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
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2014
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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2014
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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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
ππ = kΞ΄ [πΉπππ πππ: 2.1 (π)]
Where Ξ΄ 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 Newtonβs II law.
πππ‘ πΉππππ = πππ π Γ πππππππππ‘πππ
ππ β π (πΏ + π₯) = π οΏ½ΜοΏ½ [equation of motion ]
π οΏ½ΜοΏ½ = ππ β ππΏ β ππ₯ (:β mg = k Ξ΄)
π οΏ½ΜοΏ½ = βππ₯
π οΏ½ΜοΏ½ + ππ₯ = 0
π₯ + ( π /π) οΏ½ΜοΏ½ = 0 βββββ (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
π₯ = π΄ π ππ οΏ½πΎ/π π‘ + π΅ πππ οΏ½πΎ/π π‘. Where A & B are constant which can be found from
initial conditions.
The circular frequency ππ = οΏ½ π/π
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2014
The natural frequency of vibration Ζπ = ππ2ποΏ½ R
Ζπ = 1
2 π οΏ½π /π =
12 π
οΏ½ π (ππΏ/π)οΏ½ = 1
2 π οΏ½
ππΏ
Where Ξ΄ = 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
π οΏ½ΜοΏ½ = β ππ₯
π οΏ½ΜοΏ½ + ππ₯ = 0 [equation of motion ]
οΏ½ΜοΏ½ + (π /π) π₯ = 0 ββββ(1)
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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
π₯ = π΄ π ππ οΏ½οΏ½πΎ/π οΏ½π‘ + π΅ πππ οΏ½οΏ½πΎ/ποΏ½ π‘. Where A & B are constant which can be found from
initial conditions.
The circular frequency ππ = οΏ½ π/π
The natural frequency of vibration, Ζπ = ππ2ποΏ½ R
Ζπ = 1
2 π οΏ½π /π =
12 π
οΏ½ π (ππΏ/π)οΏ½ = ππ π
οΏ½ ππΉ
Where Ξ΄ = 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 ΞΈ as
shown in fig: 2.2(c).
The body is rotated through an angle ΞΈ
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. πΌ οΏ½ΜοΏ½ = β πΎπ‘ π
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πΌ οΏ½ΜοΏ½ + πΎπ‘ π = 0
οΏ½ΜοΏ½ + οΏ½ πΎπ‘πΌοΏ½ π = 0
c) NATURAL FREQUENCY:
Equation (1) is a differential equation. The solution of which is
π₯ = π΄ π ππ οΏ½οΏ½ πΎπ‘πΌ
οΏ½ π‘ + π΅ πππ οΏ½οΏ½ πΎπ‘πΌοΏ½ π‘. Where A & B are constant which can be found from
initial conditions.
The circular frequency ππ = οΏ½ πΎπ‘πΌ
The natural frequency of vibration, Ζπ = ππ2ποΏ½
ππ = οΏ½ πΎπ‘πΌ
& ππ =1
2ποΏ½ πΎπ‘πΌ
Where πΎπ‘ = πΊπ½πΏ
; π½ = π π3
32
SPRINGS IN ARBITRARY DIRECTION
Fig shows a spring K making an angle Ξ± with the direction of motion of the mass m.
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2014
If the mass is displaced by x, the spring is deformed by an amount π₯πππ πΌ along its axis
(spring Axis). The force along the spring axis is ππ₯ πππ πΌ. The component of this force along the
direction of motion of the mass is ππ₯ πππ 2 πΌ. The equation of motion of the mass m is
ππ₯ + (ππππ 2 πΌ) οΏ½ΜοΏ½ = 0.
From the above equation it may be noted that the equivalent stiffness πΎπ of a spring making
angle-Ξ± with the axis of motion is πΎπ = πΎ πππ 2 πΌ.
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.
π. π πΎ1π + πΎ2π = πΎππ
πΎ1 + πΎ2 = πΎπ
The equivalent spring shiftness is equal to
the sum of individual spring shiftiness.
SPRINGS IN SERIES
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2014
The total deflection of the system is equal
to the sum of deflection of individual springs.
