑚 𝑡

• where 𝐶1 and 𝐶2 are arbitrary constants to be determined from the initial conditions of the system.



Important TerminologiesCritical Damping Constant• The critical damping 𝑐𝑐 is defined as the value of the damping constant 𝑐 for

which the radical in the solution becomes zero

𝑐

2𝑚

2

−𝑘

𝑚= 0 ⇒ 𝑐𝑐 = 2𝑚

𝑘

𝑚= 2 𝑘𝑚 = 2𝑚𝜔𝑛

Damping Ratio• The ratio between damping constant to the critical damping constant

• Mathematically, 𝜁 =𝑐

𝑐𝑐𝑐

2𝑚=

𝑐

𝑐𝑐.𝑐𝑐

2𝑚= 𝜁𝜔𝑛

• We can simplify the roots as

• The general solution can be written as

𝑥 𝑡 = 𝐶1𝑒−𝜁+ 𝜁2−1 𝜔𝑛𝑡 + 𝐶2𝑒

−𝜁− 𝜁2−1 𝜔𝑛𝑡

• The nature and behavior of solution depends upon the magnitude of damping• Based on the value of damping ratio, we have three different system behaviors

• Undamped(𝜁 < 1), critically damped (𝜁 = 1) and overdamped (𝜁 > 1) systems



𝑠1,2 = −𝑐

2𝑚±

𝑐

2𝑚

2

−𝑘

𝑚

𝑠1,2 = −𝜁 ± 𝜁2 − 1 𝜔𝑛 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛

Case-IUnderdamped System

• (𝜁 < 1 or 𝑐 < 𝑐𝑐 or 𝑐

2𝑚<

𝑘

𝑚)

• For this condition, (𝜁2 − 1) is negative and roots 𝑠1 and 𝑠2 can be expressed as

𝑠1 = −𝜁 + 𝑖 1 − 𝜁2 𝜔𝑛; 𝑠2 = −𝜁 − 𝑖 1 − 𝜁2 𝜔𝑛

• Solution can be written as

𝑥 𝑡 = 𝐶1𝑒−𝜁+𝑖 1−𝜁2 𝜔𝑛𝑡 + 𝐶2𝑒

−𝜁−𝑖 1−𝜁2 𝜔𝑛𝑡

= 𝑒−𝜁𝜔𝑛𝑡 𝐶1𝑒𝑖 1−𝜁2𝜔𝑛𝑡 + 𝐶2𝑒

−𝑖 1−𝜁2𝜔𝑛𝑡

= 𝑒−𝜁𝜔𝑛𝑡 𝐶1 + 𝐶2 cos 1 − 𝜁2𝜔𝑛𝑡 + 𝑖 𝐶1 − 𝐶2 sin 1 − 𝜁2𝜔𝑛𝑡

= 𝑒−𝜁𝜔𝑛𝑡 𝐶1′ cos 1 − 𝜁2𝜔𝑛𝑡 + 𝐶2′ sin 1 − 𝜁2𝜔𝑛𝑡

Let 𝐶1′ = 𝑋𝑜 sin𝜙𝑜 and 𝐶2

′ = 𝑋𝑜 cos 𝜙𝑜, we get

= 𝑒−𝜁𝜔𝑛𝑡 𝑋𝑜 sin𝜙𝑜 cos 1 − 𝜁2𝜔𝑛𝑡 + 𝑋𝑜 cos𝜙𝑜 sin 1 − 𝜁2𝜔𝑛𝑡

= 𝑋𝑜𝑒−𝜁𝜔𝑛𝑡 sin( 1 − 𝜁2𝜔𝑛𝑡 + 𝜙𝑜) = 𝑋𝑒−𝜁𝜔𝑛𝑡 cos ( 1 − 𝜁2𝜔𝑛𝑡 − 𝜙)

• Where 𝐶1′, 𝐶2′, (𝑋, 𝜙) and (𝑋𝑜 , 𝜙𝑜) are arbitrary constants to be determined from initial conditions



Case-IUnderdamped System • For initial conditions 𝑥 𝑡 = 0 = 𝑥𝑜 and ሶ𝑥 𝑡 = 0 = ሶ𝑥𝑜, 𝐶1′ and 𝐶2′ can be found:

