Download - VTU-NPTEL-NMEICT Project Progress Reportnptel.vtu.ac.in/VTU-NMEICT/MV/MODULE 2.pdfMODULE-II --- SINGLE DOF FREE VIBRATIONS VIBRATION ENGINEERING 2014 VTU-NPTEL-NMEICT Project Progress

MODULE-II --- SINGLE DOF FREE VIBRATIONS VIBRATION ENGINEERING

2014

VTU-NPTEL-NMEICT Project Progress Report

The Project on Development of Remaining Three Quadrants to NPTEL Phase-I under grant in aid NMEICT, MHRD, New Delhi

DEPARTMENT OF MECHANICAL ENGINEERING, GHOUSIA COLLEGE OF ENGINEERING,

RAMANARAM -562159

Subject Matter Expert Details

SME Name : Dr.MOHAMED HANEEF

PRINCIPAL, VTU SENATE MEMBER

Course Name:

Vibration engineering

Type of the Course

web

Module

II

Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM

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CONTENTS

Sl. No. DISCRETION

1. Lecture Notes (Single-DOF free Vibration).

2. Quadrant -2

a. Animations.

b. Videos.

c. Illustrations.

3. Quadrant -3

a. Wikis.

b. Open Contents

4. Quadrant -4

a. Problems.

b. Assignments

c. Self Assigned Q & A.


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MODULE-II SINGLE DEGREE OF FREEDOM, FREE VIBRATION

1.LECTURE NOTES

SINGLE DEGREE OF FREE VIBRATIONS Un-damped Free Vibrations of Single Degree of Freedom Systems

When the elastic system vibrate because of inherent forces and no external forces is included,

it is called free vibration. If during vibrations there is no loss of energy due to friction or

resistance it is known as undamped vibration, free vibrations which occur in absence of external

force are easy to analyse for single degree of freedom systems.

A vibratory system having mass and elasticity with single degree of freedom in the simplest

case to analyse. The determination of natural frequency to avoid resonance is essential in

machine elements.

Module-II is define the following

i. Vibration model, Equation of motion-natural frequency. ii. Energy method, Rayleigh method. iii. Principle of virtual work, damping models.

2.1. VIBRATION MODEL, EQUATION OF MOTION-NATURAL FREQUENCY

2.1. i. Spring mass system displaced vertically.

a) VIBRATION MODEL:

Consider a spring mass system as shown in fig constrained to move in a collinear manner

along with the axis of spring. The spring having stiffness is fixed at one end and carries a mass m

at its free end. The body is displaced from its equilibrium position vertically downwards. This

equilibrium position is called static equilibrium.

The free body diagram of the system is shown in fig: 2.1 (a).


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Fig (a): Spring mass system

Fig (b): Free body diagram of spring mass system

Figure: 2.1 (a)

b) EQUATION OF MOTION:

In equilibrium position, the gravitational pull mg is balanced by a spring force such that

𝑚𝑔 = kδ [𝐹𝑟𝑜𝑚 𝑓𝑖𝑔: 2.1 (𝑏)]

Where δ is the static deflection of the spring. Since the mass is displaced from its equilibrium

position by a distance x and then released, so after time t as per Newton’s II law.

𝑁𝑒𝑡 𝐹𝑜𝑟𝑐𝑒 = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

𝑚𝑔 – 𝑘 (𝛿 + 𝑥) = 𝑚 �̈� [equation of motion ]

𝑚 �̈� = 𝑚𝑔 − 𝑘𝛿 − 𝑘𝑥 (:− mg = k δ)

𝑚 �̈� = −𝑘𝑥

𝑚 �̈� + 𝑘𝑥 = 0

𝑥 + ( 𝑘 /𝑚) �̈� = 0 −−−−− (1)

Eq. (1) is differential Equation of motion for free vibration

c) NATURAL FREQUENCY

Equation (1) is a differential equation. The solution of which is

𝑥 = 𝐴 𝑠𝑖𝑛 �𝐾/𝑚 𝑡 + 𝐵 𝑐𝑜𝑠�𝐾/𝑚 𝑡. Where A & B are constant which can be found from

initial conditions.

The circular frequency 𝜔𝑛 = � 𝑘/𝑚


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The natural frequency of vibration ƒ𝑛 = 𝜔𝑛2𝜋� R

ƒ𝑛 = 1

2 𝜋 �𝑘 /𝑚 =

12 𝜋

� 𝑘 (𝑘𝛿/𝑔)� = 1

2 𝜋 �

𝑔𝛿

Where δ = static deflection

2.1. ii. Spring mass system displaced horizontally.

a) VIBRATION MODEL:

In the system shown in fig: 2.2, a body of mass m is free to move on a fixed horizontal

surface. The mass is supported on frictionless rollers. The spring of stiffness is attached to a

fixed frame at one side and to mass at other side.

Figure: 2.1 (b)


As per Newtons II law

Mass x acceleration = resultant force on mass

𝑚 �̈� = − 𝑘𝑥

𝑚 �̈� + 𝑘𝑥 = 0 [equation of motion ]

�̈� + (𝑘 /𝑚) 𝑥 = 0 −−−−(1)


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Eq. (1) is differential Equation of motion for free vibration

c) NATURAL FREQUENCY:


𝑥 = 𝐴 𝑠𝑖𝑛 ��𝐾/𝑚 �𝑡 + 𝐵 𝑐𝑜𝑠��𝐾/𝑚� 𝑡. Where A & B are constant which can be found from

initial conditions.

