Vibrations Notes

1

1.1 INTRODUCTIONVibrations or oscillations constitute one of the most important fields of study in physics as wellas engineering. The characteristic feature of vibration is its periodicity i.e., there is a movementor displacement or a variation in the value of a physical quantity that repeats over and overagain. In mechanical systems, the periodicity refers to displacement and force while in theelectrical systems, it is related to current and voltage. At the microscopic level, atoms andmolecules execute periodic vibration in the solid state. The propagation of light involves thevibrations of electric and magnetic fields while that of sound, the periodic motion of atoms andmolecules. The complex and a variety of phenomena arising out of the interaction betweenmatter and radiation can be understood in the conceptual framework of vibrations.

Vibratory motion of machines and mechanical structures is a source of noise.Consequently vibration analysis and control is an important discipline in engineering. Thedesign of jet engines, rocket engines, wheeled vehicles which are made immune to shock duringmotion needs a thorough understanding of the phenomena of vibrations. Vibration isolatorsand dampers are routinely used in industry to minimise noise pollution. Vibratory motion isalso put to use in many practical devices such as vibratory conveyors, hoppers, sieves,compactors, washing machines, dentists’ drills etc. An understanding of the vibrations of strings,membranes, air columns in pipes is necessary for the design of musical instruments. Thesimulation of vibratory motion of earth’s crust is often carried out to understand the phenomenaof earthquake and the propagation of shock waves. The science of acoustics deals with thegeneration, transmission and reception of energy in the form of vibrational waves in matter. Astudy of the acoustics of building and noise control methods forms an integral part ofarchitectural engineering.

Vibrations are classified into the following types:Free vibrations: These refer to vibrations when the system is allowed to vibrate on its

own. The system is not subjected to a periodic force.Forced vibrations: If a system is subjected to a periodic force, the resulting vibrations

are known as forced vibrations.Undamped vibration: If no energy is lost or dissipated in friction during vibration, the

vibration is termed undamped vibration.Damped vibration: This refers to vibration in which there is loss of energy during

vibration. Consequently a system which is set to vibrate will come to rest after in due course oftime.

2 PHYSICS FOR ENGINEERS

Linear/non-linear vibration: If the restoring force is proportional to displacement orwhen the frictional force is linearly proportional to velocity the vibrations are termed linear,otherwise they are non-linear.

Transient vibrations: When a system is subjected to a periodic force, it takes a finitethough a short time, for the onset of steady state vibrations. The vibrations during this interimperiod is known as transient vibrations. We will begin with the study of Simple HarmonicMotion (SHM), which is the simplest of all periodic motion.

1.2 SIMPLE HARMONIC MOTIONWe define Simple Harmonic Motion (SHM) as a motion in which the acceleration of the body(or force on the body) is directly proportional to its displacement from a fixed point and isalways directed towards the fixed point. SHM possesses the following characteristics.

(i) The motion is periodic.(ii) When displaced from the fixed point or the mean position, a restoring force acts on

the particle tending to bring it to the mean position.(iii) Restoring force on the particle is directly proportional to its displacement. The study

of SHM is of practical interest. A vast variety of deformation of physical systemsinvolving stretching, compression, bending or twisting (or combinations involvingall of these) result in restoring forces proportional to displacement and hence, leadsto SHM.

Now we will consider an example of a SHM. Let aparticle A moves along the circumference of a circle with aconstant speed v(= rω) where r is the radius of the circle and‘ω’ is its angular speed. Let the centre of the circle be O and aperpendicular AP be drawn from the particle on the diameterYY′ of the circle (Fig. 1.1).

Then as the particle moves along the circumference ofthe circle, the point P, the foot of the perpendicular vibratesalong the diameter. Since the motion of A is uniform, themotion of P is periodic. At any instant, the distance OP fromO is called the displacement. If the particle moves from X toA in time t, then

AOX = PAO = ωt = θ

i.e., OP = y = r sin ωt

Velocity = v = dydt

= rω cos ωt (1.1)

= rω(1 – sin2 ωt)1/2 = rω(1 – y2/r2)1/2

Acceleration = d y

dt

2

2 = – rω2 sin ωt

= – ω2y = – ω2 × displacement (1.2)Thus, acceleration is directly proportional to displacement and directed towards a fixed

point. Hence, the above example corresponds to SHM.It is instructive to learn how velocity and acceleration in a SHM vary with time. We

notice that when the displacement is maximum (+ r or – r), the velocity = 0, because now the

Y

A

= t

P

XX O

Y

Fig. 1.1 Simple harmonic motion

VIBRATIONS AND RESONANCE 3

point P must change its direction. But when y is maximum (+ r or – r), the acceleration is alsomaximum (– ω2r or + ω2r respectively) and is directed opposite to the displacement. Wheny = 0, the velocity is maximum (rω or – rω) and the acceleration is zero.

The time period T (the time required to complete one oscillation) is given by the followingrelation.

Angular velocity = Angle described in one revolution

Time taken for one revolution

i.e., ω = 2πT

or T = 2πω

Substituting the value of ω from eqn. (1.2), we have

T = 2π

Acceleration / Displacement

T = 2π

Acceleration per unit displacement

= 2π DisplacementAcceleration

(1.3)

The frequency n is given by 1/T.The idea of phase is very important in SHM. Phase difference between two SHMs

indicates how much the two motions are out of step with each other or by how much angle orhow much time one is ahead of the other. In general, displacement is given by

y = r sin (ωt + φ)Clearly at t = 0, y = r sin φ. ‘φ’ is called the initial phase

(Fig. 1.2.)Now let us calculate the total energy associated with

the particle executing SHM. When a body undergoes SHM,its total energy consists of potential energy and kinetic energy.The velocity and consequently kinetic energy is maximum atthe mean position. Potential energy is zero at mean positionand is maximum at the extreme position.

Let us calculate the potential energy.Potential energy = Force × DistanceThe work done in moving through dy is mω2 ydy∴ P.E. = ∫ mω2 ydy

= m yω2 2

2

= m r tω ω2 2 2

2sin

K.E. = mv mr t2 2 2 2

2 2=

ω ωcos

Fig. 1.2 Simple harmonic motion

Y

A0

P

XX O

YA is the position at t = 00


∴ T.E. = mr2 2

2ω

(cos2 ωt + sin2 ωt)

= mr mv2 2 2

2 2ω

= (1.4)

= 12

× mass × (amplitude)2 × (angular velocity)2

1.3 FREE VIBRATIONSLet us consider a body of mass m executing SHM. The equation of motion is of the form:

md x

dtKx

2

2 = − or d x

dtxn

2

22= − ω where ωn

Km

2 = (1.5)

ωn is the natural angular frequency of the simple harmonic oscillator. K is often called thespring constant. This ensures that the acceleration of the particle is always directed towards afixed point on the line and proportional to the displacement from that point. For a solution ofeqn. (1.5), consider

x = Aest so that dxdt

= sAest and d x

dt

2

2 = s2 Aest (1.6)

From eqns. (1.5) and (1.6)

s2 Aest + ω n2 Aest = 0 ⇒ Aest (s2 + ω n

2 ) = 0 ⇒ s = ± iωn

Thus eqn. (1.5) is satisfied by

x = ei tnω and x = e i tn− ω

A linear combination of these two also satisfies eqn. (1.5)

∴ x = A ei tn1

ω + A e i tn2

− ω (1.7a)

where A1 and A2 are arbitrary constants. This solution may be written as:x = A1(cos ωnt + i sin ωnt) + A2(cos ωnt – i sin ωnt)

= (A1 + A2) cos ωnt + i(A1 – A2) sin ωnt= A cos ωnt + B sin ωnt

= C sin (ωnt + φ) = C sin ωnt cos φ +C cos ωnt sin φ (1.7b)Equating coefficients of sin ωnt and cos ωnt,

A = C sin φ and B = C cos φ

∴ C = A B2 2+ and φ = tan–1 AB

The velocity of the particle at any instant is given by

v = dxdt

= Cωn cos (ωnt + φ) (1.8)

The values of C and φ depend upon the initial conditions.

Let x = x0 and dxdt

= v0 at t = 0


Substituting in eqn. (1.7b) and its derivative, we get

x0 = C sin φ ⇒ sin φ = xC0 (1.8a)

v0 = dxdt t

= 0

= Cωn cos φ ⇒ cos φ = v

C n

0

ω (1.8b)

Squaring and adding eqns. (1.8a) and (1.8b)

C xv

n

= +

02 0

2

2

1/2

ω (1.9a)

Dividing eqn. (1.8a) by (1.8b)

φ = tan–1 xv

n0

0

ω(1.9b)

Clearly C is the amplitude. The value of x repeats when t changes by 2π/ωn because

x = C sin ωπ

ωφn

nt +

+

2 = C sin (ωnt + 2π + φ) = C sin (ωnt + φ)

∴ periodic time = 2πωn

and frequency = ω

πn

2

The plot of displacement and velocity with time is shown in Fig. 1.3.

2 / nx

x0

( / ) n

Time

(a)

v

v0

Time

2 / n

(b)

Fig. 1.3 (a) Displacement and (b) velocity in undamped vibration


It is possible to arrive at eqn. (1.5) by considering the principle of conservation of energy.In free vibrations without damping, the total energy is conserved, i.e., T + U = constant, whereT and U are kinetic and potential energy respectively. They are given by :

T = 12

2

mdxdt and U =

12

Kx2 (1.10)

∴ T + U = constant or ddt

(T + U) = 0 (1.11)

From eqns. (1.10) and (1.11)

ddt

mdxdt

Kx12

12

22

+

= 0 ⇒

12

m · 2 · dxdt

d x

dt

2

212

+ · 2 · Kx dxdt

= 0

⇒dxdt

md xdt

Kx2

2 +

= 0, m

d x

dt

2

2 + Kx = 0 ⇒d x

dt

2

2 + ωn2 x = 0

The above equation is identical with eqn. (1.5).

