Mechanical Vibrations Work Book Cum Lecture Notes

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MECHANICAL VIBRATIONS WORK BOOK CUM LECTURE NOTES

(FOR SIXTH SEMESTER MECHANICAL STUDENTS) (FOR PRIVATE CIRCULATION ONLY )

JAGADEESHA T Associate Professor Mechanical Engineering Department

Name

USN

Section

ST. JOSEPH ENGINEERING COLLEGE VAMANJOOR, MANGALORE – 575 028,

CHAPTER 1- INTRODUCTION TO VIBRATION

LEARNING OBJECTIVES

] Introduction to vibration ] Terminologies used in Vibration ] Simple Harmonic Motion ] Addition of Harmonics, Principle of super position applied to SHM ] Introduction to Fourier analysis, Beats ] Problems related to SHM and Fourier analysis.

Vibration is defined as a motion which repeats after equal interval of time and is also a periodic motion. The swinging of a pendulum is a simple example of vibration. Vibration occurs in all bodies which are having mass and elasticity. They are caused due to several reasons such as presence of unbalanced force in rotating machines, elastic nature of the system, external application of force or wind loads and earthquakes. Vibrations are undesirable in most engineering systems and desirable in few cases.

A body is said to vibrate if it has periodic motion. Mechanical vibration is the study of oscillatory motions of a dynamic system. An oscillatory motion is a repeated motion with equal interval of time.

Example for useful vibration:

o General industries – crushers, jackhammer, concrete compactor, etc. o Medical and health – electric massage, high frequency vibration probe for heart disease treatment o Music – string instruments i.e. guitar etc.

Example for unwanted vibration :

o Poor ride comfort in vehicle due to road irregularities o Sea sickness when traveling on ships, boats, etc. o Earthquakes o Fatigue failures in machine and structures

Cycle : The movement of vibrating body from the mean to its extreme position in one direction then to mean , then to another extreme position and back to mean is called as M cycle of vibration, Mean position 3 2

Time period : It is the time taken to complete one cycle. It is equal to the time for the vector to rotate through 2π radians

` ωωω rad/s

Reference:

Frequency: It is the number of cycles per unit time

Amplitude : It is the maximum displacement of a vibrating body from the mean position

Phase difference : it is angle between the two rotating vectors executing simple harmonic motion of same frequency ωωω rad/s ωωω rad/s The first vector is x1= X sin ( wt ) The second vector is x2= X sin ( wt+φ)

Where φ is the phase difference between x1 and x2

Resonance : it is the frequency of the external force coincides with the natural frequency of the system , a condition known as resonance occurs. During the resonance the system undergoes dangerously large oscillations

Damping : It is the resistance offered to the motion of a vibrating body.

Periodic motion If the motion is repeated after equal intervals of time, it is called periodic motion, The simplest type of periodic motion is harmonic motion

Aperiodic motion If the motion does not repeat after equal interval of time , it is called aperiodic motion

DESCRIBING MOTIONS OF VIBRATING SYSTEMS

Periodic motions

Described as sine or cosine functions [sin (ωt) and cos(ωt)]

ω = radian frequency (rad/sec) ω = 2π f ; where f is frequency (Hz)

Period = time between two adjacent peaks or valleys; P = 1/ f

Simple harmonic motion Leads / Lags y(t) = Asin(ωt + φ) Displacement

y& (t) = Aω cos(ωt + φ) Velocity

2 y& (t) = − Aω sin(ωt + φ) Acceleration

Classification of vibrations One method of classifying mechanical vibrations is based on degrees of freedom. The number of degrees of freedom for a system is the number of kinematically independent variables necessary to completely describe the motion of every particle in the system. Based on degrees of freedom, we can classify mechanical vibrations as follows:

1.Single Degree of freedom Systems 2.Two Degrees of freedom Systems 3.Multidegree of freedom Systems 4.Continuous Systems or systems with infinite degrees of freedom

Another broad classification of vibrations is: 1. Free and forced vibrations 2. Damped and undamped vibrations.

Sometime vibration problems are classified as:

1. Linear vibrations 2. Non-linear vibrations 3. Random vibrations 4. Transient vibrations 5. Longitudinal vibrations 6. Transverse vibrations 7. Torsional vibrations

Free vibration : If a system after initial disturbance is left to vibrate on its own , the resulting vibration is known as free vibrations. Free vibration takes when a system vibrates under the action of forces inherent in the system and when the external forces are absent. The frequency of free vibration of a system is called natural frequency. Natural frequency is a property of a dynamical system

Forced vibration : Vibration that takes place under the excitation of external forces is called forced vibration. the forced vibration takes place at different forced frequencies or external frequencies

Damped vibration : If any energy is lost or dissipated during oscillations then the vibration is known as damped vibration\

Undamped vibration : if no energy is lost or dissipated during oscillations ,such vibrations are known as undamped vibration

Linear vibration : If all the basic component of a vibrating system behave linearly, the resulting vibration is known as linear vibration. The differential equations govern linear vibratory system are linear. If the vibration is linear , the principle of superposition holds and mathematical techniques of analysis are well developed.

Non linear vibration : If any of the basic components of a vibrating system behave non linearly . the resulting vibration is known as non linear vibration. The differential equations that govern non linear vibratory system are non-linear. If the vibration is non linear the principle of superposition does not hold good and techniques of analysis is well known

Deterministic vibration : If the magnitude of excitation on a vibrating system is known at any given time , the resulting vibration is known as deterministic vibration

Random vibration: If the magnitude of excitation acting on a vibratory system at a given time cannot be predicted , the resulting vibration is known as non deterministic or random vibration

5 Longitudinal vibration : Consider a body of mass m carried on one end of a slender shaft and other end being fixed. If the mass vibrates parallel to the spindle axis, it is said to be execute longitudinal vibration

Transverse vibration : If the mass vibrates perpendicular to the spindle axis , it is said to execute the transverse vibration

Torsional vibration : If the shaft gets alternatively twisted and un twisted on account of an alternate torque on the disk, it is said to execute the torsional vibration

SIMPLE HARMONIC MOTION

A Vibration with acceleration proportional to the displacement and directed toward the mean position is known as SHM, ( Simple Harmonic Motion)

Consider a spring mass system as shown in the figure along with the displacement time diagram

The equation of motion of the mass can be written as follows

From the right angled triangle OAB we have

Where x= displacement at any instant of time X = amplitude of vibration W= angular velocity or frequency in rad/second

6 The velocity of the mass m at an instant of time t is given by

V =

The acceleration of mass m is given by

Hence we can conclude that in SHM the acceleration is proportional to displacement and is directed towards mean position

observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by π/2 and π radians respectively.

x-Displacement X-amplitude

T-Periodic Time x t f-Frequency f=1/T ω=Frequency in radians per second t= time

t X

X= A sin ωt

7 Degrees of freedom: The minimum number of independent coordinates required to determine completely the positions of all the parts of a system at any instant of time is called degrees of freedom

One degree of freedom

Two degree of freedom

m m1 F1 1 x kt1 1 kt2

Rotor with inertia J 1 K m2

Rotor with Inertia J 2

Three degree of freedom m2 x 2

Infinite degree of freedom

m 1 Cantilever Beam

Continuous system ( consider the mass of the beam )

Exercise

Specify the no of degree of freedom for the following

1 2

Cantilever Beam y k1 y k2 m1 m2

Continuous system ( Neglect the mass of the beam )

3k k k 4m 2m m

m, 4 J G

m2 K1 K2

5. a b

M1 θ1

L2 θ2

Addition of SIMPLE HARMONIC MOTION ( SHM)

The addition of two simple harmonic motion having frequency yields a resultant which is simple harmonic having the same frequency.

Consider tow simple harmonic motions of x1 and x2 having the same frequency and phase difference φ as given below

x 1 === X 1 sin ( ωωω t ) x 2 = X 2 sin (ω t + φ )

Adding x = x 1 +x 2

Hence the resultant displacement is also SHM of amplitude X and phase angle θ

θθθ φφφ A

ωωωt O

Tutorial problems on Simple Harmonic Motion.

1. Add the following harmonic motion analytically and verify the solution graphically

1) X1= 3 sin (((ωωω t +++ 30))) X2= 4 cos (((ωωω t +++ 10))) ( VTU Jan 2005) 2) X1= 2 cos (((ωωω t +++ 0.5))) X2= 5 sin (((ωωω t +++ 1 ))) ( VTU July 2006) 3) X1= 10 cos ωωω t +++ πππ X2= 8 sin ωωω t +++ πππ ( VTU Dec. 2007) ((( 4))) ((( 6)))

2. A body is subjected to two harmonic motions X1= 8 cos ωωω t +++ πππ X2= 15 sin ωωω t +++ πππ what harmonic motion ((( 6))) ((( 6))) should be given to the body to bring it to equilibrium (VTU July. 2005)

3. Split the harmonic motion x = 10 sin ω t + πππ into two harmonic motions (((ωω ++ 6))) having the phase of zero and the other of 45 o

4. Show that resultant motion of harmonic motion given below is zero X1= X sin ω t X2= X sin ω t + 2πππ X3= X sin ω t + 4πππ (((ωω ))) (((ωω ++ 3))) (((ωω ++ 3)))

5. The displacement of a vibrating body is given by x = 5 sin (31.41t + πππ ). 4 Draw the variation of displacement for one cycle of vibration. Also determine the displacement of body after 0.11 second. ( repeat the problem for velocity and acceleration and draw graph using Excel and compare )

Time Displacement 0 0.025 5 0.05 0.075 0 6

0.1 4 0.125 2 0.15 -3.53 0.175 0 0 0 0.05 0.1 0.15 0.2 0.25 0.2 -2

Calculate the remaining values -4

-6 6. The Motion of a particle is represented by x = 4 sin(ω t ), sketch the variation of the displacement , velocity and acceleration and determine the max value of these quantities. Assume (ω = 5) ( Try to use MATLAB/Excel) . sketch all on same graph

BEATS

When two harmonic motions with frequencies close to one another are added. The resulting motion exhibits a phenomenon known as Beats.

