chapter 9 basic terminology in · pdf filechapter 9 basic terminology in vibrations...
TRANSCRIPT
Chapter 9 Basic Terminology in Vibrations
Oscillating Motions:
The study of vibrations is concerned with the oscillating motion of elastic bodies and the force
associated with them. All bodies possessing mass and elasticity are capable of vibrations. Most
engineering machines and structures experience vibrations to some degree and their design generally
requires consideration of their oscillatory motions. Oscillatory systems can be broadly characterized
as linear or nonlinear.
Linear systems: The principle of superposition holds and mathematical technique available for their
analysis are well developed.
Nonlinear systems: The principle of superposition doesn’t hold and the technique for the analysis of
the nonlinear systems are under development (or less well known) and difficult to apply. All systems
tend to become nonlinear with increasing amplitudes of oscillations.
There are two general classes of vibrations– the free and forced.
Free vibrations: Free vibration takes place when a system oscillates under the action of forces
inherent in the system itself due to initial disturbance, and when the external impressed forces are
absent. The system under free vibration will vibrate at one or more of its natural frequencies, which
are properties of the dynamical system established by its mass and stiffness distribution.
Forced vibrations: The vibration that takes place under the excitation of external forces is called
forced vibration. If excitation is harmonic, the system is forced to vibrate at excitation frequency. If
the frequency of excitation coincide with one of the natural frequencies of the system, a condition of
resonance is encountered and dangerously large oscillations may result, which results in failure of
major structures, i.e., bridges, buildings, or airplane wings etc. Thus calculation of natural
frequencies is of major importance in the study of vibrations. Because of friction & other resistances
vibrating systems are subjected to damping to some degree due to dissipation of energy. If the
damping is small, it has very little effect on natural frequency of the system, and hence the
calculations for natural frequencies are generally made on the basis of no damping. Damping is of
great importance in limiting the amplitude of oscillation at resonance.
Degrees of Freedom (dof): The number of independent co-ordinates required to describe the motion
of a system is termed as degrees of freedom. For example: Particle - 3 dof (positions), rigid body-6
dof (3-positions and 3-orientations) and continuous elastic body - infinite dof (three positions to each
particle of the body). If part of such continuous elastic bodies may be assumed to be rigid (or lumped)
and the system may be considered to be dynamically equivalent to one having finite dof (or lumped
169
mass systems). Large number of vibration problems can be analyzed with sufficient accuracy by
reducing the system to one having a few dof.
Vibration measurement terminology:
Peak value: Indicates the maximum stress that a vibrating part is undergoing. It also places a
limitation on the “rattle space” requirement.
Average value: Indicates a steady or static value (somewhat like the DC level of an electrical current)
and it is defined as
0
lim(1/ ) ( )
T
Tx T x t dt
→∞= ∫ (1)
where x(t) is the displacement, and T is the time span (for example time period).
Example: (i) For a complete cycle of sine wave,
( ) sinx t A tω= : [ ]2
2
00
1 cossin 1.0 1.0 0
2 2 2
A t Ax A tdt
ππ ω
ωπ π ω πω
= = = − = ∫
(ii) For half cycle of the sine wave:
[ ]00
cos1/ sin 1 ( 1) 2 / 0.637
A t Ax A tdt A A
ππ ωπ ω π
π ω πω = = = − − = = ∫ (2)
where A is the amplitude of the displacement.
Mean square value: Square of the displacement generally is associated with the energy of the
vibration for which the mean square value is a measure and is defined as
2 2
0
lim(1/ ) ( )
T
Tx T x t dt
→∞< >= ∫ (3)
Example: For a complete cycle of sine wave ( ) sinx t A tω= , we have
2 2 2 2 2 22
00
(1 cos 2 ) sin 2 sin 2lim lim lim
2 2 2 2 2 2 2 2
TT
T T T
A t A A t A A T Ax dt
T T T
ω ω ωω ω→∞ →∞ →∞
− < >= = − = − =
∫ (4)
170
Root mean square value (rms): This is the square root of the mean square value. For example: for a
complete sine wave
[ ]1/ 2 1/ 22 / 2 0.7007rmsx x A A = = = (5)
Decibel (Db): It is a unit of the relative measurement of the vibration and sound. It is defined in terms
of a power ratio:
10 1 210log ( / )bD p p=
Since power is proportion to square of amplitude of vibrations or voltages, which is easily
measurable, hence
2
10 1 2 10 1 210log ( / ) 20 log ( / )bD A A A A= = (6)
where A is the amplitude. For amplitude gain of 5 the decibel has a gain of
Db = +20 log105 = +14 (7)
In vibrations decibel is used to express relative measured values of displacements, velocities and
accelerations.
