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= =

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