A Beat Frequency is the result of two closely spaced frequencies going into and out of synchronization with one another

Let us consider tow harmonic motion of same amplitude and slightly different frequencies.

X1 = X Cos (ωωω t), X2 = X Cos ((ωωω + δδδ ) t),

Where δ is a small quantity

The addition of above two harmonics can be written as X = X1 + X2

  δδδ   δδδ   X= 2X cos t cosωωω + t   2   2  

The above equation shown graphically in Figure. The resulting motion represents cosine wave with frequency ω + πππ and with a varying amplitude 2 cos( δδδ t : (ωω 2 ) ( 2 ) whenever the amplitude reaches a maximum , it is called the beat. The frequency δ at which amplitude builds up and dies down between o and 2 X is known as beat frequency.

Beat phenomenon is found in machines , structures and electric power houses. In machines , the beat phenomenon occurs when the forcing frequency is close to the natural frequency of the system

+2X

-2X

 2πππ     ωωω 

ELEMENTS OF VIBRATION

The elements of constitute vibrating systems are 1. Mass or Inertia element - m 2. Spring - k k C 3. Damper - c

4. Excitation F(t) Voigt Model

m Elements of Vibration F(t)

Passive element Active element m, C, K F(t)

Conse rvative element Non conservative element ( Mass and Spring) ( Damper)

Mass or Inertia element : The mass or inertia element is assumed to be rigid body During vibration velocity of mass changes, hence kinetic energy can be gained or loosed. The force applied on the mass from newton second law of motion can be written as F= ma. Work done on the mass is stored in the form of kinetic energy given by KE= ½ M V 2

Combination of masses

In practical cases for simple analysis, we replace several masses by a single equivalent mass.

Case1. Translational masses connected by rigid bar.

Let the masses M1, M 2 and M 3 are attached to a rigid bar at locates 1, 2 and 3 respectively as shown in the figure. The equivalent mass meq be assumed to be located at 1 is as shown in figure (b) Let the displacement of masses M1, M2 and M3 be x1, x2, x3 and similarly the velocities of respective masses be x1, x2 and x3. We can express the velocities of masses m2 and m3 in terms of m1

2 2  l   l  M = M + M  2  + M  3  eq 1 2  l  3 l   1   1  is the required answer.

Translational and rotational masses coupled together.

Let a mass m having a translational velocity x be coupled to another mass having mass moment of inertia I o with a rotational velocity θ as in rack and pinion arrangement shown in the figure.

These two masses can be combined to obtain either a single equivalent translational mass M eq or a single equivalent mass moment of inertia J eq

Equivalent translational masses

1 2  Kinetic energy of the equivalent mass =  M eq X& eq  2 

  1 1  Kinetic energy of the two masses =  M X& 2 + J θθθ& 2  2 2 0   

 J o  Meq = m m + 2  is the required answer.  R 

Also determine the equivalent rotational mass J eq

2 Jeq = m [mR + J o ] is the required answer

Spring element : Whenever there is a relative motion between the two ends of the spring, a force is developed called spring force or restoring force. The spring force is proportional to the amount of deformation x and then F α x or F = kx. Where k is stiffness of the spring or spring constant or spring gradient.

The spring stiffness is equal to spring force per unit deromation. F  The spring stiffness k = N / m  x 

Workdone in deforming a spring is equal to the strain energy or potential energy.

Strain energy = potential energy = area of the triangle OAB

F K=stiffness F

Stiffness of beams

Cantilever beam consider a cantilever beam with an end mass shown in the figure. The mass of the beam is assumed to be negligible. The static deflection of beam at free end is given by

Similarly derive the expression for Simply supported beam and fixed support beam.

 192EI  δδδ st = 3  N / m  l  3  Wl  δδδ st =  N / m  48EI 

m m x x

Stiffness of slender bar subjected to longitudinal vibrations

For a system executing the longitudinal vibrations as shown in the figure , let us derive the expression for stiffness.

E, A l

Torsional Stiffness of bar.

It is the amount of torque required to cause a unit angular deformation in the element.

T Torsional stiffness = K = t θθθ

Combination of stiffness

Determination of equivalent spring stiffness when the springs are arranged in series,

Consider two springs of stiffness K 1 and K 2 acted upon by the force F.

F The deflection of spring k1 is x 1 = K1 The deflection of spring k 1 is x 2 =

Let these two springs be replaced by an equivalent stiffness K eq upon which a force F acts and due to which its deflection is given by F x = K eq x= x 1+x 2

Determination of equivalent spring stiffness when the springs are arranged in parallel.

Force acting on K 1 spring = F 1=k 1x Force acting on K 2 spring = F 2 Force required for an equivalent spring k eq to defined by x given by F= K eq x But F = F1 +F2

Tutorial problems on Equivalent stiffness of springs

1.Determine the equivalent stiffness for the system shown in figure.

k k

M k 3k k 2k 2k 2k

2. Determine the equivalent stiffness for the system shown in figure

2x106 N/m

3x106 N/m

3. Determine the equivalent stiffness for the system shown in figure

38 kg

1x106 N/m 2x106 N/m 60 cm 70 cm

4.Determine the equivalent stiffness for the system shown in figure

60 cm 80 cm 50 cm

5.Replace the following torsional stiffness by a single shaft having radius 4cm and find the length required for the equivalent shaft . Assume the material of given system and equivalent system is same. leqn

50 cm 60 cm R1= 3cm R2= 5cm Rreq =4cm

DAMPING

Every vibration energy is gradually converted into heat or sound. Hence the displacement during vibration gradually reduces. The mechanism by which vibration energy is gradually converted into heat or sound is known as damping.

A damper is assumed to have either mass or elasticity, Hence damping is modeled as one or more of the following types: Viscous damping; Coulomb or dry friction damping; materials or solid or hysteric damping

Viscous damping Viscous damping is most commonly used damping mechanism in vibration analysis. When the mechanical system vibrates in a fluid medium such as air, gas, water or oil, the resistance offered by the fluid to the moving body causes energy to be dissipated. In this case , the amount of dissipated energy depends on many factors such as size or shape of the vibrating body. the viscosity of the fluid, the frequency of vibration and velocity of fluid. Resistance due to viscous damping is directly proportional to the velocity of vibration

Fd ααα V

Fd = C x& Where C= damping coefficient Fd = damping force

Examples of Viscous damping 1) Fluid film between sliding surface 2) Fluid flow around a piston in a cylinder 3) Fluid flow through orifice 4) Fluid flow around a journal in a bearing

Coulomb damping or dry friction damping Here a damping force is constant in magnitude but opposite in direction to that of the motion of vibrating body. It is caused by the friction between the surfaces that are dry or have insufficient lubrication

Material or solid or hysteric damping stress When the materials are deformed energy is absorbed and dissipated by the material. The effect is due to friction between the internal planes which slip or slide as the deformation takes place. When a body having the material damping is subjected to vibration, the stress Strain strain diagram shows the hysteresis loop as shown in the figure. The area of the loop denotes the energy lost per unit volume of the body per cycle.

FOURIER SERIES

The simplest of periodic motion, happens to be SHM. It is simple to handle but the motion of many vibrating system is not harmonic (but periodic) Few examples are shown below:

X(t) X(t) τττ 2τττ 3τττ τττ 2τττ 3τττ τττ τττ

Forces acting on machines are generally periodic but this may not be harmonic for example the excitation force in a punching machine is periodic and it can be represented as shown in figure 1.3. Vibration analysis of system subjected to periodic but nonharmonic forces can be done with the help of Fourier series. The problem becomes a multifrequency excitation problem. The principle of linear superposition is applied and the total response is the sum of the response due to each of the individual frequency term.

Any periodic motion can be expressed as an infinite sum of sines and cosines terms. If x(t) is a periodic function with period t its Fourier representation is given by

a X(t) = o +++ a cos((()(ωωω t )))(+++ a cos((()ωωω t )))(+++ ...... b sin((()ωωω t ))) 2 1 1 1

∞∞∞ ao 2πππ = +++ ∑∑∑ an cos((()(nωωω t )))(+++ bn sin((()ωωω t ))) ωωω === = Fundamental frequency – (1) 2 n===1 t where a o an bn are constants

Determination of constants

To find ao Integrate both sides of equation(1) over any interval τττ . All intergrals on the RHS of the equation are zero except the one containing a o

2πππ ωωω ωωω 2 τττ ao === ∫∫∫ x((()(t))) dt = ∫∫∫ x((()(t))) dt 2πππ o τττ o

To find a n multiply equation 1 by cos (((nωωωt))) and Integrate over any interval τττ . All intergrals 2πππ ωωω ωωω 2 τττ an === ∫∫∫ x((()(t))) cos((()(nωωωt))) dt === ∫∫∫ x((()(t)))( cos((()nωωωt))) dt 2πππ o τττ o

To find bn multiply equation 1 by sin (((nωωωt))) and Integrate over any interval τττ . All intergrals 2πππ ωωω ωωω 2 τττ an === ∫∫∫ x((()(t))) sin((()(nωωωt))) dt === ∫∫∫ x((()(t)))( sin((()nωωωt))) dt 2πππ o τττ o

Find the Fourier series for the curve shown below

X(t) X(t) τττ 2τττ 3τττ τττ 2τττ 3τττ

t t

1 Represent for the periodic motion shown in the figure

X(t) πππ 2πππ

X(t) 0.35 0.25

22 CHAPTER 2 : UNDAMPED FREE VIBRATION

LEARNING OBJECTIVES

] Introduction to undamped free vibration ] Terminologies used in undamped free vibration ] Three methods to solve the undamped free vibration problems ] Problems related to undamped free vibration problems.

