Me mv-16-17 unit-1.2

Mechanical Department
COURSE NAME: MECHANICAL VIBRATIONS
Prepared By:
MD ATEEQUE KHAN
(Assistant Professor)
Mechanical Engineering Department
JIT,Barabanki,U.P. INDIA
12/31/2016 1NME-013 MA KHAN

Table of Contents
Unit-1.2
1. Single Degree Freedom System
2. Equation of motion, Newton’s method
3. D’Alembert’s principle
4. Energy method
5. Free vibration, Natural frequency
6. Equivalent systems
7. Displacement, Velocity and acceleration
8. Response to an initial disturbance
9. Torsional vibrations
10.Damped vibrations
11.Vibrations of systems with viscous damping
12.Logarithmic decrement
13.Energy dissipation in viscous damping.
14.Objective Questions
12/31/2016 2NME-013 MA KHAN

Unit-1.2: Mechanical Vibrations
Single Degree Freedom System
Degree of Freedom: Degree of freedom may be defined as the minimum number of
independent coordinates required to determine completely the position of all parts of a system
at any instant of time.
Examples of single degree of freedom systems:
Fig(1) Fig(2)
Spring mass system Torsional System
k
x
m

12/31/2016 3NME-013 MA KHAN

Equation of motion: Newton’s method
Statement: Rate of change of momentum in any direction is equal to net force applied in that direction.
Free Body Diagram:
ts
k
mo o
x
Free length
ksFs 
m
mg
mgks
kx
..
xm
m
0
..
 kxxm
12/31/2016 4NME-013 MA KHAN

Unit-1: Mechanical Vibrations
D’Alembert’s principle
12/31/2016 5NME-013 MA KHAN

Energy method
Statement: In case of undamped free vibration, sum of kinetic energy and potential energy is
constant only transformation of energy takes place.
Putting this value in equation-1 and after differentiation we get:
This is known as differential equation of motion. It is homogeneous, second order, ordinary
differential equation of motion.
cUT 
)1..(....................0)( 


UT
t
2
.
2
2.
2
1
)(
2
1
2
1
)(
2
1
kxxmUT
kxU
xmT



kxxm 
..
12/31/2016 6NME-013 MA KHAN

Solution of Differential equation of Motion: Natural Frequency;
Dividing by m:
This is a homogeneous, second order equation with constant coefficient, so we
can assume the solution as below:
By substituting for x we get:
By solving the above equation:
0
..
 kxxm
0
..
 x
m
k
x
st
Cex 
m
k
is
m
k
s
Ce
Ce
m
k
s
st
st
















0
0
0
2
2
t
m
k
it
m
k
i
eCeCx
)(
2
)(
1

































t
m
k
Xx
XB
XA
Let
t
m
k
Bt
m
k
Ax
sin
cos
sin
:
sincos
m
k
n 
12/31/2016 7NME-013 MA KHAN

Equivalent systems
1k
2k
m
m
1k 2k
m
1k
2k
21
111
kkkeq
 21 kkkeq  21 kkkeq 
12/31/2016 8NME-013 MA KHAN

Torsional vibrations
Torsional vibration is angular vibration of an object, commonly it is related to
the shaft along its axis of rotation. Torsional vibration is often a concern in
power transmission systems using rotating shafts or couplings where it can
cause failures if not controlled as per design requirement.
12/31/2016 9NME-013 MA KHAN

Damped vibrations: Viscous damping
Damped vibration system FBD
k c
m m
..
xk
.
xc
12/31/2016 10NME-013 MA KHAN

Logarithmic decrement
Logarithmic decrement
Logarithmic Decrement as a function of damping ratio
t
x
1X
2X
)sin( 
 
tXex d
tn
tn
Xex 








2
1
log
x
x
e
12/31/2016 11NME-013 MA KHAN

Me mv-16-17 unit-1.2

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