X = X1 + X2
ππππππΎπ
=ππππππΎ1
=ππππππΎ2
1πΎπ
=1πΎ1
=1πΎ2
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 Newtonβs method. The other methods which are commonly used
in vibration for determination of frequency are, (i) Energy Method (ii) Rayleighβs 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. βtβ
πππ‘
(π + π) = 0
For a spring mass system shown
πΎ.πΈ = π = Β½ ποΏ½ΜοΏ½2
π.πΈ = π = Β½ ππ₯2 π π π
(Β½ ποΏ½ΜοΏ½2 + Β½ ππ₯2) = 0
π π π
( ποΏ½ΜοΏ½2 + ππ₯2) = 0
π x x + ππ₯ x = 0
οΏ½ΜοΏ½ + (π/π) π₯ = 0
Hence the natural frequency is
ππ = οΏ½ π/π Ζπ = 1 /2π οΏ½π/π = 1/ 2π οΏ½π / πΏ
2.2. ii. Rayleighβs 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 π₯ = π΄ π ππ ππ π‘
X= displacement of the body from mean position after timeβtβ
A = Maximum. displacement from mean position to extreme
position.
Differentiating w. r. t
οΏ½ΜοΏ½ = π΄ππ πππ ππ π‘
Fig:2.2 (b)
Maximum Velocity at mean position οΏ½ΜοΏ½ = ππ π΄
Maximum kinetic energy at mean position = 1 2οΏ½ (ποΏ½ΜοΏ½2)
= 1 2οΏ½ (πππ2π΄2)
And maximum potential energy at Extreme position
π.πΈ = 12οΏ½ (ππ΄2)
W.K.T, K.E = P.E
= 1 2οΏ½ (πππ2π΄2)P
= 1 2οΏ½ (ππ΄2)
ππ2 = (π/π)
ππ = οΏ½(π/π)
Hence the natural frequency is
Ζπ = 1 /2π οΏ½π/π = 1/ 2π οΏ½π / πΏ
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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) πΏπ = 0 For static equilibrium.
πΏπ = πΏπ + πΏπinertia 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 Ξ΄ 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.
Ξ΄W = m. οΏ½ΜοΏ½. Ξ΄x
The virtual work done by inertia forces is
Ξ΄W = βkx. Ξ΄x
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Equating the two quantities above and canceling Ξ΄ x, we have the equation of motion
m. οΏ½ΜοΏ½ + kx = 0
Example: Simple Pendulum
Use the virtual work method; determine the equation of motion for the system below.
πΏπβ² = πΏπinertia + πΏππ + πΏπππ
πΏπππ = 0: Neoconservative force (external force or damping force)
πΏπinertia = βοΏ½mlοΏ½ΜοΏ½οΏ½ (lδθ)
πΏππ = β(mg sinΞΈ) (lδθ)
πΏπβ² = πΏπinertia + πΏππ + πΏπππ
πΏπβ² = βοΏ½mlοΏ½ΜοΏ½οΏ½ (lδθ) + οΏ½β(mg sinΞΈ) (lδθ)οΏ½ + 0
πΏπβ² = βοΏ½mlοΏ½ΜοΏ½ + mg sinΞΈοΏ½(lδθ)
πΏπβ² = 0; δθ(t): πππππ‘ππππ¦
π¦π₯οΏ½ΜοΏ½ + π¦π π¬π’π§π = π Nonlinear equation
If ΞΈ is small, sinΞΈ = ΞΈ
mlοΏ½ΜοΏ½ + mg ΞΈ = 0
lοΏ½ΜοΏ½ + g ΞΈ = 0
οΏ½ΜοΏ½ + π π
π = π 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.
πΉπ πΌ ππ₯ππ‘
πΉπ = ποΏ½ΜοΏ½
When βcβ 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 = Β΅ N
When Β΅ = 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 Newtonβs second law of
motion will be
παΊ = β ππ₯ β ππ
παΊ + ππ₯ = β ππ
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 Οnt + A2sin Οnt β ΞΌN / k
where in the above expression Οn = (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
β’ https://www.google.co.in/#q=animations+of+single+degree+of+free+vibration
β’ www.thirdmill.org/mission/bts.asp
β’ acoustics.mie.uic.edu/Simulation/SDOF%20Undamped.htm
β’ acoustics.mie.uic.edu/Simulation/SDOF%20Damped.htm
β’ www.brown.edu/.../vibrations_free.../vibrations_free_undamped.htm
β’ se.asee.org/proceedings/ASEE2009/papers/PR2009011ERV.PDF
β’ www.efunda.com/formulae/vibrations/sdof_free_damped.cfm
β’ https://dspace.uta.edu/bitstream/.../Deshmukh_uta_2502M_11706.pdf?...
