Modul V
Systems with One Degree of Freedom—Theory
5-1 INTRODUCTION
The one-degree-of-freedom system is the keystone for more advanced studies in
vibrations. The system is represented by means of a generalized model shown in Fig. 11. The common techniques for the analysis are discussed in this chapter.
Examples of one-degree-of-freedom systems are shown in Fig. 5-1. Though such
systems differ in appearance, they all can be represented by the same generalized
model in Fig. 5-1. The model serves (1) to unify a class of problems commonly
encountered, and (2) to bring into focus the concepts of vibration. The applications to
different types of problem will be discussed in the next chapter.
Four mathematical techniques are examined. These are (1) the energy method,
(2) Newton's law of motion, (3) the frequency response method, and (4) the
superposition theorem. Our emphasis is on concepts rather than on mathematical
manipulations.
Since vibration is an energy exchange phenomenon, the simple energy method
is first presented. In applying Newton's second law, the system is described by a
second-order differential equation of motion. If the excitation is an analytical expression,
the equation can be solved readily by the "classical" method. If the excitation is an
arbitrary function, the motion can be found using the superposition theorem. The
frequency response method assumes that the excitation is sinusoidal and examines the
system behavior over a frequency range of interest.
Note that a system will vibrate in its own way regardless of the method of
analysis. The purpose of different techniques is to find the most convenient method to
characterize the system and to describe its behavior. We treat Newton's second law and
the superposition theorem as solution of a differential equation with time as the
independent variable. The frequency response method assumes that both the excitation
and the system response are sinusoidal and of the same frequency. Hence it is a
frequency domain analysis. Note that time response is intuitive but it is more convenient
to describe a system in the frequency domain.
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v
Fig.5-1
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It should be remarked that there must be a correlation between the time and the
time domain analysis, since the motion of the mass is a time function, such as the
frequency domain analyses, since they are different methods to consider the same
problem. In fact, superposition, which is treated as a time domain technique, is the basis
for t stdy of linear systems. The convolution integral derived from superposition can be
applied in the time or the frequency domain. We are presenting only one aspect of this
very important theorem and shall not discuss methods of correlation. The mathematical
correlation of the time and frequency analyses is not new. Its implementation, however,
was not practical until the advent of computers, instrumentation, and testing techniques
in recent years.
5-2 DEGREES OF FREEDOM
The number of degrees of freedom of a vibratory system is the number of
independent spatial coordinates necessary to define its configuration. A configuration is
defined as the geometric location of all the masses of the system. If the inter-relationship
of the masses is such that only one spatial coordinate is required to define the
configuration, the system is said to possess one degree of freedom.
A rigid body in space requires six coordinates for its complete identification,
namely, three coordinates to define the rectilinear positions and three to define the
angular rotations. Ordinarily, however, the masses in a system are constrained to move
only in a certain manner. Thus, the constraints limit the degrees of freedom to a much
smaller number.
Alternatively, the number of degrees of freedom of a system can be defined as
the number of spatial coordinates required to specify its configuration minus the number
of equations of constraint* We shall illustrate these definitions with a number of
examples.
The one-degree-of-freedom systems shown in Fig. 5-1 are briefly discussed as
follows:
1. The spring-mass system in Fig. 5-l(a) has a mass m suspended from a coil spring with
a spring constant k. If m is constrained to move only in the vertical direction about its
static equilibrium position 0, only one spatial coordinate x(t) is required to define its
configuration. Hence it is said to possess one degree of freedom.
2. The torsional pendulum in Fig. 5-1(b) consists of a heavy disk J and a shaft of
negligible mass with a torsional spring constant kt. If the system is constrained to
oscillate about the longitudinal axis of the shaft, the configuration of the system can be
specified by a single coordinate θ(t}.
3. The mass-spring-cantilever system in Fig. 5-l(c) has one degree of freedom if the
cantilever is of negligible mass and the mass m is constrained to move vertically. By
neglecting the inertial effect of the cantilever and considering only its elasticity, the
cantilever becomes a spring element. Hence a simple spring-mass system is obtained
from the given mass m and an equivalent spring, constructed from the combination of
the spring k and the cantilever.
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4. The mass-pulley-spring system in Fig. 5-l(d) has one degree of freedom if it is
assumed that there is no slippage between the cord and the pulley J and the cord is
inextensible. Although the system possesses two mass elements m and J, the linear
displacement x(t) of m and the angular rotation θ of J are not independent. Thus, either
x(t) or θ(t) can be used to specify the configuration of the system.
5. A simple spring-loaded flyball governor rotating with constant angular velocity fl- is
shown in Fig. 5-l(e). If a disturbance is applied to the governor, its vibratory motion can
be expressed in terms of the angular coordinate θ(t)
.
6. The simple pendulum in Fig. 5-l(f) is constrained to move in the xy plane. Its
configuration can be defined either by the rectangular Cartesian coordinates x(t) and y(t)
