Mechanical vibration note

1
MECHANICAL VIBRATIONS
The study of vibrations is concerned with the oscillatory motion of bodies and the
forces associated with them. All bodies possessing mass and elasticity are capable of
vibrating. Thus most engineering machines and structures experiences vibration to some
degree and their design generally requires consideration of their oscillatory behavior. The
oscillatory motion of the system may be objectionable or necessary for performing a task.
The objective of the designer is to control the vibration when it is objectionable and
to enhance the vibration when it is useful. Objectionable or undesirable vibration in machine
may cause the loosening of parts, its malfunctioning or its failure. The useful vibration helps
in the design of shaker in foundries, vibrators in testing machines etc.
Causes of vibration: - The main causes of vibration are:-
1) Unbalanced forces in the machine. These forces are produced from within the
machine itself because of non-uniform material distribution in a rotating machine
element.
2) Dry friction between the two mating surfaces: This is what are known as self
exited vibration.
3) External excitations. These excitations may be periodic, random or of the nature of
an impact produced external to the vibrating system.
4) Elastic nature of the system
5) Earth quakes. These are responsible for the failure of many buildings, dams etc.
6) Winds. These may cause the vibrations of transmission and telephone lines under
certain conditions.
The effect of vibrations is excessive stresses, undesirable noise, looseness of parts and
partial or complete failure of parts. Inspite of these harmful effects the vibration phenomenon
does have, some uses also e.g. in musical instruments, vibrating screens, shakers, stress
relieving.etc.

2
Methods to reduce vibrations
Elimination or reduction of the undesirable vibrations can be obtained by one or more
of the following methods
1) Removing the cause of vibrations
2) Putting in screens if noise is the only objection
3) Resting the machinery in proper type of isolators
4) Shock absorbers
5) Dynamic vibration absorbers
Definitions: - The important terms connected with vibrations are defined below
1) Periodic motion: A motion which repeats itself after equal intervals of time is known
as periodic motion. Any periodic motion can be represented by function x (t) in the
period T. the function x(t) is called periodic function.
2) Time period: - Time taken to complete one cycle is called time period.
3) Frequency: - The number of cycles per unit time is known as frequency.
4) Natural frequency: - When no external force acts on the system after giving it an
initial displacement, the body vibrates. These vibrations are called free vibrations and
their frequency as natural frequency. It is expressed in c/s or hertz
5) Amplitude: - The max displacement of a vibrating body from its equilibrium position
is called amplitude.
6) Fundamental mode of vibration: - The fundamental mode of vibration of a system
is the mode having the lowest natural frequency.
7) Resonance: - When the frequency of external exitation is equal to the natural
frequency of a vibrating body, the amplitude of vibration becomes excessively large.
This concept is known as resonance.
8) Mechanical systems: - The systems consisting of mass shiftness and damping are
known as mechanical systems.
9) Continuous and discrete systems: - Most of mechanical systems include elastic
members which have infinite number of degree of freedom. Such systems are called
continuous systems. Continuous systems are also known as distributed systems.

3
Ex. Cantilever, Simply supported beam etc. Systems with finite number of degrees of
freedom are called discrete or lumped systems.
10) Degree of freedom: - The minimum no of independent co-ordinates required to
specify the motion of a system at any instant is known as degree of freedom of the
system. Thus a free particle undergoing general motion in space will have three
degree of freedom, while a rigid body will have six degree of freedom i.e. three
components of position and three angles defining its orientation. Further more a
continuous body will require an infinite number of co-ordinates to describe its
motion; hence its degree of freedom must be infinite.

4
11) Simple harmonic motion (SHM) A periodic motion of a particle whose acceleration
is always directed towards the mean position and is proportional to its distance from
the mean position is known as SHM. It may also be defined as the motion of a
projection of a particle moving round a circle with uniform angular velocity, on a
diameter.
12) Phase difference It is the angle between two rotating vectors representing simple
harmonic motion of the same frequency.

5
Classification of vibrations
Mechanical vibrations may broadly be classified the following types
1) Free and forced vibration
2) Linear and non linear vibration
3) Damped and un damped vibration
4) Deterministic and random vibration
5) Longitudinal, transverse and torsional vibration
6) Transient vibration
1) Free and Forced vibration
Free vibration takes place when system oscillates under the action of forces inherent
in the system itself and when external impressed forces are absent. The system under
free vibration will vibrate at one or more of its natural frequencies.
Vibration that takes place under the exitation of external forces is called forced
vibration. When the exitation is oscillating the system is forced to vibrate at the
exitation frequency. If the frequency of excitation coincides with one of the natural
frequency of the system, a condition of resonance is encountered, and dangerously
large oscillations may result.
2) Linear and Non-linear vibration
If in a vibrating system mass, spring and damper behave in a linear manner, the
vibrations caused are known as linear in nature. Linear vibrations are governed by
linear differential equations. They follow law of superposition.
On the other hand, if any of the basic components of a vibrating system behaves non-
linearly, the vibration is called non-linear. Linear vibration becomes non-linear for
very large amplitude of vibration. It does not follow the law of super- positions.
3) Damped and Undamped vibration
If the vibrating system has a damper, the motion of the system will be opposed by it
and the energy of the system will be dissipated in friction. This type of vibration is
called damped vibration.
The system having no damper is known as undamped vibration

6
4) Deterministic and Random vibration
If in the vibrating system the amount of external excitation is known in magnitude, it
causes deterministic vibration. Contrary to it the non-deterministic vibrations are
known as random vibrations.
5) Longitudinal, Transverse and Torsional vibration
Fig represents a body of mass ‘m’ carried on one end of a weightless spindle, the
other end being fixed. If the mass moves up and down parallel to the spindle and it is
said to execute longitudinal vibrations as shown in fig (1).
When the particles of the body or spindle move approximately perpendicular to the
axis of the spindle as shown in fig (2) the vibrations so caused are known as
transverse vibrations
if the spindle gets alternately twisted and untwisted on account of vibrating motion of
the suspended disc, it is said to be undergoing torsional vibrations as shown in fig(3).
6) Transient Vibration
In ideal system the free vibrations continue indefinitely as there is no damping. The
amplitude of vibration decays continuously because of damping (in a real system) and
vanishes ultimately. Such vibration in a real system is called transient vibration.

8
HARMONIC MOTION (SHM)
Oscillatory motion may repeat itself regularly as in the balance wheel of a watch, or
display considerable irregularity, as in earthquakes. When the motion is repeated in equal
intervals of time T, it is called periodic motion. The repetition time T is called the period of
the oscillation, and its reciprocal f = 1/T is called the frequency. If the motion is designated
by the time function X (t), then any periodic motion must satisfy the relationship X (t) = X
(t+T).
The simplest form of periodic motion is harmonic motion. It can be demonstrated by
a mass suspended from a light spring, as shown in fig (1). If the mass is displaced from its
rest position and released, it will oscillate up and down, by placing a light source on the
oscillating mass; its motion can be recorded on a light – sensitive film strip which is made to
move past it at constant speed.
The motion recorded on the film strip can be expressed by the equation X = A Sin (2π t/T).
Where A is the amplitude of oscillation, measured from the equilibrium position of the mass
and T is the period. The motion is repeated when t = T.

9
Harmonic motion is often represented as the projection on a straight line of a point
that is moving on a circle at constant speed as shown in fig (2) with the angular speed of the
line designated by ω, the displacement X can be written as
X = A Sin ω t ---------- (1)
Fig (2) Harmonic motion as projection of a point moving as a circle
The quantity ω is generally measured in rad/sec, and is referred to as the circular frequency.
Since the motion repeat itself in 2 π radians.
We have ω = 2π / T = 2π f
Where T is period in sec and f is frequency in cycles/sec
The velocity and acceleration of harmonic motion can be determined by differentiation of
equation (1).
i.e. x
 = A.Cos ω t. ω = A ω Sin (ω t + π / 2) ------- (2)
x

 = - A ω. Sin ω t. ω = A ω 2
Sin (ω t + π ) ------- (3)
Thus the velocity and acceleration are also harmonic with the same frequency of oscillation,
but lead the displacement by π/2 and π radians respectively.
Fig (3) shows both the time variation and vector phase relationship between displacement,
velocity and acceleration in harmonic motion. ..

10
From equations (1) and (3) x

 = - X ω2

11
Addition of harmonic motion
When we add two harmonic motions of the same frequency, we get the resultant motion as
harmonic. Let us have the harmonic motions of amplitudes A1 and A2 the same frequency ω
and phase difference φ as
X1 = A1 Sin ω t
X2 = A2 Sin (ω t + φ)
The resultant motion is given by adding the above equations
X = X1 + X2 = A1 Sin ω t + A2 Sin (ω t + φ)
= A1 Sin ω t + A2 [Sin ω t Cos φ + Sin φ Cos ω t]
= A1 Sin ω t + A2 Sin ω t Cos φ + A2 Sin φ Cos ω t
X = (A1 + A2 Cos φ) Sin ω t + A2 Sin φ Cos ω t
Let A1 + A2 Cos φ = A Cos θ
A2 Sin φ = A Sin θ
Then X = A Cosθ Sin ω t + A Sin θ Cos ω t
X = A Sin (ω t + θ)
The above equation shows that the resultant displacement is also simple harmonic motion of
amplitude A and phase θ. To find out the value of A consider
A Cosθ = A1 + A2 Cos φ ------- (I)
A Sinθ = A2 Sin φ
Squaring and adding
A2
Cos2
θ + A2
Sin2
θ = (A1 + A2 Cos φ)2
+ (A2 Sin φ)2
A2
= A1
2
+ A2
2
Cos2
φ + 2 A1 A2 Cos φ + A2
2
Sin2
φ
A2
= A1
2
+ A2
2
+ 2 A1 A2 Cos φ
A = √ A1
2
+ A2
2
+ 2 A1 A2 Cos φ
The resultant phase difference can be determine from the above equation (I)
tan θ = A2 Sin φ / (A1+A2 Cos φ)
θ = tan-1
[A2 Sin φ / (A1 + A2 Cos φ)]
The graphical method of addition of two simple harmonic motion is shown in figure.

12
Beats Phenomenon: when two harmonic motions pass through some point in a medium
simultaneously, the resultant displacement at that point is the vector sum of the
displacements due to two component motions. This upper position is called interference. The
phenomenon of beat occurs as a result of interference between two waves of slightly different
frequencies moving along the same straight line in the same direction.
Consider, that at particular time, the two wave motions are in the same phase. At this stage
the resultant amplitude of vibration will be maximum. On the other hand, when the two
resultant motions are not in phase with each other, they produce minimum amplitude of
vibration.