The circular frequency 𝜔𝑛 = � 𝑘/𝑚

The natural frequency of vibration, ƒ𝑛 = 𝜔𝑛2𝜋� R

ƒ𝑛 = 1

2 𝜋 �𝑘 /𝑚 =

12 𝜋

� 𝑘 (𝑘𝛿/𝑔)� = 𝟏𝟐 𝝅

� 𝒈𝜹

Where δ = static deflection

2.1. iii. System having a rotor of mass (Torsional Vibrations)

a) VIBRATION MODEL:

Consider a system having a rotor of

mass moment of inertia I connected to a

shaft at its end of torsional stiffness Kt , let

the rotor be twisted by an angle θ as

shown in fig: 2.2(c).

The body is rotated through an angle θ

and released, the torsional vibration will

result, the mass moment of inertia of the

shaft about the axis of rotation is usually

negligible compressed to I.

Fig: 2.2 (c) The free body diagram of general angular displacement is shown Fig: 2.2 (c)


As per Newtons II law

Mass x acceleration = resultant force on mass

The equation of motion is. 𝐼 �̈� = − 𝐾𝑡 𝜃


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𝐼 �̈� + 𝐾𝑡 𝜃 = 0

�̈� + � 𝐾𝑡𝐼� 𝜃 = 0

c) NATURAL FREQUENCY:


𝑥 = 𝐴 𝑠𝑖𝑛 �� 𝐾𝑡𝐼

� 𝑡 + 𝐵 𝑐𝑜𝑠 �� 𝐾𝑡𝐼� 𝑡. Where A & B are constant which can be found from

initial conditions.

The circular frequency 𝜔𝑛 = � 𝐾𝑡𝐼

The natural frequency of vibration, ƒ𝑛 = 𝜔𝑛2𝜋�

𝜔𝑛 = � 𝐾𝑡𝐼

& 𝑓𝑛 =1

2𝜋� 𝐾𝑡𝐼

Where 𝐾𝑡 = 𝐺𝐽𝐿

; 𝐽 = 𝜋 𝑑3

32

SPRINGS IN ARBITRARY DIRECTION

Fig shows a spring K making an angle α with the direction of motion of the mass m.


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If the mass is displaced by x, the spring is deformed by an amount 𝑥𝑐𝑜𝑠𝛼 along its axis

(spring Axis). The force along the spring axis is 𝑘𝑥 𝑐𝑜𝑠 𝛼. The component of this force along the

direction of motion of the mass is 𝑘𝑥 𝑐𝑜𝑠2 𝛼. The equation of motion of the mass m is

𝑚𝑥 + (𝑘𝑐𝑜𝑠2 𝛼) �̈� = 0.

From the above equation it may be noted that the equivalent stiffness 𝐾𝑒 of a spring making

angle-α with the axis of motion is 𝐾𝑒 = 𝐾 𝑐𝑜𝑠2 𝛼.

EQUIVALENT SHIFTNESS OF SPRING COMBINATIONS

Certain systems have more than one spring. The springs are joined in series or parallel or both.

They can be replaced by a single spring of the same shiftness as they all show the same shiftness

jointly.

SPRINGS IN PARALLEL

The deflection of individual spring is

equal to the deflection of the system.

𝑖. 𝑒 𝐾1𝑋 + 𝐾2𝑋 = 𝐾𝑒𝑋

𝐾1 + 𝐾2 = 𝐾𝑒

The equivalent spring shiftness is equal to

the sum of individual spring shiftiness.

SPRINGS IN SERIES


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The total deflection of the system is equal

to the sum of deflection of individual springs.

X = X1 + X2

𝑓𝑜𝑟𝑐𝑒𝐾𝑒

=𝑓𝑜𝑟𝑐𝑒𝐾1

=𝑓𝑜𝑟𝑐𝑒𝐾2

1𝐾𝑒

=1𝐾1

=1𝐾2

Thus when springs are connected in

series, the reciprocal of equivalent spring

shiftiness is equal to the sum of the reciprocal

of individual spring shiftiness.

2.2. ENERGY METHOD, RAYLEIGH METHOD

Other Methods of Finding Natural Frequency

The above method is called Newton’s method. The other methods which are commonly used

in vibration for determination of frequency are, (i) Energy Method (ii) Rayleigh’s Method

2.2. i. Energy Method:

Consider a spring mass system as shown in fig: 2.2 (a),

assume the system to be conservative. In a conservative

system the total sum of the energy is constant in a vibrating

system the energy is partly potential and partly kinetic. The


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K.E, T is because of velocity of the mass and Potential energy

V is stored in the spring because of its elastic deformation

As per conservation law of energy.

Fig:2.2 (c)

T + V = constant

Differentiating the above equation w.r.t. ‘t’

𝑑𝑑𝑡

(𝑇 + 𝑉) = 0

For a spring mass system shown

𝐾.𝐸 = 𝑇 = ½ 𝑚�̇�2

𝑃.𝐸 = 𝑉 = ½ 𝑘𝑥2 𝒅𝒅𝒕

(½ 𝑚�̇�2 + ½ 𝑘𝑥2) = 0

𝒅𝒅𝒕

( 𝑚�̇�2 + 𝑘𝑥2) = 0

𝑚 x x + 𝑘𝑥 x = 0

�̈� + (𝑘/𝑚) 𝑥 = 0

Hence the natural frequency is

𝜔𝑛 = � 𝑘/𝑚 ƒ𝑛 = 1 /2𝜋 �𝑘/𝑚 = 1/ 2𝜋 �𝑔 / 𝛿

2.2. ii. Rayleigh’s Method:


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Consider the spring mass system as shown. In deriving the

expression, it is assumed that the maximum K.E at mean

position is equal to the maximum P.E at the extreme position.

The motion is assumed to be SH

Then 𝑥 = 𝐴 𝑠𝑖𝑛 𝜔𝑛 𝑡

X= displacement of the body from mean position after time‘t’

A = Maximum. displacement from mean position to extreme

position.