Db = 20log10(z/z0) (8)
where z is the quantity under consideration (e.g. displacement, velocity or acceleration), z0 is the
reference value (e.g. for velocity v0 = 10-8
m/sec and acceleration a0 = 9.81 × 10-6 m/sec
2). For
example Db=20 means 10 times the reference value (i.e. Db=20=2010log 10 ), and 40 Db means 100
times the reference value (i.e. Db=40 10log 10=202
10log 10 ).
Octave: The octave is used for the relative measurement of the frequency. If two frequencies have
ratio 2:1, the frequency span is one octave.
Octave = log2 (fmax/fmin) (9)
For example:
max min
2
102
10
Octave
20 10 log (2) 1
log 3 0.477130 10 log (3) 1.585
log 2 0.3010
f f
=
= = =
171
When the upper limit of a frequency range is twice its lower limit, the frequency span is said to be one
octave.
Range Bandwidth Octave
10-20 10 1
20-40 20 1
10-30 20 1.585
Oscillatory motion: It repeat itself regularly for example pendulum of a wall clock. Display
irregularity for example earthquake. Periodic Motion: This motion repeats at equal interval of time T.
Period of Oscillatory : The time taken for one repetition is called period.
Frequency - f = 1/T, It is defined reciprocal of time period. The condition of the periodic motion is
x(t+T) = x(t) (10)
where motion is designated by time function x(t).
Harmonic motion: Simplest form of periodic motion is harmonic motion and it is called simple
harmonic motion (SHM). It can be expressed as
sin 2 /x A t Tπ= (11)
where A is the amplitude of motion, t is the time instant and T is the period of motion. Harmonic
motion is often represented by projection on line of a point that is moving on a circle at constant
speed.
Figure 1: The simple harmonic motion
From Figure 1, we have sinx pr qs A tω= = = , where x is the displacement and ω is the
circular frequency in rad/sec and expressed as 2 / 2T fω π π= = , where T is the period (sec) and f
is the frequency (cycle/sec) of the harmonic motion. The SHM repeats itself in 2π radians.
Displacement can be expressed as
172
tAx ωsin= (12)
So that the velocity and acceleration can be written as
)2/sin(cos πωωωω +== tAtAx� (13)
and
)sin(sin 22 πωωωω +=−= tAtAx�� (14)
Equation (12) to (14) are plotted in Figure 2
Figure 2: Variation of displacement, velocity and acceleration
It should be noted from equations (12-14) that when displacement is a SHM the velocity and
acceleration are also harmonic motion with same frequency of oscillation (i.e. displacement).
However, lead in phases by 900 and 1800 respectively with respect to the displacement as shown in
Figure 2. From equations (12) and (14) we find
xx 2ω−=�� (15)
In harmonic motion acceleration is proportional to the displacement and is directed towards the origin.
The Newton’s second law of motion states that the acceleration is proportional to the force. Hence for
a spring (linear), we write
kxFs = (16)
where FS is the spring force and k is the stiffness of the spring. It executes harmonic motion as force
is proportional to the displacement.
Exponential form: From Euler’s equation, we have
iθ cos sine jθ θ= + (17)
173
Figure 3
A rotating vector as shown in Figure 3 can be expressed as
jθ jωtz Ae Ae= = (18)
where A is the magnitude, θ is the orientation and j = 1− is the imaginary number. Equation (18)
can also be written as
cos j sin jz A t A t x yω ω= + = + with cos and sinx A t y A tω ω= = (19)
where z is the complex sinusoid, x is the real component and y is the imaginary component.
Differentiating equation (20) with respect to time gives
jωtz j Aeω=� and
2 jωt 2z Ae zω ω= − = −�� (21)
From equation (20), we can write
1
2( )x z z= + (22)
where z is the complex conjugate of z. We can also write
Re( ) Re[ ] cosi tx z Ae A tω ω= = = (23)
where Re( )z is the real part of quantity z. The exponential form of the harmonic motion offers
mathematical advantages over the trigonometric form.
