CHAPTER 12 TWO-DEGREE- OF-FREEDOM-SYSTEMS
Introduction to two degree of freedom systems:
•
The vibrating systems, which require two coordinates to describe its motion, are called two-degrees-of –freedom systems.
•
These coordinates are called generalized coordinates when they are independent of each other and equal in number to the degrees of freedom of the system.
•
Unlike single degree of freedom system, where only one co-ordinate and hence one equation of motion is required to express the vibration of the system, in twodof systems minimum two co-ordinates and hence two equations of motion are required to represent the motion of the system. For a conservative natural system, these equations can be written by using mass and stiffness matrices.
•
One may find a number of generalized co-ordinate systems to represent the motion of the same system. While using these co-ordinates the mass and stiffness matrices may be coupled or uncoupled. When the mass matrix is coupled, the system is said to be dynamically coupled and when the stiffness matrix is coupled, the system is known to be statically coupled.
•
The set of co-ordinates for which both the mass and stiffness matrix are uncoupled, are known as principal co-ordinates. In this case both the system equations are independent and individually they can be solved as that of a singledof system.
•
A two-dof system differs from the single dof system in that it has two natural frequencies, and for each of the natural frequencies there corresponds a natural state of vibration with a displacement configuration known as the normal mode.
Mathematical terms associated with these quantities are eigenvalues and eigenvectors.
•
Normal mode vibrations are free vibrations that depend only on the mass and stiffness of the system and how they are distributed. A normal mode oscillation is defined as one in which each mass of the system undergoes harmonic motion of same frequency and passes the equilibrium position simultaneously.
•
The study of two-dof- systems is important because one may extend the same concepts used in these cases to more than 2-dof- systems. Also in these cases one can easily obtain an analytical or closed-form solutions. But for more degrees of
209

freedom systems numerical analysis using computer is required to find natural frequencies (eigenvalues) and mode shapes (eigenvectors).
The above points will be elaborated with the help of examples in this lecture.
Few examples of two-degree-of-freedom systems
Figure 1 shows two masses m and m
1 2
with three springs having spring stiffness k k
1
,
2
and k
3
free to move on the horizontal surface. Let x and x
1 2 be the displacement of mass m
1
and m
2
respectively.
k
1 m
1 x
1 k
2 m
2 x
2 k
3
Figure 1
As described in the previous lectures one may easily derive the equation of motion by using d’Alembert principle or the energy principle (Lagrange principle or Hamilton’s principle) m x
1 1
Using d’Alembert principle for mass m
1
, from the k x
1 1
(
2 1
− x
2
) free body diagram shown in figure 1(b) m x
+
( k
1
+
2
)
1
− k x
2
=
0 (1) m
1 and similarly for mass m
2 m x
2 2
− k x
1 1
+
( k
2
+
)
3 2
=
0 (2)
Important points to remember
•
Inertia force acts opposite to the
direction of acceleration, so in both the free body diagrams inertia forces are shown towards left.
•
For spring k
2
, assuming x
2
> x
1
,
2
(
2
− x
1
) m x m
2 2
2 k x
3 2
Figure 1 (b), Free body diagram
210

The spring will pull mass m
1
towards right by k x
2 2
− x
1
) and it is stretched by x
2
− x
1
(towards right) it will exert a force of k x
2 2
− x
1
) towards left on mass m
2
. Similarly assuming x
1
> x
2
, the spring get compressed by an amount x
2
− x
1
and exert tensile force of (
2 1
− x
2
) . One may note that in both cases, free body diagram remain unchanged .
Now if one uses Lagrange principle,
The Kinetic energy = T
=
1
2
Potential energy = U
=
1
2 k x
1 1
2 +
1
2
So, the Lagrangian
2 +
1
2 m x
2
2 and (3)
(
2 1
− x
2
)
2 +
1
2 k x
2
3 2
(4)
L T U
1
2 m x
1 1
2 +
1
2 m x
2
2 2
−
⎠ ⎝
1
2 k x
1 1
2 +
1
2
(
2 1
− x
2
)
2 +
1
2 k x
3 2
2
(5)
The equation of motion for this free vibration case can be found from the Lagrange principle d dt
⎛
⎝
∂
∂ q
L k
⎞
⎠
−
∂
∂ q
L k
=
0 , (6) and noting that the generalized co-ordinate q
1
= x
1
and q
2
= x
2 which yields m x
1 1
+
( k
1
+
)
2 1
− k x
2 2
=
0 m x
2 2
− k x
1 1
+
( k
2
+
)
3 2
=
0
(7)
(8)
Same as obtained before using d’Alembert principle.
Now writing the equation of motion in matrix form
⎡
⎣ m
1
0
0 m
2 x x
2
+
⎡
⎣ k
1
+ k
− k
2
2 k
2
− k
+
2 k
3
⎤
⎦ x
1
0
⎝ ⎠
. (9)
Here it may be noted that for the present two degree-of-freedom system, the system is dynamically uncoupled but statically coupled.
211