Differentiating w. r. t

�̇� = 𝐴𝜔𝑛 𝑐𝑜𝑠 𝜔𝑛 𝑡

Fig:2.2 (b)

Maximum Velocity at mean position �̇� = 𝜔𝑛 𝐴

Maximum kinetic energy at mean position = 1 2� (𝑚�̇�2)

= 1 2� (𝑚𝜔𝑛2𝐴2)

And maximum potential energy at Extreme position

𝑃.𝐸 = 12� (𝑘𝐴2)

W.K.T, K.E = P.E

= 1 2� (𝑚𝜔𝑛2𝐴2)P

= 1 2� (𝑘𝐴2)

𝜔𝑛2 = (𝑘/𝑚)

𝜔𝑛 = �(𝑘/𝑚)


ƒ𝑛 = 1 /2𝜋 �𝑘/𝑚 = 1/ 2𝜋 �𝑔 / 𝛿


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2.3. PRINCIPLE OF VIRTUAL WORK, DAMPING MODELS

2.3. i. PRINCIPLE OF VIRTUAL WORK:

The virtual work method is another scalar method besides the work and energy method. It is

useful especially for systems of interconnected bodies of higher DOF.

The principle of virtual work states that If a system in equilibrium under the action of a set of

forces is given a virtual displacement, the virtual work done by the forces will be zero. In other

words

(1) 𝛿𝑊 = 0 For static equilibrium.

𝛿𝑊 = 𝛿𝑊 + 𝛿𝑊inertia for dynamic equilibrium.

(2) Virtual displacements should satisfy the displacement boundary conditions. It will be

shown that these conditions are not crucial. Virtual displacement: imaginary (not real)

displacement

Example: Use the virtual work method; determine the equation of motion for the system below.

Draw the system in the displaced position x and place the forces acting on it, including

inertia and gravity forces. Give the system a small virtual displacement δ x and determine the

work done by each force. Using the fact that virtual work done by external forces equals virtual

work done by inertia forces, we then obtain the equation of motion for the system.

δW = m. �̈�. δx

The virtual work done by inertia forces is

δW = −kx. δx


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Equating the two quantities above and canceling δ x, we have the equation of motion

m. �̈� + kx = 0

Example: Simple Pendulum

Use the virtual work method; determine the equation of motion for the system below.

𝛿𝑊′ = 𝛿𝑊inertia + 𝛿𝑊𝑐 + 𝛿𝑊𝑛𝑐

𝛿𝑊𝑛𝑐 = 0: Neoconservative force (external force or damping force)

𝛿𝑊inertia = −�ml�̈�� (lδθ)

𝛿𝑊𝑐 = −(mg sinθ) (lδθ)


𝛿𝑊′ = −�ml�̈�� (lδθ) + �−(mg sinθ) (lδθ)� + 0

𝛿𝑊′ = −�ml�̈� + mg sinθ�(lδθ)

𝛿𝑊′ = 0; δθ(t): 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦

𝐦𝐥�̈� + 𝐦𝐠 𝐬𝐢𝐧𝛉 = 𝟎 Nonlinear equation

If θ is small, sinθ = θ

ml�̈� + mg θ = 0

l�̈� + g θ = 0

�̈� + 𝐠𝒍

𝛉 = 𝟎 Linear equation

2.3. DAMPING MODELS

DAMPED FREE VIBRATION OF SINGLE DEGREE OF FREEDOM SYSTEMS

In general, all physical systems are associated with one or the other type of damping. In

certain cases the amount of damping may be small and in other cases large. When damped free

vibrations takes place, the amplitude of vibration gradually becomes small and finally is


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completely lost. The rate, at which the amplitude decays, depends upon the type and amount of

damping in the system. The aspects we are primary interested in damped free vibrations are 1)

the frequency of damped oscillations 2) the rate of decay

Different Models of Damping

Damping is associated with energy dissipation. There are several types of damping. Four of

which are important types which are discussed here.

1) Viscous damping

2) Coulomb damping

3) Structural damping or solid damping

4) Slip or Interfacial damping

Viscous Damping: Viscous damping is encountered by bodies moving at moderate speed

through a liquid. This type of damping leads to a resisting force proportional to the velocity. The

damping force.

𝐹𝑑 𝛼 𝑑𝑥𝑑𝑡

𝐹𝑑 = 𝑐�̇�

When ‘c’ is the constant of proportionality and is called viscous damping Co-efficient with the

dimension of N-s/m.

Coulomb Damping: - This type of damping arises from sliding of dry surfaces. The friction

force is nearly constant and depends upon the nature of sliding surface and normal pressure

between them as expressed by the equation of kinetic friction.

F = µ N

When µ = co- efficient of friction

N = normal force


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Now coming towards damping force, if we analyze above expression, we can deduce result that

it only depends upon normal force irrespective of displacement, velocity of the body. Therefore

in order to findout mathematical solution of single degree of freedom with coulomb damping, we

consider this reciprocatory motion into two cases.

In this case, when the displacement x of the body is positive and dx/dt will be positive or

displacement x is negative, dx/dt will still be positive. Such kind of condition can only be

fulfilled if the body moves from left side to the right side. Therefore Newton’s second law of

motion will be

𝑚ẍ = – 𝑘𝑥 – 𝜇𝑁

𝑚ẍ + 𝑘𝑥 = – 𝜇𝑁

The above equation is second order nonhomogeneous differential equation. In order to verify and

to make calculations easier, we will assume that the system exhibit harmonic motion. Therefore

x(t) = A1cos ωnt + A2sin ωnt – μN / k

where in the above expression ωn = (k / m)1/2 and A1 and A2 are constant and there values can be

find out by using initial conditions.

Solid or Structural Damping:-

Solid damping is also called structural damping and is due to internal friction within the material

itself. Experiment indicates that the solid damping differs from viscous damping in that it is

independent of frequency and proportional to maximum stress of vibration cycle. The

independence of solid damping frequency is illustrated by the fact that all frequencies of

vibrating bodies such as bell are damped almost equally.