174
Undamped Free Vibration: The undamped free vibration executes the simple harmonic motion as
shown in Figure 4.
Figure 4: Simple harmonic motion
Harmonic motions have the following form
x = cos( )nX tω φ−
where X is the amplitude, nω is the circular frequency in rad/sec ( fT ππ 2/2 == ), T is the time
period in sec, f is the frequency in cycle/sec and φ is the phase. X and φ depend upon initial
conditions.
175
Chapter 10 Free Vibration
Objective of the present section will be to write the equation of motion of a system and evaluate its
natural frequency, which is mainly a function of mass, stiffness, and damping of the system from its
general solution. Damping has little influence on the natural frequency and may be neglected in its
calculation. The system can thus be considered as conservative and principle of conservation of
energy offers another approach to the calculation of the natural frequency. The effect of damping is
mainly evident in diminishing of the vibration amplitude at or near the resonance.
Vibration Model:
The basic vibration model consists of a mass, spring (stiffness) and damper a shown in Figure 1.
Figure 1: Spring-damper vibration model.
Figure 2: The static equilibrium position
The inertia force model is
176
iF mx= �� (1)
where m is the mass in kg, x�� is the acceleration in m/sec2 and Ft is the inertia force in N. The linear
stiffness force model is
FS = kx (2)
where k is the stiffness (N/m), x is the displacement and Fs is the spring force. The damping force
model for viscous damping is
xcFd�= (3)
where c is the damping coefficient in N/m/sec, x� is the velocity in m/sec and Fd is the damping force.
Choose of x in positive downward direction as positive. Also velocity, �x , acceleration, ��x , and force,
F, are positive in downward direction. From Newton’s second law, we have
∑= Fxm �� or )( xkmgxm +∆−=��
From Figure 2, we have mgk =∆ , so the above equation becomes
0=+ kxxm �� (4)
The choice of the static equilibrium position as reference for x axis datum has eliminated the force due
to the gravity. Equation (4) can be written as
02 =+ xx nω�� or xx n
2ω−=�� (5)
with 2 /n k mω = (6)
From equation (5), it can be observed that: (i) The motion is harmonic (i.e. the acceleration is
proportional to the displacement) (ii) It is homogeneous, second order, linear differential equation (in
the solution two arbitrary constants should be there) (iii) Function cos ntω and sin ntω both satisfy
equation (3). The general solution can be written as
tBtAx nn ωω cossin += (7)
where A and B are two arbitrary constants, which depend upon initial conditions i.e. x(0) and )0(x� .
Equation (5) can be differentiated to give
tBtAx nnnn ωωωω sincos −=� (8)
On application of boundary conditions in equation (7) and (8), we get
177
x(0) = B and nAx ω=� or (0) /ω= � nA x and )0(xB = (9)
On substituting equation (9) into equation (8), we get
( (0) / )sin (0)cos cos( )n n n nx x t x t X tω ω ω ω φ= + == −� (10)
with, )0(cos xX =φ , nxX ωφ /)0(sin �= ,
where X is the amplitude, nω is the circular frequency and φ is the phase. The sine & cosine
functions repeat after 2π radians (i.e. Frequency × Time period = 2π)
πω 2=∴ Tn (11)
The time period can be written as
kmT /2π= (12)
The natural frequency can be written as
mkTf /)2/1(/1 π== (13)
since we have
∆=⇒=∆ // gmkmgk ∆=∴ /)2/1( gf π (14)
T, f, nω are dependent upon mass & stiffness of the system, which are properties of the system.
Above analysis is valid for all kind of SDOF system including beam or torsional members. For
torsional vibrations the mass may be replaced by the mass moment of inertia and stiffness by stiffness
torsional spring. For stepped shaft an equivalent stiffness can be taken or for distributed mass an
equivalent lumped can be taken.
1 2
1 2
1;
1/ 1/k k k k
k k= = +
+
3
3
3
3
Pl P EIk
EI lδ
δ= ∴ = =
Figure 3
178
Example 1: Obtain equation of motion of a simple pendulum.
