LECTURE MODULE

in ENGINEERING VIBRATION
Engr. G Sainz Belonio

TERM: PRELIM

LESSON 1: INTRODUCTION TO VIBRATION

Vibrations are the fluctuations of a mechanical system about an equilibrium position. In order
for vibration to occur, the mechanical system must be subjected to a restoring force or
restoring moment, that continually pulls the system toward its equilibrium position.
Vibrations are initiated when energy is imparted to the mechanical system by an external
source.

The study of vibration is concerned with the oscillation of masses and the associated forces.
Oscillations occur because of the way masses and the elasticity of materials combine. When
vibrating, the mass of a material possesses momentum which tries to stretch and squash the
material. This is analogous to a mass suspended on a spring.

All bodies possess mass and in so doing also possess a degree of elasticity. It can be said that
everything will vibrate to some extent. The vibration is at a fairly high frequency and is
perceived as sound, but dies away quickly. Example, hold a ruler over the edge of a desk and
flick the end. The oscillations are so slow that they can be seen. The ruler oscillates because
there is energy within the mass, trying to move it. The material itself, however, provides the
“springiness” which allows the mass to vibrate.

Examples of Common Occurrences of Vibrations in Engineering Systems:

(1) Vibrations of a structure induced by an earthquake lead to large stresses and can result in structural
failure.
(2) Vibrations resulting from the rotating unbalance in a helicopter blade can lead to the pilot’s losing
control of the helicopter.
(3) Vibrations of machine tools lead to improper machining of a part.
(4) Vibrations of compressors and pumps increase the noise level in the surroundings.
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(5) Vibrations in the suspension system of an automobile protects the passengers and the automobile
from rough roads.
(6) Vibrations of cushioning materials used in packaging fragile items to prevent leakage when moved
or even dropped.

CAUSES OF VIBRATION
The main causes of vibration are the following listed below:
(1) Unbalanced Force
The unbalanced force maybe due to mass unbalance, such as in an eccentrically mounted rotor, or
it may be due to variable inertia forces in the machinery, which does not move uniformly, like the
crank and connecting rod motion or the cam-follower system.
(2) Dry Friction Between Mating Surfaces
When two parts rub against each other, they are definitely producing self-excited vibrations.
(3) Misalignment
When two shafts are connected through coupling or a shaft is supported on a bearing,
misalignment may result. The misalignment between the shaft axes give rise to vibrations.
(4) Wind- Induced Vibration
Winds may cause vibration of transmission lines and communication cables which result in faults,
service interruptions and sometimes structural damage. A bridge can also collapse due to this.
(5) External Excitations
Those excitations result due to some external fluctuating force acting on the system. Vibrations
due to external may be periodic, random or of the nature of impact load. Vibrations caused by
rough road to a person driving an automobile is an example of external excitation.

EFFECTS OF VIBRATION
The following are adverse effects of uncontrolled vibrations:
(1) Excessive stresses and force levels may be set up as a result of vibration and in extreme cases
may lead to part failure.
(2) There is increased wear of parts and unsatisfactory equipment performance.
(3) Loosening of fasteners such as bolts and nuts may result.
(4) In metal machining processes, vibration may cause chatter, resulting in poor surface finish. Tool
life will be affected and machinery will more frequent maintenance.

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(5) Undesirable noise is a result of vibration. The vibration and noise generated by engines cause
annoyance to people and sometimes, damage to property.

CLASSIFICATIONS OF VIBRATION
(1) Free and Forced Vibration
Free Vibration
This takes place when energy within the mass-elastic system creates internal forces which
push and pull the mass into oscillatory motion. A system which is allowed to vibrate in this
natural way will do so at one of its natural frequencies. A natural frequency is a particular “speed”
of oscillation and is a product of the amount of mass and the elastic distribution within the
vibrating system. It is the frequency at which system naturally prefers to vibrate. Example of free
vibrating system was considered earlier with the ruler.

Forced Vibration
This takes place when forces from outside the mass-elastic system pushes or pull the mass. Often
these external forces are themselves oscillating and cause the system to oscillate at the same speed.
An excellent example of forced vibration is when the fuel ignites within an internal combustion
engine, and the subsequent oscillations of the engine impart forces to the vehicle chassis. These
are perceived as vibration and engine noise within the passenger compartment. Another good
example is that of a washing machine. When set to spin, the washing may congregate in lumps
within the drum. As the drum spins at high speed, the mass of washing imparts a force on the
drum, due to the centrifugal force, which rotates as the drum rotates. This rotating force is passed
to the washing machine body as an oscillating force.

