Vibrations

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MECHANICAL VIBRATIONS

WORK

WORK BOOK

BOOK CUM

CUM LECTURE NO

LECTURE NOTES

TES

(FOR SIXTH SEMESTER MECHANICAL STUDENTS)

((

FOR PRIVAT FOR PRIVATE CIRCULATION OE CIRCULATION ONLY NLY

))

JAGADEESHA

JAGADEESHA T

T

Associate Professor

Mechanica

Mechanical

l Engineering Department

Engineering Department

Name

USN USN Section Section

ST. JOSEPH ENGINEERING COLLEGE

VAMANJOOR, MANGALORE – 575 028,

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M M

CHAPTER 1- INTRODUCTION TO VIBRATION

LEARNING OBJECTIVES



 Introduction to vibrationIntroduction to vibration



 Terminologies used in VibrationTerminologies used in Vibration



 Simple Harmonic MotionSimple Harmonic Motion



 Addition of Harmonics, Principle of super position applied to SHMAddition of Harmonics, Principle of super position applied to SHM



 Introduction to Fourier analysis, BeatsIntroduction to Fourier analysis, Beats



 Problems related to SHM and Fourier analysis.Problems related to SHM and Fourier analysis.

Vibration

is defined as a motion which repeats after equal interval of time and isis 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. 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 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, 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 elastic nature of the system, external application of force or wind loads and earthquakes. Vibrations are undesirable in most engineering systems and desirable earthquakes. Vibrations are undesirable in most engineering systems and desirable in few cases.

in few cases. A

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

motion is a repeated motion with equal interval of time.

Example for useful vibration: Example for useful vibration:

o

o General industries – crushers, jackhammer, concrete compactor, etc.General industries – crushers, jackhammer, concrete compactor, etc. o

o Medical and health – electric massage, high frequency vibration probe for Medical and health – electric massage, high frequency vibration probe for heart disease treatment

heart disease treatment o

o Music – string instruments i.e. guitar etc.Music – string instruments i.e. guitar etc.

Example for unwanted vibration Example for unwanted vibration::

o

o Poor ride comfort in vehicle due to road irregularitiesPoor ride comfort in vehicle due to road irregularities o

o Sea sickness when traveling on ships, Sea sickness when traveling on ships, boats, etc.boats, etc. o

o EarthquakesEarthquakes o

o Fatigue failures in machine and structuresFatigue failures in machine and structures

Cycle

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

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Time period

: : It is the time taken It is the time taken to complete one to complete one cycle. It is eqcycle. It is equal to the time for thual to the time for thee vector to rotate through 2π radians

vector to rotate through 2π radians

``

Frequency:

It is the number of cycles per unit timeIt is the number of cycles per unit time

Amplitude

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

from the mean position

Phase difference

: it is angle between the two rotating vectors executing simple: it is angle between the two rotating vectors executing simple harmonic motion of same f

harmonic motion of same frequencyrequency The

The first first vector vector is is x1= x1= X sX sin in ( ( wt wt )) The

The second second vector vector is is x2= x2= X X sin sin ( ( wt+φ)wt+φ)

Where φ is the phase difference between x1 and x2 Where φ is the phase difference between x1 and x2

Resonance

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

resonance the system undergoes dangerously large oscillations

Damping

: It is the : It is the resistance offered to the motion of a vibrating body.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 If the motion is repeated after equal intervals of time, it is called periodic motion, The simplest type of periodic motion is harmonic motion

simplest type of periodic motion is harmonic motion

Aperiodic motion

If the motion does not repeat after equal interval of

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

rad/s rad/s Reference: Reference: rad/s rad/s

ω

rad/s rad/s ω ω ω ω ω ω ω ω

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DESCRIBING MOTIONS OF VIBRATING SYSTEMS

Periodic motions Periodic motions

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

ω = radian frequency (rad/sec) ω = radian frequency (rad/sec) ω = 2π

ω = 2πf f ; where; where f f is frequency (Hz)is frequency (Hz)

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

Simple harmonic motion Simple harmonic motion

(

t t

)

== A A

(

ω +ω t t +φ φ

))

y

y sinsin DisplacementDisplacement

(

t t

)

== A Aω ω

(

ω ω t t ++

))

φ φ y

y_&& coscos VelocityVelocity

(

t t

)

== −− A Aω ω

(

ω ω t t ++

))

φ φ y

y&& 22sinsin AccelerationAcceleration

Classification of vibrations

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

as follows:

