VIBROMETER

1. CHAPTER-5
VIBRATION MEASURING INSTRUMENTS
Topics covered:
Theory of Vibration measuring instruments
Displacement measuring instrument (Vibrometer)
Velocity measuring instrument (Velometer)
Acceleration measuring instrument (Accelerometer)
Numerical Problems/Discussions
Theory of Vibration measuring instruments
It is well known that the dynamic forces in a vibratory system depend on the
displacement, velocity and acceleration components of a system:
Spring force  displacement
Damping force  velocity
Inertia force  acceleration
Therefore, in vibration analysis of a mechanical system, it is required to measure the
displacement, velocity and acceleration components of a system. An instrument,
which is used to measure these parameters, is referred as vibration measuring
instrument or seismic instrument. A simple model of seismic instrument is shown in
Fig.5.1. The major requirement of a seismic instrument is to indicate an output, which
represents an input such as the displacement amplitude, velocity or acceleration of a
vibrating system as close as possible.
m-seismic mass
c-damping coefficient of seismic unit
K-stiffness of spring used in seismic unit
x-absolute displacement of seismic mass
y-base excitation (assume SHM)
K
x
m
c
z
Machine
y=Y sint
Frame
Scale
Fig.5.1 Seismic instrument

2. VTU e-learning Course ME65 Mechanical
Vibrations
Dr. S. K. Kudari, Professor Session: I&II 03-04/04/07
Deptt. Mech. Engg.,
B. V. B. College of Engineering and Technology, Hubli - 580031.
2
z=(x-y) displacement of seismic mass relative to frame
To study the response of the system shown in Fig.5.1, we shall obtain the equation of
motion of seismic mass:
0
)
(
)
( 



 y
x
K
y
x
c
x
m 


 (1)
y
m
Kz
z
c
z
m 



 


 (2)
Considering base excitation to be SHM:
y(t) =Y sint (3)
t
Y
m
Kz
z
c
z
m 
 sin
2


 

 (4)
The above equation represents a equation of motion of a forced vibration with
F
Y
m 
2

Solution of governing differential equation is:
zc is the complimentary solution, which nullifies after some time. The total solution is
thus, only steady state solution zp
Let, the steady state solution of Eqn.(4) is:
)
sin(
)
( 
 
 t
Z
t
z (5)
Eqn.(5) has to satisfy Eqn.(4). Substitute Eqn.(5) in (4) and draw force polygon as
already studied in forced vibration. The amplitude of steady state vibration is:
2
2
2
2
)
(
)
( 


c
m
K
Y
m
Z


 (6)
devide above equation by K
2
2
2
2
)
2
(
)
1
( r
r
Y
r
Z



 (7)
substitute eqn.(5.7) in eqn.(5.5)
)
sin(
)
2
(
)
1
(
)
(
2
2
2
2






 t
r
r
Y
r
t
z (8)
the phase angle is:







 
2
1
tan



m
K
c
(9)







 
2
1
1
2
tan
r
r

 (10)
The variation of non-dimensional amplitude (Z/Y) with respect to frequency ratio (r)
is shown in Fig.5.2
)
(
)
(
)
( t
z
t
z
t
z p
c 


3. VTU e-learning Course ME65 Mechanical
Vibrations
Dr. S. K. Kudari, Professor Session: I&II 03-04/04/07
Deptt. Mech. Engg.,
B. V. B. College of Engineering and Technology, Hubli - 580031.
3
Displacement measuring instrument (Vibrometer)
It is an instrument used to measure the displacement of a vibrating system.
In Eqn.(8) if,
1
)
2
(
)
1
( 2
2
2
2


 r
r
r

(11)
then,
)
sin(
.
)
( 
 
 t
Y
t
z (12)
Eqn.(11) is the condition for vibrometer.
Acceleration measuring instrument (Accelerometer)
It is an instrument used to measure the acceleration of a vibrating system. The
response of the seismic mass is given by Eqn.(8). Double differentiating the Eqn.(8),
we get.
))
sin(
(
)
2
(
)
1
(
)
( 2
2
2
2
2
2




 




 t
Y
r
r
r
t
z (13)
))
sin(
(
)
2
(
)
1
(
1
)
( 2
2
2
2
2




 




 t
Y
r
r
t
z n (14)
In above equation if
1
)
2
(
)
1
(
1
2
2
2


 r
r 
(15)
Then,
)
sin(
)
( 2
2



 


 t
Y
t
z n (16)
we have acceleration component of base excitation:
Eqn.(15) is the condition for accelerometer.
Fig.5.2 Plot of equation 7
0 1 2 3 4
0
1
2
3
4
=0.0
=0.1
=0.2
=0.3
=0.4
=0.5
=0.707
=1
Z/Y
/n
(r)

4. VTU e-learning Course ME65 Mechanical
Vibrations
Dr. S. K. Kudari, Professor Session: I&II 03-04/04/07
Deptt. Mech. Engg.,
B. V. B. College of Engineering and Technology, Hubli - 580031.
4
Numerical problems
Problem-1
A seismic instrument is mounted on a machine running at 1000 rpm. The natural
frequency of the seismic instrument is 20 rad/sec. the instrument records relative
amplitude of 0.5 mm. Compute the displacement, velocity and acceleration of the
machine. Neglect the damping in seismic instrument.
Given data
n=20 rad/s, =0
Speed of the machine (N) = 1000 rpm
60
)
1000
(
2
60
2 

 

N
=104.72 rad/s
Frequency ratio
23
.
5
20
72
.
104



n
r


For seismic instrument
2
2
2
2
)
2
(
)
1
( r
r
r
Y
Z




For the given system damping is neglected
mm
Z
Y 48
.
0
042
.
1
5
.
0
042
.
1



042
.
1
23
.
5
1
23
.
5
2
2



Y
Z
Displacement of the machine:
mm
Z
Y 48
.
0
042
.
1
5
.
0
042
.
1



Velocity of the machine:
.Y = (104.72) 0.48 = 50.26 mm/s
Acceleration of the machine:
2
.Y = (104.72)2
0.48 = 5263.81 mm/s2
Problem-2
A seismic instrument has natural frequency of 6 Hz. What is the lowest frequency
beyond which the amplitude can be measured within 2% error. Neglect damping
Given data
n = 6 Hz,  =0 and error = 2%

5. VTU e-learning Course ME65 Mechanical
Vibrations
Dr. S. K. Kudari, Professor Session: I&II 03-04/04/07
Deptt. Mech. Engg.,
B. V. B. College of Engineering and Technology, Hubli - 580031.
5
Damping is neglected for given system
2
2
1 r
r
Y
Z


02
.
0
Error 


Y
Y
Z
Z = Y+0.02 Y = 1.02 Y
2
2
1
02
.
1
r
r
Y
Z



2
2
02
.
1
02
.
1 r
r 

r= 0.7034
The lowest frequency beyond which the amplitude can be measured within 2% error
is:
 =r. n
 = (0.7034) 6
 = 4.22 Hz
Summary
Seismic instruments are used to measure the displacement, velocity and acceleration
components of a vibratory system. Basic theory of Seismic instruments is based on
forced vibration considering the vibratory system under base excitation. A single
Seismic instrument can be sued as vibrometer, velometrer and accelerometer using
suitable calibration.