TWO DEGREE OF FREEDOM
SYSTEM
INTRODUCTION

Systems that require two independent coordinates to
describe their motion;


Two masses in the system X two possible types of motion of
each mass.
Example: motor pump system.
There are two equations of motion for a 2DOF system,
one for each mass (more precisely, for each DOF).
 They are generally in the form of couple differential
equation that is, each equation involves all the
coordinates.

EQUATION OF MOTION FOR FORCED VIBRATION
 Consider
a viscously damped two degree of
freedom spring-mass system, shown in Fig.5.3.
Figure 5.3: A two degree of freedom spring-mass-damper
system
EQUATIONS OF MOTION FOR FORCED VIBRATION
m1x1  (c1  c2 ) x1  c2 x2  (k1  k2 ) x1  k2 x2  F1
m2 x2  c2 x1  (c2  c3 ) x2  k2 x1  (k2  k3 ) x2  F2
4
The application of Newton’s second law of motion
to each of the masses gives the equations of
motion:
(5.1)
(5.2)
Both equations can be written in matrix form




as
[m]x (t )  [c]x (t )  [k ]x (t )  F (t )
(5.3)
where [m], [c], and [k] are called the mass,
damping, and stiffness matrices, respectively,
and are given by
EQUATIONS OF MOTION FOR FORCED VIBRATION
m1 0 
[ m]  

m
0
2

5
c1  c2  c2 
[c ]  

c

c
c

3
2
2

k1  k 2  k 2 
[k ]  

k

k
k

3
2
2

And the displacement and force vectors are given
respectively:

 x1 (t ) 
F1 (t ) 

x (t )  
F (t )  


 x2 (t )
F2 (t )
It can be seen that the matrices [m], [c], and [k]
are all 2 x 2 matrices whose elements are known
masses, damping coefficient and stiffnesses of the
system, respectively.
EQUATIONS OF MOTION FOR FORCED VIBRATION
oFurther,
these matrices can be seen to be
symmetric, so that,
[m]T  [m],
[c]T  [c],
[k ]T  [k ]
where the superscript T denotes the transpose of
the matrix.
oThe
solution of Eqs.(5.1) and (5.2) involves four
constants of integration (two for each equation).
Usually the initial displacements and velocities of
the two masses are specified as
x1(t = 0) = x1(0) and x1( t = 0) = x 1(0),
x2(t = 0) = x2(0) and x2 (t = 0) = x2(0).
6
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
7
By setting F1(t) = F2(t) = 0, and damping
disregarded, i.e., c1 = c2 = c3 = 0, and the equation
of motion is reduced to:
m1x1 (t )  (k1  k 2 ) x1 (t )  k2 x2 (t )  0
(5.4)
m2 x2 (t )  k 2 x1 (t )  (k 2  k3 ) x2 (t )  0
(5.5)
Assuming that it is possible to have harmonic
motion of m1 and m2 at the same frequency ω and
the same phase angle Φ, we take the solutions as
x1 (t )  X 1 cos(t   )
x2 (t )  X 2 cos(t   )
(5.6)
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
Substituting into Eqs.(5.4) and (5.5),
 m   (k  k )X  k X cos(t   )  0
 k X  m   (k  k )X cos(t   )  0
2
1
2
1
2
2
2
2
1
2
2
8
1
3
2
(5.7)
Since Eq.(5.7)must be satisfied for all values of
the time t, the terms between brackets must be
zero. Thus,
 m 
1
2


 (k1  k2 ) X 1  k2 X 2  0

 k2 X 1  m2  (k2  k3 ) X 2  0
2
(5.8)
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
9
which represent two simultaneous homogenous
algebraic equations in the unknown X1 and X2. For
trivial solution, i.e., X1 = X2 = 0, there is no
solution. For a nontrivial solution, the determinant
of the coefficients of X1 and X2 must be zero:


  m1 2  (k1  k 2 )
det 
k 2


0
2
m1  (k1  k 2 ) 
k 2


or
(m1m2 )  (k1  k2 )m2  (k2  k3 )m1
4


 (k1  k2 )( k2  k3 )  k22  0
(5.9)
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
which is called the frequency or characteristic
equation. Hence the roots are:
1  (k1  k 2 )m2  (k 2  k3 )m1 
 ,  

