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MENG 475: Applied Vibration Measurement, Analysis, and Control
Fall 2010
INTRODUCTION
Definitions
Oscillatory motion is a pattern of motion where a system moves back and forth across some
equilibrium position. The motion may or may not have a repeating pattern.
Periodic motion is a specific form of oscillatory motion where the motion pattern repeats itself
with a uniform time interval. This time interval is known as the period or periodic time. Its
reciprocal is the frequency.
Harmonic motion is a special type of periodic motion where the motion pattern is described by a
pure sine or cosine function. This motion is also known as simple harmonic motion.
The number of degrees of freedom of a system is the number of independent coordinates
necessary to determine the configuration (state) of a system at any given time.
In this chapter we will review the vibration of single degree of freedom systems. Please refer to
the textbooks by Ogata (System Dynamics) and Rao (Mechanical Vibrations).
Free vibration
The system shown in Fig. 1(a) is an undamped single degree of freedom system undergoing free
vibration. By free vibration we mean the system is disturbed from equilibrium (i.e. given some
initial conditions) and is left to vibrate freely without any external forces acting on it.
x
k
k
m
m
x
(a)
(b)
Figure 1. Single degree of freedom system.
The equation of motion is given by:
(1)
mx  kx  0
Note that the equation of motion is the same if the mass is undergoing vertical oscillation as in
Fig. 1(b). To get the response, we can assume a solution in the form:
(2)
x  Cest
where C and s are constants to be determined. Substituting Eq. (2) into Eq. (1) yields:
1
 ms
2
 k  Cest  0
(3)
For nontrivial solution, we must have
(4)
ms2  k  0
which is known as the characteristic equation. The solution of Eq. (4) is:
k
s  i
(5)
m
Letting
k
n 
(6)
m
Then we get
(7)
s  in
The solution of Eq. (1) then becomes:
x(t )  C1eint  C2e int
(8)
By using the identities:
(9)
eit  cost  i sin t
the solution becomes:
x(t )   C1  C2  cos nt  i  C1  C2  sin nt
(10)
The constants C1 and C2 must be of the nature to make the RHS expression of x(t) real. It follows
that C1 and C2 must be complex conjugates, i.e.
(11)
C1  A  iB , C2  A  iB
Therefore Eq. (10) becomes:
(12)
x(t )  2 A cos nt  2B sin nt
which can be conveniently written as:
(13)
x(t )  A1 cos nt  A2 sin nt
The constants A1 and A2 are to be determined from the initial conditions x0 and x0 . In this
regards, we have:
x(0)  x0  A1
(14)
x(0)  x0  A2n
Hence
x
(15)
x(t )  x0 cos nt  0 sin nt
n
Another solution
By observing the equation of motion:
mx  kx  0
(16)
x  n2 x  0
(17)
and rewriting it in the form:
We can seek a solution in the form:
or:
x(t )  A cos nt
x(t )  B sin nt
2
Both expressions satisfy the differential equation (17). The general solution must contain 2
arbitrary constants, therefore the solution becomes:
(18)
x(t )  A cos nt  B sin nt
which is the same as Eq. (13). Note that Eq. (18) is equivalent to:
x(t )  X sin nt   
(19)
To convert between the 2 forms, i.e. to get the constants X and , substitute t = 0 and equate the
displacements and velocities in both equations. This yields:
x(0)  A  X sin 
x(0)  B n  X n cos 
Thus
 A
,   tan 1  
B
The solution indicates that when the mass is disturbed from equilibrium, it will oscillate with a
frequency n which is called the natural frequency.
X  A2  B2
Another solution
From MENG 375, we can also get the response using Laplace Transform. Consider the equation
of motion:
(20)
mx  kx  0
Taking Laplace Transform (See Table 2-1 in Ogata) gives:
(21)
m  s 2 X (s)  sx0  x0   kX (s)  0
Rearranging:
X (s) ms 2  k   msx0  mx0
(22)
msx0  mx0
ms 2  k
msx
mx
 2 0  2 0
ms  k ms  k
x

s
 x0 2
 0 2 n 2
2
s  n n s  n
Taking inverse Laplace Transform gives:
x
x(t )  x0 cos nt  0 sin nt
(23)
Hence
X ( s) 
n
which is identical to Eq. (15). Note that the Laplace Transform was taken without assuming zero
initial conditions. Notice that the same solution was obtained using the 3 approaches.
3
Damped vibration
Figure 2 shows a spring-mass-damper system. The damper (or dashpot) provides viscous
damping.
x
k
m
c
Figure 2. Spring-mass-damper system.
The equation of motion is given by:
mx  cx  kx  0
(24)
and can be written in the standard form as:
x  2n x  n2 x  0
(25)
where
c
 