13
Again after some time the two motions are in phase and produce maximum amplitude and
then minimum amplitude. This process goes on repeating and the resultant amplitude
continuously keeps on charging from Maximum to Minimum. This phenomenon is known as
beat.
Let us consider two waves of the same amplitude A and slightly different frequencies ω1 and
ω 2 .If X1 and X2 are the displacements of these waves at any time t, then
X1 = A Sin ω 1t, X2 = A Sin ω 2t
The resultant displacement X at any time is given by adding the above two equations
X = X1 + X2 = A Sin ω 1t + A Sin ω 2t
= A (Sin ω 1t + Sin ω 2t)
= A. 2 Sin (ω 1 + ω 2 / 2) t. Cos (ω 1 - ω 2 / 2) t --- (1)
X = B Sin (ω 1 + ω 2 / 2) t ----- (2)
Where B = 2A Cos (ω 1 - ω 2 / 2) t
Equation (2) represents SHM whose amplitude is B. The Maximum value of B in 2A and
Minimum is Zero. The frequency of beat is (ω 1 - ω 2 / 2π) Hertz.
Differentiating equation (1) with respect to time we get
dx / dt = 2 A Sin (ω 1 + ω 2 / 2) t [-Sin (ω 1 - ω 2 / 2) t (ω 1 - ω 2 / 2)]
+ 2 A Cos (ω 1 - ω 2 / 2) t [Cos (ω 1 + ω 2 / 2) t (ω 1 + ω 2 / 2)]

14
Where dx / dt is called the slope of the beat.
If a particle is subjected to two different harmonic motions given by
X1 = a Sin ω 1 t and X2 = b Sin ω 2 t
Then the Maximum amplitude Amax = (a + b)
And Minimum amplitude Amin = (a – b)
FOURIER THEOREM
J. Fourier, a French mathematician developed a periodic function in terms of series of Sines
and Cosines. With the help of this mathematical series known as Fourier series, the vibration
results obtained experimentally can be analyzed analytically. If X(t) is a periodic function
with period T, the Fourier series can be written as
X (t) = a0 / 2 + a1 Cosω t + a2 Cos2ω t + a3 Cos3ω t + ………..
+ b1 Sin ω t + b2 Sin2ω t + b3 Sin3ω t + ………..
X (t) = a0 / 2 + (an Cos n ω t + bn Sin n ω t ---------- (1)
Where ω = 2π/T is the Fundamental frequency and a0, a1, a2 ….b0, b1, b2… are constants Co –
efficient
Determination of ao
Integrate both sides of equation (1) over any interval of length T = 2π / ω. All the integrats on
the RHS of the above equation are Zero except the one containing ao that is
0∫2π/ ω
X(t) dt = 0∫2π/ ω
ao dt = ao 2π/ω
ao = ω /2π 0∫2π/ ω
X(t) dt ------ (2)
Determination an: To find an multiply both sides of equation (1) by Cos n ωt and integrate
over any interval of time T = 2π/ω
0∫2π/ ω
X (t) Cos (n ωt) dt = 0∫2π/ ω
an Cos2
(n ωt) dt
= 0∫2π/ ω
an(1+ Cos 2n ωt /2)dt
= an 0∫2π/ ω
½ dt + an 0∫2π/ ω
Cos2n ωt/2 dt
= an /2 [t] 0
2π/ ω
+ an/2 [Sin 2nωt / 2nω] 0
2π/ ω
= an /2 [2 π/ω] + an/4nω [Sin 4nπt - 0]

15
= an . π/ω
an = ω/ π 0∫2π/ ω
X(t) Cos (n ωt)dt
Similarly bn = ω/ π 0∫2π/ ω
X (t) Sin (n ωt) dt
The above analysis is known as Harmonic analysis
Work done by a Harmonic Force
Let a harmonic force F = Fo Sin ωt is acting on a vibrating body having motion
x = Xo Sin (ωt – φ). The work done by the force during a small displacement dx is F
dx. So the work done in one cycle
ω = 0∫T
F dx / dt .dt
= 0∫T
[Fo Sin ω t. d / dt Xo Sin (ω t – φ)] dt
= 0∫T
Fo Sin ω t. X1 ω Cos (ω t – φ)] dt
= Xo Fo ω 0∫T
Sin ω t. Cos (ω t – φ)] dt
= Xo Fo ω 0∫T
Sin ω t. [Cos ω t Cosφ + Sin ω t Sinφ)]
= Xo Fo ω 0∫T
(Sin ω t Cos ω t) Cosφ + Sin2
ω t Sinφ)]
= Xo Fo ω 0∫T
Sin [(ω t + ω t) /2 ] Cosφ + (1 – Cos 2 ω t) / 2)] Sinφ
ω = Xo Fo ω 0∫T
[Sin 2 ω t Cosφ / 2 + Sinφ (1 – Cos 2 ω t) / 2] dt
Putting T = 2π/ ω we get
W = π Fo Xo Sinφ
In the above equation if φ = 0, the Work done will be zero. It means force and
displacement should not be in phase to get the work done.

16
PROBLEMS
1. The motion of a particle is represented by the equation x = 10 Sin ωt. Show the
relative positions and magnitudes of the displacements, velocity and acceleration
vectors at time t = 0, for the case when (i) ω = 2 rad/ s (ii) ω = 0.5 rad/ s.
2. A harmonic motion has an amplitude of 0.05m and a frequency of 25 Hz. Find the
time period, Maximum velocity and Maximum acceleration.
3. A harmonic motion has an amplitude of 0.2 cm and a period of 0.15 sec. Determine
the maximum velocity and acceleration.
4. An accelerometer indicates that a structure is vibrating harmonically at 82 cps with a
maximum acceleration of 50g. Determine the amplitude of vibration.
5. A harmonic motion uses a frequency of 10cps and its maximum velocity is 4.57 m/s.
Determine its amplitude, its period and its maximum acceleration.
6. Show that the resultant motion of three harmonic motions given below is zero.
X1 = a sinωt, X2 = a sin (ωt + 2π /3), X3 = a sin (ωt + 4 π / 3).
7. A harmonic motion is given by the equation (t) = 5sin (15t - π/4) cm where phase
angle is in radians and t in seconds. Find i) period of motion ii) Frequency iii)
Maximum displacement, velocity and acceleration.
8. The rectilinear motion of a point is given by α = -9X where α and X are the
acceleration and displacement of SHM and the amplitude is 2 units. Find (i) The
period and frequency (ii) Displacement, velocity and acceleration after 21.5 sec.
9. A harmonic motion is described as X(t) = X Cos (100t + ψ) mm the initial conditions
.
are X (0) = 4mm and X (0) = 1m/s. (i) Find the constant X and ψ (ii) Expressing X
(t) in the form X = A Cos ω t + B Sin ω t. Find the constants A and B.
10. Given X (t) = X Cos (100t + ψ) = A Cos 100t + B Sin 100t. Find A, B, X and ψ for
the following cases.
(i) X (0.1) = - 8.796 mm and X (0.2) = 10.762 mm
(ii) X (0.1) = - 8.796 mm and x
 (0.1) = 621.5 mm/s

17
. .
(iii) X (0.1) = - 8.796 mm and X (0.2) = -10.76 x 104
mm/s2
11. A body describes simultaneously two motions X1 = 3Sin40t X2 = 4Sin41t. What is
the maximum and minimum amplitude of combined motion and what is the beat
frequency.
12. A body has two simultaneous motions represented by X1 = A sin ω1t and X2 = A sin
ω2t . Where ω1 > ω2 . Find out the slope of beat phenomenon curve at t = π / (ω1 - ω2)
also find the condition for this slope to become zero.
13. Add the following harmonic motions analytically and check the solution graphically.
X1 = 4Cos (ωt + 100
) X2 = 6Sin (ωt + 600
).
14. Add the following motions analytically and check the result by graphically
X1 = 2 Cos (ωt + 0.5), X2 = 5 Sin (ωt + 1.0).
15. Split the harmonic motion X = 10 Sin (ωt + π/6) into two harmonic motions one
having a phase angle of zero and the other of 450
.
16. A body is subjected to two harmonic motions.
X1 = 15 Sin (ωt + π / 6) X2 = 8 Cos (ωt + π / 6).
What harmonic should be given to the body to bring it to equilibrium?
17. Split up the harmonic motion X = 10Sin (ωt + π / 6) into two harmonic motions one
having a phase of zero degree and another of 46.
18. A particle is under the influence of two harmonic motions y1 = 0.03 Sin (14t + 68.80
)
and y2 = 0.02 Sin (10t + 59.60
). Determine the resulting amplitude and phase angle.

18
PROBLEMS ON WORKDONE BY A HARMONIC FORCE
1. A force Po Sin ωt acts on a displacement Xo Sin (ωt – π/3). If Po = 100N, Xo = 0.02m,
ω = 2π rad/s. Find the work done during (i) the first cycle (ii) the first second (iii) the
first quarter second.
2. A force Po Sin ωt acts on a displacement Xo Sin (ωt – π/6). Where Po = 25N,
Xo = 0.05m and ω = 20π rad/s. What is the work done during (i) the first second
(ii) the first 1 / 40 second.
PROBLEMS ON FOURIER SERIES
i. Represent the period motion given in figure by harmonic series.
ii. A periodic motion observed on the oscilloscope is illustrated in figure. Represent this
motion by harmonic series.

19
iii. Represent the periodic motions given in figure by harmonic motion.
iv. Represent the periodic motions given in figure by harmonic series.

20
COMPLEX METHOD OF REPRESENTING HARMONIC VIBRATION.
Let there be a vector V in the X – Y plane represented by a complex number
V = a + ib i = √-1
r = √a2
+ b2
is the modulus
And θ = tan-1
(b/a) is argument
V = a + ib = r (Cos θ + iSin θ)
V = reiθ
by Euler formula
For a particular vibrating particle, r in the amplitude and ω its circular frequency, then the
displacement of the particle can be written as
X = r (Cos ωt + iSin ωt) = reiωt
x

 = ωr (-Sin ωt + iCos ωt)
= iωr (Cos ωt + iSin ωt)
= iωreiωt
This is known as velocity vector.
x

 = iωr [-Sin ωt + iCos ωt)
i2
ωr [Cos ωt + iSin ωt)
x

 = i2
ωr . eiωt
acceleration vector

21
PROBLEMS.
1. Represent 17 e-i3.74
in rectangular form
2. Represent 3 + i6 in exponential form
3. Represent the following complex number in exponential terms
(i) 3 + 7J (ii) -5 + 4J.
4. Represent the following complex number in rectangular form (i) 5eJo.1
(ii) 14e-2.8J

22
Undamped 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.
Equation of Motion and Natural Frequency of Vibration of a Simple Spring Mass
System

23
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 /2π
ƒn = 1 /2 π √k /m = 1/2 π √ k / kδ/g = 1 / 2 π √g / δ
Where δ = static deflection

24
Spring mass system in horizontal position
In the system shown in fig 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.
As per Newtons II law
Mass x acceleration = resultant force on mass
m x

 = - kx
m x

 + kx = 0 = x

 + k /m x = 0
: ωn = √ k/m and ƒn = 1 / 2 π √ k/m = 1 /2 π √g /δ

25
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
Energy Method: Consider a spring mass system; 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
T + V = constant
Differentiating the above equation w.r.t.1
t1
d (T+V) = 0
dt
For a spring mass system shown
K.E = T = ½ m x
 2
P.E = V = ½ kx2
: d ( ½ m x
 2
+ ½ kx2
) = 0
dt
d (m x
 2
+ kx2
) = 0
dt

26
m x

 x
 + kx x
 = 0
x

 + k/m x = 0
Hence ωn = √ k/m ƒn = 1 /2π √ k/m = 1/ 2π √g / δ
Rayleigh’s 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 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
x
 = Aωn cos ωn t
Maximum Velocity at mean position x
 = ωn A

27
Maximum kinetic energy at mean position = 1/2m x
 2
= 1/2 m ωn
2
A2
And maximum potential energy at Extreme position
P.E = 1 /2 kA2
K.E = P.E
= 1/2m ωn
2
A2
= 1/2 kA2
ωn
2
= k/m
ωn = √k/m
: ƒn = 1/2 π √ k/m =1/2 π √g /d
Torsional Vibrations
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.