𝐶1′ = 𝑥𝑜 and

𝐶2′ =

ሶ𝑥𝑜 + 𝜁𝜔𝑛𝑥𝑜

(1 − 𝜁2𝜔𝑛

• and hence the solution becomes

𝑥 𝑡 = 𝑒−𝜁𝜔𝑛𝑡 𝑥𝑜 cos 1 − 𝜁2𝜔𝑛𝑡 +ሶ𝑥𝑜 + 𝜁𝜔𝑛𝑥𝑜

(1 − 𝜁2𝜔𝑛

sin 1 − 𝜁2𝜔𝑛𝑡

• The constants (𝑋, 𝜙) and (𝑋𝑜 , 𝜙𝑜) can be expressed as

𝑋 = 𝑋𝑜 = 𝐶1′ 2 + 𝐶2

′ 2 =𝑥𝑜2𝜔𝑛

2 + ሶ𝑥𝑜 + 2𝑥𝑜 ሶ𝑥𝑜𝜁𝜔𝑛

(1 − 𝜁2𝜔𝑛

𝜙𝑜 = tan−1(𝐶1′

𝐶2′) = tan−1(

𝑥𝑜𝜔𝑛 (1 − 𝜁2

ሶ𝑥𝑜 + 𝜁𝜔𝑛𝑥𝑜)

𝜙 = tan−1(𝐶2′

𝐶1′) = tan−1(

ሶ𝑥𝑜 + 𝜁𝜔𝑛𝑥𝑜

𝑥𝑜𝜔𝑛 (1 − 𝜁2)



Case-IUnderdamped System • The motion described by the equation

𝑥 𝑡 = 𝑒−𝜁𝜔𝑛𝑡 𝑥𝑜 cos 1 − 𝜁2𝜔𝑛𝑡 +ሶ𝑥𝑜 + 𝜁𝜔𝑛𝑥𝑜

(1 − 𝜁2𝜔𝑛

sin 1 − 𝜁2𝜔𝑛𝑡

• is a damped harmonic motion of angular frequency 𝜔𝑑, the frequency of damped vibration. Mathematically, its given by

𝜔𝑑 = (1 − 𝜁2𝜔𝑛

• Factor 𝑒−𝜁𝜔𝑛𝑡 exponentially decays the amplitude with time,



Case-IICritically damped System

• (𝜁 = 1 or 𝑐 = 𝑐𝑐 or 𝑐

2𝑚=

𝑘

𝑚)

• In this case the two roots are equal

𝑠1 = 𝑠2 = −𝑐𝑐2𝑚

= −𝜔𝑛

• Because of repeated roots, the solution of the EOM is given by𝑥 𝑡 = 𝐶1 + 𝐶2𝑡 𝑒

−𝜔𝑛𝑡

• Application of initial conditions 𝑥 𝑡 = 0 = 𝑥𝑜 and ሶ𝑥 𝑡 = 0 = ሶ𝑥𝑜, givesC1 = 𝑥𝑜 and 𝐶2 = ሶ𝑥𝑜 + 𝜔𝑛𝑥𝑜

• and the solution becomes𝑥 𝑡 = 𝑥𝑜 + ሶ𝑥𝑜 + 𝜔𝑛𝑥𝑜 𝑡 𝑒−𝜔𝑛𝑡



Case-IIOverdamped System

• (𝜁 > 1 or 𝑐 > 𝑐𝑐 or 𝑐

2𝑚>

𝑘

𝑚)

• In this case the roots are real and distinct and are given by

𝑠1 = −𝜁 + 𝜁2 − 1 𝜔𝑛 < 0

𝑠2 = −𝜁 − 𝜁2 − 1 𝜔𝑛 < 0

• the solution of the EOM is given by

𝑥 𝑡 = 𝐶1𝑒−𝜁+ 𝜁2−1 𝜔𝑛𝑡 + 𝐶2𝑒

−𝜁− 𝜁2−1 𝜔𝑛𝑡

• For the initial conditions 𝑥 𝑡 = 0 = 𝑥𝑜 and ሶ𝑥 𝑡 = 0 = ሶ𝑥𝑜, the constants can be obtained as

C1 =𝑥𝑜𝜔𝑛(𝜁 + 𝜁2 − 1 + ሶ𝑥𝑜

2𝜔𝑛 𝜁2 − 1

C2 =−𝑥𝑜𝜔𝑛(𝜁 − 𝜁2 − 1 − ሶ𝑥𝑜

2𝜔𝑛 𝜁2 − 1



Graphical representation



Logarithmic Decrement• It represents the rate at which the amplitude of free-damped vibration decreases• It is given by the natural logarithm of the ratio of any two successive amplitudes• Let 𝑡1and 𝑡2denote the times corresponding to two consecutive amplitudes,

measured one cycle apart for an underdamped system

• we can form the ratio

𝑥1𝑥2

=𝑋𝑒−𝜁𝜔𝑛𝑡1 cos(𝜔𝑑𝑡1 − 𝜙)