(2) Undamped and Damped Vibration

If during oscillation, no energy is lost or dissipated in friction or by other resisting element, the
vibration is known as undamped vibration.

If a damper element is present in the system, energy of the system will be dissipated by friction
and other resistances and the resulting vibration is called damped vibration.

(3) Deterministic and Random Vibration

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 If the magnitude of excitation (force or motion) acting on a vibratory system is known at a given
time, the excitation is known as deterministic.

If the value or magnitude of excitation acting on a system can’t be predicted at a given time, the
excitation is called as random.

(4) Steady and Transient Vibration

A steady state vibration is one in which the motion is repeated exactly in each cycle.
Transient vibration is one in which amplitude of vibration decays continuously because of
damping present in the system.

(5) Linear and Non-linear Vibration

If elements of a vibrating system, example, spring, mass and damper behave in a linear manner,
the vibration is known as linear vibration. Linear differential equations are formed in linear
vibrations.

If any of the basic elements of a vibrating system behave non-linearly, the vibration is called non-
linear vibration.

(6) Rectilinear and Torsional Vibration

Rectilinear translation appears in two forms, longitudinal and transverse vibration. Longitudinal
vibration is the axial extension or compression of bars and wires and include the stretching and
compression of coil springs.
Transverse vibration include motion of beams perpendicular to their center lines. The amplitude of
rectilinear vibration is measured in mm or fraction of an inch.

Torsional vibration happens in motion of rotation or twisting, as in shafts and their amplitude is
measured in radians or degrees.

IMPORTANT TERMS AND DEFINITIONS

(1) Oscillation: The movement of a mass, to and fro, between two extreme points.

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 (2) Periodic Motion: Any motion that repeat itself after equal intervals of time is called periodic
motion.

(3) Simple Harmonic Motion (SHM): It is the simplest type of periodic motion in which
acceleration is always directed towards the mean position and is proportional to displacement.

(4) Cycle: The interval of time within which the motion sequence repeats itself is called a cycle.

(5) Period (T): Time taken to complete one cycle of oscillatory motion measured in seconds.

(6) Frequency (f): Number of cycles of motion per unit time. It is usually measured in cycles per

1 ω
second. f = = where: f – frequency in Hertz (Hz)
T 2π
T = time period in seconds
(7) Amplitude (A): The maximum height of an oscillation measured from the axis of the sine
wave to the top of the peak. Often amplitude refers to the displacement but it may also refer to
values of velocity, acceleration or force.

(8) Natural Frequency(ω ¿: After giving initial displacement from the equilibrium position, if the
system is left to vibrate on its own without any external force, the frequency with which it
vibrates is known as its natural frequency.

(9) Resonance: When the frequency of external excitation equals with one of the natural
frequencies of the system, the phenomenon of resonance occurs. At resonance, the system
vibrates with large amplitude resulting in high stresses.

(10) Degree of Freedom: The number of independent spatial coordinates required to describe
the motion of a system is termed as degree of freedom.

ASSIGNMENT # 1 - PRELIMS
Instructions: Essay type questions. Answer and explain in your own understanding.
(1) What do you understand about the term “vibration” as far as engineering viewpoint is concerned?

5
 (2) Why engineering vibration is important in the study of mechanical engineering?
(3) Cite three (3) situations that vibration can occur.

LESSON 2: HARMONIC MOTION

The simplest form of periodic motion is harmonic motion, and is demonstrated by a mass
suspended on a spring, oscillating with an up-and-down motion. If the amplitude of this
motion is plotted against time, a sine wave is produced, as shown below.