Free vibrations are oscillations about a systems equilibrium position that occur in the absence of an external excitation force. If during vibrations there is no loss of energy, it is known as undamped vibration. The first step in solving a vibration problem is setting up the differential equation of motion

Consider spring mass system which is assumed to move only along the vertical direction as shown below. Let m be the mass of the block and k be the stiffness of the spring. When block of mass m is attached to spring , the deflection of spring will be ∆ , known as static deflection. In the static equilibrium position, the free body diagram of forces acting on the mass is shown in Figure(b). Hence mg= kA Once the system is disturbed, the system executes vibrations.

Let at any instant of time t, the mass is displaced from the equilibrium position x, the different forces acting on the system are shown in figure (d)

From Newton’s second law of motion ∑ F = ma Inertia force ( disturbing force) = restoring force m x&& = − k ( ∆ + x ) + mg

m x&& + k ( x ) = 0 k or x&& + ( x ) = 0 m equation 2 is the differential equation of motion for spring mass system shown in 2 figure. Comparing equation (2) with the equation of SHM x&& + ω ( x ) = 0 since the vibrations of the above system are free( without the resistance of external forces) we can write

k ω = rad / sec n m

1 m time period , τ = = 2 π f n k from the equation(1) mg = k∆ k q = m ∆

Difference between the translation ( rectilinear) and rotational system of vibration.

Translatory Rotational

kt k k ω = rad / sec n m ω = rad/sec n

In the analysis the disturbing and restoring In the analysis FORCES are considered In the analysis MASS Moment of Inertia (J) is considered

Linear stiffness K , in N/m is considered

Problems

1.A mass of 10kg when suspended from a spring causes a static deflection of 1cm . Find the natural frequency of system.

2. A spring mass system has a spring stiffness K N/m and a mass of m Kg. It has a natural frequency of vibration 12 Hz. An extra 2kg mass coupled to it. then the natural frequency reduces by 2 Hz. find K and m.

25 3. A steel wire of 2mm diameter and 30m long. It is fixed at the upper end and carries a mass of m kg at its free end. Find m so that the frequency of longitudinal vibration is 4 Hz.

4. A spring mass system has a natural period of 0.2 seconds. What will be the new period, if the spring constant is 1) increased by 50% 2) decreased by 50%,

5. A spring mass system has a natural frequency of 10 Hz when the spring constant is reduced by 800 N/m, the frequency is altered by 45%, Find the mass and spring constant of the original system.

6. Determine the natural frequency of system shown in fig is which shaft is supported in SHORT bearings.

7. Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings.

l where l is the length of bearing and E – young’s modulus and I is moment of Inertia.

7. Determine the natural frequency of system shown in fig is which shaft is supported in LONG bearings.

8. A light cantilever beam of rectangular section( 5 cm deep x 2.5cm wide) has a mass fixed at its free end. Find the ratio of frequency of free lateral vibration in vertical plane to that in horizontal.

9.. Determine the natural frequency of simple pendulum

10. Homogeneous square plate of size l and mass m is suspended from the mid point of one of its sides as shown in figure, Find the frequency of vibration.

26 11. A compound pendulum which is rigid body of mass m and it is pivoted at O. The point of pivot is at distance d from the centre of gravity. It is free to rotate about its axis. Find the frequency of oscillation of such pendulum.

12. A connecting rod shown in fig is supported at the wrist pin end. It is displaced and allowed to oscillate. The mass of rod is 5kg and centre of gravity is 20 cm from the pivot point O. If the frequency of oscillation is 40 cycles/minute, calculate the mass moment of inertia about its C.G.

13. A semi circular homogenous disc of radius r and mass m is pivoted freely about its centre as shown in figure. Determine the natural frequency of oscillation.

14.A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass of 0.575 kg at the middle is found to have natural frequency of 30 rad/sec. Determine young’s modulus of elasticity of beam.

15. A spring mass system, k1 and m have a natural frequency f1. Determine the value of k2 of another spring in terms of k1 which when placed in series with k1 lowers the natural frequency to 2/3 f1.

COMPLETE SOLUTION OF SYSTEM EXECUTING SHM

The equation of motion of system executing SHM can be represented by m x&& + k ( x ) = 0 ------(1) 2 dx 2 dx + ω = 0 dt2 dt

The general solution of equation (1) can be expressed as

X = A cos( ω t) + B sin (ω t) ------(2)

Where A and B are arbitrary constant which can be determined from the initial conditions of the system. Two initial conditions are to be specified to evaluate these constants. x=x 0 at t=0 and x& = V o at t=0. substituting in the equation (2)

V o Is the required complete x = x 0 cos( ω t) + sin (ω t) ω solution

Energy method

In a conservative system the total energy is constant. The differential equation as well as natural frequency can be determined by the principle of conservation of energy. For free vibration of undamped system at any instant of time is partly kinetic and partly potential. The kinetic energy T is stored in the mass by virtue of its velocity where as the potential energy U is stored in the form of strain energy in elastic deformation or work done in a force field such as gravity.

The total energy being constant T+U = constant. Its rate of change

28 d Is given by []T + U = 0 dt From this we get a differential equation of motion as well as natural frequency of the system.

Determine the natural frequency of spring mass system using energy method.

Determine the natural frequency of system shown in figure.

θ l k

Determine the natural frequency of the system shown in figure. Is there any limitation on the value of K. Discuss?

θ k l

Determine the natural frequency of system shown below. Neglect the mass of ball.

k m

a l

A string shown in figure is under tension T which can be assumed to remain constant for small displacements. Find the natural frequency of vertical vibrations of spring.

T T m

An acrobat 120kg walks on a tight rope as shown in figure. The frequency of vibration in the given position is vertical direction is 30 rad/s. Find the tension in the rope.

8m 36m

A manometer has a uniform bore of cross section area A. If the column of liquid of length L and Density ρ is set into motion as shown in figure. Find the frequency of the resulting oscillation.

Find the expression for natural frequency of system shown in the figure. Neglect the mass of the cantilever beam. Study the special case i) k=Infinity ii) I = infinity.

Determine the expression for the natural frequency of a system shown in figure. The two discs are keyed to a common shaft and have a combined mass moment of inertia about centre of oscillation O. It is assumed that the card attached to mass m does not stretch and is always under tension.

Determine the natural frequency of a system shown

Determine the expression for the natural frequency of the system shown in figure. Assume that the wires connecting the masses do not stretch and are always in tension.

M3 M1

k2 k1

Determine the natural frequency of spring mass system taking the MASS OF SPRING (m s ) into account.

RAYLEIGH’S METHOD.

This is the extension of energy method. Here we can determine the natural frequency of a conservative system without finding the equation of motion. The natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system.

From the principle of conservation of energy we have,

T+U = Constant

T= K.E and U =P.E

We can write the above equation as

T1+U1 = ------(1)

Where 1 and 2 represents tow instances of time.

Let 1 be the time when the mass is passing through static equilibrium position

Let 2 be the time corresponding to the mass displacement of the mass, At this instant the velocity f the mass is zero and hence

Substituting in equation (1) we get

( ) Max = ( ) Max

Above equation leads directly to natural frequency.

Determine the natural frequency of spring mass system using Rayleigh’s method.

Home work .

Determine the natural frequency of pendulum using Rayleigh’s method.

A Cylinder of mass m and moment of inertia J o is to role without slipping but is sustained by spring K as shown in figure. Determine the natural frequency of oscillation.

Determine the natural frequency of system shown in figure. If the cylinder is free to roll without slipping.

K K

Determine the natural frequency of system shown where in cylindrical disc rolls over the inextensible spring.

A cylinder of radius r rolls without slipping on cylindrical portion of radius R. Derive the equation of motion and hence determine the natural frequency of small oscillations about the lowest point shown.

Repeat the above problem – Instead of cylinder assume sphere of radius with r rolls without slipping on concave surface as shown above.

35 The torsional pendulum has a natural frequency of 5 Hz. What length of steel wire of diameter 2 mm should be used for pendulum. The inertia of mass fixed at the free end is 0.0098 kgm 2. Take g= 0.83x10 11 N/m 2

Determine the torsional natural frequency of the system shown in figure. Neglect the mass moment of inertia of the shaft. ( Figure not proportionate)

Kt2 0.6m Kt1 0.8m 0.4m

20cm 20cm 20cm 20cm

Determine the natural frequency of simple pendulum considering the mass of the rod into account.

Using the energy method, find the natural frequency of the system shown in figure. The chord may be assumed inextensible in the spring mass pulley system.