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Slip or Interfacial Damping

Energy of vibration is dissipated by microscopic slip on the interfaces of machine parts in

contact under fluctuating loads. Microscopic slip also occurs on the interfaces of the machine

elements having various types of joints. The amount of damping depends amongst other things

upon the surface roughness of the mating parts, the contact pressure and the amplitude of

vibration. This type of damping is essentially of a non-linear type.

QUADRANT-2 Animations

• https://www.google.co.in/#q=animations+of+single+degree+of+free+vibration

• www.thirdmill.org/mission/bts.asp

• acoustics.mie.uic.edu/Simulation/SDOF%20Undamped.htm

• acoustics.mie.uic.edu/Simulation/SDOF%20Damped.htm

• www.brown.edu/.../vibrations_free.../vibrations_free_undamped.htm

• se.asee.org/proceedings/ASEE2009/papers/PR2009011ERV.PDF

• www.efunda.com/formulae/vibrations/sdof_free_damped.cfm

• https://dspace.uta.edu/bitstream/.../Deshmukh_uta_2502M_11706.pdf?...

• www.vibrationdata.com/matlab.htm

• www.vibrationdata.com/animation.htm

• www.acs.psu.edu/drussell/demos.html

• facultad.bayamon.inter.edu/.../Chapter%202%20Free%20vibration%20o...

• web.itu.edu.tr/~gundes/2dof.pdf


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http://www.brown.edu/.../vibrations_free.../vibrations_free_undamped.htm

%E2%80%A2%09https://www.google.co.in/#q=animations+of+single+degree+of+free+vibration

www.thirdmill.org/mission/bts.asp%E2%80%8E

http://acoustics.mie.uic.edu/Simulation/SDOF%20Undamped.htm

http://acoustics.mie.uic.edu/Simulation/SDOF%20Damped.htm%E2%80%8E

http://se.asee.org/proceedings/ASEE2009/papers/PR2009011ERV.PDF%E2%80%8E

www.efunda.com/formulae/vibrations/sdof_free_damped.cfm

https://dspace.uta.edu/bitstream/.../Deshmukh_uta_2502M_11706.pdf

www.vibrationdata.com/matlab.htm%E2%80%8E

www.vibrationdata.com/animation.htm%E2%80%8E

www.acs.psu.edu/drussell/demos.html%E2%80%8E

facultad.bayamon.inter.edu/.../Chapter%202%20Free%20vibration%20o...%E2%80%8E

http://web.itu.edu.tr/%7Egundes/2dof.pdf%E2%80%8E


2014

Videos www.youtube.com/watch?v=vBLuOXBfzJk

www.youtube.com/watch?v=PfxlTw3BS7g

www.youtube.com/watch?v=bDa8Ghm9aRw

ww.youtube.com/watch?v=_Rn68hC4rlc

www.youtube.com/watch?v=JZWf2sdKhS8

www.youtube.com/watch?v=PsXLBphWNuE

www.youtube.com/watch?v=DErLaGaJ1d0

www.youtube.com/watch?v=RKfZ081epsM

www.youtube.com/watch?v=MUWI-yi9Y2s

www.youtube.com/watch?v=DdkIai5oQtU www.youtube.com/watch?v=V_Lj4Pun_WM

Illustrations 1. Derive an expression for an Equation of Motion and Natural Frequency of Vibration of a

Simple Spring Mass System .


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http://www.youtube.com/watch?v=PsXLBphWNuE

http://www.youtube.com/watch?v=DErLaGaJ1d0

http://www.youtube.com/watch?v=MUWI-yi9Y2s

www.youtube.com/watch?v=vBLuOXBfzJk%E2%80%8E

www.youtube.com/watch?v=PfxlTw3BS7g%E2%80%8E

www.youtube.com/watch?v=bDa8Ghm9aRw

www.youtube.com/watch?v=_Rn68hC4rlc%E2%80%8E

www.youtube.com/watch?v=JZWf2sdKhS8

www.youtube.com/watch?v=RKfZ081epsM%E2%80%8E

www.youtube.com/watch?v=DdkIai5oQtU%E2%80%8E

www.youtube.com/watch?v=V_Lj4Pun_WM%E2%80%8E


2014

Consider a spring mass system as shown in fig constrained to move in a collinear manner along

with the axis of spring. The spring having stiffness is fixed at one end and carries a mass m at its

free end. The body is displaced from its equilibrium position vertically downwards. This

equilibrium position is called static equilibrium.

The free body dia of the system is shown in fig.

In equilibrium position, the gravitational pull mg is balanced by a spring force such that

mg = kδ.

Where δ is the static deflection of the spring. Since the mass is displaced from its equilibrium

position by a distance x and then released, so after time t as per Newton’s II law.

Net Force = mass x acceleration

mg – k (δ + x) = m x

m x = mg - kδ - kx (:- mg = k δ)

m x = -kx

m x + kx = 0

x + k /m x = 0 ----- (1)

Equation (1) is a differential equation. The solution of which is x = A sin √K/m t +

B cos√K/m t. Where A and B are constant which can be found from initial conditions.

The circular frequency ωn = √ k/m

The natural frequency of vibration ƒn = ωn /2

ƒn = 1 /2 π √k /m = 1/2 π √ k / kδ/g = 1 / 2 π √g / δ

Where δ = static deflection 2) Explain a) Energy method b) Rayliegh Ritz method of finding Natural Frquency.

