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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
VTU-NPTEL-N
MEICT Proj
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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.
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
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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).
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
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2014
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
๐๐ = kฮด [๐น๐๐๐ ๐๐๐: 2.1 (๐)]
Where ฮด 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 Newtonโs II law.
๐๐๐ก ๐น๐๐๐๐ = ๐๐๐ ๐ ร ๐๐๐๐๐๐๐๐๐ก๐๐๐
๐๐ โ ๐ (๐ฟ + ๐ฅ) = ๐ ๏ฟฝฬ๏ฟฝ [equation of motion ]
๐ ๏ฟฝฬ๏ฟฝ = ๐๐ โ ๐๐ฟ โ ๐๐ฅ (:โ mg = k ฮด)
๐ ๏ฟฝฬ๏ฟฝ = โ๐๐ฅ
๐ ๏ฟฝฬ๏ฟฝ + ๐๐ฅ = 0
๐ฅ + ( ๐ /๐) ๏ฟฝฬ๏ฟฝ = 0 โโโโโ (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
๐ฅ = ๐ด ๐ ๐๐ ๏ฟฝ๐พ/๐ ๐ก + ๐ต ๐๐๐ ๏ฟฝ๐พ/๐ ๐ก. Where A & B are constant which can be found from
initial conditions.
The circular frequency ๐๐ = ๏ฟฝ ๐/๐
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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2014
The natural frequency of vibration ฦ๐ = ๐๐2๐๏ฟฝ R
ฦ๐ = 1
2 ๐ ๏ฟฝ๐ /๐ =
12 ๐
๏ฟฝ ๐ (๐๐ฟ/๐)๏ฟฝ = 1
2 ๐ ๏ฟฝ
๐๐ฟ
Where ฮด = 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)
b) EQUATION OF MOTION:
As per Newtons II law
Mass x acceleration = resultant force on mass
๐ ๏ฟฝฬ๏ฟฝ = โ ๐๐ฅ
๐ ๏ฟฝฬ๏ฟฝ + ๐๐ฅ = 0 [equation of motion ]
๏ฟฝฬ๏ฟฝ + (๐ /๐) ๐ฅ = 0 โโโโ(1)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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Eq. (1) is differential Equation of motion for free vibration
c) NATURAL FREQUENCY:
Equation (1) is a differential equation. The solution of which is
๐ฅ = ๐ด ๐ ๐๐ ๏ฟฝ๏ฟฝ๐พ/๐ ๏ฟฝ๐ก + ๐ต ๐๐๐ ๏ฟฝ๏ฟฝ๐พ/๐๏ฟฝ ๐ก. Where A & B are constant which can be found from
initial conditions.
The circular frequency ๐๐ = ๏ฟฝ ๐/๐
The natural frequency of vibration, ฦ๐ = ๐๐2๐๏ฟฝ R
ฦ๐ = 1
2 ๐ ๏ฟฝ๐ /๐ =
12 ๐
๏ฟฝ ๐ (๐๐ฟ/๐)๏ฟฝ = ๐๐ ๐
๏ฟฝ ๐๐น
Where ฮด = 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 ฮธ as
shown in fig: 2.2(c).
The body is rotated through an angle ฮธ
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)
b) EQUATION OF MOTION:
As per Newtons II law
Mass x acceleration = resultant force on mass
The equation of motion is. ๐ผ ๏ฟฝฬ๏ฟฝ = โ ๐พ๐ก ๐
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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๐ผ ๏ฟฝฬ๏ฟฝ + ๐พ๐ก ๐ = 0
๏ฟฝฬ๏ฟฝ + ๏ฟฝ ๐พ๐ก๐ผ๏ฟฝ ๐ = 0
c) NATURAL FREQUENCY:
Equation (1) is a differential equation. The solution of which is
๐ฅ = ๐ด ๐ ๐๐ ๏ฟฝ๏ฟฝ ๐พ๐ก๐ผ
๏ฟฝ ๐ก + ๐ต ๐๐๐ ๏ฟฝ๏ฟฝ ๐พ๐ก๐ผ๏ฟฝ ๐ก. Where A & B are constant which can be found from
initial conditions.
The circular frequency ๐๐ = ๏ฟฝ ๐พ๐ก๐ผ
The natural frequency of vibration, ฦ๐ = ๐๐2๐๏ฟฝ
๐๐ = ๏ฟฝ ๐พ๐ก๐ผ
& ๐๐ =1
2๐๏ฟฝ ๐พ๐ก๐ผ
Where ๐พ๐ก = ๐บ๐ฝ๐ฟ
; ๐ฝ = ๐ ๐3
32
SPRINGS IN ARBITRARY DIRECTION
Fig shows a spring K making an angle ฮฑ with the direction of motion of the mass m.