36 A rod of length l and depth H having a mass m rests on a semicylidrical surface os radius r. It is tipped slightly and let go. Find the frequency of oscillation

A thin bar of l having a mass m rests on a semicylidrical surface radius r. It is tipped slightly and let go. Find the frequency of oscillation l

37 CHAPTER 3

DAMPED FREE VIBRATION OF SINGLE DEGREE FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to damped free vibration ] Terminologies used in damped free vibration ] Three different types of damping ] Concept of critical damping and its importance ] Study of response of viscous damped system for cases of under damping, critical and over damping. ] Concept of logarithmic decrement ] Problems related to damped free vibration problems.

Viscously damped free vibration of single degree freedom sytem

Consider a single degree freedom system with viscous damping as shown in figure, For a viscous damped system, the damping force is

where c is a constant of proportionality. It is called damping coefficient or coefficient of viscous damping

Symbolically. it is designated by a dashpot, as shown in Fig. 3.

x x& x

38 At any instant of time, let the mass be displaced by x from the equilibrium position. Then different forces acting on system is shown in the free body diagram

By Newton’s second law of motion, mx&& = − xc& − kx = O mx&& + xc& + kx = O ------(1) or d 2 x dx M + C + kx = 0 dt 2 dt Assuming the solution of equation as

−Ct 2m The first term in the above equation e is an exponentially decaying function of time. The behavior of the terms in the parenthesis however depends on whether numerical value within the radicals is positive , zero or negative.

 c  2 k CASE 1: if   − = 0  2m  m

 c  2 k CASE 2: if   > = 0  2m  m

 c  2 k CASE 3: if   < = 0  2m  m

40 Critically damped coefficient

Critically damped coefficient is defined as the value of the damping coeffient for which radical in equation

 c   c  2 k S =   ±   − becomes zero. 2,1  2m   2m  m

Damping Ratio ( Damping factor) ( ξ )

It is defined as the ratio of damping coefficient to the critical damping coefficient

Note : The damping coefficient c is defined as the actual amount of damping while the damping coefficient is a constant depending upon the mass and stiffness of the system. The roots of characteristics equation can be now written as

Note : if c=c c or ξ =1 then system is critically damped

if c> cc or ξ >1 then system is over damped

I if c< cc or ξ <1 then system is under damped

Take home quiz Derive General solution of different damped system 1. Critically damped system 2. Under damped system 3. Over damped system

Logarithmic Decrement

It is used to determine the amount of damping present in system. This measure The rate of decay of free vibration. The larger the damping the greater will be rate of decay. Logarithmic decrement is defined as the natural logarithm of ratio of any two successive amplitudes .   x1 δ = ln   x 2 

Expression for logarithmic decrement ( δ ) Consider a damped free vibration which is expressed by

()ξω t 2 x = Xe n sin ω 1− ξ t + φ ( n ) x is the displacement at any instant of time. X and φ are arbitrary constants

2πξ δ = 1− ξ 2

42 Tutorial problems

1. Derive an expression for damping ratio in terms of logarithmic decrement

1  x   o  2. Show that logarithmic decrement δ = ln  where x o is the initial amplitude x n n  x n  is amplitude after n cycles.

3. A mass of 1 kg is to be supported on a spring having stiffness. K= 9800 N/m. The damping constant c= 5.9 Ns/m Determine the natural frequency of system. Find also logarithmic decrement and amplitude after 3 cycles if initial displacement is 0.3 cm.

4. A vibrating system is defined by the following parameters. M = 3kg , k= 100 N/m. damping coefficient = 3 Ns/m. determine Damping factor, natural frequency , damped vibration, logarithmic decrement , ratio of consecutive amplitude, no of cycles after which the original amplitude is reduced to 20%.

5.A single degree damped vibrating system ,a suspended mass of 18kg makes 15 oscillations in 0.03 seconds. The amplitude decreases to 0.25 of the initial value after 5 oscillations. Determine the stiffness of spring, Logarithmic decrement , damping factor and damping coefficient.

6. A damped vibration record of a spring mass system shows the following data. Amplitude at the end of 2 nd cycle is 9mm. amplitude at eh end of 3 rd cycle is 6mm. amplitude at the end of 4 th cycle is 4mm. stiffness of spring =8000 N/m weight =20N. Find the logarithmic decrement, damping force at unit velocity, periodic time of vibration.

7. For the system shown in figure . the characteristics of dashpot is such that the constant force of 49 N is applied to the piston it velocity is found to be constant at 0.12 m/s. Determine the value of C, would you expect the complete system to be periodic or aperiodic.

8. For a torsional system shown in the figure.f1 is natural frequency in air and f2 is the natural frequency when immersed in oil. Determine the damping coefficient.

Disc of mass moment of Inertia J Viscous fluid

43 9. A thin plate of area A and weigh w is fixed to the end of the spring and allow to oscillate in a viscous fluid as shown in figure. It f 1 is the frequency of oscillation in air and f 2 in liquid. Show that 2πω µ = f 2 −f 2 where the damping force on the plate gA 1 2

Fd = µ2AV , V being the velocity of vibration.

10. A torsional pendulum has a natural frequency of 175 cycles/second vibrating in vacuum. The mass moment of inertia is 2 kgm 2. It is immersed in oil and it is observed that the natural frequency is 142 cycles/second. Find the damping torque. If the disc is replaced by 3.4 deg. When in oil, find the displacement at the end of the first complete cycle.

11. An automobile can be modeled as a mass placed on 4 shock M 200kg=200kg absorbers each consisting of a spring and a damper. Such that kg kg each spring is equally loaded. Determine the stiffness damping constant of each shock absorber. So that natural frequency is 2Hz and system is critically damped. The mass of vehicle is 200kg.

12. Write the differential equation of motion for the system shown in the figure below . Determine the natural frequency of the damped oscillation and critical damping coefficient.

Determine the differential equation of motion and find the critical coefficient and damped natural frequency for system shown in figure.

a a

2a C k

44 Derive the equation of motion and find the critical damping coefficient and damped natural frequency for the system shown below:

k M C

Write the differential equation of motion and determine the expression for critical damping coefficient and damped natural frequency of system shown in figure.

Set up a differential equation of motion for system shown in figure and determine the following: Undamped natural frequencies, critical damping coefficient , damping ratio, damped natural frequency,

45 A mass of 300N is resting on two springs of stiffness 3000 N/m each and a dashpot of damping coefficient 150 Ns/m as shown in figure. If the initial velocity of 10cm/s is given to mass at its equilibrium position what will be the displacement from the equilibrium position at the end of first second. Derive the formula used.

W= 3000 N

A gun barrel of mass 600kg has a recoil spring of stiffness 294 kn/m. if the barrel recoils 1.3m on firing determine i) The initial recoil velocity of barrel ii) The critical damping coefficient of dashpot which is engaged at the end of recoil stroke. Iii) Time required for the barrel to return at the end of recoil stroke. Iv) time required for the barrel to return to a position 5cm from the initial position.

FIRING

46 CHAPTER 4

FORCED VIBRATION

LEARNING OBJECTIVES

] Introduction to forced free vibration ] Terminologies used in forced vibration ] Concept of magnification factor ] Study of reciprocating and rotating unbalance. ] Concept of base excitation and support excitation ] Concept of Transmissibility ratio ] Problems related to forced vibration.

The vibration that takes place under the excitation of external forces are called forced vibration. When a system is subjected to harmonic excitation, it is forced to vibrate at the same frequency as that of excitation. Most of the mechanical systems are prone to steady state forced vibration which cause fatigue failure. The vibration problem also becomes very important when any of excitation frequencies coincide with one of natural frequencies of a system. Under this condition resonance occurs during which large amplitude of vibrations are observed. To avoid resonance external frequency (operating speed) may be changed or properties of the system may be changed to alter the natural frequency. In some cases, sufficient amount of damping may be provided to avoid large amplitude during resonance. Thus the problem of force vibration is very important in mechanical design.

FORCED VIBRATIONS OF SINGLE DOF SYSTEMS UNDER HARMONIC EXCITATION

F = Fo sin[ω t] ` F = F sin ω t o [ ] x x& x&&

Consider a spring mass system having a viscous damping excited by a harmonic force F = Fo sin[ω t] as shown in figure. The various forces acting on the system is shown in figure Then different forces acting on system is shown in the free body diagram

By Newton’s second law of motion,

mx&& = − xc& − kx + Fo sin(ω )t = O ------(1) mx&& + xc& + kx = Fo sin(ω )t or d 2 x dx M + C + kx = F sin(ω )t dt 2 dt o

Equation (1) represents the differential equation of motion of the system. The solution to this equation consists of two parts namely complementary function and a particular integral.

The first of solution is also termed as transient solution of homogeneous solution and can be obtained from the homogeneous differential equation. mx&& + xc& + kx = 0

The second part of solution, particular integral is also termed as steady state solution ad is of form x = Xsin(ωt − φ) . Where X= steady state amplitude , φ is phase difference between external force and displacement . x = Xsin(ωt − φ)

The vector representation of the shown

The force polygon for the above vector diagram can be written as follows

Cωx m ω 2 x

Fo KX

(ωt − φ) O φ ωt

  ω  2ξ  −1  ωn  φ = tan  2    ω   is the governing equation. 1−      ωn  

50 Zero frequency deflection (X st ) It is deflection observed at zero frequency ( when the external frequency, ω = 0

 F o     k  F o x st =   =  ()1 − 0 2 + 0 2  k  

Magnification factor (M) It is defined as the ratio of steady state amplitude to zero frequency deflection x st .

X F M = = o X ar k

  A plot of magnification factor versus frequency ratio  ω  for varying values of  ωn  damping factor (ξ) is called frequency plot of a harmonically excite vibrating system.