β’ www.vibrationdata.com/matlab.htm
β’ www.vibrationdata.com/animation.htm
β’ www.acs.psu.edu/drussell/demos.html
β’ facultad.bayamon.inter.edu/.../Chapter%202%20Free%20vibration%20o...
β’ web.itu.edu.tr/~gundes/2dof.pdf
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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 = kΞ΄.
Where Ξ΄ 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 Newtonβs II law.
Net Force = mass x acceleration
mg β k (Ξ΄ + x) = m x
m x = mg - kΞ΄ - kx (:- mg = k Ξ΄)
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 βK/m t +
B cosβK/m t. Where A and B are constant which can be found from initial conditions.
The circular frequency Οn = β k/m
The natural frequency of vibration Ζn = Οn /2
Ζn = 1 /2 Ο βk /m = 1/2 Ο β k / kΞ΄/g = 1 / 2 Ο βg / Ξ΄
Where Ξ΄ = 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. βtβ
πππ‘
(π + π) = 0
For a spring mass system shown
πΎ.πΈ = π = Β½ ποΏ½ΜοΏ½2
π.πΈ = π = Β½ ππ₯2 π π π
(Β½ ποΏ½ΜοΏ½2 + Β½ ππ₯2) = 0
π π π
( ποΏ½ΜοΏ½2 + ππ₯2) = 0
π x x + ππ₯ x = 0
οΏ½ΜοΏ½ + (π/π) π₯ = 0
Hence the natural frequency is
ππ = οΏ½ π/π Ζπ = 1 /2π οΏ½π/π = 1/ 2π οΏ½π / πΏ
b) Rayleighβs 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 π₯ = π΄ π ππ ππ π‘
X= displacement of the body from mean position after timeβtβ
A = Maximum. displacement from mean position to extreme
position.
Differentiating w. r. t
οΏ½ΜοΏ½ = π΄ππ πππ ππ π‘
Fig:2.2 (b)
Maximum Velocity at mean position οΏ½ΜοΏ½ = ππ π΄
Maximum kinetic energy at mean position = 1 2οΏ½ (ποΏ½ΜοΏ½2)
= 1 2οΏ½ (πππ2π΄2)
And maximum potential energy at Extreme position
π.πΈ = 12οΏ½ (ππ΄2)
W.K.T, K.E = P.E
= 1 2οΏ½ (πππ2π΄2)P
= 1 2οΏ½ (ππ΄2)
ππ2 = (π/π)
ππ = οΏ½(π/π)
Hence the natural frequency is
Ζπ = 1 /2π οΏ½π/π = 1/ 2π οΏ½π / πΏ
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.
πΉπ πΌ ππ₯ππ‘
πΉπ = ποΏ½ΜοΏ½
When βcβ 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 = Β΅ N
When Β΅ = 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 Ξ΄ 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.
Ξ΄W = m. οΏ½ΜοΏ½. Ξ΄x
The virtual work done by inertia forces is
Ξ΄W = βkx. Ξ΄x
Equating the two quantities above and canceling Ξ΄ x, we have the equation of motion
m. οΏ½ΜοΏ½ + kx = 0
Example: Simple Pendulum
Use the virtual work method; determine the equation of motion for the system below.
πΏπβ² = πΏπinertia + πΏππ + πΏπππ
πΏπππ = 0: Neoconservative force (external force or damping force)
πΏπinertia = βοΏ½mlοΏ½ΜοΏ½οΏ½ (lδθ)
πΏππ = β(mg sinΞΈ) (lδθ)
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πΏπβ² = πΏπinertia + πΏππ + πΏπππ
πΏπβ² = βοΏ½mlοΏ½ΜοΏ½οΏ½ (lδθ) + οΏ½β(mg sinΞΈ) (lδθ)οΏ½ + 0
πΏπβ² = βοΏ½mlοΏ½ΜοΏ½ + mg sinΞΈοΏ½(lδθ)
πΏπβ² = 0; δθ(t): πππππ‘ππππ¦
π¦π₯οΏ½ΜοΏ½ + π¦π π¬π’π§π = π Nonlinear equation
If ΞΈ is small, sinΞΈ = ΞΈ
mlοΏ½ΜοΏ½ + mg ΞΈ = 0
lοΏ½ΜοΏ½ + g ΞΈ = 0
οΏ½ΜοΏ½ + π π
π = π 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
= π€π
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 = -mglΞΈ since ΞΈ is small sin ΞΈ β ΞΈ
According to Newtonβs second law
Accelerating torque = Restoring torque
i.e., πΌπ Μ = -mglΞΈ
i.e, οΏ½ΜοΏ½+ ππππΌ ΞΈ = 0
i.e, οΏ½ΜοΏ½+ ππππ(π2+ π2
ΞΈ = 0
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β΄οΏ½ΜοΏ½+ ππ(π2+ π2
ΞΈ = 0
β΄ ππ= οΏ½ πππ2+π2
πππ/π ππ
Hence natural frequency ππ = 12π
ππ = 12ποΏ½ πππ2+π2
π»π§.
