Free Vibrations of Systems with Two
Degrees of Freedom
1.
Introduction
The number of degrees of freedom (DOF) of a vibrating system can be defined as the
number of independent coordinates required to completely specify the configuration
of that system at any instant. It is often possible to simplify complex systems into
acceptable single degree of freedom (SDOF) models, though there are times when this
cannot be done. Then it becomes necessary to consider multiple degree of freedom
(MDOF) models, in which there are more than one resonant condition, each with its
characteristic mode shape. We will first consider systems with two degrees of
freedom (TDOF). The methods developed for TDOF systems can be readily extended
to MDOF systems. The principal differences between SDOF and TDOF systems are
greater than the differences between TDOF and ten DOF systems, for instance.
In practice, with the use of appropriate simplifying assumptions, many real life
systems may often be reduced to representative TDOF models. Some common TDOF
models are presented below.
1.1 The Two-mass, Two-spring Translational Model
The model, illustrated in Fig. 1.1, can be used, for example, to model a vehicle trailer
mass (m2 ) , its suspension (k 2 ) , the mass of the axle (m1 ) , and the tyre stiffness (k1 ) .
y2
m2
k2
y1
m1
k1
Fig. 1.1
Free Vibrations of Two Degree of Freedom Systems
A slightly different model is achieved if three, rather than two springs, are used, the
third spring being located at the top of the model with its (the third spring’s) upper end
fixed.
1.2 The Single-mass, Two-spring Model with Both Rotation and Translation
The single-mass, two-spring model with both rotation and translation is illustrated in
Fig. 1.2 below. It may be used, for instance, to model a vehicle body mounted on
front and rear suspension springs, if the vehicle wheels are assumed to be completely
rigid and the damping effects of the shock absorbers are neglected. The pitching (θ )
and bouncing ( y ) motions can then be analysed by treating the body as rigid and
neglecting the ‘unsprung’ mass of the vehicle.
y
θ
G
m
k2
k1
Fig. 1.2
1.3 The Two-rotor, Two-shaft Model with a Single Axis of Rotation
The two-rotor, two-shaft model with a single axis of rotation is illustrated in Fig. 1.3
below.
K1
J1
θ1
K2
J2
θ2
Fig. 1.3
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Free Vibrations of Two Degree of Freedom Systems
The shafts are assumed to be linearly elastic in the torsional mode. This model may
be used to represent the dynamics of mechanical drive transmissions. Where a gear
ratio is included in the system, appropriately modified values of rotor inertia and shaft
stiffness must be used, as will be seen later. In the context of mathematical analysis,
this model is analogous to that in Fig. 1.1 above.
1.4 The Two-rotor, Two-spring Model with Coplanar Rotors and Parallel Axes
of Rotation.
The two-rotor, two-spring model with coplanar rotors and parallel axes of rotation is
illustrated in Fig. 1.4 below. This model may be used to model a belt drive, for
example.
k2
θ1
r1
J1
r2
θ2
J2
k1
Fig. 1.4
A mode is a description of motion. There are various qualifications of modes, such as
the first mode, the second mode, and a coupled mode. All these describe a particular
manner of motion. Usually the number of natural frequencies of a system is equal to
the number of DOF, each such frequency being associated with a principal or normal
mode of free vibration. In a principal mode, which is also known as a natural mode:
•
All the components of the system vibrate with the same frequency, which is a
natural frequency of the system.
•
All the components of the system reach their extreme positions at the same
instant in time. This means that all the components of the system are either in
phase or anti-phase (1800 out of phase). When two components of a system are
in anti-phase, a transition point or zone will occur somewhere, whose
displacement will be zero. Such a point or zone is known as a node.
•
If the amplitude of one co-ordinate is conveniently made equal to unity, then the
mode is said to be normalised, or it is simply referred to as a normal mode.
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Free Vibrations of Two Degree of Freedom Systems
When external excitation is applied to MDOF systems, forced vibrations will be
induced and resonance will occur if the frequency of excitation coincides with any of
the natural frequencies of the system. Consequently, the first and most important
concern in the study of the vibrations of MDOF systems is to estimate the natural
frequencies of the systems in question. For this purpose, the effects of damping need
not be considered.