or by the angular rotation θ(t). The (x,y) coordinates, however, are not independent.
They are related by the equation of constraint
x2+y2 = L2 ……….(5-1)
where the length L of the pendulum is assumed constant. Thus, if x(t) is chosen
arbitrarily, y(t) is determined from Eq. (5-1).
Several systems with two degrees of freedom are shown in Fig. 5-2.
Fig.5-2
1. The two-spring-two-mass system of Fig. 5-2(a) possesses two degrees of freedom if
the masses are constrained to move only in the vertical direction. The two spatial
coordinates defining the configuration are x1(t) and x2(t).
2. The spring-mass system shown in Fig. 5-2(b) was described previously as a onedegree-of-freedom system. If the mass m is allowed to oscillate along the axis of the
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spring as well as to swing from side to side, the system possesses two degrees of
freedom.
3. The pendulum in space in Fig. 5-2(c) can be described by the θ(t) and φ(t)
coordinates as well as by the x(t), y(t),and z(t) coordinates. The latter are related by the
equation of constraint
x2 + y2 + z2 = L2.
Thus, this pendulum has only two degrees of freedom.
5-3 EQUATION OF MOTION— ENERGY METHOD
5-3.1 Rectilinear System (Gerak Translasi)
The equation of motion of a conservative system can be established from energy
considerations. If a conservative system in Fig. 5-3 is set into motion, its total
mechanical energy is the sum of the kinetic energy and the potential energy. The kinetic
energy T is due to the velocity of the mass, and the potential energy U is due to the
strain energy of the spring by virtue of its deformation. Since the system is conservative,
the total mechanical energy is constant and its time derivative must be zero. This can be
expressed as
T+U= (total mechanical energy) = constant (5-2)
Fig.5-3
d
(T  U )  0 ……(5-3)
dt
To derive the equation of motion for the spring –mass system of Fig.5-3.assume that the
displacement x(t) of the mass m is measured from its static equilibrium position. Let x(t)
be positive in the downward direction. Since the spring element is of negligible mass, the
kinetic energy T of the system is
T = 1/2m x 2 …..(5-4)
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The corresponding potential energy of the entire system is the algebraic sum of (1) the
strain energy of the spring and (2) that due to the change in elevation of the mass. The
net potential energy of the system about the static equilibrium is
….(5-5)
Substituting Eqs. (5-4) and (5-5) into Eq. (5-3) gives
Since the velocity x(t) in the equation cannot be zero for all values of time, clearly
(5.6)& (5-7)
where   k / m . The equation of motion of the system can be expressed as shown in
Eq. (5-6) or (5-7).
2
n
It can be shown that the solution of Eq. (5-7) is of the form
…(5-8)
where A2 and A2 are arbitrary constants to be evaluated by the initial conditions x(0) and
x(0). It is apparent that can in Eq. (5-7) is the circular frequency of the harmonic motion
x(t). Since the components of the solution are harmonic of the same frequency, their
sum is also harmonic and can be written as
…(5-9)
where
angle.
A
A A
2
1
2
2
is the amplitude of the-motion and ψ= tan-1 A1/A2 is the phase
Equation (5-9) indicates that once this system is set into motion it will vibrate with simple
harmonic motion, and the amplitude A of the motion will not diminish with time. The
system oscillates because it possesses two types of energy storage elements, namely,
the mass and the spring. The rate of energy interchange between these elements is the
natural frequency fn of the system.*
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Note that the natural frequency is a property of the system. It is a function of the values
of m and fc and is independent of the amplitude of oscillation or the manner by which the
system is set into motion. Evidently, only the amplitude A and the phase angle i// are
dependent on the initial conditions.
Disini penulis akan memberikan solusi lengkap bagi pers.(5-6)&(5-7) diatas sebagai
berikut:
Substitusikan pers.(5-3) dan(5-4) ke (5-2) menghasilkan
d 1
1
( mx 2  kx 2 )  (mx  kx) x  0
dt 2
2
x  0
maka:
mx  kx  0 …….(5-10)
Karena
Atau
x   n 2 x  0
2
Disini
n  k /
Dengan mengambil
x  e t
Sehingga
m
solusi (5-10) menjadi
2
2
…..(5-11)
n
   0
   j n , jadi
x1  e j nt ,
x 2  e  j nt
Kedua jawaban diatas dapat memenuhi persamaan (5-11) dan bila kedua jawaban tsb
dikalikan oleh A1 dan B1 dan ditambahkan ,maka
x  A1 x1  B1 x2  A1e jnt  B1e  jnt
…..(5-12)
Karena
e  j nt  cos  n t  j sin  n t ,pers.(5-8) menjadi
x  A1 (cos  n t  j sin  n t )  B1 (cos  n t  j sin  n t )
 ( A1  B1 ) cos  n t  j ( A1  B1 ) sin  n t
…….(5-13)
 C cos  n t  D sin  n t
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A
C 2  D2
C
cos  
C
Bila diambil
sin  
 D2
C
2
C 2  D2
D
tan   
C
Maka pers.(5-13) menjadi
x  C 2  D2 .
C
cos  n t  C 2  D 2 .
C 2  D2
 A cos  cos  n t  A sin  sin  n t
D
C 2  D2
sin  n t
 A cos( n t   )
…….(5-14)
Atau dapat saja ,bila diambil
A
C 2  D2
cos  
sin  
D
C 2  D2
C
C 2  D2
C
tan  
D
Maka
x  A(sin  cos  n t  cos  sin  n t )
 A sin(  n t   )
……(5.15)
5.3.2. Rotational System (Gerak Rotasi)
Hal yg sama dpt diturunkan gerak pendulum putar (Gamb.5-2.b),yaitu:
J
d 2
  k …….(5.16)
dt 2
Juga utk simple pendulum Gamb.5-2.f),yaitu:
Example 5-1
Determine the equation of motion of the simple pendulum shown in Fig.5-f).
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Solution;
Assume (1) the size of the bob is small as compared with the length L of the pendulum
and (2) the rod connecting the bob to the hinge point 0 is of negligible mass. The mass
moment of inertia of the bob of mass m about O is
J 0  ( J c. g .  mL2 )  mL2 …(i)
where Jcg is the mass moment of inertia of m about its mass center. If the bob is
sufficiently small in size, then Jcg= mL2.
The angular displacement θ(t) is measured from the static equilibrium position of the
pendulum. The kinetic energy of the system is T = \JflQi -5inL202. The corresponding
potential energy is T= mgL(l— cos ψ), where L(l-cos 8) is the change in elevation of the
pendulum bob. Substituting these energy quantities gives
mL2  mgL sin   0
 