28
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. The free body diagram of general angular displacement is shown
The equation of motion is.
. .
I θ = - Kt θ
. .
I θ + Kt θ = 0
θ + Kt θ / I = 0
: ωn = √Kt /I and ƒn = 1/2 π √ kt/I
Where Kt = GJ /L J = π d
3
/32
Show That for Finding The Natural Frequency of Spring Mass System the Mass of The
Spring Can Be Taken Into Account By Adding One –Third Its Mass To The Main
Mass.
Solution: 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

29
The K.E of the element dy of the spring is equal to 1/2 (ρdy) (y/ L x
 x)2
x
dy
y

k
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
∫0y2
dy
= 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
= 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

30
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.

31
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
Ke = K1 + K2
The equivalent spring shiftness is equal to the sum of individual spring shiftness.
SPRINGS IN SERIES

32
The total deflection of the system is equal to the sum of deflection of individual springs.
X = X1 + X2
Force = Force = Force
Ke K1 K2
1 = 1 + 1
Ke K1 K2
Thus when springs are connected in series, the reciprocal of equivalent spring shiftness is
equal to the sum of the reciprocal of individual spring shiftness.

33
SPRINGS IN ARBITRARY DIRECTION
Fig shows a spring K making an angle α with the direction of motion of the mass m.
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 + (K cos2
α) x

 = 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.

34
PROBLEMS
1. An unknown mass m kg attached to the end of an unknown spring K has a natural frequency
of 94 cpm, When a 0.453 kg mass is added to m, the natural frequency is lowered to 76.7
cpm. 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.
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. Find the natural frequency of the system shown in fig (3) K= 2x105
N/m, m= 20kg

35
6. 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.
7. Find the natural frequency of the system shown in fig (5). K1= K2=K3=K4=K5=K6=K = 1000
N/m.

36
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.
9. Find the equivalent spring constant of the system shown in fig (2) in the direction of the load P.

37
PROBLEMS ON UNDAMPED FREE VIBRATIONS
1. A U tube, open to atmosphere at both ends contains a column length L of a certain liquid.
Find the natural frequency of the liquid column.
2. A simple U tube monometer filled with liquid is shown in fig (1).Calculate the frequency of
resulting motion. If the minimum length of a monometer tube is 0.15m.
3. A electric motor is supported by 4 springs each having spring constant K as shown in fig
(2).If the Moment of inertia of the motor about the central axis of rotation is Jo find its natural
frequency of oscillations.
4. An electric motor is supported by six springs of stiffness K each as shown in Fig (3).The
Moment of inertia of the motor is I. Determine the natural frequency of the system.

38
5. A semi circular homogeneous disk of radius r and mass m is pivoted freely about its centre as
shown in fig (4).Determine its natural frequency of oscillation for small displacements.
6. A homogeneous square plate of side L and mass m is suspended from the mid point of one of
the sides as shown in fig (5).Find the natural frequency of the system.
7. Show that the equation of motion of two simple spring- mass systems shown in fig (1) & fig
(2) - are the same and are equal to that of the system as shown in fig (3).

39
8. A circular cylinder of mass m and radius r is connected by a spring of stiffness ‘K’ the
cylinder rolls on a horizontal plane surface while the other end of the spring is fixed with the
spring parallel to the plane surface. Find the natural frequency.
OR
A cylinder of mass ‘m’ and mass Moment of inertia Jo is free to roll without slipping, but is
retained by spring K as shown in fig (6).Find the natural frequency of oscillation.

40
9. Find the frequency of oscillation for the system shown in Fig (7). When the roller rolls
without slipping.
10. A circular cylinder of mass m and radius r is connected by a spring of shiftness K on an
inclined plane as shown in fig (8).If it is free to roll on the rough surface which is horizontal
without slipping. Determine its natural frequency.
11. A circular cylinder of mass 4 kg and radius 15cm is connected by a spring of shiftness
4000N/m as shown in fig (9). It is free to roll on horizontal rough surface without slipping,
determine the natural frequency

41
12. Find the natural frequency of the single degree freedom system shown in figure. The
rod AOB of the system is light, straight and stiff.
13. Find the natural frequency of motion for the system shown in fig (10). If the cylinder rolls
without slipping
14. Determine the natural frequency of the spring mass pully system as shown in fig.

42
15. The mass m is hanging from a cord attached to the circular homogeneous disc of mass M and
radius R as shown in fig. the disc is restrained from rotating by a spring attached at radius r
from the centre. If the mass is displaced downward from the rest position, determine the
frequency of oscillation.
16. For small angles of oscillations find the frequency of oscillations of the system shown in fig.

43
17. Find the natural frequency of system shown in fig. the cord may be assume inextensible in
the spring mass pully system no ship. Use energy method.
18. Determine the differential equation of motion of the system shown in fig and hence find the
natural frequency of the system.

44
19. Derive the differential equation of motion for a spring controlled simple pendulum as shown
in fig. the spring is in its unstretched position when the pendulum rod is vertical.
20. Find the natural frequency of the system shown in fig.

45
21. Use Rayleigh’s method to find the natural frequency of the semi-circular shell of
mass m and radius r which rolls from side to side without slipping as shown in fig
below.
22. The cylinder of mass m and radius r rolls without slipping on a circular surface of
radius R as shown in fig below. Determine the frequency of oscillation when the
cylinder is displaced slightly from its equilibrium position. Use the energy method.
23. A homogeneous sphere of radius r and mass m is free to roll without slipping on a
spherical surface of radius R. if the motion of the sphere is restricted to a vertical
plane as shown in figure. Determine the natural frequency of oscillation of the sphere.
Fig Q.21 & Q.22

46
24. The mass m is attached to one end of a weightless stiff rod which is rigidly connected
to the center of a homogeneous cylinder of radius r as shown in figure. If the cylinder
rolls without slipping, what is the natural frequency of oscillation of the system?

47
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
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 Types 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) Colomb 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 dx
dt
Fd = c x

When ‘c’ is the constant of proportionality and is called viscous damping Co-efficient with
the dimension of N-s/m.
Colomb 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

48
When µ = co- efficient of friction
N = normal force
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.
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.
EXPRESSION FOR AMPLITUDE OF FREE VIBRATION WITH VISCOUS DAMPING:
Consider a spring- mass – damper system as shown in fig .For free vibration with viscous
damping

49
From the free body diagram using Newton’s II law of motion, the equation of motion can be
written as
m x

 = - kx - c x

m x

 + c x
 + kx=0 -- (1)
Equation (1) is the differential equation assuming the solution of the form
x = est
 x
 = sest
, x

 = s2
est
where s is the constant to be determined. Substituting the above
values in equation (1) we get
ms2
est
+ csest
+ kest
= 0
(ms2
+ cs + k) est
= 0
s2
+ (c/m) s + k/m = 0
This is the quadrant equation, the solution of which is
s1,2 = -(c/m) ± √ (c/m)2
– 4x 1x (k/m)
2 x 1
s1,2 = - c/ 2m ± √ (c/2m)2
– (k/m)
The general solution may be written as
X = A e
s1t
+ B e
s2t
Where A & B are arbitrary integration constant which depends upon how the motion is
started.
Substitute equation (2) in equation (3)
X = A e
(-c/2m) + √(c/2m)2 – k/m) t +
B e (-c/2m) - √(c/2m)2 – k/m) t
The behavior of the system depends upon whether the radial of equation (4) is real,
imaginary or zero. When the damping term (c/2m)2
> (k/m) the exponents in the above
3
2
4

50
equation are real numbers and no oscillation are possible. We refer to this case as over
damped
When the damping term (c/2m)2
< k/m, the exponents are imaginary, this case is known as
under damped.
In the limiting case between oscillatory and nonoscillatory motion (c/2m)2
= k/m, and the
radical is zero. The damping corresponding to this case is called critical damping.
Critical Damping and Damping factor
Critical damping co-efficient Cc is defined as that damping as Co- efficient which makes the
radical zero i.e (Cc /2m)2
– k/m = 0
(Cc / 2m)2
= k/m
Cc / 2m = √ k/m
Cc / 2m = ωn
Cc = 2mωn
The damping factor denoted by ζ, is defined as the ratio of damping Co efficient to critical
damping Co efficient i.e.
It is a non dimensional relation
The term C / 2m = C / Cc x Cc/2m = ζ ωn
Substituting this in Equation (4) we get
X = Ae
(-ζ ω
n+√ (ζ ω
n
) ²- ω
n
² ) t
+ Be
(-ζ ω
n
-√ (ζ ω
n
) ²- ω
n
² ) t
X = Ae
(-ζ +√ ζ ² - 1) ω
n
t
+ Be
(-ζ - √ (ζ ² - 1) ω
n
t
5

51
It is clear from equation (5)
Three cases of interest exists when
i) ζ >1
ii) ζ =1
iii) ζ <1
Case (i) when ζ >1 (large damping or over damping):
In this case the radical is real and always less than ζ . Hence the two roots S1 & S2 are
Negative therefore
X = Ae
(-ζ + √ζ ² - 1) ω
n
t
+ Be
(-ζ - √(ζ ² - 1) ω
n
t
The displacement became the sum of two decaying components. The above equation
represents the case with very large damping, the motion is not periodic and no vibration takes
place. This type of vibration is also called as aperiodic vibration.
Case (ii) when ζ <1 (light damping or under damping):
The radical in this case is imaginary and definite oscillatory motion takes place, in this case
the displacement is oscillatory with diminishing amplitudes.
x =A
(-ζ + i√ 1- ζ ² ) ω
n
t
+ Be
(-ζ – i √ 1- ζ ² ) ω
n
t