1.4 DAMPED VIBRATIONS1.4.1 IntroductionIn many practical systems, the vibrational energy is gradually converted to heat or sound. Dueto the reduction in energy, the response, such as the displacement of the system graduallydecreases. The mechanism by which the vibrational energy is gradually converted into heat orsound is known as damping. Although the energy loss due to damping may be small,consideration of damping becomes important for an accurate prediction of the vibrationalresponse of the system. Damping is modelled as one or more of the following types:

Viscous DampingWhen a mechanical system vibrates in a fluid such as air, gas, water and oil, the resistance

offered by the fluid to the moving body causes energy to be dissipated. The amount of dissipatedenergy depends on many factors such as the size and shape of the vibrating body, the viscosityof the fluid, the frequency of vibration and the velocity of the vibrating body. In viscous damping,the damping force is proportional to the velocity of the vibrating body. Typical examples ofviscous damping include

(a) fluid films between sliding surfaces(b) fluid flow around a piston in a cylinder(c) fluid flow through an orifice and(d) the fluid flow around a journal in a bearing.

Coulomb or Dry Friction DampingHere the damping force is constant in magnitude but opposite in direction to the motion

of the vibrating body. It is caused by friction between rubbing surfaces that are either dry orhave insufficient lubrication.

Material or Solid or Hysteretic DampingEnergy is absorbed or dissipated when a material is deformed under periodic stress.

The effect is due to friction between the internal atomic planes, which slip or slide as the


deformation takes place. When a body having material damping is subjected to a periodicstress, a periodic strain results which has a phase lag. The stress (σ)-strain (ε) curve shows ahysteresis loop as shown in Fig. 1.4b. The area of the loop denotes energy loss per unit volumeof the body per cycle due to damping. This phenomenon is very much similar to ferromagnetichysteresis where phase lag is observed between the magnetic induction (B) and the appliedfield (H). In a ferromagnetic material, the application of magnetic field leads to domain growthand rotation. This involves some sort of friction between the adjacent domains.

Bs

Br

HcH 0

(a) (b)

B

0

Fig. 1.4 Hysteretic damping (a) ferromagnetic hysteresis (b) elastic hysteresis

1.4.2 Free Vibration with Viscous DampingHere the viscous force is proportional to velocity and acts opposite to the direction of velocity.Generally a damped oscillator is represented by a spring and a dashpot. (Fig. 1.5). The eqn. ofmotion of a body subject to viscous drag is of the form

md x

dt

2

2 = − cdxdt

– Kx ⇒d x

dt

2

2 + cm

dxdt

+ Km

x = 0 (1.12)

Kx cx.

m

+x

O

+x

m

Kc

System(a)

Free body diagram(b)

Fig. 1.5 One dimensional damped oscillator (a) system and (b) free body diagram


(c/m) denotes the frictional force per unit mass per unit velocity. For convenience, denote

Km

= ωn2 and

cm

= 2ξωn (1.13)

ξ is a measure of damping. It is related to factors like quality factor (Q), logarithmic decrement(δ), relaxation time, loss coefficient and specific damping capacity, which are used to characterisea damped oscillator.

From eqns. (1.12) and (1.13) we have

d x

dt

2

2 + 2ξωn dxdt

+ ωn2 x = 0 (1.14)

Let x = Aest be the solution of eqn. (1.14)

Thendxdt

= Asest and d x

dt

2

2 = As2est (1.15)

From eqns. (1.14) and (1.15)

Aest (s2 + 2ξωns + ω n2 ) = 0 ⇒ s2 + 2ξωns + ω n

2 = 0

This is a quadratic equation in s. The solutions are given by

s = ωn − ± −

ξ ξ2 1

The general solution to eqn. (1.12) is

x = A e A en nt t1

12

12 1( ) ( )− + − − − −+ξ ξ ω ξ ξ ω (1.16)Critical Damping Constant:

The critical damping constant (cc) is defined as the value of the damping constant forwhich radical is eqn. (1.16) is zero. i.e.,

ξ = 1 ⇒ cc = 2mωn = 2 mK [from eqn. (1.13)] (1.17)

From eqns. (1.13) and (1.17)

ξ = =ccc

Damping constantCritical damping constant

⇒ c = ξcc = 2ξmωn = 2ξ mK (1.18)

The solution given by eqn. (1.16) can be analysed under the following three cases:Case (i) Underdamped system

ξ < 1 or c < cc or c < 2mωn or c < 2 mK

For this condition (ξ2 – 1) is negative and the two roots can be expressed as

s1 = ( )− + −ξ ξ ωi n1 2 ; s2 = ( )− − −ξ ξ ωi n1 2

Hence the solution eqn. (1.16) takes the form

x = A e i tn1

1 2( )− + −ξ ξ ω + A e i tn2

1 2( )− − −ξ ξ ω = e A e A en n nt i t i t− − − −+

ξω ξ ω ξ ω1

12

12 2

(1.19)

Following the procedure in simplifying as in the case of eqn. (1.7)

x(t) = Ce nt−ξω sin 1 2− +

ξ ω φnt = Ce ntξω sin (ωdt + φ) (1.20)

where ωd = ωn 1 2− ξ (1.21)


C and φ can be evaluated by knowing the initial conditions and are given by eqn. (1.8).On comparing eqn. (1.20) with eqn. (1.7b) we find that eqn. (1.20) corresponds to a dampedharmonic motion of angular frequency ωd given by eqn. (1.21). The amplitude decreasesexponentially with time given by:

Ce nt−ξω

A sketch of displacement with time for various ξ is given in Fig. 1.6. Clearly ωd < ωn. Thevariation of ωd with ξ is shown in Fig. 1.7.

x = 0.01

= 0.05x

x

x

= 0.1

= 0.2

t

t

t

t

Fig. 1.6 Time variation of displacement as afunction of damping ratio ξ < 1 (underdamping)

Case (ii) Critically Damped System

ξ = 1 or c = cc = 2mωn = 2 mK

For this case the eqn. (1.16) takes the form

x e A Ant= +−ξω ( )1 2

Clearly the displacement decreases exponentially with time (Fig. 1.8). This solutiondoes not provide any more information. On substituting this solution in eqn. (1.14) we get backthe condition that ξ = 1. Hence let us assume that the expression within the radical sign is notequal to zero but a small quantity h. Later we will analyse the solution in the limit h → 0.

i.e., assume i hξ2 1− =

d

n

1

0 1

d n= 1 – 2

Fig. 1.7 Variation of (ωd/ωn) with ξ

Fig. 1.8 Time variation of displacement with(a) ξ = 1 (critical damping) and (b) ξ > 1(slightly greater than critical damping)

x(t)

Criticallydamped( = 1)

Slightlyoverdamped( > 1)

Time


Substituting the above expression in eqn. (1.19), we get

∴ x = A e h tn1

( )− +ξ ω + A e h tn2

( )− −ξ ω = e A e A en n nt h t h t− −+ξω ω ω1 2

≈ e nt−ξω [A1(1 + hωnt + ···) + A2(1 – hωnt + ···)]

≈ e nt−ξω [(A1 + A2) + hωnt(A1 – A2)] = e nt−ξω [P + Qt] (1.22)Where P = A1 + A2 and Q = hωn (A1 – A2)It is clear from eqn. (1.22) that as t increases, the factor P + Qt increases but the factor

e nt−ξω decreases. Thus the displacement x first increases due to the factor (P + Qt) and thenexponentially decreases to zero as it increases. The displacement is not periodic (Fig. 1.8).

Case (iii) Overdamped System

ξ > 1 or c > cc or c > 2mωn > 2 mK

For this condition (ξ2 – 1) is positive and the two roots are:

s1 = ( )− + −ξ ξ ω2 1 n < 0

s2 = ( )− − −ξ ξ ω2 1 n < 0

They are real. Hence x decays exponentially withtime. Note that the exponent factor is greater inmagnitude than in case (ii). Here the mass movesto equilibrium more slowly. Further, there is nooscillation in this case as well (Fig. 1.9). In criticaldamping the mass returns to rest in shortestpossible time without overshooting. This propertyof critical damping is used in many practicalapplications. For example, large guns havedashpots with critical damping value so that theycan return to their original position after recoilin the minimum time without vibrating. If thedamping provided were more than critical value,some delay would occur before the next firing.

1.4.3 Energy of a Weakly Damped Oscillator

The amplitude of the oscillator is given by eqn. (1.20) to be Ce nt−ξω

From eqn. (1.4) the energy of oscillator

E = 12

× mass × (amplitude)2 × (angular frequency)2

= 12

mω n2 C2e nt−2ξω = E0e nt−2ξω (1.23)

Thus the energy decays with time as shown in Fig. (1.10).

x(t)

Time

Highlyoverdamped

( >> 1)

Fig. 1.9 Time variation of displacement withfor ξ >> 1 (overdamping)


x

t

(a)

E0

E(t)

(b)

E = E e0–2n

t

Fig. 1.10 Time variation of (a) amplitude and (b) energy of an underdamped oscillator

1.4.4 Logarithmic Decrement, Relaxation Time, Specific Damping Capacity, LossCoefficient and Q-Factor

Several parameters such as logarithmic decrement, relaxation time, specific damping capacity,loss coefficient and quality factor are used to describe the characteristics of damped oscillator.