Ans)

a) Energy Method:






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Fig:2.2 (c)

T + V = constant


𝑑𝑑𝑡

(𝑇 + 𝑉) = 0


𝐾.𝐸 = 𝑇 = ½ 𝑚�̇�2


(½ 𝑚�̇�2 + ½ 𝑘𝑥2) = 0

𝒅𝒅𝒕

( 𝑚�̇�2 + 𝑘𝑥2) = 0

𝑚 x x + 𝑘𝑥 x = 0

�̈� + (𝑘/𝑚) 𝑥 = 0


𝜔𝑛 = � 𝑘/𝑚 ƒ𝑛 = 1 /2𝜋 �𝑘/𝑚 = 1/ 2𝜋 �𝑔 / 𝛿

b) Rayleigh’s Method:


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position.



Fig:2.2 (b)



= 1 2� (𝑚𝜔𝑛2𝐴2)


𝑃.𝐸 = 12� (𝑘𝐴2)

W.K.T, K.E = P.E

= 1 2� (𝑚𝜔𝑛2𝐴2)P

= 1 2� (𝑘𝐴2)

𝜔𝑛2 = (𝑘/𝑚)

𝜔𝑛 = �(𝑘/𝑚)


ƒ𝑛 = 1 /2𝜋 �𝑘/𝑚 = 1/ 2𝜋 �𝑔 / 𝛿

3.Eriefly Explain different models of damping.

Ans) Different Models of Damping




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5) Viscous damping

6) Coulomb damping





damping force.


𝐹𝑑 = 𝑐�̇�


dimension of N-s/m.




F = µ N


N = normal force

4) Explian the principle of Virtual Work.And derive an expression for equation of Motion for a

spring Mass System and Simple Pendulam.


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Ans)





δW = m. �̈�. δx


δW = −kx. δx


m. �̈� + kx = 0








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ml�̈� + mg θ = 0

l�̈� + g θ = 0

�̈� + 𝐠𝒍


5) Determine the natural frequency of a compound pendulum.

Solution:

Figure below shows a compound pendulum in the displaced position.

Let m = Mass of the rigid body

= 𝑤𝑔

l = Distance of point of suspension from G

O = Point of suspension

G = Centre of gravity

I = Moment of inertia of the body about O

= mk2 + ml2 = m(k2 + l2)

k = Radius of gyration of the body

If OG is displaced by an angle,

Restoring torque = -mglθ since θ is small sin θ ≈ θ

According to Newton’s second law

Accelerating torque = Restoring torque

i.e., 𝐼𝜃 ̈ = -mglθ

i.e, �̈�+ 𝑚𝑔𝑙𝐼 θ = 0

i.e, �̈�+ 𝑚𝑔𝑙𝑚(𝑘2+ 𝑙2

θ = 0


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∴�̈�+ 𝑔𝑙(𝑘2+ 𝑙2

θ = 0

∴ 𝜔𝑛= � 𝑔𝑙𝑘2+𝑙2

𝑟𝑎𝑑/𝑠𝑒𝑐

Hence natural frequency 𝑓𝑛 = 12𝜋

𝜔𝑛 = 12𝜋� 𝑔𝑙𝑘2+𝑙2

𝐻𝑧.

6) Determine the natural frequency of a spring mass system where the mass of the spring is also

to be taken into account.

Solution: Figure shows a spring mass system If the mass of the spring is taken into account then, let x = Displacement of mass 𝑥 ̇ = Velocity of the free end of the spring at the instant under consideration. m' = Mass of spring wire per unit length l = Total length of the spring wire. Consider an elemental length dy at a distance ‘y’ measured from the fixed end. Velocity of the spring wire at the distance y from the fixed end = 𝑥 ̇ �𝑦

𝑙�

Kinetic energy of the spring element dy = 12

(𝑚′𝑑𝑦) ��̇� 𝑦𝑙�2

7) A block of mass 0.05 kg is suspended from a spring having a stiffness of 25 N/m.

The block is displaced downwards from its equilibrium position through a distance of 2 cm and released with an upward velocity of 3 cm/sec. Determine (i) Natural Frequency (ii) Period of Oscillation (iii) Maximum Velocity (iv) Maximum acceleration (v) Phase angle.

Data: m = 0.05 kg; k = 25 N/m; x(0) = x0 = 2 cm �̇�R0 = ν0 = 3cm/sec.

Solution:

The differential equation of the motion is given by

�̈� + 𝑘𝑚𝑥 = 0

The general solution for the above differential equation is,

x(t) = A cos ωn t + B sin ωn t = X cos (ωn t - φ)

When t = 0, x(0) = x0 = A = 2cm


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∴ �̇� R(0) = ν0 = B ωn; ∴ B = 𝜗0𝜔𝑛

ωn = �𝑘𝑚

= � 250.05

= 22.36 𝑟𝑎𝑑/𝑠𝑒𝑐

Maximum amplitude of vibration X = √𝐴2 + 𝐵2 = �𝑥02 + 𝜗02

𝜔𝑛2

= �22 + 32

22.362 = 2.0045𝑐𝑚

i) Natural Frequency 𝑓𝑛 = 12𝜋

𝜔𝑛 = 12𝜋

× 22.36 = 3.56 𝐻𝑧

ii) Period of oscillation T = 1𝑓𝑛

= 13.56

= 0.28 sec.

iii) Maximum Velocity �̇� Rmax = X ωn = 2.0045 × 22.36 = 44.82 cm/sec.

iv) Maximum Acceleration �̈� Rmax = X 𝜔𝑛2 = �̇� Rmax .ωn = 44.82 × 22.36 = 1002.2 cm/sec2

v) Phase angle φ = tan-1 � 𝜗0𝜔𝑛𝑥0

� = 𝑡𝑎𝑛−1 � 322.36×2

� = 3.838𝑜

8) An oscillating system with a natural frequency of 3.98 Hz starts with an initial displacement of x0 = 10 mm and an initial velocity of �̇�R0 = 125 mm/sec. Calculate all the vibratory parameters involved and the time taken to reach the first peak.

