On the sinusoidal vibration of automobiles by ogunmiluyi ifeoluwa (olimilove)

ON THE SINUSOIDAL VIBRATION OF AUTOMOBILES
1
BY
OGUNMILUYI, IFEOLUWA MICHEAL
MATRICULATION NUMBER: 2009/1842
A PROJECT SUBMITTED TO THE DEPARTMENT OF MATHEMATICS,
COLLEGE OF NATURAL SCIENCES,
FEDERAL UNIVERSITY OF AGRICULTURE, ABEOKUTA.
IN PARTIAL FULFILMENT OF THE REQUIREMENT FOR THE
AWARD OF BACHELOR OF SCIENCES (B.Sc.) DEGREE IN
MATHEMATICS
JANUARY, 2014.

CERTIFICATION
This is to certify that this project was carried out by Ogunmiluyi Ifeoluwa Micheal, with
matriculation number 20091842, in the Department of Mathematics, College of Natural Sciences, Federal
University of Agriculture, Abeokuta, Ogun State, Nigeria.
………………………... ..…………………
Ogunmiluyi Ifeoluwa Date
2
Student
………………………… ...…………………
Dr. I.O Abiala Date
(Supervisor)
………………………… ….……………….
Dr. B.I Olajuwon Date
(Head of Department)

DEDICATION
This project work is dedicated to ALMIGHTY God (the creator of heaven and the earth) for his love
and mercy, who in His infinite mercy brought me thus far.
3

ACKNOWLEDGEMENTS
First and foremost, I would like to thank my greatest teacher of all, God. I know that I am here and
that I am able to write all of this for a reason. I will do my best in never forgetting what a great fortune I
have had in just being here, and that it comes with a lesson and a responsibility. I hope I am doing the work
you have planned me to do.
I would like to thank my supervisor, Dr. I.O. Abiala, for being like a father to me and giving
invaluable suggestions to improvement of my project and for his patience throughout my project and also,
those first few days of uncertainty that you pulled with me are ones that I will not never forget, thank you for
believing in me, even if it’s only was for a few moments.
I would like to thank my friends and colleagues that I have met in this home far away from home
called Abeokuta. Specially, David, Moyo, Seyi (my editor) and Peter, who, even though have reduced me to
a fifth wheel in our relationship, have blossomed into a partnership that will not be forgotten. Whatever
happens with you too, do know that, throughout these couples of years, our relationship has provided me
with an impressively beautiful site to see, as it is when five friends fall in love with each other. You guys
have been more than friends.
I would like to thank honourable senior, Jinadu Ayo, though our relationship was born in a very odd
way, but I would not have expected otherwise, as both of us are odd in our own beautifully weird world. For
guiding me and helping my shortcomings. I have become a better man because of the mirror you held up for
me. Thank you.
A special thanks goes to my sister and my siblings, Ibukun and Joshua and also my dearest cousin,
Oluwadarasimi. I cannot but appreciate the effort of DLCF FUNAAB family, departmental mates and my
castle mates. Thank you, I love you all.
4

Finally, my parents: Pastor A.I Ogunmiluyi and Mrs R.T Ogunmiluyi. They gave me my name, they
gave me my life, and everything else in between. I pride myself in having words for everything, but they
truly shut me up when it comes down to describing how much I love them and appreciate the efforts they
have put into giving me the life I have now. They are the reason I did this; they are the reason I thrive to be
better. Their pride for me is my main goal in life. As I have always taught and hoped; when they lay in their
death bed they would think, “I am proud of my son.” Thank you, thank you, thank you.
5

TABLE OF CONTENTS
6
PAGES
Certification ii
Declaration iii
Acknowledgement iv
Table of contents vi
Nomenclature viii
Abstract ix
CHAPTER ONE
1.1 Introduction 1
1.2 Historical development of Sinusoidal vibration 3
1.3 Definition of key terms 5
CHAPTER TWO
2.1 Literature review 7
CHAPTER THREE
3.1 Introduction 10
3.2 Problem formulation 11
3.3 Problem analysis 14
3.4 Application or economic importance of sinusoidal vibration 21

7
CHAPTER FOUR
4.1 Introduction 23
4.2 Method of solution to the problem 24
4.3 Numerical solution 28
CHAPTER FIVE
5.1 Conclusion 44
5.2 Recommendation 44
5.3 References 46

NOMENCLATURE
m Mass of the automobiles measured in kilogram.
c This is the damping constant with unit force/unit velocity.
f This is the natural cyclic frequency of vibration (1/T ).
T The natural period of frequency of the vibration.
k Spring constant or stiffness.
x The displacement of the automobile which is measured in meter (m).
dx = & This is defined as the rate of change of displacement (x) and it’s often called the velocity
which is measured in m/s.
f This is called the phase angle and they are arbitrary constant to be determined by initial condition.
w = k This is the square of the stiffness ‘k’ per the mass ‘m’ and is called the natural frequency of
the circular vibration and is measured in rad/sec.
F This is the external applied force on the automobile which is measured in Newton per meters (N/m).
z = c This is the fraction of critical damping.
8
x
dt
m
w
d = This the static displacement.
k st
c c
c mw c = 2 This is the critical damping coefficient.
d w
This is the damped natural frequency.
g This is the gravitational acceleration, g = 32.16
W This is the weight of member or structure (F)
2
2
dt
d x This is the acceleration of the automobile which is measured in m/s2.

ABSTRACT
This project work attempt to express the problem of sinusoidal vibration of automobiles as the
general equation of motion which is a second order linear differential equation comprising of the inertial
force, damping force, stiffness force and the external force and also explained the level of the damping force
acting on the automobile such as under damping, critical damping and over damping. Also, this project
attempt to express solution to the problem using the step-by-step integration with the central difference
method and Excel. It was discovered that increase in the time step leads to increase in the response.
9

CHAPTER ONE
10
1.1 INTRODUCTION
Sinusoidal in relation to dynamics can be defined as having a magnitude that varies as the sine of an
independent variable. There are few machines that will vibrate in pure sinusoidal fashion.
Vibration is the motion of a particle or a body or system of connected bodies released from a position
of equilibrium. Most vibrations has much disadvantages in machines and structures because they have the
tendency to produce increased stresses, energy losses(damping), caused added wear, induce fatigue and
create passenger discomfort in vehicles. Also in rotating parts like gear needs to be given a lot attention to
when balancing in order to prevent damage from vibrations.
It also occurs when a system is released from its state of being balance. The system tends to return to
this balanced position under the action of restoring forces (such as it is well known in simple pendulum).
The system keeps moving back and forth across its position of equilibrium. The combination of elements
with intention of accomplishing a goal is called a system. For example, an automobile is a system whose
elements are the wheels, suspension, car body, brake and so on.
Also, Vibration can be defined as “the cyclical change in the position of an object as it moves
alternately to one side and the other of some reference or datum position” (Macinante, 1984). Vibration of
rigid bodies can be rectilinear (or translational), rotational, or a combination of the two. Rectilinear vibration
refers to a point whose path of vibration is a straight line, and rotational vibration refers to a rigid body
whose vibration is angular about some reference line. Additionally, vibration of flexible bodies can be
described by flexural or other elastic vibrations such as longitudinal, tension and compression, and torsional
or twisting.
Many types of engines, compressors, pumps, and other machinery that run continuously generate a
form of periodic vibration. If a motion is periodic, its velocity and acceleration are also periodic.
The three terms used in describing vibration are amplitude, frequency, and type. Thus, a vibration is
said to be sinusoidal if it corresponds to a sinusoidal function of time.