(26)
2mn
is the damping ratio. To get the response due to initial conditions, we can assume a solution in
the form:
(27)
x  Cest
Substituting Eq. (27) into (25) gives:
(28)
 s2  2n s  n2  Cest  0
For nontrivial solution, we must have:
s 2  2n s  n2  0
(29)
s  n  n  2  1
(30)
whose roots are:
and the general solution becomes:
 

 2 1 t
 

 2 1 t
(31)
x(t )  C1e
 C2e
The nature of the roots s1 and s2, and hence the behavior of the system, depends on the value of
 . For   1 (underdamping) the response is given by:
n
n
n
n
x(t )  e nt C sin d t  
(32)
where d  n 1   2 is the frequency of damped vibration and C and  are constants to be
determined from the initial conditions (see section 2.6 in Rao). Note: as an application, we can
4
estimate the damping ratio  by observing the rate of decay of free oscillations using the
“logarithmic decrement”. For   1 (critical damping) the response is:
x(t )   A  Bt  e nt
(33)
For   1 (overdamping) the response is:
  

 2 1 t
  

 2 1 t
x(t )  Ae
 Be
(34)
In this case there is no vibration. When the system is disturbed from its equilibrium, it creeps
back to it very slowly. Note: the same solutions can be obtained by using Laplace Transform, see
Ogata section 8.3.
n
n
n
n
Example: free vibration with Coulomb damping
Coulomb damping or dry friction arises when bodies slide with respect to one another. The
friction force is given by F   N and acts in a direction opposite to velocity. For the system
shown in Fig. 3,
x
k = 200 N/m
m= 10 kg
 = 0.7
Figure 3. System with Coulomb damping.
the equation of motion is given by:
(35)
mx  kx  mg sgn( x)  0
where sgn(y) is known as the signum function whose value = 1 for y > 0, -1 for y < 0 and 0 for y
= 0. Use MATLAB to determine the response of the system shown if it is given an initial
condition x0 = 5m and x0  0 .
Solution: in order to solve the 2nd order differential equation, we can rewrite it as two 1st order
differential equations (ENGR 313). Let
x1  x
x2  x
It follows that:
x1  x2
k
x2   x1   g sgn( x2 )
m
5
We can then use “ode23” in MATLAB to solve for the response. The m-files are listed below.
%
Free vibration with Coulomb damping
clear; close all;
tspan = 0:0.05:10; % time span
x0 = [5;0];
% initial conditions
[t,x] = ode23('odefun1',tspan,x0);
plot(t,x(:,1));
xlabel('Time [s]')
ylabel('Displacement [m]')
function f = odefun1(t,x)
f = zeros(2,1);
f(1) = x(2);
f(2) = -200/10*x(1)-0.7*9.81*sign(x(2));
The response is shown in Fig. 4. Observe that the amplitude drops linearly, whereas it reduces
exponentially for a viscously damped system. Notice also that the final position of the mass is
usually displaced from equilibrium and represents a permanent displacement.
5
4
3
Displacement [m]
2
1
0
-1
-2
-3
-4
-5
0
1
2
3
4
5
Time [s]
6
7
8
9
Figure 4. System response with Coulomb friction.
6
10
Forced vibration
In forced vibration, the system is acted upon by an external force which remains to be applied
while the system is moving. From MENG 375, the response of a system to an impulse, step and
ramp inputs can be determined. The response to a sinusoidal input is also of interest.
Harmonic force
Figure 5 shows a single degree of freedom system subjected to a harmonic force.
x
k
F0 cos t
m
c
Figure 5. Harmonic force acting on a single degree of freedom system.
The equation of motion is:
mx  cx  kx  F0 cos t
(36)
Note that F0 is the amplitude of the driving force and  is the driving (or forcing) frequency, not
to be confused with n. Equation (36) is a non-homogeneous, 2nd order differential equation.
This will have two solutions: the homogeneous (F0 = 0) and the particular (the periodic force),
with the total response being the sum of the two responses. The homogeneous solution is the free
vibration problem from the last section. Note that the homogeneous solution dies out with time,
and the steady-state solution prevails as long as the forcing function is present. This is why we
are interested in the steady state response. We will assume that the particular solution is of the
form:
x p (t )  A1 sin t  A2 cos t
(37)
Thus the particular solution is a steady-state oscillation having the same frequency  as the
exciting force and a phase angle, as suggested by the sine and cosine terms. Taking the
derivatives and substituting into Eq. (36) we get:
(38)
 k  m 2   A1 sin t  A2 cos t   c  A1 cos t  A2 sin t   F0 cos t
Equating the coefficients of the sine and the cosine terms, we get two equations:
c A1  k  m 2 A2  F0