A x

M m

A p
θ x
GGGG

ANALYSIS:
The height of the sine wave (x) or the displacement of the particle can be given by the equation,
x = A sin θ = A sin ω t
where : θ = ω t
x = height of the sine wave at any instant in time (mm)
A = maximum height of the sine wave or displacement (mm)
θ = angle projected by the height of sine wave (x) and amplitude (A)
ω = natural frequency in rad/sec
t = elapsed time in seconds

Illustrative Problem 1:
A body is suspended on a spring and oscillates at a frequency of 2 Hz. If the maximum

6
 displacement amplitude is measured as 10 mm, determine:
(a) The frequency in radian per second.
(b) The displacement x after an elapsed time of 4 seconds

Approach:
Given: f = 2 Hz A = 10 mm

Solution:
(a) Finding for the natural frequency in rad/sec (ω):
1 ω
f= =
T 2π
ω = 2πf = (2) (π) (2) = 12.56 rad/sec (answer)

(b) Finding for the displacement (x):

At t = 4 sec,
x = A sin θ = A sin ω t
x = (10 mm) sin [ (12.56 rad /sec)( 4 sec) ] = 7.7 mm (answer)

Illustrative Problem 2:
A harmonic motion has an amplitude of 0.06 m and a frequency of 20 Hz. Find:
(a) The time period.
(b) The maximum velocity
(c) The maximum acceleration

Approach:
Given: A = 0.06 m f = 20 Hz

Solution:
(a) Finding for the time period in sec (T):
1 ω
f= =
T 2π

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 1 1
T= = = 0.05 sec (answer)
f 20
(b) Finding for the maximum velocity (v):
dx
v= = A sin ω t
dt

LESSON 3: SIMPLE HARMONIC MOTION

Simple Harmonic Motion – is defined as the periodic motion of a body, or point, the acceleration of which
is always towards a fixed point lying on its path and proportional to its displacement from that point.

Spring – Mass System:

The preceding discussion has described the importance of predicting natural frequency but this
needs to be done in a way which takes account of the physical aspects of the vibrating system. For a
simple mass-elastic system there would be mass (m) and spring stiffness (k).

Spring Stiffness
Spring stiffness (k) is the “strength” of the spring. Due to high loads, the spring stiffness, say in a vehicle
suspension, will have a high value of k, while the spring in a ball-point pen will have a low value of k.

The value of k is the force required to stretch or compress a spring over a certain distance. Double the
force and the distance will also be doubled. For a uniform open coiled spring this relationship is largely
linear.

The value of k for any spring or elastic component can be determined by considering the expression
below.

force applied F
k= =
distance∨extension done d

The spring stiffness k is usually the slope of the line and can be shown by hanging mass on an elastic
band.

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 Force (F) k
y
x

Extension (d)

Calculation of Displacement, Velocity, Acceleration and Force on a Vibrating System:

p
A A
GGGG θ
x t

ANALYSIS:
1. Displacement of the particle (x):
x = A sin θ = A sin ω t where : θ = ω t

x = displacement at any instant in time

A = maximum height of the sine wave or amplitude
θ = angle projected by the height of sine wave
ω = natural frequency in rad/sec
t = elapsed time in seconds

2. Velocity of the particle can be solved by the equation,

dx
v = dt = ω x cos ω t

3. Acceleration of the particle can be solved by the equation,

dv
a = dt = - ω 2 x sin ω t

Note: if the maximum velocity and maximum acceleration of the particle is required respectively,
then x = A or use the amplitude instead of displacement x.
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 dx
(a) Maximum velocity: v= = ω A cos ω t = ω A
dt

dv
(b) Maximum acceleration: a = = - ω 2 A sin ω t = ω 2 A
dt

4. External force applied to create vibration (F):

F=mxa

5. Spring stiffness (k)

force applied F
k= =
distance∨extension done d
k
k = m ω2 or ω=
√ m

ILLUSTRATIVE PROBLEM 1:
A body of mass 2.5 kg describes SHM of amplitude 500 mm and periodic time of 2.5 sec.
Determine:
(a) Its velocity and acceleration at a displacement of 250 mm from the center.
(b) Its maximum velocity and acceleration.
(c) The maximum force acting on the body.

Approach:
Given: m = 2.5 kg A = 500 mm t = 2.5 sec

Solution:
(a) velocity and acceleration at a displacement of 250 mm from the center:
Extremity from the center = 250 mm
2π
v = ω x cos ω t where: ω = 2 π /t = 2.5 = 2.51 rad/sec

x = A – extremity = 500 mm – 250 mm = 250 mm

for the velocity:

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 v = (2.51 rad/sec) (0.025 m) cos (2.51 rad/sec) (2.5 sec)
v = 1.09 m/sec (answer)

for the acceleration:

a = - ω 2 x sin ω t
a = - (2.51)2(0.025) sin (2.51)(2.5)
a = 1.58 m/sec2 (answer)