All the frequency terms starts from magnification factor M=1, what ever may be value of damping ratio (ξ) . The amplitude of vibration at any particular frequency depends on the value of damping. The amplitude will be higher for lesser damping and lesser for higher damping. The amplitude of vibration at resonance will be very high. The   amplitude tends to zero as  ω  tends to infinity.  ωn 

  ζ = 0 ζ = 0  ω  2ξ −1  ωn  φ = tan  2    ω   1−      ωn   Magnification ζ = 0.25 Factor M = X/X ζ = 0.375 st ζ = 0. 5 ζ = 0.707

ζ = 1 ζ = 2

Frequency Ratio r = (ω/ω n)

ζ = 0.25 Plot of phase angle φ versus frequency ratio ζ = 0.5    ω  is shown in the ζ = 0.707  ωn  figure. The phase angle ζ = 1.0 φ is zero for all the values of damping ratio Phase Angle, φ, ζ = 2.0   at  ω  =0. The phase  ωn  angle φ is 90 for all the values of damping ratio   at  ω  =1. i.e at  ωn  resonance. The phase angle between o to 90 0.5 1.0 1.5 2.0 2.5 3.0 deg for all the values of   Frequency Ratio r = (ω/ω n)  ω  = 0 to 1 and  ωn    between 90 deg to 180 deg for all the values of  ω  >1  ωn 

Class work Derive an expression for frequency at peak amplitude

2 ω = ωp = ωn 1− 2ξ is the required equation.

52 Derive expression amplitude at resonance ( In dimensional form)

In dimensional form

We know that steady state amplitude

   F  x =  o  Equation in dimensional form  2 2 2   ()k − m ω + ()C ω 

k ω = ω = We know that at resonance n m

      F o x R =   ,  2   k  2   k − m  + ()C ω n    m  

     F o  F o x R =   x R =    2   ()C ω n   ()C ω n 

Class work Derive an expression for amplitude at resonance ( in non dimensional form)

53 Tutorial Problems on Forced Vibration

1.A weight of 60N is suspended by a spring stiffness 1.2 kN/m is forced to vibrate by a harmonic force of 10N. assuming viscous damping of 0.086 kN-s/m. Find i) resonance frequency ii) amplitude at resonance iii) phase angle at resonance iv) frequency corresponding to peak amplitude.

2. A periodic torque having a maximum value of 0.5888Nm at a frequency of 4 rad/s is impressed upon a flywheel suspended from a wire. The wheel has mass moment of inertia 0.12 kg/m 2 and the wire has stiffness 1.176 Nm/rad. A viscous dashpot applies damping couple of 0,784 Nm an angular velocity of 2 rad/s. calculate i)maximum angular displacement from rest position ii) maximum couple applied to dashpot. Iii) angle by which the angular displacement lags the torque.

A mass of 10.19 kg is suspended from a spring stiffness 20 N/mm. viscous damping causes the amplitude to decrease to 1/10 th of initial value in 4 complete cycle. If a periodic force of 20 cos(50 t) is applied to the mass. Find the amplitude of forced vibration. What would be the amplitude of oscillation if the applied force coincides with the natural period. m=10.19 kg

F = 30 cos[50 t]

54 ROTATING UNBALANCE

Unbalance in rotating machinery is a common source of vibration excitation. Consider a machine of mass m having mounted on a foundation of stiffness k and damping coefficient C as shown in figure. The unbalance is represented by an eccentric rotating mass m with eccentricity (e) and having a angular velocity ( ω). let x be the displacement of non rotating mass from the static equilibrium position. The displacement of rotating mass m from the static equilibrium is (x + e sin(ω t)

Applying Newton’s law of motion , we have

  ωωω  2ξξξ    −1 ωωωn φφφ = tan  2    ωωω   1−      ωωωn  

55 RECIPROCATING UNBALANCE

Unbalance in reciprocating machinery is a common source of vibration excitation. Consider a machine of mass M having mounted on a foundation of stiffness K and damping coefficient C as shown in figure. The unbalance is represented by a reciprocating mass (m) having crank radus (e) and connecting rod of length (l) Let ωωω be angular velocity of the crank . Let x be the displacement of non reciprocating mass from the static equilibrium position at any instant of time. The displacement of reciprocating mass m from the static equilibrium position is

e x + e sinωωωt + sin(2ωωωt) + ...... l

  ωωω  2ξξξ  −1  ωωωn  φφφ = tan  2    ωωω   1−         ωωωn  

56 Note : The equations obtained in reciprocating unbalance is same as the rotating unbalance and having the following differences. Rotating Unbalance Reciprocating Unbalance Here mass of the rotating element Here mass of reciprocating element is is considered considered

Here the eccentricity (e) which is the Here the eccentricity e is the crank radius distance between the centre of rotation and crank radius = stroke length/2 and C.G

The frequency plot of non dimensional factor( MX versus frequency ratio ( me)    ω  for various values of (ξξξ )is shown in figure. This frequency plot is same for  ωn  the rotating and reciprocating unbalance with the following points are noted from the diagram.

1. Damping ratio plays very important role during resonance in controlling the amplitude.   2. When  ω  =0 then (MX )=0 for all the values of damping ratio (ξξξ )  ωn  me   3. When  ω  is higher than (MX ) tends to unity  ωn  me   4. At very high speeds i.e at very high frequency ratio  ω  there is no effect  ωn  of damping

The frequency plot of phase angle versus the frequency ratio is given as shown in figure. The explanation is same as in forced harmonic vibration.

57 VIBRATION ISOLATION

Vibratory forces and motions generated by machine and other caused are unavoidable . Howeever their effect on the other dynamical systems can be minimized by proper isolator design. An isolation system is either to protect o delicate machine from excessive vibration transmitted to it from its supporting structure or to prevent vibratory forces and motions generated by machines from being transmitted to its surroundings. The effectiveness of isolation may be measured in terms of forces and motion transmitted. The first type is known as force isolation and the second type is known as motion isolation.

Force isolation (Force Transmissibility) When vibratory forces generated by machine are isolated the effectiveness of forced isolation can be made by defining the term transmissibility. The term transmissibility is defined as the ratio of force transmitted to the foundation to that of the external impressed force.

Force transmitted to foundation F Force Transmissibility = = tr External Impressed force Fo

Consider viscous damped spring mass system as shown in figure. Upon this system an external harmonic force Fo Sin (ω t ) is applied. The difference forces acting on system are inertia forces , damping force, spring force and external force. All these forces are represented on force polygon shown below.

2 2

¡ ¢ = (kx) + (cωx)

This is the expression for force transmissibility in dimensional form. To represent same in the non dimensional form divide numerator and denominator by k.

2  ω  1+ 2ξ  ω Ftr  n  TR = = 2 F 2 2 o   ω    ω  1−    + 2ξ  ω ω   n    n 

PHASE DIFFERNECE BETWEEN EXTERNAL FORCE AND DISPLACEMENT

Out of these four forces the spring force and damping force are common forces acting on mass as well as on foundation. Hence the force transmitted to the foundation F tr is the vector sum of these  ω   2ξ  −1  Cω  −1  ωn  φ = tan   = tan   K − mω2   2   ω   1−      ωn   The phase difference between force transmitted to the foundation Ftr to the displacement.

Cωx cω −1  Cωx  −1  ω  tan α = = or α = tan   = tan 2ξ  Kx k  Kx   ωn 

The phase difference between external force F o and force transmitted to foundationis β

 ω   2ξ  −1  ω  −1  ω  = = n -   β φ − α tan  2  tan 2ξ     ω    ω   n 1−      ωn  

Displacement Transmissibility

It is the ratio of amplitude of steady state vibration of machine to that of the base motion.

Forced vibration due to support motion

Consider a spring mass damper system subjected to harmonic excitation y = Ysin(ωt) at the base as shown in figure. Let x be the absolute displacement of the mass at any instant of time t. The different forces acting on system are shown in free body diagram.

Applying Newton’s Law of motion. Ma = ∑ F m x&& = − x(c & − y& ) − k x( − y )

m x&& + x(c & − & )y + k x( − )y = 0

Frequency plot of transmissibility versus frequency ratio.

The plot of transmissibility versus frequency ratio is as shown in figure. The following points can be noted from it.

  i) The value of transmissibility is unity at frequency ratio  ω  =0 and 2  ωn  ii) For an undamped system (ξ) =0 transmissibility tends to ∞ at resonance iii) The value of transmissibility is less than unity for the value of frequency ratio greater than 2

  iv) For frequency ratio  ω  < 2 , smaller damping ratio lead to larger values of  ωn    transmissibility on the other hand for frequency ratio  ω  > 2 smaller damping  ωn  ratio leads to smaller value of transmissibility.   The frequency plot of transmissibility (TR) versus frequency ratio  ω  can be  ωn  broadly divided into two zones

  1  ω  = 0 to 2  ωn 

  2.  ω  = 2 to ∞  ωn 

CASE 1

  When  ω  = 0 to 2  ωn  In this zone , the value of transmissibility is always greater than 1 for all values of damping ratio. In this zone, the transmissibility decreases with increase in damping. Hence we can conclude that in this zone, damping is useful in isolating vibration forces being transmitted to the support

CASE 2

  When  ω  = 2 to ∞  ωn  In this zone, the value of transmissibility is less than 1 for all values of damping ratio. In this zone, the transmissibility increases with increase in damping. Hence we can conclude that in this zone, damping will adversely affect in isolating vibratory forces.