1.5 Free Natural Mode Vibrations of TDOF Systems
The following is an outline of a systematic procedure to be followed in the
mathematical analysis of TDOF models that undergo free vibrations:
(a) Consider the system in a configuration that is displaced from equilibrium, and
choose the most suitable parameters (co-ordinates) for specifying of this
configuration and the formulating the equations of motion. See Fig. 1.5.
x2
x1
k1
k2
m2
m1
Fig. 1.5
(b) Write down the equations of motion for each mass or inertia, whichever is
applicable, in terms of the chosen parameters. For the above system, by applying
Newton’s second law of motion, the equations are found to be the following:
m1&x&1 = − k1 x1 + k 2 ( x2 − x1 )
m2 &x&2 = − k 2 ( x2 − x1 )
(1)
(c) Assume solutions to the equations of motion that will represent the principal
modes. Since damping will not have been considered, all parts of the system will
either be in-phase or in anti-phase. Note that with no damping, the sine of any
phase angle will be zero. Thus, simple harmonic solutions of the following form
will suffice:
xi = X i sin ωnt ,
for i = 1,2
Hence:
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Free Vibrations of Two Degree of Freedom Systems
&x&i = −ωn 2 X i sin ωnt ,
for i = 1,2
(2)
(d) In one approach to solving the differential equations of motion, the above
assumed solutions may be substituted into the equations of motion and the
amplitude ratio X 1 X 2 be eliminated from the equations so obtained. Thus:
− m1 X 1ωn = −k1 X 1 + k 2 ( X 2 − X 1 )⎫⎪
⎬
2
⎪⎭
− m2 X 2 ωn = −k 2 ( X 2 − X 1 )
2
(3)
or:
⎛
X1 2
X
X ⎞⎫
ωn = − k1 1 + k 2 ⎜⎜1 − 1 ⎟⎟⎪
X2
X2
X 2 ⎠⎪
⎝
⎬
⎛
X1 ⎞
2
⎪
⎟⎟
− m2 ωn = − k 2 ⎜⎜1 −
⎪
X2 ⎠
⎝
⎭
− m1
(3a)
or:
⎫
− m1uωn 2 = −k1u + k 2 (1 − u )⎪
⎪⎪
− m2 ωn 2 = −k 2 (1 − u )
⎬
⎪
X
⎪
u= 1
⎪⎭
X2
(3b)
Each of equations (3) may now be re-arranged to give expressions of the
amplitude ratio as follows:
X1
k2
⎫
=
2⎪
X 2 k1 + k 2 − m1ωn ⎪
⎬
2
X 1 k 2 − m2 ωn
⎪
=
⎪⎭
X2
k2
(4)
The amplitude ratio may now be eliminating between equations (4) to obtain the
following:
k2
k1 + k 2 − m1ωn 2
k 2 − m2 ωn 2
=
k2
(5)
(e) Through manipulation of equation (5), a frequency equation may now be
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Free Vibrations of Two Degree of Freedom Systems
obtained, the roots of which will lead to the values of the natural frequencies
(ωn ). In the present case, the frequency equation is as follows:
m1m2 ωn 4 − [k 2 (m1 + m2 ) + k1m2 ]ωn 2 + k1k 2 = 0
or:
m1m2 ωn 4 − [m2 (k1 + k 2 ) + m1k 2 ]ωn 2 + k1k 2 = 0
As can be seen in the above expression, in the case of TDOF systems, the
2
frequency equation is a quadratic in ωn and may be solved to yield two positive
roots, which give the values of the two natural frequencies of the system. Note
that negative frequencies have no physical meaning.
(f) The modal shapes can then be found by substituting each value of ωn into any of
the earlier obtained expressions for the amplitude ratios.
Note that any complex periodic function may be analysed into simpler harmonics by
use of the Fourier series. It so happens to be that any arbitrary free vibration of a
MDOF system must consist of separate vibrations in the principal modes, combined in
their proper proportions and phasing.