1.Single Degree of freedom Systems 1.Single Degree of freedom Systems 2.Two Degrees of freedom Systems 2.Two Degrees of freedom Systems 3.Multidegree of freedom Systems 3.Multidegree of freedom Systems

4.Continuous Systems or systems with infinite

4.Continuous Systems or systems with infinite degrees of freedomdegrees of freedom Another broad classification of vibrations is:

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

1. Free and forced vibrations

Leads / Lags Leads / Lags

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Sometime vibration problems are classified as: Sometime vibration problems are classified as: 1. Linear vibrations 1. Linear vibrations 2. Non-linear vibrations 2. Non-linear vibrations 3. Random vibrations 3. Random vibrations 4. Transient vibrations 4. Transient vibrations 5. Longitudinal vibrations 5. Longitudinal vibrations 6. Transverse vibrations 6. Transverse vibrations 7. Torsional vibrations 7. Torsional vibrations

Free vibration

: If a system after initial disturbance is left to vibrate on its own , the: 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 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 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 forces are absent. The frequency of free vibration of a system is called natural frequency. Natural frequency is a property of

frequency. Natural frequency is a property of a dynamical systema dynamical system

Forced vibration

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

frequencies or external frequencies

Damped vibration

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

vibration is known as damped vibration\

Undamped vibration

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

vibrations are known as undamped vibration

Linear vibration

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

superposition holds and mathematical techniques of analysis are well developed.developed.

Non linear vibration

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

analysis is well known

Deterministic vibration

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

Random vibration:

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

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Longitudinal vibration

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

axis, it is said to be execute longitudinal vibration

Transverse vibration

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

said to execute the transverse vibration

Torsional vibration

: : If the If the shaft gshaft gets alternativets alternatively twisely twisted and ted and un twisted un twisted onon account of an a

account of an alternate torque on the lternate torque on the disk, it is said disk, it is said to execute the tto execute the torsional vibrationorsional vibration

SIMPLE HARMONIC MOTION

A

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

the mean position is known as SHM, ( Simple Harmonic Motion) Consider a

Consider a spring masspring mass syss system as stem as shown in hown in the figure the figure along with along with the the displacementdisplacement time diagram

time diagram

The equation of motion of the mass can be written as follows The equation of motion of the mass can be written as follows From the right angled triangle OAB we have

From the right angled triangle OAB we have

Where Where

x= displacement at any instant of time x= displacement at any instant of time X = amplitude of vibration

X = amplitude of vibration W= angular velocity or f

W= angular velocity or frequency in rad/secondrequency in rad/second

m m

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The velocity of the mass m at an instant of t

The velocity of the mass m at an instant of time t is given byime t is given by V =

V =

The acceleration of mass m is given by The acceleration of mass m is given by

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

displacement and is directed towards mean position

observing the equations 2 and 3 the velocity and acceleration are harmonic with the observing the equations 2 and 3 the velocity and acceleration are harmonic with the same frequency but lead a displacement vector by

same frequency but lead a displacement vector by π/2 and π π/2 and π radians respectively.radians respectively.

tt

x

X= A sin X= A sin ωωtt x-Di x-Displacementsplacement X-amplitude X-amplitude T-Periodic Time T-Periodic Time f-Frequency f-Frequency f=1/T f=1/T ω ω=Frequency in=Frequency in radians per second radians per second t= time

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Rotor with inertia J Rotor with inertia J₁₁ k

k11

k k22

Rotor with Inertia J Rotor with Inertia J22

Degrees of freedom:

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

is called degrees of freedom

One degree of freedom

Two degree of freedom

Three degree of freedom

Infinite degree of freedom

m m11 m m22 m m11 m m22 m m33 Cantilever Beam Cantilever Beam Continuous system Continuous system

( consider the mass of the beam ) ( consider the mass of the beam ) m m11 m m11 K K11 m m22 K K x x11 x x22 F F11

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K K11 KK22 m, m, a a bb

Exercise

Specify the no of degree of freedom for the following Specify the no of degree of freedom for the following

1

22

33

44

55

..