Example 2.
Consider a lathe machine, which can be modeled as a rigid bar with its center of mass not coinciding with its geometric center and supported by two springs, k k
1 2
. k
1 k
Figure 2 k
2
( c
− l
1
θ c
) l
1
C x c
J
θ e l
2
G
( c
+ e c
θ
θ c
) k x
2 c
+ l
2
θ
Figure 3: Free body diagram of the system c
)
In this example, it will be shown, how the use of different coordinate systems lead to static and or dynamic coupled or uncoupled equations of motion. Clearly this is a twodegree-of freedom system and one may express the co-ordinate system in many different ways. Figure 3 shows the free body diagram of the system where point G is the center of mass. Point C represents a point on the bar at which we want to define the co-ordinates of l
1 l
2 x is the linear displacement of point C and c
θ c the rotation about point C, the equation of motion of this system can be obtained by using d’Alember’s principle. Now summation of all the forces, viz. the spring forces and the inertia forces must be equal to zero leads to the following equation. mx c
+ me
θ c
+
( c
− l
1
θ c
)
+
2
( c
+ l
2
θ c
)
=
0 (10)
Again taking moment of all the forces about point C
J
G
θ c
+
( mx c
+ me
θ c
)
−
( c
− l
1
θ c
) l
1
+
( c
+ l
2
θ c
) l
2
=
0 (11)
Noting J c
=
J
G
+ me
2
, the above two equations in matrix form can be written as
⎡
⎣ m me me
J c
⎤ ⎛ ⎞
⎥
θ
+
⎡
⎣ k
1
+ k
2 k l
2 2
− k l
1 1 k l
− k l k l
2 + k l
2
⎤
⎦
θ c =
(12)
⎝ ⎠
212

⎡
⎣ m
0
Now depending on the position of point C, few cases can are studied below.
Case 1 : Considering e
=
0 , i.e., point C and G coincides, the equation of motion can be written as
0
J
G
⎤
⎦
⎛ ⎞
θ
+
1
(
−
1
θ
) x
⎡
⎣ k
1
+ k
2 k l
2 2
− k l
1 1 k l
2 2
− k l
1 1 k l
1 1
2 + k l
2
2 2
(
+
2
θ
)
θ
⎤
⎦
θ x
=
(13)
⎝ ⎠ ⎝ ⎠
So in this case the system is statically coupled and if k l
= k l , this coupling disappears,
θ
vibrations.
Case 2 : If, k l
2 2
= k l
1 1
, the equation of motion becomes
⎡
⎣ m me me
J c
⎤ ⎛ ⎞
⎥
θ
+
⎡
⎣ k
1
+
0 k
2
0 k l
1 1
2 + k l
2 2
2
⎤
⎦
θ c =
⎝ ⎠
. (14)
Hence in this case the system is dynamically coupled but statically uncoupled.
Case 3: If we choose l
1
=
0 , i.e. point C coincide with the left end, the equation of motion will become
⎡
⎣ m me x me J c
⎤ ⎛ ⎞
⎥
θ c +
⎡
⎣ k
1
+ k
2 k l
2 2 k l
2 2 k l
2
2 2
⎤
⎦
θ c =
. (15)
Here the system is both statically and dynamically coupled.
Normal Mode Vibration
Again considering the problem of the spring-mass system in figure 1 with m
1
= m , m
2
=
2 m , k
1
= k
2
= k
3
= k , the equation of motion (9) can be written as mx
1
+
(
1
− x
2
)
+ kx
1
=
0
2 mx
2
−
(
1
− x
2
)
+ kx
2
=
0
(16)
213

We define a normal mode oscillation as one in which each mass undergoes harmonic motion of the same frequency, passing simultaneously through the equilibrium position.
For such motion, we let x
1
=
A e , x
2
=
A e (17)
Hence,
(2 k
− m
ω
2
) A
1
− kA
2
=
0
− kA
1
+ k
− m
ω
2
(2 2 ) A
2
=
0
(18) or, in matrix form
⎡
⎣
2 k
− m
ω
− k
2 − k
2 k
−
2 m
ω
2
⎤
⎦
⎛
A
1 =
0
⎝ A
2
⎠ ⎝ ⎠
(19)
Hence for nonzero values of A
1
and A
2
(i.e., for non-trivial response)
2 k
− m
ω
2
− k
− k
2 k
−
2 m
ω
2
=
0 . (20)
Now substituting
ω
2 = λ
, equation 6.1. yields
λ
2 − k
(3 ) m
λ +
3
2 k
( ) 0 m
2 =
(21)
Hence,
λ
1
=
(
3
−
1
2 2 k
3) m
= k
0.634
m
and
λ
2
=
(
3
+
1
2 2 k
3) m
= k
2.366
m
So, the natural frequencies of the system are
ω
1
= λ
1
= k
0.634
and m
ω
2
= k
2.366
m
Now from equation (1)., it may be observed that for these frequencies, as both the equations are not independent, one can not get unique value of A
1 and A
2
. So one should find a normalized value. One may normalize the response by finding the ratio of A
1
to A
2
.
From the first equation (19) the normalized value can be given by
A
1
A
2
= k
2 k
− m
ω
2
= k
2 k
− m
λ
(22) and from the second equation of (19), the normalized value can be given by
A
A
2
1
=
2 k
−
2 k m
ω
2
=
2 k
−
2 m
λ k
(23)
Now, substituting
ω
2
λ
1
= =
1
0.634
k m
in equation (22) and (23) yields the same values, as both these equations are linearly dependent. Here,
214