1.6 Worked Examples
Worked Example 1
Illustrated in the figure below is a possible model of a belt drive. Derive the
differential equations of motion of the system in the coordinates shown on the figure
and express the equations so derived in matrix notation.
k
J1
r1
θ1
J2
r2
θ2
k
Figure for Worked Example 1
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Free Vibrations of Two Degree of Freedom Systems
If in the above figure, r2 = 2r1 = 2r and J 2 = 4 J1 = 4 J , determine the natural
frequencies of the system. Moreover, suppose that the modal fractions are denoted as
follows:
u1 =
A11
A
; u 2 = 12
A21
A22
Show that the following relationship holds true:
u1u 2 = −
J2
J1
Solution
The figure below gives the free-body diagrams of the two rotors.
k (r1θ1 − r2θ2 )
J1
r1
θ1
J2
r2
θ2
k (r2 θ 2 − r1θ1 )
Free-body Diagrams
By applying Newton’s second law of motion to each rotor in turn, one obtains the
following equations:
J1&θ&1 = k (r2 θ 2 − r1θ1 )r1 − k (r1θ1 − r2 θ 2 )r1 ⎫
⎬
J 2&θ& 2 = k (r1θ1 − r2 θ 2 )r2 − k (r2 θ 2 − r1θ1 )r2 ⎭
(6)
The above equations may be rearranged into the following standard form:
2
J1&θ&1 + 2kr1 θ1 − 2kr1r2 θ 2 = 0⎫⎪
⎬
2
J 2&θ& 2 − 2kr1r2 θ1 + 2kr2 θ 2 = 0⎪⎭
(7)
Furthermore, the above equations may be re-written in matrix notation, as follows:
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Free Vibrations of Two Degree of Freedom Systems
⎡ J1
⎢0
⎣
0 ⎤ ⎧&θ&1 ⎫ ⎡ 2kr12
⎨ ⎬+ ⎢
J 2 ⎥⎦ ⎩&θ&2 ⎭ ⎣− 2kr1r2
− 2kr1r2 ⎤ ⎧ θ1 ⎫ ⎧0⎫
⎬=⎨ ⎬
2 ⎥⎨
2kr2 ⎦ ⎩θ 2 ⎭ ⎩0⎭
(8)
Given that r2 = 2r1 = 2r and J 2 = 4 J1 = 4 J , equations (7) above can be reduced into
the following:
J&θ&1 + 2kr 2 θ1 − 4kr 2 θ 2 = 0⎫⎪
⎬
J&θ& 2 − kr 2 θ1 + 2kr 2 θ 2 = 0⎪⎭
(9)
In the natural modes, the displacements and accelerations of the rotors may be
described by the following equations:
θi (t ) = Ai sin ωt ⎫
&θ& (t ) = −ω2 θ (t )⎬
i
i
⎭
(10)
Substituting equations (10) into equations (9) and eliminating the harmonic term
yields the following:
(2kr
)
− ω2 J A1 − 4kr 2 A2 = 0⎫⎪
⎬
− kr 2 A1 + 2kr 2 − ω2 J A2 = 0⎪⎭
2
(
)
(11)
Equations (11) may now be expressed in matrix notation, as follows:
(
⎡ 2kr 2 − ω2 J
⎢
2
⎣ − kr
)
⎤ ⎧ A1 ⎫ ⎧0⎫
⎥⎨ ⎬ = ⎨ ⎬
2kr 2 − ω2 J ⎦ ⎩ A2 ⎭ ⎩0⎭
(
− 4kr 2
)
(12)
In another approach to solving the differential equations of motion, as opposed to the
elimination of amplitude ratios, Cramer’s rule may be applied to equation (12) to yield
the following frequency equation:
(2kr
)
⎫
0
=
⎪
2kr 2 − ω2 J
− kr 2
⎪
⎪
2
2kr 2 − ω2 J − 4k 2 r 4 = 0⎬
⎪
4k 2 r 4 − 4ω2 Jkr 2 + ω4 J 2 − 4k 2 r 4 = 0⎪
ω2 ω2 J 2 − 4 Jkr 2 = 0⎪⎭
2
− ω2 J
(
(
(
− 4kr 2
)
)
(13)
)
The corresponding natural frequencies and amplitude ratios (modal fractions) are as
follows:
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Free Vibrations of Two Degree of Freedom Systems
⎫
⎪
⎪
⎬
4kr 2
A12
=
; u2 =
= −2⎪
⎪⎭
J
A22
ωn1 = 0; u1 =
ωn 2
A11
=2
A21
(14)
It can be seen from equations (14) that:
u1u 2 = − 4 = −
J2
J1
Note in equation (14) that double subscripts are used in the notation for the amplitudes
of the rotors. The first subscript denotes the rotor while the second subscript denotes a
natural mode of vibration. Since the system in question has two DOF, it also has two
natural modes vibration.