Sinusoidal vibration which is also known as simple harmonic motion is the simplest form of
vibration, in which a body moves around an equilibrium position in a periodic (changes with time) and
smooth way or it can be defined as a type of vibration that smoothly changes with time. The best known
example of sinusoidal vibration is the simple pendulum where a ball is attached to a spring and its displaced
from its equilibrium with time.
Now relating sinusoidal vibration to automobiles, the terms used that changes with time are
displacement (distance), velocity (speed) and acceleration. Sinusoidal motion often occurs in our day to day
activities. When riding a Mazda car, each compartment moves in a circular manner as it changes with time
and when tracking the height of the compartment the motion is clearly sinusoidal.
The motion of any vehicle depends upon all the forces and moments that act upon it. These forces
and moments are caused by interaction of the vehicle with the surrounding medium(s) such as air or water
(fluid static and dynamic forces), gravitational attraction (gravity forces), Earth’s surface (support, ground,
or landing gear forces), and also for ship or aircrafts (propulsion forces).
Another important parameter to discuss when describing vibration is damping. Structural damping
occurs as material layers slide over one another during vibration. It is important to remember that damping is
one of the most difficult phenomena to model in vibrating systems. In fact, in the twenty years from 1945 to
1965, 2000 papers were published in the area of damping technology. Damping is usually best estimated
experimentally. Although damping mechanisms in real systems are rarely viscous, the nice analytical
properties of vibrating systems with viscous damping are worth exploiting if possible. In fact, the concept of
equivalent viscous damping is in wide use within the noise & vibration engineering community. If a system
is initially displaced at a certain distance and then released, such as a pendulum, it will vibrate about a
certain datum line for a finite amount of time before coming to rest. The amplitude of the motion decays,
and the cause of this decay in motion, or dissipation of energy, is referred to as damping. It is present
naturally, and if a system is not being forced to vibrate by an external source, its motion will eventually
decay because of the intrinsic damping that is present. Damping can also be introduced into a system as a
means of controlling the vibrations.
11

1.2 HISTORICAL DEVELOPMENT OF SINUSOIDAL VIBRATION
The critical aspects of our knowledge about vibrations, we owe to Galileo Galileo, who in 1610 gave
the concept of mass, and also Hooke joined with Marriott in 1660 propound the Hooke’s law and also Isaac
Newton in 1665 who gave the laws of motion and the various equations to each ones, with Leibnitz in 1684,
they gave the calculus and Newton who in 1687 declare the laws of motion acceptable and valid.
Far back in 1100-1165, Hibat Allah Abu’l-Barakat al-Baghdaadi discovers that force is proportional
to acceleration rather than speed, which is now a fundamental law in classical mechanics under which there
is sinusoidal vibration. Later, Newton also developed calculus which is necessary to perform the
mathematical calculations involved in classical mechanics. However, it was Gottfried Leibniz who
independently of Newton, developed a calculus with the notation of the derivative and integral which are
used to this day, but classical mechanics retains the Newton’s dot notation for time derivatives which is
applied in sinusoidal vibration which is also respective of time. With the help of Hooke’s law in 1660, the
restoring force was used and Taylor’s in 1713 when coming up with the dawn of vibration analysis also used
the Hooke’s law and deduced an expression for the resultant force. Therefore, the experimental work of
Isaac Newton in 1665 and Leibnitz in 1684 led to the general equation of motion of second-order non-homogenous
12
linear equation.
The foundations of vibration theory for continuous media were established between 1733 and 1735
by Daniel Bernoulli and Leonard Euler. These two mathematical scientists by 1734 finally achieved the
fourth-order equation using an infinite series approach. The solutions were given by Eigen value equations
for several kinds of end conditions, which are common knowledge today. In 1739, Euler had another
discovery of the generality of the exponential method for the solution of differential equations with
constants coefficients. This method is the basic method used by analysts today to solve problems involving
differential equations of linear systems. He also solves the ordinary differential equation for a forced
harmonic oscillator and notices the resonance phenomenon.

Another method that is used in solving the second-order differential equation is the finite difference
which is a numerical method for approximating the solutions to differential equations using finite difference
equations to approximate derivatives. But the possible and likely sources of error in finite difference
methods are round-off error which is the loss of precision due to computer rounding of decimal quantities
and truncation error which is the difference between the exact solution of the finite difference equation and
the exact quantity assuming perfect arithmetic.
Looking back at the short history of finite difference method. The finite difference method (FDM)
was first developed by A. Thom in the 1920s under the title “the method of square” to solve nonlinear
hydrodynamic equations which is in his book, ‘A. Thom and C. J. Apelt, Field Computations in Engineering
and Physics. London: D. Van Nostrand, 1961’. After this, there was some history in the 1930s and further
development of the finite difference method. Although some ideas may be traced back further, we begin the
fundamental paper by Courant, Friedrichs and Lewy (1928) on the solutions of the problems of
mathematical physics by means of finite differences. A finite difference approximation was first defined for
the wave equation and the CFL stability condition was shown to be necessary for convergence. Error bounds
for difference approximations of elliptic problems were first derived by Gerschgorin (1930) whose work was
based on a discrete analogue of the maximum principle for Laplace’s equation. This approach was pursued
through the 1960s and various approximations of elliptic equations and associated boundary conditions were
analysed. The finite difference theory for general initial value problems and parabolic problems then had an
intense period of development during 1950s and 1960s, when the concept of stability was explored in the lax
equivalence theorem and the Kreiss matrix lemmas. Independently of the engineering applications, a number
of papers appeared in the mathematical literature in the mid-1960s which were concerned with the
construction and analysis of finite difference schemes by the Rayleigh-Ritz procedure with piecewise linear
approximating functions.
Beginning in the mid-1950s, efforts to solve continuum problems in elasticity using small, discrete
"elements" to describe the overall behaviour of simple elastic bars began to appear. Argyris (1954) and
13

Turner, et al. (1956) were the first to publish use of such techniques for the aircraft industry. Actual coining
of the term "finite element" appeared in a paper by Clough (1960).
The early use of finite elements lay in the application of such techniques for structural-related
problems. However, others soon recognized the versatility of the method and its underlying rich
mathematical basis for application in non-structural areas. Sienkiewicz and Cheung (1965) were among the
first to apply the finite element method to field problems (e.g., heat conduction, irrotational fluid flow, etc.)
involving solution of Laplace and Poisson equations, Gangadharan, et al. 2008 applied finite element
method to model the vehicle/track system and used Power Spectral Density (PSD) of track irregularities as
input to the system.
The underlying mathematical basis of the finite element method first lies with the classical
Rayleigh-Ritz and variational calculus procedures introduced by Rayleigh (1877) and Ritz (1909). These
theories provided the reasons why the finite element method worked well for the class of problems in which
variational statements could be obtained (e.g., linear diffusion type problems).
In finite difference method, three forms are commonly considered, these are; the forward, backward
and central differences. So also, the Newton’s series which also consist of the terms of the Newton forward
difference equation named after Isaac Newton, in essence, it is the Newton interpolation formula which was
published in his Principia Mathematica in 1687, namely the discrete analogue of the continuum Taylor
expansion
14
1.3 DEFINITIONS OF KEY TERMS
Damping: Dissipation of oscillatory or vibratory energy, with motion or with time.
Critical damping ( ) c c : This is that value of damping that provides most rapid response to a step function
without overshoot.
Damping ratio: This is a fraction of c c .

Displacement: Specified change of position, or distance, usually measured from the mean position or
position of rest. Usually applies to uniaxial, less often to angular motion.
Harmonic: A sinusoidal quality having a frequency that is an integral multiple (x2, x3, etc) of a fundamental
(x1) frequency.
Phase: A periodic quality, the fractional part of a period between a reference time (such as when
displacement = zero) and a particular time of interest; or between two motions or electrical signals having
the same fundamental frequency.
Stiffness: The ratio of force (or torque) to deflection of a spring-like element.
15
Velocity ( ) x
v
& : Rate of change of displacement with time, usually along a specified axis; it may refer to
angular motion as well as to uniaxial motion.
Vibration: Mechanical oscillation or motion about a reference point of equilibrium.
Natural Frequency(w): The frequency of an undamped system’s free vibration; also, the frequency of any of
the normal modes of vibration. Natural frequency drops when damping is present.
Free Vibration: Free vibration occurs without force, similar after a reed is plucked.
Datum: This is a fixed starting point of a scale or operation from which inferences can be drawn from.
Datum Line: This is a standard on comparison or point of reference.

CHAPTER TWO
16
2.1 LITERATURE REVIEW
Vibrations occur in many spheres of our life. For example, any unbalance in machines with rotating
parts such as fans, ventilators, centrifugal separators, washing machines, lathes, centrifugal pumps, rotary
presses, and turbines, can cause vibrations. For these machines, vibrations are generally undesirable.
Buildings and structures can experience vibrations due to operating machinery; passing vehicular (consisting
of vehicles), air, and rail traffic; or natural phenomena such as earthquakes and winds. Pedestrian bridges
and floors in buildings also experience vibrations due to human movement on them. In structural systems,
the fluctuating stresses due to vibrations can result in fatigue failure. Vibrations are also undesirable when
performing measurements with precision instruments such as an electron microscope and when fabricating
micro-electro-mechanical system
The study of vibration started in the sixteen century and since then, it has become a subject of intense
research. The behaviour of the solution is studied in a sufficiently small neighbourhood of a given solution,
for example, in a neighbourhood of stationary point or a periodic solution. The summary of some literature
review pertaining to some research on sinusoidal vibration, basically on damping will be presented in this
section.
In spite of a large amount of research, understanding of damping mechanisms is quite primitive. A
major reason for this is that, by contrast with inertia and stiffness forces, it is not in general clear which state
variables are relevant to determine the damping forces. Moreover, it seems that in a realistic situation it is
often the structural joints which are more responsible for the energy dissipation than the (solid) material.
There have been detailed studies on the material damping like (Bert, 1973) and also on energy dissipation
mechanisms in the joints (Earls, 1966, Beards and Williams, 1977). But here difficulty lies in representing
all these tiny mechanisms in different parts of the structure in a unified manner. Even in many cases these
mechanisms turn out be locally non-linear, requiring an equivalent linearization technique for a global
analysis (Bandstra, 1983). A well-known method to get rid of all these problems is to use the so called