52
The general solution of above equation can be obtained as follows:
x = Ae
(-ζ + i√ 1- ζ ²) ωnt
+ Be
(-ζ – i √ 1- ζ ²) ωnt
x= e
- ζ ω
n
t
{A [cos (√1- ζ
²
) ωnt + i sin(√1- ζ
²
) ωnt] + B [cos(√1- ζ
²
) ωnt -i sin(√1- ζ
²
) ωnt)] }
x= e
- ζ ω
n
t
{ (A+B) cos ((√1- ζ
²
) ωnt) + (A-B) i sin√ (1- ζ
²
) ωnt) }
x= e
- ζ ω
n
t
{C1 Cos(√1- ζ
²
) ωnt) + C2 Sin √ (1- ζ
²
) ωnt.
Where, C1= A+B & C2 = i (A-B)
x = e
- ζ ω
n
t
( C1 Cos ωdt + C2 Sin ωdt)
Where ωd = (√1- ζ
²
) ωn is damped frequency
The term ωd is the circular frequency of damped oscillation in radians/sec and damped time
period.
Td = 2π/ωd
And damped frequency, fd = ωd /2π
Equation (6) is the solution of the differential equation for small damping this can be written
as x = X e
- ζ ω
n
t
Sin (ωdt + Ф)
6

53
= X e
- ζ ω
n
t
Sin [(√1- ζ
²
) ωnt + Ф ]
Where X = √C1
2
+ C2
2
tan Ф = (C1 / C2)
Case (iii): ζ = 1[Critical damping or initial damping]: ζ = 1, represents the transition
between oscillatory and nonoscillatory condition. The amount of damping corresponds to
this case is referred to as critical or initial damping. Since the radical of equation (1) is zero
for critical damping, the two roots ζ1 & ζ2 are equal and alike.
X = (A+B)e
- ζ ω
n
t
X = Ce
- ζ ω
n
t
Where C= A+B = constant
The above equation contains only one arbitrary constant. The solution lacks the required
number of independent constant to represent general solution. In this case a function of the
form te - ωnt
will found to satisfy the differential equation. When substituting the general
solution can be written as
X = (A+Bt)e
- ζ ω
n
t
Logarithmic Decrement: A convenient way of determining the amount of damping in a
system is to measure the rate of decaying oscillations. This can be best expressed by a term
called ‘logarithmic Decrement’.
It is defined as the natural logarithmic ratio of any two successive amplitudes. It is denoted
by δ
Consider a damped free vibration express by the equation.
X = e
- ζ ω
n
t
( C1 Cos ωdt + C2 Sin ωdt) and the oscillatory motion as shown in fig

54
The logarithmic decrement δ = ln (x1/x2) = ln (e
- ζ ω
n
t
1/ e
- ζ ω
n
t
2)
δ = ln e-
ζ ω
n
(t
1
-t
2
)
δ = ln e
ζ ω
n
(t
2
-t
1
)
δ = ln e
ζ ω
n
T
Where T= t2-t1 , is the period of oscillation = 2π/ωd = 2π/(√1- ζ2
) ωn
δ = ln e
ζ ω
n
x 2π / (√1- ζ2) ω
n
δ = 2π ζ / √1- ζ2
For small values of ζ
δ = 2π ζ

55
PROBLEMS ON UNDER DAMPED SYSTEM
1) Find the equation of the motion for the system shown in fig. when (a) ζ = 1 ii) ζ = 0.3 and
iii) ζ = 2. If the mass m is displaced by a distance of 3cm and released.
Fig (1)
2) An over damped system shown in figure (i) has a spring stiffness of 15N/mm, mass 10kg
and damping co – efficient 1.5Ns/mm. It is at rest in its static equilibrium position when it
receives an impulse force acting to the right that creates an initial instantaneous velocity of
25m/sec. Determine (i) An expression for displacement in terms of t. (ii) Maximum
displacement of mass from initial position. (iii) Time required for the mass to attain the
position of Maximum displacement
3) A simple spring damp system has a mass of 2 kg. The spring constant is 1000N/m and
damping co – efficient has a value of 15 N-s/m. The system is initially at rest when a
velocity of 10cm/sec is imported to it. Find (i) The subsequent displacement and velocity of
mass (ii) Displacement and velocity of mass when time t equal to 1sec.
4) A gun barrel having mass 500 kg is designed with the following data. Initial recoil velocity
= 36m/s. Recoil distance on firing = 1.5m. Calculate (i) Spring constant (ii) Critical
damping co – efficient and iii) The time required for the barrel to return to a position 0.12m
from its initial position.

56
5) A gun barrel having mass 600 kg has a recoil spring of stiffness 294000 N/m. if the barrel
recoils 1.3m on spring determine a) the initial recoil velocity of the barrel b) The critical
damping co-efficient of the dish pot which is engaged at the end of the recoil spring. c) The
time required for the barrel to a return to a position 5cm from the initial position.
6) The vibrating system consists mass of 2 kg and spring stiffness 3 KN/m viscously damped
such that ratio of any two consecutive amplitudes is 1.00 to 0.98. determine
(i) Natural frequency
(ii) Logarithmic decrement
(iii) Damping factor
(iv) Damping Co - efficient
7) A vibrating system of 4.534 kg, a spring of stiffness 3500 N/m and a dash pot having
damping Co – efficient 1.243 N-s/m. Find
(i) Damping factor
(iii) Ratio of any two successive amplitudes
8) A mass of single degree damped vibrating system is 75 kg and makes 24 free oscillation in
14sec and disturb from its equilibrium position. The amplitude of vibration reduces to 0.25
of its initial value after 5 oscillation. Determine
(i) Stiffness of spring
(iii) Damping factor
9) A vibrating system in a vehicle is to be designed with the following parameters
K= 100N/m, C= 2N-S/m m= 1 kg. Calculate 1) the decrease of amplitude from its starting
value after 3 complete oscillation and 2) the frequency of oscillation.

57
10) For the system shown in fig, the characteristic of the dash pot is such that when a constant
force of 49N is applied to the piston its velocity is found to be constant at 0.12m/s.
1) Determine the value of C 2) would you expect the complete system to-
periodic or a periodic.
11) A damper offers resistance 0.05N at constant velocity 0.04m/s. the damper is used with K=
9N/m.Determine the damping frequency of the system when the mass of the system is 10kg.
12) A vibrating system is defined by the following parameters m = 3 kg K= 100N/m C= 3N-
S/m. determine a) damping factor b) the natural frequency of damped vibrations c) the ratio
of two consecutive amplitudes e) The number of cycles after which the original amplitude is
reduced to 20%.
13) The damped vibration record of a spring mass dash pot system shows the following data
Amplitude on second cycle = 1.2cm
Amplitude on third cycle = 1.05cm
Spring constant = 80N/cm
Mass of spring = 2kg
Determine the damping constant, causing the viscous damping.

58
14) Free vibration records of 1 tone machine mounted on an isolar is shown. Identify the type of
isolator and its characteristics i.e. the spring.
15) A machine of mass 100kg is mounted on springs and is fitted with a dashpot to damp out
vibrations. There are 4 springs each of stiffness 7.5 N/mm and it is found that the amplitude
of vibration diminishes from 3.84cm to 0.64cm in 2 complete oscillations. Assuming that
the damping force varies as the velocity, determine the resistance of dashpot at unit velocity
and compare the frequency of damped vibration with the frequency when dash pot is not in
operation.
16) A spring mass damper system makes 10 damped oscillations per second. Its amplitude
reduces to 10% of its initial value in 50 cycles. Determine (i) Damped frequency (ii)
Logarithmic decrement (iii) Damping factor.
17) A mass of 10kg is kept on two slabs of isolators placed one over the other one of the
isolator is of rubber having a stiffness of 3KN/m and damping co-efficient of 100N-S/m
while the other isolator is of felt with stiffness of 12KN/m and damping co-efficient of
300N-S/m.if the system is set in motion in vertical direction. Determine the damped and
undamped natural frequency of the system.
18) The disc of a torsional pendulum has a moment of inertia of 600 kg-cm2
and is immersed in
a viscous fluid. The brass shaft attached to it is of 10cm dia and 40cm long. When the
pendulum is vibrating, the observed amplitudes on the same side of the rest portion for
successive cycles are 90
, 60
, and 40
determine.

59
a) Logarithmic decrement
b) Damping torque at unit level
c) The periodic time of vibration
Assume for brass shaft G= 4.4x 1010
N/m2
19) A torsional pendulum when immersed in oil indicates its natural frequency as 200Hz. But
when it was put to vibrate in vacuum having no damping its natural frequency was observed
as 250 Hz. Find the value of damping factor of oil.
20) The torsional pendulum with a disc of moment of inertia 0.05k-m2
is immersed in a viscous
fluid is shown in fig. during vibrations of pendulum, the observed amplitudes on the same
side of the neutral axis for successive cycles are found to decay 50% of the initial value.
Determine 1) logarithmic decrement.
2) Damping torque/unit velocity.
3) The periodic time of vibration.
4) The frequency when the disc is removed from fluid.
Assume G= 4.5x1010
N/m2
for material of shaft, d= 0.1m and L= 0.5m.

60
21) Derive equation of motion for the system shown in fig. if m= 1.5 kg K= 4900N/m a= 60mm
& b= 140mm. determine the value of C for which the system is critically damped. Also find
the natural frequency.
22) Setup the differential equation of motion for the system shown in fig. Determine the
expression for critical damping co-efficient. Also determine the value of critical damping
when m = 1.5kg, k=4900N/m, a = 0.06m and l = 0.14m.
23) Determine suitable expression for frequency of motion of the damped vibratory system
shown. Find the critical damping co-efficient when a= 0.1m b= 0.13m K= 4900N/m.
M= 1.5kg.

61
24) Write the differential equation of the motion for the system shown in figure. Determine the
natural frequency of damped oscillation and the critical damping co – efficient. Determine
the damped frequency and critical damping co – efficient for a = 100mm b = 150mm, the
mass m may be taken as 2.5kg. the damping co – efficient is 120 Ns/m and the spring
constant is 6000 N/m.
25) Write the differential equations of the motion for the system shown in fig and the natural
frequency of damped vibrations and the damping ratio for the system.

62
26) The single pendulum is pivoted at point O as shown in fig. if the mass of the rod is
negligible, for small oscillations; find the damped natural frequency of the pendulum.

63
Forced Vibration
Energy Dissipated by Damping
When a system is undergoing steady state forced vibrations with viscous damping, energy is
continuously being absorbed by the dash pot. This energy absorbed per cycle can be
determined from the work done/ cycle by a harmonic force on a harmonic motion.
We know that
Work done = π Fo Xo sin φ
Where Xo = amplitude of vibratory motion
Fo= amplitude of vibratory force
φ = the angle by which the motion lags the force
For the case under study X0 = X Fo = C ωX
φ = -900
(Vector Diagram)
There fore work done / cycle = π (c ωX). X (-1)
= - π c ωX2
Or the energy dissipated per cycle is given
Energy dissipated / cycle = π c ωX2
Problem
1) Determine the power required to vibrate a spring – mass – dashpot system with an
amplitude of 15mm and at a frequency of 100 Hz. The system has a damping factor
of 0.05 and a damped natural frequency of 22 Hz as found out from the vibration
record. The mass of the system is 0.5 kg.