(b) for maximum velocity and acceleration:

Maximum velocity: v = ω A = (2.51 rad/sec) (0.5 m) = 1.255 m/sec

Maximum acceleration: a = ω 2 A = (2.51)2 = (0.5 m) = 3.15 m/sec2

(c) The maximum force acting on the body:

F = m . amax = (2.5 kg) (3.15 m/sec2) = 7.875 kg – m/sec2 or 7.875 N

ILLUSTRATIVE PROBLEM 2:
A block of mass 10 kg vibrates with Simple Harmonic Motion at a frequency of 100 Hz. The
total displacement of the mass has been measured at 4 mm. Determine:
(a) The elapsed time when the mass is 0.5 mm from the extremity of its stroke.
(b) The velocity of the mass when it is 0.5 mm from the extremity of its stroke.
(c) The acceleration of the mass when it is 0.5 mm from the extremity of its stroke.
(d) The maximum force acting on the block.

Approach:
Given: m = 10 kg, f = 100 Hz, total displacement of mass = 4 mm
therefore, A = 2 mm =amplitude

Solution:
(a) Solving for the elapsed time (t):
x = A sin ω t

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 where: x = A – extremity of the stroke = 2 mm – 0.5 mm = 1.5 mm
x
ω t = sin-1 ( )
A
x
sin−1 ( )
t= A where: ω = 2 π f = 2 π (100) = 628.3 rad/sec
ω

1.5
sin−1 ( )
hence, t= 2 = 0.00135 sec (answer)
628.3

(b) Solving for the velocity of the mass (v):

dx
v= = ω x cos ω t = (628.3 rad/sec) (0.002 m) cos (628.3 rad/sec)(0.00135 sec)
dt

v = 0.831 m/sec (answer)

(c) Solving for the acceleration of the mass (a):

dv
a = dt = - ω 2 x sin ω t

a = - (628.3)2(0.002) sin (628.3)(0.00135)

a = 592.2 m/sec2 (answer)

(d) Solving for the maximum force acting on the mass (F):

F = m . amax

where: amax = ω 2 A = (628.3)2 rad2/sec2 (0.002 m) = 789.5 m/sec2

F = (10 kgm) (789.5 m/sec2) = 7895 kgm -m/sec2 or 7895 N

ASSIGNMENT # 1:
Instructions: Essay type questions. Answer and explain in your own understanding.

(1) What do you understand about the term “vibration” as far as engineering viewpoint is concerned?
12
 (2) Why engineering vibration is important in the study of mechanical engineering?

(3) Cite three (3) situations that vibration can occur and explain the situation.

ASSIGNMENT # 2:
Instructions: Solve the following problems below. Show pertinent solutions.

1. A harmonic motion has frequency of value of 15 cps and its maximum velocity is 5 m/s. Calculate
the following:
(a) Amplitude (ans. 0.053 m)
(b) Time period of oscillation (ans. 0.067 sec)
(c) Maximum acceleration (ans. 466.1 m/sec2)

2. A harmonic moving particle has an amplitude of 3.5 cm and a frequency of 5 Hz. Find the
following:
(d) the time period
(e) the maximum velocity
(f) the maximum acceleration

3. A harmonic motion is given by x = 6 sin (20 t + 60o) m, where t is in seconds and phase angle in
degrees. Find:
(a) Frequency and time period of oscillation (ans. f = 20 rad/sec, t = 0.314 sec)
(b) Maximum velocity and acceleration (ans. v = 120 m/s, a = 2400 m/sec2)

ASSIGNMENT # 3:
Instructions: Solve the following problems below. Show pertinent solutions.

4. A body of mass 20 kg moves with simple harmonic motion whose amplitude is 150 mm and
frequency of vibration is 5 Hz. Determine:
(a) The periodic time
(b) The circular frequency of the motion.
(c) The maximum values of velocity and acceleration.
(d) The displacement at which velocity and acceleration are half of the maximum values.
(e) The maximum force which acts on the body.

5. A body of mass 50 kg describes SHM along a straight line. The time for one complete oscillation
is 10 sec and the amplitude of the motion is 1.3 m. Determine:
(a) The maximum force acting on the body.
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 (b) The velocity of the body when it is 1 m from the amplitude position.
(c) The time taken for the body to travel 0.3 m from the amplitude position.