viscous damping. This approach was first introduced by (Rayleigh, 1877) via his famous dissipation
function, a quadratic expression for the energy dissipation rate with a symmetric matrix of coefficients, the
damping matrix.
A further idealization, also pointed out by Rayleigh, is to assume the damping matrix to be a linear
combination of the mass and stiffness matrices. Since its introduction this model has been used extensively
and is now usually known as ‘Rayleigh damping’, proportional damping or classical damping. With such a
damping model, the modal analysis procedure, originally developed for undamped systems, can be used to
analyse damped systems in a very similar manner. (Rayleigh, 1877) has shown that undamped linear
systems are capable of so-called natural motions. This essentially implies that all the system coordinates
execute harmonic oscillation at a given frequency and form a certain displacement pattern. The oscillation
frequency and displacement pattern are called natural frequencies and normal modes, respectively. Thus,
any mathematical representation of the physical damping mechanisms in the equations of motion of a
vibrating system will have to be a generalization and approximation of the true physical situation.
As (Scanlan, 1970) has observed, any mathematical damping model is really only a crutch which
does not give a detailed explanation of the underlying physics. Free oscillation of an undamped single
degree of frequency (SDOF) like sinusoidal vibration system never dies out and the simplest approach to
introduce dissipation is to incorporate an ideal viscous dashpot in the model. The damping force is assumed
to be proportional to the instantaneous velocity and the coefficient of proportionality which is known as the
dashpot-constant or viscous damping constant. The loss factor, which is the energy dissipation per radian to
the peak potential energy in the cycle, is widely accepted as a basic measure of the damping.
This dependence of the loss factor on the driving frequency has been discussed by (Crandall, 1970)
where it has been pointed out that the frequency dependence, observed in practice, is usually not of this
form. In such cases one often resorts to an equivalent ideal dashpot. Theoretical objections to the
approximately constant value of damping over a range of frequency, as observed in aero elasticity problems,
have been raised by (Naylor, 1970). Dissipation of energy takes place in the process of air flow and
coulomb-friction dominates around the joints. This damping behaviour has been studied by many authors in
17

some practical situations, for example by (Cremer and Hecki, 1973). (Earls, 1966) has obtained the energy
dissipation in a lap joint over a cycle under different clamping pressure. (Beards and Williams, 1977) have
noted that significant damping can be obtained by suitably choosing the fastening pressure at the interfacial
slip in joints.
18

CHAPTER THREE
19
3.1 INTRODUCTION
In this chapter, we shall introduce and formulate the problem on the sinusoidal vibration of
automobiles using the concept of second-order linear differential equation with the aid of Newton’s second
law of motion which states that when an applied force acts on a mass, the rate of change of momentum is
equal to the applied force which is the product of the mass and the acceleration. We shall be working on the
force ‘ F ’ applied to an automobile, the displacement ‘ x ’, the velocity ‘ v ’, the acceleration ‘ a ’, the mass ‘
m’ of the automobile, the damping constant ‘ c ’ and the stiffness ‘ k ’.
When undergoing this project, Hooke’s law which states that force in the spring is proportional to
displacement from its equilibrium position where ‘ k ’ is the spring stiffness and the stiffness ‘ k ’ is
measured in Newton’s per meter (N/m), will be introduced in order to generate the stiffness ‘ k ’. The typical
stiffness parameters for a car: k = 17000 N/m for spring and k = 180000 N/m for a tire. For a truck the
stiffness can be 10 times the magnitude for a car.
However, the purpose and focus of this chapter is to explain how the second-order linear differential
equation
m&x&(t)+ cx&(t)+ kx(t) = F(t) (3.1)
was formulated where the ‘m’ is the mass, ‘ x ’ is the displacement, ‘ k ’ is the stiffness, ‘ c ’ is the damping
constants, ‘ t ’ is the time of the vibration or motion and ‘ F ’ is the external force applied to the automobile.
Also, in this chapter, the analysis of the problem will be explained and the economic importance of
the problem will also be examined.

20
3.2 PROBLEM FORMULATION
Sinusoidal vibration is also known as a single degree of freedom (SDOF) where the shape of the
system can be represented in terms of a single dynamic coordinate x(t). From the diagram below;
Figure 1
Where m = the mass of the automobile, k = stiffness, x(t) = displacement and c = the damping which
leads to loss of energy in the vibrating system.
We shall now formulate the problem which is the second-order linear differential equation.
Let us consider when the object moves in a positive displacement x(t) with a positive velocity x&(t)
under an external force F(t). The accelerations, velocities and displacements in a system produce forces
when multiplied respectively by mass, damping and stiffness. For mass and inertia, the acceleration between
mass ‘m’ and acceleration ‘ &x& ’ is given by Newton’s second law which states that when a force acts on a
mass, the rate of change of momentum is equal to the applied force which is the product of mass and the
acceleration.
Mathematically;
F
÷ø
d m dx
= dt
dt
ö
æ
çè

df of the straight line between these extremes, where ıı and ıı
1 kx in the figure (b), where - kx is the restoring force where k is the
21
This can also be written/expressed as;
F = m&x&
(This is a second-order linear differential equation)
Mass (m)
Force (F) Acceleration (&x&)
Initial force (m&x&)
Stiffness of the system.
The stiffness can be determined by any of the standard methods of static structure analysis. Using the
Hooke’s law which states that force in the spring is proportional to displacement from its equilibrium
position as in figure (a). The slope
dx
represent small changes in f and x respectively, is the stiffness ‘ k ’. The potential energy at any value of ‘
1 or 2
x ’ is the shaded triangle, Fx
2
2
stiffness and is measured in Newton’s per meter and ‘m’ is the mass of the automobile and is measured in
kilograms.
E, a F elastic limit
F slope
df Area= 2
dx
1 kx
2
L x
(a) (b)

m d x (3.22)
22
Damping in a system
Damping is different, in that it dissipates energy, which is lost from the system. This happens
because nature has built in our system a retarding property, which implicitly acts against motion from the
advent of the motion and brings it to a stop, this is known as the damping of a system. The damping may be
deliberately added to a system or structure to reduce unwanted oscillations. Examples are discrete units,
usually using fluids, such as vehicle suspension dampers and viscoelastic damping layers on panels. In
vehicle suspension dampers, such a device typically produces a damping force, F , in response to closure
velocity, x& , by forcing fluid through a hole or opening in the system. This is inherently a square-law rather
than a linear effect, but can be made approximately linear by the use of a special valve, which opens
progressively with increasing flow. The damper is then known as an automotive damper. Then the force and
velocity are related by:
F cx d = & (3.21)
Where d F is the external applied damping force and x& the velocity at the same point. The quantity c is
called the damping constant having the dimensions force/unit velocity. Equation (3.21) is called the damping
force.
Thus;
å[inertial force + damping force - restoring force] = External force at anytime t
Therefore, from the second law of motion, the equation of motion for automobile system having a degree of
two (2) at any time t is express as;
2
+ c dx
+ kx =
0 2
dt
dt
Equation (3.22) is a homogenous and a second-order differential equation where the external forces
neglected. If an external force is considered, we have;
2
m d x + + = 2
kx F(t)
c dx
dt
dt
(3.23)

m d x (3.31)
23
3.3 PROBLEM ANALYSIS
The purpose of this section is to analyse the problem that will be solved in the chapter three of this
project. We shall analyse the general equation which is the second-order linear differential equation under
two conditions, which deals with the presence of damping force.
CASE 1(without damping)
For a system oscillating without an external force and damping being applied, then equation (3.23)
becomes;
m&x&+ kx = 0
From the equation, mıı= ma = F where ‘a’ is the acceleration of the system, and since the acceleration is
2
dt
a = d x , it follows that;
describe as the change of velocity with time. This implies that 2
Acceleration - restoring force = 0
2
+ kx =
dt
This gives, 0 2
The auxiliary equation is; ml2 + k = 0
From equation (3.1), let x be the instantaneous displacement varying with time, then;
Instantaneous displacement x(t) = x sin 2pft
Differentiating this and the result is called instantaneous velocity which is varying with time, then;
Instantaneous velocity (v) = x& = 2pfx cos 2pft
Also differentiating the velocity to get the instantaneous acceleration;
Instantaneous acceleration (a) = v& = &x& = -4p 2 f 2 x sin 2pft
Since w = 2pf , then, the above equations changes to;
Instantaneous displacement, velocity and acceleration are x sinwt, xw coswt ,and -w 2 x sinwt respectively.
Since x(t) = xsinwt and &x&(t) = -w 2 xsinwt
This implies,
m(-w 2 x sinwt) + k(x sinwt) = 0