⎛
⎝
A
A
2
1
⎞
⎠
=
1
=
0.732
1
(24) and similarly for
ω
2
λ
2
= =
2 k
2.366
m
⎛
⎝
A
A
2
1
⎞
⎠
=
2
=
−
2.73
1
(25)
It may be noted
•
Equation (19) gives only the ratio of the amplitudes and not their absolute values, which are arbitrary.
•
If one of the amplitudes is chosen to be 1 or any number, we say that amplitudes ratio is normalized to that number.
•
The normalized amplitude ratios are called the normal modes and designated by
φ n x .
From equation (24) and (25), the two normal modes of this problem are:
φ
1
=
⎧
0.731
⎩ 1.00
⎭
φ
2 x
=
⎧
⎩
−
2.73
1.00
In the 1 st
normal mode, the two masses move in the same direction and are said to be in phase and in the 2 nd
mode the two masses move in the opposite direction and are said to be out of phase. Also in the first mode when the second mass moves unit distance, the first mass moves 0.731 units in the same direction and in the second mode, when the second mass moves unit distance; the first mass moves 2.73 units in opposite direction.
When the system is disturbed from its initial position, the resulting free-vibration of the system will be a combination of the different normal modes. The participation of different modes will depend on the initial conditions of displacements and velocities. So for a system the free vibration can be given by x
= φ
1
A sin(
ω ψ
1 1
)
+ φ
2
B sin(
ω
2 t
+ ψ
2
) (27)
215

Here A and B are part of participation of first and second modes respectively in the resulting free vibration and
ψ
1 and
ψ
2
are the phase difference. They depend on the initial conditions. This is explained with the help of the following example.
Example : Let us consider the same spring-mass problem (figure 4) for which the natural frequencies and normal modes are determined. We have to determine the resulting free vibration when the system is given an initial displacement x
1
(0)
=
5, x
2
=
and initial velocity x
1
(0)
= x
2
(0)
=
0 . x
1 x
2 k m k
2 m k
Figure 4
Solution:
Any free vibration can be considered to be the superposition of its normal modes. For each of these modes the time solution can be expressed as: x
⎨ ⎬ x
2
1
=
⎧
⎩
0.731
1 x
1
⎩ ⎭
2
=
⎧
⎩
−
2.731
⎨ ⎬ ⎨ x
2
1.00
sin
ω
1 t sin
ω
2 t
The general solution for the free vibration can then be written as: x
⎨ ⎬
⎩ ⎭
=
A ⎨
0.731
1.00
⎭ sin(
ω ψ
1 t
+ +
1
) B
⎧
⎩
−
1
2.73
sin(
ω
2 t
+ ψ
2
) where A and B allow different amounts of each mode and
ψ
1 and
ψ
allows the two
2 modes’ different phases or starting values.
Substituting:
216

⎧
⎩ x x
2
1
(0)
(0) ⎭
=
5
1
=
A ⎨
0.731
⎫
1 sin
ψ
1
+
B
⎧
⎩
−
1
2.731
⎭ sin
ψ
2 x
1
(0)
⎩ x
2
(0) ⎭
=
⎩ ⎭
= ω
1
A
⎧
0.731
⎩ 1 ⎭ cos
ψ ω
1
+
2
B
⎧
⎩
−
1
2.731
cos
ψ
2 cos
ψ
1
= cos
ψ
2
=
0 =>
ψ
1
= ψ
2
= 0
90
Substituting in 1 st
set:
5
⎨ ⎬
1
=
A ⎨
0.731
1
⎬ +
B ⎨
−
2.731
1 ⎭
0.731A-2.731B= 5 A= 2.233
A+B = 1 B=-1.233
Hence the resulting free vibration is x
1
⎩ ⎭
=
2.233
⎧
0.731
⎩ 1.00
⎭ cos
ω
1 t
−
1.233
⎧
⎩
−
2.731
1.000
cos
ω
2 t
Normal modes from eigenvalues
The equation of motion for a two-degree-of freedom system can be written in matrix form as
M x
+
K x
=
0 (28) where M and K are the mass and stiffness matrix respectively; is the vector of generalized co-ordinates. Now pre-multiplying M
−
1
in both side of equation 6.2. one may get
I x
+ −
1
M K x
=
0 or, I x
+
A x
=
0
Here A
= −
1
M K is known as the dynamic matrix. Now to find the normal modes,
(29)
(30) x
1
=
X e , x
2
=
X e
[
A
− λ ] =
0
, the above equation will reduce to
(31)
217