64
FORCED VIBRATION OF SINGLE DEGREE OF FREEDOM SYSTEM
Forced vibration: in case of free vibration the system once disturbed from its equilibrium
position executes vibration because of its elastic properties. This system will come to rest
sooner or later, depending upon the amount of damping. In case of forced vibration there is
an impressed force on the system which keeps it vibrating. The vibration of air compressor
Internal. Combustion engine, machine tools and various machinery are all examples of forced
vibrations. This are in general three types of forcing functions: - i) Periodic forcing function
ii) Impulsive type of forcing function iii) Random forcing function.
Periodic forces can be further divided into harmonic and non harmonic. Harmonic periodic
forces are commonly encountered in vibration system. Impulse type of forces gives rise to
transient vibrations. The random forcing function gives rise to vibration such as earthquake
etc.
EXPRESSION FOR AMPLITUDE AND PHASE ANGLE FOR SPRING MASS
DASHPOT SYSTEM SUBJECTED TO HARMONIC FORCING FUNCTIONS

65
Consider a single degree of freedom system with viscous damping excited by a harmonic
force to F0 sin ωt as shown in figure. From the free body diagram applying Newton’s II law
of motion we get
m x

 = -c x
 - kx + F0 sinωt
Equation (1) a differential equation
m x

 + c x
 + kx = F0 sin ωt 1
The complete solution of the equation (1) consists of a complementary function and a
particular integral.
i.e x = xc + xp
Where xc is a complementary solution
and xp is a particular integral
The complimentary solution is the solution of the homogeneous equation m x

 + c x
 + kx =0
i.e x

 + c/m x
 + k/m x = 0
The complementary solution is
xc = e
- ζ ω
n
t
{C1 Cos ωdt + C2 Sin ωdt }
The particular solution. xp is the solution for steady state vibration
let xp = x sin (ωt -φ)
x
 p = x ω Cos (ω t - φ)
= x ω Sin (ω t - φ + п/2)

66
x

 p = -x ω2
Sin (ω t - φ)
= x ω2
Sin (ω t - φ + п)
Substituting xp, x
 p and x

 p in
Equation (1) we get
m x ω2
Sin (ωt - φ + п) + c x ω Sin (ωt - φ + п/2) + k x sin (ωt - φ) = F0 Sin ωt
F0 Sin ωt - k x sin (ωt - φ) – cx Sin (ωt - φ + п/2) – mx ω2
sin (ωt - φ + п )] = 0
Representing by vectors
From the right angle triangle OAD.

67
OA2
= OB2
+ AB2
FO = (cωx)2
+ (kx - m ω2
x)2
F0
2
= X2
[(cω)2
+ (k - m ω2
)2
]
Or X2
= FO
2
/ (cω)2
+ (k - m ω2
)2
X= FO / √ (cω)2
+ (k - m ω2
)2
And phase angle tan θ = c ω/( K - m ω2
)
Or θ = tan
-1
[(c ω/ (K - m ω2
)]
X= FO/k / √ (cω /k)2
+ (1 - m ω2
/k)2
But Cω / K = Cω / k. Cc / Cc. 2m / 2m
= C / Cc. Cc / 2m. 2m / k. ω = ξ . ωn . 2/ ωn
2
. ω
= 2 ξ . ω /ωn
And mω2
/k = ω2
/ ωn
2
X= FO/k / √ [1-(ω/ωn
2
]2
+ [2ξ. ω/ωn)2
tan θ = 2ξ (ω/ωn) / [1-(ω/ωn)2
]
i.e θ = tan-1
[2 ξ ω/ωn /(1 - (ω/ωn)2
]
Let, Xs = Fo/k
X = Xs/√ [1-(ω/ωn]2
+ [2ξ . ω/ωn]2

68
Where Xs may be defined as zero frequency definition of the spring mass systems under a
steady force F0. The above equation gives the particulars solution of equation (1) therefore
the total or complete solution is given by
X = Xc + Xp
e
- ζ ω
n
t
{C1 Cos ωdt + C2 Sin ωdt } + Xs /√ [1-( ω/ωn]2
+ [2ξ . ω/ωn)2
This is the complete solution to an under damped system subjected to sinusoidal
excitation. It is found to contain two arbitrary constants C1 & C2. Which has to be determined
from this complete solution from the two initial conditions.
The first part of the compete solution, i.e. complementary function is seen to decay with time
and vanishes ultimately. This part is called as ‘Transient Vibrations’. The second part i.e. the
particular solution is seen to be a sinusoidal vibration with a constant amplitude and is called
‘steady state vibration’
STEADY STATE VIBRATION or MAGNIFICATION FACTOR or FREQUENCY
RESPONSE CURVE
For all practical systems subject to harmonic excitation, the transient vibration die out
within a matter of short time, living only the steady state vibration thus it is important to
know the steady state behaviour of system when subjected to different excitation frequencies.
By the behaviour of the system we mean its steady state amplitude and the phase lag. The
ratio of the steady state amplitude to the zero frequency deflection. i.e. X/Xs is defined as the
magnification factor and is denoted by M.F. it is the factor by which the zero frequency
deflection is to be multiplied to get the amplitude.
M.F. = X/Xs = 1/√ [1-(ω/ωn]2
+ [2ξ. ω/ωn]2
θ = tan-1
[2 ξ ω / ω n / (1 - (ω / ω n)2
)]

69
The dimensionless plot of magnification versus frequency ratio and phase lag versus
frequency ratio for different values of damping factor are shown in fig (1) & fig (2). The
curves of fig (1) drawn with magnification versus frequency ratio are known as frequency
response curves.
The curve between phase lag and frequency ratio is known as phase frequency
response curve. The following characteristics can be noted from the above equations and fig
(1) and fig (2).
Fig (1) Magnification factor v/s frequency ratio for different amounts of damping

70
i) At Zero frequency the magnification ratio is unity and damping does not have any effect on it
ii) Damping reduces the magnification ratio for all value of frequency
ii) Maximum value of amplitude occurs a little towards left of resonant frequency
iv) At resonance frequency the phase angle is 900
v) The phase angle increases for decreasing value of damping above resonance
vi) The amplitude of vibration is infinite at resonance frequency at zero damping factor
vii) The amplitude ratio is below unity for all values of damping which are more than 0.7
viii) The variation in phase angle is because of damping. Without damping it is either 1800
or 00
.
Fig (2) Phase lag v/s frequency ratio for different amounts of damping

71
Solution by Complex Algebra (Complex Frequency response)
The Harmonic force can be represented in complex form as,
F(t) = F0 (Cos ωt + i sin ωt)
= F0 eiωt
Hence the equation of motion is,
m x

 + c x
 + kx = F0 eiωt
The response is,
x = Xei(ωt-φ)
; x
 = i ω X ei(ωt-φ)
; x

 = (i ω)2
X ei(ωt-φ)
--------- (1)
Substituting these values in the above equation we get
mX (i ω)2
ei(ωt-φ)
+ c (i ω X) ei(ωt-φ)
+ kX ei(ωt-φ)
= F0 eiωt
i.e., (-m ω2
+ icω + k) X ei(ωt-φ)
= F0 eiωt
Therefore Xe –iφ
= F0 / (-m ω2
+ icω + k) = F0 / [(k -m ω2
) + icω]
Using equation (1) we obtain
x = F0 eiωt
/ [(k -m ω2
) + icω] (x = X ei(ωt-φ)
= X eiωt
e –iφ
Considering only the real part of the above equation for the solution of the system with harmonic
excitation
x = Re F0 eiωt
/ [(k -m ω2
) + icω]
It is know convenient to introduce the complex frequency response H (ω) defined as the output
divided by the input,
Therefore H (ω) = X e –iφ
/ F0 = 1 /[(k -m ω2
) + icω]
The frequency response from the above equation is,
H (ω) = xmax / x0 = 1 / √ (k -m ω2
)2
+ (cω)2
And φ = tan-1
(cω / k -m ω2
)

72
VIBRATION ISOLATION and TRANSMISSIBILITY
Vibration forces generated by machines and engines are often unavoidable and are
undesirable hence should be eliminated or atleast reduced for example, the inertia force
developed in a reciprocating engines or unbalanced forces produced in any rotating
machinery should be isolated from the foundation so that adjoining structure is not set into
heavy vibrations. The method by which vibrations can be reduced by the use of isolators is
called vibration isolation. Another example may be the isolation of delicate instruments from
their support which may be subjected to certain vibrations.
In either case the effectiveness of isolation may be measured in terms of the force or
motion transmitted to that in existance. The first type is known as force isolation and second
type is known as motion isolation. The lesser the force or motion transmitted, the greater is
said to be the isolation.
In this connection, we discuss the ways of reducing the transmitted vibrations by
vibration isolation which is obtained by placing properly chosen isolator materials between
the body and the supporting structure. The isolating materials may be pads of rubber, felt or
cork or metallic springs. All these isolating materials are elastic and possess damping
properties.
TRANSMISSIBILITY OR FORCE TRANSMISSIBILITY
The terms transmissibility or force transmissibility in the case of force excited system
is defined as “the ratio of the force transmitted to the foundation to that impressed upon the
system. Imagine a mass ‘m’ supported on the foundation by means of an isolator having an
equivalent stiffness and damping co-efficient k and c respectively and excited by a force Fo
sinωt as shown in fig(1) Under steady state conditions, force acting on the mass can be
represented by means of a vector diagram as shown in fig (2).