C A iB
24
m(w 2 xsinwt) = k(x sinwt)
m
k
= k x t =
2 ( sin w
)
m x t
( sin w
)
w
m
w 2 = k
m
w = k
From the auxiliary equation ml2 + k = 0
m
l2 = - k
l = ±i k
m
l = ±iw
Since equation (3.31) is a linear, homogenous second-order ordinary differential equation, the
solution is of the form,
x(t) = Celt
Where C and λ are constant and t is the time.
l = ±iw
x(t) = C eiwt + C e-iwt 1 2 (3.32)
To change the complex component of x(t) to a real component, we make use of Euler’s formula,
e±q = cosq ± sinq
Equation (3.32) now becomes;
( ) (cos sin ) (cos sin ) 1 2 x t = C wt + i wt +C wt - i wt
(C C )coswt i(C C )sinwt 1 2 1 2 = + + - (3.33)
Since x(t) is not yet a real-valued function because of the component i , then ( ) 1 2 C +C and ( ) 1 2 i C -C must
be real-valued. Using the complex conjugate pair;
-
C =
A iB
2 1
+
= (3.34)
2 2

Substituting equation (3.34) back into equation (3.33) to have;
ù
é
ù
A iB A iB wt i A iB -
æ A +
iB sinwt
2 2
é - - -
+ úû
25
cos
ö
ö
æ -
+
2 2 úû
êë
÷ø
çè
-
+ úû
é
êë
÷ø
çè
+
A iB A B wt i A iB A iB sinwt
2
cos
ù
é + + -
2 úû
êë
ù
êë
A cos wt i 2
iB ð sinwt
2
2
2
ù
úû
+ é úû
êë
ù
é
êë
ð Acoswt + iBsinwt
Recall that i2 = -1;
x(t) = Acoswt + Bsinwt (3.35)
Therefore, the general solution of the undamped vibration is Acoswt + Bsinwt
Let sinf 1 A = C and cosf 1 B = C (3.36)
Hence, f 2 2f
A2 = C sin and B = C cos
1
2 2 2
1
Therefore, f 2 2f
A2 + B2 = C sin + C cos
1
2 2
1
2 ( 2f 2f )
1 = C sin + cos
Recall that; sin 2 + cos2 = 1
ð 2
A2 + B2 = C
1
Hence, the amplitude; 2 2
1 C = A + B (3.37)
Substituting equation (3.36) into (3.35), we have;
x(t) C sinf coswt C cosf sinwt 1 1 = +
C (sinf coswt cosf sinwt) 1 = +
Recall from trigonometry,
sin(A + B) = cos Asin B + cosBsin A
x(t) = C sin(wt +f ) 1 (3.38)
Also, from equation (3.36),

w = k is the natural frequency of circular vibration and T is the natural period of frequency which is
2p T = and f is the natural cyclic frequency of vibration denoted as;
d = and since
26
f = A f =
and B
sin cos
C
C
1 1
ð
A
C
1
B
1
tan sin
f f
cos
C
= = f
C
A 1
B
1
C
´ = ı
ıı
× ıı
ı
tanf = A
B
ð f = tan-1 ( A
) B
Where f is the phase angle.
m
w
k
m
= 1 = =
T
f
w
1
2
2
p p
Where w = mg and the static displacement
w
st k
kg g
st
= = 1 =
g
m
1
p w p d
w
2
2
From equation (3.35);
x(t) = Acoswt + Bsinwt
We consider two cases to have the general equation;
For case 1: Suppose that the mass is pulled down to the point 0 x and then released at time t = 0 i.e.
( ) x 0 = x0 and x&(0) = 0, then;
x(t) = Acoswt + Bsinwt
x&(t) = -Aw sinwt + Bw coswt
At (0) , (0) 0 0 0 x = x x& = and t =

27
x(0) = Acos0 + sin 0
x(0) = A
0 A = x
x&(0) = -Aw sin 0 + Bw cos0
0 = 0 + Bw
B = 0
Therefore, the motion of the system for case 1 is governed by;
x(t) x coswt 0 =
For case 2: Suppose that we have an impulse impacts initial velocity 0 v to the mass and also
( ) ( ) 0 x t = 0 and x& 0 = v , then;
x(0) = Acos0 + Bsin 0
0 = A + 0
A = 0
x&(0) = -Aw sin 0 + Bw cos0
v0 = 0 + Bw
0 v
w
B =
Therefore, the motion of the system for case 2 is governed by;
v
( ) t
x t = 0 sin
w
w
Therefore, the amplitude;
2 2
1 C = A + B
2
ö
x v
= + æ
2 0
0 ÷ø
çè
w

CASE 2(with damping)
For a system oscillating with damping and no external force being applied, then equation (3.23) becomes;
m&x&+ cx& + kx = 0 (3.39)
= l + l (3.3.10)
28
From the auxiliary equation;
ml2 + cl + +k = 0
Since it is an homogenous equation, then the solution will be of the form;
x(t) C e 1t C e 2t
1 2
The form of the solution of equation (3.39) depends upon whether the damping coefficient is equal to,
greater than or less than the critical damping coefficient c C and where C is the damping coefficient.
C m k c = 2 = 2 w = 2
m mk
m
t = C is the fraction of critical damping.
The ratio
c C
Considering each cases,
Case 1: When = (t = 1) c C C (critical damping)
In this case, it is called the critical damping and the solution has no oscillation, then the solution is of the
form;
ct
-
( ) ( ) m
x t = A +
Bt e 2
x(t)1 critical damping
0 1 2 3 t
C
-1 free vibration of a system with = 1
c C

Case 2: When < (t < 1) C Cc (under-damping)
In this case, it is also called the less than critical damping or under-damping and the solution is of the form;
ct
= 2 sinw + cosw -
x(t) e (A t B t) d d
29
m
ct
- Ce t d
Or = 2 m
sin(w +f )
= -twt cosw
Ce t d
Where f is the phase angle and d w
is the damped natural frequency which is related to the undamped
natural frequency ω by;
w =w 1-t 2 d
When (t < 1), the solution consist of two actors, the first on decreasing exponential and the second a sine
wave. The combined result is exponentially decreasing sine wave lying in the space between the exponential
curve on both sides of the phase angle axis in the figure below;
C
Free vibration of a system with < 1
c C
The smaller the damping constant C , the flatter will be the exponential curve and the more cycles will it
take for the vibration to be eliminated.
Case 3: When > (t > 1) c C C (overdamping)
In this case, it is called the greater-than-critical damping or the over-damping and the solution is of the form;
ct
( ) m ( t t )
x t - = e 2 Ae w t 2 - 1 + Be - w t
2 -
1

When (t >1), the motion is not oscillating but rather a creeping back to the original position, this is due to
the fact that whent > 1, then C is large.
C
Free vibration of a system with > 1
30
c C
3.4 APPLICATION OR ECONOMIC IMPORTANCE OF THE PROBLEM
The economic importance of sinusoidal vibration in relation to the problem is numerous. Some of them are;
1. It is the most important and central point of physics. Anything that oscillates produces motion that is
partly or almost sinusoidal.
2. Another important use of sinusoidal vibration is that it is an Eigen-function of linear systems. This
means that it is important for the analysis of filters such as reverberators (objects that is repeated several
times as it bounces off different surfaces), equalizers, certain (but not all) ``effects'', etc.
3. From the point of view of computer music research, is that the human ear is a kind of spectrum
analyser. That is, the cochlea of the inner ear physically splits sound into its (near) sinusoidal components.
This is accomplished by the basilar membrane in the inner ear: a sound wave injected at the oval window
(which is connected via the bones of the middle ear to the ear drum), travels along the basilar membrane
inside the coiled cochlea. The membrane starts out thick and stiff, and gradually becomes thinner and more
compliant toward its apex (the helicotrema). A stiff membrane has a high resonance frequency while a thin,
compliant membrane has a low resonance frequency (assuming comparable mass density, or at least less of a

difference in mass than in compliance). Thus, as the sound wave travels, each frequency in the sound
resonates at a particular place along the basilar membrane. The highest frequencies resonate right at the
entrance, while the lowest frequencies travel the farthest and resonate near the helicotrema. The membrane
resonance effectively ``shorts out'' the signal energy at that frequency, and it travels no further. Along the
basilar membrane there are hair cells which feel the resonant vibration and transmit an increased firing rate
along the auditory nerve to the brain.
Thus, the ear is very literally a Fourier analyser for sound, albeit nonlinear and using analysis
parameters that are difficult to match exactly. Nevertheless, by looking at spectra (which display the amount
of each sinusoidal frequency present in a sound), we are looking at a representation much more like what the
brain receives when we hear.
4. It also bring about the concept of phase (i.e starting a system in motion that changes with time),
which is used in some advanced diagnostic techniques and the basic concept used in rotor balancing. For
example, in balancing a rotor and understanding what is happening, one must definitely understand
sinusoidal vibration and phase.
5. One of the major applications of sinusoids in Science and Engineering is the study of harmonic
motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the
motion of an object on a spring, to the response of an electronic circuit.
6. It’s appropriate for fatigue testing of products that operate primarily at a known speed (frequency)
under in-service conditions.
7. It helps in detecting sensitivity of a device to a particular excitation frequency.
8. It also helps in detecting resonances, natural frequencies, modal damping, and mode shapes.
9. It’s appropriate for calibration of vibration sensors and control systems.
10. For sinusoidal waveforms, it is easy to convert between acceleration, velocity and displacement.
11. Any vibration waveform, no matter how complex, can be decomposed into sinusoidal components.
This fact is the base of frequency analysis, perhaps the most known tool for vibration diagnostics.
31