4m

2m

m

3k

k

Cantilever Beam Cantilever Beam Continuous system Continuous system

( Neglect the mass of the beam ) ( Neglect the mass of the beam )

k k11 _k_k₂₂ m m11 mm22 y y y y m2 m2

θ

1

1 θ

θ

2

L1 L1 L2 L2 M1 M1 M2 M2

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O

A

B

Addition of SIMPLE HARMONIC MOTION ( SHM)

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

which is simple harmonic having the same frequency.same frequency. Consider tow simple harmonic motions of

Consider tow simple harmonic motions of x x ₁₁ andand x x ₂₂ having the same frequencyhaving the same frequency and phase difference φ as

and phase difference φ as given belowgiven below

)) (( sin sin 1 1 1 1 X X t t x x ======== ω ω ωωω ω ω ω x x₂₂ == X X ₂₂ sinsin ((ω ω t t ++ φ φ )) Adding Adding x = xx = x11 +x+x22

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

θ

ω

ωtt

φ

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-6 -6 -4 -4 -2 -2 0 0 2 2 4 4 6 6 0 0 00..005 5 00..1 1 0..1015 5 00..2 2 00..2255

Tutorial problems on Simple Harmonic Motion. Tutorial problems on Simple Harmonic Motion.

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

graphically

1)

1) X1= 3 X1= 3 sinsin

((((((( (

ω ω ω ω ωωω ω tt ++++++++ 3030

))))))))

X2= 4 cosX2= 4 cos

((((((( (

ω ω ωωω ω ω ω tt ++++++++ 1010

))))))))

( VTU Jan 2005)( VTU Jan 2005) 2)

2) X1= 2 X1= 2 coscos

((((((( (

ωω ωω ω ω ω ω tt ++++++++ 0.50.5

))))))))

X2= 5 sinX2= 5 sin

((((((( (

ω ω ωωω ω ω ω tt ++++++++ 11

))))))))

( VTU July 2006)( VTU July 2006) 3) 3) X1= 10 X1= 10 coscos 4 4 tt πππ π π π π π ω ω ω ωω ω ω ω ++++++++ X2= 8 sinX2= 8 sin 6 6 tt πππ π π π π π ω ω ω ωω ω ω

ω ++++++++ ( ( VTU VTU Dec. Dec. 2007)2007)

2. A body is subjected to two harmonic motions

X1= 8 cos X1= 8 cos

))))

6 6 tt π π πππ π π π ω ω ω ω ω ω ω ω ++++++++ X2= 15 sinX2= 15 sin

))))

6 6 tt π π πππ π π π ω ω ω ωω ω ω

ω ++++++++ what harmonic motionwhat harmonic motion should

should be be given given to to the the body body to to bring bring it it to to equilibrium equilibrium (VTU (VTU July. July. 2005)2005)

3. Split the harmonic motion

x = 10 sinx = 10 sin ω ω ωωω ω ω ω tt ++++++++ π π π π πππ π ₆₆

))))

into two harmonic motionsinto two harmonic motions having the phase of zero and the other of 45

having the phase of zero and the other of 45oo

4. Show that resultan

4. Show that resultant motion of harmon

t motion of harmonic motion

ic motion given below is zer

given below is zero

o

X1= X sin

((((((( (

ωωω ω ω ω ω ω tt

))))))))

X2= X sinX2= X sin

))))

3 3 2 2 tt π π πππ π π π ω ω ω ωω ω ω ω ++++++++ X3= X sinX3= X sin

))))

3 3 4 4 tt πππ π π π π π ω ω ω ω ω ω ω ω ++++++++

5. The displacement of a vibrating body is given by x = 5 sin (31.41t +

4 4 π π π π π π π π

_).

Draw the variation of displacement for one cycle of vibration. Also determine

the displacement o

the displacement of body

f body after 0.11 secon

after 0.11 second.

d. ( repeat the

( repeat the problem for

problem for velocity

velocity

and acceleration

and acceleration and draw graph

and draw graph using Excel

using Excel and compare )

and compare )

Time

Displacement

Time

Displacement

00

0.025

5

0.025

5

0.05

0.075

0

0.075

0

0.1

0.125

0.15 -3.53

0.175

0

0.175

0

0.2

0.2 Calculate the remaining values

Calculate the remaining values

6. The Motion of a particle is

represented by x = 4

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BEATS

When two

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

resulting motion exhibits a phenomenon known as Beats. A Beat

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

of synchronization with one another

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

frequencies.