E N D OF P R E L I M
MIDTERMS:

LESSON 4: SYSTEMS HAVING LINEAR OSCILLATION

Consider a spring – mass system as shown in the figure (picture) below. The mass (m) is
connected at the lower end of a weightless spring of stiffness (k), while the upper end of spring is
attached to a rigid support. The mass is constrained to move in vertical direction only.

Due to weight W of mass m, the spring will deflect by an amount ∆st, called static
deflection. Figure (b), which is at the static equilibrium position, the downward
gravitational force on the mass is balanced by upward spring force.

Analysis:
(1) Weight of mass, W:
W = m g = k (∆st)

(2) Force acting on the spring, F:

F=ma

(3) Spring stiffness (k)

force applied F
k= =
distance∨extension done d
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 k
k = m ω2 ω=
√ m
(4) Natural frequency:
k
ω=
√ m
, rad/sec

g
ω=
√ ∆ st
ω 1 k
f = =
2π 2π √ m
, cycle/sec (Hertz)

ILLUSTRATIVE PROBLEM 1:
A mass of 10 kg when suspended from a spring, causes a static deflection of 0.01 m. Find the
natural frequency of the system in Hertz.

Given: m = 10 kg ∆st = 0.01 m

Solution: Solving for the natural frequency, f

ω
f=
2π
g 9.81m/ sec 2
where: ω=
√ ∆ st
=
√ 0.01 m
= 31.3 rad/sec

31.3
hence, f = 2 π = 4.98 Hz (answer)

ILLUSTRATIVE PROBLEM 2:
A spring mass system has spring stiffness of k (N/m) and mass of m (kg). The natural
frequency of the system is 12 Hz. When an extra 2 kg mass is coupled to mass m, the natural
frequency reduces by 2 hertz. Find the value of k and m.

Given:
Original system: k = spring stiffness m = mass f1 = 12
New system: new mass = m + 2
new frequency = 12 – 2 = 10 Hz

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 Solution:
(a) Solving for the original mass (m):
1 k
For original system, f1 =
2π √ m
= 12

1 k
For new system, f2 =
2π √
m+2
= 12 – 2 = 10

Dividing the f1 and f2:

1 k
10
12
=
2 π m+2
1 k
√
2π m √
Squaring both sides:
100 m
144
= m+ 2
m = 4.54 kg (answer)

(b) Solving for the original spring index, k:

1 k
f1 =
2π √ m
1 k
12 =
2π √ 4.54
k = 25,809.4 N/m (answer)

LESSON 5: SYSTEMS HAVING ANGULAR OSCILLATION

SIMPLE PENDULUM
This is an example of a system having single degree of freedom and possessing undamped
free vibrations.
A simple pendulum consists of a concentrated mass (m) carried at one end of a light
inextensible string or wire or rod, the other end of which is pivoted.
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 o o

θ L
L
m
θ
m mg sin θ mg cos θ
W = mg
Let the pendulum be displaced from vertical mean position by an angular deflection θ, as shown
in the figure.
 The component of the weight, mg sin θ act as a restoring force which tend to restore the pendulum
bob or mass, to its equilibrium position and the corresponding restoring torque is mg sin θ L,
where L = length of the pendulum rod or string.
 The inertia torque about the pivot point O is equal to is I α , where I = mass moment of inertia
and α = angular acceleration.
 From the Alembert’s Principle: Inertia Torque + Restoring Torque = 0
I α + mg sin θ L = 0
Analysis:
(1) Mass moment of inertia, I:
I = m L2

(2) Natural frequency:

g
ω = 2π f =
√ L
rad/sec

ω 1 g
f = =
2π 2π √ L
cycle/sec (Hertz)

(3) Time period of oscillation, T:

1 L
T=
f
= 2π
√ g

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 (4) Time of one beat, t:
The movement of the bob from one end to the other constitutes half an oscillation is called
a beat or a swing.
1 L
t=
2
T=π
√ g

(5) Number of beats, n:

T g
n=
π √ L

ASSIGNMENT # 1- MIDTERMS
A closed coiled helical spring undergoes a static deflection of 30 mm when a mass of 2.5 kg is placed
on its lower end. The mass is then pulled downwards through a further distance of 20 mm and
released so that it oscillates about the static equilibrium position. Neglecting air resistance, energy
losses in the spring material and the mass of the spring,
(a) Determine the periodic time and natural frequency of vibration.
(b) Determine the maximum velocity and acceleration of the mass.