CHAPTER FOUR
m d x (4.1)
32
4.1 INTRODUCTION
From the problem analysed in the previous chapter, the method to be used to solve the problem is the
step-by-step integration coupled with the central difference method which is among the forms of the finite
difference method and it will be solved numerically. The finite difference techniques are based upon the
approximations that permit replacing differential equations by finite difference equations. These finite
difference approximations are algebraic in form, and the solutions are related to grid points.
Thus, a finite difference solution basically involves three steps: these are;
1. Dividing the solution into grids of nodes.
2. Approximating the given differential equation by finite difference equivalence that relates the solutions to
grid points.
3. Solving the difference equations subject to the prescribed boundary conditions and/or initial conditions.
In finite difference method, three forms are commonly considered, these are; the forward, backward and
central differences
Similarly, solutions and examples to this model are critically considered. For instance, it is a simple
matter to choose m, c and k so that the equation
2
+ c dx
+ kx =
0 2
dt
dt
is a valid linear equation. However, if one needs to specify the nature of the equation above, the settling time
and the peak time, then there may be a choice of m, c and k that will satisfy the equation.

4.2 METHOD OF SOLUTION TO THE PROBLEM
The central difference method combine with the direct integration techniques which is otherwise
known as the step-by-step integration will be used to solve the problem.
For the forward difference [df (x)]= f (x + h)- f (x) (4.21)
For the backward difference [df (x)]= f (x)- f (x - h) (4.22)
The central difference is the summation of (4.21) and (4.22), we have;
[df (x)]= [f (x + h)- f (x)]+ [f (x)- f (x - h)]
= f (x + h)- f (x)+ f (x)- f (x - h)
33
[df (x)]= f (x + h)- f (x - h)
1
Taking the average of the central difference = [ f (x + h)- f (x - h)]
2
Now using the central difference to solve;
m&x&(t)+ cx&(t)+ kx = F(t) (4.23)
d 1
From [ f (x)] = [ f (x + h)- f (x - h)]
2
For this system, f = x, x = t, h = Dt
d 1
=> [ x(t)] = [x(t + Dt)- x(t - Dt)]
2
x(t) = x(t)
Taking the derivative of [dx(t)] with respect to Dt

úû
m x t t x t x t t c x t t x t t
+ = mx t t cx t - D
t
+ =
mx t t mx t
+ =
P m 2 2 2
and T m
R m
= k
- = ÷ø
34
( ) ( ) ( )
x t x t + D t - x t - D
t
t
D
=
2
&
( ) ( ) ( ) ( )
x t + D t - x t + x t x t - D
t
2
2
D
t
&& =
Substituting this into (4.23), we have;
( ) ( ) ( ) ( + D ) - ( - D
) k[x(t)] F(t)
t
+ D - + - D
t
ù
é
êë
D
ù
+ úû
é
êë
D
2
2
2
Opening the bracket;
( ) ( ) ( ) ( ) ( ) kx(t) F(t)
t
cx t + D
t
t
mx t t
t
mx t
t
t
D
-
D
+
D
- D
+
D
-
D
+ D
2 2
2
2 2 2
Collecting like terms;
( ) ( ) ( ) ( ) ( ) kx(t) F(t)
t
cx t - D
t
t
mx t t
t
cx t + D
t
t
t
D
-
D
-
D
- D
+
D
+
D
2
+ D
2 2 2
2 2
ö
k x F(t)
x 2
m
t
æ -
D
t
c
x m
t
t
c
m
t
ö
æ
t t t t t = ÷ø
çè
- ÷ø
çè
D
-
D
ö
+ ÷ø
æ
çè
D
+
D 2 +D 2 -D 2
2 2
(4.24)
ö
æ -
D
æ
ö
æ
Let ÷ø
çè
ö
çè
D
-
D
= ÷ø
çè
D
+
D
t
t
c
t
t
c
t
2
2
,
2
=> Pxt t Rxt t Txt F(t) + + = +D -D
Making Pxt t +D subject of formula, we have;
=> ( ) Pxt t F t Txt Rxt t +D -D = - - (4.25)
This implies that to get xt t +D , we need to have xt and xt t -D .
From the boundary condition, at t = 0 , then = 0 = 0 xt and x&t
To have xt t -D , by the Taylor’s series expansion of degree two;

= - D + -D (4.26)
1 , 1 2
2 2 =
a 1
35
D 2
t t t t t x x tx& t &x&
2
= - D + -D
At = 0, xt we have;
x x tx t x t & &&
0
2
D
0 0 0 2
After getting 0 x and x -Dt , then (4.25) becomes;
( ) t t t Px F t Tx Rx +D -D = - - 0
From equation (4.24), let us take integration constant;
c
a d t
t
c
t
b
t
2
, 2 2 ,
2
D
= =
D
=
D
=
D
=
Then (4.26) becomes;
0 0 0 x t = x - Dtx& + d&x& -D
Then (4.24) becomes;
(ma bc)x (ma cb)x (k cm)x F(t) t t t t t + + - + - = +D -D (4.27)
Also, from (4.27), we can now have three forms of matrix, namely;
(i) Mass matrix: This is a sparse matrix, that is, it is primarily populated with zero (Stoer and Bulirsch,
2002). This is; P = ma + bc
(ii) Stiffness matrix: This is a band matrix in which the non-zero elements are clustered near the diagonal.
This is; T = -(cm- k ) = k - cm
(iii) Damping matrix: This is a symmetric matrix which is equal to its transpose that is aij = a ji . This is;
R = ma + cb

36
Let Pxt t F(t) = +D
ð x =
P -1
F(t) t +D t
Therefore, the effective force vector is;
( ) ( ) F t F t Tx Rx t -D = - - 0
All the above expression can be summarise under the following algorithm;
A. Initial computation
1. Form stiffness [K], mass [M] and damping [C] matrices.
2. Initialize [ ] [ ] [ ] x0 , x&0 and &x&0
3. Select time step Dt and calculate integration constants;
a , 2 , 1
c
1 , 1
2 = =
c a d
t
b
t
2
D
=
D
=
4. Calculate [ ] [ ] [ ] [ ] 0 0 0 x t = x - Dt x& + d &x& -D
5. Form effective mass matrix [P] = a[m]+ b[c]
B. For each time step;
1. Calculate effective force vector at time t;
[ ] [ ] [ ] [ ] t t t F F T x R x -D = - - 0
2. Solve the displacement at time t + Dt
[ xt ] =
[ P -1
][ ] +D t Ft When solving most problems under structural dynamics, the following should be put in place, the
initial condition of the general equation of motion for dynamics system for the displacement and velocity at
t = 0 . After this, the next step of direct integration comes to place. In direct integration procedure, it
requires the value of the previous time xt , before getting xt t +D and also to get xt , we must also have xt t -D .