74
Out of these forces acting on the mass, spring force kx, the dashpot force cωx are two
common forces acting on the mass and also on the structure (foundation). Therefore the force
transmitted to the foundation is the vector sum of these two forces acting in the direction
opposite to that of the mass as shown in figure (3). These forces are 900
out of phase with
each other and their vector sum Ftr is the force transmitted to the foundation
Fo Ftr
cx
kx


Fig(3)
Ftr = √(kx)2
+ (cωx)2
= x√ k2
+ (cω)2
= x k√1+(cω/k)2
= x k√1+(2ξω/ωn)2
[cω/k = 2ξω/ωn]
But x = F0/k / √ [1-(ω/ωn)2
]2
+ (2ξω/ωn)2
Ftr = F0/k . k √ 1+ (2ξω/ωn)2
/ √ [1-(ω/ωn)2
] 2
+ (2ξω/ωn)2
Ftr/ F0 = √ 1+ (2ξω/ωn)2
/ √1-(ω/ωn)2
] 2
+ (2ξω/ωn)2
------------- (1)
Where Ftr/Fo = Є (transmissibility)

75
The angle through which the transmitted force lags the impressed force is given by (θ-α)
whose
α = tan-1
(cωx/kx) = tan-1
(2ξω/ωn)
φ = tan-1
[(2 ξω/ωn)/(1-(ω/ωn)2
]
(φ -α) = tan-1
[(2ξω/ωn) / (1-(ω/ωn)2
)] – tan-1
(2ξω/ωn) --------- (2)
Equation (1) & (2) gives the transmissibility and phase lag of transmitted force from the
impressed force and are plotted in fig (1) and (2) respectively with different damping factors.
ω/ωn
Fig (1)
Transmissibility v/s frequency ratio for different amounts of damping

76
Fig (2)
Phase angle v/s frequency ratio for different amounts of damping
1. The area of vibration isolation starts when transmissibility is less than unity and ω/ωn > √2.
Thus when the frequency ω of the exciting force is given, the isolation mounts can be
designed such that ω/ωn > √2
2. Damping is important especially at resonance otherwise transmissibility will be
excessively large.
3. In the vibration isolation region ω/ωn >√2, hence in transmissibility equation the absence
of damping can be written as
Є = 1/ (ω/ωn)2
– 1
4. Unity value of transmissibility occurs at two points where ω/ωn is zero and √2 for all
values of damping.
5. Under ideal operating condition transmissibility must be zero hence ω/ωn should be as
large as possible.

77
Show that providing damping in vibration isolation is not useful when the frequency
ratio ω/ωn is more than √2.
Consider the curve of transmissibility V/s frequency ratio ω/ωn as shown in fig.
ω/ωn
all the curves passes to a particular point and at that point the transmissibility Є = 1
we have
Є = √ 1+ (2ξω/ωn)2
/ √ [1-(ω/ωn)2
]2
+ (2ξω/ωn)2
Take Є = 1 in the above equation
1 = √ 1+ (2ξω/ωn)2
/ √ [1-(ω/ωn)2
]2
+ (2ξω/ωn)2
√ [1-(ω/ωn)2
]2
+ (2ξω/ωn)2
= √ 1+ (2ξω/ωn)2
[1-(ω/ωn)2
]2
+ (2ξω/ωn)2
= 1+ (2ξω/ωn)2

78
1+ (ω/ωn)4
- 2(ω/ωn)2
= 1
(ω/ωn)4
- 2(ω/ωn)2
= 0
(ω/ωn)2
[(ω/ωn)2
-2] = 0
Since (ω/ωn)2
# 0
(ω/ωn)2
-2 = 0
(ω/ωn)2
= 2
ω/ωn = √2
Hence it is shown that vibration isolation is possible in the range of ω/ωn > √2

79
PROBLEMS
1) A 1000kg machine is mounted on four identical springs of total shiftness K= 1.95x
107
N/m and having negligible damping. The machine is subjected to a harmonic
external force of magnitude of 490N and frequency of 180 rpm.
i) Determine the amplitude of machine.
ii) The maximum force transmitted to the foundation.
2) A refrigerator unit has a mass of 30kg is to be supported by 3 springs of shiftness ‘K’.
The unit operates at 580 rpm, what should be the spring constant k if only 10% of
shaking force of the unit to be transmitted to the supporting structure.
3) A machine of mass 200kg supported on spring of total shiftness 16000N/m has an
unbalanced rotating mass which results in a disturbing force of 800N at a speed of
3000rpm. Assuming the damping factor of 20% of critical damping. Determine
i) The amplitude of motion of machine due to unbalance
ii) The transmissibility
iii) Force transmitted to foundation.
4) A mass of 10kg is suspended from one end of a spiral spring, the other end of which
is fixed, the shiftness of spring is 4N/mm. damping causes the vibration to decrease
1/10 of its initial value in four complete oscillation. A periodic force of magnitude
150cos.25t is applied to the mass. Find the amplitude of forced oscillation. What
would be the amplitude if period of applied force coincides with the natural period of
vibration of the system?
5) A machine having a mass of 10kg and supported on a spring of total stiffness
778KN/m has unbalanced force of 392N at a speed of 3000rpm. Assume the damping
factor of 0.2. determine
a. The amplitude of motion of machine
b. Transmissibility
c. Force transmitted to foundation

80
6) A body of mass 70kg is suspended from a spring which deflects 20mm under the
load. It is subjected to a damping affect adjusted to a value 0.23 times that required
for critical damping. Find the natural frequency of the underdamped vibration and
damped vibration and ratio of successive amplitudes for the damped vibration.
If the body is subjected to the periodic force of 700N and of frequency equal to 0.78
time the natural undamped frequency. Find the amplitude of forced vibration and
phase angle with respect to the disturbance force.
7) A machine of mass one ton is acted upon by an external force of 2450N at frequency
of 1500 rpm. To reduce the effect of vibration isolators of rubber having static
deflection of 2mm under the machine load and an estimated damping factor ξ = 0.2
assumed. Determine
(a) The force transmitted to the foundation
(b) The amplitude of vibration of machine
(c) Phase angle.
8) A mass of 100kg is suspended on a spring having a scale of 19600N/m, and is acted
upon by a harmonic force of 39.2 N at the underdamped natural frequency. The
damping may be considered to be viscous with a co-efficient of 98 N-sec/m
Determine
(a) The under damped natural frequency
(b) The amplitude of the vibration of the mass, and
(c) The phase difference between the force and the displacement.
9) A huge machinery is mounted on a bed plate which is supported on four elastic
members, each having a stiffness of 3.92 x 106
N/m. the total mass to be supported is
1ton. It is estimated that the total damping force exerted on the system is 20% of the
critical and is of viscous nature.
When the speed of the rotation of the machine is 2000rpm, the amplitude of vertical
motion of the bed plate is 0.06mm, calculate the total maximum force transmitted
through each mounting to the foundation.

81
10) A compressor unit of mass 200kg is mounted rigidly on a concrete bed having a mass
of 500kg. The disturbing force whose frequency is the same as the compressor speed
and which is sinusoidal has maximum value of 294N. If the compressor speed is
1000rpm, determine the stiffness of rubber pads to be used beneath the concrete bed
such that the force transmitted is 0.5% of the disturbing force. Neglect damping.
11) A Flywheel of mass moment of inertia 0.1 Kgm2
is suspended from a thin wire of
stiffness 1.2 N-m/rad. A periodic torque having a maximum value of 0.6Nm at a
frequency of 4rad/sec is impressed upon the flywheel. A viscous dashpot applies
damping couple of 0.8Nm at an angular velocity of 2rad/sec. Determine
(i) Maximum angular displacement
(ii) Maximum couple applied to the dashpot
(iii) Critical damping co – efficient
(iv) Angle by which the angular displacement lags the torque

82
ROTATING UNBALANCE
Unbalance in rotating machine is a common source of vibration excitation. We
consider here a spring mass system constrained to move in vertical direction and excited by
rotating machine that is unbalance as shown in fig. The unbalance is represented by an
eccentric mass m, with the eccentricity e which is rotating with angular velocity ω rad/sec. let
x be the displacement of the non rotating mass (M-m). From the static equilibrium position
the displacement of the mass m is x + e sinωt.
The equation of motion is then (M-m) x

 + m d2
/dt2
(x + e sinωt) = - kx-c x

M x

 - m x

 + m x

 - me ω2
sin ωt = - kx-c x

M x

 + c x
 + kx = me ω2
sin ωt
This is similar to the general equation of motion for forced vibrations

83
M x

 + c x
 + kx = Fo sin ωt
Where F0 = me ω2
Hence the study state solution is
X = (FO/k) /√ [(1-(ω / ωn)2
]2
+ [2 ζ ω/ ωn]2
And the phase angle
tan Ф = (2 ζ ω/ ωn )/ [(1-( ω/ ωn)2
]
The above equation can be further reduced to the following form
X = (meω2
/k) / √ [(1-(ω/ ωn)2
]2
+ [2 ζ ω/ ωn]2
X = (meω2
/ωn
2
M) / √ [(1-(ω/ ωn)2
]2
+ [2 ζ ω/ ωn]2
MX /me = (ω2
/ωn
2
) / √ [(1-(ω/ ωn)2
]2
+ [2 ζ ω/ ωn]2
And the phase angle
tan Ф = (2 ζ ω/ ωn )/ [(1-( ω/ ωn)2
]

84
Problems on Rotating Mass
12) A machine of 75 kg is mounted on spring of shiftness 1182 N/m with a damping
factor of 0.2. A piston within the machine has a mass of 2kg and has a reciprocating
motion with a stroke of 76mm and a speed of 3000rpm. Assume the motion of piston
to be S.H.M. Determine
i) The amplitude of motion of machine
ii) Phase angle with respect to disturbing force
iii) The transmissibility
iv) Force transmitted to the foundation
v) Phase angle of the transmitted force with respect to distributing force.
13) A single cylinder vertical petrol engine of total mass of 300kg is mounted upon
a steel chasis frame and cases a vertical static deflection of 2mm. the reciprocating
parts of engine has a mass of 20kg and moves through a vertical stroke of 150mm
with SHM. A dashpot is provided whose damping resistance is directly proportional
to the velocity and amounts to 1500Ns/m. considering that steady state vibration is
reached. Determine 1) amplitude of vibration when driving shaft of engine rotates at
480 rpm. ii) The speed of driving shaft at which the resonance occurs.
14 A vibrating body of mass 150kg supported on springs of total stiffness 1MN/m
having rotating unbalance of 525N at a speed of 6000rev/min. if damping factor is
0.3. Determine
i) The amplitude caused by the unbalanced and its phase angle
ii) The transmissibility
iii) The actual force transmitted and its phase angle.

85
15) A small rotor driving a compressor weighs 300N causes each of the rubber isolation
to deflect by 4mm, the motor runs at a constant speed of 1800 rpm. The compressor
piston has a 50mm stroke. The piston and reciprocating parts weighs 5N and can be
assumed to perform SHM. Assume damping ratio for rubber to be 0.25. Determine
the amplitude of vertical motion at the operating speed.
16) A vibrating system as shown in figure is displaced for vibrational characteristics. The
total mass of the system is 25kg at a speed of 1000rpm. The system and eccentric
mass have a phase difference of 900
and the corresponding amplitude is 15mm.
The eccentric unbalance mass of 1kg has a radius of 40mm. Determine
a. The natural frequency of the system
b.The damping factor
c. The amplitude at 1500rpm
d. Phase angle at 1500rpm

86
17) A counterrotating eccentric mass exciter shown in figure is used to determine the
vibrational characteristics of a structure of mass of 181.4 kg, at a speed of 900 rpm. A
stroboscops shows the eccentric masses to be at the top at the instant the structure is
moving upward through its static equilibrium position and the corresponding
amplitude is 21.6mm.s if the unbalance of each wheel of the exciter is 0.0921kgs – m.
determine
i) The natural frequency of the structure
ii) The damping factor of the structure
iii) The amplitude at 1200 rpm
iv) The angular position of the eccentric at the instant the structure is moving upward
through its equilibrium position.