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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If the mass is displaced by x, the spring is deformed by an amount ๐ฅ๐๐๐ ๐ผ along its axis
(spring Axis). The force along the spring axis is ๐๐ฅ ๐๐๐ ๐ผ. The component of this force along the
direction of motion of the mass is ๐๐ฅ ๐๐๐ 2 ๐ผ. The equation of motion of the mass m is
๐๐ฅ + (๐๐๐๐ 2 ๐ผ) ๏ฟฝฬ๏ฟฝ = 0.
From the above equation it may be noted that the equivalent stiffness ๐พ๐ of a spring making
angle-ฮฑ with the axis of motion is ๐พ๐ = ๐พ ๐๐๐ 2 ๐ผ.
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.
๐. ๐ ๐พ1๐ + ๐พ2๐ = ๐พ๐๐
๐พ1 + ๐พ2 = ๐พ๐
The equivalent spring shiftness is equal to
the sum of individual spring shiftiness.
SPRINGS IN SERIES
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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The total deflection of the system is equal
to the sum of deflection of individual springs.
X = X1 + X2
๐๐๐๐๐๐พ๐
=๐๐๐๐๐๐พ1
=๐๐๐๐๐๐พ2
1๐พ๐
=1๐พ1
=1๐พ2
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 Newtonโs method. The other methods which are commonly used
in vibration for determination of frequency are, (i) Energy Method (ii) Rayleighโs 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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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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. โtโ
๐๐๐ก
(๐ + ๐) = 0
For a spring mass system shown
๐พ.๐ธ = ๐ = ยฝ ๐๏ฟฝฬ๏ฟฝ2
๐.๐ธ = ๐ = ยฝ ๐๐ฅ2 ๐ ๐ ๐
(ยฝ ๐๏ฟฝฬ๏ฟฝ2 + ยฝ ๐๐ฅ2) = 0
๐ ๐ ๐
( ๐๏ฟฝฬ๏ฟฝ2 + ๐๐ฅ2) = 0
๐ x x + ๐๐ฅ x = 0
๏ฟฝฬ๏ฟฝ + (๐/๐) ๐ฅ = 0
Hence the natural frequency is
๐๐ = ๏ฟฝ ๐/๐ ฦ๐ = 1 /2๐ ๏ฟฝ๐/๐ = 1/ 2๐ ๏ฟฝ๐ / ๐ฟ
2.2. ii. Rayleighโs Method:
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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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 ๐ฅ = ๐ด ๐ ๐๐ ๐๐ ๐ก
X= displacement of the body from mean position after timeโtโ
A = Maximum. displacement from mean position to extreme
position.
Differentiating w. r. t
๏ฟฝฬ๏ฟฝ = ๐ด๐๐ ๐๐๐ ๐๐ ๐ก
Fig:2.2 (b)
Maximum Velocity at mean position ๏ฟฝฬ๏ฟฝ = ๐๐ ๐ด
Maximum kinetic energy at mean position = 1 2๏ฟฝ (๐๏ฟฝฬ๏ฟฝ2)
= 1 2๏ฟฝ (๐๐๐2๐ด2)
And maximum potential energy at Extreme position
๐.๐ธ = 12๏ฟฝ (๐๐ด2)
W.K.T, K.E = P.E
= 1 2๏ฟฝ (๐๐๐2๐ด2)P
= 1 2๏ฟฝ (๐๐ด2)
๐๐2 = (๐/๐)
๐๐ = ๏ฟฝ(๐/๐)
Hence the natural frequency is
ฦ๐ = 1 /2๐ ๏ฟฝ๐/๐ = 1/ 2๐ ๏ฟฝ๐ / ๐ฟ
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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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) ๐ฟ๐ = 0 For static equilibrium.
๐ฟ๐ = ๐ฟ๐ + ๐ฟ๐inertia 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 ฮด 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.
ฮดW = m. ๏ฟฝฬ๏ฟฝ. ฮดx
The virtual work done by inertia forces is
ฮดW = โkx. ฮดx
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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Equating the two quantities above and canceling ฮด x, we have the equation of motion
m. ๏ฟฝฬ๏ฟฝ + kx = 0
Example: Simple Pendulum
Use the virtual work method; determine the equation of motion for the system below.