Tension in spring is given by
T = mgcosθ (a)
The moment balance about centre of rotation O is given as
0 ( sin )I M mg lθ θ= = −∑�� or 2 sinml mglθ θ= −�� or ( / ) sin 0g lθ θ+ =�� (b)
which is the equation of motion of the pendulum.
Figure 4
Alternatively the equation of motion can be derived by the energy method. The kinetic and potential
energy can be written as the kinetic energy and potential energy of the pendulum are given by
1 2
2T Iθ= � and )cos1( θ−= mglU
From the principle of conservation of energy, we have
( ) 0d
T Udt
+ =
On substituting energy terms, we get
(1 cos ) 0 or [ sin ] 0d
I mgl I mgldt
θθ θ θθ θ θ+ − = + =��� ��� �
For pendulum I = ml2 hence above equation becomes
2[ (sin )] 0ml mglθ θ θ+ =� ��
Since θ� cannot be zero, hence
179
( / ) sin 0g lθ θ+ =��
Above equation is non-linear because of the term sinθ . For small amplitude of oscillation above
equation reduces to (i.e.sinθ θ≈ )
0)/( =+ θθ lg��
from which the natural frequency of the pendulum can be obtained as
Tflg nn /22/2 ππωω ===
lgn /=ω rad/sec or l
gf
π21
= cycle/sec
The time period is given as 1
2 /T l gf
π= = sec
Spring Mass system
The loss of potential energy of m due to position x is cancelled by the work done by the equilibrium
force of the spring, (k∆) in position x = 0.
Figure 5
Energy method:
In a conservative system (i.e. with no damping) the total energy is constant, and differential EOM can
also be established by the principle of conservation of energy. For the free vibration of undamped
system: Energy=(partly kinetic+ partly potential energy). Kinetic energy T is stored in mass by virtue
of its velocity. Potential energy U is stored in the form of strain energy in elastic deformation or work
done in a force field such as gravity, magnetic field etc.
180
T + U = constant (15)
Hence
0)( =+UTdt
d (16)
Example 2: Obtain the equation of motion of a spring mass system by energy method.
Solution: For a spring mass system, we have
2 2 21 1 12 2 2
; ( )T mx U kx mgx k x kx= = + − + ∆ =��
From equation (12)
02221
21 =+ xxkxxm ���� or 0)( =+ xkxxm ��� since 0≠x� hence 0)( =+ kxxm ��
Our interest is to find natural frequency of the system, writing equation (11) for two positions i.e.
T1 + U1 = T2 + U2 = constant (17)
where, 1 & 2 represents two instants of time. Let 1 represents at static equilibrium position and let U1
= 0, choosing the reference point of potential energy and 2 represents corresponds to maximum
displacement of mass and at this position velocity of mass will be zero and hence T2 = 0. Equation
(14) reduces to
T1 + 0 = 0 + U2 (18)
If mass is undergoing harmonic motion then T1 & U2 are maximum values.
Tmax=Umax (19)
For spring mass system, having harmonic motion at the natural frequency, ptax cos= ,
sinn nx tω ω= −� , 2 2cosn n nx a t xω ω ω= − = −�� ,
21max 2
U ka= , 21
max 2( )nT m aω= − ,
0maxattnx aω
== −�
Equation (15) will directly give natural frequency. So,
/n k mω =
The undamped free response can be written as
[cos sin ] [cos sin ] ( )cos ( )sin
cos sin
n ni t i t
n n n n n n
n n
x Ae Be A t i t B t i t A B t i A B t
a t b t
ω ω ω ω ω ω ω ω
ω ω
−= + = + + − = + + −
= +
where a & b are another constants to be determined from initial conditions.
181
Damped System: Vibration system may encounter damping of type (i) internal molecular friction (ii)
sliding friction and (ii) fluid resistance. Generally mathematical model of such damping is quite
complicated and not suitable for vibration analysis. Simplified mathematical model (such as viscous
damping or dash-pot) have been developed which leads to manageable mathematical model. The force
displacement curve with damping will enclose an area, referred as the hysteresis loop, that is
proportional to the energy lost per cycle.A mathematical model of damping in which force is
proportional to displacement i.e., Fd = cx is not possible because with cyclic motion this model will
encounter an area of magnitude equal to zero. So dissipation of energy is not possible with this model.