2
m1m2

2
1
2
2
2

1  (k1  k 2 )m2  (k 2  k3 )m1 
 

2 
m1m2


1/ 2
 (k1  k 2 )( k 2  k3 )  k 
 4

m1m2


2
2
(5.10)
The roots are called natural frequencies of the
system.
10
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
To determine the values of X1 and X2, given ratio
X 2( 2)  m122  (k1  k 2 )
k2
r2  ( 2) 

X1
k2
 m222  (k 2  k3 )
11
X 2(1)  m112  (k1  k 2 )
k2
r1  (1) 

X1
k2
 m212  (k 2  k3 )
(5.11)
The normal modes of vibration corresponding to
ω12 and ω22 can be expressed, respectively, as
 (1)  X 1(1)   X 1(1) 
 ( 2 )  X 1( 2 )   X 1( 2 ) 
X   (1)    (1)  and X   ( 2 )   
( 2) 
 X 2  r1 X 1 
 X 2  r2 X 1 
(5.12)
which are known as the modal vectors of the system.
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
The free vibration solution or the motion in time
can be expressed itself as
12
(1)
(1)



x
(
t
)
X
 (1)
 1
  1 cos(1t  1 ) 
x (t )   (1)    (1)
  first mode
 x2 (t ) r1 X 1 cos(1t  1 )
 x1( 2 ) (t )  X 1( 2 ) cos(2t  2 ) 
 ( 2)
x (t )   ( 2 )   
  second mode
( 2)
 x2 (t ) r2 X 1 cos(2t  2 )
(5.17)
Where the constants X 1(1) , X 1(2) , and 2 are
determined by the initial conditions. The initial
conditions are
1
x1 (t  0)  X 1(i )  some constant,
x1 (t  0)  0,
x2 (t  0)  ri X 1(i ) ,
x2 (t  0)  0
Thus the components of the vector can be
expressed as
x1 (t )  x1(1) (t )  x1( 2 ) (t )  X 1(1) cos(1t  1 )  X 1( 2 ) cos(2t  2 )
x2 (t )  x2(1) (t )  x2( 2 ) (t )
 r1 X 1(1) cos(1t  1 )  r2 X 1( 2 ) cos(2t  2 )
(5.15)
where the unknown constants can be determined
from the initial conditions:
13
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
The resulting motion can be obtained by a linear
superposition of the two normal modes, Eq.(5.13)



x(t )  c1x1 (t )  c2 x2 (t )
(5.14)
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
x1 (t  0)  x1 (0),
x2 (t  0)  x2 (0),
x1 (t  0)  x1 (0),
x2 (t  0)  x2 (0)
(5.16)
Substituting into Eq.(5.15) leads to
x1 (0)  X 1(1) cos 1  X 1( 2 ) cos 2
x1 (0)  1 X 1(1) sin 1  2 X 1( 2) sin 2
x2 (0)  r1 X 1(1) cos 1  r2 X 1( 2) cos 2
x2 (0)  1r1 X 1(1) sin 1  2 r2 X 1( 2) sin 2
(5.17)
The solution can be expressed as
 r x (0)  x2 (0) 
X 1(1) cos 1   2 1
,
r2  r1


  r x (0)  x2 (0) 
X 1(1) sin 1   2 1
,
 1 (r2  r1 ) 
  r x (0)  x2 (0) 
X 1( 2 ) cos 2   1 1

r

r
2
1


 r x (0)  x2 (0) 
X 1( 2 ) sin 2   1 1


(
r

r
)
2 2
1

 14
FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
from which we obtain the desired solution

 X
(1)
1
 
cos 1  X
2
(1)
1
sin 1

2 1/ 2
15
X
(1)
1

1 
 r2 x1 (0)  x2 (0) 
2

r2 x1 (0)  x2 (0) 

(r2  r1 ) 
12

2
X
( 2)
1

 X
( 2)
1
 
cos 2  X
2
( 2)
1
sin 2
1/ 2

2 1/ 2

1 
r1 x1 (0)  x2 (0) 
2



  r1 x1 (0)  x2 (0) 

(r2  r1 ) 
22

2
1/ 2
(1)


1 (0)  x2 (0) 
X
1
1   r2 x
1 sin 1
1  tan  (1)
  tan 

X
cos


[
r
x
(
0
)

x
(
0
)
1
2
 1
 1 2 1

( 2)