 k  m  A  c A
2
1
2
This leads to a solution of:
A1 
F0 c
 k  m    c 
2
2
2
,
7
A2 
0

F0 k  m 2

 k  m    c 
2
2
2
(39)
(40)
Aside: The equation
x p (t )  A1 sin t  A2 cos t
can also be written as:
x p (t )  X cos t   
To convert between the 2 forms, i.e. to get the constants X and , substitute t=0 and
equate the displacements and velocities in both equations. This yields:
A2  X cos  , A1  X sin 
Thus
A
X  A12  A22 , tan   1
A2
The solution can then be written in the form:
x p (t )  X cos t   
where
F0
 c 
X
,   tan 1 
2 
2
2
 k  m 
k  m 2   c 

(41)
(42)

using n and  from before and introducing X0 and r to be
k
c
c
c
c
n 
,  

,  2n
m
cc 2mn 2 mk m
F
X 0  0  deflection under static force F0
k
r

 frequency ratio
n
we can write
M
X
1


Xo
2 2
2
    

1       2
  n    n 


where M is the magnification factor (amplitude ratio) and
 2 r 
  tan 1 
2 
1 r 
The steady state response is shown in the Fig. 6.
8
1
1  r    2 r 
2
2
(43)
2
(44)
Figure 6. Response (magnitude and phase) to a harmonic force.
9
Complex Analysis
Using the fact that the complex exponential is periodic, we can look at the real component of
mx  cx  kx  F0eit
(45)
we can assume a solution of the form:
(46)
x p (t )  Xeit  
This leads to the equation:
(47)
m 2  ci  k Xeit    F0eit


which can be written as:


 k  m 2  ic   F0 ei

 X
This can be represented vectorially as:
(48)
Im
F0
X
c

from which we can get:
Re
k  m 2
F0

X
 k  m    c 
X
F0
2
2
2
(49)
so
 k  m    c 
2
2
(50)
2
and
 c 
2 
 k  m 
  tan 1 
(51)
which is the same solution (see Eq. 42) we had for the cosine case before, with a lot less work.
10
Base Motion
Figure 7 shows a base-excited system.
x
m
c
k
y(t) = Y sin t
Figure 7. System subjected to base excitation.
The equation of motion of the system is:
mx  c( x  y )  k  x  y   0
I. Relative Motion
Sometimes we are concerned with the relative motion of the mass with respect to the base.
(Example: accelerometer and the velocity meter). In this case, we can define z  x  y . Our
equation of motion becomes:
mx  c( x  y )  k  x  y   0
If the periodic input is in the form:
y  Y sin t
the equation of motion becomes:
mz  cz  kz  my  m 2Y sin t
(52)
(53)
(54)
(55)
P0
which is identical to the previous case (harmonic force), with F0 replaced by P0. The steady state
solution can be written as:
z (t )  Z sin t   
(56)
where
Z
1
(57)

2
2
Z0
2
1  r  2 r

 

But
Z0 
P0 m 2Y

 Yr 2
k
k
(58)
hence the solution is:
Z

Y
r2
1  r    2 r 
2
2
(59)
2
and
 2 r 
2 
 1 r 
  tan 1 
11
(60)
Figure 8 shows the response to base excitation.
Figure 8. Response (relative motion) to base excitation.
12
Application: vibration measuring instruments
For negligible damping (   1 ), we have
Z
r2