๐ฟ๐โฒ = ๐ฟ๐inertia + ๐ฟ๐๐ + ๐ฟ๐๐๐
๐ฟ๐๐๐ = 0: Neoconservative force (external force or damping force)
๐ฟ๐inertia = โ๏ฟฝml๏ฟฝฬ๏ฟฝ๏ฟฝ (lฮดฮธ)
๐ฟ๐๐ = โ(mg sinฮธ) (lฮดฮธ)
๐ฟ๐โฒ = ๐ฟ๐inertia + ๐ฟ๐๐ + ๐ฟ๐๐๐
๐ฟ๐โฒ = โ๏ฟฝml๏ฟฝฬ๏ฟฝ๏ฟฝ (lฮดฮธ) + ๏ฟฝโ(mg sinฮธ) (lฮดฮธ)๏ฟฝ + 0
๐ฟ๐โฒ = โ๏ฟฝml๏ฟฝฬ๏ฟฝ + mg sinฮธ๏ฟฝ(lฮดฮธ)
๐ฟ๐โฒ = 0; ฮดฮธ(t): ๐๐๐๐๐ก๐๐๐๐ฆ
๐ฆ๐ฅ๏ฟฝฬ๏ฟฝ + ๐ฆ๐ ๐ฌ๐ข๐ง๐ = ๐ Nonlinear equation
If ฮธ is small, sinฮธ = ฮธ
ml๏ฟฝฬ๏ฟฝ + mg ฮธ = 0
l๏ฟฝฬ๏ฟฝ + g ฮธ = 0
๏ฟฝฬ๏ฟฝ + ๐ ๐
๐ = ๐ 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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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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.
๐น๐ ๐ผ ๐๐ฅ๐๐ก
๐น๐ = ๐๏ฟฝฬ๏ฟฝ
When โcโ 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 = ยต N
When ยต = co- efficient of friction
N = normal force
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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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 Newtonโs second law of
motion will be
๐แบ = โ ๐๐ฅ โ ๐๐
๐แบ + ๐๐ฅ = โ ๐๐
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 ฯnt + A2sin ฯnt โ ฮผN / k
where in the above expression ฯn = (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.
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
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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
โข https://www.google.co.in/#q=animations+of+single+degree+of+free+vibration
โข www.thirdmill.org/mission/bts.asp
โข acoustics.mie.uic.edu/Simulation/SDOF%20Undamped.htm
โข acoustics.mie.uic.edu/Simulation/SDOF%20Damped.htm
โข www.brown.edu/.../vibrations_free.../vibrations_free_undamped.htm
โข se.asee.org/proceedings/ASEE2009/papers/PR2009011ERV.PDF
โข www.efunda.com/formulae/vibrations/sdof_free_damped.cfm
โข https://dspace.uta.edu/bitstream/.../Deshmukh_uta_2502M_11706.pdf?...
โข www.vibrationdata.com/matlab.htm
โข www.vibrationdata.com/animation.htm
โข www.acs.psu.edu/drussell/demos.html
โข facultad.bayamon.inter.edu/.../Chapter%202%20Free%20vibration%20o...
โข web.itu.edu.tr/~gundes/2dof.pdf
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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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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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 = kฮด.
Where ฮด 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 Newtonโs II law.
Net Force = mass x acceleration
mg โ k (ฮด + x) = m x
m x = mg - kฮด - kx (:- mg = k ฮด)
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 โK/m t +
B cosโK/m t. Where A and B are constant which can be found from initial conditions.
The circular frequency ฯn = โ k/m
The natural frequency of vibration ฦn = ฯn /2
ฦn = 1 /2 ฯ โk /m = 1/2 ฯ โ k / kฮด/g = 1 / 2 ฯ โg / ฮด
Where ฮด = static deflection 2) Explain a) Energy method b) Rayliegh Ritz method of finding Natural Frquency.
Ans)
a) 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. โtโ
๐๐๐ก
(๐ + ๐) = 0
For a spring mass system shown
๐พ.๐ธ = ๐ = ยฝ ๐๏ฟฝฬ๏ฟฝ2
๐.๐ธ = ๐ = ยฝ ๐๐ฅ2 ๐ ๐ ๐
(ยฝ ๐๏ฟฝฬ๏ฟฝ2 + ยฝ ๐๐ฅ2) = 0
๐ ๐ ๐
( ๐๏ฟฝฬ๏ฟฝ2 + ๐๐ฅ2) = 0
๐ x x + ๐๐ฅ x = 0
๏ฟฝฬ๏ฟฝ + (๐/๐) ๐ฅ = 0
Hence the natural frequency is
๐๐ = ๏ฟฝ ๐/๐ ฦ๐ = 1 /2๐ ๏ฟฝ๐/๐ = 1/ 2๐ ๏ฟฝ๐ / ๐ฟ
b) Rayleighโs 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 ๐ฅ = ๐ด ๐ ๐๐ ๐๐ ๐ก
X= displacement of the body from mean position after timeโtโ
A = Maximum. displacement from mean position to extreme
position.
