UNIT-5 VIBRATION NOTES- FREE -LONGITUDINAL VIBRATION

4(ii) Derive an expression for the natural frequency of single degrees of freedom system.
We know that the kinetic energy is due to the motion of the body and the potential energy
is with respect to a certain datum position which is equal to the amount of work required to move
the body from the datum position. In the case of vibrations, the datum position is the mean or
equilibrium position at which the potential energy of the body or the system is zero.
In the free vibrations, no energy is transferred to the system or from the system.
Therefore the summation of kinetic energy and potential energy must be a constant quantity
which is same at all the times. In other words,
We know that kinetic energy,
Comparing equations,
A vertical shaft of 5 mm diameter is 200 mm long and is supported in long bearings at its
ends. A disc of mass 50 kg is attached to the centre of the shaft. Neglecting any increase in
stiffness due to the attachment of the disc to the shaft, find the critical speed of rotation

and the maximum bending stress when the shaft is rotating at 75% of the critical speed.
The centre of the disc is 0.25 mm from the geometric axis of the shaft. E = 200 GN/m
2
.
Solution. Given : d = 5 mm = 0.005 m ; l = 200 mm = 0.2 m ; m = 50 kg ; e = 0.25 mm =
0.25 × 10–3
m ; E = 200 GN/m2
= 200 × 109
N/m2
Critical speed of rotation
We know that moment of inertia of the shaft,
Since the shaft is supported in long bearings, it is assumed to be fixed at both ends. We
know that the static deflection at the centre of the shaft due to a mass of 50 kg,
Maximum bending stress
Let σ = Maximum bending stress in N/m2, and
N = Speed of the shaft = 75% of critical speed = 0.75 Nc . . . (Given)
When the shaft starts rotating, the additional dynamic load (W1) to which the shaft is
subjected, may be obtained by using the bending equation,
We know that for a shaft fixed at both ends and carrying a point load (W1) at the centre,
the maximum bending moment

6.(i) A shaft 50 mm diameter and 3 metres long is simply supported at the ends and
carries three loads of 1000 N, 1500 N and 750 N at 1 m, 2 m and 2.5 m from the left
support. The Young's modulus for shaft material is 200 GN/m
2
. Find the frequency
of transverse vibration.
Solution. Given :2
= 200 × d = 50 mm = 0.05 m ;109
N/m2
l = 3 m, W1 = 1000 N ; W2 = 1500
N ; W3 = 750 N;
E = 200 GN/m
The shaft carrying the loads is shown in Fig. 23.13
We know that moment of inertia of the shaft,

6.(ii) Calculate the whirling speed of a shaft 20 mm diameter and 0.6 m long
carrying a mass of 1 kg at its mid-point. The density of the shaft material is 40 Mg/m
3
,
and Young’s modulus is 200 GN/m
2
. Assume the shaft to be freely supported.

