Two-Degrees-of-Freedom Systems
One-degree-of-freedom systems allow basic concepts, such as
frequency, damping, initial conditions, resonance and phase, etc.,
to be introduced and appreciated. However, real systems have
infinite number of degrees-of-freedom. On many occasions, these
systems can be approximated as having finite degrees-offreedom, the simplest of which has two degrees-of-freedom.
Multi-degrees-of-freedom systems afford a venue for
introducing eigenvectors (modes), matrices and vibration
absorption.
Free Vibration
A mass-spring system:
k2
k1
m1
k3
m2
Free-body diagrams
x1
k1x1
m1
x2
k2(x1-x2)
k2(x1-x2)
m2
k3x2
Equations of motion
m1x1  k1 x1  k 2 ( x1  x2 )
m2 x2  k3 x2  k 2 ( x1  x2 )
m1x1  (k1  k 2 ) x1  k 2 x2  0
or m x  k x  (k  k ) x  0
2 2
2 1
2
3
2
1
(1)
In matrix form as
m1 0   x1  k1  k2  k2   x1  0
  x   0
 0 m  x     k
k

k
2
2
3  2 
 
2  2 


(2)
or
Mx  Kx  0
Following the procedure for one-degree-of-freedom systems,
assume the solution in the form of
 x  a 
x   1    1  sin  t
 x2  a2 
This leads to
 k2 
m 0   a1  0
k  k
2 1
( 1 2


)    



 0 m2  a2  0
  k 2 k 2  k3 
(3)
The condition for a non-trivial solution of (K   2M)a  0 to exist is
k1  k2  m1 2
det 
 k2


0
2
k2  k3  m2 
 k2
(4)
or
det( K   2M)  0
(5)
Notice  in equation (5) is the natural frequency (frequencies)
2
This is known in linear algebra as an (generalized) eigenvalue
problem. For matrices of very low orders (22, 33, 44), ‘hand’
calculation is feasible. For higher-orders matrices, suitable
algorithms and computer programs must be used.
For the above simple eigenvalue problem, the determinant
method gives (characteristic equation)
(k1  k2  m1 2 )( k2  k3  m2 2 )  k22  0
or
m1m2 4  [m1 (k2  k3 )  m2 (k1  k2 )] 2  k1k2  k1k3  k2 k3  0
This quadratic equation has two roots of 12 and  22 . Suppose
m1  m , m2  2m and k1  k2  k3  k . The above characteristic
equation become
2m 2 4  6mk 2  3k 2  0
k
k
The two analytic solutions are 1  0.7962 m and 2  1.5382 m .
Equation (3) allows the ratio of the two amplitudes to be found as
a1
k
2k  2m 2


a2 2k  m 2
k
Substitution of one eigenvalue a time into equation (3) yields
(1)
 a1 
For 1 :  a   0.731
 2
3
 a1 
For 2 :  a 
 2
( 2)
 2.73
Mode shapes
0.73
a1
1
a1
1
a2
a2
-2.73
Proportionality
A mode shape is represented by a vector, called eigenvector
mathematically. The absolute magnitude of individual elements
of an eigenvector is indefinite  an eigenvector multiplied by a
scalar is still an eigenvector. But the relative proportion of the
individual elements of an eigenvector is fixed. For this reason, an
eigenvector is normally normalized (scaled according to a rule)
and then becomes unique. Normalisation helps to graphically
display a mode, compare modes and facilitate computation.
Orthogonality
≠𝟎 𝒊=𝒌
≠𝟎 𝒊=𝒌
𝐱 𝒊𝐓 𝐌𝐱 𝒌 = {
𝐱 𝒊𝐓 𝐊𝐱 𝒌 = {
=𝟎 𝒊≠𝒌
=𝟎 𝒊≠𝒌
Recall orthogonality between two vectors.
Normalisaton
(1) Let the maximum element be one and the other elements
xk
 
x
(k  1, 2, ......, n)
k
scaled as
x
max i
i
4
T
(2) Multiply a factor to an eigenvector so that x i Mx k  δ ik
(mass-normalisation)
(3) Scale an eigenvector so that
2
 xi  1 .
i
A normalised eigenvector is called a normal eigenvector.
The two-degree-of-freedom system may vibrate in any one of
the two modes. In general, the motion of the system is a linear
combination of all modes as
(1)
( 2)
 x1 (t ) 
0.731

  A1 
 sin( 0.796 t  1 )
 1 
 x2 (t )
 x1 (t ) 


 x2 (t )
 2.73
 A2 
 sin( 1.538 t  2 )
 1 
Factors A1 and A2, phase angles 1 and  2 , like in one-degree-offreedom systems, should be determined by the initial conditions.
Initial Conditions
For the linear system, given initial conditions as
 x1 (0)  2
 x1 (0)  0

    and 
 
 x2 (0) 4
 x2 (0) 0
Substituting them into
 x (t ) 
0.731
 2.73
x(t )   1   A1 
 sin( 0.796 t  1 )  A2 
 sin( 1.538 t   2 )
x
(
t
)
1
1




 2 
leads to
5
2 
0.731
 2.73

A
sin


A
 

 sin  2
1
1
2
4
1
1
 




0
0.731
 2.73

A

cos


A

 

 cos  2
1 1
1
2 2
0
1
1
 




The four unknowns can be determined and the solution is
 x (t )  2.732
 0.732
x(t )   1   
cos(
0
.
796
t
)



 cos(1.538 t )
 0.268 
 x2 (t ) 3.732
4
2
x1(t)
0
x2(t)
0
10
20
30
-2
-4
For a system of n degrees-of-freedom, M and K are nn square
matrices. There usually exist n natural frequencies i and n
eigenvectors ψ i , by solving equation (5). The motion can be
written as
n
x(t )   Ai ψ i sin( it  i )
i 1
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Rotational Systems
1
k1
2
k2
J1
1
k11
k3
J2
 J1 0  1   K1  K 2
 0 J        K
2
2   2  

2
k2(1-2)
J1
k32
J2
 K 2  1  0
  
K 2  K 3   2  0
Coupled Pendulum
a
l
 ka2  1  0
1 0 1  ka2  mgl
ml 
    
     
2
2
0
1


ka
ka

mgl

 2  
  2  0
2
m
Two natural frequencies and modes can be found as
g
1 
,
l
1
ψ1    ,
1
g
a2 k
2 
2 2
l
l m
7
1
ψ2   
 1
m
Damped Vibration
c
k2
k1
k3
m1
m2
m1 0   x1   c  c  x1  k1  k2  k2   x1  0
  x   0
 0 m  x    c c   x     k
k

k
 2  
2
2
3  2 
 
2  2  

 x1 (t )   a1 
It is no longer valid to assume that  x (t )  a  sin t due to
 2   2
presence of damping. In general, the solution can be assumed as
 x1 (t )   a1 

    exp( t ) , where
x
(
t
)
 2  a2 
 is complex. This leads to
 k2 
k  k
 c  c 2 m1 0   a1  0
( 1 2


 
)    




k 2  k3 
 c c 
 0 m2  a2  0
  k2
Suppose m1  m , m2  2m and k1  k2  k3  k . The above equation
becomes
2m 24  3cm3  6mk2  2ck  3k 2  0
Introduce a new variable  
k
c
, 
,   z
m
2m
. A new equation
appears as
2 z 4  6z 3  6 z 2  4z  3  0
Two roots at   5% : z1, 2  0.0014  i0.80 ; z3, 4  0.0736  i1.54 ;
k
k
eigenvalues: 1, 2  (0.0014  i0.80) m ; 3, 4  (0.0736  i1.54) m .
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