(ii)
g
sin   0
L
The equation of motion of the simple pendulum is as shown above. If it is further
assumed that the amplitude of oscillation is small, then sin θ = θ and Eq. (ii) becomes
g
L
    0
….(iii)
This is of the same form as Eq. (5-7) and the solution follows. The frequency of
oscillation of a simple pendulum is ωn= Vg/L.
Atau lebih lengkapnya dapat penulis uraikan sbb:
Dalam hal ini (1) ukuran bandul cukup kecil dibanding panjang lengan L dan (2) massa
lengan L dpt diabaikan .
Momen inertia dari massa bandul m thd titik O :
Disini J CG :
: moment inertia massa m thd titik beratnya . Bila bandul cukup kecil ,
2
maka JCG<< mL .
Dgn metoda energy,bila (t) diukur dari posisi keseimbangan static ,maka energy kinetik
system tsb
T 
1
1
J o  mL2 2 , dan
2
2
Dan potential energy
U  mg ( L  L cos )  mgL(1  cos )
Karena d(T+U)=0,
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mL2  mgL sin  .  0
( mL2  mgL sin  )  0
atau : mL2  mgL sin   0
Frekkuensi oscillasi pendulum :
n 
mgL

mL2
g
L
Example 5-2
Sebuah silinder dgn massa m dan jari2 R1menggelinding tanpa slip pada permukaan
lengkung ber-jari2 R. Turunkanlah persamaan gerak system tsb dgn energy method.
Jawab:
Kinetic energy yg ada pada silinder adalah karena gerak translasi dan rotasi. Ambil
gerak rotasi pusat massa sbb:
x  xo sin   OO sin 
Kecepatan translasi bagi pusat massa silinder adalah
x  OO cos  x max  OO  ( R  R1 )
Karena saat sudut rotasi =,sudut rotasi silinder =1
Maka net displacement sudut rotasi silinder = 1-,jadi kecepatan sudut rotasi silinder


=1  
Karena silinder menggelinding tanpa sslip
R=R11 atau 1=R/R1
Kecepatan sudut dpt ditulis:
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
1
 R

R
         (  1)
R1
 R1


Total energy kinetic:
T 

1
m R  R1 
2

2

 R
 
1

J  

1
R
 .......( 4)
2
1

 

disini
1
2
mR1 adalah momen inersia silinder thd sumbu longitudinal .Dan energy
2
potential:
U  mgR  R  R1  cos    mgR  mg R  R1  cos  ......(5)
d
T  U   0
Karena....
dt
2
   
m 2  R
2  


 m R  R1    
R1 
 1    mg R  R1  sin   0
2
 R1
 
2

 
1
2 
2 R
mR  R1    mR1 
 1    mg R  R sin    0
2
R

 1
 
J 


2
R
1
2
2
mR  R1    m 1 2 R  R1    mg{R  R1  sin  }}  0
2
R1
3
R  R1 2   mg{R  R1 sin  }  0......  0
2
3
2
 R  R1    mg R  R1  sin   0
2
Karena   sin  ,maka pers.gerak:
3
R  R1 2   mg R  R1   0.......(6)
2


2g
n  

 3R  R1  
1/ 2
......(7)
Soal-soal latihan 5:
3 5-1Sebuah piringan digantungkan pada beam berdiameter 0.50 cm dan panjangnya
2m(lihat Gamb.(i)).Piringan diberikan sudut penyimpangan  dan dilepas dan
membuat 10 osilasi dalam 30.2 sec.
Hitunglah I momen inertia polar piringan.
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(i)
(ii)
Gambar.
5-2.Hitunglah frekuensi natural system pada Gambar (ii) diatas.
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