FORCED VIBRATION
4.1 INTRODUCTION:
When a system is subjected continuously to time varying disturbances, the
vibrations resulting under the presence of the external disturbance are referred to as forced
vibrations.
Forced vibration is when an alternating force or motion is applied to
a mechanical system. Examples of this type of vibration include a shaking washing machine
due to an imbalance, transportation vibration (caused by truck engine, springs, road, etc), or
the vibration of a building during an earthquake. In forced vibration the frequency of the
vibration is the frequency of the force or motion applied, with order of magnitude being
dependent on the actual mechanical system.
When a vehicle moves on a rough road, it is continuously subjected to road
undulations causing the system to vibrate (pitch, bounce, roll etc). Thus the automobile is
said to undergo forced vibrations. Similarly whenever the engine is turned on, there is a
resultant residual unbalance force that is transmitted to the chassis of the vehicle through the
engine mounts, causing again forced vibrations of the vehicle on its chassis. A building
when subjected to time varying ground motion (earthquake) or wind loads, undergoes forced
vibrations. Thus most of the practical examples of vibrations are indeed forced vibrations.
4.2 CAUSES RESONANCE:
Resonance is simple to understand if you view the spring and mass as energy storage
elements – with the mass storing kinetic energy and the spring storing potential energy. As
discussed earlier, when the mass and spring have no force acting on them they transfer energy
back and forth at a rate equal to the natural frequency. In other words, if energy is to be
efficiently pumped into both the mass and spring the energy source needs to feed the energy in at
a rate equal to the natural frequency. Applying a force to the mass and spring is similar to
pushing a child on swing, you need to push at the correct moment if you want the swing to get
higher and higher. As in the case of the swing, the force applied does not necessarily have to be
high to get large motions; the pushes just need to keep adding energy into the system.
The damper, instead of storing energy, dissipates energy. Since the damping force
is proportional to the velocity, the more the motion, the more the damper dissipates the
energy. Therefore a point will come when the energy dissipated by the damper will equal the
energy being fed in by the force. At this point, the system has reached its maximum
amplitude and will continue to vibrate at this level as long as the force applied stays the
same. If no damping exists, there is nothing to dissipate the energy and therefore
theoretically the motion will continue to grow on into infinity.

4.3 FORCED VIBRATION OF A SINGLE DEGREE-OF-FREEDOM SYSTEM:
We saw that when a system is given an initial input of energy, either in the
form of an initial displacement or an initial velocity, and then released it will, under the right
conditions, vibrate freely. If there is damping in the system, then the oscillations die away. If
a system is given a continuous input of energy in the form of a continuously applied force or
a continuously applied displacement, then the consequent vibration is called forced
vibration. The energy input can overcome that dissipated by damping mechanisms and the
oscillations are sustained.
We will consider two types of forced vibration. The first is where the ground to
which the system is attached is itself undergoing a periodic displacement, such as the vibration
of a building in an earthquake. The second is where a periodic force is applied to the mass, or
object performing the motion; an example might be the forces exerted on the body of a car by
the forces produced in the engine. The simplest form of periodic force or displacement is
sinusoidal, so we will begin by considering forced vibration due to sinusoidal motion of the
ground. In all real systems, energy will be dissipated, i.e. the system will be damped, but often
the damping is very small. So let us first analyze systems in which there is no damping.
4.4 STEADY STATE RESPONSE DUE TO HARMONIC OSCILLATION:
Consider a spring-mass-damper system as shown in figure 4.1. The equation of
motion of this system subjected to a harmonic force can be given by
(4.1)
where, m , k and c are the mass, spring stiffness and damping coefficient of the system, F is the
amplitude of the force, w is the excitation frequency or driving frequency.
Figure 4.1 Harmonically excited system

Figure 4.2: Force polygon
The steady state response of the system can be determined by solving equation(4.1) in many
different ways. Here a simpler graphical method is used which will give physical
understanding to this dynamic problem. From solution of differential equations it is known
that the steady state solution (particular integral) will be of the form
(4.2)
As each term of equation (4.1) represents a forcing term viz., first, second and third terms,
represent the inertia force, spring force, and the damping forces. The term in the right hand
side of equation (4.1) is the applied force. One may draw a close polygon as shown in figure
4.2 considering the equilibrium of the system under the action of these forces. Considering a
reference line these forces can be presented as follows.
Spring force = (This force will make an angle with the
reference line, represented by line OA).
Damping force = (This force will be perpendicular to the spring
force, represented by line AB).
Inertia force = (this force is perpendicular to the damping force
and is in opposite direction with the spring force and is represented by line BC) .
Applied force = which can be drawn at an angle with respect to the reference
line and is represented by line OC.
From equation (1), the resultant of the spring force, damping force and the inertia force will be the
applied force, which is clearly shown in figure 4.2.
It may be noted that till now, we don't know about the magnitude of X and which can be easily computed
from
Figure 2. Drawing a line CD parallel to AB, from the triangle OCD of Figure 2,