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 π₯ Μ = 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 βyβ measured from the fixed end. Velocity of the spring wire at the distance y from the fixed end = π₯ Μ οΏ½π¦
ποΏ½
Kinetic energy of the spring element dy = 12
(πβ²ππ¦) οΏ½οΏ½ΜοΏ½ π¦ποΏ½2
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 οΏ½ΜοΏ½R0 = Ξ½0 = 3cm/sec.
Solution:
The differential equation of the motion is given by
οΏ½ΜοΏ½ + πππ₯ = 0
The general solution for the above differential equation is,
x(t) = A cos Οn t + B sin Οn t = X cos (Οn t - Ο)
When t = 0, x(0) = x0 = A = 2cm
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β΄ οΏ½ΜοΏ½ R(0) = Ξ½0 = B Οn; β΄ B = π0ππ
Οn = οΏ½ππ
= οΏ½ 250.05
= 22.36 πππ/π ππ
Maximum amplitude of vibration X = βπ΄2 + π΅2 = οΏ½π₯02 + π02
ππ2
= οΏ½22 + 32
22.362 = 2.0045ππ
i) Natural Frequency ππ = 12π
ππ = 12π
Γ 22.36 = 3.56 π»π§
ii) Period of oscillation T = 1ππ
= 13.56
= 0.28 sec.
iii) Maximum Velocity οΏ½ΜοΏ½ Rmax = X Οn = 2.0045 Γ 22.36 = 44.82 cm/sec.
iv) Maximum Acceleration οΏ½ΜοΏ½ Rmax = X ππ2 = οΏ½ΜοΏ½ Rmax .Οn = 44.82 Γ 22.36 = 1002.2 cm/sec2
v) Phase angle Ο = tan-1 οΏ½ π0πππ₯0
οΏ½ = π‘ππβ1 οΏ½ 322.36Γ2
οΏ½ = 3.838π
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 οΏ½ΜοΏ½R0 = 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; οΏ½ΜοΏ½R0 = Ξ½0 = 125 mm/sec. Solution:
The differential equation of the motion is given by
οΏ½ΜοΏ½ + πππ₯ = 0
The general solution for the above differential equation is,
x(t) = A cos Οn t + B sin Οn t = X cos (Οn t - Ο)
When t = 0, x(0) = x0 = A = 10 mm
β΄ οΏ½ΜοΏ½ R(0) = Ξ½0 = B Οn; β΄ B = π0ππ
Frequency π = 12πππ
i.e, 3.98 = 12πππ
β΄ ππ = 25 rad/sec
i) Maximum amplitude of vibration X = βπ΄2 + π΅2 = οΏ½π₯02 + π02
ππ2 = οΏ½102 + 1252
252 = 11.18 ππ.
ii) Period of oscillation T = 1ππ
= 13.98
= 0.251 π ππ.
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iii) Maximum Velocity οΏ½ΜοΏ½ Rmax = X Οn = 11.18 Γ 25 = 279.5 mm/sec
iv) Maximum Acceleration οΏ½ΜοΏ½ Rmax = X ππ2 = οΏ½ΜοΏ½ Rmax Οn = 279.5 Γ 25 = 6987.5 mm/sec
v) Phase angle Ο = tan-1 οΏ½ π0πππ₯0
οΏ½ = π‘ππβ1 οΏ½ 12525Γ10
οΏ½ = 26.565π
vi) Time taken to reach the first peak = πππ
=26.565Γ π180
25= 0.018546 π ππ.
vii) Lead angle Ξ¨ = tan-1 (πππ₯0π0
= π‘ππβ1 οΏ½25 Γ10125
οΏ½ = 1.107 ππππππ = 63.435π
QUADRANT-3
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Wikis
β’ en.wikipedia.org/wiki/Vibration
β’ wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp...