X1

X1 =

= X

X Cos

Cos

( ( ))

ω ω ωωω ω ω ω tt

, ,

X2

X2 =

= X

X Cos

Cos

( (

ω ω +ωωω ω ω ω +δ δ δδδ δ δ δ

))

tt

))

,,

Where δ is a small quantity Where δ is a small quantity

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

























+













t

X

2

2 cos

cos

2

2 cos

cos

2

2 δ

δ

ω

δ

The above equation shown graphically in Figure. The resulting motion represents The above equation shown graphically in Figure. The resulting motion represents cosine wave with frequency

cosine wave with frequency

2 2 π π π π π π π π ω ω ω ω ω ω ω

ω ++ and with a varying amplitude 2 cos(and with a varying amplitude 2 cos(

2 2 δ δ δ δ δ δ δ δ _tt

_::

whenever the amplitude reaches a maximum , it is called the beat. The frequency whenever the amplitude reaches a maximum , it is called the beat. The frequency δδ at which amplitude builds up and dies down between o and 2 X is known as beat at which amplitude builds up and dies down between o and 2 X is known as beat frequency.

frequency.

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

natural frequency of the system

-2X -2X       22π π πππ π π π

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ELEMENTS OF VIBRATION

The elements of constitute vibrating systems are The elements of constitute vibrating systems are

1.

1. Mass Mass or Inertia or Inertia element element - m- m 2. 2. Spring Spring - - kk 3. 3. Damper Damper - - cc 4. Excitation F(t) 4. Excitation F(t)

Mass or Inertia element

::

The mass or inertia element is assumed to be rigid body During vibration velocity of The mass or inertia element is assumed to be rigid body During vibration velocity of mass changes, hence kinetic energy can be gained or

mass changes, hence kinetic energy can be gained or loosed. The force applied onloosed. The force applied on the mass from newton second law of motion can be written as F= ma. W

the mass from newton second law of motion can be written as F= ma. W ork done onork done on the mass is stored in the form of

the mass is stored in the form of kinetic energy given by KE= ½ M Vkinetic energy given by KE= ½ M V22

Combination of masses

In practical cas

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

equivalent mass.

Case1.

Case1. Translational m

Translational masses connected

asses connected by r

by rigid bar.

igid bar.

Let the masses M

Let the masses M11, , MM22 and Mand M33 are attached to a rigid bar at locates 1, 2 and 3are 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 respectively as shown in the figure. The equivalent mass meq be assumed to be located at 1 is as shown in figure (b)

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 Let the displacement of masses M1, M2 and M3 be x1, x2, x3 and similarly the velocities o

velocities of respective masf respective masses be ses be x1, x2 x1, x2 and x3. and x3. We can We can express express the velocities the velocities of of masses m2 and m3 in terms of m1

masses m2 and m3 in terms of m1

Elements of Vibration

Passive element Passive element m, C, K m, C, K Active element Active element F(t) F(t) Conservative element Conservative element ( Mass and Spring) ( Mass and Spring)

Non conservative element Non conservative element

( Damper) ( Damper)

C

k

m

F(t)

Voigt Model Voigt Model

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is the required answer. is the required answer.

Translational and rotational masses

Translational and rotational masses coupled together.

coupled together.

Let a mass

Let a mass m having a translational m having a translational velocity x be cvelocity x be coupled to another mass houpled to another mass havingaving mass

mass moment moment of of inertia inertia IIoo with a rotational velocity θ as in rack and pinionwith a rotational velocity θ as in rack and pinion

arrangement shown in the figure. arrangement shown in the figure.

These two masses can be combined to obtain either a single equivalent translational These two masses can be combined to obtain either a single equivalent translational mass M

mass Meqeq or a single equivalent mass moment of inertia Jor a single equivalent mass moment of inertia Jeqeq 2 2 1 1 3 3 3 3 2 2 1 1 2 2 2 2 1 1













+













+

=

l

M

l

M

_eq_eq

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Equivalent

Equivalent translational translational massesmasses

Kinetic energy of the equivalent mass =

_











22 2 2 1 1 eq eq eq eq X X M M

&&

Kinetic energy of the two masses = Kinetic energy of the two masses =

            + + ₀₀ 22 2 2 2 2 1 1 2 2 1 1 θ θ θ θ θ θ θ θ && && J J X X M M M Meqeq= m= m













+ + ₂₂ R R J J m

m oo

is the required answer.

Also determine the equivalent rotational mass J Also determine the equivalent rotational mass Jeqeq

J

Jeqeq= m= m mRmR ++ J J oo

2

_{is the required answer}

Spring element

::

Whenever there is a relative motion between the two ends of the spring, a force is 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 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 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.

spring or spring constant or spring gradient.

The spring stiffness is equal to spring force per unit deromation. The spring stiffness is equal to spring force per unit deromation.