Tutorial problems on Forced Vibration

1. Machine of mass 75kg is mounted on spring of stiffness 12000 N/m with an assumed damping factor ξ =0.2 piston within the machine of mass 2 kg has a reciprocating motion with a stroke of 7.5 cm and speed of 15 cycles/second. Assuming the motion of piston to be harmonic, determine amplitude of machine ( X), Transmissibility and force transmitted to foundation, phase angle of transmitted force w.r.t to exciting force.

2. Motor weighs 600N is mounted on a simple beam that has a spring rate of 400 N/cm. The motor armature base 160 N and has an eccentricity of 0.005cm. What will be the amplitude of motion when it runs at 1760 RPM. Neglect damping, the deflection of armature shaft and weight of the beam.

3. A Heavy machine of mass 300 kg is supported on resilient foundation the static deflection of foundation due to weight of machine is found to be 75mm. It is observed that the machine vibrates with an amplitude of 10mm, when the base of the foundation is subjected to harmonic oscillation at the undampeds natural frequency of system with an amplitude of 2.5mm. Find the damping constant of foundation. Ii) dynamic force amplitude on the base iii) amplitude of the displacement of machine relative to base.

4. An air compressor of mass 450 kg operates at a constant speed of 1750 RPM. The reciprocating masses are 10 kg , the crank radius is 100mm. If the damper for mounting introduces a damping factor 0.15 i) specify the springs of mounting such that only 20% of the unbalanced force is transmitted to the foundation. ii) determine the magnitude of the transmitted force.

• Repeat the above problem by assuming that there is no damping in system.

5. A body of a car having mass 1500 kg and is mounted on 4 equal springs which deflect through 22.5 cm under the weight of the body. A total damping forces the 4 shock absorbers is 46 N at a velocity of 1cm/sec. The car is placed with all the four wheels on a test platform which is move up and down at resonant speed. With an amplitude of 2.5cm. Find the amplitude of the car body on its springs assuming its C.G. to be in the centre of wheel base.

6.A 75 kg machine which is mounted on springs of stiffness 11.6x10 5 N/m with the damping of ξ = 0.2 . A 2kg piston within the machine as reciprocating motion with a stroke of 0.08m and speed of 3000 RPM. Assuming the motion of piston to be harmonic , determine the amplitude of vibration of machine.

7. The spring of an automobile are compressed 10cm under its own weight . Find the critical speed when a trailer is traveling over road with a people approximaterly by a sine wave of amplitude 7.5 cm and wave length 15m. What will be the amplitude of vibration at 64km/hour.

8. An eccentric mass excites us used to determine vibratory characteristics of a structure of mass 25kg as given in figure. At a speed of 1000 RPM , the eccentric mass to beat top position at the instant the structure was moved upwards through its static equilibrium position and corresponding amplitude was 20mm. The eccentric mass is 0.5kg at an eccentricity of 10mm. Determine i) Undamped natural frequency of the system ii) Damping ratio iii) Steady state amplitude and phase angle at 1500 RPM. iv) Force transmitted to the ground at 1200 RPM.

Eccentric mass

M=25kg

C K

64 CHAPTER 5

VIBRATION MEASURING INSTRUMENTS WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to Vibrometer and accelerometer ] Terminologies used ] Concept of critical speed and its importance in engineering practice ] Derivation of critical speeds with and without air damping ] Discussion of speeds above and below critical speeds ] Problems related to whirling of whirling of shafts

Support motion

Consider a spring mass damper system as shown in figure. Let it excited by motion of support, let y be the harmonic displacement of support.

From Newton’s second law of motion

m x&& = − c(x& − y& ) − k (x − y) m x&& + c(x& − y& ) + k (x − y) = 0 − − − − − − − − − − − − − − − − − − (1)

Let z be the relative displacement of the mass with respect to support then Z=x-y

 ω   2ξ   ω  φ = tan −1  n   2   ω   1−      ωn  

66 From the frequency plot shown following points can be noted. Z • At higher frequency ratio the amplitude ratio is almost equal to unity. Then Y relative amplitude z and support amplitude Y are equal. Z • When is equal to 1 or Z=Y. it means that the mass will not be having any Y displacement or zero displacement. • For higher frequency ratios, the amplitude ratio will not have any effect.

VIBRATION MEASURING INSTRUMENTS.

It measure displacement velocity and acceleration of a vibrating body. These are also known as seismic instruments. This instrument have mass spring and dashpot. The construction of vibrating measuring instrument is as shown in above figure. Seismic mass m is permanent magnet that is moving relative to coil fixed to the casing thus the voltage is generated due to coils cutting the magnetic flux lines. The output of electrical signal of instrument will be proportional to the quantity which is to be measured.

There are two types of vibration measuring system. 1. Vibrometer (seismometer) – an instrument with low natural frequency 2. Accelerometer

VIBROMETER

This is a device to measure the displacement of vibrating body. This is designed wit low natural frequency. We know for vibration measuring instrument

Z  r 2  =   2 2 Y  ()1 − r 2 + ()2 ξ r    when the natural frequency of instrument is low then frequency ratio  ω  tends to  ωn  higher values for undamped system. Z  r 2  Z = which implies = 1 or Z= Y.  2  Y  ()1 − r 2  Y so relative amplitude Z is equal to the amplitude of vibrating body Y for every values Z   of damping ratio. This is shown in frequency plot of versus ω  ω  Y  n  One disadvantage of vibrometer is its large size because it is an instrument with low natural frequency. The seismic mass remains stationary while the frame moves with an vibrating body. These instruments can be used to measure velocity and acceleration by incorporating differentiators.

ACEELEROMETER

Accelerometer is used to measure the acceleration of vibratory body. The natural frequency of accelerometer is high compare to the frequency which is to be measured.

We know for a vibration measuring instrument 2  ω    Z  ω  = n Y 2 2 2   ω    ω  1−    + 2ξ  ω ω   n    n 

ω2 Y where Z = .f 2 ω n where f is a factor which remains constant for the useful range of accelerometer.  1    Where f =   with r =  ω  2 2  ω   ()1 − r 2 + ()2 ξ r  n

The equation (1) ω2 Y is the acceleration of the vibrating body. It is clearly seen that 1 the acceleration is multiplied by factor . To keep the value of factor f=1, for the 2 ω n   higher range of  ω  , ξ should be high, then amplitude Z becomes proportional to  ωn  amplitude of acceleration to be measured. i.e. Z> ω2 Y or Z α acceleration where

ωn is constant

From the equation (2) the following plot can be drawn. The figure shows the response of the accelerometer . It is seen that for ξ =0.7 there is complete linearity for accelerometer i.e. f=1 for frequency ratio less than 0.25. Since the natural frequency of accelerometer is high , it is very light in construction. Using integration circuits, one can get the display of velocity and displacement . the same instrument can be used to measure the velocity and displacement by incorporating integrator

TUTORIAL PROBLEMS ON VIBRATION MEASURING INSTRUMENTS

1.A vibrometer has a natural frequency of 4 rad/s and ξ 0.2 is attached to a structure that perform the harmonic motion. If the difference of maximum and minimum recorded value is 8mm. Find the amplitude of motion of vibrating structure when the frequency is 40 rad/second.

2. A damping in a vibrometer is observed to be negligible it gave a relative displacement of 0.1cm when used on vibrating structures. The natural frequency of vibrometer is 300 RPM and the vibrating unit was running at 120 cycles/minutes. Find the magnitude of displacement and acceleration of the vibrating units.

3.A device used to measure torsional acceleration consists of a ring having a mass moment of inertia 0.049 kg/m2 is connected to shaft by a spiral spring having constant 0.98 Nm/rad and a viscous damper having a constant of 0.11 Nm-s/rad . When the shaft vibrates with a frequency of 15 cycles/minute. The relative amplitude between ring and shaft is found to be 2 deg. What is the maximum acceleration of the shaft.

4.A commercial vibration pick-up has a natural frequency of 5.75 Hz and damping factor of 0.65. What is the lowest frequency beyond which the amplitude can be measured with in (a) 1% error (b) 2% error.

5..An accelerometer is made with a crystal of natural frequency 20 kHz. The damping ratio of accelerometer is found to be 0.71. Determine the upper cut off frequency of the accelerometer for 1% accuracy.

6.A vibrometer having a natural frequency of 4 rad/sec and ζ = 0.2 is attached to a structure that executes harmonic motion. If the difference between the maximum and minimum recorded value is 8 mm, find the amplitude of vibration of structure when its frequency is 40 rad/sec.

7. A vibrometer has a natural frequency of 10 cps and has a damping ratio of 0.7. It is used, by mistake, to measure vibrations of a fan base at an exciting frequency of 180 rpm. The measured vibration velocity of the fan base is 3 mm/s. What is the actual velocity of the fan base?

8. A seismic instrument is fitted to measure the vibration characteristics of a machine running at 120 rpm. If the natural frequency of the instrument is 5 Hz and if it shows 0.004 cm determine the displacement, velocity and acceleration assuming no damping

9.A vibrometer indicates 2 percent error in measurement and its natural frequency is 5 Hz. If the lowest frequency that can be measured is 40 Hz, find the value of damping factor

70 CHAPTER 5 ( CONTD..)

WHIRLING OF SHAFTS

LEARNING OBJECTIVES

] Introduction to whirling of shafts ] Terminologies used ] Concept of critical speed and its importance in engineering practice ] Derivation of critical speeds with and without air damping ] Discussion of speeds above and below critical speeds ] Problems related to whirling of whirling of shafts

Introduction

Critical speed occurs when the speed of the rotation of shaft is equal to the natural frequency of the lateral vibration of shafts, At this speed shaft starts to vibrate violently in the transverse direction. Whirling is defined as the rotation of the plane created by bent shaft and the line of centre of bearings.