where X
= { x
1 x
2
}
T
2
.
From equation (31) it is apparent that the free vibration problem in this case is reduced to that of finding the eigenvalues and eigenvectors of the matrix A .
Example: Determine the normal modes of a double pendulum.
Solution
Kinetic energy of the system =
T
=
1
2 m l
2
θ
1 1 1
2 +
1
2
(
2
θ
2 1 1
2 + l
2
θ
2 2
2 +
2 l l
θ θ
1 2 1 2 cos(
θ
2
− θ
1
))
Potential energy of the system =
U
= m g
1
=
− θ
1
)
+ m g l
2
{
1 g
{
( m
1
+
2
− θ
1
)
+
− θ
1
)
+ l
2
− θ
2
)
}
− θ
2
)
}
θ
1
θ
2 l
θ
1 1 l
θ
θ θ
2
−
1 l
θ
1 1
So Lagrangian of the system =
L T U
Figure 5
=
1
2 m l
2
θ
1 1 1
2 +
1
2
(
θ
2 1 1
+ l
θ
2 2
+
2 l l
1 2 cos(
θ θ
2
− −
1
)) g
{
( m
1
+ m
2
− θ
1
)
+ m
2
−
So using Lagrange principle, and assuming small angle of rotation, the equation of
θ
2
)
} motion can be written in matrix form as
⎡
( m
1
+
)
2 1
2 m l l
2 1 2
⎤
⎥ m l l
2 1 2 m l
2 2
2 θ
+
⎡
( m
1
+
)
2 1
0
0 m l g
2 2
⎤ ⎛
⎦ ⎝
θ
θ
2
⎞
=
⎠
Now considering a special case when m
1
= m
2
= m and l
1
= =
2 l , the above equation becomes
⎡
⎢
⎣
2 ml
2 ml
2 ml
2 ml
2
⎤
⎥ or, ml
2
⎡
⎣
2 1
1 1
⎤
⎦
θ
θ
+
⎡
2 mlg
0
0 mlg
⎤
⎥
θ
θ
+ mlg
⎡
⎣
2 0
0 1
⎤
⎦
θ
θ
=
0
⎜ ⎟ ⎜ ⎟
Now A
=
1 ml 2
⎛
⎝
−
1
−
1
1 2
⎞ ⎡
⎟ mlg
⎣
2 0
⎢
0 1
⎤
⎦
= l g
⎛
⎜
−
2
−
1
2 2
⎞
⎠
To find eigenvalues of A,
218

A
− λ
Or, 4
I 0
2 g l
− λ
−
2 g l
2
−
4 l g
λ λ
2 −
2
− g l
2 g l
− λ
=
0
2
=
0
Or,
λ
2 −
4 g l
λ +
2 g
⎝ ⎠
2
=
0
Or,
λ =
4 g l
±
⎜ 4 l g
⎟
2
−
8 g
⎜ ⎟
2
2
=
(2
±
2) g l
Hence natural frequencies are
ω
1
= g
0.7653
, l
ω
2
=
1.8478
g l
The normal modes can be determined from the eigenvalues.
The corresponding principal modes are obtained as g
θ
θ
λ λ
1
= l
(2 2 2) g l g
=
1
2
θ
θ
λ λ
2
= l
(2 2 2) g l
= −
1
2
It may be noted that while in the first mode
Both the pendulum moves in the same direction,
In the second mode they move in opposite direction
Figure 6
One may solve the same problem by taking x
1 and x as the generalized coordinates.
2
Here x
1
is the horizontal distance moves by mass m
1
and x is the distance move by
2 mass m
2
. Figure 7 show the free body diagram of both the masses.
T
1 y
T
2
θ
1
θ
2 y m x x m x
2 2
θ
2 m g m g
1
Figure 7
T
2 x
219

From the free body diagram of mass m
2
,
T
2
T
2 cos
θ
2 sin
θ
2
= m
2 g
= − m
2 x
2
Also from the free body diagram of mass m
1
,
T
1
T
1 cos
θ
1 sin
θ
1
−
T
2
−
T
2 cos
θ
2 sin
θ
2
+
= m g
1 m x
1 2
=
0
Assuming
θ
1
and
θ
2
to be small, sin
θ
1
= tan and sin
θ
2
=
= =
1 tan x l
1
/
θ
2
= θ
2
=
( x
2
− x
1
) / l
Hence
T
2
= m g , and T
1
=
( m
1
+
) m x
1 1
+
⎛
⎜
⎛
⎝
( m
1
+ l
1 m g
2
⎞
⎠
+ m g l
2
2
⎞
⎟ x
1
+
− m g
2 l
2 x
2
=
0 m x
2 2
+ m g
2
⎛
⎝ x
2
− x
1 l
2
⎞
⎠
=
0
Hence in matrix form
⎡
⎣ m
0
1
0 m
2
⎥ x x
1 ⎜ ⎟
+
⎢
⎣
⎡
⎢
⎢
⎢
⎛
⎜
⎛
⎝
( m
1
+ l
1 m g
2
⎞
⎠
+ m g l
2
2
− m g
2 l
2
⎞
⎟
− m g
2 m g l l
2
2
2
⎥
⎦
⎥
⎥
⎤
⎥ x
1 =
⎝ ⎠
Considering the case in which m
1
= m
2
= m and l
1
= =
2 l , the above equation becomes
⎡
1 0
⎣ 0 1
⎤
⎦ x
1
⎝ ⎠
+ l g
⎡
⎣ −
3
−
1
1 1
⎤
⎦
=
⎛
⎝
0
0
⎞
⎠
220