F F



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Workdone in deforming a spring is equal to the strain energy or potential energy. Workdone in deforming a spring is equal to the strain energy or potential energy. Strain energy = potential energy = area of the triangle

Strain energy = potential energy = area of the triangle OABOAB

Stiffness of beams

Cantilever beam

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

free end is given by

Similarly derive the expression for

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

F F x x K=stiffness K=stiffness F F k k m m x x m m x x m m N N EI EI Wl Wl st st // 48 48 3 3       = = δ δ δ δδ δ δ δ m m x x m m N N l l I I E E st st // 192 192 3 3       = = δ δ δ δδ δ δ δ

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Stiffness of slender bar

Stiffness of slender bar subjected to longitudinal vibrations

subjected to longitudinal vibrations

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

derive the expression for stiffness.

Torsional

Torsional Stiffness

Stiffness of

of bar.

bar.

It is the amount o

It is the amount of torque required to caf torque required to cause a unit angulause a unit angular r deformation in thedeformation in the element. element. Torsional stiffness = K Torsional stiffness = Ktt == θ θ θ θθ θ θ θ T T

Combination of stiffness

Determination of equivalent spring stiffness when the

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

series,

Consider two springs of stiffness K

Consider two springs of stiffness K11 and Kand K22 acted uponacted upon

by the force F. by the force F.

The deflection of spring k1 is x The deflection of spring k1 is x11 ==

1 1 K K F F The deflection of spring k

The deflection of spring k11 is xis x22==

Let these two springs be

Let these two springs be replaced by an equivalentreplaced by an equivalent stiffness K

stiffness Keqeq upon which a force F acts and due toupon which a force F acts and due to

which its deflection is given by which its deflection is given by

F F m m E, A E, A ll

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60

60 cm cm 70 cm70 cm

Determination of equivalent spring stiffness when the

Determination of equivalent spring stiffness when the springs are arranged in

springs are arranged in

parallel.

Force acting on K

Force acting on K11 spring = Fspring = F11=k=k11xx

Force acting on K

Force acting on K22 spring = Fspring = F22

Force required for an equivalent spring k

Force required for an equivalent spring keqeq to defined by x givento defined by x given

by F= K by F= Keqeq xx

But F = F1 +F2 But F = F1 +F2

Tutorial problems on Equivalent stiffness of springs Tutorial problems on Equivalent stiffness of springs

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

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

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

k k 2k 2k k k 3k 3k 2k 2k k k M M k k _2k_2k 38 kg 38 kg N/m N/m 10 10 2 2 x x 66 N/m N/m 10 10 1 1 x x 66 22 x x 101066 N/m N/m N/m N/m 10 10 3 3 66 x x M M

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60 cm

60 cm ₈₀_{80 cm}_cm ₅₀_{50 cm}_cm

50 cm

50 cm _{60 cm}_{60 cm}

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

5.Replace the following torsional stiffness by a

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

system and equivalent system is same.

R1=

R1= 3cm 3cm R2= R2= 5cm5cm

DAMPING

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

energy is gradually converted into heat or sound is known as damping.sound is known as damping. A

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

damping; materials or solid or hysteric damping

Viscous damping

Viscous damping is most commonly used damping mechanism in vibration analysis. 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 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 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 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

such as size or shape of the vibrating body. the viscosity of the fluid, the frequency of fluid, the frequency of vibration and velocity of fluid. Resistance due to viscous damping is directly vibration and velocity of fluid. Resistance due to viscous damping is directly proportional

proportional to the to the velocity velocity of vibrationof vibration

d d F F α α α ααα α α V V d d F F ==C C x x &&

Where C= damping coefficient Where C= damping coefficient F

Fdd= damping force= damping force

Examples of Viscous damping

1)

1) Fluid Fluid film between film between sliding sliding surfacesurface 2)

2) Fluid flow Fluid flow around a around a piston in piston in a cylinder a cylinder

R

R reqreq =4cm=4cm

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stress stress

Strain Strain

Coulomb damping or dry friction damping

Here a damping force is constant in magnitude but opposite in direction to that of the 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

motion of vibrating body. It is caused by the friction between the surfaces that are dryfriction between the surfaces that are dry or have insufficient lubrication

or have insufficient lubrication

Material or solid or hysteric damping

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

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τ τ τττ τ τ τ 22τττ τ τ τ τ τ 33τττ τ τ τ τ τ τ τ τττ τ τ τ X(t) X(t)

FOURIER SERIES

The simplest of periodic motion, happens to be SHM. It is simple to

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

shown below:

Forces acting on machines are generally periodic but this may not be harmonic for 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 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 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 periodic but nonharmonic forces can be done with the help of Fourier series. The problem becomes a multifrequency excitation problem. The principle of linear 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 superposition is applied and the total response is the sum of the response due to each of the individual frequency term.

each of the individual frequency term.