87
FORCED VIBRATI ON DUE TO EXCITATION OF SUPPORT
In case of locomotives or vehicles the wheels acts as base or support for the system.
The wheels can move vertically up and down during the motion of vehicle, at the same time
there is relative motion between the wheels and chasis. Hence chasis is having relative
motion to the wheels and wheels relative to road surface. The amplitude of vibration in case
of support motion depends on speed of vehicle and nature of road surface. In vibratory
system where the support is put to excitation absolute and relative motion becomes
important.
Amplitude motion or amplitude (motion or displacement transmissibility)
Consider a spring mass system as shown in fig. It can be noted that the support is not
fixed but is movable in y direction as shown in fig
Fo
m x
k c
y
m
..
mx
k(x-y)
. .
c(x-y)

88
If x is the absolute motion of the mass m then the equation of the motion can be written as
m x

 = -c ( x
 - y
 ) – k(x-y)
m x

 + c x
 + kx = c y
 + ky
The support is subjected to harmonic vibration y = Y sin ωt
y
 = Yω cos ωt
Substituting this in above equation we get
m x

 + c x
 + kx = k Y sin ωt + c Yω cos ωt
m x

 + c x
 + kx = Y √k2
+ cω2
[k/√k2
+ cω2
. sin ωt + c ω/√k2
+ cω2
. cos ωt]
m x

 + c x
 + kx = Y √k2
+ cω2
[cosα . sin ωt +. Sinα cos ωt]
Where cos α = k /√k2
+ cω2
and sin α = c ω/√k2
+ cω2
m x

 + c x
 + kx = Y √k2
+ cω2
sin (ωt + α)
Hence the study state solution can be written as follows
Fo = Y √k2
+ cω2
Fo = Y /k√1 + (cω /k)2
Fo = Y /k√1 + (2ζ ω/ωn)2
But x = F0/k / √ [1-(ω/ωn)2
+ (2ξω/ωn)2
x = y √ 1+ (2ξω/ωn)2
/ √ [1-(ω/ωn)2
+ (2ξω/ωn)2

89
x/y= √ 1+ (2ξω/ωn)2
/ √ [1-(ω/ωn)2
]2
+ (2ξω/ωn)2
And the phase angle
φ = tan-1
[2 ξω/ωn/(1-(ω/ωn)2
)]
α = tan-1
(cωx/kx) = tan-1
(2ξω/ωn)
(φ -α) = tan-1
[2 ξ(ω/ωn)/1-(ω/ωn)2
] – tan-1
(2ξω/ωn)
The ratio X/y is sometimes called as displacement transmissibility
Relative amplitude:
If z represents the relative motion of the mass with respect to support then we have
z = x – y or x = y + z
Substituting for x in the equation
m x

 + c x
 + kx = c y
 + ky
We get
m z

 + c z
 + kz = - m y


But we have y = Y sin ωt
y
 = Yω cos ωt
Substituting this in the above equation we get
m z

 + c z
 + kz = - mω2
Y sinωt
Hence the study state relative amplitude and the phase lag between the excitation and
the relative displacements are given by
Z/y = (ω/ωn)2
/ √ [1-(ω/ωn)2
]2
+ (2ξω/ωn)2
φ = tan-1
[2 ξω/ωn/1-(ω/ωn)2
]

90
Problems on support motion
18) Determine the relative amplitude of the end of the cantilever read with respect to
the base for the system shown in fig. The base is performing a harmonic motion
y = 0.8 sin 10t cms in a direction perpendicular to the read. The natural frequency
of the system is twice the disturbing frequency.
19) A 3kg mass is suspended in a box by a spring as shown in fig. the box is put on a
platform having vibration y = 0.8 sin 6t cms. Determine the absolute amplitude of
mass given k = 6000 N/m.
k
m

91
20) A mass of 1.93 kg is suspended in a box by vertical spring whose constant
K = 100N/m. the box is placed on the top of shave table producing a vibration
x = 0.09 sin 8t. Find the absolute amplitude of the mass assume no damping.
21) A mass of 1.75 kg is attached to a vertical spring of stiffness 9000N/m. The spring
mass is enclosed in a box and kept on a vibrating table which produces a harmonic
vibration of x = 2.25 sin 3t where x is in mm. Find the absolute amplitude of the
mass.
22) The support of a spring mass system is vibrating with an amplitude of 5mm and a
frequency of 1150cps. If the mass is 0.9kg and the spring has a stiffness of 1960N/m.
determine the amplitude of vibration of mass. What amplitude will result if a damping
factor of 0.2 is included in a system.
23) The spring of an automobile trailer are compressed by 0.1m under its own weight .
Find the critical speed when the trailer is traveling over a road. With a profile
approximated by a sine wave of amplitude 0.08m and wave length 0f 14 m . What
will be the amplitude of vibration at 60km/hr.
24) An automobile trailer which moves over the road surface making approximately
sinusoidal profile with a wave length of 8m and amplitude of 0.06m. The trailer is
pulled on the road surface with a velocity of 60km/hr. calculate the critical speed of
trailer if the vibration amplitude is 15mm. for the trailer mass of 50kg.
25) A vibrating body is supported by six isolators each having stiffness 32000 N/m and
six dash pots having damping factor as 400N s/m. The vibrating body is to be isolated
by a rotating device having an amplitude of 0.06mm at 600 rpm take m = 30kg.
Determine i) amplitude of vibration of the body ii) dynamic load on each isolator due
to vibration.

92
26) A radioset of 20kg mass must be isolated from a machine vibrating with amplitude of
0.5 mm at 500 rpm. The set is mounted on four isolators each have a spring scale of
31400N/m and damping factor of 392N-s/m.
e. what is the amplitude of vibration of radio
f. What is the dynamic load in each isolator due to vibration?
27) An instrument of mass 50kg is located in an aeroplane cabin which vibrates at 2000
c.p.m with amplitude of 0.1mm. Determine the stiffness of four steel spring required
as supports for the instrument to reduce its amplitude to 0.005mm also calculate the
maximum total load for which each spring must.

93
VIBRATION MEASURING INSTRUMENTS
The primary purpose of the vibration measuring instrument is to give an output signal which
represents, as closely as possible, the vibration phenomenon. This phenomenon may be
displacement, velocity or acceleration of the vibrating system and accordingly the
instruments which reproduce signals proportional to these are called vibrometers (or
vibration pick-ups), velocity pick-ups or accelerometers.
Seismic Instrument
Fig shows the schematic of a seismic instrument which is used to measure any of the
vibration phenomenon. It consists of a frame in which the seismic mass m is supported by
means of spring k and dashpot c. the frame is mounted on a vibrating body and vibrates along
with it. The system reduces to a spring mass dashpot system having base or support
excitation. Consider the vibrating body (base) to have a sinusoidal motion
y = Y sin ωt
Fig Seismic instrument

94
Then the steady state relative amplitude Z of the seismic mass with respect to the frame can
be obtained. And the phase difference between the exciting motion and the relative motion
can also be obtained.
Imagine that a scale is fixed on the frame and a pointer on the seismic mass. Then the
amplitude of the motion of the mass over the scale represents the relative motion z having an
amplitude Z. this motion is also harmonic.
Frequency measuring instruments.
One of the methods of measuring the frequency of vibration of a system is by means of
Frahm’s Read Tachometer. It consists of a system having a number of reeds fixed over it in
the form of cantilevers carrying small masses at their free ends as shown in fig. The natural
frequencies of the set of these reeds is adjusted to give a definite series of known frequencies.
When this instrument is attached to the body where frequency of vibration is to be measured,
the reed whose natural frequency is nearest to the excitation frequency vibrates near resonant
condition and has a large amplitude of vibration. The frequency of vibrating body is then
given by the known frequency of reed vibrating with maximum amplitude.

95
Fig Frahm’s reed tachometer
The accuracy of the instruments depends upon the difference between the natural frequencies
of the successive reeds. The smaller the difference, more accurate is the instrument and vice-
versa. Of course, with a more accurate instrument of this type, the range of frequencies that
can be measured will be smaller.
Seismometer – instrument with low natural frequency.
When the natural frequency ωn of the instrument is low in comparison to the vibration
frequency ω to be measured, the ratio ω/ωn approaches a large number, and the relative
displacement Z approaches Y regardless of the value of damping ζ, as indicated in fig,
The mass m then remains stationery while the supporting case moves with the vibrating
body. Such instruments are called seismometers.

96
One of the disadvantages of seismometers is its large size. Since Z = Y, the relative
motion of the seismic mass must be of the same order of magnitude as that of the vibration to
be measured.
The relative motion z is usually converted to an electric voltage by making the
seismic mass a magnet moving relative to coils fixed in the case, as shown in fig. since the
voltage generated is proportional to the rate of cutting of the magnetic field, the output of the
instrument will be proportional to the velocity of the vibrating body. Such instruments are
called velometers. A typical instrument of this kind may have a natural frequency between
1 Hz to 5Hz and a useful frequency range of 10Hz to 2000Hz. The sensitivity of such
instruments may be in the range of 20 mV/cm/s to 350 mV/cm/s, with the maximum
displacement limited to about 0.5 cm peak to peak.
Both the displacement and acceleration are available from the velocity type
transducer by means of integrator or the differentiator provided in most signal conditioner
units.

97
Accelerometer – instrument with high natural frequency.
When the natural frequency of the instrument is high compared to that of the vibration to be
measured, the instrument indicates acceleration. Examination of equation
Z = y (ω/ωn)2
/√ [1-(ω/ωn)2
+ (2ξω/ωn)2
shows the factor.
√ [1-(ω/ωn)2
+ (2ξω/ωn)2
approaches unity for ω/ωn 0.
So that Z = ω2
y/ωn
2
= acceleration / ωn
2
Thus Z becomes proportional to the acceleration of the motion to be measured with a factor
1/ωn
2
.
Several different accelerometers are in use today. The seismic mass accelerometer is
often used for low-frequency vibration, and the supporting springs may be four electric strain
gauge wires connected in a bridge circuit. A more accurate variation of this accelerometer is
one in which the seismic mass is servo-controlled to have zero relative displacement; the
force necessary to accomplish this becomes a measure of the acceleration. Both of these
instruments require an external source of electric power.

98
The piezoelectric properties of crystals like quartz or barium titanate are utilized in
accelerometers for higher frequency measurements. The crystals are mounted so that under
acceleration they are either compressed or bent to generate an electric charge. Figure shows
one such arrangement. The natural frequency of such accelerometers can be made very high,
in the 50,000-Hz range, which enables acceleration measurements to be made upto 3000 Hz.
The size of the crystal accelerometer is very small, approximately 10mm in diameter and
height.
Problems
1) A vibrometer has a period of free vibration of 2 seconds. It is attached to a machine
with a vertical harmonic frequency of 1Hz. If the vibrometer mass has amplitude of
2.5 mm relative to the vibrometer frame what is the amplitude of vibration of
machine?
2) A commercial type vibration pick up has a natural frequency of 5.75 Hz and a
damping factor of 0.65. What is lowest frequency beyond which the amplitude can be
measured within one percent error.