Viscously damped free vibration:
Figure 6 Damping in proportional to displacement (no energy dissipation)
Viscous damping force is expressed as,
xcFd�= (20)
c is the constant of proportionality. From Newton’s second law, we have
xmF ��=∑
From free body diagram
xmxckxtF ��� =−−)( or )(tFkxxcxm =++ ��� (21)
Solution of equation (21) will be in two parts (i) homogeneous solution and (ii) particular solution.
(i) Homogeneous Solution (Free damped motion)
0=++ kxxcxm ��� (22)
Let us assume a solution of the form
stex = (23)
dF
0 x
182
where s is a constant (can be a complex number). So that stsex =� and
stesx 2=�� , substituting in
equation (22), we get,
2( ) 0stms cs k e+ + =
So the condition that equation (23) is a solution for all values of t, we get, as characteristic equation
02 =++m
ks
m
cs (24)
m
k
m
c
m
cs −
±−=2
2,122
(25)
Hence the general solution is given by the equation
tsts
BeAex s 2+= (26)
Substituting (21) into (22), we get
+= −−−− mtkmcmtkmctmc
BeAeeX/)2/(/)2/()2/(
22
(27)
The term outside the bracket in RHS is an Exponentially decaying function. The term
−2( / 2 ) / can have three cages.c m k m
(i)
2
2
c k
m m
<
: Exponents in above equation are real number and no oscillation possible. It is called
overdamped system.
(ii)
2
2
c k
m m
<
: Exponent in above equation are imaginary number
2
2
k ci
m m
± −
.
We can write
22 2
2 cos sin2 2
k ci t
m m k c k ce t t
m m m m
± − = − ± −
Hence,
2 2
2 ( )cos ( )2 2
ct
m k c k cx e A B t i A B t
m m m m
= + − = − −
Let φcos)( XBAa =+= and φsin)( XBAib =−= , so that we have
183
2
2 cos2
ct
m k cx Xe t
m mφ
= − −
; 1 2 2tan ( / )b a a bφ −= = + (28)
Oscillations are possible (decaying type) and it is called the undamped system.
(iii) Linearity case between oscillatory and non-oscillatory motion.
2
2
c k
m m
=
. Damping
corresponding to this case is called critical damping, cc :
kmmmkmc nc 22/2 === ω (29)
Any damping can be expressed in terms of critical damping by a non-dimensional number ζ called
the damping ratio
ccc /=ζ (30)
EOM can be expressed in terms of ζ and nω as
)(1
tFm
xm
kx
m
cx =++ ���
Noting the equation (29), it can be written as
)(1
2 2 tFm
xxx nn =++ ωζω ��� (31)
This form of equation is useful in identification of natural frequency and lamping of system. Useful in
modal summation of MDOF system. The roots of characteristic equation can be written as
2
1,22 2
c c ks
m m m
= − ± −
( )2 2
n n nζω ζω ω= − ± − 2( 1) nζ ζ ω = − ± −
(32)
with n
nc
m
m
m
c
m
cζω
ωζζ2
2=== .
Depending upon value of damping ratio we can have the following cases (i) 1>ζ , overdamped
condition (ii) 1<ζ , underdamped condition (iii) 1=ζ , critical damping (iv) 0=ζ , undamped
system. Complex plane with ζ along the horizontal axis (of equation 32) and s along the vertical axis
for 0=ζ , is
n
±=ω
2,1, 110 22,1 =+=
n
s
ω.
184
for 10 << ζ , 22,1
1 ζζω
−±−= is
n
where 1s and 2s are complex conjugate points on a circular arc,
( ) 11)(2
222,1 =−+−= ζζω n
s
for 1=ζ , 012,1 ×±−= i
s
nω
and for 1>ζ , 122,1 −±−= ζζ
ω n
s, always real numbers.
In complex plane various points are as follows: A and B, 0=ζ ; undamped system, C and D,
10 << ζ ; underdamped system or between A and E and between B and E; underdamped system, E,
1=ζ , criticadamping and F and G, 1>ζ , overdamped system.
