TWO DEGREE OF FREEDOM

SYSTEM
 INTRODUCTION
THE FIRST STEP IN ANALYZING MULTIPLE DEGREES OF
FREEDOM (DOF) IS TO LOOK AT 2 DOF

 DOF: Minimum number of coordinates to specify the

position of a system
 Many systems have more than 1 DOF
 Examples of 2 DOF systems
 car with sprung and unsprung mass (both heave)
 elastic pendulum (radial and angular)
 motions of a ship (roll and pitch)
EQUATION OF MOTION FOR FORCED VIBRATION
 Considera viscously damped two degree of
freedom spring-mass system, shown below.
EQUATIONS OF MOTION FOR FORCED VIBRATION

Both equations can be written in matrix form as

where [m], [c], and [k] are all square and

symmetric ([m]=[m]T) called the mass, damping,
and stiffness matrices, respectively, and are given
by m 0  c  c c 
[ m]   1 [c ]  
1 2 2
   
 0 m2  c 2 c 2 c3

k1  k 2  k 2 
[k ]   
  k 2 k 2  k 3
EQUATIONS OF MOTION FOR FORCED VIBRATION
And the displacement and force vectors are given
respectively:
  x1 (t )    F1 (t ) 
x (t )    F (t )   
 2 
x (t )  2 
F (t )
 FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
By setting F1(t) = F2(t) = 0, and damping
disregarded, i.e., c1 = c2 = c3 = 0, the equation of
motion is reduced to:

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
 FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
After substitution,

Thus,

For a nontrivial solution,

 
  m1 2  (k1  k2 ) k 2 
det  0
 k 2  2

m1  (k1  k2 ) 
 FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
The above equation is called frequency or characteristic
equation because its solution yields the frequencies of the
system. The roots of the equation are given by.

Roots are called natural frequencies of the system.

Q: Let m1=m2=m and k1=k2=k3=k. Then find natural
frequencies of the system. 8
 FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
Solving the following equation of motion for X1 and X2

To determine the values of X1 and X2, given ratio

Above vectors are known as the modal vectors of the

system. They describe the relative amplitude of vibration of
each mass for each of the natural frequencies.
 FREE VIBRATION ANALYSIS OF AN UNDAMPED
SYSTEM
The free vibration solution or the motion in time
can be expressed itself as

Where the constants X 1(1) , X1(2) , and 2 are

1

determined by the initial conditions.

 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)

Let c1 = c2 = 1, thus the components of the vector

can be expressed as

where the unknown constants can be determined

from the initial conditions:
Initial conditions:

substituting the initial conditions:

The following unknowns can be identified:

12
The solution can be expressed as
 r x (0)  x2 (0)    r x (0)  x2 (0) 
X 1(1) cos1   2 1 , X 1( 2) cos2   1 1 
 r2  r1   r 2  r1 
  r2 x1 (0)  x2 (0)   r1 x1 (0)  x2 (0) 
X 1 sin 1  
(1)
, X 1 sin 2  
( 2)

  (
1 2 r  r1 )  
 2 2 1  ( r  r )

13
from which we obtain the desired solution
 IMPORTANT EXPRESSIONS (FREE UNDAMPED
VIBRATION ANALYSIS)

  m1 2  (k1  k2 )  k 2 
det  0
 k 2 
m1  (k1  k2 ) 
2


 r x (0)  x2 (0)    r x (0)  x2 (0) 

X 1(1) cos1   2 1 , X 1( 2) cos2   1 1 
 r2  r1   r 2  r1 
  r x (0)  x2 (0)   r x (0)  x2 (0) 
X 1(1) sin 1   2 1 , X 1( 2) sin 2   1 1 
 1 (r2  r1 )  
 2 2 1 
( r  r )
 EXAMPLE 1: FREE VIBRATION RESPONSE OF A
TWO DEGREE OF FREEDOM SYSTEM
Find the free vibration response of the system shown
below with k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1 and c1 =
c2 = c3 = 0 for the initial conditions x1 (0)  1, x1 (0)  x2 (0)  x2 (0).