𝑋𝑒−𝜁𝜔𝑛𝑡2 cos(𝜔𝑑𝑡2 − 𝜙)• But 𝑡2 = 𝑡1 + 𝜏𝑑, where 𝜏𝑑 = 2𝜋/𝜔𝑑 is the period of damped vibration.• Hence cos 𝜔𝑑𝑡2 − 𝜙 = cos(2𝜋 + 𝜔𝑑𝑡1 − 𝜙) = cos(𝜔𝑑𝑡1 − 𝜙)



Logarithmic Decrement• The ratio

𝑥1𝑥2

=𝑋𝑒−𝜁𝜔𝑛𝑡1 cos(𝜔𝑑𝑡1 − 𝜙)

𝑋𝑒−𝜁𝜔𝑛𝑡2 cos(𝜔𝑑𝑡2 − 𝜙)• can be written as

𝑥1𝑥2

=𝑒−𝜁𝜔𝑛𝑡1

𝑒−𝜁𝜔𝑛(𝑡1+𝜏𝑑)= 𝑒𝜁𝜔𝑛𝜏𝑑

• The logarithmic decrement 𝛿 can be obtained as

𝛿 = ln𝑥1𝑥2

= 𝜁𝜔𝑛𝜏𝑑 = 𝜁𝜔𝑛

2𝜋

1 − 𝜁2𝜔𝑛

=2𝜋𝜁

1 − 𝜁2=2𝜋

𝜔𝑑.𝑐

2𝑚

• For small damping, 𝛿 can be approximated as

𝛿 ≅ 2𝜋𝜁



Logarithmic Decrement• The logarithmic decrement is

dimensionless and is actually another form of dimensionless damping ratio 𝜁

• Once 𝛿 is known, 𝜁 can be found by

𝜁 =𝛿

2𝜋 2 + 𝛿2

• When using linear approximation

𝜁 =𝛿

2𝜋• If damping in the system is unknown, we

can determine it experimentally by measuring any two consecutive displacement 𝑥1 and 𝑥2




Torsional System with Viscous Damping• Viscous damping torque is given by

𝑇 = 𝑐𝑡 ሶ𝜃• Equation of motion for the given single DOF

system can be derived as𝐽𝑜 ሷ𝜃 + 𝑐𝑡 ሶ𝜃 + 𝑘𝑡𝜃 = 0

• Solution of the above equation can be determined exactly in the same way as was done in translational vibrations.

• For an undamped case, the frequency of the damped vibrations is given by

𝜔𝑑 = 1 − 𝜁2𝜔𝑛

• where,

𝜔𝑛 =𝑘𝑡

𝐽𝑜and 𝜁 =

𝑐𝑡

𝑐𝑡𝑐=

𝑐𝑡

2𝐽𝑜𝜔𝑛=

𝑐𝑡

2 𝑘𝑡𝐽𝑜



An underdamped shock absorber is to be designed for a

motorcycle of mass 200 kg as shown in Fig. (a). When the shock

absorber is subjected to an initial vertical velocity due to a road

bump, the resulting displacement-time curve is to be as

indicated in Fig. (b). Find

Example 2.11

a) the necessary stiffness and damping constants of the shock absorber if the damped period of vibration is to be 2s and the amplitude is to be reduced to one-fourth in one half cycle (i.e., 𝒙𝟏.𝟓 = 𝒙𝟏/𝟒)

b) the minimum initial velocity that leads to a maximum displacement of 250 mm. (Assume that the max displacement envelop is given by 𝒙 =

𝑿𝒆−𝜻𝝎𝒏𝒕 𝟏 − 𝜻𝟐 and 𝒔𝒊𝒏 𝝎𝒅𝒕 = 𝟏 − 𝜻𝟐) at 𝒕𝟏


Review Examples: 2.10, 2.12; Practice Problems: 2.97, 2.98, 2.101-2.106, 2.109, 2.110, 2.119, 2.121, 2.122, 2.124, 2.127-2.130

Graphical Representation of Characteristic Roots


Roots of Characteristic Equation• The free vibration of a SDOF spring-mass-damper system is given by

𝑚 ሷ𝑥 + 𝑐 ሶ𝜃 + 𝑘𝜃 = 0• whose characteristic equation can be written as

𝑚𝑠2 + 𝑐𝑠 + 𝑘 = 0• or

𝑠2 + 2𝜁𝜔𝑠 + 𝜔𝑛2 = 0

• The roots of the characteristic equation, called the characteristic roots, help us in understanding the behavior of the system

• The roots are given by

𝑠1 = 𝑠2 =−𝑐 ± 𝑐2 − 4𝑚𝑘

2𝑚• or,

𝑠1 = 𝑠2 = −𝜁𝜔𝑛 ± 𝑖𝜔𝑛 1 − 𝜁2

• The roots given by above equations can be plotted in a complex plane, alsoknown as the s-plane, by denoting the real part along the horizontal axis andthe imaginary part along the vertical axis.