Logarithmic DecrementThe logarithmic decrement represents the rate at which the amplitude of a free damped

vibration decreases. It is defined as the natural logarithm of the ratio of any two successiveamplitudes. Let x1 and x2 be the two consecutive amplitudes. Let t1 and t2 denote the timecorresponding to two consecutive amplitudes (Fig. 1.11).

x x1

dt

x2

t1 t2

Fig. 1.11 Logarithmic decrement

xx

1

2 =

Ce t

Ce t

n

n

td

td

−

−

+

+

ξω

ξω

ω φ

ω φ

1

2

1

2

sin

sinwhere ωd is given by eqn. (1.21) (1.24)

But t2 = t1 + τd where τd = 2πω d

(1.25a)

∴ sin (ωdt2 + φ) = sin ω πω

φdd

t12

+

+

= sin (ωdt1 + 2π + φ) = sin (ωdt1 + φ) (1.25b)


From eqns. (1.24) and (1.25)

xx

e n d1

2= ξω τ

Taking natural logarithms and making use of eqn. (1.21)

∴ lnxx

1

2

= δ = ξωnτd = ξωn

2πω d

=

2

1 2

πξ

ξ− ≈ 2πξ for ξ << 1 (1.26)

Figure 1.12 shows the plot of logarithmic decrement as a function of ξ.

0.2 0.4 0.6 0.8 1.00

2

4

6

8

10

12

= 2

= 2

1 –

2

= ccc

= Damping factor

Loga

rithm

icde

crem

ent

Fig. 1.12 Logarithmic decrement as a function of ξ

In general if x1 and xm + 1 denote the amplitudes corresponding to t1 and tm + 1 = t1+ mτd

where m is an integer, then

xxm

1

1+ =

xx

1

2 ×

xx

2

3 ×

xx

3

4 × ··· ×

xx

m

m + 1 = ( )e n d mξω τ = em n dξω τ = emδ

Taking natural logarithms

lnx

xm

1

1+

= mδ ⇒ δ =

1 1

1mx

xmln

+

(1.27)

To determine the number of cycles m elapsed for a 50% reduction in amplitude, we havefrom eqns. (1.26) and (1.27)

δ ≈ 2πξ = 1m

ln 2 = 0 693.

m⇒ mξ =

0 6932.

π = 0.110

This is the eqn. of a rectangular hyperbola and is shown in Fig. 1.13.


0.05 0.10 0.15 0.200

1

2

3

4

5

6

Num

ber

ofcy

cles

for

50%

redu

ctio

nin

ampl

itude

ccc

= = Damping factor

Fig. 1.13 Number of cycles for 50% reduction in amplitude

Relaxation Time

The amplitude is given by x(t) = Ce nt−ξω

Relaxation time is defined as the time taken by the amplitude to decrease (1/e)th of itsoriginal value

i.e., xx o( )( )τ =

1e

= Ce

C

n−ξω τ = e n−ξω τ ⇒ τ =

1ξωn

(1.28)

From eqns. (1.28) and (1.26)

τ ⇒1

ξω n =

1

2δπ

ω n

= 2 1πω δn

=

Tδ

(1.29)

Relaxation time is inversely proportional to logarithmic decrement.Natural time period = Relaxation time × Logarithmic decrement

Specific Damping Capacity, Loss Coefficient and Q-factorThe rate of decay of energy with time can be obtained by differentiating eqn. (1.23)

i.e.,dEdt

= – E0 2ξωne nt−2ξω = – 2ξωnE

∴ loss of energy in one cycle = dEdt

× time period

= ∆E = – 2ξωnE × 2πω d

(1.30)

From eqn. (1.21) and (1.30)

∆EE

= specific damping capacity = 4πξω

ωn

d =

4

1 2

πξ

ξ− ≈ 4πξ


Quantity ∆E/E is called the specific damping capacity and is useful in comparing thedamping capacity of materials. Another coefficient known as the loss coefficient is also usedfor comparing the damping capacity.

Loss coefficient = Energy dissipated per radian

Total energy =

∆E

E2π

= 1

2π∆EE

= 2ξ ≈

δπ

(1.31)

Q-factor = Q = 2π Energy stored

Energy dissipated per cycle = 2π

EE∆

= 2π ×

14πξ

= 1

2ξ = πδ

(1.32)

Note that Q-factor is the reciprocal of the loss coefficient.

1.5 STEADY STATE RESPONSE OF AN OSCILLATOR UNDER THE AC-TION OF A PERIODIC FORCE

1.5.1 IntroductionIn many situations the response of an oscillator to a periodic force has to be analysed both inthe fields of physics and engineering. An important idea associated with the response of anoscillator to the periodic force is resonance i.e., when the natural frequency of the oscillator isequal to the frequency of the periodic force. The differential equation which describes thisphenomenon is used in the analysis of mechanical vibrations as well as a.c. circuits. The samemathematical formalism can be used to understand the response of electrons and ions to anelectromagnetic field. In geophysics it is used to understand the phenomenon of tides whichare the result of moon’s periodic motion about the earth. Resonance is also encountered inmany phenomena such as nuclear magnetic resonance, Mossbauer effect, infrared absorptiondielectric dispersion and microwave absorption.

1.5.2 Undamped OscillatorWe consider an undamped system subjected to a periodic force of frequency ω i.e., F(t) = F0 sin ωt. The equation of motion is given by

md x

dt

2

2 = F0 sin ωt – Kx ⇒ md x

dt

2

2 + Kx = F0 sin ωt (1.33)

Since the exciting force F(t) is periodic, x(t) is also periodic and has the same frequencyi.e., x(t) = A sin ωt (1.34a)

∴dxdt

= Aω cos ωt and d x

dt

2

2 = – Aω2 sin ωt (1.34b)

From eqns. (1.34a), (1.34b) and (1.33)– mAω2 sin ωt + KA sin ωt = F0 sin ωt

⇒ A sin ωt[K – mω2] = F0 sin ωt

⇒ A = F

K mF K

n

st

n

02

02 2

1 1( )

/−

=

−

=

−

ω ω

ω

δ

ωω

( ω n2 = K/m) (1.35a)


δst has the units of displacement and is given by (F0/K). Itdenotes the deflection of mass under force F0 and issometimes referred to as static deflection because F0 isconstant. The quantity (A/δst) represents the ratio of thedynamic to static amplitude and is called magnificationfactor or amplitude ratio. It is given by eqn. (1.35b). Itsvariation with the frequency of the driving force is shown inFig. 1.14

A

stδ =

1

12

−

ω

ωn

(1.35b)

It is instructive to consider the variation of the amplitude ratio for the following twocases:

Case (i) When ω < ωn , the amplitude ratio ( A/δst ) is positive and the displacement is inphase with the force.

Case (ii) When ω > ωn, the amplitude ratio (A/δst) is negative and the solution is expressedas

x = – A sin ωt and A = δ

ωω

st

n

−

2

1

In this case there is a phase difference of π between the displacement and the force.Case (iii) When ω = ωn, the amplitude ratio (A/δst) becomes infinite and the phenomenon

is known as resonance. But in practical situations, amplitude ratio does not become infinitesince damping is invariably present.

1.5.3 Damped OscillatorThe response of a damped oscillator is of practical interest since most oscillatory motions inreal life are damped. Some examples of damping are described below:

Resistive damping: In simple LCR circuit, the damping effect is produced by resistance.Energy is dissipated due to Joule heating and is referred to as Resistive damping.

Electromagnetic Damping: A galvanometer consists of a current carrying coil mountedon an axis in a magnetic field. The radial field produced by a properly shaped permanentmagnet results in a deflection which is proportional to the current. The steady current in thecoil gives rise to a torque which is proportional to the current. The coil comes to rest in aposition when this electromagnetic torque is balanced by the elastic torque due to the stiffnessof the suspension. A suspended coil is subjected to various mechanical damping processes.The viscosity of the atmosphere will produce a damping torque which is proportional to theangular velocity. There is also an electromagnetic source of damping. When the coil rotatestowards its new equilibrium position, the magnetic field will induce a voltage proportional tothe instantaneous angular speed. According to Lenz’s law the induced voltage will reduce thecurrent flowing in the coil by an amount proportional to the angular velocity and inverselyproportional to the resistance of the circuit. Thus the electromagnetic torque will also be reduced.

A/st

3

2

1

0

–1

–2

–3

1 2 3 4

r = / n

Fig. 1.14 Magnification factor foran undamped oscillator


Collision damping: Whenever an electron in a metal or in atmosphere is subjected to anelectromagnetic field, the oscillatory motion of the electron is damped by collision with otherelectrons. This is often referred to as collision damping.

Radiation damping: An electron subjected to an oscillatory motion experiences a periodicacceleration. An electric charge in acceleration emits electromagnetic radiation and thus theelectron loses energy. This is known as radiation damping.