A l
⎡
⎣ −
3
−
1
1 1
⎤
⎦
Hence and A
− λ
I 0
3 g l
− λ − g l
− g g l l
− λ
=
0 or ,
λ
1
λ
2 − l g
λ +
=
(
2
−
2
)
2 g
⎝ ⎠
2 l g and
=
λ
2
0
=
(
2
+
2
)
Same as those obtained by taking
θ
1 g l and
θ
as the generalized coordinates.
2
Now
⎛
⎝
X
X
1
2
⎞
λ λ
1
=
3 g l l g
− λ
1 g
⎛
⎝
X
X
1
2
⎞
λ λ
2
=
3 g l l
− λ
2
=
1
3 2 2
=
1
3 2 2
=
1
2.4142
=
0.4142
= −
1
0.4142
= −
2.4142
1
1
-0.4142
2.4142
Figure 8
The different modes are as shown in the above figure.
Example Determine the equation of motion if the double pendulum is started with initial conditions x
1
(0)
= x
2
(0)
=
0.5, x
1
(0)
= x
2
(0)
=
0.
Solution:
The resulting free vibration can be considered to be the superposition of the normal modes. For each of these modes, the time solution can be written as
221

x
1
⎝ ⎠
1
=
⎛
⎝
X
X
1
2
⎞
⎠
1 sin
ω
1 t x
1
⎝ ⎠
2
=
⎛
⎝
X
X
1
2
⎞
⎠
2 sin
ω
2 t
The general solution for the free vibration can be written as x
⎜ ⎟
=
A
⎛
⎜
0.4142
1
⎞
⎠ sin(
ω ψ
1 t
+ +
1
) B
⎛
⎝
−
2.4142
1
⎞
⎠ sin(
ω
2 t
+ ψ
2
) where A and B are the amounts of first and second mode’s participation and
ψ
1 and
ψ
2 are the starting values or phases of the two modes. Substituting the initial conditions in the above equation
⎛
0.5
⎞ ⎛
⎝ 0.5
⎠
=
A
⎝
0.4142
1
⎞
⎠ sin
ψ
1
+
B
⎛
⎝
−
2.4142
1
⎞
⎠ sin
ψ
2 and
0
⎝ ⎠
=
A
ω
1
⎛
⎝
0.4142
1
⎞
⎠ cos
ψ
1
+
B
ω
2
⎛
⎝
−
2.4142
1
⎞
⎠ cos
ψ
2
For the second set of equations to be satisfied,
Hence A
=
0.6035
cos
ψ
1
= cos
ψ
2
=
0 , so that
ψ
1
= ψ
2
=
90
0
. and B
= −
0.1036
. So the equation for free vibration can be given by x
1
⎝ ⎠
=
0.6035
⎛
⎝
0.4142
1
⎞
⎠ cos
ω
1 t
−
0.1036
⎛
⎝
−
2.4142
1
⎞
⎠ cos
ω
2 t
Damped-free vibration of two-dof systems
Consider a two degrees of freedom system with damping as shown in figure k
1 x
1 k
2 x
2 k
3 c
1 m
1 c
2 m
2 c
3
Figure 9
Now the equation of motion of this system can be given by
⎡
⎣ m
1
0
0 m
2 x x
2
+
⎡
⎣ k
1
+ k
2
− k
2 k
2
− k
+
2 k
3
⎤
⎦ x
⎜ ⎟
+
⎡
⎣ c
1
+ c
2
− c
2 c
2
− c
+
2 c
3
⎤
⎦ x
1 =
⎝ ⎠
(32)
222

As in the previous case, here also the solution of the above equations can be written as x
1
=
A e
1 st
and x
2
=
A e
2 st
(33)
A A
1 2
Substituting (33) in (32) , one may write
⎡
⎣ m s
1
2 +
( c
−
1
+
2
) c s k
2
−
2
+ +
1 k
2
− c s k
2
−
2 m s
2
2 +
( c
2
+ c s
3 k
2 k
3
⎤
⎦
⎛
⎝
A
A
2
1
⎠
=
0
⎝ ⎠
(34)
Now for a nontrivial response i.e., for non-zero values of A
1 and A
2
, the determinant of their coefficient matrix must vanish. Hence m s
2 +
( c
1
+
2
)
+ + k
2
− c s
2
− k
2
− c s
− k
2 m s
2
2 +
( c
2
+ c s
3
)
+ +
2 k
3
=
0 or, ( m s
1
2 +
( c
1
+ c s
2
+ +
1 k
2
)( m s
2
2 +
( c
2
+
3
)
+ k
2
+ k
3
)
+
( c s
2
+ k
2
)
2 =
0
(35)
(36) which is a fourth order equation in s and is known as the characteristic equation of the system. This equation is to be solved to get four roots. The general solution of the system can be given by x
1
=
A e
11
+
A e
12
+
A e
13
+
A e
14 x
2
=
A e
21
+
A e
22
+
A e
23
+
A e
24
Here
(37)
1 i
,
=
1, 2, 3, 4 are four arbitrary constants to be determined from the initial conditions and the coefficients
2 i
,
=
1, 2, 3, 4 are related to A
2 i and can be determined from equation (34) as
A
A
1 i
2 i
= m s i
2 +
( c c s
1
2
+ i
+
) k
2 i k k
2
(38)
For a physical system with damping, the motion will die out with time. For a stable system, all the four roots must be either real negative numbers or complex number with negative real parts. It may be recalled that, if the roots contain complex conjugate numbers, the motion will be oscillatory.
Example: Find the response of the system as shown in figure 9 considering m
1
= m
2
= m , k
1
= k
2
= k
3
= k and c
1
= c
3
=
0 and c
2
= c .
Solution.
In this case the characteristics equation becomes
( ms
2 + + k ms
2 cs 2 ) ( cs
+ k )
2 =
0
223