Data: f = 3.98 Hz; x0 = 10 mm; �̇�R0 = ν0 = 125 mm/sec. Solution:


�̈� + 𝑘𝑚𝑥 = 0



When t = 0, x(0) = x0 = A = 10 mm

∴ �̇� R(0) = ν0 = B ωn; ∴ B = 𝜗0𝜔𝑛

Frequency 𝑓 = 12𝜋𝜔𝑛

i.e, 3.98 = 12𝜋𝜔𝑛

∴ 𝜔𝑛 = 25 rad/sec

i) Maximum amplitude of vibration X = √𝐴2 + 𝐵2 = �𝑥02 + 𝜗02

𝜔𝑛2 = �102 + 1252

252 = 11.18 𝑚𝑚.

ii) Period of oscillation T = 1𝑓𝑛

= 13.98

= 0.251 𝑠𝑒𝑐.


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iii) Maximum Velocity �̇� Rmax = X ωn = 11.18 × 25 = 279.5 mm/sec

iv) Maximum Acceleration �̈� Rmax = X 𝜔𝑛2 = �̇� Rmax ωn = 279.5 × 25 = 6987.5 mm/sec

v) Phase angle φ = tan-1 � 𝜗0𝜔𝑛𝑥0

� = 𝑡𝑎𝑛−1 � 12525×10

� = 26.565𝑜

vi) Time taken to reach the first peak = 𝜋𝜔𝑛

=26.565× 𝜋180

25= 0.018546 𝑠𝑒𝑐.

vii) Lead angle Ψ = tan-1 (𝜔𝑛𝑥0𝜗0

= 𝑡𝑎𝑛−1 �25 ×10125

� = 1.107 𝑟𝑎𝑑𝑖𝑎𝑛 = 63.435𝑜

QUADRANT-3


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Wikis

• en.wikipedia.org/wiki/Vibration

• wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp...

• Wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp...

• apmr.matelys.com/BasicsMechanics/SDOF/index.html

• www.structuralwiki.org › Home › Topics

• petrowiki.org/Basic_vibration_analysis

• vibrationdata.com/python-wiki/index.php?title=Runge-Kutta_ODE...

• en.wikipedia.org/wiki/Energy_functional

Open Contents:

• Mechanical Vibrations, S. S. Rao, Pearson Education Inc, 4th edition, 2003. • Mechanical Vibrations, V. P. Singh, Dhanpat Rai & Company, 3rd edition, 2006. • Mechanical Vibrations, G. K.Grover, Nem Chand and Bros, 6th edition, 1996 • Theory of vibration with applications ,W.T.Thomson,M.D.Dahleh and C

Padmanabhan,Pearson Education inc,5th Edition ,2008 • Theory and practice of Mechanical Vibration : J.S.Rao&K,Gupta,New Age International

Publications ,New Delhi,2001

QUADRANT-4 Problems


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http://www.structuralwiki.org/en/Main_Page

http://www.structuralwiki.org/en/StructuralWiki:Topics

http://en.wikipedia.org/wiki/Vibration

http://wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp...%E2%80%8E

http://wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp...%E2%80%8E

http://apmr.matelys.com/BasicsMechanics/SDOF/index.html%E2%80%8E

www.structuralwiki.org

http://petrowiki.org/Basic_vibration_analysis%E2%80%8E

http://vibrationdata.com/python-wiki/index.php?title=Runge-Kutta_ODE...

http://en.wikipedia.org/wiki/Energy_functional%E2%80%8E


2014

1. An instrument panel of natural period 0.1 second, is excited by a step function 0.5 cm

magnitude for a period of 0.075 second. Determine the response of the system.

Solution:

Natural frequency = 10.1

= 10 𝐻𝑧

Also fn = 12𝜋𝜔𝑛

i.e, 10 = 12𝜋𝜔𝑛

∴ 𝜔𝑛 = 20𝜋 𝑟𝑎𝑑𝑖𝑎𝑛/ sec = 62.832 𝑟𝑎𝑑𝑠𝑒𝑐

.

With the reference to translated equilibrium position for the first part,

x(t) = X cos ωn t; ∴ x(0) = 0.5 = X

�̇� R(t) = ν0 = 0; ∴ x(t) = 0.5 cos 62.832 t

With reference to the original mean equilibrium position of the mass

x = x(t) - x(0) = 0.5 cos 62.832t – 0.5

= 0.5[cos 62.832t-1]

∴ �̇� = (0.5×62.832)(-sin 62.832t)

i.e., �̇�62.832

= −0.5 sin 62.832𝑡

i.e., �̇�𝜔𝑛

= −0.5 sin 62.832𝑡

At the end of first part, t = 0.075 sec

∴ x(0.075) = 0.5 [cos (20π × 0.075) – 1] = -0.5 cm

�̇�0.075𝜔𝑛

= -0.5 sin (20π × 0.075) = 0.5 cm = 𝜗0.075𝜔𝑛

∴ For the second part with t’ as time,

x = A cos + 𝐵 sin𝜔𝑛𝑡′

= X cos (𝜔𝑛𝑡′- φ)

X = √𝐴2 + 𝐵2 = �𝑥(0.075)2 + �𝜗0.075

𝜔𝑛� = �(−0.5)2 + 0.52 = 0.7071

φ = tan-1 �𝜗0.075𝜔𝑛

𝑥0.075� = 𝑡𝑎𝑛−1 �−0.5

0.5� = 0.7854 𝑟𝑎𝑑 = 2.3562 𝑟𝑎𝑑𝑖𝑎𝑛

∴ x = 0.7071 cos (62.832t’ – 2.3562) cm.


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2. The solution to the differential for single degree freedom motion is given by x =

X cos (100t + φ) with initial condition �̇�R(0) = 1250mm/sec and x(0) = 0.25 mm find the

values of X and φ, and express the given equation in the form x = A

sin ωn t + B cos ωn t.