We consider a damped oscillator subject to a periodic force of frequency ω.i.e.,F(t) = F0 sin ωt. The equation of motion is given by

md x

dt

2

2 = F0 sin ωt – cdxdt

– Kx ⇒ md x

dt

2

2 + cdxdt

+ Kx = F0 sin ωt (1.36)

Under steady state conditions x(t) is also expected to be periodic. i.e.,x(t) = A sin (ωt – φ) (1.37a)

so thatdxdt

= Aω cos (ωt – φ) andd x

dt

2

2 = – Aω2 sin (ωt – φ) (1.37b)

From eqns. (1.36), (1.37a) and (1.37b)m[– Aω2 sin (ωt – φ)] + c[Aω cos (ωt – φ)] + KA sin (ωt – φ)

= F0 sin ωt = F0 sin [(ωt – φ) + φ]⇒ A [(K – mω2)] sin (ωt – φ) + cω cos (ωt – φ)]

= F0 [sin (ωt – φ) cos φ + cos (ωt – φ) sin φ] (1.38)Equating coefficients of sin (ωt – φ) and cos (ωt – φ) in eqn. (1.38)

Acω = F0 sin φ (1.39a) A(K – mω2) = F0 cos φ (1.39b)

Squaring and adding eqns. (1.39a) and (1.39b)

F02 = A2 [(K – mω2)2 + c2 ω2] ⇒ A =

F

K m c0

2 2 2 2 1/2[( ) ]− +ω ω

= F K

mK

cK

0

2 2 2 2

2

1/2

1

/

−

+

ω ω(1.40a)

Dividing eqn. (1.39a) by eqn. (1.39b)

tan φ = c

K mω

ω( )− 2 (1.40b)

The A and φ for various values of Q is shown in Fig. 1.15.It is convenient to recast eqn. (1.40a) in a simplified form by making the following

substitutions:

ωn = Km

; c = 2mξωn ; δst = FK

0 ; r = ω

ω n(1.41a)


0.4 0.8 1.2 1.6 2.0 ( / ) n

Q=1

Q = 3

Q = 10

Q = 30

0

2

4

6

8

10

12

14

16

18

20

22

A

(a)

0.4 0.8 1.2 1.6 2.0 ( / ) n

0

20°

40°

60°

80°

100°

120°

140°

160°

180°

Q = 30

Q = 10

Q = 3

Q = 1

Q =

(b)

0 0

Fig. 1.15 Frequency dependence of amplitude and phase for various values of Q

A

r rstδ ξ=

− +

1

1 22 2 2( ) ( ) (1.41b)

φ = tan–1 2

1 2

rrξ

−

= tan–1 2

1ξ

rr−

(1.41c)

Equation (1.41) gives the amplitude and phase of the forced vibration. Depending on r,the following cases are possible.

Case (i) r << 1 i.e., when the driving frequency is very much less than the naturalfrequency.

In this case (A/δst) ~ 1, since terms r2, r4 can be neglected. Note that r can lie onlybetween 0 and 1. The amplitude of vibration is independent of frequency of external force. Itdepends only on the magnitude of the applied force F0.

Further φ ~ tan– 1(0) ~ 0. i.e., the displacement and the force are in phase.Case (ii) r = 1, when the frequency of the applied force is equal to the natural frequency of

the oscillator.In this case A = δst /2ξ = Qδst. i.e., the amplitude is proportional to the applied force and

the quality factor.Further φ = tan– 1(∞) = π/2. i.e., the amplitude and the applied force are out of phase by

90°.Case (iii) r >> 1, when the frequency of the applied force is very much greater than the

natural frequency.

Assuming that only r4 is the dominant term, A ~ δst/r2 ~ δst ωn

2 / ω2. i.e., the amplitudedecreases inversely as the square of the frequency of the applied force. It is also proportional tothe magnitude of the force.

Further φ = tan–1 (– 0) = – π. i.e., the displacement and force are out of phase by 180°.


1.5.4 Amplitude ResonanceEquation (1.41a) clearly indicates that amplitude is a function of frequency. It is interesting tofind for what r, A will be maximum. Clearly A is maximum when the denominator is least i.e.,when

ddr

(1 – r2)2 + (2rξ)2 = 0

⇒ 2(1 – r2)(– 2r) + 8rξ2 = 0 ⇒ 4r [2ξ2 – (1 – r2)] = 0 or 1 – r2 = 2ξ2 or r = 1 2 2− ξ

i.e., amplitude is maximum for angular frequency

ω = ωn 1 2 2− ξ (1.42a)

which is neither the natural frequency ωn nor the frequency ωd the damped oscillator.

1.5.5 Velocity ResonanceThe solution to eqn. (1.36) is given by

x = A sin (ωt – φ) where A and φ are given by eqn. (1.41)

v = velocity = dxdt

= Aω cos (ωt – φ) ⇒ vmax = Aω

i.e., vmax = Aω = δ ω

ξst

r r

.

( )1 42 2 2 2− + =

δ ω

ξst n r

r r

. .

( )1 42 2 2 2− + =

δ ω

ξ

st n

rr

14

22−

+

(1.42b)

The above expression is greatest when r = 1 i.e., ω = ωn and its value is δ ω

ξst n

2

Note that the amplitude resonance and velocity resonance occur at frequencies differentfrom each other. For ξ << 1 (~ 0.05), both the resonances are at ω = ωn.

1.5.6 Power Absorbed by a Driven OscillatorWhenever an oscillator is driven by an external force, energy is absorbed by the oscillator. Theenergy absorbed by the oscillator is equal to the energy dissipated due to damping. The rate ofenergy absorption or power absorbed is a function of driving frequency. It is maximum atresonance i.e., when the frequency of the periodic force is equal to that of the natural frequencyof the oscillator. The power absorbed by the oscillator is given by:

Power = Viscous force × VelocityFrom eqn. (1.36)

Viscous force = cdxdt

∴ power = cdxdt

2

From eqn. (1.37b)

x(t) = A sin ωt anddxdt

= Aω cos ωt assuming φ = 0 at t = 0

Power (∆P) absorbed by the oscillator in one cycle is given by:

∆P = c A t dt cA tdt[ cos ] cos//

ω ω ω ωπ ωπ ω 2 2 2 2

0

2

0

2=


= +

= × =cA

t dt dtcA

c A2 2

0

2

0

2 2 22

22

12ω ω ω πω

π ωπ ωπ ω

cos//

From eqn. (1.37b)Power = No. of cycles/sec × Average power per cycle

= ωπ2

× πcωA2 = 12

cA2ω2 (1.43)

The energy absorbed by the oscillator is equal to the energy dissipated due to damping.From eqns. (1.43) and (1.41a)

P cAc

r rst= =

− +

12 2 1 2

2 22 2

2 2 2ωδ ω

ξ( ) ( )

=− +

=−

+

c r

r r

c

rr

st n st n2 1 2 21

14

2 22

2 2 22 2

22

δ ωξ

δ ωξ

( ) ( ) (1.44)

A plot of P for various Q-factors is shown in Fig. 1.16.

1.0 2.0 ( / ) n

0

2

4

6

8

10

12

Q = 30

Q = 10

Q = 3

Q = 1

Pow

er(A

bsor

bed)

Fig. 1.16 Frequency dependence of mean power absorbed by an oscillator for various values of Q

Clearly this is maximum at r = 1 or ω = ωn. The frequency of an undamped oscillator(ωd) = ωn (1 – ξ2)1/2. Hence the power resonance occurs at a frequency different from the ωd .

Pmax = c

st n21

42 2

2δ ωξ

(1.45)


From eqns. (1.44) and (1.45)

PP

rr

=−

+

max 4

14

2

22

ξ

ξ (1.46)

1.5.7 Resonance, Quality Factor and BandwidthLet us find the value of r for which

PP

= max

2

⇒ 12

4

14

2

22

=−

+

ξ

ξr

r

⇒1 2

rr−

= 4ξ2 or

1r

r−

= ± 2ξ 1 – r2 = ± 2rξ or r2 ± 2rξ – 1 = 0

Since r > 0

r1 = ωω

ξ ξ124 1 2

2n=

+ −and r

n2

224 1 2

2= =

+ +ωω

ξ ξ

⇒ 2(r2 – r1) = 4ξ or (ω2 – ω1) = (∆ω) = 2ξωn ⇒ω

ξn

∆ω= 1

2 = Q (1.47)

The frequencies ω1 and ω2 corresponding to r1 and r2 are known as half power points(Fig. 1.17). ω2 ~ ω1 = ∆ω is known as the bandwidth. It is possible to arrive at an expression forthe bandwidth by considering the expression for the amplitude. For small values of damping(ξ < 0.05), from eqn. (1.41a), ωd ≈ ωn.

AQ

stn

δ ξω ω

≈ ≈

=

12

Pmax

P2max

1 n 2

Fig. 1.17 Bandwidth and half power points


A/st

12

Q =

Q2

Bandwidth

r1 r21.0 / n

Half power points

Fig. 1.18 Bandwidth and half power points

Thus Q is also equal to the amplitude ratio at resonance (Fig. 1.18). The points r1 and r2

where the amplification factor falls to A/ 2 as called half-power points because the powerabsorbed by the damped oscillator responding harmonically at a given frequency, is proportionalto the square of amplitude. The difference between the frequencies associated with the halfpower points r1 and r2 is called the bandwidth. To find the values of r1 and r2 , from eqn.(1.41a), we have

Q

r r21

1 2

12 22 2 2

=− +

=( ) ( )ξ ξ

⇒ ( ) ( )1 2 82 2 2 2− + =r rξ ξ

⇒ r r4 2 2 24 2 1 8 0+ − + − =( ) ( )ξ ξ

This is a quadratic in r2

⇒ r12 = 1 – 2ξ2 – 2ξ(1 + ξ2)1/2 and r2

2 = 1 – 2ξ2 + 2ξ(1 + ξ2)1/2

where ω = ω1 at r1 and ω = ω2 at r2

Making the approximation

(1 + ξ2)1/2 ≈ 112

2+

ξ , we get

rn

12 1

22 21 2 2 1

12

1 2=

= − − +

≈ −ωω

ξ ξ ξ ξ( )

rn

22 2

22 21 2 2 1

12

1 2=

= − + +

≈ +ωω

ξ ξ ξ ξ( )

ω22 – ω1

2 = (ω2 + ω1)(ω2 – ω1) = (r22 – r1

2)ωn2 ≈ 4ξωn

2

But ω1 + ω2 = 2ωn

∴ ∆ω = ω2 – ω1 ≈ 2ξωn

∴ Q = 1

2 2 1ξω ω

ω ω≈ ≈

−n n

∆ω(1.48)


Sharpness of ResonanceWe have seen that the amplitude of forced vibration is maximum when the frequency of

the applied force has the value ω = ωn(1 – 2ξ2)1/2. If the frequency changes from this value, theamplitude falls. When the fall in amplitude for a small departure from the resonance conditionis very large, the resonance is said to be sharp. On the other hand if the fall in amplitude issmall, the resonance is termed as flat. Thus, the term sharpness of resonance means the rateof fall in amplitude with the change of forcing frequency on each side of the resonance frequency.Figure 1.15 shows the variation of amplitude with forcing frequency at different amounts ofdamping or Q-factor. Clearly smaller the damping, sharper the resonance and larger thedamping flatter the resonance. As can be seen from the figure, larger the Q, smaller the band-width and sharper the resonance. In fact Q-factor is inversely proportional to the bandwidthas shown in eqns. (1.47) and (1.48).