( ms
2 + + k
2 −
( cs
+ k )
2 =
0
,
2 4 +
2 mcs
3 +
(4 mk
+ − 2 2 c s
+
(4 kc
−
2 )
+
4 k
2 − k
2 =
0
,
2 4 +
2 mcs
3 +
4 mks
2 +
2 kcs
+
3 k
2 =
0
,
2 or ms ms
2 +
2 cs
+ k
+ k ms
2 +
2 cs
+
3 )
=
0 or ms
2 + k ms
2 +
2 cs
+
3 )
=
0 or ms
2 + k ms
2 +
2 cs
+
3 )
=
0
Hence the roots are s
1,2
= ± i k m s m c
2
− k m
So the system has a pair of complex conjugate
SEMI-DEFINITE SYSTEMS
The systems with have one of their natural frequencies equal to zero are known as semidefinite or degenerate systems. One can show that the following two systems are degenerate systems. x
1 k m
1
Figure 10 m
2 x
2
I
1
θ
From figure 10 the equation of motion of the system is
⎡
⎣ m
0
1
0 m
2 x x
+
⎡ k
−
− k k k
⎤ x x
1 =
⎛
0
⎞
⎜ ⎟ ⎜ ⎟
1
Figure 11
θ
2
I
2
(39)
Assuming the solution x
1
=
A e and x
2
=
A e (40)
⎡
⎣ k
− m
1
ω
− k
2 − k k
− m
2
ω 2
⎤
⎦
⎛
⎝
A
A
2
1
⎠
=
0
⎝ ⎠
So for non-zero values of A A
1
,
2
,
(41)
224

k
− m
1
ω
2
− k
(
− k
=
0 k
− m
2
ω
2
1
ω
2
)( k
− m
2
ω
2
)
− k
2 = k 2 − k m
1
+ m
2
)
ω
2 + m m
1 2
ω
4 − k 2 =
0
ω
2 or, ( m m
2
ω
2 −
(
1
+ m
2
))
=
0
(42)
(43)
(44)
(45)
⇒ ω
1
=
0, and,
ω
2
=
(
1
+ m
2
) m m
1 2
(46)
Hence, the system is a semi-definite or degenerate system. Corresponding to the first mode frequency, i.e.,
ω =
1
0, A
1
=
A
2
.
So the system will have a rigid-body motion. For the second mode frequency
A
A
1
2
= k
− k m
1
ω
2
= km m
1 2 km m
2
−
(
1
+ m
2
)
=
− km m
1 2
1
= − m
2 m
1
(47) amplitude ratio is inversely proportional to the mass ratio the system.
Similarly one may show for the two-rotor system,
θ
1
θ
2
= −
I
I
1
2 (48) the ratio of angle of rotation inversely proportional to the moment of inertia of the rotors.
Forced harmonic vibration, Vibration Absorber
Consider a system excited by a harmonic force F
1 sin
ω t expressed by the matrix equation
⎡
⎣ m m
11
21 m m
12
22 x x
2
⎢ k
11 k
12 x k k x
21 22 2
1
⎝ ⎠ sin
ω t (49)
Since the system is undamped, the solution can be assumed as x
⎜ ⎟ ⎜ x
2
X
X
1
2
⎞
⎠ sin
ω t (50)
Substituting equation (50) in equation (49), one obtains
⎡
⎢
⎣ k
11
− m
11
ω k
21
− m
21
ω
2
2 or ,
⎡
⎢
⎣ k
11
− m
11
ω k
21
− m
21
ω
2
2 k
12
− m
12
ω k
22
− m
22
ω
2
2
⎤
⎥ ⎜
⎦
⎛
⎝
X
X
1
2
⎞
⎠ sin
ω t k
12
− m
12
ω
2 k
22
− m
22
ω
2
=
⎤
⎥ ⎜ ⎟ ⎜ ⎟
⎦
⎛
⎝
X
X
1
2
⎞ ⎛ ⎞
⎠
=
F
⎝ ⎠ sin
ω t
(51)
225