Any periodic motion can be expressed as an infinite sum of

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

If x(t) is a periodic function with period t its Fourier representation is given byperiod t its Fourier representation is given by X(t) =

X(t) = coscos

((((((( ( ))))))) )

tt coscos

((((((( ( ))))))) )

tt ... sinsin

((((((( ( ))))))))

tt 2 2 aa11 ω ω ωωω ω ω ω aa11 ω ω ωωω ω ω ω bb11 ω ω ω ωωω ω ω a a_o_o + + + +++ + + + + + +++ + + + + + +++ + + =

= coscos

((((((( (

tt

))))))) )

sinsin

((((((( ( ))))))))

tt

2 2 11 ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω _n_n n n n n o o b b n n a a a a + + + +++ + + + + + +++ + +

∑

∞ ∞ ∞ ∞∞∞ ∞ ∞ = = = === = = t t π π π ππ π π π ω ω ω ωω ω ω

ω ======== 22 = Fundamental frequency – (1)= Fundamental frequency – (1) where a

where aooaann bbnn are constantsare constants

τ τ τττ τ τ τ 22τττ τ τ τ τ τ 33τ τ τ τττ τ τ τ τ τττ τ τ τ X(t) X(t)

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X(t) X(t) 0.35 0.35 0.250.25 Determination of constants Determination of constants To find a

To find aoo Integrate both sides of equation(1) over Integrate both sides of equation(1) over any intervalany interval τττ τ τ τ τ τ . All intergrals on. All intergrals on

the RHS of the equation are zero except the one containing a the RHS of the equation are zero except the one containing aoo

((((((( ( ))))))))

t

dt

x

a

o o o o

2

2 2

∫∫∫∫∫∫∫∫

=

ω ω ω ω ω ω ω ω π π π ππ π π π

π

ω

= =

x

((((((( ( ))))))))

t

dt

o o

2

2 ∫∫∫∫∫∫∫∫

τ τ τττ τ τ τ

τ

τττ

τ

To find a

To find ann multiply equation 1 by cosmultiply equation 1 by cos

((((((( (

nnω ωω ω ω ω ωω t t

))))))))

and and Integrate Integrate over over any any intervalinterval τττ τ τ τ τ τ . All. All

intergrals intergrals

((((((( ( ))))))) ) ((((((( (

t t

))))))))

dt dt x x a a o o n n coscos nn tt 2 2 2 2 ω ω ω ωω ω ω ω π π π π π π π π ω ω ω ωω ω ω ω ω ω ωωω ω ω ω π π π ππ π π π

∫∫∫∫∫∫∫∫

= = = = = = = = x x

((((((( ( ))))))) ) ((((((( (

t t

))))))))

dt dt o o tt n n cos cos 2 2 ω ω ω ωω ω ω ω τ τ τττ τ τ τ τ τ τττ τ τ τ

∫∫∫∫∫∫∫∫

= = = === = = To find b

To find bnn multiply equation 1 by sinmultiply equation 1 by sin

((((((( (

nnω ωω ω ω ω ωω t t

))))))))

and and Integrate ovIntegrate over any er any intervalinterval τττ τ τ τ τ τ . All. All

intergrals intergrals

((((((( ( ))))))) ) ((((((( (

t t

))))))))

dt dt x x a a o o n n sinsin nn tt 2 2 2 2 ω ω ω ωω ω ω ω π π π π π π π π ω ω ω ωω ω ω ω ω ω ωωω ω ω ω π π π ππ π π π

∫∫∫∫∫∫∫∫

= = = = = = = = x x

((((((( ( ))))))) ) ((((((( (

t t

))))))))

dt dt o o tt n n sin sin 2 2 ω ω ω ω ω ω ω ω τ τ τττ τ τ τ τ τ τττ τ τ τ

∫∫∫∫∫∫∫∫

= = = === = =

Find the Fourier series for the curve shown below Find the Fourier series for the curve shown below

τ τ τττ τ τ τ 22τττ τ τ τ τ τ 33τ τττ τ τ τ τ t t X(t) X(t) E E τ τ τττ τ τ τ 22τττ τ τ τ τ τ 33τ τ τττ τ τ τ t t X(t) X(t) π π π ππ π π π 2 2 t t X(t) X(t)

π

1 1

Represent for the periodic motion Represent for the periodic motion shown in the figure

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CHAPTER 2 : UNDAMPED FREE VIBRATION

LEARNING OBJECTIVES



 Introduction to undamped free vibrationIntroduction to undamped free vibration



 Terminologies used in undamped free vibrationTerminologies used in undamped free vibration



 Three methods to solve the Three methods to solve the undamped free vibration problemsundamped free vibration problems



 Problems related to undamped free Problems related to undamped free vibration problems.vibration problems.