However, in the present case, one of the natural mode frequencies happens to be zero
and therefore the system has only one non-trivial natural frequency. Such a system is
referred to as semi-definite and can usually be readily reduced into an equivalent
single DOF system, as we shall see in due course.
Worked Example 2
For the system illustrated in the figure below, derive the differential equations of
motion and express them in matrix notation. Solve for the two natural frequencies and
the corresponding modal fractions.
θ1
2J
J
K
θ2
2K
Figure for Worked Example 2
Solution
With the use of free-body diagrams and the application of Newton’s second law of
motion to each of the inertias in turn, the equations of motion are found to be the
following:
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Free Vibrations of Two Degree of Freedom Systems
J&θ&1 + 3Kθ1 − 2 Kθ2 = 0⎫
⎬
2 J&θ&2 − 2 Kθ1 + 2 Kθ2 = 0⎭
These equations may be re-written in matrix notation as follows:
0 ⎤ ⎧ &θ&1 ⎫ ⎡ 3K
⎨ ⎬+
2 J ⎥⎦ ⎩&θ&2 ⎭ ⎢⎣− 2 K
⎡J
⎢0
⎣
− 2 K ⎤ ⎧ θ1 ⎫ ⎧0⎫
⎨ ⎬=⎨ ⎬
2 K ⎥⎦ ⎩θ 2 ⎭ ⎩0⎭
To solve for the two natural frequencies we assume solutions of the following form:
θi (t ) = Θi cos ω t
⎫⎪
i = 1, 2
&θ& (t ) = −ω2Θ cos ω t ⎬⎪
i
i
⎭
Substitution of the above solutions into the earlier derived differential equations of
motion yields the following results:
− ω2 JΘ1 + 3KΘ1 − 2 KΘ 2 = 0⎫⎪
⎬
− 2ω2 JΘ 2 − 2 KΘ1 + 2 KΘ 2 = 0⎪⎭
The above may be expressed in matrix notation as follows:
(
⎡ 3 K − ω2 J
⎢
⎣ − 2K
)
⎤ ⎧ Θ1 ⎫ ⎧0⎫
⎥⎨ ⎬ = ⎨ ⎬
2 K − ω2 J ⎦ ⎩Θ 2 ⎭ ⎩0⎭
(
− 2K
)
On the application of Cramer’s rule, the following is readily obtained:
(
)(
)
2 3 K − ω2 J K − ω 2 J − 4 K 2 = 0
The above equation may be re-arranged and simplified into the following:
J 2 ω4 − 4 KJω2 + K 2 = 0⎫
⎪
2
⎬
K
K
⎛ ⎞
⎛ ⎞
ω4 − 4⎜ ⎟ω2 + ⎜ ⎟ = 0⎪
⎝J⎠
⎝J⎠
⎭
The above is a quadratic in ω2 and can be solved, with the use of the quadratic
formula, to yield the following:
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Free Vibrations of Two Degree of Freedom Systems
⎫
⎪
⎪
⎬
K ⎪
= 0.5176
J ⎪⎭
ωn1 = 3.732
K
K
; ωn1 = 1.932
J
J
ωn 2 = 0.268
K
; ωn 2
J
2
2
To determine the modal fractions, recall that:
− ω2 JΘ1 + 3KΘ1 − 2 KΘ 2 = 0⎫⎪
⎬
− 2ω2 JΘ 2 − 2 KΘ1 + 2 KΘ 2 = 0⎪⎭
Thus, with the substitution of appropriate values of the natural frequencies in any of
the above equations, the modal fractions are found to be the following:
Θ11
⎫
= −2.732⎪
Θ 21
⎪
⎬
Θ12
= 0.732 ⎪
⎪⎭
Θ 22
Worked Example 3
For the two degree-of-freedom system illustrated in the figure below, derive the
differential equations of motion and express them in matrix notation. Further, derive
the frequency equation.