Y 1 r2
Case 1: r  1  Z  Yr 2  Y
(61)
2
n2
At “low” frequencies, Z is proportional to 2Y, i.e. proportional to acceleration. In this range the
accelerometer works. It must have a high n such that  n is small .
Case 2: r  1  Z  Y
At “high” frequencies, the ratio between Z and Y is one, i.e. Z is equal to the displacement. In
this range the vibrometer works. It must have a low n such that  n is high.
13
I. Absolute Motion
If we are concerned with the absolute motion, we can write the equation of motion as:
mx  cx  kx  ky  cy
Here it becomes more convenient to assume the base motion having the form1:
y  Yeit
We seek a solution in the form:
x  Xeit  
Substituting into the equation of motion yields:
m 2  ci  k Xeit    k  ic Yeit

from which

X  i
k  ic
e 
Y
k  m 2  ic

(62)
(63)
(64)
(65)
(66)

hence
X

Y
1   2 r 

1 r2

2
2
  2 r 
(67)
2
and

2 r 3
  tan 1 
 1  r 2   2 r 

2




The response is shown in Fig. 9.
1
We can also assume a periodic base input of the form y = Y sin t, see page 16.
14
(68)
Figure 9. Response (absolute motion) to base excitation.
Note: the motion transmitted is less than 1 for r  2 . Hence for vibration isolation, we must
have  n  2 , i.e. n must be small compared to .
15
The above solution can also be obtained if we assume a periodic input of
y  Y sin t
the equation of motion can be written as
mx  cx  kx  kY sin t  cY cos t
mx  cx  kx  A sin t   
where
2
 c 
A  Y k 2   c  ,   tan 1  
 k 
(see page 8)
But the above equation is in the same format now as the forced vibration one, with an
amplitude of A instead of F0 and an additional phase . So we can find (see page 2)
Y k 2   c 
x  X sin t    1  
 k  m 
2
2
2
c 
2
sin t    1 
2
 c 
2 
 k  m 
We can write the ratio of the amplitudes as
1  tan 1 
X

Y
k 2   c 
2
 k  m    c 
2
2
2

1   2 r 
2
1  r    2 r 
2
2
2
using the same definition for  and r as before, we can also write the total solution as
x p (t )  X sin t   
where




mc 3
2 r 3
1



  tan 1 

tan
 1   4 2  1 r 2 
 k  k  m 2   c 2 




which is the same solution obtained using complex analysis.
16
Force Transmitted
To find the force on the base during base motion, we have:
F  k ( x  y )  c  x  y   mx
but x is known from before, so F is given by:
F  m 2 X sin t     FT sin t   
(69)
(70)
where FT is the amplitude or maximum value of the force transmitted to the base. Hence:
FT  m Y
2
1   2 r 
2
1  r    2 r 
2
2
(71)
2
from which we get:
FT
 r2
kY
1  (2 r ) 2
1  r 2    2 r 
2
2
(72)
In this relationship, kY represents the force on the mass if it remained stationary, while the base
moved with only the spring attached. This ratio is called the force transmissibility.
The force transmitted can also be calculated for the case of a harmonic force (Chapter 9). In this
case:
FT  kx  cx  kx  icx
(73)
so the force amplitude is:
FT 
 kX   cX 
2
2
Thus, the transmissibility or transmission ratio of the isolator (Tr) can be calculated to be:
17
(74)
Tr 
FT

F0

k 2   2c 2
k  m 2

2
  2c 2

1  (2 r )2

1 r2

2
  2 r 
2
(75)
This has the same amplitude as the base motion ratio found earlier. If Tr is less than one, then
the system behaves like a vibration isolator, i.e. the ground receives less force than the input
force. (See figure 3.15)
Rotating Unbalance
Having a rotating unbalance is a common problem in machinery. In this case, we have the
following equation of motion:
Mx  cx  kx  me2 sin t
If we denote the particular solution xp(t) as:
x p  t   X sin t   
then the displacement amplitude is:
X
1

2
2
X0
1  r 2  2 r

 

(76)
(77)
(78)
where
X0 
me 2 mer 2

k
M
(79)
then
XM

me
r2
1  r    2 r 
2
2
(80)
2
and
 2 r 
2 
 1 r 
  tan 1 
18
(81)
The force transmitted because of the rotating imbalance is
F  kx  cx
The transmissibility can be analyzed as before to give:
F
F
F
Tr 


2
F0 me
mer 2n 2
thus
F
 r2
men 2
(82)
(83)
1  (2 r ) 2
1  r 2    2 r 
2
19
2
(84)