β’ Wikis.controltheorypro.com/Single_Degree_of_Freedom,_Free_Undamp...
β’ apmr.matelys.com/BasicsMechanics/SDOF/index.html
β’ www.structuralwiki.org βΊ Home βΊ Topics
β’ petrowiki.org/Basic_vibration_analysis
β’ vibrationdata.com/python-wiki/index.php?title=Runge-Kutta_ODE...
β’ en.wikipedia.org/wiki/Energy_functional
Open Contents:
β’ Mechanical Vibrations, S. S. Rao, Pearson Education Inc, 4th edition, 2003. β’ Mechanical Vibrations, V. P. Singh, Dhanpat Rai & Company, 3rd edition, 2006. β’ Mechanical Vibrations, G. K.Grover, Nem Chand and Bros, 6th edition, 1996 β’ Theory of vibration with applications ,W.T.Thomson,M.D.Dahleh and C
Padmanabhan,Pearson Education inc,5th Edition ,2008 β’ 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 π»π§
Also fn = 12πππ
i.e, 10 = 12πππ
β΄ ππ = 20π ππππππ/ sec = 62.832 ππππ ππ
.
With the reference to translated equilibrium position for the first part,
x(t) = X cos Οn t; β΄ x(0) = 0.5 = X
οΏ½ΜοΏ½ R(t) = Ξ½0 = 0; β΄ 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 β 0.5
= 0.5[cos 62.832t-1]
β΄ οΏ½ΜοΏ½ = (0.5Γ62.832)(-sin 62.832t)
i.e., οΏ½ΜοΏ½62.832
= β0.5 sin 62.832π‘
i.e., οΏ½ΜοΏ½ππ
= β0.5 sin 62.832π‘
At the end of first part, t = 0.075 sec
β΄ x(0.075) = 0.5 [cos (20Ο Γ 0.075) β 1] = -0.5 cm
οΏ½ΜοΏ½0.075ππ
= -0.5 sin (20Ο Γ 0.075) = 0.5 cm = π0.075ππ
β΄ For the second part with tβ as time,
x = A cos + π΅ sinπππ‘β²
= X cos (πππ‘β²- Ο)
X = βπ΄2 + π΅2 = οΏ½π₯(0.075)2 + οΏ½π0.075
πποΏ½ = οΏ½(β0.5)2 + 0.52 = 0.7071
Ο = tan-1 οΏ½π0.075ππ
π₯0.075οΏ½ = π‘ππβ1 οΏ½β0.5
0.5οΏ½ = 0.7854 πππ = 2.3562 ππππππ
β΄ x = 0.7071 cos (62.832tβ β 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 + Ο) with initial condition οΏ½ΜοΏ½R(0) = 1250mm/sec and x(0) = 0.25 mm find the
values of X and Ο, and express the given equation in the form x = A
sin Οn t + B cos Οn t.
Data: οΏ½ΜοΏ½R0 = Ξ½0 = 1250 mm/sec; x(0) = x0 = 0.25 mm
Solution:
Given equation x = X cos (100t + Ο) = X cos (Οn t + Ο)
β΄Οn = 100rad/sec
Also x(t) = A sin Οn t + B cos Οn t
β΄ x(t) = x0 = 0 + B
β΄ B = x0 = 0.25 mm
οΏ½ΜοΏ½ R(t) = A Οn cos Οn t - B Οn sin Οn t
β΄ οΏ½ΜοΏ½ R(0) = Ξ½0 = A Οn
i.e., 1250 = 100 A
β΄ A = 12.5 mm
Maximum amplitude X = βπ΄2 + π΅2 = οΏ½οΏ½π02
ππ2οΏ½ + π₯02
= β12.52 + 0.252 = 12.5025 ππ.