Differentiating w. r. t
๏ฟฝฬ๏ฟฝ = ๐ด๐๐ ๐๐๐ ๐๐ ๐ก
Fig:2.2 (b)
Maximum Velocity at mean position ๏ฟฝฬ๏ฟฝ = ๐๐ ๐ด
Maximum kinetic energy at mean position = 1 2๏ฟฝ (๐๏ฟฝฬ๏ฟฝ2)
= 1 2๏ฟฝ (๐๐๐2๐ด2)
And maximum potential energy at Extreme position
๐.๐ธ = 12๏ฟฝ (๐๐ด2)
W.K.T, K.E = P.E
= 1 2๏ฟฝ (๐๐๐2๐ด2)P
= 1 2๏ฟฝ (๐๐ด2)
๐๐2 = (๐/๐)
๐๐ = ๏ฟฝ(๐/๐)
Hence the natural frequency is
ฦ๐ = 1 /2๐ ๏ฟฝ๐/๐ = 1/ 2๐ ๏ฟฝ๐ / ๐ฟ
3.Eriefly Explain different models of damping.
Ans) 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.
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5) Viscous damping
6) Coulomb damping
7) Structural damping or solid damping
8) 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.
๐น๐ ๐ผ ๐๐ฅ๐๐ก
๐น๐ = ๐๏ฟฝฬ๏ฟฝ
When โcโ 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 = ยต N
When ยต = co- efficient of friction
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)
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 ฮด 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.
ฮดW = m. ๏ฟฝฬ๏ฟฝ. ฮดx
The virtual work done by inertia forces is
ฮดW = โkx. ฮดx
Equating the two quantities above and canceling ฮด x, we have the equation of motion
m. ๏ฟฝฬ๏ฟฝ + kx = 0
Example: Simple Pendulum
Use the virtual work method; determine the equation of motion for the system below.
๐ฟ๐โฒ = ๐ฟ๐inertia + ๐ฟ๐๐ + ๐ฟ๐๐๐
๐ฟ๐๐๐ = 0: Neoconservative force (external force or damping force)
๐ฟ๐inertia = โ๏ฟฝml๏ฟฝฬ๏ฟฝ๏ฟฝ (lฮดฮธ)
๐ฟ๐๐ = โ(mg sinฮธ) (lฮดฮธ)
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๐ฟ๐โฒ = ๐ฟ๐inertia + ๐ฟ๐๐ + ๐ฟ๐๐๐
๐ฟ๐โฒ = โ๏ฟฝml๏ฟฝฬ๏ฟฝ๏ฟฝ (lฮดฮธ) + ๏ฟฝโ(mg sinฮธ) (lฮดฮธ)๏ฟฝ + 0
๐ฟ๐โฒ = โ๏ฟฝml๏ฟฝฬ๏ฟฝ + mg sinฮธ๏ฟฝ(lฮดฮธ)
๐ฟ๐โฒ = 0; ฮดฮธ(t): ๐๐๐๐๐ก๐๐๐๐ฆ
๐ฆ๐ฅ๏ฟฝฬ๏ฟฝ + ๐ฆ๐ ๐ฌ๐ข๐ง๐ = ๐ Nonlinear equation
If ฮธ is small, sinฮธ = ฮธ
ml๏ฟฝฬ๏ฟฝ + mg ฮธ = 0
l๏ฟฝฬ๏ฟฝ + g ฮธ = 0
๏ฟฝฬ๏ฟฝ + ๐ ๐
๐ = ๐ Linear equation
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
= ๐ค๐
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 = -mglฮธ since ฮธ is small sin ฮธ โ ฮธ
According to Newtonโs second law
Accelerating torque = Restoring torque
i.e., ๐ผ๐ ฬ = -mglฮธ
i.e, ๏ฟฝฬ๏ฟฝ+ ๐๐๐๐ผ ฮธ = 0
i.e, ๏ฟฝฬ๏ฟฝ+ ๐๐๐๐(๐2+ ๐2
ฮธ = 0
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โด๏ฟฝฬ๏ฟฝ+ ๐๐(๐2+ ๐2
ฮธ = 0
โด ๐๐= ๏ฟฝ ๐๐๐2+๐2
๐๐๐/๐ ๐๐
Hence natural frequency ๐๐ = 12๐
๐๐ = 12๐๏ฟฝ ๐๐๐2+๐2
๐ป๐ง.
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 ๐ฅ ฬ = 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 โyโ measured from the fixed end. Velocity of the spring wire at the distance y from the fixed end = ๐ฅ ฬ ๏ฟฝ๐ฆ
๐๏ฟฝ
Kinetic energy of the spring element dy = 12
(๐โฒ๐๐ฆ) ๏ฟฝ๏ฟฝฬ๏ฟฝ ๐ฆ๐๏ฟฝ2
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 ๏ฟฝฬ๏ฟฝR0 = ฮฝ0 = 3cm/sec.