(1) Oscillatory motion: ( 0.1<ζ , underdamped case) General Solution equation (23) becomes:
[ ]tsitsit nnn BeAeexωωζω 22
11 −−−− += (33)
[ ]tBAitBAe nn
tn ωζωζζω 22 1sin)(1cos)( −−+−+= − (34)
( )φωζζω +−= −tXe n
tn 21 (35)
where φsinXC = and φcosXD = and 22 DCX += , )/(tan 1 DC−=φ , C & D and X, φ are
arbitrary constants to be determined from initial conditions, x(0) and )0(x� . From equation (34), we
have
{ } { }
2 2
2 2
2 2 2 2
cos 1 sin 1
( ) cos 1 sin 1
sin 1 1 cos 1 1
n
n
n
t
n n
t
n n n
t
n n n n
x e C t D t
x e C t D t
e C t D t
ζω
ζω
ζω
ζ ω ζ ω
ζω ζ ω ζ ω
ζ ω ζ ω ζ ω ζ ω
−
−
−
= − + −
= − − + −
+ − − − + − −
�
�
On application of initial conditions, we get
Cx =)0(
185
and
nn DCx ωζζω 21sin −+−=� ; 21
)0()0(
ζω
ζω
−
+=
n
n xxD
�
Hence, equation (34), becomes
−+−
−
+= −
txtxx
ex nn
n
ntn ωζωζζω
ζωζω 22
21cos)0(1sin
1
)0()0(� (36)
Equation (36) indicates that the frequency of damped system is equal to,
212
ζωπ
ω −== n
d
dT
(37)
From (36), we have
{ }
2
2
21 1 2
sin 1sin 1
(0) (0) lim (0) 1 (0) (0) lim (0)1 1
n n
nt tn
n n
nn
dt
t dx e x x x e x x x
d
d
ω ω
ζ ζ
ζ ωζ ω ζζω ζω
ω ζ ω ζζ
− −
→ →
− −= + + × = + +
− −
� �
{ } ( )
+
−−
−−−+=
→
−)0(
)2(1)2/1(
)2(1)2/1(1coslim)0()0(
2
2/122
1x
ttxxe
n
nn
n
tn
ζζω
ωζζωζζω
ζ
ω�
{ }[ ])0()0()0( xtxxe n
tn ++= − ζωω� (38)
(2) Non-oscillatory motion (ζ >1 over damped case). Two roots remain real with one increases and
another decreases. The general solution becomes
tt nn
BeAexωζζωζζ
−−−
−+−
+=11
22
(39)
so that
( ) ( ) t
n
t
n
nn
BeAexωζζωζζ
ωζζωζζ
−−−
−+−
−−−+−+−=1
21
222
11�
On application of initial conditions, we have
BAx +=)0(
and
( ) ( ) BAx nn ωζζωζζ 11)0( 22 −−−+−+−=�
or, ( ) ( ) [ ]AxBAx nn −−−−+−+−= )0(11)0( 22 ωζζωζζ�
186
which gives
( )12
)0(1)0(
2
2
−
−+−+=
ζζ
ωζζ xxA
n�
and ( )
12
)0(1)0(
2
2
−
−−−−−=
ζζ
ωζζ xxB
n�
(40)
(3) Critically damped systems: )0.1( =ζ We obtained two roots 1 2 ns s ω= = − . Two terms in
solution combines to give one constant. Hence the general solution will be
tneBtAx
ω−+= )( (41)
so that,
)())((tt
nnn eBeBtAxωωω +−+=�
On application of initial conditions, we get
Ax =)0( and )0()0()0( xxBBAx nn ωω +=⇒+−= ��
The necessary and sufficient conditions for crossing once can be obtained as
[ ]{ } t
nnetxxxxωω −++= )0()0()0( � (42)
0)0()0( <+ xx nω� or )0()0( xx nω−<�
(0) 0x <� is a necessary condition for crossing once but sufficient conditions is given by equation (37).
Logarithmic Decrement:
Rate of decay of free vibration is a measure of damping present in a system. Greater is the decay,
larger will be the damping. Damped (free) vibration, general equation of the response is given as
( ) ( )φωζζω +−= −tXex n
tn 21sin
Defining a term logarithmic decrement which is defined as the natural logarithm of the rato of any
two successive amplitudes.