Solution: For the given data,

 m1 2  k1  k 2 k2   X 1  0
     
 k 2 m2  k 2  k3   X 2  0
2


or   10 2
 35  5   X 1  0
     (E.1)
 -5   5  X 2  0
2
 SOLUTION OF THE EXAMPLE 1
By setting the determinant of the coefficient
matrix in Eq.(E.1) to zero, we obtain the
frequency equation,
10  85  150  0
4 2
(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 (E.4)
 X 2  2
 ( 2)  X 1( 2 )   1  ( 2 )
X   ( 2)     X 1 (E.5)
 X 2   5 17
The free vibration responses of the masses m1
and m2 are given by (see Eq.5.15):
x1 (t )  X 1(1) cos(1.5811t  1 )  X 1( 2 ) cos(2.4495t  2 ) (E.6)
x2 (t )  2 X 1(1) cos(1.5811t  1 )  5 X 1( 2) cos(2.4495t  2 )

18
(E.7)

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  1.5811 X 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)
The solution of Eqs.(E.8) and (E.9) yields
5 2
X1 cos1  ;
(1)
X 1 cos2 
( 2)
(E.12)
7 7
while the solution of Eqs.(E.10) and (E.11) leads
to (1)
X 1 sin 1  0, X 1 sin 2  0
( 2)
(E.13)

Equations (E.12) and (E.13) give

5 2
X (1)
1  , X ( 2)
1  , 1  0, 2  0 (E.14)
7 7

19
Thus the free vibration responses of m1 and m2
are given by
5 2
x1 (t )  cos1.5811t  cos 2.4495t (E.15)
7 7
10 10
x2 (t )  cos1.5811t  cos 2.4495t (E.16)
7 7

20
 GENERALIZED COORDINATES, COORDINATE
COUPLING AND PRINCIPAL COORDINATES
Generalized coordinates are sets of n coordinates
used to describe the configuration of the system.
 The lathe can be simplified to be represented by a
2DoF with the bed considered as a rigid body with
two lumped masses representing the headstock and
tailstock assemblies. The supports are represented
by two springs.

21
The following set of coordinates can be used to describe
the system:
1. the deflection at each extremity of the lathe x1(t) and x2(t)
2. the deflection at the center of gravity x(t) and the rotation θ(t)
3. the deflection at extremity A x1(t) and the rotation θ(t)

22
•Equations of motion Using x(t) and θ(t).
Using the FBD, in the vertical direction and about the
C.G. respectively:

In matrix form:

 As each eqn. contains

both x and θ the system
is coupled-Elastic or
static coupling.
 Whenever a displacement
or torque is applied thru
the C.G. the resulting
23
motion will contain both
translation and rotation.
•Equations of motion Using x(t) and θ(t).
Using the FBD, in the vertical direction and about the
C.G. respectively:

In matrix form:

 The system is uncoupled

(equations independent)
only when k1l1 = k2l2
 Only then can pure
translation or rotation be
generated by a
displacement or torque 24
thru the C.G.
•Equations of motion Using x(t) and θ(t).
When the deflection y(t) at point P located at distance e to
the left of the C.G. and the rotation θ(t)
 As each eqn. contains both y, y’’, θ and θ’’ the system is
coupled with both elastic (static) and mass
(dynamic) coupling
 When k1l’1 = k2l’2 , the system is dynamically coupled
only → the inertial force my’’ produced by vertical
motion will induce a rotational motion (my’’e) and vice-
versa.
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             (5.26)
 21 m22  x2  c21 c22   x2  k 21 k 22   x2  0
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
dynamically. These uncoupled equations can be
solved independently.
27
4. System has elastic (static) coupling if the
stiffness matrix is not diagonal.
5. System has damping or velocity (dynamic)
coupling if the damping matrix is not diagonal.
6. System has mass or inertial (dynamic) coupling
if the mass matrix is not diagonal.