Free vibrations are oscillations about a systems equilibrium position that occur in the 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, 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 it is known as undamped vibration. The first step in solving a vibration problem is setting up the differential equation of motion

setting up the differential equation of motion

Consider spring mass system which is assumed to move only along the vertical Consider spring mass system which is assumed to move only along the vertical direction as shown below

direction as shown below. Let m be the mass of . Let m be the mass of the block and k be the stiffness the block and k be the stiffness of of the spring. When block of mass m is attached to spring , the deflection of spring will the spring. When block of mass m is attached to spring , the deflection of spring will be

be ∆∆ , known as static deflection. In the static equilibrium position, the free body, 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

diagram of forces acting on the mass is shown in Figure(b). Hence mg= kA Once the system is disturbed,

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Let at any instant of time t, the mass is displaced from the equilibrium position x, the 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)

different forces acting on the system are shown in figure (d) From Newton’s second law of motion

From Newton’s second law of motion

∑

F F == mama

Inertia force Inertia force

( disturbing force) = restoring force ( disturbing force) = restoring force

mg mg x x k k x x m m

&&

== −− ((∆∆ ++ )) ++ 0 0 )) (( == + + k k xx x x m m

&&

or

++ (( xx )) == 00 m m k k x x

&&

equation 2 is the differential equation of motion for spring mass system shown in equation 2 is the differential equation of motion for spring mass system shown in figure. Comparing equation (2) with the equation of SHM

figure. Comparing equation (2) with the equation of SHM x x

&&

++ ω ω 22 (( xx )) == 00

since the vibrations of the above system are free( without the resistance of external since the vibrations of the above system are free( without the resistance of external forces) we can write

forces) we can write

sec sec // rad rad m m k k n n == ω ω

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time period , time period , k k m m 2 2 f f 1 1 n n π π τ τ == ==

from the equation(1) mg = k∆ from the equation(1) mg = k∆

∆ ∆ = = qq m m k k

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

Translatory Rotational

In the analysis the disturbing and restoring In the analysis the disturbing and restoring FORCES are considered

FORCES are considered

In the analysis In the analysis

In the analysis MASS Moment of Inertia In the analysis MASS Moment of Inertia (J) is considered

(J) is considered Linear stiffness K , in N/m is

Linear stiffness K , in N/m is consideredconsidered

Problems Problems

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

Find the natural frequency of system.system. 2. A spring mass

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

natural frequency reduces by 2 Hz. find K and m.

k k tt

sec

//

rad

n n

=

ω

k

m m

sec

//

rad

m

k

n n

=

ω

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

vibration is 4 Hz.

4. A spring mass system has a natural period of 0.2 seconds. What will be the new 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%,

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 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 is reduced by 800 N/m, the frequency is altered by 45%, Find the mass and spring constant of the original system.

constant of the original system.

6. Determine the natural frequency of

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

in SHORT bearings.

7

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

in LONG bearings.

where l is the length of bearing and E –

where l is the length of bearing and E – young’s modulus and I is moment of Inertia.young’s modulus and I is moment of Inertia. 7. Determine the natural frequency of

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

in LONG bearings.

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

free lateral vibration in vertical plane to that in horizontal.in horizontal. 9.. Determine the natural frequency of simple pendulum 9.. Determine the natural frequency of simple pendulum

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

frequency of vibration.

ll

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

axis. Find the frequency of oscillation of such pendulum.requency of oscillation of such pendulum.

12. A connecting rod shown in fig is supported at the wrist pin end. It is displaced 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 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 the pivot point O. If the frequency of oscillation is 40 cycles/minute, calculate the mass moment of inertia about its C.G.

mass moment of inertia about its C.G.

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

of oscillation.

14.A simply supported beam of square cross 14.A simply supported beam of square cross section 5mmx5mm and length 1m carrying a mass section 5mmx5mm and length 1m carrying a mass of 0.575 kg at the

of 0.575 kg at the middle is found to have natural fmiddle is found to have natural f requency of 30 rad/sec. Determinerequency of 30 rad/sec. Determine young’s modulus of elasticity of

young’s modulus of elasticity of beam.beam.