From the previous module of free-vibration it may
be recalled that
H
• Natural frequency
• Critical
damping
• Damping factor or damping ratio
ence,
or
As the ratio is the static deflection of the spring, is known
as the magnification factor or amplitude ratio of the system 4.5 FORCED
VIBRATION WITH DAMPING:

In this section we will see the behaviour of the spring mass damper model when we add
a harmonic force in the form below. A force of this type could, for example, be generated by a
rotating imbalance.
If we again sum the forces on the mass we get the following ordinary differential equation:
The steady state solution of this problem can be written as:
The result states that the mass will oscillate at the same frequency, f, of the applied force, but
with a phase shift φ.
The amplitude of the vibration ―X‖ is defined by the following formula.
Where ―r‖ is defined as the ratio of the harmonic force frequency over the
undamped
natural frequency of the mass–spring–damper model.
The phase shift , φ, is defined by the following formula.
The plot of these functions, called "the frequency response of the system", presents one of the most
important features in forced vibration. In a lightly damped system when the forcing frequency
nears the natural frequency ( ) the amplitude of the vibration can get extremely high. This
phenomenon is called resonance (subsequently the natural frequency of a system is often referred

to as the resonant frequency). In rotor bearing systems any rotational speed that excites a resonant
frequency is referred to as a critical speed.
If resonance occurs in a mechanical system it can be very harmful – leading to
eventual failure of the system. Consequently, one of the major reasons for vibration analysis is to
predict when this type of resonance may occur and then to determine what steps to take to prevent
it from occurring. As the amplitude plot shows, adding damping can significantly reduce the
magnitude of the vibration. Also, the magnitude can be reduced if the natural frequency can be
shifted away from the forcing frequency by changing the stiffness or mass of the system. If the
system cannot be changed, perhaps the forcing frequency can be shifted (for example, changing
the speed of the machine generating the force).
The following are some other points in regards to the forced vibration shown in the frequency response
plots.
At a given frequency ratio, the amplitude of the vibration, X, is directly proportional to the
amplitude of the force F0 (e.g. if you double the force, the vibration doubles)
With little or no damping, the vibration is in phase with the forcing frequency when the frequency ratio r <
1 and 180 degrees out of phase when the frequency ratio r > 1
When r ≪ 1 the amplitude is just the deflection of the spring under the static force F0. This
deflection is called the static deflection δst. Hence, when r ≪ 1 the effects of the damper and the mass are minimal.
When r ≫ 1 the amplitude of the vibration is actually less than the static deflection δst. In this region the force generated by the mass (F = ma)
is dominating because the acceleration seen by the mass increases with the frequency. Since the deflection seen in the spring, X, is reduced in
this region, the force transmitted by the spring (F = kx) to the base is reduced. Therefore the mass–
spring–damper system is isolating the harmonic force from the mounting base – referred to as vibration isolation. Interestingly, more
damping actually reduces the effects of vibration isolation when r ≫ 1 because the damping force (F = cv) is also transmitted to the base.
4.6 ROTATING UNBALANCE FORCED VIBRATION:
One may find many rotating systems in industrial applications. The unbalanced
force in such a system can be represented by a mass m with eccentricity e , which is rotating with angular
velocity as shown in Figure 4.1.