Solution:
The differential equation of the motion is given by
๏ฟฝฬ๏ฟฝ + ๐๐๐ฅ = 0
The general solution for the above differential equation is,
x(t) = A cos ฯn t + B sin ฯn t = X cos (ฯn t - ฯ)
When t = 0, x(0) = x0 = A = 2cm
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โด ๏ฟฝฬ๏ฟฝ R(0) = ฮฝ0 = B ฯn; โด B = ๐0๐๐
ฯn = ๏ฟฝ๐๐
= ๏ฟฝ 250.05
= 22.36 ๐๐๐/๐ ๐๐
Maximum amplitude of vibration X = โ๐ด2 + ๐ต2 = ๏ฟฝ๐ฅ02 + ๐02
๐๐2
= ๏ฟฝ22 + 32
22.362 = 2.0045๐๐
i) Natural Frequency ๐๐ = 12๐
๐๐ = 12๐
ร 22.36 = 3.56 ๐ป๐ง
ii) Period of oscillation T = 1๐๐
= 13.56
= 0.28 sec.
iii) Maximum Velocity ๏ฟฝฬ๏ฟฝ Rmax = X ฯn = 2.0045 ร 22.36 = 44.82 cm/sec.
iv) Maximum Acceleration ๏ฟฝฬ๏ฟฝ Rmax = X ๐๐2 = ๏ฟฝฬ๏ฟฝ Rmax .ฯn = 44.82 ร 22.36 = 1002.2 cm/sec2
v) Phase angle ฯ = tan-1 ๏ฟฝ ๐0๐๐๐ฅ0
๏ฟฝ = ๐ก๐๐โ1 ๏ฟฝ 322.36ร2
๏ฟฝ = 3.838๐
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 ๏ฟฝฬ๏ฟฝR0 = 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; ๏ฟฝฬ๏ฟฝR0 = ฮฝ0 = 125 mm/sec. Solution:
The differential equation of the motion is given by
๏ฟฝฬ๏ฟฝ + ๐๐๐ฅ = 0
The general solution for the above differential equation is,
x(t) = A cos ฯn t + B sin ฯn t = X cos (ฯn t - ฯ)
When t = 0, x(0) = x0 = A = 10 mm
โด ๏ฟฝฬ๏ฟฝ R(0) = ฮฝ0 = B ฯn; โด B = ๐0๐๐
Frequency ๐ = 12๐๐๐
i.e, 3.98 = 12๐๐๐
โด ๐๐ = 25 rad/sec
i) Maximum amplitude of vibration X = โ๐ด2 + ๐ต2 = ๏ฟฝ๐ฅ02 + ๐02
๐๐2 = ๏ฟฝ102 + 1252
252 = 11.18 ๐๐.
ii) Period of oscillation T = 1๐๐
= 13.98
= 0.251 ๐ ๐๐.
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iii) Maximum Velocity ๏ฟฝฬ๏ฟฝ Rmax = X ฯn = 11.18 ร 25 = 279.5 mm/sec
iv) Maximum Acceleration ๏ฟฝฬ๏ฟฝ Rmax = X ๐๐2 = ๏ฟฝฬ๏ฟฝ Rmax ฯn = 279.5 ร 25 = 6987.5 mm/sec
v) Phase angle ฯ = tan-1 ๏ฟฝ ๐0๐๐๐ฅ0
๏ฟฝ = ๐ก๐๐โ1 ๏ฟฝ 12525ร10
๏ฟฝ = 26.565๐
vi) Time taken to reach the first peak = ๐๐๐
=26.565ร ๐180
25= 0.018546 ๐ ๐๐.
vii) Lead angle ฮจ = tan-1 (๐๐๐ฅ0๐0
= ๐ก๐๐โ1 ๏ฟฝ25 ร10125
๏ฟฝ = 1.107 ๐๐๐๐๐๐ = 63.435๐
QUADRANT-3
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Wikis
โข en.wikipedia.org/wiki/Vibration
โข wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp...
โข Wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp...
โข apmr.matelys.com/BasicsMechanics/SDOF/index.html
โข www.structuralwiki.org โบ Home โบ Topics
โข petrowiki.org/Basic_vibration_analysis
โข vibrationdata.com/python-wiki/index.php?title=Runge-Kutta_ODE...
โข en.wikipedia.org/wiki/Energy_functional
Open Contents:
โข Mechanical Vibrations, S. S. Rao, Pearson Education Inc, 4th edition, 2003. โข Mechanical Vibrations, V. P. Singh, Dhanpat Rai & Company, 3rd edition, 2006. โข Mechanical Vibrations, G. K.Grover, Nem Chand and Bros, 6th edition, 1996 โข Theory of vibration with applications ,W.T.Thomson,M.D.Dahleh and C
Padmanabhan,Pearson Education inc,5th Edition ,2008 โข 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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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 ๐ป๐ง
Also fn = 12๐๐๐
i.e, 10 = 12๐๐๐
โด ๐๐ = 20๐ ๐๐๐๐๐๐/ sec = 62.832 ๐๐๐๐ ๐๐
.