3000 1200
3000 1200
3000 1200
37
4.3 NUMERICAL SOLUTION
Example 1: Find the displacement x by central difference method at time step 0.04 and;
ù
úû
500sin12.5
é
= úû
êë
ù
é
-
êë
-
700 2800
ù
= úû
é
-
êë
-
ù
= úû
50 0
é
=
êë
t
t
m c k and F
200sin12.5
1200 51000
,
2800 12300
,
0 100
Solution
ù
é
=
1
x
Let the displacement be úû
êë
2
x
x
From initial boundary condition, i.e. t = 0, then = 0, = 0 xt x&t and also F = 0 since sin12.5(0) = 0
ù
úû
500sin12.5
é
= úû
êë
ù
é
êë
ù
úû
é
-
êë
-
ù
+ úû
é
êë
100 2800
ù
úû
é
-
êë
-
ù
+ úû
é
êë
ù
úû
50 0
é
êë
t
t
x
x
x
&
x
x
&&
x
200sin12.5
1200 51000
2800 12300
0 100
1
2
1
2
1
2
&
&&
ù
úû
é
= úû
êë
ù
0
é
êë
ù
úû
é
-
êë
-
ù
+ úû
é
êë
100 2800
ù
úû
é
-
êë
-
ù
+ úû
é
êë
ù
úû
50 0
é
êë
0
0
0
1200 51000
0
0
2800 12300
0 100
x
&&
&&
1
x
2
ù
0
é
= úû
é
x
&&
&&
1
ð úû
êë
ù
êë
= 0
t x
2 0
From the question, Dt = 0.04
625
1 1
2 2 = =
D
0.04
=
t
a
1 = 1
=
D
( ) 12.5
2 0.04
2
=
t
b
c = 2 ´ a = 2´ 625 = 1250
0.0008
d = 1 =
1250
For mass matrix P;

8750 35000
ù
ù
é
= úû
0
38
32150 0
8750 35000
35000 153750
50 0
700 2800
2800 12300
40000 35000
ù
úû
é
-
êë
ù
-
é
é
-
ma
bc
= + =
ù
úû
é
-
êë
-
= úû
êë
é
= úû
-
=
ù
úû
êë
ù
êë
=
35000 216250
12.5
0 62500
0 100
625
P ma bc
For stiffness matrix T;
ù
úû
3000 1200
ù
62500 0
1200 51000
é
= úû
êë
ù
50 0
é
êë
é
-
=
úû
êë
-
=
0 12500
0 100
1250
k
cm
ù
úû
59500 1200
é
- -
êë
- -
= - =
1200 74000
T k cm
For damping matrix R;
ù
úû
32150 0
é
é
êë
22500 35000
-
=
úû
é
-
êë
-
ù
- úû
êë
= - =
35000 91250
35000 153750
0 62500
R ma cb
From the other initial condition at Dt = 0.04 ,
D
x = x - D tx -
t x t
& &&
( ) ( )
úû
êë
ù
é
êë
0.04 0.04
= - -
-D
-
0
0
0
2
2
2
0.04 0 0
0
2
0 0
x x x
&

8650000000 1225000000
216250 35000
6 6
- -
29.12 ´ 10 4.71 ´
10
6 6
ù
39
To get P-1 ;
adj P
40000 35000
ù
úû
- =
é
-
êë
-
=
35000 216250
1
P
P
P
Determinant of P= P = (40000´ 216250)- (- 35000´ -35000)
( ) ( )
= -
7425000000
=
To get adj P , we need the cofactors, 11 12 21 22 c ,c ,c ,c
( ) +
( )
( ) ( )
1 216250 216250
= - =
+
( ) +
( )
( 1) (40000) 40000
1 35000 35000
ù
35000 40000
1 35000 35000
= - - =
2 2
22
2 1
21
1 2
c
12
1 1
11
= - =
úû
é
êë
Þ =
= - - =
+
c
c
adj P
c
- -
4.71 10 5.39 10
6
1
10
29.12 4.71
4.71 5.39
216250 35000
ù
35000 40000
7425000000
-
-
ù
´ úû
é
é
=
êë
êë
úû
´ ´
=
úû
é
êë
P =
To now get the displacement by Dt = 0.04 ;
1. At Dt = 0.04
Ft Ft Tx Rx t -D = - - 0

0
é
+ úû
29.12 4.71 6
é
´ ´ úû
239.713
29.12 239.713 4.71 95.885
t
500sin12.5 59500 1200
7.432
3
é
ù
= -
500sin 12.5 0.08 59500 7.432 1200 1.646
3
é
- ´ + - ´
= -
40
[ ]0.04 [ ]0.04 0 -0.04 F = F - Tx - Rx
Where w = 12.5rad / sec,t = 0.04
( )
( )
500sin 12.5 0.04
239.713
ù
úû
é
é
=
êë
ù
úû
êë
ù
0
é
+ úû
êë
ù
êë
´
´
=
95.885
0
0
200sin 12.5 0.04
ù
úû
êë
ù
é
= úû
x
1
x
-
êë
ù
é
êë
95.885
10
4.71 5.39
2 0.04
( ´ ) + ( ´
)
( ) ( )
3
6
10
7.432
é
=
1.646
10
4.71 239.713 5.39 95.885
-
-
ù
´ úû
êë
ù
´ úû
é
êë
´ + ´
=
2. At Dt = 0.08
[ ] [ ] 0.08 0.08 0.04 0 F = F - Tx - Rx
ù
úû
0
é
êë
ù
úû
é
êë
22500 35000
-
ù
- ´ úû
êë é
úû
é
- -
- - -
úû
êë
ù
êë
0
35000 91250
10
1.646
1200 74000
t
200sin12.5
( )
( )
úû
( ) úû
( )
é
ù
( ) ( ) ´ - êë
ù
êë
- ´ + - ´
ù
- úû
é
êë
´
´
0
0
10
1200 7.432 74000 1.646
200sin 12.5 0.08
420.735
é
+ úû
ù
úû
é
=
864.9142
é
=
êë
ù
úû
0
444.1792
é
- úû
êë
ù
êë
ù
êë
299.0164
0
130.7224
168.294

ù
864.9142
29.12 4.71
é
´ ´ úû
é
= úû
299.0164
10
4.71 5.39
( 29.12 ´ 864.9142 ) + ( 4.71 ´
299.0164
)
( ) ( )
t
500sin12,5 ù
é
0.02680
úû
´ - 59500 0.02680 1200 0.00569
é
- ´ + - ´
1200 0.0268 74000 0.00569
500sin 12.5 ´
0.12
200sin 12.5 0.12
ù
22500 ´ 7.432 + 35000 ´
1.646
224.83
1875.3455
29.12 4.71
é
´ ´ úû
542.7965
10
4.71 5.39
29.12 ´ 1875.3455 + 4.71 ´
542.7965
41
0.02680
ù
úû
é
é
=
êë
ù
´ úû
êë
´ + ´
=
úû
êë
ù
êë
ù
é
êë

-
-
0.00569
10
4.71 864.9142 5.39 299.0164
6
6
x
1
x
2 0.08
3. At Dt = 0.12
[ ] [ ] 0.12 0.12 0.08 0.04 F = F - Tx - Rx
10 3
7.432
é
1.646
22500 35000
ù
é
35000 91250
ù
0.00569
59500 1200
é
- -
- - -
úû
1200 74000
200sin12.5
ù
êë
úû
êë
-
- úû
êë
úû
êë
ù
é
=
êë
t
( )
( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
10 3
35000 7.432 91250 1.646
ù
- ´ úû
é
êë
´ + - ´
ù
- úû
êë
- ´ + - ´
- úû
é
êë
´
=
ù
úû
1601.428
498.7475
é
+ úû
2100.1755
é
- úû
êë
ù
é
=
é
=
êë
ù
úû
é
êë
ù
úû
êë
ù
êë
224.83
109.9225
652.719
109.9225
_
453.22
199.4990
1875.3455
ù
úû
é
=
êë
542.7965
ù
( ) ( )
( ) ( )
0.05717
ù
úû
é
é
=
êë
ù
´ úû
êë
´ + ´
=
úû
êë
ù
êë é
ù
= úû
é
êë

-
-
0.01176
10
4.71 1875.3455 5.39 542.7965
6
6
x
1
x
2 0.12

0.16 0.16 0.12 0.08 F F Tx Rx
59500 1200
59500 0.05717 1200 0.01176
é
- ´ + - ´
1200 0.05717 74000 0.01176
22500 0.0268 35000 0.00569
802.15
3055.5255
701.916
10
29.12 4.71
4.71 5.39
29.12 3050.64 4.71 701.916
42
4. At Dt = 0.16
[ ] [ ]
úû
( )
é
ù
( ) êë
ù
úû
é
êë
22500 35000
-
0.05717
ù
- úû
é
êë
ù
úû
é
- -
- - -
úû
êë
ù
500sin 12.5 0.16
é
êë
´
´
=
= - -
0.0268
0.00569
35000 91250
0.01176
1200 74000
200sin 12.5 0.16
( ) ( )
( ) ( )
( ) ( )
( ) ( )úû
ù
ù
é
êë
´ + ´
´ + - ´
454.6487
-
ù
úû
êë
- ´ + - ´
- úû
é
=
êë
35000 0.0268 91250 0.00569
181.8595
3403.0262
é
+ úû
454.6487
3857.6755
é
- úû
3055.5255
ù
úû
é
=
é
=
é
=
êë
é
- úû
ù
úû
êë
ù
êë
ù
úû
êë
ù
êë
ù
êë
701.916
802.15
418.7875
1120.7035
418.7875
938.844
181.8595
ù
é
´ ´ úû
é
= úû
( ´ ) + ( ´
)
( ) ( )
0.09213
ù
úû
é
é
=
êë
ù
´ úû
êë
´ + ´
=
úû
êë
ù
êë
ù
é
êë