Figure 4.1 : Vibrating system with rotating unbalance
Figure 4.2. Freebody diagram of the system
Let x be the displacement of the nonrotating mass (M-m) from the static equilibrium
position, then the displacement of the rotating mass m is
From the freebody diagram of the system shown in figure 4.2, the equation of motion is
(4.1)
or (4.2)
This equation is same as equation (1) where F is replaced by . So from the force polygon
as shown in figure 4.3

(4.3)
or (4.4)
(4.5)
Figure 4.3: Force polygon
(4.6)
and (4.7)
So the complete solution becomes
(4.8)
4.7 VIBRATION ISOLATION AND TRANSMISSIBILITY:
When a machine is operating, it is subjected to several time varying forces because of which it

tends to exhibit vibrations. In the process, some of these forces are transmitted to the foundation – which
could undermine the life of the foundation and also affect the operation of any other machine on the same
foundation. Hence it is of interest to minimize this force transmission. Similarly when a system is subjected to
ground motion, part of the ground motion is transmitted to the system as we just discussed e.g., an automobile
going on an uneven road; an instrument mounted on the vibrating surface of an aircraft etc. In these cases, we
wish to minimize the motion transmitted from the ground to the system. Such considerations are used in the
design of machine foundations and in order to understand some of the basic issues involved, we will study
this problem based on the single d.o.f model discussed so far. we get the expression for force transmitted to
the base as follows:
4.7.1 Vibration Isolators:
Consider a vibrating machine; bolted to a rigid floor (Figure 2a).The force transmitted to the
floor is equal to the force generated in the machine. The transmitted force can be decreased by adding
a suspension and damping elements (often called vibration isolators) Figure 2b , or by adding what is
called an inertia block, a large mass (usually a block of cast concrete), directly attached to the
machine (Figure 2c).Another option is to add an additional level of mass (sometimes called a seismic
mass, againa block ofcast concrete) and suspension(Figure2d).
Figure 2.Vibration isolation systems: a) Machine bolted to a rigid foundation
Supported on isolation springs, rigid foundation c) machine attached to an inertial block. d)
Supported on isolation springs, non-rigid foundation (such as a floor); or machine on isolation
springs, seismic mass and second level of isolator springs
When oscillatory forces arise unavoidably in machines it is usually desired to

prevent these forces from being transmitted to the surroundings. For example, some
unbalanced forces are inevitable in a car engine, and it is uncomfortable if these are
wholly transmitted to the car body. The usual solution is to mount the source of vibration
on sprung supports. Vibration isolation is measured in terms of the motion or force
transmitted to the foundation. The lesser the force or motion transmitted the greater the
vibration isolation
Suppose that the foundation is effectively rigid and that only one direction of
movement is effectively excited so that the system can be treated as having only one degree of
freedom.
4.8 RESPONSE WITHOUT DAMPING:
The amplitude of the force transmitted to the foundations is Where k is the Stiffness
of the support and x(t) is the displacement of the mass m.
The governing equation can be determined by considering that the total forcing on the machine
is equal to its mass multiplied by itsacceleration(Newton’s second law)

4.9 SOLVED PROBLEMS
Derive the relation for the displacement of mass from the equilibrium position of
the damped vibration system with harmonic forcing.
Consider a system consisting of spring, mass and damper as shown in Fig. 23.19. Let
the system is acted upon by an external periodic (i.e. simple harmonic) disturbing force,
Fx F cos ω .t
where F = Static force, and
= Angular velocity of the periodic disturbing force.
When the system is constrained to move in vertical guides, it has only one degree of
freedom. Let at sometime t, the mass is displaced downwards through a distance x from
its mean position.
The equation of motion may be written as,
This equation of motion may be solved either by differential equation method or by
graphi-cal method as discussed below :
Differential equation method
The equation (i) is a differential equation of the second degree whose right hand side is
some function in t. The solution of such type of differential equation consists of two parts ; one
part is the complementary function and the second is particular integral. Therefore the solution
may be written as
x = x1 + x2
where x1 = Complementary function, and x2 = Particular integral.
The complementary function is same as discussed in the previous article, i.e.

x1 Ce−
at
cos (ω d t− θ ) . . . (ii) where C and θ are constants. Let
us now find the value of particular integral as discussed below :
Let the particular integral of equation (i) is given by

UNIT-5 VIBRATION NOTES- FREE -LONGITUDINAL VIBRATION

A mass of 10 kg is suspended from one end of a helical spring, the other end being
fixed. The stiffness of the spring is 10 N/mm. The viscous damping causes the
amplitude to decrease to one-tenth of the initial value in four complete oscillations. If a
periodic force of 150 cos 50 t N is applied at the mass in the vertical direction, find the
amplitude of the forced vibrations. What is its value of resonance ?