( )( )( )
++−
+−==
+−
−
φωζ
φωζδ
ζω
ζω
dn
Tt
n
t
TtXe
tXe
x
x
dn
n
1
2)(
2
2
1
1sin
1sinlnln
1
187
Since ( ){ } { }tTt ndn ωζωζ 2
1
2 1sin1sin −=+− ; where dT = damped period, πω 2=ddT where
nωζ21− = damped natural frequency. We have damped period
21/2/2 ζωπωπ −== nddT ,
hence
21
2
ζ
πζδ
−= (43)
and since 11 2 ≅−ζ
πζδ 2≈ (44)
Experimental determination of natural frequency and damping ratio:
Natural frequency: d
dT
πω
2= rad/sec, dT can be obtained from displacement-time free vibration
oscillations.
Damping ratio: 2
1ln2
1
x
x
πζ = , where 1x and 2x are two consecutive amplitudes in free vibration
displacement time curve.
Example 1 : A flywheel weight = 311.36 N which swings as pendulum about a knife-edge. The
period of oscillations T = 1.22 sec. Find mass moment of inertia about center of gravity of the
flywheel.
Solution: The kinetic energy and potential energy are given as
21
2T Iω= and
1
2U mg=
For a compound pendulum the period is given by
2l c
Tg
π+
=
T = Time period = 1.22 sec., l = the distance between centers of rotation and gravity = 1.1524 m, g =
acceleration due to 9.81 m/sec2, c = the distance between the centers of gravity and percussion =
2
24
T gl
π−
( )22
1.22 9.810.1524
4 π×
= −×
= 0.2174 m
We have: IG= m2
GK = m(l.c)311.36
0.1524 0.21749.81
= × × = 10.315 kg m2
188
Q.2. For a SDOF system J is mass- moment of inertia, m is mass suspended by thread. Coordinate
required is the location of mass, x, then flywheel will have rotation 1
parameter
radius
x
rθ = = . Kinetic
energy is due to the mass of the block in m and the polar mass moment of inertia of flywheel J.
2 21 1
2 2T mx Jθ= + �� 2 2
1
1 1( / )
2 2mx J x r= +� � (1)
The potential energy is given as (no gravity effect since static equilibrium position as reference for
axes chosen)
21
2U ky=
1 1
x y
r rθ = = 1
2
ry x
r
=
Hence the potential energy can be written as
2
2
1
1
2
rk x
r
=
2 2 2
2 1
1( / )
2k r r x= (2)
For the conservative system, we have
T + U = Constant ( ) 0d
T Udx
∴ + =
22
1
2
2
22
1
2 )/(2
1)/(
2
1
2
1xrrkxrJxm ++ �� = constant
Differentiating with respect to time and equating to zero i.e.
0)2)(/(2
1)2()/(
2
1)2(
2
1 2
1
2
2
2
1 =++ xxrrkxxrJxxm �������� or 0)]/([)/( 2
1
2
2
2
1 =++ xrrkxrJxm ����
or [ ] 0)]/([)/( 2
1
2
2
2
1 =++ xrrkxrJm ��
or [ ] 0)/(
)/(2
1
2
1
2
2 =+
+ xrJm
rrkx��
For harmonic motion 02 =+ xx nω��
22 2 1
2 2
1 1
( / )
( ) /n
k r r
mr J rω∴ =
+
Jmr
krn +=∴
2
1
2
2ω
Example 3. A rotor has ω = 1200 rpm, unbalance mass = 0.015 kg, radius at which unbalance is
there = 0.05m.
2
0F m rω∴ = = unbalance force amplitude
22 1200
0.015 0.0560
π × = × ×
=11.843 N
189
f = 18 cycle/sec, 2
60
nπω = =125.66 rad/sec, nω = 2 fπ =113.1 rad/sec
125.661.111
113.1n
Rωω
∴ = = = , 10.0=ζ damping ratio
90.1]11.11.02[)11.11(
1
]2[)1(
1
2222220
=××+−
=+−
=RRF
Xk
ζ
2 2 2, = 100 113.1 =1279161 N/mn n
Kk m
mω ω= = ×
0176.01279161
843.1190.190.1 0 =×
=×
=k
FX mm
0
2
1
2
1 4.43)111.1(1
111.110.02tan
1
2tan −=⇒
−××
=
−
= −− φζ
φR
R(lag behind)
01.90 1.90 11.8430.0176mm
1279161
FX
K
× ×= = =
ii) for n=1080 rpm =>ω =113.097 0
2.23XK
F∴ =
R 0.99976n
ωω= =
---------------------------------*********************--------------------------------