QUIZ 1 - MIDTERMS
An elastic spring of stiffness 0.4 KN/m is suspended vertically with a load attached to its lower end.
When displaced, the load is seen to oscillate with a periodic time of 1.27 seconds. Determine:
(a) the magnitude of the load
(b) the acceleration of the load when it is 25 mm from the equilibrium position
(c) the tension in the spring when the load is 25 mm from the equilibrium position

ASSIGNMENT # 2 - MIDTERMS
A simple pendulum of length 1.25 m is allowed to swing freely with amplitude of 150 mm.
Determine:
(a) the circular frequency of the system
(b) the periodic time of the oscillations
(c) the maximum velocity
(d) the maximum acceleration
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QUIZ 2 - MIDTERMS
A simple pendulum is made from light cord 900 mm long and a concentrated mass of 2.25 kg.
(a) What will be the periodic time of the oscillations when the pendulum is given a small
displacement and allowed to swing freely?
(b) What will be the stiffness of the spring which will have the same periodic time when carrying
the same mass?

QUIZ # 3 - MIDTERMS
The length of a close-coiled helical spring extends by a distance of 25 mm when a mass of 0.5 kg is
placed on its lower end. An additional mass of 1.5 kg is then added and displaced so that the system
oscillates with amplitude 50 mm. Determine
(a) the circular frequency of the system
(b) the natural frequency of the oscillations
(c) the maximum velocity
(d) the maximum acceleration

END OF MIDTERMS

TERM: FINALS

LESSON 7: VIBRATION ON EQUIVALENT SPRINGS

In many practical applications, springs are used in combination. The combination types are series and
parallel. These springs can be replaced by a single equivalent spring.

A. SPRINGS IN SERIES
The figure shows a system having two springs of stiffness k1 and k2 connected in series. If these are
replaced by a single equivalent spring, then the total static deflection of the body in two cases under
the same load can be the same.

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 Let: δ 1 = deflection of the first spring k1 = stiffness of the first spring
δ 2 = deflection of the second spring k2 = stiffness of the second spring

Analysis:

(a) Static deflection of equivalent spring, δ :

δ = δ1 + δ2

(b) Spring constant of equivalent spring, k e:

1 1 1
= +
ke k1 k2

(c) Natural frequency:

ke
ω=
√ m
, rad/sec

ω
f= , hertz
2π

(d) Time period of vibration, t:

1
t= , sec
f

B. SPRINGS IN PARALLEL
The figures (a) and (b) below shows the two cases with springs in parallel.

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Analysis:

(a) Static deflection of equivalent spring, δ :

δ = δ1 = δ2

(b) Spring constant of equivalent spring, k e:

ke = k1 + k2

(c) Natural frequency:

ke
ω=
√ m
, rad/sec

ω
f= , hertz (Hz)
2π

(d) Time period of vibration, t:

1
t= , sec
f

SAMPLE PROBLEM 1:
Find the natural frequency of the spring-mass system shown below. Given k1 = 2000 N/m,
k2 = k3 = 1000 N/m, m = 5 kg.
Figure:

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Given: k1 = 2000 N/m
k2 = k3 = 1000 N/m
m = 5 kg.

Required: Natural frequency in rad/sec and in Hertz

Solutions:
(a) Solving for the natural frequency in rad/sec:
ke
ω=
√ m
, rad/sec

Finding for the equivalent spring stiffness ke :

in figure (c), the spring equivalent becomes parallel, thus

ke = k1 + k2 + k3 = 2000 + 1000 + 1000 = 4000 N/m

m
4000 N /m 4000(kg− )/m
then, ω =
√
5 kg
=
5 kg
sec 2 = 28.28 rad/sec (answer)

(b) Solving for the natural frequency in hertz:

22
 ω 28.28 rad /sec
f = = = 4.5 cycle/sec (Hertz) (answer)
2π 2 π rad /cycle

SAMPLE PROBLEM 2:

A spring system is shown in the figure below. It is given that k1 = 10 N/m, k2 = 5 N/m, k3 = 10 N/m,
k4 = k5 = 20 N/m. Find the mass (m) if the system has a natural frequency of 2.5 Hz.