Now X cos (Οn t + Ο) = A sin Οn t + B cos Οn t (given)
i.e., X cos Οn t.cosΟ - X sin Οn t sinΟ = A sin Οn t + B cos Οn t
β΄A = -X sin Ο; B = X cos Ο
β΄ tan Ο = -π΄π΅
β΄ Phase angle Ο = tan-1 οΏ½β π΄π΅οΏ½ = π‘ππβ1 οΏ½
βπ0πππ₯0οΏ½ = π‘ππβ1 οΏ½β12.5
0.25οΏ½ = β1.55 ππππππ
= 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 βyβ from the support as shown. If Ο is the mass per unit length of the spring in equilibrium condition, then the mass of the spring ms= ΟL and the mass of the element dy is equal to Οdy. 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 = Β½ 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 Β½ m x 2
The K.E of the element dy of the spring is equal to 1/2 (Οdy) (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 + β« 0 Β½ (Ο dy) (y/L x )2 = 1/2 m x 2 + Β½ Ο x 2 / L2 β«0y2dy = 1/2 m x 2 + Β½ Ο x 2 / L2 [y3 / 3] L 0 = 1/2 m x 2 + Β½ Ο x 2 / L2 [L3/3] = 1/2 m x 2 + Β½ Ο L /3 x 2
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= 1/2 m x 2 + Β½ 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 Β½ k (2x) ( x ) + Β½ (2 x ) ( x ) [m + ms /3] = 0 kx + (m + ms/3) x = 0 Or (m + ms /3) x + kx = 0 f n = 1/2Ο β k / (m + ms /3) Οn = βk / (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 β 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. βtβ
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πππ‘
(π + π) = 0
For a spring mass system shown
πΎ.πΈ = π = Β½ ποΏ½ΜοΏ½2
π.πΈ = π = Β½ ππ₯2 π π π
(Β½ ποΏ½ΜοΏ½2 + Β½ ππ₯2) = 0
π π π
( ποΏ½ΜοΏ½2 + ππ₯2) = 0
π x x + ππ₯ x = 0
οΏ½ΜοΏ½ + (π/π) π₯ = 0
Hence the natural frequency is
ππ = οΏ½ π/π Ζπ = 1 /2π οΏ½π/π = 1/ 2π οΏ½π / πΏ
d) Rayleighβs 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 π₯ = π΄ π ππ ππ π‘
X= displacement of the body from mean position after timeβtβ
A = Maximum. displacement from mean position to extreme
position.
Differentiating w. r. t
οΏ½ΜοΏ½ = π΄ππ πππ ππ π‘
Fig:2.2 (b)
Maximum Velocity at mean position οΏ½ΜοΏ½ = ππ π΄
Maximum kinetic energy at mean position = 1 2οΏ½ (ποΏ½ΜοΏ½2)
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= 1 2οΏ½ (πππ2π΄2)
And maximum potential energy at Extreme position
π.πΈ = 12οΏ½ (ππ΄2)
W.K.T, K.E = P.E
= 1 2οΏ½ (πππ2π΄2)P
= 1 2οΏ½ (ππ΄2)
ππ2 = (π/π)
ππ = οΏ½(π/π)
Hence the natural frequency is
Ζπ = 1 /2π οΏ½π/π = 1/ 2π οΏ½π / πΏ
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.
πΉπ πΌ ππ₯ππ‘
πΉπ = ποΏ½ΜοΏ½
When βcβ 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 = Β΅ N
When Β΅ = 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 Ξ΄ 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.
Ξ΄W = m. οΏ½ΜοΏ½. Ξ΄x
The virtual work done by inertia forces is
Ξ΄W = βkx. Ξ΄x
Equating the two quantities above and canceling Ξ΄ x, we have the equation of motion
m. οΏ½ΜοΏ½ + kx = 0
Example: Simple Pendulum
Use the virtual work method; determine the equation of motion for the system below.