1.6 VIBRATION ISOLATION1.6.1 IntroductionMachines such as motors, fans and compressors produce a vibratory force at a particular or arange of frequencies. These are often a source of irritating noise that propagates through airand the ground. The vibratory force generated by such machines often leads to loosening offasteners, excessive wear of bearings, formation of cracks and structural as well as mechanicalfailures. Electronic malfunctioning through fracture of solder joints and abrasion of insulationaround conducting wires can also occur. Also there are many practical situations in which adelicate machinery has to be isolated from vibratory impacts that it is subjected to. All vehiclesare provided with shock absorbers so that when they move on a rough and bumpy surface, thejerky motion is not communicated to the engine and the passengers. In all these cases oneemploys vibration isolation techniques, which reduce the undesirable effects of vibration. Thedesign of a vibration isolation system is based on the theory of forced vibration. The vibrationisolation system is said to be active or passive depending on whether the external power isrequired for the isolator to perform its function or not. A passive isolator consists of a resilientmember (stiffner or spring) and an energy dissipator (dampner). Examples of passive isolatorsinclude metal springs, cork, felt, pneumatic springs and elastomer (rubber) springs. An activeisolator is composed of a servomechanism with a sensor, signal processor and an actuator. Aservomechanism essentially senses the vibratory motion and if it exceeds a preset value,activates an actuator for corrective action.

1.6.2 Basic TheoryVibration isolation is used in two types of situations as shown in Figs. 1.19 and 1.20.

Case (i):Consider for instance a heavy machine. To reduce the vibratory motion of a heavy machine

that is communicated to the floor, it is mounted on a rigid base. In this case, efficiency ofisolation is defined by force transmittability.

TFF

rr r

T= = = +− +

Amplitude of the force transmitted to the baseAmplitude of the force exerted by the machine 0

2 2

2 2 2 2

1 41 4

ξξ( )

Case (ii):This corresponds to the design of a shock absorber where a delicate machinery has to be

insulated from the vibrations transmitted from the ground. Here one defines displacementtransmittibility


Motor Pump

Base

(a)

Vibratingmachine

m F0

x(t)

Kc

FF

T

0T =

Rigid baseor foundation

FT (b)

Fig. 1.19 Force transmittibility (a) typical machine mounting (b) conceptual diagram

xT

m v km/hr

cK2

K2

x0

Road surface

One cycle

(a)

xT

x(t)

mDelicate

instrumentor machine

xx

T

0T =

cK

Base(Package) x0

(b)

Fig. 1.20 Displacement transmittibility (a) Shock absorber (b) Conceptual diagram

Txx r r

T= = =− +

Amplitude of transmitted displacement to the machineryAmplitude of applied displacement 0 2 2 2 2

1

1 4( ) ξ

For a simple mass-spring system the force and displacement transmittibilities are equaland are given by the above equations. Figure 1.21 shows how T varies with frequencies forvarious amounts of damping. At frequencies below resonance T = 1, indicating that the massand the base move in effect together as if rigidly connected. As the resonant frequency isapproached, the transmittability increases greatly, indicating an amplification of the vibrationbeing transmissied through the structure. The maximum transmittibility at resonance dependson the amount of damping. Above resonance, the transmittibility falls to 1 until at a frequencycorresponding to r = 1.414. At higher frequencies T is less than 1, indicating that the vibrationsare being alternated or isolated as they travel through the structure. In this region, well aboveresonance, it can be seen that the amount of damping affects the transmittibility. A lightly


damped system has a lower T value at the same transmitted frequency ratio than a moreheavily damped system.

10

5

3

2

1.0

0.5

0.3

0.2

0.10

0.05

0.03

0.02

0.01

= 0

0.050.05

0.10.1

0.20.2

0.50.51.01.0

= 1.0

= 1.0

0.50.5

0.20.2

0.10.1

0.050.05

=

0

=0

0.1 0.2 0.3 0.5 1.0 2 3 5 1099

90

80

50

200

Am

plifi

catio

nIs

olat

ion

effic

ienc

y(%

)

Tran

smitt

ibili

ty

r = ( / ) n

Fig. 1.21 Frequency dependence of transmittibility for various values of ξ

A machine such as a motor, fan or engine produces a vibratory force at a particularfrequency ω, related to its rotational speed. The machine and the isolator form a mass-springsystem with a resonating frequency ωn, which is related to the mass of the machine and thestiffness of the isolator. The isolator must be selected such that ωn is low enough to achieve avalue of r(= ω/ωn) high enough to produce the required degree of isolation i.e., r >> 1. Thedamping of the isolator should in theory be as small as possible to achieve the best reductionsat a given frequency ratio. However, in practice, damping can be useful since in many machinesthe vibratory force passes through the resonance frequency of the system during run up or rundown. The damping helps to limit the vibration amplitude of the machine as it passes throughthe resonance speed.

1.6.3 Vibration AbsorbersA machine or system will experience excessive vibration if it is acted upon by a force whoseexcitation frequency nearly coincides with the natural frequency of the machine or system. Insuch cases, the vibration of the machine or system can be reduced by using a vibrationneutralizer or dynamic vibration absorber. This vibration absorber is another spring masssystem. It is designed such that the natural frequency of the resulting system is away from theexcitation frequency (Fig. 1.22).


Motor Generator

Vibrationabsorber

(a)

F sin t0

Machine (m )1

K2

c2

x (t)1

Isolator(K /2)1

m2 x (t)2

Isolator(K /2)1

Dynamic vibration absorber

(b)

Fig. 1.22 Vibration absorber

1.6.4 Vibration Analysis and ControlEngineers routinely carry out vibration analysis of machines and structures. This is essentialfor optimal efficiency of operation. The measurement of the natural frequencies of a structureor a machine is useful in selecting the operation speeds of nearby machinery to avoid resonantconditions. This also helps in the design of vibration isolation systems. In addition it helps inunderstanding the operations of machines or structures under specific vibrational environmentsuch as road surface conditions, fluctuating wind velocities, random variation of ocean wavesand ground vibrations due to earthquake. Figure 1.23 shows the basic features of vibrationalmeasurement. The motion of the vibrating body is converted into an electrical signal by atransducer or pick-up. Transducers measure displacement, velocity and acceleration. Theoutput signal is usually amplified and the data presented on a display unit for visual inspection.The vibration measuring instrument is called a vibrometer.

Vibratingmachine orstructure

Vibrationtransduceror pickup

Signalconversioninstrument

Display unit,recorder orcomputer

Dataanalysis

Fig. 1.23 Basic vibration measurement scheme

1.7 LCR CIRCUIT ANALYSIS—ANALYTICAL APPROACH1.7.1 IntroductionThe analytical approach that we have adopted in analysing the free, damped and forcedvibrations can be adopted to analyse LCR circuits as well. This brings out the similarity betweenthe mechanical and electrical oscillations.


1.7.2 Free Oscillation in an LC CircuitFigure 1.24 shows a simple circuit consisting of a pure capacitorand a pure inductor. The capacitor is initially charged using abattery and then allowed to discharge across the inductor. Atany time

The potential across the plate VC = qC

The potential across the inductor VL = – L didt

= – L d q

dt

2

2

The negative sign indicates that the voltage opposes the increase of current.On applying Kirchhoff’s second law

Ld qdt

qC

2

2 0

+ = or d q

dt LCq

2

21

0+ = ⇒d q

dt LCq

2

21= − (1.49)

Comparing eqns. (1.5) and (1.49) we get

ωn = 1

LC or T = 2π LC (1.50)

Thus the magnitudes of charge on the plate varies periodically with a time

T = 2π LCA pure LC circuit is an electrical analogue of the undamped simple pendulum. Just as

in the case of simple pendulum where the energy alternates twice the potential to kineticenergy, here the energy is alternately stored in the capacitor as electric field and inductor asmagnetic field.

1.7.3 Damped Oscillations in a Series LCR CircuitFigure 1.25 shows a circuit consisting of a capacitor,inductor and a resistor. The capacitor is initially chargedand then allowed to discharge across the inductor andthe resistor. At any instant of time

Voltage across the capacitor = VC = qC

Voltage across the inductor = – L didt

= – L d q

dt

2

2

Voltage across the resistor = iR = R dqdt

Applying Kirchhoff’s second law

Ld q

dtR

dqdt

qC

2

2 0+ + =. ⇒d q

dt

RL

dqdt

qLC

2

2 0+ + =. (1.51)

Comparing eqns. (1.51) and (1.14) we get

ωn2 =

1LC

and 2ξωn = RL

⇒ ξ = R

LR C

Ln2 2ω= (1.52)

Fig. 1.24 LC circuit

L

C+++

+–––

–

i

Fig. 1.25 LCR circuit

C L R

i

–––

–+++

+


Thus the variation of q and i(= dq/dt) is same as that of displacement and velocity withtime and is shown in Fig 1.3. Note also that there is a phase lag of 90° between the chargeoscillations and current oscillations. Here also one can differentiate between underdampedand overdamped cases.

Overdamping : This refers to the case when

ξ > 1 ⇒R C

L2 > 1 ⇒ R > 2

LC

In this case charge on the plate decays with time. It is non-oscillatory and does notquickly decay with time.