X
sin 2 
r1 x1 (0)  x2 (0) 
1 
2  tan 1  1( 2 )

tan



X
cos


[

r
x
(
0
)

x
(
0
)
2
2
 1
 2 1 1

(5.18)
EXAMPLE 5.3:FREE VIBRATION RESPONSE OF A
TWO DEGREE OF FREEDOM SYSTEM
16
Find the free vibration response of the system
shown in Fig.5.3(a) with k1 = 30, k2 = 5, k3 = 0, m1
= 10, m2 = 1 and c1 = c2 = c3 = 0 for the initial
x1 (0)  1, x1 (0)  x2 (0)  x2 (0).
conditions
Solution: For the given data, the eigenvalue
problem, Eq.(5.8), becomes
 m1 2  k1  k 2
  X 1  0
k 2

    
2
k2
m2  k 2  k3   X 2  0

or
 10 2  35
 5   X 1  0
(E.1)

    
2
-5
  5  X 2  0

EXAMPLE 5.3 SOLUTION
By setting the determinant of the coefficient
matrix in Eq.(E.1) to zero, we obtain the
frequency equation,
4
2
10  85  150  0
(E.2)
from which the natural frequencies can be found
as
12  2.5,
22  6.0
1  1.5811, 2  2.4495
(E.3)
The normal modes (or eigenvectors) are given by
 (1)  X 1(1)  1  (1)
X   (1)     X 1
 X 2  2
 ( 2 )  X 1( 2 )   1  ( 2 )
X   ( 2)     X 1
 X 2   5
(E.4)
(E.5)
17
EXAMPLE 5.3 SOLUTION
The free vibration responses of the masses m1
and m2 are given by (see Eq.5.15):
(E.6)
x2 (t )  2 X 1(1) cos(1.5811t  1 )  5 X 1( 2) cos(2.4495t  2 )
(E.7)
18
x1 (t )  X 1(1) cos(1.5811t  1 )  X 1( 2) cos( 2.4495t  2 )
By using the given initial conditions in Eqs.(E.6)
and (E.7), we obtain
x1 (t  0)  1  X 1(1) cos 1  X 1( 2) cos 2
(E.8)
x2 (t  0)  0  2 X 1(1) cos 1  5 X 1( 2) cos 2
(E.9)
x1 (t  0)  0  1.5811X 1(1) sin 1  2.4495 X 1( 2) sin 2 (E.10)
x2 (t  0)  3.1622 X 1(1)  12.2475 X 1( 2 ) sin 2
(E.11)
EXAMPLE 5.3 SOLUTION
The solution of Eqs.(E.8) and (E.9) yields
5
2
(1)
( 2)
X 1 cos 1  ;
X 1 cos 2 
7
7
(E.12)
while the solution of Eqs.(E.10) and (E.11) leads
to
(1)
( 2)
X 1 sin 1  0, X 1 sin 2  0
(E.13)
Equations (E.12) and (E.13) give
X
(1)
1
5
 ,
7
X
( 2)
1
2
 ,
7
1  0, 2  0
(E.14)
19
EXAMPLE 5.3
SOLUTION
Thus the free vibration responses of m1 and m2
are given by
5
2
x1 (t )  cos 1.5811t  cos 2.4495t
7
7
10
10
x2 (t )  cos 1.5811t  cos 2.4495t
7
7
(E.15)
(E.16)
20
TORSIONAL SYSTEM
Figure 5.6: Torsional system with discs mounted on a shaft
Consider a torsional system as shown in Fig.5.6.
The differential equations of rotational motion for
the discs can be derived as
21
TORSIONAL SYSTEM
2 2
t2
2
1
t3 2
t2
which upon rearrangement become
J11  (kt1  kt 2 )1  kt 2 2  M t1
J 22  kt 21  (kt 2  kt 3 ) 2  M t 2
(5.19)
For the free vibration analysis of the system,
Eq.(5.19) reduces to
J11  (kt1  kt 2 )1  kt 2 2  0
J 22  kt 21  (kt 2  kt 3 ) 2  0
(5.20)
22
J11  kt11  kt 2 ( 2  1 )  M t1
J   k (   )  k   M
EXAMPLE 5.4:NATURAL FREQUENCIES OF
TORSIONAL SYSTEM
Find the natural frequencies and mode shapes
for the torsional system shown in Fig.5.7 for J1 =
J0 , J2 = 2J0 and kt1 = kt2 = kt .
Solution:
The differential equations of motion,
Eq.(5.20), reduce to (with kt3 = 0,
kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):
J 01  2kt1  kt 2  0
2 J 02  kt1  kt 2  0
(E.1)
Fig.5.7:
Torsional system
23
EXAMPLE 5.4 SOLUTION
Rearranging and substituting the harmonic
solution:
i (t )  i cos(t   ); i  1,2
(E.2)
gives the frequency equation:
2 4 J 02  5 2 J 0 kt  kt2  0
(E.3)
The solution of Eq.(E.3) gives the natural
frequencies
kt
1 
(5  17 ) and
4J 0
kt
2 
(5  17 ) (E.4)
4J 0
24
EXAMPLE 5.4 SOLUTION
The amplitude ratios are given by