The excessive vibrations associated with critical speeds may cause permanent deformation resulting in structural damage. Example: The rotor blades of a turbine may come in contact with stator blades. Larger shaft deflections produce larger bearing reactions,which may lead to bearing failure. The amplitude build up is a time dependent phenomenon and therefore, it is very dangerous to continue to run the shaft at it critical speed.

This phenomenon results from the following causes: 1.Mass unbalance 2.hystersis damping in shaft 3.Gyroscope forces 4.Fluid friction in bearing.

Bearing centre

Undeflected Position O O C X C Deflected Position

G G

71 Bearing centre line

ωωω

Bearing Bearing

ωωω Bent up shaft axis Rotor or Disc

The whirling motion of a shaft consists of two components of motion as shown a. Spinning of the shaft along with rotor about the bent up shaft axis. b. Rotation of plane A made by the centre line of the bearings and bent up-shaft, about the centre line of the bearings.

Bearing centre line

ωωω

Plane A Plane A

Bearing Bearing Rotation of plane A

ωωωplane A Bent up shaft axis Rotor or Disc

The rotation of plane A, which is generally referred as whirling, may take place in the same sense as that of spinning of the shaft or in the opposite sense. Further the speed of whirling may or may not be equal to the speed of spinning of the shaft. When the whirling speed is equal to the speed of rotation of shaft it is called “synchronous whirl ”.

Critical speed of a shaft with a single rotor (with out damping): Consider a shaft on which a rotor in symmetrically located between two bearings. The expression for the deflection of the shaft in terms of frequency ratio and eccentricity can be obtained as follows based on the following assumptions. 1. Shaft is light and flexible. 2. Gravity effects are negligible. 3. Friction at shaft centre is small. 4. Damping due to air is neglected.

Let m: mass of the disc. ω: Angular rotation of the disc (uniform angular velocity of shaft) e: eccentricity of the disc: radial distance of the mass centre of the disc from its geometric centre- G. K: Stiffness of the shaft in transverse direction C: Geometric centre of the disc. G: C.G of disc (mass centre) X: Lateral deflection of the shaft centre from 0. (OC) (deflection of the geometric centre of the disc). ωc: Critical speed of the shaft.

The rotor, (disc) is in equilibrium under the action of two forces. Centrifugal force, which acts radially outwards through G = m ω2 (x + e) Restoring force which act radially inwards through C = KX ∴For equilibrium restoring force = Centrifugal force

X/e = 1/[(1/r 2) –1]

Discussions: The relation between X/e and ωc/ω can be plotted as shown below in Figure 5.13.

+ ve

r < 1

X/e

ωωωn 1.0 ωωω

- ve r >1

Relation between X/e and ωωωn /ωωω

Case (i): When ωωω =ωωωn (r =1) • Forcing frequency coincides with the natural frequency of transverse vibration of the shaft. X/e – approaches infinity i.e., the deflection of geometric centre of the disc tends to infinity. • The disk has a tendency to fly out, if the damping is insufficient. There will be severe vibrations of the shaft thereby producing huge bearings reactions. • At ω = ωn, the above undesirable effects would occur and therefore ωωω = ωωωn = ωωωc is called the critical speed of the shaft.

Case (ii): ωωω < ωωωc, r < 1 ω <<< ωn, r < 1 ∴X/e = is positive. The deflection x and eccentricity ‘e’ are in the same sense. This condition of disc is referred as “ Heavy side outside ” i.e.,. The disc rotates with heavy side outside. Thus C will lie between O and G. Positive sign indicates that X is in phase with CF.

O · ·C

· G

Disk with Heavy side outside

Case (iii): When ωωω > ωωωc, r > 1 ω >>> ωn X/e = negative, the deflection x and the eccentricity e are in opposite sense. This condition of the disc is referred as “ Heavy side inside ”. G falls between O and C. Negative sign indicates that X is out of phase with CF.

O · · G ·C

Disk with Heavy side inside

When ωωω is very large, ωωω/ωωωn = r ∞∞∞ G tends to coincide with O. The disc tends to rotate about its mass centre and hence vibrations are very minimum. This is the principle used for stabilization of aircrafts at high speeds.

Dynamic force transmitted to the bearings . 2 2 Fd = KX ωn = K/m, K = m ωn at the critical speed 2 = m ω nX = m ω2X, at ω Note: 2 1. ω < ωn, r < 1, X/e is positive, F d = m ω (X + e) 2 2. ω > ωn, r > 1, X/e is negative, F d = m ω (X – e) If the shaft is vertical dynamic load on each bearing F B = F d/2 If the shaft is horizontal dynamic load on each bearing = F B = (mg/2 + F d/2)

75 Tutorial problems on Critical speeds with out damping

1. A rotor has a mass of 12 Kg and is mounted midway on a horizontal shaft of 24 mm φ supported at the ends by two bearings. The bearings are 1 m apart. The shaft rotates at 1200 rpm. The mass centre of the rotor is 0.11 mm away from the geometric centre of the rotor due to certain manufacturing errors. Determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings if E = 200 GN/m 2.

2. A rotor having a mass of 5 Kgs is mounted midway on a 10 mm diameter horizontal shaft supported at the ends by two bearings. The bearing span is 400 mm. Due to certain manufacturing in accuracies the CG of the disc is 0.02 mm away from the geometric centre of the rotor. If the system rotates at 3000 rpm determine the amplitude of steady state vibrations and dynamic force transmitted to the bearings. Neglect damping. Take E = 1.96 * 10 5 N/mm 2

3. shaft of 14 mm φand the length 1.2 m is held in long bearings . It carries a rotor of 16 Kgs at its midspan. The eccentricity of the mass centre of the rotor from the shaft centre is 0.4 mm. The shaft is made of steel for which E = 200 GN/m 2 and permissible stress is 70 MPa. Determine the critical speed of the shaft and range of speed over which it is unsafe to run the shaft. Assume the shaft is mass less, (a) When the shaft is horizontal (b) When the shaft if vertical.

4. A shaft of 25 mm diameter is freely supported on bearings 750 mm apart carries a rotor of 20 Kgs, midway between the bearings. Determine the critical speed of the shaft, if the shaft material has a density of 8000 Kg/m3 and E = 2.1 * 10 5N/mm 2.

5. A 6mm diameter shaft is mounted on two bearings 50 cm apart which acts as simply supports. Midway between these bearings a block of CI weighing 5N is tightly shrunk on the shaft so that its C G is 2.5 cm from the centre line. The shaft rotates at 700 RPM. Neglect the weight of the shaft , assume that the elastic limit of the shaft is not exceeded. Determine the maximum bending and minimum bending stresses on the surface of shaft, if the shaft is horizontal. ASSUME NO DAMPING take E=200 GPA .

Critical Speed of a Shaft with a Single Rotor with Damping In engineering applications, rotors are subjected to air-resistance or structural damping. However, for analytical purposes, equivalent viscous damping may be considered with a viscous damping coefficient C and damping ratio ζ.

It has been shown that, in a viscously damper system subjected to forced vibrations the displacement lags behind the forcing function by an angle φ which is given by 2 tan φ = 2 ξr/(1-r ), r = ω/ωn ξ = damping ratio.

Due to damping the points O, C and G no longer remain collinear and take up the configuration given below as shown in Figure 5.16. The point C is pulled back due to damping. Thus the rotor will be in equilibrium under the action of the following forces.

76 a. Centrifugal force = me ω2, due to the eccentricity of mass of the rotor. b. Spring force = KX. c. Centrifugal force mX ω2 due to whirling. d. Damping force C ωX.

The above forces are shown both in magnitude and direction as given below in Figure 5.16.

2 me ω2 me ω Sin φ

me ω2 G ) φ ) φ

. 2 O KX C mX ω2 (K -mω )x me ω2 Cos φ X CωX CωX

Resolve these forces in horizontal and vertical direction and for equilibrium.

X/e = r 2/ √[(1-r2)2 + (2ζr) 2] and tan φφφ = C ωωω/(K-mωωω2) = 2 ξξξr/(1-r2)

These expressions are very much similar to frequency response curve of single DOF system subjected to harmonic excitation due to rotating unbalance. The frequency response curves are as shown in Figure 5.17 and 5.18.

ζ = 0

ζ = 0.1

X/e ζ = 0.15 ζ = 0.25 ζ = 0.5

ζ = 1.0

r = ω/ ωn

ζ = 0.25 ζ = 0 ζ = 0.5

ζ = 0.707

ζ = 1.0

Phase Angle, φ, ζ = 2.0

0.5 1.0 1.5 2.0 2.5 3.0

Frequency Ratio r = (ω/ω n) Figure 5.18

Discussions a. When ωωω<<< ωωωn, r <<< 1. tan φ ∞, φ 90 0 Disc rotates with heavier side outside i.e., G outside C as shown in figure (a)

G φ C O

(a) φφφ < 90

b. When ωωω = ωωωn, r = 1 tan φ ∞, φ 90 0

G φ C O

(b) φφφ = 90

Resonance occurs: Deflection X is maximum. As damping increases deflection reduces. Severe lateral vibrations occurs.

c. ωωω>>>> ωωωn, r >>> 1 90 0 < φ < 180 0

φ G C O

Disc rotates with heavy side inside.

d. When φφφ = 180 0

Irrespective of amount of damping, the point G approaches O. The system tends to be more stable and it is the desirable conditions.