-
-
0.01817
10
4.71 3055.5255 5.39 701.916
6
6
x
1
x
2 0.16
5. At Dt = 0.2
[ ] [ ] 0.2 0.2 0.16 0.12 F = F - Tx - Rx

( )
ù
é
0.09213
( ) úû
4104.8501
úû
29.12 ´ 4104.8501 + 4.71 ´
646.9804 ´ - 59500 1200
59500 0.12258 1200 0.02282
ù
2708.875
43
0.05717
ù
é
êë
ù
úû
é
êë
22500 35000
-
ù
- úû
êë
úû
59500 1200
é
- -
- - -
úû
êë
ù
500sin 12.5 0.2
é
êë
´
´
=
0.01176
35000 91250
0.01817
1200 74000
200sin 12.5 0.2
1697.925
ù
úû
é
- úû
êë
ù
5503.539
299.2361
é
+ úû
êë
ù
é
=
êë
927.85
1455.136
119.6944
1697.925
ù
úû
5802.7751
é
- úû
êë
ù
é
=
êë
927.85
1574.8304
4104.8501
ù
úû
é
=
êë
646.9804
ù
úû
29.21 4.71 6
é
´ ´ úû
êë
ù
é
= úû
x
1
x
-
êë
ù
é
êë
646.9804
10
4.71 5.39
2 0.2
( ) ( )
( ) ( )
10 6
4.71 4104.8501 5.39 646.9804
ù
é
êë
´ + ´
=
é
=
0.02282
ù
úû
êë
0.12258
6. At Dt = 0.24
[ ] [ ] 0.24 0.24 0.2 0.16 F = F - Tx - Rx
( )
é
0.09213
ù
( ) úû
êë
ù
úû
é
êë
22500 35000
-
0.12258
ù
- úû
é
êë
ù
úû
é
- -
- - -
úû
êë
ù
500sin 12.5 0.24
é
êë
´
´
=
0.01817
35000 91250
0.02282
1200 74000
200sin 12.5 0.24
( ) ( )
( ) ( )
( ) ( )
( ) ( )úû
ù
é
êë
22500 ´ 0.09213 + 35000 ´
0.01817
´ + - ´
ù
- úû
é
- ´ + - ´
êë
- ´ + - ´
ù
- úû
é
=
êë
35000 0.09213 91250 0.01817
1200 0.12258 74000 0.02282
70.56
28.224
úû
7320.894
é
- úû
êë
ù
é
+ úû
êë
ù
é
=
êë
1566.5375
1835.776
70.56
28.224

4682.579
ù
29.12 4682.579 4.71 297.4625 -
é
=
0.02366
= úû
44
ù
úû
é
- úû
êë
ù
7391.454
é
=
êë
2708.875
1566.5375
1864
é
=
297.4625
ù
úû
êë
4682.579
úû
29.12 4.71 6
é
´ ´ úû
êë
ù
é
= úû
x
1
x
-
êë
ù
é
êë
297.4625
10
4.71 5.39
2
( ) ( )
( ) ( )
6
10
´ + ´
4.71 4682.579 5.39 297.4625
ù
´ úû
é
êë
´ + ´
ù
êë
0.13776

The table below gives the value of displacement and effective force for 25 time steps when Dt = 0.04
for central difference method and plotted below using Excel;
S/No Time step F1 F2 x1 x2
1 0.04 239.713 95.885 0.007432 0.001636
2 0.08 864.9142 299.0164 0.02680 0.00569
3 0.12 1875.3455 542.7965 0.05717 0.01176
4. 0.16 3055.5255 701.916 0.09213 0.01817
5. 0.20 4104.8501 646.9804 0.12258 0.02282
6. 0.24 4682.579 297.4625 0.13776 0.02366
7. 0.28 4492.9704 -361.9796 0.12953 0.01921
8. 0.32 3423.9858 -1237.0095 0.09388 0.00946
9. 0.36 1521.6719 -2163.4475 0.03426 -0.00449
10. 0.40 -889.7801 -2905.5079 -0.03960 -0.01985
11. 0.44 -3346.4902 -3266.3406 -0.11283 -0.03337
12. 0.48 -5307.3867 -3085.9716 -0.16909 -0.04163
13. 0.52 -6296.636 -2422.5145 -0.19477 -0.04271
14. 0.56 -6049.9987 -1143.1902 -0.18156 -0.03466
15. 0.60 -4498.247 324.5495 -0.12987 -0.01943
16. 0.64 -1957.7019 1797.0826 -0.04854 0.00047
17. 0.68 1113.8026 2955.6279 0.04635 0.02118
18. 0.72 4065.0002 3447.1512 0.13461 0.03773
19. 0.76 6232.8204 3233.8477 0.19673 0.04679
20. 0.80 7140.2979 2411.2443 0.21928 0.04663
21. 0.84 6599.1931 921.8543 0.19651 0.03605
22. 0.88 4669.7599 -716.2985 0.13261 0.01813
23. 0.92 1791.0999 -2262.6259 0.04150 -0.00376
24. 0.96 -1421.8235 -3322.7421 -0.05705 -0.02461
25. 1.00 -4259.3179 -3698.4643 -0.14145 -0.03990
45

Example 2: Find the displacement x by central difference method at time step 0.05 and;
3000 1200
3000 1200
3000 1200
46
ù
úû
500sin12.5
é
= úû
êë
ù
é
-
êë
-
700 2800
ù
= úû
é
-
êë
-
ù
= úû
50 0
é
=
êë
t
t
m c k and F
200sin12.5
1200 51000
,
2800 12300
,
0 100
Solution
ù
é
=
1
x
Let the displacement be úû
êë
2
x
x
From initial boundary condition, i.e. t = 0, then = 0, = 0 xt x&t and also F = 0 since sin12.5(0) = 0
ù
úû
500sin12.5
é
= úû
êë
ù
é
êë
ù
úû
é
-
êë
-
ù
+ úû
é
êë
100 2800
ù
úû
é
-
êë
-
ù
+ úû
é
êë
ù
úû
50 0
é
êë
t
t
x
x
x
&
x
x
&&
x
200sin12.5
1200 51000
2800 12300
0 100
1
2
1
2
1
2
&
&&
ù
úû
é
= úû
êë
ù
0
é
êë
ù
úû
é
-
êë
-
ù
+ úû
é
êë
100 2800
ù
úû
é
-
êë
-
ù
+ úû
é
êë
ù
úû
50 0
é
êë
0
0
0
1200 51000
0
0
2800 12300
0 100
x
&&
&&
1
x
2
ù
0
é
= úû
é
x
&&
&&
1
ð úû
êë
ù
êë
= 0
t x
2 0
From the question, Dt = 0.05
400
1 1
2 2 = =
D
0.05
=
t
a
1 = 1
=
D
( ) 10
2 0.05
2
=
t
b
c = 2´ a = 2´ 400 = 800
0.00125
d = 1 =
800
For mass matrix P;

7000 28000
ù
ù
0
47
20000 0
7000 28000
28000 123000
50 0
700 2800
2800 12300
27000 28000
ù
úû
é
-
êë
-
é
é
-
ma
bc
= + =
ù
úû
é
-
êë
-
ù
= úû
êë
-
=
ù
úû
é
= úû
êë
ù
êë
=
28000 163000
10
0 40000
0 100
400
P ma bc
For stiffness matrix T;
ù
úû
3000 1200
ù
1200 51000
é
= úû
êë
ù
50 0
é
êë
é
-
=
úû
êë
-
=
40000 0
0 80000
0 100
800
k
cm
ù
úû
37000 1200
é
- -
êë
- -
= - =
1200 29000
T k cm
For damping matrix R;
ù
úû
é
20000 0
13000 28000
é
êë
-
=
úû
é
-
êë
-
ù
- úû
êë
= - =
28000 83000
28000 123000
0 40000
R ma cb
From the other initial condition at Dt = 0.05 ,
D
x = x - D tx -
t x t
& &&
( ) ( )
úû
é
= úû
êë
ù
é
êë
0.05 0.05
= - -
-D
-
0
0
0
2
2
2
0.05 0 0
0
2
0 0
x x x
&