28
 EXAMPLE 2
Determine the principal coordinates for the
spring-mass system shown below:

Find 1 and 2
 
  m1 2  (k1  k2 ) k 2 
det  0
 k 2  2

m1  (k1  k2 ) 

29
 SOLUTION OF THE EXAMPLE 2
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  (E.1)
 m   m 

We define a new set of coordinates such that

30
  k 
q1 (t )  B1 cos t  1 
 m 
 3k 
q2 (t )  B2 cos t  2  (E.2)
 m 
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 (E.3)
m
From Eqs.(E.1) and (E.2), we can write

x1 (t )  q1 (t )  q2 (t )
x2 (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 )] (E.5)
2
 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 
m               (5.27 )
 12 m22   x2  c21 c22   x2  k 21 k 22   x2   F2 

Consider the external forces to be harmonic:

F j (t )  F j 0 eit , j  1,2 (5.28)

where ω is the forcing frequency. We can write the

steady-state solutions as
x j (t )  X j eit , j  1,2 (5.29)
33
 FORCED VIBRATION ANALYSIS
Substitution of Eqs.(5.28) and (5.29) into
Eq.(5.27) leads to
( 2 m11  ic11  k11 ) ( 2 m12  ic12  k12 )   X 1 
  
( m12  ic12  k12 ) (2 m22  ic22  k 22 )  X 2 
2

 F10 
  (5.30)
 F20 

We defined as in section 3.5 the mechanical

impedance Zre(iω) as

Z rs (i)   mrs  icrs  krs , r, s  1,2

2
(5.31)
 FORCED VIBRATION ANALYSIS
And write Eq.(5.30) as:
 
Z (i )X  F0 (5.32)

Where,

 Z11 (i ) Z12 (i ) 

Z (i )     Impedance matrix
 Z12 (i ) Z 22 (i )
  X1 
X  
X 2 
  F10 
F0   
 F20 
 FORCED VIBRATION ANALYSIS
Eq.(5.32) can be solved to obtain:
 
X  Z (i ) F0
1
(5.33)
where the inverse of the impedance matrix is
given
 Z 22 (i ) -Z12 (i )
Z (i )
1

1
  (5.34)
Z11 (i ) Z 22 (i )  Z12 (i )  Z12 (i ) Z11 (i ) 
2

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 )  (5.35)
Z11 (i ) Z 22 (i )  Z122 (i ) 36
 EXAMPLE 3
(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.

37
 SOLUTION OF THE EXAMPLE 3
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     (E.1)
  2    2   0 

We assume the solution to be as follows.

x j (t )  X j cost; j  1,2 (E.2)
Eq.(5.31) gives
Z rs (i)   2 mrs  icrs  krs , r, s  1,2 (5.31)
Z11 ( )  Z 22 ( )  m 2  2k , Z12 ( )  k (E.3)
38
 SOLUTION OF THE EXAMPLE 3
Hence, from Eq.(5.35)

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 )  (5.35)
Z11 (i ) Z 22 (i )  Z122 (i )

( 2 m  2k ) F10 ( 2 m  2k ) F10

X 1 ( )   (E.4)
( m  2k )  k
2 2 2
(m 2  3k )(m 2  k )
kF10 kF10
X 2 ( )   (E.5)
(m  2k )  k
2 2 2
(m 2  3k )(m 2  k )
 SOLUTION OF THE EXAMPLE 3

( 2 m  2k ) F10 ( 2 m  2k ) F10

X 1 ( )   (E.4)
( m  2k )  k
2 2 2
(m 2  3k )(m 2  k )
kF10 kF10
X 2 ( )   (E.5)
(m  2k )  k
2 2 2
(m 2  3k )(m 2  k )

Eqs.(E.4) and (E.5) can be expressed as

    2 
2    F10
  1  
X 1 ( )  (E.6)
   2    2      2 
k  2      1    
 1   1     1  
 SOLUTION OF THE EXAMPLE 3
F10
X 2 ( )  (E.7)
          
2 2 2

k  2      1    
 1   1     1  

There are two

resonance conditions
for the system: one at
1 and another at 2.
At all other values of
 the amplitudes of
vibration are finite.

Fig.5.14: Frequency response curves

41