φ G C O

(d) φφφ = 180 0

Fig shows the phase at different rotational speeds.

G φ G φ φ φ G G C C C C

O O O O

0 0 0 (a) φφφ < 90 (b) φφφ = 90 (c) 90 < φφφ < 180 (d) φφφ = 180

Tutorial problems on Critical speeds with out damping

1.A disc of mass 4 kg is mounted midway between two bearings which are 480 mm apart, on a horizontal steel shaft 9 mm in diameter. The CG of the disc is displaced by 3 mm from its geometric centre. Equivalent viscous damping at the centre of the disc is 49 Ns/m. If the shaft rotates at 760 rpm determine (a) the maximum stress in the shaft. (b) What is the power required to drive the shaft at this speed. (c) Also compare the maximum bending stress with the dead load stress in the shaft. Also find the power required to drive the shaft at this speed.

2 single rotor weighing 100 N is mounted between midway between the bearings on a steel shaft 1cm in diameter . The bearing span is 40cm. It is known that C G of rotor is 0.25mm from its geometric centre . If the system rotates at 1000 RPM and damping ratio is estimated to be 0.05. Find 1. amplitude of vibration 2. Dynamic load transmitted to bearings 3. Maximum stress induced in the shaft, when the shaft is supports vertically

Neglect the weight of the shaft. Assume the shaft to be simply supported and take E = 196 GPa

3 A turbine disc has a mass of 40 kg with operating speed of 6000 RPM, is mounted at the centre of the shaft of span 0.8m and diameter 5cm. The disc has an unbalance of 10 -3 kg-m. Assume the bearings to be rigid and the end conditions to be simply supported . Damping is represented by an equivalent viscous damping ratio 0.08. Find the critical speed , whirling amplitude at the critical speed, Maximum bending stress in the shaft at the operating speed ( assume the shaft to be horizontal) , Power required to direve the disc at a speed of 25 % more than critical speed. Take E =196 GPa.

4. A vertical shaft 20mm diameter rotates in LONG BEARINGS and disc of mass 15kg is attached to the mid span of the shaft. The span of shaft between the bearings is 1m, The mass centre of the disc is 0.005mm from the axis of the shaft. Neglecting the mass of the shaft and taking the deflection as per beam fixed at both ends , determine the critical speed of rotation. Determine the range of speed over which the stress in the shaft due to bending will exceed 125 MPa. Take E=200 GPa.

80 CHAPTER 6

TWO DEGREES OF FREEDOM SYSTEM

LEARNING OBJECTIVES

] Introduction to Two degrees of freedom system ] Terminologies used – principle modes and normal modes of vibrations ] Concept of co ordinate coupling ] Generalized and principal co ordinates ] Free vibration in terms of initial conditions. ] Problems on Geared system ] Forced oscillations – Harmonic excitation applications ] Vehicle suspension ] Dynamic vibration absorber ] Dynamics of reciprocating engines

The general rule for computation of number of degrees of freedom can be stated as follows

Number of degrees of freedom = {No of Masses in the system}{Noof possible type of motion of each mass} = 2 x 1 = 2 ( for the examples from 1 to 5) = 1x 2 = 2 ( for the examples from 6 to 7)

Figure 3

K K1 t 1 Figure 2 Figure 1

m J1 1 θ1 x1 X1 X2 Kt2 K 2 K J2 m1 m2 θ2

m 2 • • • • x2 Figure 4 Figure 7 81 X1 X2

K1 K2 K3

m1 m2

• • • •

Figure 5

Figure 7

Figure 6

Two degrees of freedom will vibrate in two different modes called as principal modes. If masses m1 and m2 shown in figure (1) are vibrating in phase , such mode of vibration is called first principal mode. When the masses m1 and m2 are vibrating in out of phase, such mode of vibration is called second mode of vibration.

There are two equations of motions for a two degree of freedom system, one for each mass ( more precisely for each DOF0. They are generally in the form of coupled differential equation. i.e each equation involves all the coordinates . After assuming harmonic solution to the equation of motion, it gives two natural frequencies. During free vibration at one of natural frequencies, there exists a definite value of ratio of the amplitude, which is referred to as principal mode ( the ratio of amplitude of vibration in a particular mode is known as principal mode) In the principal mode if one of amplitude is taken as unity, then principal mode is referred at as normal mode.

UNDAMPED FREE VIBRATION OF TWO DEGREES OF FREEDOM

1.write the equation of motion for the following fig. Determine the mode shape , natural frequency , no of modes in different modes. ( Figure 1)

Figure 2

Figure 1

2. Determine the natural frequencies and draw mode shapes and locate the nodes for the systems shown in Figure 2.

3. Find the natural frequencies of system shown in figure, Take J1 = 06 kgm2, J2 = 1.34 kgm2, G=0.83x1011 N/m2

K1θ1 J1θ&& 1

Disc-1 θ1

J1 θ1

K (θ − θ ) 2 2 1 J2θ&& 2 Kt2

θ2 J2 Disc-2 θ2

Free body diagram

Definite and semi-definite systems A system, which is free from both the ends, is referred as semi-definite system. Ex: Train, Turbo-generator, etc.

X1 X2

m1 m2

• • • •

x2 > x 1

X1 X2

KX KX 2 2 m1 m2

KX 1 KX 1

K (X –X ) 2 1 m1 m2 K (X –X ) 2 1

For the linear system shown below (i) derive the equations of motion, (ii) setup the frequency equation and obtain the fundamental natural frequencies (iii) obtain the modal vectors and modal matrix (iv) draw mode shapes of the system.

Class work

1. A electric train made of 2 cars 200kg is attached by a coupling of stiffness 40x106 N/m. Determine the natural frequency of the system.

85 2. Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends. The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm. The length of the shaft is 3 m and its diameter = 10 cm. Modulus of rigidity for shaft material of the shaft G = 0.83 * 1011 N/m2

Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm

( hint for the part II is given below)

J1 J2

K K t1 t2

GENERAL SOLUTION OF TWO DEGREE FREEDOM SYSTEM .

The two degree system performs free harmonic vibration only when certain amplitude ratios are satisfied. For a general disturbance not satisfying amplitude ratios, the motion cannot be harmonic. The general solution includes the response containing harmonics as well as second natural frequencies of the system. Since the differential euation for mass m1 and mass m2 both are of second order, there will be two constants for each differential equations, hence the general solution can be written as

For the system shown mass m1 is displaced 0.01 m from its static equilibrium position and released. Determine the resulting displacements x1 and x2 of masses m1 and m2 , if m1=m2=m. Also determine the displacement of masses after 0.1 second. If k=10 KN/m and m=10kg.

Find the natural frequency and principal modes for system shown in figure Take m1=m2=m and l1=l2=l

87 Determine the natural frequencies of the system shown below. Take m1=m2=m.

Derive the differential equation of motion for system shown. Assume a chord passing over cylinder do not slip. Determine an expression for natural frequencies of system .

Derive the equation of motion for the double pendulum shown in the figure and hence derive the frequency equation by considering m1=m2=m and l1=l2=l also find the amplitude ratio an draw the mode shapes.

UNDAMPED DYNAMIC VIBRATION ABSORBER. ( Forced harmonic vibration of two degree freedom system)

When a single degree freedom system having large amplitudes of vibration under external excitation at resonance, the same can be eliminated by coupling a properly designed auxiliary spring mass system to the main system. This is the principle of undamped dynamic vibration absorber. Here the excitation is finally transmitted to the auxiliary system, bringing the main system to rest.

Consider a two degree freedom system as shown in Figure. The spring mass system k1, m1 is called the main system which is excited by an external F = F sinωt harmonic force, 1 0 K1x1

F = Fsinωt m1 mx 1 0 1 && 1

K2(x 2-x1)

m 2 m2x&& 2

Force equilibrium diagram of the system

From Force equilibrium diagram of mass m1: m x + K x − K (x − x ) = F sinωt 1 && 1 1 1 2 2 1 0 m x + (K + K )x − K x = F sinωt 1 && 1 1 2 1 2 2 0 (137) From Force equilibrium diagram of mass m2 ; m2x&& 2 + K2(x2 − x1) = 0 m2x&& 2 − K2x1 + K2x2 = 0 (138) It is possible to have pure harmonic free vibration for both the masses. Therefore, Let us assume. x1 = A1sinωt x2 = A2sinωt where ω is the forcing frequency on the system .

F K o 2 x 2 = 4 2 [m1m 2ω − k( 1m 2 + k 2 m 2 + k 2 m1 )ω + k1k 2 ]

F [k − m ω2 ] o 2 2 x1 = 4 2 [m1m 2ω − k( 2 m1 + k 2 m1 + k1m 2 )ω + k1k 2 ]

The amplitude of vibration of main system can be reduced to zero, if the numerator of equation (2) is zero

F (K − m ω2 ) = 0 0 2 2

2 K 2 − m2ω = 0

K ω2 = 2 m 2

K ω = 2 rad s/ m2 This shows that the amplitude of vibration of main mass can be made equal to zero if the operating frequency is equal to the natural frequency of the absorber system. The addition of a vibration absorber to main system is not much meaningful unless ω = ωn the main system is operating under resonance. Then 1 but absorber to be ω = ωn ω = ωn = ωn effective 2 we can write 1 2 When this condition is fulfilled the absorber is known as tuned absorber.