15. A spring mass system, k1 and m have a natural frequency f1. Determine the 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 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.

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COMPLETE SOLUTION OF SYSTEM EXECUTING SHM COMPLETE SOLUTION OF SYSTEM EXECUTING SHM The equation of motion of

The equation of motion of system executing SHM can be represented bysystem executing SHM can be represented by

0 0 )) (( == + + k k xx x x m m

&&

---(1)---(1)

0

2

2 =

=

+

dt

dx

dt

dx

ω

The general solution of equation (1)

The general solution of equation (1) can be expressed ascan be expressed as ---(2) ---(2)

Where A and B are arbitrary constant which can be determined from the initial 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 conditions of the system. Two initial conditions are to be specified to evaluate these constants. x=x

constants. x=x00at t=0 andat t=0 and

x

&&

= V= Voo at t=0. substituting in the equation (2)at t=0. substituting in the equation (2)

Energy method Energy method

In a conservative system the total energy is constant. The differential equation as 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 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 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

and partly potential. The kinetic energy T is stored in the mass by virtue of its the mass by virtue of its velocityvelocity where as the potential energy U is stored in the form of strain energy in elastic 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.

deformation or work done in a force field such as gravity.

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

t)

((

sin

B

t)

cos(

A

X

=

ω

+

ω

t)

((

sin

V

t)

cos(

x

oo 0 0

ω

+

ω

=

Is the required completeIs the required complete

solution solution

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k k ll

k

Is given by Is given by

[ [

T

U

]]

0

0 dt

dt

d

=

+

From this we get a differential equation of motion as well as

From this we get a differential equation of motion as well as natural frequency of thenatural frequency of the system.

system.

Determine the natural frequency of spring

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

Determine the natural frequency of system shown in figure. Determine the natural frequency of system shown in figure.

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

on the value of K. Discuss?

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

m m ll k k m m a a θ θ m m θ θ

m

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A string shown in figure is under tension T whic

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

An

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

rope.

A manometer has a uniform bore of cross sec

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

length L and Density ρ is set into motion as shown in figure. Find t he frequency of he frequency of the resulting oscillation.

the resulting oscillation.

m

ll

aa

T

36m 36m 8m 8m

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k

R R k k m m r r

Find the expression for natural frequency of system shown in the figure. Neglect the 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

mass of the cantilever beam. Study the special case i) k=Infinity ii) I = infinity.= infinity.

Determine the expression for the natural frequency of a system shown in figure. The 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 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 inertia about centre of oscillation O. It is assumed that the card attached to mass m does not stretch and is

does not stretch and is always under tension.always under tension.

Determine the natural frequency Determine the natural frequency of a system shown

of a system shown

m m ll

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Determine the expression for the natural frequency of the system shown in figure. Determine the expression for the natural frequency of the system shown in figure. Assume that the wires connecting the mas

Assume that the wires connecting the masses do not stretch and are alwayses do not stretch and are always ins in tension.

tension.

Determine the natural frequency of spring mass system taking the MASS OF Determine the natural frequency of spring mass system taking the MASS OF SPRING (m

SPRING (mss ) into account.) into account.

M M33 M M11

M2

k2

k1

m m

k

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RAYLEIGH’S METHOD.

This is the extension of energy method. Here we can determine the natural 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 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 natural frequency is a function of the rate of change of Kinetic energy and potential energy of the system.

energy of the system. From the principle of

From the principle of conservation of energy we have,conservation of energy we have, T+U = Constant

T+U = Constant T=

T= K.E K.E and and U U =P.E=P.E

We can write the above equation as We can write the above equation as T1+U1

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

Where 1 and 2 1 and 2 represents tow represents tow instances of instances of time.time.

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

instant the velocity f the mass is zero and hence

Substituting in equation (1) we get Substituting in equation (1) we get

(

( ))MaxMax = = ( ( ))MaxMax

Above equation leads directly to natural

Above equation leads directly to natural frequency.frequency.

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

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r r B B K K KK

Home work

..

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

A Cylinder Cylinder of of mass mass m m and and moment moment of of inertia inertia JJoo is to role without slipping but isis to role without slipping but is

sustained by spring K as shown in figure. Determine the natural frequency of sustained by spring K as shown in figure. Determine the natural frequency of oscillation.

oscillation.

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

roll without slipping.

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

roll without slipping.

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

the inextensible spring.

A

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

oscillations about the lowest point shown.shown.

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

without slipping on concave surface as shown above.

r r

K K