Data: �̇�R0 = ν0 = 1250 mm/sec; x(0) = x0 = 0.25 mm

Solution:

Given equation x = X cos (100t + φ) = X cos (ωn t + φ)

∴ωn = 100rad/sec

Also x(t) = A sin ωn t + B cos ωn t

∴ x(t) = x0 = 0 + B

∴ B = x0 = 0.25 mm

�̇� R(t) = A ωn cos ωn t - B ωn sin ωn t

∴ �̇� R(0) = ν0 = A ωn

i.e., 1250 = 100 A

∴ A = 12.5 mm

Maximum amplitude X = √𝐴2 + 𝐵2 = ��𝜗02

𝜔𝑛2� + 𝑥02

= √12.52 + 0.252 = 12.5025 𝑚𝑚.

Now X cos (ωn t + φ) = A sin ωn t + B cos ωn t (given)

i.e., X cos ωn t.cosφ - X sin ωn t sinφ = A sin ωn t + B cos ωn t

∴A = -X sin φ; B = X cos φ

∴ tan φ = -𝐴𝐵

∴ Phase angle φ = tan-1 �− 𝐴𝐵� = 𝑡𝑎𝑛−1 �

−𝜗0𝜔𝑛𝑥0� = 𝑡𝑎𝑛−1 �−12.5

0.25� = −1.55 𝑟𝑎𝑑𝑖𝑎𝑛

= 1.5908 radian = 91.146o

Hence the given equation is,

x = 12.5 sin 100t + 0.25 cos 100t

= 12.5025 cos (100t + 1.5908) = 12.5025 cos (100t + 91.146o).


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3. Determine the natural frequency of an Spring Mass System where the Mass of the Spring is

also to be taken in to account

consider a spring mass system as shown in fig. let L be the length of the spring under equilibrium condition. Consider an element dy of the spring at a distance ‘y’ from the support as shown. If ρ is the mass per unit length of the spring in equilibrium condition, then the mass of the spring ms= ρL and the mass of the element dy is equal to ρdy. At any instant, let the mass be displaced from the equilibrium position through a distance x, then the P.E of the system is P.E = ½ kx2 The K.E of vibration of the system at this instant consists of K.E of the main mass plus the K.E of the spring. The K.E of the mass is equal to ½ m x 2

The K.E of the element dy of the spring is equal to 1/2 (ρdy) (y/ L x x)2

x

dy

y

lk

m

Therefore the total K.E of the system is given by L K.E = 1/2 m x 2 + ∫ 0 ½ (ρ dy) (y/L x )2 = 1/2 m x 2 + ½ ρ x 2 / L2 ∫0y2dy = 1/2 m x 2 + ½ ρ x 2 / L2 [y3 / 3] L 0 = 1/2 m x 2 + ½ ρ x 2 / L2 [L3/3] = 1/2 m x 2 + ½ ρ L /3 x 2


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= 1/2 m x 2 + ½ ms /3 x 2 K.E = 1/2 x 2 [m + ms /3] We have by energy method P.E + K.E = Constant

1/2 kx2 +1/2 x 2 [m + ms / 3 ] = Constant Differentiating the above equation ½ k (2x) ( x ) + ½ (2 x ) ( x ) [m + ms /3] = 0 kx + (m + ms/3) x = 0 Or (m + ms /3) x + kx = 0 f n = 1/2π √ k / (m + ms /3) ωn = √k / (m + ms /3) Hence the above equation shows that for finding the natural frequency of the system, the mass of the spring can be taken into account by adding one – third its mass to the main mass.

4) Explain a) Energy method b) Rayliegh Ritz method of finding Natural Frquency.

Ans)

c) Energy Method:








Fig:2.2 (c)

T + V = constant



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𝑑𝑑𝑡

(𝑇 + 𝑉) = 0


𝐾.𝐸 = 𝑇 = ½ 𝑚�̇�2


(½ 𝑚�̇�2 + ½ 𝑘𝑥2) = 0

𝒅𝒅𝒕

( 𝑚�̇�2 + 𝑘𝑥2) = 0

𝑚 x x + 𝑘𝑥 x = 0

�̈� + (𝑘/𝑚) 𝑥 = 0


𝜔𝑛 = � 𝑘/𝑚 ƒ𝑛 = 1 /2𝜋 �𝑘/𝑚 = 1/ 2𝜋 �𝑔 / 𝛿

d) Rayleigh’s Method:








position.



Fig:2.2 (b)




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= 1 2� (𝑚𝜔𝑛2𝐴2)


𝑃.𝐸 = 12� (𝑘𝐴2)

W.K.T, K.E = P.E

= 1 2� (𝑚𝜔𝑛2𝐴2)P

= 1 2� (𝑘𝐴2)

𝜔𝑛2 = (𝑘/𝑚)

𝜔𝑛 = �(𝑘/𝑚)


ƒ𝑛 = 1 /2𝜋 �𝑘/𝑚 = 1/ 2𝜋 �𝑔 / 𝛿

1. Eriefly Explain different models of damping.

Ans) Different Models of Damping



9) Viscous damping

10) Coulomb damping





damping force.


𝐹𝑑 = 𝑐�̇�


dimension of N-s/m.


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F = µ N


N = normal force

6) Explian the principle of Virtual Work.And derive an expression for equation of Motion for a

spring Mass System and Simple Pendulam.

Ans)





δW = m. �̈�. δx


δW = −kx. δx


m. �̈� + kx = 0




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ml�̈� + mg θ = 0

l�̈� + g θ = 0

�̈� + 𝐠𝒍


7) Determine the natural frequency of a compound pendulum.

Solution:

Figure below shows a compound pendulum in the displaced position.