r2 

( 2)
2
( 2)
1
(5  17 )
 2
4
25
(21)
(5  17 )
r1  (1)  2 
1
4
(E.5)
Equations (E.4) and (E.5) can also be obtained by
substituting the following in Eqs.(5.10) and (5.11).
k1  kt1  kt ,
k 2  kt 2  kt ,
m1  J1  J 0 , m2  J 2  2 J 0 and
k3  0
COORDINATE COUPLING AND PRINCIPAL
COORDINATES
Generalized coordinates are sets of n coordinates
used to describe the configuration of the system.
•Equations of motion Using x(t) and θ(t).
26
COORDINATE COUPLING AND PRINCIPAL
COORDINATES
mx  k1 ( x  l1 )  k2 ( x  l2 )
(5.21)
and the moment equation about C.G. can be
expressed as
J 0  k1 ( x  l1 )l1  k 2 ( x  l2 )l2
(5.22)
Eqs.(5.21) and (5.22) can be rearranged and
written in matrix form as
27
From the free-body diagram shown in Fig.5.10a,
with the positive values of the motion variables
as indicated, the force equilibrium equation in
the vertical direction can be written as
COORDINATE COUPLING AND PRINCIPAL
COORDINATES
 (k1l1  k 2l2 )   x  0
 
2
2  
(k1l 1  k 2l 2 )   0
(5.23)
28
m 0   x  (k1  k 2 )
0 J      (k l  k l )

0   
11
2 2

The lathe rotates in the vertical plane and has
vertical motion as well, unless k1l1 = k2l2. This is
known as elastic or static coupling.
•Equations of motion Using y(t) and θ(t).
From Fig.5.10b, the equations of motion for
translation and rotation can be written as
my  k1 ( y  l1 )  k2 ( y  l2 )  me
COORDINATE COUPLING AND PRINCIPAL
COORDINATES
J P  k1 ( y  l1 )l1  k2 ( y  l2 )l2  mey
(5.24)
These equations can be rearranged and written
in matrix form as
m
me

me  y (k1  k 2 )
   

J P    (k1l1  k 2l2 )
(k 2l2  k1l1)   y  0
 
2
2  
(k1l1  k 2l 2 )   0
(5.25)
If k1l1  k2l2 , the system will have dynamic or
inertia coupling only.
Note the following characteristics of these
systems:
29
COORDINATE COUPLING AND PRINCIPAL
COORDINATES
1. In the most general case, a viscously damped
two degree of freedom system has the
equations of motions in the form:
m11 m12  x1  c11 c12   x1  k11 k12   x1  0
  
  
  
m



 21 m22  x2  c21 c22   x2  k21 k22   x2  0
(5.26)
2. The system vibrates in its own natural way
regardless of the coordinates used. The choice of
the coordinates is a mere convenience.
3. Principal or natural coordinates are defined as
system of coordinates which give equations of
motion that are uncoupled both statically and
30
dynamically.
EXAMPLE 5.6:PRINCIPAL COORDINATES OF
SPRING-MASS SYSTEM
Determine the principal coordinates for the
spring-mass system shown in Fig.5.4.
31
EXAMPLE 5.6 SOLUTION
Approach: Define two independent solutions as
principal coordinates and express them in terms
of the solutions x1(t) and x2(t).
The general motion of the system shown is
 k