163000 28000
6 6
- -
45.06 ´ 10 7.74 ´
10
6 6
ù
48
To get P-1 ;
adj P
27000 28000
ù
úû
- =
é
-
êë
-
=
28000 163000
1
P
P
P
Determinant of P= P = (27000´163000)- (- 28000´ -28000)
( ) ( )
4401000000 784000000
= -
3617000000
=
To get adj P , we need the cofactors, 11 12 21 22 c ,c ,c ,c
( ) +
( )
( ) ( )
1 163000 163000
= - =
+
( ) +
( )
( 1) (27000) 27000
1 28000 28000
ù
28000 27000
1 28000 28000
= - - =
2 2
22
2 1
21
1 2
c
12
1 1
11
= - =
úû
é
êë
Þ =
= - - =
+
c
c
adj P
c
- -
7.74 10 7.46 10
6
1
10
45.06 7.74
7.74 7.46
163000 28000
ù
28000 27000
3617000000
-
-
ù
´ úû
é
é
=
êë
êë
úû
´ ´
=
úû
é
êë
P =
To now get the displacement by Dt = 0.05 ;
1. At Dt = 0.05
t t t F F Tx Rx -D = - - 0

é
+ úû
0
45.06 7.74 6
é
´ ´ úû
292.5486
45.06 292.5486 7.74 117.0195 6
= -
ù
37000 0.01409 1200 0.00314
é
- ´ + - ´
49
[ ]0.05 [ ]0.05 0 -0.05 F = F - Tx - Rx
Where w = 12.5rad / sec,t = 0.05
( )
( )
500sin 12.5 0.05
292.5486
ù
úû
é
é
=
êë
ù
úû
êë
ù
é
+ úû
êë
ù
êë
´
´
=
117.0195
0
0
0
200sin 12.5 0.05
ù
úû
êë
ù
é
= úû
x
1
x
-
êë
ù
é
êë
117.0195
10
7.74 7.46
2 0.05
( ´ ) + ( ´
)
( ) ( )
0.01409
ù
úû
é
=
êë
ù
´ úû
é
êë
´ + ´
0.00314
10
7.74 292.5486 7.46 117.0195
2. At Dt = 0.1
[ ] [ ] 0.1 0.1 0.5 0 F = F - Tx - Rx
ù
úû
é
êë
ù
úû
13000 28000
é
êë
-
0.01409
ù
- úû
êë é
úû
37000 1200
é
- -
- - -
úû
êë
ù
t
500sin12.5
é
=
êë
0
0
28000 83000
0.00314
1200 29000
t
200sin12.5
( )
( )
úû
úû
( ) ( )
é
ù
( ) ( ) - êë
ù
êë
- ´ + - ´
ù
- úû
é
êë
500sin 12.5 ´
0.1
´
=
0
0
1200 0.01409 29000 0.00314
200sin 12.5 0.1
525.098
474.4923
é
+ úû
999.5903
ù
úû
é
=
é
=
êë
ù
úû
0
é
- úû
êë
ù
êë
ù
êë
297.7649
0
107.968
189.7969

999.5903
ù
é
´ ´ úû
é
= úû
297.7649
10
45.06 7.74
7.74 7.46
( 45.06 999.5903 ) ( 7.74 297.7649
)
( ) ( )
é
ù
37000 0.04735 1200 0.00996
é
- ´ + - ´
ù
1200 0.04735 29000 0.00996
500sin 12.5 0.15
200sin 12.5 0.15
13000 ´ 0.01409 + 28000 ´
0.00314
271.09
_
1969.8549
é
´ ´ úû
402.5772
10
45.06 7.74
7.74 7.46
45.06 ´ 1969.8549 + 7.74 ´
402.7965
50
´ + ´
0.04735
ù
úû
é
é
=
êë
ù
´ úû
êë
´ + ´
=
úû
êë
ù
êë
ù
é
êë

-
-
0.00996
10
7.74 999.5903 7.46 297.7649
0.6
6
x
1
x
2 0.1
3. At Dt = 0.15
[ ] [ ] 0.15 0.15 0.1 0.05 F = F - Tx - Rx
0.01409
ù
úû
é
êë
ù
úû
13000 28000
é
êë
-
0.04735
ù
- úû
êë
úû
37000 1200
é
- -
- - -
úû
êë
ù
é
=
êë
0.00314
28000 83000
0.00996
1200 29000
t
500sin12.5
t
200sin12.5
( )
( )
( ) ( )
( ) ( )
´
( ) ( )
( ) ( )úû
ù
é
êë
´ + - ´
- úû ù
êë
- ´ + - ´
- úû
é
êë
´
=
28000 0.01409 83000 0.00314
1763.902
ù
úû
477.0429
é
+ úû
2240.9449
é
- úû
êë
ù
é
=
é
=
êë
ù
úû
é
êë
ù
úû
êë
ù
êë
271.09
133.9
536.4772
133.9
345.66
190.8172
1969.8549
ù
úû
é
=
êë
402.5772
ù
é
= úû ù
( ) ( )
( ) ( )
0.09188
ù
úû
é
é
=
êë
ù
´ úû
êë
´ + ´
=
úû
êë
ù
êë
é
êë

-
-
0.01825
10
7.74 1969.8549 7.46 402.7965
6
6
x
1
x
2 0.15

The table below gives the value of displacement and effective force for 20 time steps when Dt = 0.05
for central difference method and plotted below using Excel;
S/No Time step F1 F2 x1 x2
1. 0.05 292.5486 117.0195 0.01409 0.00314
2. 0.10 999.5903 297.7649 0.04735 0.00996
3. 0.15 1969.8549 402.5772 0.09188 0.01825
4. 0.20 2826.2661 260.0804 0.12936 0.02382
5. 0.25 3117.7599 -208.5596 0.13887 0.02258
6. 0.30 2530.8653 -937.8683 0.10678 0.01259
7. 0.35 1056.6069 -1709.6984 0.03438 -0.00458
8. 0.40 -953.5581 -2228.2189 -0.06021 -0.02400
9. 0.45 -2243.7111 -2123.3685 -0.11754 -0.03321
10. 0.50 -2950.6916 -1416.8938 -0.14392 -0.03341
11. 0.55 -2628.298 -495.3304 -0.12226 -0.02404
12. 0.60 -1277.029 600.457 -0.05290 -0.00540
13. 0.65 780.4684 1400.5798 0.04601 0.01649
14. 0.70 2873.42 1691.3668 0.14257 0.03486
15. 0.75 4281.9507 1272.3655 0.20279 0.04263
16. 0.80 4452.8854 272.2338 0.20275 0.03650
17. 0.85 3249.5802 -1023.3334 0.13851 0.01752
18. 0.90 1004.24 -2102.7267 0.02898 -0.00791
19. 0.95 -1547.2188 -2746.2527 -0.09097 -0.03246
20. 1.00 -3593.2629 -2531.7384 -0.18151 -0.04672
51

52
0.25
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-0.15
-0.2
-0.25
0.2 0.4 0.6 0.8 1
Displacement
Time Steps
x1
x2
Figure 3: Displacement against time step of 0.05
0.25
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-0.15
-0.2
-0.25
0.2 0.4 0.6 0.8 1
Displacement
Time Step
x1
x2
Figure 2: Displacement against time step of 0.04

CHAPTER FIVE
53
5.1 CONCLUSION
From the above analysed problem and numerical examples, by using central difference method, it
shows that it is through the general equation of motion (equation 3.23) that displacement (xt t ) +D can be
calculated as explained in the steps of central difference method, and it reveals that the sinusoidal vibration
of automobiles is a function of its inertial force, damping force, stiffness force and the external force which
combines to form a second order linear differential equation called the general equation of motion. And also
by using one of the methods to solve the problem, I conclude that the plotted graph (Figure 2 &3) are
conditionally stable because if exceeded, the displacement, velocity and acceleration grows without limit.
5.2 RECOMMENDATION
I recommend that;
1. A course that deals with the fundamental of dynamics should be introduced to the students to aid their
knowledge about dynamics.
2. A good and appropriate time step should be selected for meaningful and proper evaluation and
analysis of result based on direct integration method.
3. In a dynamics system, this method of step-by-step integration using central difference method should
be used to solve problems because it is advantageous and more accurate in predicting the response of
the dynamics system.
4. The damping is not high because if it is too high, the solution may not undergo core overflow.

54
5.3 REFERENCES
1. Balakumar Balachandran and Edward B. Magrab (2009). Vibrations, second edition.
2. Boyd D. Schimel, Jow-Lian Ding, Michael J. Anderson and Walter J. Grantham (1997),
Dynamic Systems Laboratory Manual, School of Mechanical and Materials Engineering,
Washington State University.
3. Sondipon Adhikar (2000). Damping Models for Structural Vibration. Trinity College, Cambridge.
September.
4. Indrajit Chowdhury & Shambhu P. Dasgupta (2009), Dynamics of Structure and Foundation – A
Unified Approach 1. Fundamentals, CRC Press/Balkema
5. Ankush Jalhotra, (2009), Study of vibration characteristics of different materials by sine sweep test.
M.Eng. Thesis, Mechanical Engineering department, Thapar University, Patiala, India.
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publications.

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