With the reference to translated equilibrium position for the first part,
x(t) = X cos ฯn t; โด x(0) = 0.5 = X
๏ฟฝฬ๏ฟฝ R(t) = ฮฝ0 = 0; โด 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 โ 0.5
= 0.5[cos 62.832t-1]
โด ๏ฟฝฬ๏ฟฝ = (0.5ร62.832)(-sin 62.832t)
i.e., ๏ฟฝฬ๏ฟฝ62.832
= โ0.5 sin 62.832๐ก
i.e., ๏ฟฝฬ๏ฟฝ๐๐
= โ0.5 sin 62.832๐ก
At the end of first part, t = 0.075 sec
โด x(0.075) = 0.5 [cos (20ฯ ร 0.075) โ 1] = -0.5 cm
๏ฟฝฬ๏ฟฝ0.075๐๐
= -0.5 sin (20ฯ ร 0.075) = 0.5 cm = ๐0.075๐๐
โด For the second part with tโ as time,
x = A cos + ๐ต sin๐๐๐กโฒ
= X cos (๐๐๐กโฒ- ฯ)
X = โ๐ด2 + ๐ต2 = ๏ฟฝ๐ฅ(0.075)2 + ๏ฟฝ๐0.075
๐๐๏ฟฝ = ๏ฟฝ(โ0.5)2 + 0.52 = 0.7071
ฯ = tan-1 ๏ฟฝ๐0.075๐๐
๐ฅ0.075๏ฟฝ = ๐ก๐๐โ1 ๏ฟฝโ0.5
0.5๏ฟฝ = 0.7854 ๐๐๐ = 2.3562 ๐๐๐๐๐๐
โด x = 0.7071 cos (62.832tโ โ 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 + ฯ) with initial condition ๏ฟฝฬ๏ฟฝR(0) = 1250mm/sec and x(0) = 0.25 mm find the
values of X and ฯ, and express the given equation in the form x = A
sin ฯn t + B cos ฯn t.
Data: ๏ฟฝฬ๏ฟฝR0 = ฮฝ0 = 1250 mm/sec; x(0) = x0 = 0.25 mm
Solution:
Given equation x = X cos (100t + ฯ) = X cos (ฯn t + ฯ)
โดฯn = 100rad/sec
Also x(t) = A sin ฯn t + B cos ฯn t
โด x(t) = x0 = 0 + B
โด B = x0 = 0.25 mm
๏ฟฝฬ๏ฟฝ R(t) = A ฯn cos ฯn t - B ฯn sin ฯn t
โด ๏ฟฝฬ๏ฟฝ R(0) = ฮฝ0 = A ฯn
i.e., 1250 = 100 A
โด A = 12.5 mm
Maximum amplitude X = โ๐ด2 + ๐ต2 = ๏ฟฝ๏ฟฝ๐02
๐๐2๏ฟฝ + ๐ฅ02
= โ12.52 + 0.252 = 12.5025 ๐๐.
Now X cos (ฯn t + ฯ) = A sin ฯn t + B cos ฯn t (given)
i.e., X cos ฯn t.cosฯ - X sin ฯn t sinฯ = A sin ฯn t + B cos ฯn t
โดA = -X sin ฯ; B = X cos ฯ
โด tan ฯ = -๐ด๐ต
โด Phase angle ฯ = tan-1 ๏ฟฝโ ๐ด๐ต๏ฟฝ = ๐ก๐๐โ1 ๏ฟฝ
โ๐0๐๐๐ฅ0๏ฟฝ = ๐ก๐๐โ1 ๏ฟฝโ12.5
0.25๏ฟฝ = โ1.55 ๐๐๐๐๐๐
= 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 โyโ from the support as shown. If ฯ is the mass per unit length of the spring in equilibrium condition, then the mass of the spring ms= ฯL and the mass of the element dy is equal to ฯdy. 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 = ยฝ 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 ยฝ m x 2
The K.E of the element dy of the spring is equal to 1/2 (ฯdy) (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 + โซ 0 ยฝ (ฯ dy) (y/L x )2 = 1/2 m x 2 + ยฝ ฯ x 2 / L2 โซ0y2dy = 1/2 m x 2 + ยฝ ฯ x 2 / L2 [y3 / 3] L 0 = 1/2 m x 2 + ยฝ ฯ x 2 / L2 [L3/3] = 1/2 m x 2 + ยฝ ฯ L /3 x 2
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2014
= 1/2 m x 2 + ยฝ 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 ยฝ k (2x) ( x ) + ยฝ (2 x ) ( x ) [m + ms /3] = 0 kx + (m + ms/3) x = 0 Or (m + ms /3) x + kx = 0 f n = 1/2ฯ โ k / (m + ms /3) ฯn = โk / (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 โ third its mass to the main mass.