The mass of an electric motor is 120 kg and it runs at 1500 r.p.m. The armature mass
is 35 kg and its C.G. lies 0.5 mm from the axis of rotation. The motor is mounted on
five springs o f negligible damping so that the force transmitted is one -eleventh of
the impressed force. Assume that the mass of the motor is equally distributed among
the five springs.
Determine : 1. stiffness of each spring; 2. dynamic force transmitted to the ba se at the
operating speed; and 3. natural frequency of the system.

What do you understand by transmissibility? Describe the method of finding the
transmissibility ratio from unbalanced machine supported with foundation.
A little consideration will show that when an unbalanced machine is installed on the
foundation, it produces vibration in the foundation. In order to prevent these vibrations or to
minimize the transmission of forces to the foundation, the machines are mounted on springs
and dampers or on some vibration isolating material, as shown in Fig. 23.22. The
arrangement is assumed to have one degree of freedom, i.e. it can move up and down only.
It may be noted that when a periodic (i.e. simple harmonic) disturbing force F cos ω t
is applied to a machine of mass m supported by a spring of stiffness s, then the force is
transmitted by means of the spring and the damper or dashpot to the fixed support or
foundation.
The ratio of the force transmitted (FT) to the force applied (F) is known as the isolation
factor or transmissibility ratio of the spring support.
We have discussed above that the force transmitted to the foundation consists of the fol-
lowing two forces :
1.Spring force or elastic force which is equal to s. xmax, and
2.Damping force which is equal to c. ω .xmax.
Since these two forces are perpendicular to one another, as shown in Fig.23.23,
therefore the force transmitted,

A machine has a mass of 100 kg and unbalanced reciprocating parts of mass
2 kg which move through a vertical stroke of 80 mm with simple harmonic motion.
The machine is mounted on four springs, symmetrically arranged with respect to
centre of mass, in such a way that the machine has one degree of freedom and can
undergo vertical displacements only.
Neglecting damping, calculate the combined stiffness of the spring in order that
the force transmitted to the foundation is 1 / 25 th of the applied force, when the speed
of rotation of ma-chine crank shaft is 1000 r.p.m.
When the machine is actually supported on the springs, it is found that the
damping reduces the amplitude of successive free vibrations by 25%. Find : 1.
force transmitted to foundation at 1000 r.p.m., 2. the force transmitted to the
the
foundation at reso nance, and 3. the amplitude of the forced vibration of the machine
at resonance.

6.(i) Derive the relation for magnification factor in case of forced vibration.

6.(ii) A single cylinder vertical petrol engine of total mass 300 kg is mounted upon a steel
chassis frame and causes a vertical static deflection of 2 mm. The reciprocating parts of the
engine has a mass of 20 kg and move through a vertical stroke of 150 mm with simple

harmonic motion. A dashpot is provided whose damping resistance is directly
proportional to the velocity and amounts to 1.5 kN per metre per second.
Considering that the steady state of vibration is reached ; determine : 1. the amplitude of
forced vibrations, when the driving shaft of the engine rotates at 480 r.p.m., and 2. the
speed of the driving shaft at which resonance will occur.
Solution : Given. m = 300 kg; δ = 2 mm = 2 × 10–3 m ; m1 = 20 kg ; l = 150 mm
= 0.15 m ; c = 1.5 kN/m/s = 1500 N/m/s ; N = 480 r.p.m. or ω 2 480 / 60 = 50.3 rad/s

UNIT-5 VIBRATION NOTES- FREE -LONGITUDINAL VIBRATION

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