Roots of Characteristic EquationThe following observations can be made by examining characteristic equations• The exponent of a larger real negative number (such as 𝑒−2𝑡) decays faster than

the exponent of a smaller real negative number (such as 𝑒−𝑡 ), the roots lying farther to the left in the s-plane indicate that the corresponding responses decay faster than those associated with roots closer to the imaginary axis.



Roots of Characteristic EquationThe following observations can be made by examining characteristic equations• If the roots have positive real values of s that is, the roots lie in the right half of

the s-plane the corresponding response grows exponentially and hence will be unstable.



Roots of Characteristic EquationThe following observations can be made by examining characteristic equations• If the roots lie on the imaginary axis (with zero real value), the corresponding

response will be naturally stable.• If the roots have a zero imaginary part, the corresponding response will not

oscillate.



Roots of Characteristic EquationThe following observations can be made by examining characteristic equations• The response of the system will exhibit an oscillatory behavior only when the

roots have nonzero imaginary parts.• The larger the imaginary part of the roots, the higher the frequency of

oscillation of the corresponding response of the system.

Parameter Variations and Root Locus Representation


Interpretation of 𝝎𝒏, 𝝎𝒅, 𝜻 and 𝝉 in the s-plane• Although the roots 𝑠1and 𝑠2 appear as complex conjugates, we consider only

the roots in the upper half of the s-plane• Root 𝑠1 is plotted as point A with the real value as 𝜁𝜔𝑛 and the complex value

as 1 − 𝜁2𝜔𝑛, so that the length of OA is 𝜔𝑛

• Thus the roots lying on the circle ofradius 𝜔𝑛 correspond to the samenatural frequency of the system

• Different concentric circles representsystems with different naturalfrequencies



Interpretation of 𝝎𝒏, 𝝎𝒅, 𝜻 and 𝝉 in the s-plane• Horizontal line passing through A corresponds to damped natural frequency• Lines parallel to real axis denote systems having different 𝜔𝑑

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

sin 𝜃 =𝜁𝜔𝑛

𝜔𝑛= 𝜁 ⇒ 𝜃 = sin−1 𝜁

• Radial lines passing through the origincorrespond to different damping ratios



Interpretation of 𝝎𝒏, 𝝎𝒅, 𝜻 and 𝝉 in the s-plane• Therefore, when 𝜁 = 0, we have no damping (𝜃 = 0), and the damped natural

frequency will reduce to the undamped natural frequency• Similarly, when 𝜁 = 1, we have critical damping and the radical line lies along

the negative real axis.



Interpretation of 𝝎𝒏, 𝝎𝒅, 𝜻 and 𝝉 in the s-plane

• The time constant of the system, is defined as, 𝜏 =1

𝜁𝜔𝑛

• The distance DO or AB represents the reciprocal of the time constant, 𝜁𝜔𝑛 =1

𝜏

• Different lines parallel to the imaginary axis denote reciprocals of different time constants

Reading Assignment:Section 2.8.2: Root Locus and Parameter VariationsExample: 2.13


• Stability is one of the most important characteristics for any vibrating system

• Definition of stability depends on the kind of system or the point of view

• Following are the terms related to stability for linear time-invariant systems (the parameters m, c, and k do not change with time)

• A system is stable (marginally stable in controls literature) if its free-vibration response neither decays nor grows, but remains constant or oscillates as time approaches infinity.

• A system is asymptotically stable if its free-vibration response approaches zero as time approaches infinity

• A system is unstable if its free-vibration response grows without bound as time approaches infinity

• It is evident that an unstable system whose free-vibration response grows without bounds can cause damage to the system, adjacent property, or human life.

• Usually, dynamic systems are designed with limit stops to prevent their responses from growing with no limit.

Stability of Systems


Consider a uniform rigid bar, of mass m and length l, pivoted at one end and

connected symmetrically by two springs at the other end. Assuming that the

springs are unstretched when the bar is vertical

Example 2.18

a) derive the equation of motion of the system for small angular displacements (u) of the bar about the pivot point

b) investigate the stability behavior of the system.

Stability of Systems

Practice Problem: 2.115

Haider

Sticky Note

Submission Date: 16th Nov, 2015

Free Vibrations

Documents

Transcript of Free Vibrations