4) Explain a) Energy method b) Rayliegh Ritz method of finding Natural Frquency.
Ans)
c) 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
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. โtโ
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๐๐๐ก
(๐ + ๐) = 0
For a spring mass system shown
๐พ.๐ธ = ๐ = ยฝ ๐๏ฟฝฬ๏ฟฝ2
๐.๐ธ = ๐ = ยฝ ๐๐ฅ2 ๐ ๐ ๐
(ยฝ ๐๏ฟฝฬ๏ฟฝ2 + ยฝ ๐๐ฅ2) = 0
๐ ๐ ๐
( ๐๏ฟฝฬ๏ฟฝ2 + ๐๐ฅ2) = 0
๐ x x + ๐๐ฅ x = 0
๏ฟฝฬ๏ฟฝ + (๐/๐) ๐ฅ = 0
Hence the natural frequency is
๐๐ = ๏ฟฝ ๐/๐ ฦ๐ = 1 /2๐ ๏ฟฝ๐/๐ = 1/ 2๐ ๏ฟฝ๐ / ๐ฟ
d) Rayleighโs Method:
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 ๐ฅ = ๐ด ๐ ๐๐ ๐๐ ๐ก
X= displacement of the body from mean position after timeโtโ
A = Maximum. displacement from mean position to extreme
position.
Differentiating w. r. t
๏ฟฝฬ๏ฟฝ = ๐ด๐๐ ๐๐๐ ๐๐ ๐ก
Fig:2.2 (b)
Maximum Velocity at mean position ๏ฟฝฬ๏ฟฝ = ๐๐ ๐ด
Maximum kinetic energy at mean position = 1 2๏ฟฝ (๐๏ฟฝฬ๏ฟฝ2)
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= 1 2๏ฟฝ (๐๐๐2๐ด2)
And maximum potential energy at Extreme position
๐.๐ธ = 12๏ฟฝ (๐๐ด2)
W.K.T, K.E = P.E
= 1 2๏ฟฝ (๐๐๐2๐ด2)P
= 1 2๏ฟฝ (๐๐ด2)
๐๐2 = (๐/๐)
๐๐ = ๏ฟฝ(๐/๐)
Hence the natural frequency is
ฦ๐ = 1 /2๐ ๏ฟฝ๐/๐ = 1/ 2๐ ๏ฟฝ๐ / ๐ฟ
1. Eriefly Explain different models of damping.
Ans) 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.
9) Viscous damping
10) Coulomb damping
11) Structural damping or solid damping
12) 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.
๐น๐ ๐ผ ๐๐ฅ๐๐ก
๐น๐ = ๐๏ฟฝฬ๏ฟฝ
When โcโ is the constant of proportionality and is called viscous damping Co-efficient with the
dimension of N-s/m.
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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 = ยต N
When ยต = co- efficient of friction
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)
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 ฮด 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.
ฮดW = m. ๏ฟฝฬ๏ฟฝ. ฮดx
The virtual work done by inertia forces is
ฮดW = โkx. ฮดx
Equating the two quantities above and canceling ฮด x, we have the equation of motion
m. ๏ฟฝฬ๏ฟฝ + kx = 0
Example: Simple Pendulum
Use the virtual work method; determine the equation of motion for the system below.