99
3) A vibrometer is used to fixed the displacement, velocity and acceleration of machine
running at 120 rpm. If the natural frequency of the instrument is 5 Hz and it shows
0.004cm. What are the three readings .Assume no damping.
4) A seismic instrument with a natural frequency of 6 Hz is used to measure the
vibration of a machine operating at 120 rpm. The relative displacement of the seismic
mass as read from the instrument is 0.05 mm. determine the amplitude of machine.
Neglect damping.
5) A device used to measure torsional acceleration consists of a ring having a moment of
inertia of 0.049 kg-m2
connected to a shaft by a spiral spring having a scale of 0.98
N-m/rad, and a viscous damper having a constant of 0.11 N-m/rad,
When the shaft vibrates with a frequency of 15 cpm the relative amplitude between
the ring and the shaft is found to be 20
. What is the maximum acceleration of the
shaft?
6) An instrument for measuring acceleration records 30 oscillation/sec. the natural
frequency of the system is 800 cycles/sec. what is the acceleration of the machine part
to which the instrument is attached if the amplitude recorded is 0.02mm? What is the
amplitude of vibration of the machine part?
7) A vibrometer with natural frequency of 2 Hz and with negligible damping is attached
to a vibrating system which performs a harmonic motion. Assuming the difference
between the maximum and minimum recorded values as 0.6 mm, determine the
amplitude of motion of the vibrating system when its frequency is (i) 20 Hz (ii) 4 Hz.

100
Critical or Whirling or Whipping speed of shafts.
In actual practice, a rotating shaft carries different mounting and accessories in the
form of gears, pulley etc. this mounting creates centrifugal forces in the shaft during rotation
there by producing an eccentricity between the centre of gravity of the mountings and the
axis of rotation. This process is cumulative and ultimately the shaft fail at high speed due to
high centrifugal forces.
The speed, at which the shaft tends to vibrate in transverse direction so that additional
deflection of the shaft from the axis of rotation becomes infinite, is known as critical or
whipping or whirling speed of a shaft.
To prove that the whirling speed of a rotating shaft same as the transverse vibrations
Consider a shaft carrying a disk, rotating about the axis of rotation at a uniform speed of ω
rad/sec as shown in fig.
Let m = mass of the disk in kg
e = initial distance of centre of gravity of the disk from the axis of rotation
r = additional deflection of centre of gravity of disk when the shafts starts rotating.
K= stiffness of the shaft

101
The centrifugal force acting radially outward and causes deflection, is given by
Fc = m ω2
(r+ e)
Again, the force resisting the deflection = k r
For equilibrium the centrifugal force must be equal to the resisting force
m ω2
(r + e) = k r
m ω2
r + m ω2
e = k r

102
k r - m ω2
r = m ω2
e
r (k- mω2
) = m ω2
e
Or r = m ω2
e/(k- mω2
)
= ω2
e/(k/m - ω2
)
But ωn = √k/m
r = ω2
e/( ωn
2
- ω2
)
From the above expression it is evident that when ω > ωn the value of r is – ve. Hence to have
the value of r always +ve we take
r = + ω2
e/(ωn
2
- ω2
)
r = + (ω/ ω n )2
e / [1 – (ω/ ωn) 2
]
Or r = + e/ [(ωn / ω)2
– 1]
Replace ωn by ωc
r = + e / [(ωc / ω)2
– 1]
From the above expression it can be seen that when ω = ωc the value of r becomes infinite.
At this speed resonance will occur the natural frequency of vibration ωn is equal to the speed
of the shaft. i.e ωn = ω
The critical speed ωc = ω = ωn = √k/m

103
if nc is the critical speed in r.p.s
2 π nc = √k/m
nc = 1/2 π √k/m
nc = fn ---------- Natural frequency of transverse vibration
Hence the critical speed is the same as the natural frequency of transverse vibration but its
units will be r.p.s.
The above equation r = + (ω/ ω n )2
e / [1 – (ω/ ωn) 2
] is the equation for amplitude
of vibration when damping is not considered. If damping is taken into account the
equation reduces to r = + (ω/ ω n )2
e / √[1 –(ω/ ωn) 2
]2
+ (2ζ ω/ ωn)2

104
Problems on Whirling or Critical speed of shaft
1) A rotor of mass 12kg is mounted in the middle of 25mm diameter shaft supported
between two bearings placed at 900mm from each other. The rotor is having 0.02mm
eccentricity. If the system rotates at 3000 rpm. Determine the amplitude of steady state
vibrations and the dynamic force on the bearings take E = 2 x 1011
N/m2
.
2) A disc of mass 4kg is mounted midway between bearings which may be assumed to be
simple supports. The beam of span is 480 mm. the steel shaft which is horizontal is 9mm
in diameter. The centre of gravity of the disc is displaced 3mm from the geometric centre.
The equivalent viscous damping at the centre of the disc shaft may be taken as 49-Ns/m.
if the shaft rotates at 760 rpm, find the Maximum stress in the shaft and compare it with
dead load stress in the shaft. Also find the power required to drive the shaft.

105
3) A vertical shaft 12.5 mm in diameter rotates in spherical bearings with a span of 0.9m
and carries a disc of mass 10kg midway between the two bearings. The mass centre of the
disc is 0.25 mm away from the geometric axis. If the stress in the shaft is not to exceed
10.3 x 107
N/m2
. Determine the range of speed within which it is unsafe to run the shaft.
Neglect the mass of the shaft and the damping in the system.
4) A rotor of weight 10kg is mounted mid way on a 20mm diameter horizontal shaft
supported at the ends by two bearings. The bearing span is 800mm because of certain
manufacturing defects; the centre of gravity of the disc is 0.1 mm away from the
geometric centre of gravity of the rotor. If the system rotates at 3000rpm determine, the
amplitude of the steady state vibration and the dynamic force transmitted to the bearing.
E = 2 x 106
kgs/cm2
.
5) The following data relate to a shaft held in long bearings
Length of shaft = 1.2m
Diameter of shaft = 14mm
Mass of a rotor at midpoint = 16kg
Eccentricity of centre of mass of rotor from centre of rotor = 0.4mm
Modulus of elasticity of shaft material = 200GN/m2
Permissible stress in shaft material = 70 x 106
N/m2
Determine the critical speed of the shaft and the range of speed over which it is unsafe to
run the shaft. Neglect the mass of shaft.

106
INFLUENCE COEFFICIENTS
An influence coefficient, denoted by α12, is defined as the static deflection of the
system at position 1 due to a unit force applied at position 2 when the unit force is the only
force acting. The influence coefficient is therefore a convenient method to keep account of
all the induced deflections due to various applied forces, and to set up the differential
equations of motion for the system.
It can be shown that the following expression is true:
αij = αji
Where
αij = deflection at position i due to a unit force applied at position j,
αji = deflection at position j due to a unit force applied at position i,
This is Maxwell’s reciprocal theorem.
An influence co efficient αij, on the other hand, can be interpreted to mean the
angular displacement at coordinate i due to a unit torque applied at coordinate j for rotational
motions.
MATRIX ITERATION
This is an iterative procedure that leads to the principal modes of vibration of a
system and its natural frequencies.
Displacements of the masses are estimated, from which the matrix equation of the
system is written. The influence coefficients of the system are substituted into the matrix
equation which is then expanded. Normalization of the displacement and expansion of the
matrix is repeated. This process is continued until the first mode repeat itself to any desired
degree of accuracy.
For the next higher modes and natural frequencies, the orthogonality principle is used
to obtain a new matrix equation that is free from any lower modes the iterative procedure is
repeated.

107
THE STODOLA METHOD
The stodola method may be set up in the following tabular form as follows, assuming
an arbitrary set of values for the fundamental principal mode, the inertia force acting on each
mass is equal to the product of the assumed deflection and the square of the natural frequency
as shown in row 2. The spring force in row 3 is equal to the total inertia force acting on each
spring. Row 4 is obtained by dividing row 3, term by term, by their respective spring
constants. The calculated deflections in row 5 are found by adding the deflections due to the
springs, with the mass near the fixed end having the least deflection and so on. The
calculated deflections are then compared with the assumed deflections. This process is
continued until the calculated deflections are proportional to the assumed deflections. When
this is true the assumed deflections will represent the configuration of the fundamental
principal mode of vibration of the system.
k1 m1 k2 m2 K3 M3
1.Assumed deflection
2.Inertia force
3.Spring force
4.Spring deflection
5.Calculated deflection

108
THE HOLZER METHOD
Begin the Holzer tabulation with the column of position, indicating the masses of the
system. The second column is for the values of the different masses of the system; this
information is given. The third column is the product of mass and frequency squared.
Displacement comes next, and is obtained from the preceding row minus the total
displacement at the end of the same row. Column five is just the product of columns three
and four. The total inertia force is inserted in column six. It is equal to the sum of the total
inertia force in the preceding row plus the inertia force on the same row. The rest are plainly
evident.
An initial displacement, usually equal to unity for convenience, is assumed. If the
assumed frequency happens to be one of the natural frequencies of the systems, the final total
inertia force on the system should be zero. This is because the system is having free
vibration. If the final total inertia force is not equal to zero, the amount indicates the
discrepancy of the assumed frequency.
Position m1 m1ω2
xi m1xiω2
i
∑mixiω2
1
kij
i
∑mixiω2/ kiij
1
Assumed frequency, ω =

109
THE DUNKERLEY’S METHOD
The Dunkerley’s equation for multi dot system is given by
1/ ωn
2
=1/ ω1
2
+ 1/ ω2
2
+ ω3
3
+ …………..
Where ωn = fundamental natural frequency of the system.
ω1, ω2, ω3 etc are the natural frequency of the system with each mass acting separately at its point of
application, in the absence of other masses, using influence co-efficients Prof. Dunkerley has
suggested the following empirical equation.
1/ ωn
2
= α11m1 + α22m2 + α33m3 + …………..
Dunkerley’s equation is used to determine the fundamental natural frequency of the system.
RAYLEIGH’S METHOD
This is the energy method to find the frequency. This method is used to find the natural
frequency of the system when transverse point loads are acting on the beam or shaft. Good estimate
of fundamental frequency can be made by assuming the suitable deflection curve for the fundamental

110
mode. The maximum kinetic energy is equated to maximum potential energy of the system to
determine the natural frequency.
ωn = √ g ∑ mi yi / ∑ mi yi
2
where,
y1= m1 g α11 + m2 g α12
y2 = m1 g α12 + m2 g α22

Mechanical vibration note

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Mechanical vibration note

Similar to Mechanical vibration note (20)

More from Mohammed Imran

More from Mohammed Imran (20)

Recently uploaded

Recently uploaded (20)

Mechanical vibration note