⎛
⎝
X
X
1
2
⎞
⎠ ⎣ k
11
− m
11
ω k
21
− m
21
ω
2
2
= k
12
− m
12
ω k
22
− m
22
ω
2
2
⎤
⎥
⎦
−
1
⎡
⎣ k
22
− m
22
ω
− k
21
+ m
21
ω
2
2 k
11
− m
11
ω
2 k
21
− m
21
ω
2
− k
12
+ m
12
ω 2 k
11
− m
11
ω 2
⎤ k
12
− m
12
ω
2 k
22
− m
22
ω
2
Hence
X
1
=
( k
22
− m
22
ω
Z
2
)
F
, where
[
Z
ω ] = ⎢
⎡
⎣ k
11
− m
11 k
21
− m
21
ω
ω
2
2 k
12
− m
12
ω k
22
− m
22
ω
2
2
⎤
⎦
(52)
(53)
X
2
=
( k
21
− m
21
ω
2
)
F
Z
(54)
Example Consider the system shown in figure 12 where the mass m
1 is subjected to a force F sin
ω t . Find the response of the system when m
1
= m
2 k
1
= k
2
= k
3 k
1 m
1 x
1
F sin
ω t k
2 m
2 x
2 k
3
Figure 12
Solution:
The equation of motion of this system can be written as
⎡
⎣ m
1
0
0 x m
2
⎥ x
1 ⎜ ⎟
+
⎡
⎣ k
1
+ k
− k
2
2
− k
2 k
2
+ k
3
⎤
⎦ x
1 x
2
F sin
ω t
0
⎞
⎠
⎡
⎣ m
0
0 m
⎤
⎦ x
⎜ ⎟ +
⎡
⎣
2 k
− k
− k 2 k
⎤
⎦ x
1 =
⎛
⎝
F sin
ω t
0
⎞
⎠
So assuming the solution
226

[ x
⎜ ⎟ ⎜ x
2
X
X
1
2
⎞
⎠ sin
ω t and proceeding as explained before
Z
]
⎣
2 k
− m
ω
2
− k
− k
2 k
− m
ω
2
⎤
⎦
Z
ω =
(
2 k
− m
ω
2
)
2 − k
2 = m
2
ω
4 −
4 mk
ω
2 +
3 k
2 = m
2
(
ω
4 −
4 k m
ω
2 +
2 k
3 ) m 2 or, Z
= m
2
(
ω
2 − k m
)(
ω
2 − k
3 ) m
= m
2
(
ω
2 − ω ω
1
)(
2 − ω
2
2
) where,
ω
1
2 = k m and
ω
2
2 = k
3 m
are normal mode frequencies of this system.
Hence,
X
1
= m
2
(
2 k
− m
ω
2
)
F
(
ω 2 − ω ω 2 − ω
2
2
)
X
2
= m
2 kF
(
ω
2 − ω ω
1
2 − ω
2
2
)
So it may be observed that the system will have maximum vibration when
ω ω
1 or ,
ω ω
2
.
Also it may be observed that X
1
=
0, when
ω
2 =
.
Tuned Vibration Absorber
Consider a vibrating system of mass m
1
, stiffness k
1
, subjected to a force F sin
ω t . As studied in case of forced vibration of single-degree of freedom system, the system will have a steady state response given by x
=
F sin
ω t m (
ω
2 n
− ω
2
)
, where
ω n
= k
1
/ m
1
(55) which will be maximum when
ω ω n .
Now to absorb this k
1
F sin
ω t vibration, one may add a secondary spring and mass system as shown in figure 13. k
2 m
1 x
1 m
2
Figure 13 x
2
227

The equation of motion for this system can be given by
⎡
⎣ m
1
0
0 m
2 x x
2
⎢ k
1
+ k
2
− k
2 x
1
− k
2 k
2 x
2
⎝
F sin
ω t
0
⎞
⎠
(56)
Comparing equation (49) and (56), m
11
= m m
1
,
12
=
0, m
21
=
0, m
22
= m k
2, 11
= +
1 k
2
, k
12
= − k
2
, k
21
= − k
2
, and k
22
= k .
Hence,
Z
= k
1
+ −
2 m
1
ω
2
− k
2
− k
2 k
2
− m
2
ω
2
= k k
1 2
− m k
1 2
ω
2 − k m
1 2
ω
2 − k m
2 2
ω
2 + m m
1 2
ω
4
= m m (
1 2 1
)(
2
) (57)
2 where
λ
1 and
λ
2
are the roots of the characteristic equation Z
ω =
0 of the freevibration of this system., which can be given by
λ
1,2
=
0.5
⎧
⎨
⎛
⎜
⎩
⎝ k
1
+ k
2
+ k
2 m
1 m
2 m
1
⎞
±
⎛
⎟ ⎜ k
1
+ k
2
+ k
2 m
1 m
2 m
1
⎞ 2
−
4 k k
1 2 m m
2
⎬
⎭
(58)
Now from equation (53) and (54)
X
1
=
( k
22
− m
22
ω
Z
2
)
F
=
( k
2
−
Z m
2
ω
ω
2
)
F
, (59)
X
2
=
− k F
2
Z
(60)
From equation (59), it is clear that, X
1
=
0, when
ω 2 = k
2 m
2
.
Hence, by suitably choosing the stiffness and mass of the secondary spring and mass system, vibration can be completely eliminated from the primary system. For
ω
2 = k
2 m
2
,
Z
ω = k k
− m k k
2 m
2
− k m
2 k
2 m
2
− k m
2 k
2 m
2
+ m m
2 k
2 m
2 k
2 m
2
(61)
= k k
1 2
− m
1 k
2 m
2
2
− k k
1 2
− k
2
2
+ m
1 k
2
2 m
2
= − k
2
2 and X
2
=
−
− k F
=
F k
2
2 k
2
(62)
228