Underdamping : This refers to the case when

ξ < 1 ⇒R C

L2 < 1 ⇒ R < 2

LC

In this case charge on the plate oscillates with an angular frequency

ωd = ωn 11

14

14

22 2

2− = − = −ξLC

R CL LC

R

L(1.53)

The energy stored in the capacitor decays with time as shown in Fig. 1.10

1.7.4 Forced Oscillations in a Series LCR CircuitFig. 1.26 shows a series LCR circuit driven by analternating applied voltage

V = V0 sin ωt.Once again applying Kirchhoff’s second law

Ld q

dtR

dqdt

qC

V t2

2 0+ + =. sin ω

⇒d q

dt

RL

dqdt

qLC

VL

t2

20

0+ + =. sin ω (1.54)

Comparing eqn. (1.54) with eqn. (1.36) following analogy is possible

Table 1.1 Correspondence between physical quantitiesin mechanical and electrical oscillations

Displacement x → Charge q

Velocity dx/dt → Current dq/dt

Mass m → Inductance L

Damping coefficient c → Resistance R

Spring constant K → Reciprocal of capacitance 1/C

Force amplitude Fo → Voltage amplitude V0

Driving frequency ω → Oscillator ω

ξω

= cm n2 → ξ = R C

L2

C L R

V sin t0

Fig. 1.26 Driven LCR circuit


Comparing with eqn. (1.41a) we have

δst = V0C; ωn = 1

LC; r =

ωω

ξn

R CL

; =2

(1.55a)

∴q

CV LC R C

0

0 2 2 2 2 2

1

1=

− +( )ω ω (1.55b)

tan φ = RC

LC

RX XC L

ωω1 2−

=−

(1.55c)

where XC and XL are the impedances due to capacitor and inductor and are given by

XC = 1

Cω and XL = Lω

The variation of current with frequency can be obtained as follows:

q = q0 sin (ωt – φ)

idqdt

= = q0ω cos (ωt – φ) = i0 cos (ωt – φ) (1.56)

From eqns. (1.55) and (1.56)

iV C

L R C

V

CL R

V

X X RC C L

00

2 2 2 2 20

22

02 21 1

=− +

=

−

+

=− +

ω

ω ωω

ω( ) ( )

(1.57)

Power absorption:

From eqn. (1.45), (1.55) and table 1.1

Pmax = C R

CVLC R

LC

VR

VRst n2

14 2

1 12

2 22 0

22

02 2

. . . ( ) .δ ωξ

=

= = rms

(1.58)

Also from eqn. (1.46)

PP

rr

P

rr

P

RCLR

P

X XR

C L

=−

+=

−

+

=−

+=

−

+

max max max max. 4

14 1

2 21

11 1

2

22

2 2 2

ξ

ξξ ξ ω

ω(1.59)

Q-factor

From eqns. (1.47) and (1.52)

Q = 1

21

ξω

= =n

RLC∆ω

(1.60)

1.7.5 Resonance in a Series LCR Circuit—Phasor AnalysisConsider and a.c. circuit consisting of resistance R, inductance L, and capacitance C connectedin series as shown in Fig. 1.27.


VR

I

VL

I

I

VC

R L C

VR VL VC

II

B

D

O

C

V = RR I A

V = XC CI

I

V = ZI

V = XL LI(V – V ) = (X – X )L C L CI

F

Z

R X

=(X

–X

)L

C

(a) (b) (c)V

Fig. 1.27 Series RLC circuit(a) circuit diagram (b) phasor diagram (c) impedence triangle

Let V be the r.m.s. value of applied voltage and I be the r.m.s. value of current.∴ Voltage drop across R, VR = IR (in phase with I)

Voltage drop across L, VL = IXL (leading I by 90°)Voltage drop across C, VC = IXC (lagging I by 90°)

The applied voltage is the vector sum of VR, VL and VC

The phasor diagram is drawn as shown in Fig. 1.27b

(OA = VR = IR, OB = VL = IXL, OC = VC = IXC

Assuming that VL > VC . OD = OB – OC

= VL – VC = IXL – IXC = I(XL – XC)= IX

OF = IZFrom triangle OAF

V = V V VR L C2 2+ −( )

= ( ) ( )IR IX IXL C2 2+ −

= I R X XL C× + −2 2( ) = IZ

where Z = R X XL C2 2+ −( ) is the impedance of the circuit.

In the phasor diagram the current I lags behind the applied voltage by an angle φ suchthat

tan φ = AFOA

I X XIR

X XR

XR

L C L C=−

=−

=( ) ( )

Where X = (XL – XC ) is the net reactance of the circuit.

∴ Phase angle, φ = tan– 1 (X/R)

Impedance triangle of the circuits is shown in Fig. 1.27c.


As in the previous cases, the power in the RLC series circuits can be obtained as

P = VI cos φ (= I2R)

Note that power factor = cos φ = R

R X XL C2 2+ −( )

Power is consumed only in the resistance (R) of the circuit.If VC > VL, then the current will lead the applied voltage V, by an angle ‘φ’ such that

φ = tan– 1 (XC – XL)/R and power factor = cos φ = R R X XC L/ ( )2 2+ −

If VL = VC (i.e., XL = XC ) then power factor = cos φ = unity. In this situation current ismaximum and corresponds to the series resonance condition.

The frequency of the resonance is given by

XL = XC

Lω = 1

Cω i.e., ω2 1=

LC i.e., f

LCr2

21

4=

π

or fr = 1

2π LC

Resonance Curve and Q-FactorA plot of current versus frequency in the LCR circuits is known as resonance curve. The shapeof such a curve for various values of R is shown in Fig. 1.28a. For smaller values of R, thecurrent frequency curve is sharply peaked, but for large values of R the curve is flat. Thevariation of Z, power factor and I are separately shown in Fig. 1.32b.

I

R verysmall

R verylarge

I

f0 f

cos = 1

Im = V/R

Z = R

I

f0

fResonant frequency,

Z

LaggingP.F.

Lead

ing

P.F.

(a) (b)

Fig. 1.28


The ability of a reactive circuit to store energy is expressed in terms of the quality factoror Q-factor. It is a figure of merit which enables us to compare different coils. The Q-factor isdefined as the ratio of the energy stored in the coil to the energy dissipated on the circuitacross the resistance.

Thus Q = 2π Energy stored

Energy dissipated per cycleThus larger the Q-factor, greater is the storing ability for a given dissipation.In the case of an inductor the energy stored is I2XLt. The energy dissipated is I2Rt

∴ QL = I Lt

I Rt

LR

2

2ω ω

=

= 1R

LC

at resonance ω =

1LC

In the case of a capacitor

Q = I C t

I RtC

R CR

2

2

1 1 1( / ) ( / )ω ωω

= =

= 1R

LC

at resonance ω =

1LC

The Q-factor varies from 5 to 100 for inductive circuits and from about 1400 to 10000 forcapacitive circuits. When a series circuit is in resonance the energy stored in the capacitor isequal to that stored in inductor. In one quarter cycle the inductor stores energy while in thenext quarter cycle it is stored in the capacitor. The energy flows back and forth betweeninductance and capacitance. The only loss of energy is due to the energy loss in the resistance.Thus if the resistance is small, the energy oscillation continues for a long time even thoughthere is no supply of energy from an external source.

In the case of an LCR circuit, Q-factor may also be defined as equal to the voltagemagnification in the circuit at resonance. We have seen that at resonance, the current ismaximum

i.e., Imax = VR

Voltage across the coil = Imax · XL

Supply voltage = V = Imax R

∴ Voltage magnification = I XI R

XR

LR

L Lmax

max .= =

ω

∴ Q-factor = ω πLR

f LR

r= 2

But the resonant frequency = fr = 1

2π LC

∴ Q = 1R

LC


Voltage across capacitor = Imax · XC

∴ Voltage magnification = I XI R

XR CR

C Cmax

max= = 1

ω

∴ Q-factor = 1 1

11

ωCR LC CR RLC

= =/ .

The Q-Factor of a Series LCR Circuit and Selectivity

At resonance the reactance is zero. Hence the current at resonance is Ir = VR

.

Consider two frequencies on either side of fr where the reactance is equal to resistance

(Fig. 1.29). Hence at these frequencies Z = R R R2 2 2+ = . .

The current I at these frequencies is

I = V

R2 .

Power dissipation at these frequencies

P1 = V

R2

2

.

R =

VR

2

2 = P2

A

B C

I = ER

I I I1 2= = 0.072

Cur

rent

f1 f2 f3f

Fig. 1.29

The power dissipation at resonant frequency is

Pr = VR

2

R = VR

2

Thus the power dissipation at f1 and f2 is half that at resonance. Hence the points B andC are called half power points.


At f1 which is below fr

XL – XC = – R

i.e., ω1L – 1

1ω C = – R or

1

1ω C – ω1L = R (1.61)

the circuit is capacitive.At f2 which is above fr

XL – XC = R

i.e., ω2L – 1

2ω C = R (1.62)

the circuit is inductive.From (1.61) and (1.62), we get

(ω2 – ω1) L + 1 2 1

1 2Cω ω

ω ω−

= 2R

or (ω2 – ω1) LC

+

1

1 2ω ω = 2R

Dividing by L

(ω2 – ω1) 11 2

1 2+

=

ω ω LCRL

(1.63)

As f1 and f2 are close together we can write to a first degree of approximation

ω1L = 1

2ω C

Then ω1ω2LC = 1. Then (1.63) becomes

(ω2 – ω1) (1 + 1) = 2RL

(ω2 – ω1) = RL

or f2 – f1 = R

L2π(1.64)

But Qr = ω πr rL

Rf LR

= 2

or fr = Q R

L2π(1.65)

From (1.64) and (1.65)

f ff Qr

−=1 1

(1.66)


This is identical to eqn (1.48).f2 – f1 is called bandwidth B

B = f2 – f1 = fr/Q (1.67)If the bandwidth is small, the resonance is to be sharp.Suppose the applied a.c. voltage has a number of frequency components. The LCR circuit

will give maximum response to that component frequency that is equal or nearly equal to itsresonant frequency. Thus the circuit exhibits selectivity. For this reason the circuit is called anacceptor circuit. Equation (1.67) shows that smaller the bandwidth greater the Q-factor. Hencea decrease in bandwidth results in better selectivity. The selectivity also increases as the ratioL/R is increased or the product CR is decreased.