Figure:

Given:
k1 = 10 N/m k2 = 5 N/m k3 = 10 N/m
k4 = k5 = 20 N/m f = 2.5 Hz

Required: mass (m) attached to the spring system

ke
Solution: ω =
√ m
, rad/sec

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 ω 1 ke
f =
2π
= 2π √ m

Finding for the equivalent spring stiffness of the spring system, ke:

Note: Springs k1 , k2 and k3 are connected in series, so they can be replaced by a single
equivalent spring ka where the stiffness can be,
1 1 1 1 1 1 1
= + +
ka k1 k2 k3
= 10 + 5 + 10

4
ka = = 2.5 N/m
10

Springs k4 and k5 are connected in parallel, so they can be replaced by a spring equivalent
kb,

kb = k4 + k5 = 20 + 20 = 40 N/m

Referring to figure (b), the spring stiffness ka and kb are in parallel now, and can be
reduced to a single equivalent stiffness ke:

ke = ka + kb = 2.5 + 40 = 42.5 N/m

1 ke
finally, from the equation f = 2 π
√ m
1 42.5
2.5 = 2 π
√ m

m = 0.17 kg (answer)

LESSON 8: TORSIONAL VIBRATIONS

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Consider a disc or rotor suspended on an elastic rod as shown in the figure. The rod is rigidly fixed at
Its base and rigidly attached to the center of the disc.

If the disc is given an angular twist about its vertical axis and then released, torsional vibrations will
occur in the system.

Let θ = angular twist of the disc from its equilibrium position

T = torque required to produce the twist
I = moment of inertia
G = modulus of rigidity of the material (or rod)
J = polar moment of inertia of the rod
d = diameter of the rod
kt = torsional spring constant
L = length of the rod

Analysis:

(e) Angular Twist of the Disc, θ

TL
θ=
JG

(f) Torsional Spring Constant, kt:

T GJ
kt = =
θ L

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 (g) Natural frequency:

kt
ω=
√ I
, rad/sec

ω
f= , hertz
2π

(h) Time period of vibration, t:

1
t= , sec
f
SAMPLE PROBLEM:

Find the torsional stiffness natural frequency of torsional oscillations for the system shown in the
figure
below. Neglect the mass of steel shaft. Take G = 83 Gpa.
Figure:

Given: d1 = 0.05 m L1 = 2 m
d2 = 0.025 m L2 = 1 m
I = 15 kg-m2
Required:
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 (a) Torsional stiffness
(b) Natural frequency in rad/sec and hertz

Solution:
(a) Solving for the natural stiffness, kt:
T GJ
kt = =
θ L

Since rods 1 and 2are connected in series, so the equivalent torsional stiffness k te:
1 1 1
= +
k te k 1 k 2

T G J1 π d 14
where: k1 = = , J1 = = π ¿ ¿ = 6.136 x 10-7m4
θ L1 32
( 83 x 109 ) N /m2 (6.136 x 10−7 m 4)
k1 = = 25,464.4 N/m
2m

T G J2 π d 14
k2 = = , J2 = = π ¿ ¿ = 3.835 x 10-8 m4
θ L2 32

( 83 x 109 ) N /m2 (3.835 x 10−8 m4 ) = 3183.05 N/m

k2 =
1m

1 1 1
then, = +
k te 25,464.4 3183.05

kte = 2.83 x 103 N/m (answer)

(b) Solving for the natural frequency:

kt
ω=
√ I
, rad/sec

N kg−m/ sec 2
ω=
√ 2.80 x 10 3
m
x

15 kg−m 2
N = 13.73 rad/sec (answer)

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 ω 13.73
f= = = 2.18 Hz (answer)
2π 2π

ASSIGNMENT – FINALS
A mass m is suspended from a spring system shown in the figure. The following data are given: k 2 = k3 = 500
N/m and k1 = 1000 N/m, m = 10 kg.
(a) Determine the equivalent spring stiffness.
(b) Determine the natural frequency in rad/sec and in Hertz.
Figure:

QUIZ – FINALS
1. Find the equivalent spring constant of the system shown below if m = 3 kg , k = 12 N/m. Find also the natural
frequency in rad/sec and hertz.

Figure:

2. Determine the natural frequency of the system shown in the figure given that k 1 = 2000 N/m, k2 = k3 = 1500
N/m and mass m = 5 kg.

Figure:

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END OF FINAL TERM

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