Centrifugal Pendulum Vibration Absorber
The tuned vibration absorber is only effective when the frequency of external excitation equals to the natural frequency of the secondary spring and mass system. But in many cases, for example in case of an automobile engine, the exciting torques are proportional to the rotational speed ‘n’ which may vary over a wide range. For the absorber to be effective, its natural frequency must also be proportional to the speed. The characteristics of the centrifugal pendulum are ideally suited for this purpose.
Placing the coordinates through point O’, parallel and normal to r, the line r rotates with angular velocity ( ) r j
ˆ i
ˆ
R
O
O
′
The acceleration of mass m a m
⎣ R
θ
2 cos
φ +
R
θ sin
φ − r (
θ φ
Since the moment about O
′
is zero,
)
2
⎤ i
ˆ + ⎡
R
θ
2 sin
φ +
R
θ cos
φ + r (
θ φ
)
2
⎤⎦ j
ˆ (63)
M
O
′
=
⎣
θ
2 sin
φ +
R
θ cos
φ + r (
θ φ
Assuming
φ
to be small, cos
φ =
1, sin
)
2 r
=
0
, so
(64)
φ +
R r
= −
R
+ r r
θ
(65) oscillation of frequency
ω
, one may write
θ = nt
+ θ ω
0 sin t
θ = nt
+ ωθ
0 cos
ω t
≅ n
θ = − θ ω
0
2 sin
ω t
Substituting the above equations in equation (65) yields,
(66)
(67)
(68)
229

φ +
R n
2 r
⎞
φ =
⎛
R
+ r r
⎠ 0 sin
ω t
Hence the natural frequency of the pendulum is
(69)
ω n
= n
R r and its steady-state solution is
φ =
( R
+ r ) / r
− ω
2 +
( Rn
2 r
ω θ
0 sin
ω t
(70)
(71)
It may be noted that the same pendulum in a gravity field would have a natural frequency of g
. So it may be noted that for the centrifugal pendulum the gravity r field is replaced by the centrifugal field Rn
2
.
Torque exerted by the pendulum on the wheel
With the component of a m
equal to zero, the pendulum force is a tension along , given by m times the component of a m
.
T
=
(
R cos
φ i
ˆ +
R sin
φ j
ˆ
)
× m
⎡ −
R
θ
2 cos
φ +
R
θ sin
φ − r (
θ φ
)
2
= − mR
φ
⎣
−
R
ω θ
0 sin
ω t sin
φ −
Rn
2 − rn
2 − r
φ
2 −
2 r
θφ
⎦
Now assuming small angle of rotation
T
= − ( + )
2
φ i
ˆ
(72)
(73)
Now substituting the (73) in (72),
T
=
−
(
+
)
2 2 r n / r
( Rn
2 r
− ω
2
ω θ
0 sin
ω t
= −
⎡
⎣ 1
−
(
+ r )
2 r
ω
2
/ Rn
2
⎤
⎦
θ =
J eff
θ
(74)
Hence the effective inertia can be written as
J eff
= −
⎡
⎣ 1
−
(
+ r )
2 r
ω
2
/ Rn
2
⎤
⎦
= −
1
−
(
+ r )
2
( )
2 n
(75) which can be
∞
at its natural frequency. This possesses some difficulties in the design of the pendulum. For example to suppress a disturbing torque of frequency equal to four times the natural speed n , the pendulum must meet the requirement
ω
2 =
(4 )
2 = 2 n n R / r . Hence, as the length of the pendulum r
=
R /16 becomes very
230

small it will be difficult to design it. To avoid this one may go for Chilton bifilar design.
Exercise problems
1.
In a certain refrigeration plant, a section of pipe carrying the refrigerant vibrated violently at a compressor speed of 232 rpm. To eliminate this difficulty, it was proposed to clamp a cantilever spring mass system to the pipe to act as an absorber. For a trial test, for a 905 gm. Absorber tuned to 232 cpm resulted in two natural frequencies of 198 and 272 cpm.
If the absorber system is to be designed so that the natural frequencies lie outside the region 160 to 320 cpm, what must be the weight and spring stiffness?
2.
Derive the normal modes of vibration of a double pendulum with same length and mass of the pendulum.
3.
Develop a matlab code for determination of free-vibration of a general two-degree of freedom system.
4.
Derive the equation of motion for the double pendulum shown in figure p1 in terms of
θ
1 and
θ
2 using Lagrange principle. Determine the natural frequencies and mode shapes of the systems. If the system is started with the following initial conditions: x
1
(0) = x
2
(0) = X, v
1
(0) =v
2
(0)=0, (v
1 and v
2
are velocity) determine the equation of motion. If the lower mass is given an impulse F
0
δ
(t), determine the response in terms of normal modes.
θ
1
L
1
m
1
L
2
x
1
θ
2
m
2
x
2
Figure P1
5. A centrifugal pump rotating at 500 rpm is driven by an electric motor at 1200 rpm through a single stage reduction gearing. The moments of inertia of the pump impeller and the motor are 1600 kg.m
2
and 500 kg.m
2 respectively. The lengths of the pump shaft and the motor shaft are 450 and 200 mm, and their diameters are 100 and 50 mm respectively. Neglecting the inertia of the gears, find the frequencies of torsional oscillations of the system. Also determine the position of the nodes.
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