1.7.6 Resonance in a Parallel LCR CircuitWe will consider the practical case of a coil in parallel with a capacitor as shown in Fig. 1.30.Such a circuit is said to be in electrical resonance when the reactive (or wattless) component ofline current becomes zero. The frequency at which this happens is known as resonant frequency.

ILR L

Z

IC CI

IC

90°IL Lcos

VL

L Lsin L

L

R

ZXL

(a) (b) (c)

Fig. 1.30 RLC parallel circuit(a) circuit diagram (b) phasor diagram (c) impedance triangle

The vector diagram for the circuit is shown in Fig. 1.30b.Net reactive or wattless component = IC – IL sin φL

As at resonance its value is zero.∴ IC – IL sin φL = 0 or IL sin φ = IC

Now IL = VZ

; sin φL = XZ

L ; IC = VXC

Hence the condition for resonance becomes

VZ

XZ

VX

L

C= or XL XC = Z2

Now XL = ωL and XC = 1

ωC

ωω

LC

= Z2 or LC

= Z2

orLC

= R2 + XL2


or LC

= R2 + (2πfr L)2

or (2πfr L)2 = LC

– R2

2πfr = 1 2

2LCR

L− or

fr = 1

21 2

2π LCR

L−

This is the resonant frequency and is given in Hz if R is in ohms, L is in henrys and C isin farads. If R is negligible

fr = 1

2π LC which is same as for series resonance.

Because the wattless current is zero, the circuit current is minimum and is

Imin = IL cos φ = VZ

RZ

or Imin = VR

Z2

Putting the value Z2 = L/C, we get

Imin = VR

L CV

L CR/ /=

The denominator (L/CR) is known as the equivalent or dynamic impedance of the parallelcircuit at resonance. It should be noted that this resistance is resistive only. Since current isminimum at resonance, L/CR must then represent the maximum impedance of the circuit.

Current at resonance is minimum. Hence such a circuit is sometimes known as rejectorcircuit because it rejects (or takes minimum current of) that frequency to which it resonates.

Q-Factor of a Parallel CircuitIt is defined as the ratio of the current through the coil or capacitor to the main current

or as the current magnification at resonance.Then Q-factor = IC/INow IC = V/XC = V/1/ωC = VωC

Imin = V

L CR( / )

Q = I

Ic

min = VωC/

VL CR

LR

f LR

r

/= =ω π2

Now at resonant frequency when R is negligible,

fr = 1

2π LC


∴ Q = 2 1

21π

πL

R LC RLC

. .=

It should be noted that in series circuits, Q-factor gives the voltage magnification, whereasin parallel circuits, it gives the current magnification.

Note that Q is also given by IL/I and that can also be shown to be equal to 1R

LC

.

REFERENCES1. A.P. French, Vibration and Waves, Arnold-Heinemann India, New Delhi, 1973.2. H.J. Pain, The Physics of Vibrations and Waves, John Wiley & Sons, New York, 2003.3. S.S. Rao, Mechanical Vibrations, Pearson Education, New Delhi, 2004.4. I.G. Main, Vibrations and Waves in Physics, Cambridge University Press, Cambridge,

1995.5. W.H. Erickson and N.H.Bryant, Electrical Engineering Theory and Practice, John Wiley

& Sons, New York, 1967.6. Vincent Del Toro, Principles of Electrical Engineering, Prentice Hall of India Pvt. Ltd.,

New Delhi 1987.7. H. Alex Romanowitz, Introduction to Electric Circuits, John Wiley & Sons, New York,

1971.8. B.L. Theraja, Fundamentals of Electric Engineering, Niraja Construction and

Development Pvt. Ltd., New Delhi, 1988.

SOLVED EXAMPLES1. A massless spring, suspended from a rigid support, carries a flat disc of mass 100 g at its

lower end, it is observed that the system oscillates with a frequency 10Hz and the am-plitude of the damped oscillations reduces to half its undamped value in one minute.Calculate the resistive force constant and the relaxation time of the system.Solution: The amplitude of the damped oscillator at any instant t is given by

A = Ce nt−ξω

since AC

= 12

for t = 1 minute = 60s

12

= e en n− −=( )ξω ξω60 60

∴ ξωn = ln 260

= 1.16 × 10–2 rad/s

The resistive force constant = c = 2mξωn

= 2 × (100 × 10–3) × (1.16 × 10–2)= 2.32 × 10–3 newton/s/meter

Relaxation time τ = 1 1

116 10 2ξωn=

× −. / secred = 86.96 s.

2. A massless spring of spring constant 10 N/m is suspended from a rigid support and carriesa mass of 0.1 kg at its lower end. The system is subjected to a resistive force c(dx/dt).


where c is the resistive force constant and dx/dt is the velocity. It is observed that thesystem performs damped oscillatory motion and its energy decays to 1/e of its initialvalue in 50 s.(a) What is the value of resistive force constant c ?(b) What is the natural angular frequency of the oscillator ?(c) What is the damping ratio ξ and Q-factor ?(d) What is the percentage change in frequency due to damping ?Solution:

(a) m = 0.1 kgThe decay of the energy of the damped oscillator is given by

E(t) = E e n0

2− ξω

where E0 is the initial energy.

E tE e( )

0

1= = e–1 = e n−100ξω

⇒ ξωn = 10–2/sHence the resistive force constant = c

= m · 2ξωn = 0.1 × 2 × 10–2 = 2 × 10–3 newton/s/meter (1)(b) Since K = 10N/m, the angular frequency ω in the absence of damping is

ωn = Km

=100 1.

= 10 rad/s (2)

(c) From (1) and (2)

ξ = 1 10

10

2× − = 1 × 10–3 and Q =

12

1

2 10 3ξ=

× − = 500

The angular frequency of damped oscillation is

ωd = ωn 1 2− ξ

= 10 1 1 10 3 2− × −( )

≈ 10 radian/s (3)

(d) The fractional change in frequency is given by ω ω

ωωω

n d

n

d

n

−−

1

= 1 −

ωω

d

n = 1 – (1 – ξ2)1/2 ≈ 1 1

12

2− −

ξ ≈ 12

2ξ ≈ 12

1 10 3 2× × −( )

≈ 0.5 × 10–6

∴ percentage change in frequency = 0.5 × 10–6 × 102 = 5 × 10–5 (4)3. An object of mass 0.1 kg is hung from a spring whose spring constant is 100Nm–1. A

resistive force c (dx/dt) acts on the object where dx/dt is the velocity in meters persecond and c = 1 Nsm–1. The object is subjected to a harmonic driving force of the form


F0 cos ωt where F0 = 2N and ω = 50 radian/s. In the steady state what is the amplitude ofthe oscillations and the phase relative to applied force ?Solution:

m = 0.1 kg ; K = 100 N/m ; c = 1 Nsm–1 ; F0 = 2N ; ω = 50 rad/s

ωn = Km

= =1000 1

1000.

= 31.6 rad/s

r = ω

ωn= 50

31.6 = 1.58 ; δst

Fk

= =0 2100

= 2 × 10–2 m

ξ = c

m n21

2 0.1 31.6ω=

× × = 0.158

A = δ

ξst

r r( ) ( )1 2

2 102 2 2

2

− += ×

− + × ×

−

(1 0.158 ) (2 1.58 0.158)2 2 = 1.26 × 10–2 m

tan φ = 2

1 2r

r

ξ−

=× ×

−= −2 1.58 0.158

1 1.58132 ⇒ φ = °161.7

i.e., the oscillations lag behind the applied force by 161.7°.4. A series circuit consists of a resistance of 15 ohms, an inductance of 0.08 henry and a condenser

of capacity 30 microfarads. The applied voltage has a frequency of 500 radians/s. Does thecurrent lead or lag the applied voltage and by what angle ?Solution:Here ω = 500 radian/s, L = 0.08 H

R = 15 ohm, and C = 30 × 10–6 F Lω = 0.08 × 500 = 40 ohm

1 1

30 10 5006Cω=

× ×− = 66.7 ohm

tan φ = L

CR

ωω

− 1

= 40 66.7

1526.715

− = − = – 1.78

φ = – 60.65°The current leads the applied voltage by 60.65°.

5. A series circuit consists of a resistance, inductance and capacitance. The applied voltageand the current at any instant are given by

E = 141.4 cos (3000t – 10°)I = 5 cos (3000t – 55°)

The inductance is 0.01 henry. Calculate the values of the resistance and capacitance.Solution:

The E.M.F. is ahead of the current by 55° – 10° = 45°∴ φ = 45°

tan φ = tan 45° = 1


Also tan φ = L

CR

ωω

−

1

= 1

LC

ωω

− 1 = R

Also Impedance, Z = R LC

221

+ −

ωω

Z = R R R2 2 2+ = = 1.414 R

But Z = EI

0

0= 141.4

5 = 28.28

∴ 1.414 R = 28.28i.e., R = 20 ohm

LC

ωω

− 1 = 20

ω = 3000 radian/sL = 0.01 henry

Lω = 0.01 × 3000 = 30 ohm

∴ 30 – 1

Cω = 20

i.e.,1

Cω = 10 ohm

C = 1

101

10 3000ω=

×

C = 33.33 × 10–6 FC = 33.33 µF.