x1
x2
k
m1
m2
Figure for Worked Example 3
Moreover, show that the differential equations of motion of the system may be rewritten as a single equation in the coordinate
q = ( x2 − x1 )
and equivalent mass:
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Free Vibrations of Two Degree of Freedom Systems
meq =
m1m2
m1 + m2
Solution
With the use of free-body diagrams and Newton’s second law of motion, the system’s
differential equations of motion are found to be the following:
m1&x&1 + kx1 − kx2 = 0 ⎫
⎬
m2 &x&2 − kx1 + kx2 = 0⎭
The above equations may be expressed in the concise matrix notation as follows:
− k ⎤ ⎧ x1 ⎫ ⎧0⎫
⎨ ⎬=⎨ ⎬
k ⎥⎦ ⎩ x2 ⎭ ⎩0⎭
⎡m1 0 ⎤ ⎧ &x&1 ⎫ ⎡ k
⎢ 0 m ⎥ ⎨ &x& ⎬ + ⎢− k
⎣
2 ⎦⎩ 2 ⎭ ⎣
To obtain the system’s frequency equation, we assume harmonic solutions of the
following form:
xi (t ) = X i cos ω t
⎫⎪
⎬ i = 1, 2
&x&i (t ) = −ω2 X i cos ω t ⎪⎭
By substituting these solutions into the differential equations of motion, one obtains
the following:
− ω2 m1 X 1 + kX 1 − kX 2 = 0⎫⎪
⎬
− ω2 m2 X 2 − kX 1 + kX 2 = 0⎪⎭
The above may be expressed in matrix notation as follows:
(
⎡ k − ω2 m1
⎢
⎣ −k
)
⎤ ⎧ X 1 ⎫ ⎧0⎫
−k
⎥⎨ ⎬ = ⎨ ⎬
k − ω2 m2 ⎦ ⎩ X 2 ⎭ ⎩0⎭
(
)
Application of Cramer’s rule to the above equation leads to the following result:
(k − m ω )(k − m ω ) − k = 0⎫⎪
⎬
ω [m m ω − k (m + m )] = 0⎪⎭
2
2
1
2
2
2
2
1
2
1
2
Hence the frequency equation as follows:
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Free Vibrations of Two Degree of Freedom Systems
⎡ 2 k (m1 + m2 )⎤ 2
⎢ω − m m
⎥ω = 0
⎣
⎦
1 2
Now, recall that the differential equations of motion can be expressed as follows:
m1&x&1 + kx1 − kx2 = 0 ⎫
⎬
m2 &x&2 − kx1 + kx2 = 0⎭
If the first of the above equations is multiplied by m2 and the second multiplied by
m1 , the following equations are readily obtained:
m1m2 &x&1 + km2 x1 − km2 x2 = 0⎫
⎬
m1m2 &x&2 − km1 x1 + km1 x2 = 0⎭
Now, if the first of the above equations is subtracted from the second, the following
are obtained:
m1m2 (&x&2 − &x&1 ) + k (m1 + m2 )( x2 − x1 ) = 0⎫
⎪
⎛ m1m2 ⎞
⎪
⎜⎜
⎟⎟q&& + kq = 0⎬
⎝ m1 + m2 ⎠
⎪
meq q&& + kq = 0⎪⎭
This completes the proof.