Let m = Mass of the rigid body

= 𝑤𝑔

l = Distance of point of suspension from G

O = Point of suspension

G = Centre of gravity

I = Moment of inertia of the body about O


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= mk2 + ml2 = m(k2 + l2)

k = Radius of gyration of the body

If OG is displaced by an angle,

Restoring torque = -mglθ since θ is small sin θ ≈ θ

According to Newton’s second law

Accelerating torque = Restoring torque

i.e., 𝐼𝜃 ̈ = -mglθ

i.e, �̈�+ 𝑚𝑔𝑙𝐼 θ = 0

i.e, �̈�+ 𝑚𝑔𝑙𝑚(𝑘2+ 𝑙2

θ = 0

∴�̈�+ 𝑔𝑙(𝑘2+ 𝑙2

θ = 0

∴ 𝜔𝑛= � 𝑔𝑙𝑘2+𝑙2

𝑟𝑎𝑑/𝑠𝑒𝑐

Hence natural frequency 𝑓𝑛 = 12𝜋

𝜔𝑛 = 12𝜋� 𝑔𝑙𝑘2+𝑙2

𝐻𝑧.

8) Determine the natural frequency of a spring mass system where the mass of the spring is also

to be taken into account.

Solution: Figure shows a spring mass system If the mass of the spring is taken into account then, let x = Displacement of mass 𝑥 ̇ = Velocity of the free end of the spring at the instant under consideration. m' = Mass of spring wire per unit length l = Total length of the spring wire. Consider an elemental length dy at a distance ‘y’ measured from the fixed end. Velocity of the spring wire at the distance y from the fixed end = 𝑥 ̇ �𝑦

𝑙�

Kinetic energy of the spring element dy = 12

(𝑚′𝑑𝑦) ��̇� 𝑦𝑙�2

9) A block of mass 0.05 kg is suspended from a spring having a stiffness of 25 N/m.

The block is displaced downwards from its equilibrium position through a distance of 2 cm and released with an upward velocity of 3 cm/sec. Determine


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(i) Natural Frequency (ii) Period of Oscillation (iii) Maximum Velocity (iv) Maximum acceleration (v) Phase angle.

Data: m = 0.05 kg; k = 25 N/m; x(0) = x0 = 2 cm �̇�R0 = ν0 = 3cm/sec.

Solution:


�̈� + 𝑘𝑚𝑥 = 0



When t = 0, x(0) = x0 = A = 2cm

∴ �̇� R(0) = ν0 = B ωn; ∴ B = 𝜗0𝜔𝑛

ωn = �𝑘𝑚

= � 250.05

= 22.36 𝑟𝑎𝑑/𝑠𝑒𝑐

Maximum amplitude of vibration X = √𝐴2 + 𝐵2 = �𝑥02 + 𝜗02

𝜔𝑛2

= �22 + 32

22.362 = 2.0045𝑐𝑚

vi) Natural Frequency 𝑓𝑛 = 12𝜋

𝜔𝑛 = 12𝜋

× 22.36 = 3.56 𝐻𝑧

vii) Period of oscillation T = 1𝑓𝑛

= 13.56

= 0.28 sec.

viii) Maximum Velocity �̇� Rmax = X ωn = 2.0045 × 22.36 = 44.82 cm/sec.

ix) Maximum Acceleration �̈� Rmax = X 𝜔𝑛2 = �̇� Rmax .ωn = 44.82 × 22.36 = 1002.2 cm/sec2

x) Phase angle φ = tan-1 � 𝜗0𝜔𝑛𝑥0

� = 𝑡𝑎𝑛−1 � 322.36×2

� = 3.838𝑜

Assignment 1. An unknown mass m kg attached to the end of an unknown spring K has a natural frequency of

94 HZ, When a 0.453 kg mass is added to m, the natural frequency is lowered to 76.7 HZ.

Determine the unknown mass m and the spring constant K N/m.

2. An unknown mass is attached to one end of a spring of shiftness K having natural frequency of 6

Hz. When 1kg mass is attached with m the natural frequency of the system is lowered by 20%.

Determine the value of the unknown mass m and stiffness K.


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3. Find the natural frequency of the system shown in fig (1).given K1 = K2 = 1500 N/m K3 = 2000

N/m and m= 5 kg.

4. A mass is suspended from a spring system as shown in fig (2). Determine the natural frequency of

the system. Given k1= 5000N/m, K2=K3= 8000N/m and m= 25 kg.

5. Consider the system shown in fig (4). If K1= 20N/cm, K2= 30N/cm K4= K5= 5N/cm. Find the

mass m if the systems natural frequency is 10 Hz.


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7. Find the natural frequency of the system shown in fig (5). K1= K2=K3=K4=K5=K6=K = 1000

N/m.

8. A mass m guided in x-x direction is connected by a spring configuration as shown in fig (1).

Set up the equation of mass m. write down the expression for equivalent spring constant.


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9. Find the equivalent spring constant of the system shown in fig (2) in the direction of the load P.

10. Determine the natural frequency of spring mass system taking the mass of the spring in to

account.

11. Drive the differential equation for an undamped spring mass system using Newton’s method.

12. Derive the equation of motion of a simple pendulum having an angular displacement of

𝜃.

13. Show that the frequency of undamped free vibration of a spring mass system is given

by 𝐹𝑛 = 1/ 2𝜋 �𝑔 𝛿

.

14. Show that the natural frequency of undamped free vibration of a spring mass system is

given by 𝜔𝑛 = 1/ 2𝜋 �𝑘 𝐽

.

15. Using the energy method derive the differential equation of motion of an undamped free

vibration and show that frequency 𝜔𝑛 = �𝑘 𝑚

16. Using the Rayleigh method derive the differential equation of motion of an undamped

free vibration and show that frequency 𝜔𝑛 = �𝑘 𝑚

17. Derive the natural frequency of torsional vibrations.


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