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๐ฟ๐โฒ = ๐ฟ๐inertia + ๐ฟ๐๐ + ๐ฟ๐๐๐
๐ฟ๐๐๐ = 0: Neoconservative force (external force or damping force)
๐ฟ๐inertia = โ๏ฟฝml๏ฟฝฬ๏ฟฝ๏ฟฝ (lฮดฮธ)
๐ฟ๐๐ = โ(mg sinฮธ) (lฮดฮธ)
๐ฟ๐โฒ = ๐ฟ๐inertia + ๐ฟ๐๐ + ๐ฟ๐๐๐
๐ฟ๐โฒ = โ๏ฟฝml๏ฟฝฬ๏ฟฝ๏ฟฝ (lฮดฮธ) + ๏ฟฝโ(mg sinฮธ) (lฮดฮธ)๏ฟฝ + 0
๐ฟ๐โฒ = โ๏ฟฝml๏ฟฝฬ๏ฟฝ + mg sinฮธ๏ฟฝ(lฮดฮธ)
๐ฟ๐โฒ = 0; ฮดฮธ(t): ๐๐๐๐๐ก๐๐๐๐ฆ
๐ฆ๐ฅ๏ฟฝฬ๏ฟฝ + ๐ฆ๐ ๐ฌ๐ข๐ง๐ = ๐ Nonlinear equation
If ฮธ is small, sinฮธ = ฮธ
ml๏ฟฝฬ๏ฟฝ + mg ฮธ = 0
l๏ฟฝฬ๏ฟฝ + g ฮธ = 0
๏ฟฝฬ๏ฟฝ + ๐ ๐
๐ = ๐ Linear equation
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
= ๐ค๐
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 = -mglฮธ since ฮธ is small sin ฮธ โ ฮธ
According to Newtonโs second law
Accelerating torque = Restoring torque
i.e., ๐ผ๐ ฬ = -mglฮธ
i.e, ๏ฟฝฬ๏ฟฝ+ ๐๐๐๐ผ ฮธ = 0
i.e, ๏ฟฝฬ๏ฟฝ+ ๐๐๐๐(๐2+ ๐2
ฮธ = 0
โด๏ฟฝฬ๏ฟฝ+ ๐๐(๐2+ ๐2
ฮธ = 0
โด ๐๐= ๏ฟฝ ๐๐๐2+๐2
๐๐๐/๐ ๐๐
Hence natural frequency ๐๐ = 12๐
๐๐ = 12๐๏ฟฝ ๐๐๐2+๐2
๐ป๐ง.
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 ๐ฅ ฬ = 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 โyโ measured from the fixed end. Velocity of the spring wire at the distance y from the fixed end = ๐ฅ ฬ ๏ฟฝ๐ฆ
๐๏ฟฝ
Kinetic energy of the spring element dy = 12
(๐โฒ๐๐ฆ) ๏ฟฝ๏ฟฝฬ๏ฟฝ ๐ฆ๐๏ฟฝ2
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 ๏ฟฝฬ๏ฟฝR0 = ฮฝ0 = 3cm/sec.
Solution:
The differential equation of the motion is given by
๏ฟฝฬ๏ฟฝ + ๐๐๐ฅ = 0
The general solution for the above differential equation is,
x(t) = A cos ฯn t + B sin ฯn t = X cos (ฯn t - ฯ)
When t = 0, x(0) = x0 = A = 2cm
โด ๏ฟฝฬ๏ฟฝ R(0) = ฮฝ0 = B ฯn; โด B = ๐0๐๐
ฯn = ๏ฟฝ๐๐
= ๏ฟฝ 250.05
= 22.36 ๐๐๐/๐ ๐๐
Maximum amplitude of vibration X = โ๐ด2 + ๐ต2 = ๏ฟฝ๐ฅ02 + ๐02
๐๐2
= ๏ฟฝ22 + 32
22.362 = 2.0045๐๐
vi) Natural Frequency ๐๐ = 12๐
๐๐ = 12๐
ร 22.36 = 3.56 ๐ป๐ง
vii) Period of oscillation T = 1๐๐
= 13.56
= 0.28 sec.
viii) Maximum Velocity ๏ฟฝฬ๏ฟฝ Rmax = X ฯn = 2.0045 ร 22.36 = 44.82 cm/sec.
ix) Maximum Acceleration ๏ฟฝฬ๏ฟฝ Rmax = X ๐๐2 = ๏ฟฝฬ๏ฟฝ Rmax .ฯn = 44.82 ร 22.36 = 1002.2 cm/sec2
x) Phase angle ฯ = tan-1 ๏ฟฝ ๐0๐๐๐ฅ0
๏ฟฝ = ๐ก๐๐โ1 ๏ฟฝ 322.36ร2
๏ฟฝ = 3.838๐
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.
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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MODULE-II --- SINGLE DOF FREE VIBRATIONS VIBRATION ENGINEERING
2014
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.
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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MODULE-II --- SINGLE DOF FREE VIBRATIONS VIBRATION ENGINEERING
2014
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.
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
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MODULE-II --- SINGLE DOF FREE VIBRATIONS VIBRATION ENGINEERING
2014
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 Newtonโs method.
12. Derive the equation of motion of a simple pendulum having an angular displacement of
๐.
13. Show that the frequency of undamped free vibration of a spring mass system is given
by ๐น๐ = 1/ 2๐ ๏ฟฝ๐ ๐ฟ
.
14. Show that the natural frequency of undamped free vibration of a spring mass system is
given by ๐๐ = 1/ 2๐ ๏ฟฝ๐ ๐ฝ
.
15. Using the energy method derive the differential equation of motion of an undamped free
vibration and show that frequency ๐๐ = ๏ฟฝ๐ ๐
16. Using the Rayleigh method derive the differential equation of motion of an undamped
free vibration and show that frequency ๐๐ = ๏ฟฝ๐ ๐
17. Derive the natural frequency of torsional vibrations.
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