Evidently, the system that we have dealt with in this worked example is a two DOF
system. However, it has only one non-trivial natural frequency, the other natural
frequency being zero, as can be seen from its frequency equation. We have already
seen that such systems are referred to as semi-definite and they can usually be reduced
to equivalent systems of fewer DOF, through the manipulation of coordinates.
1.7 Example Problems
1.
In systems with free or natural vibrations, what is meant by “natural mode”?
You may use a TDOF spring-mass system to explain your answer.
The tractor-trailer combination of Fig. Q1 moves at constant advance velocity in
a straight line on perfectly plane and level ground.
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Free Vibrations of Two Degree of Freedom Systems
Hitch
Trailer
Tractor
Fig. Q1
Given that the mass of the trailer with its load is m1 , the mass of the tractor is
m2 , the hitch is linearly elastic in the longitudinal direction with stiffness k , and
all frictional forces are negligible:
(a) Model the system as a two degree-of-freedom spring-mass system and
derive the equations of motion.
(b) Determine the natural mode frequencies and the corresponding amplitude
ratios in terms of the given quantities.
2.
Figure Q2, illustrates a torsional system. Derive the frequency equation. In the
particular case where J1 = 2 J 2 = 2 J and K1 = K c = K 2 = K , determine the
ratios of amplitudes corresponding to the two natural modes.
K1
J1
Kc
J2
K2
Fig. Q2
3.
Referring again to Fig. Q2, given that K1 = K 2 = 0 and K c = K , solve for the
natural frequencies and the amplitude ratios in terms of the given quantities.
A turbine-electric generator unit may be dynamically modelled as two rotational
inertias J1 = 4 kgm2 (turbine) and J 2 = 2 kgm2 (generator), and a linearly
elastic shaft with K = 40 kNm/rad and negligible inertia. Determine the nontrivial natural frequencies and the corresponding ratios of amplitudes.
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Free Vibrations of Two Degree of Freedom Systems
4.
In an oscillating or vibrating system, define the terms degree of freedom and
node.
A shaft of 700 mm length and uniform circular cross-section connects two
flywheels with moments of inertia of 0.3 kgm2 and 0.5 kgm2. The torsional
stiffness of the shaft is 300 kNm/rad. Determine:
(a) The non-trivial natural frequency of torsional oscillations.
(b) The corresponding ratios of amplitudes.
(c) The location of the node from the flywheel with greater inertia.
5.
A shaft of 50 mm diameter and 600 mm length is attached at one of its ends to a
flywheel weighing 1764 N with a radius of gyration of 250 mm and at its other
end to a generator armature whose mass moment of inertia is half that of the
flywheel. Determine the non-trivial natural frequency of vibration of the system.
The elastic shear modulus of the shaft material is 80 × 109 N/m2.
gravitational acceleration may be assumed to be 9.8 m/s2.
6.
The
In Fig. Q6, the motor is connected to the flywheel by a shaft of 16 mm diameter
and 900 mm length. The torsional stiffness of the shaft is 530 Nm/rad. The mass
moments of inertia of the motor and the flywheel are 2.7 and 6.3 kgm2,
respectively. Assuming shaft inertia to be negligible, determine the non-trivial
natural mode and mode shape. Where along the shaft would you locate a
coupling, if necessary, and why?
J1
K
J2
Linearly Elastic Shaft
Motor Inertia
Flywheel
Fig. Q6
7.
For the system illustrated in Fig. Q7, derive the frequency equation. Given that
m1 = m2 = 227 kg, k1 = 263 kN/m, and k 2 = 131.5 kN/m, solve for the natural
modes.
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Free Vibrations of Two Degree of Freedom Systems
x2
x1
k2
k1
m2
m1
Fig. Q7
8.
In an open V-belt drive, the pitch diameter of the smaller sheave is half that of
the larger sheave. The moment of inertia of each sheave about its own axis of
rotation may be estimated by use of the following equation:
mr 2
J=
2
where m is the mass, and r is the pitch radius of the particular sheave.