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πΏπβ² = πΏπinertia + πΏππ + πΏπππ
πΏπππ = 0: Neoconservative force (external force or damping force)
πΏπinertia = βοΏ½mlοΏ½ΜοΏ½οΏ½ (lδθ)
πΏππ = β(mg sinΞΈ) (lδθ)
πΏπβ² = πΏπinertia + πΏππ + πΏπππ
πΏπβ² = βοΏ½mlοΏ½ΜοΏ½οΏ½ (lδθ) + οΏ½β(mg sinΞΈ) (lδθ)οΏ½ + 0
πΏπβ² = βοΏ½mlοΏ½ΜοΏ½ + mg sinΞΈοΏ½(lδθ)
πΏπβ² = 0; δθ(t): πππππ‘ππππ¦
π¦π₯οΏ½ΜοΏ½ + π¦π π¬π’π§π = π Nonlinear equation
If ΞΈ is small, sinΞΈ = ΞΈ
mlοΏ½ΜοΏ½ + mg ΞΈ = 0
lοΏ½ΜοΏ½ + g ΞΈ = 0
οΏ½ΜοΏ½ + π π
π = π 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
= π€π
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
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MODULE-II --- SINGLE DOF FREE VIBRATIONS VIBRATION ENGINEERING
2014
= mk2 + ml2 = m(k2 + l2)
k = Radius of gyration of the body
If OG is displaced by an angle,
Restoring torque = -mglΞΈ since ΞΈ is small sin ΞΈ β ΞΈ
According to Newtonβs second law
Accelerating torque = Restoring torque
i.e., πΌπ Μ = -mglΞΈ
i.e, οΏ½ΜοΏ½+ ππππΌ ΞΈ = 0
i.e, οΏ½ΜοΏ½+ ππππ(π2+ π2
ΞΈ = 0
β΄οΏ½ΜοΏ½+ ππ(π2+ π2
ΞΈ = 0
β΄ ππ= οΏ½ πππ2+π2
πππ/π ππ
Hence natural frequency ππ = 12π
ππ = 12ποΏ½ πππ2+π2
π»π§.
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 π₯ Μ = 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 βyβ measured from the fixed end. Velocity of the spring wire at the distance y from the fixed end = π₯ Μ οΏ½π¦
ποΏ½
Kinetic energy of the spring element dy = 12
(πβ²ππ¦) οΏ½οΏ½ΜοΏ½ π¦ποΏ½2
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
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MODULE-II --- SINGLE DOF FREE VIBRATIONS VIBRATION ENGINEERING
2014
(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 οΏ½ΜοΏ½R0 = Ξ½0 = 3cm/sec.
Solution:
The differential equation of the motion is given by
οΏ½ΜοΏ½ + πππ₯ = 0
The general solution for the above differential equation is,
x(t) = A cos Οn t + B sin Οn t = X cos (Οn t - Ο)
When t = 0, x(0) = x0 = A = 2cm
β΄ οΏ½ΜοΏ½ R(0) = Ξ½0 = B Οn; β΄ B = π0ππ
Οn = οΏ½ππ
= οΏ½ 250.05
= 22.36 πππ/π ππ
Maximum amplitude of vibration X = βπ΄2 + π΅2 = οΏ½π₯02 + π02
ππ2
= οΏ½22 + 32
22.362 = 2.0045ππ
vi) Natural Frequency ππ = 12π
ππ = 12π
Γ 22.36 = 3.56 π»π§
vii) Period of oscillation T = 1ππ
= 13.56
= 0.28 sec.
viii) Maximum Velocity οΏ½ΜοΏ½ Rmax = X Οn = 2.0045 Γ 22.36 = 44.82 cm/sec.
ix) Maximum Acceleration οΏ½ΜοΏ½ Rmax = X ππ2 = οΏ½ΜοΏ½ Rmax .Οn = 44.82 Γ 22.36 = 1002.2 cm/sec2
x) Phase angle Ο = tan-1 οΏ½ π0πππ₯0
οΏ½ = π‘ππβ1 οΏ½ 322.36Γ2
οΏ½ = 3.838π
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
VTU-NPTEL-N
MEICT Proj
ect
Page 37 of 40
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
VTU-NPTEL-N
MEICT Proj
ect
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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
VTU-NPTEL-N
MEICT Proj
ect
Page 39 of 40
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 Newtonβs method.
12. Derive the equation of motion of a simple pendulum having an angular displacement of
π.
13. Show that the frequency of undamped free vibration of a spring mass system is given
by πΉπ = 1/ 2π οΏ½π πΏ
.
14. Show that the natural frequency of undamped free vibration of a spring mass system is
given by ππ = 1/ 2π οΏ½π π½
.
15. Using the energy method derive the differential equation of motion of an undamped free
vibration and show that frequency ππ = οΏ½π π
16. Using the Rayleigh method derive the differential equation of motion of an undamped
free vibration and show that frequency ππ = οΏ½π π
17. Derive the natural frequency of torsional vibrations.
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 40 of 40