 3k




x1 (t )  B1 cos
t  1   B2 cos
t  2 
 m

 m

 k

 3k

x2 (t )  B1 cos
t  1   B2 cos
t  2 
 m

 m

(E.1)
We define a new set of coordinates such that
32
EXAMPLE 5.6 SOLUTION
 3k

q2 (t )  B2 cos
t  2 
 m

33
 k

q1 (t )  B1 cos
t  1 
 m

(E.2)
Since the coordinates are harmonic functions,
their corresponding equations of motion can be
written as
k
q1   q1  0
m
 3k 
q2   q2  0
m
(E.3)
EXAMPLE 5.6 SOLUTION
From Eqs.(E.1) and (E.2), we can write
x2 (t )  q1 (t )  q2 (t )
34
x1 (t )  q1 (t )  q2 (t )
(E.4)
The solution of Eqs.(E.4) gives the principal
coordinates:
1
q1 (t )  [ x1 (t )  x2 (t )]
2
1
q2 (t )  [ x1 (t )  x2 (t )]
2
(E.5)
FORCED VIBRATION ANALYSIS
The equations of motion of a general two degree
of freedom system under external forces can be
written as
m11 m12  x1  c11 c12   x1  k11 k12   x1  F1 
  
  
     (5.27)
m



 12 m22  x2  c21 c22   x2  k21 k22   x2  F2 
Consider the external forces to be harmonic:
it
Fj (t )  Fj 0e , j  1,2
(5.28)
where ω is the forcing frequency. We can write the
steady-state solutions as
it
x j (t )  X j e , j  1,2
(5.29)
35
FORCED VIBRATION ANALYSIS
Substitution of Eqs.(5.28) and (5.29) into
Eq.(5.27) leads to
We defined as in section 3.5 the mechanical
impedance Zre(iω) as
Z rs (i )   2 mrs  icrs  k rs , r , s  1,2
(5.31)
36
( 2 m11  ic11  k11 ) ( 2 m12  ic12  k12 )   X 1 

 
2
( m12  ic12  k12 ) (2 m22  ic22  k 22 )  X 2 
 F10 
 
(5.30)
 F20 
FORCED VIBRATION ANALYSIS
And write Eq.(5.30) as:
(5.32)
Where,
 Z11 (i ) Z12 (i ) 
Z (i )  
 Impedance matrix

 Z12 (i ) Z 22 (i )
  X1 
X  
X 2 
  F10 
F0   
 F20 
37
 
Z (i)X  F0
FORCED VIBRATION ANALYSIS
Eq.(5.32) can be solved to obtain:


1
X  Z (i) F0
(5.33)
where the inverse of the impedance matrix is
given
Z (i )
1
Z 22 (i ) -Z12 (i )
1

 Z (i ) Z (i )  (5.34)
2
Z11 (i ) Z 22 (i )  Z12 (i )  12
11

Eqs.(5.33) and (5.34) lead to the solution
Z 22 (i ) F10  Z12 (i ) F20
X 1 (i ) 
Z11 (i ) Z 22 (i )  Z122 (i )
 Z12 (i ) F10  Z11 (i ) F20
X 2 (i ) 
Z11 (i ) Z 22 (i )  Z122 (i )
(5.35)
38
EXAMPLE 5.8:STEADY-STATE RESPONSE OF
SPRING-MASS SYSTEM
Find the steady-state response of system shown in
Fig.5.13 when the mass m1 is excited by the force
F1(t) = F10 cos ωt. Also, plot its frequency response
curve.
39
EXAMPLE 5.8 SOLUTION
The equations of motion of the system can be
expressed as
m 0 x1  2k -k   x1  F10 cos t 

0 m x   -k 2k   x   
0

 2  
 2  

(E.1)
We assume the solution to be as follows.
x j (t )  X j cos t ;
j  1,2
(E.2)
Eq.(5.31) gives
Z11 ( )  Z 22 ( )  m 2  2k ,
Z12 ( )  k
(E.3)
40
EXAMPLE 5.8 SOLUTION
Hence,
(E.4)
41
( 2 m  2k ) F10
( 2 m  2k ) F10
X 1 ( ) 

2
2
2
( m  2k )  k
(m 2  3k )( m 2  k )
kF10
kF10
X 2 ( ) 

2
2
2
(m  2k )  k
(m 2  3k )(  m 2  k )
(E.5)
Eqs.(E.4) and (E.5) can be expressed as
    2 
2     F10
  1  
X 1 ( ) 
   2    2      2 
k  2      1    
 1   1     1  
(E.6)
EXAMPLE 5.8 SOLUTION
X 2 ( ) 
F10
          
k  2      1   
 1   1     1 
2
2
2



Fig.5.14: Frequency response curves
(E.7)
42