Model the drive as a two-rotor, two-spring vibratory system with coplanar rotors
and parallel axes of oscillation. Derive expressions for the non-trivial natural
mode frequency and amplitude ratio.
9.
For the system illustrated in Fig. Q9, derive the equations of motion and express
them in matrix notation. Given that m1 = m2 = 1.75 kg, k1 = k3 = 3500 N/m,
and k 2 = 1750 N/m, solve for the two natural modes.
x2
x1
k2
k1
k3
m2
m1
Fig. Q9
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Free Vibrations of Two Degree of Freedom Systems
10. In Fig. Q10 the pulley rotates about a fixed centroidal axis and may be
considered to be a rigid solid cylinder of homogeneous material. As shown, the
mass of the pulley is equal to that of the suspended mass. There can be no sliding
between the cord connecting the two springs and the pulley. Determine the
normal mode frequencies of the system.
k
θ
r
m
k
y
m
Fig. Q10
11. For the system shown in Fig. Q11, show that the values of the two natural
frequencies of vibration are given by the following equation:
⎡ (k + k ) (k + k ) ⎤ 2 ⎡ k k + k 2 k3 + k3k1 ⎤
4
ωn − ⎢ 1 2 + 2 3 ⎥ ωn + ⎢ 1 2
⎥=0
m
m
m
m
⎣
⎦
⎣
⎦
1
2
1 2
x1
x2
k2
k1
k3
m2
m1
Fig. Q11
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Free Vibrations of Two Degree of Freedom Systems
If m1 = m2 = m and k1 = k2 = k3 = k , obtain expressions for the values of the
two natural frequencies and the corresponding amplitude ratios.
12. For the system illustrated in Fig. Q12 below, derive the differential equations of
motion and express them in matrix notation. Given that J1 = J 2 = 2 kgm 2 ,
K1 = K 2 = 400 Nm/rad and K c = 200 Nm/rad , solve for the two natural
frequencies.
K1
J1
J2
Kc
K2
Fig. Q12
13. For the system illustrated in Fig Q13, overleaf, derive the differential equations
of motion and express them in matrix notation. Solve for the two natural
frequencies and the corresponding modal fractions. Sketch the mode shapes.
x1
x2
k
k
k
k
2m
2m
k
Fig. Q13
14. In Fig. Q14, an automobile of mass 2000 kg pulls a trailer of mass 500 kg on
horizontal level ground. If the stiffness of the hitch is 150 kN/m, model the
system as a two-DOF system and determine, from first principles, the non-trivial
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Free Vibrations of Two Degree of Freedom Systems
natural frequency. Further, in the non-trivial natural mode, show that the
amplitude ratio will be:
X2
m
=− 1
X1
m2
Fig. Q14
Further Reading
1.
DEN HARTOG, J. P. (1985). Mechanical Vibrations. Dover Publications
Incorporated, New York.
2.
MEIROVITCH, LEONARD (1986).
Edition. McGraw-Hill.
3.
RAO, SINGIRESU S. (1986).
Publishing Company.
4.
SRIVIVASAN, P. (1990). Mechanical Vibration Analysis. Tata McGraw-Hill
Publishing Company Limited.
5.
STEIDEL, ROBERT F. (1989). An Introduction to Mechanical Vibrations,
3rd Edition. John Wiley and Sons.
6.
WARBURTON, G. B. (1964). The Dynamical Behaviour of Structures.
Pergamon Press and The MacMillan Company.
7.
THOMSON, W. T. and MARIE DILLON DAHLEH (1998).
Vibration with Applications, 5th Edition. Prentice Hall.
8.
TSE, FRANCIS S., IVAN E. MORSE and ROLLAND T. HINKLE (1983).
Mechanical Vibrations, theory and Applications, First Indian Edition. CBS
Publishers and Distributors, Delhi.
9.
www.mame.mu.oz.au/dynamics/14lec.pdf
Elements of Vibration Analysis, 2nd
Mechanical Vibrations.
Addison Wesley
Theory of
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