6. A series RLC circuit with a resistance of 50 Ω, a capacitance of 25 µF and an inductanceof 0.15H is connected across 230-V, 50-Hz supply. Determine (i) impedance (ii) current(iii) power factor and (iv) power consumption of the circuit.Solution:

XL = 2πfL = 2π × 0.15 = 47.1 ΩXC = 1/2πfC = 106/ωπ × 50 × 25 = 127.3 Ω

Net X = XL – XC = 47.1 – 127.3 = 80.2 Ω (capacitive)

(i) Z = R X2 2 2 250 80 2+ = + . = 94.4 Ω

(ii) I = V/Z = 230/94.4 = 2.44 A(iii) p.f. = cos φ = R/Z = 50/94.4 = 0.53 (lead)(iv) power consumed = VI cos φ = 230 × 2.44 × 0.53 = 297 W.


7. A coil of insulated wire of resistance of 8 Ω and inductance 0.03 H is connected to an a.c.supply at 240-V, 50-Hz. Calculate:(i) the current, p.f. and the power

(ii) the value of capacitance which when connected in series with the above coil, causesno change in the values of the current and power taken from the supply.

Solution:(i) XL = 314 × × 0.03 = 9.42 Ω

Z = 8 9 422 2+ . = 12.36 Ω

I = 240/12.36 = 19.4 A W = I2R = 19.42 × 8 = 3011 Wp.f. = R/Z = 8/12.36 = 0.65 (lag)

(ii) If the circuit is to draw the same current and at the same power factor, then the totalreactance of the RLC circuit must be 9.42 Ω. This can be achieved by selecting acapacitor which should not only neutralize the inductive reactance of 9.42 Ω of thecoil but must add a further capacitive reactance of 9.42 Ω. In other words, capacitormust have a reactance of 2 × 9.42 = 18.84 Ω.

1/ωC = 18.84 or C = 1/314 × 18.84 = 169 µF8. A resonant circuit consists of a 4 µF capacitor in parallel with an inductor of 0.25 H

having a resistance of 50 Ω. Calculate the frequency of resonance.Solution:

The resonance frequency is

fLC

R

L0

2

21

21= −

π

It is seen that f0 is dependent on R. When R is large, f0 is reduced. In a series circuitf0 = 1/2π LC . Obviously, f0 does not depend on R.

f0 = 1

210

4 0 2550

0 25

4 2

2π ×−

. . = 156 Hz.

9. A coil of resistance 30 Ω and inductance 20 mH is connected in parallel with a variablecapacitor across a supply of 25 V and frequency 1000/π Hz. The capacitance of the ca-pacitor is then varied until the current taken from the supply is a minimum (i.e., untilthe overall p.f. of the circuit is unity). For this condition, find(i) the capacitance of the circuit

(ii) the value of the current.Solution:

XL = 2πfL = 2π × (1000/π) × 20 × 10–3 = 40 Ω

Z = 30 402 2+ = 50 Ω

(i) At resonance Z2 = L/CC = L/Z2 = 20 × 10–3/502 = 8 × 10–6 F = 8 µF.


(ii) Imin = V

L CR/Dynamic impedance = 20 × 10–3/8 × 10–6 × 30 = 83.3 Ω

Imin = 25/80.3 = 0.3 A.

QUESTIONS 1. Define (a) free vibrations (b) forced vibrations (c) transient vibrations (d) damped vibrations and

(e) linear vibrations.2. Show that the average energy of a weakly damped harmonic oscillator decays exponentially with

time.3. Establish the equation of motion of a damped oscillator subjected to a resistive force that is

proportional to the first power of its velocity. If the damping is less than critical, show that themotion of the system is oscillatory with its amplitude decaying exponentially with time.

4. Discuss the theory of vibration isolator.5. Discuss the theory of electrical oscillations in a series (a) LC and (b) LCR circuit by setting up

the relevant differential equations.6. Discuss the theory of forced electrical oscillations in a series LCR circuit by setting up the relevant

differential equation.7. Find the expression for current in the case of a series LCR circuit. Describe the frequency

dependence of current, and hence discuss the concept of resonance.8. Find the expression for current in the case of parallel LCR circuit. Describe the frequency

dependence of current.9. What are acceptor and rejector circuits ?

10. What do you understand by the term ‘quality factor’ and the ‘relaxation time’?11. What is Q-factor ? Obtain an expression for the same.

PROBLEMS1. Prove that in simple harmonic motion the average potential energy equals the average kinetic

energy when the average is taken with respect to time over one period of motion and that eachaverage is equal to (1/4) Ka2 where K is the spring constant and a is the amplitude. But whenthe average is taken with respect to position over one cycle, the average potential energy is equalto (1/6) Ka2 and the average kinetic energy is equal to (1/3) Ka2. Explain why the two results aredifferent.

2. A simple pendulum consists of a rod of mass m and length l which is pivoted at the upper endand carries a mass M at the other end. Using energy consideration determines the frequency ofthe pendulum if (a) m << M and (b) m is comparable with M.

3. A massless spring suspended from a rigid support carries a flat disc of mass 100g at its lowerend. It is observed that the system oscillates with a frequency of 10Hz and the amplitude of thedamped oscillations reduces to half its undamped value in one minute. Calculate the (a) theresistive force constant (b) relaxation time (c) the quality factor and (d) the spring constant.

(Ans. 0.0023 Nsm–1, 86.58 s, 2720, 394.8 Nm–1)4. The viscous force on sphere of radius a moving with a velocity v in a medium of coefficient of

viscosity η is 6πηav. Determine the effects of air viscosity on the amplitude and period of simplependulum consisting of a aluminium bob of radius 0.5cm suspended by means of a 1m longthread. Take the density of aluminium to be 2.65 g/cc and η to be 1.78 × 10–4 gcm–1 s–1.

5. According to classical electromagnetic theory, an electron orbiting around the nucleus can bethought of as executing simple harmonic oscillation. It is subjected to an acceleration whichresults in the emission of radiation. It radiates energy at the rate of (ke2ω4A2/3c3) watts, wherek = 9 × 109 Nm2 c–2, e is the charge on the electron, c is the velocity of light, A is the amplitude of


oscillation or the radius of the orbit and ω is the angular frequency. If the emitted radiation hasa wavelength of 600 nm. Calculate the Q value of the oscillator and the radiation life time i.e.,the time required for the energy to fall to e–1 of the original value. (Ans. 5 × 107, 16.9 ns)

6. For a mass-spring system, the spring constant = 10 N/m, m = 10 kg and resistive force constant= 8 Ns/m. Determine the motion of the mass when it is given a velocity of .068 m/s at t = 0.

7. An oscillator with small damping has mass 5 g and a force constant of 2 N/m. If the Q for theoscillator is 200 and the system oscillate in energy resonance with an applied force (a) what isthe critical damping constant cc (b) what is the frequency of the applied force (c) what frequencyof the applied force will produce amplitude resonance ? (d) what is the amplitude of the displace-ment and velocity oscillations of the oscillator if the frequency of applied force is 90% of thenatural frequency of the free oscillator ? Express the answer as fraction of the amplitude atenergy resonance (e) what is the power delivered to the oscillator by the impressed force ex-pressed as the fraction of the power at resonance (e) what is the bandwidth of the oscillator.

(Ans. 5 × 10–3 kg/s, 31.8Hz, ~ 31.8Hz, 0.0237, 0.000561 1rad/s)

8. A weakly damped harmonic oscillator is driven by a force F0 cos ωt, whose amplitude F0 is keptconstant but its angular frequency is varied. It is experimentally observed that the amplitude ofthe steady state oscillations is 0.1 mm at very low ω and attains a maximum value of 10 cmswhen ω = 100 rad/s. Calculate (a) the Q value of the system (b) the time during which the energyof the oscillator falls to 1/e of its initial value and (c) half-width of the power resosnance.

(Ans. 1000, 10 s, 0.05 rad/s)

9. A vibrating system of natural frequency 500 Hz is forced to vibrate with a periodic force ofamplitude 10–1 N/kg in the presence of damping coefficient of 10–5 N s/m. Calculate the maximumamplitude of the vibration of the system. (Ans. 3.15 × 10–2 m)

10. The energy of a piano string of frequency 256 Hz reduces to half its initial value in 2 s. Whatis the Q-factor of the string ? (Ans. 4643)

11. The quality factor of a sonometer wire of frequency 500 Hz is 5000. In what time will its energyreduce to 1/e of its value because of damping? (Ans. 1.59s)

12. An alternating EMF, E = E0 sin ωt is applied across a parallel combination of R, L and C asshown in Fig. 1.31. Calculate the current in each branch and express the total as a sine function.

Ans. sin ( tanI ER

LC

t R C= + − × + − −

1 2 1 21 1

ωω

ω ωω

E sin t0

I

R

IR IL

LIC

C

Fig. 1.31

13. A resistance of 10 ohms is connected in series with an inductance of 0.5 henry. What capacitanceshould be put in series with the combination to obtain the maximum current ? What will be thepotential difference across the resistance, inductance and capacitance? The current is being sup-plied by 200 volts and 50 cycles per second mains. (Ans. 200 V ; 3142 V ; 3142 V)

14. A 60 cycles AC circuit has resistance of 2 ohms and inductance of 10 millihenries. What is the


power factor ? What capacitance placed in the circuit will make the power factor unity ?(Ans. 0.4687 ; 703 µF)

15. A series circuit consisting of a condenser of capacitative reactance 30 ohms, a noniductive resist-ance of 44 ohms and a coil of inductive reactance 90 ohms and resistance 36 ohms is connected to200 V, 60 hertz line. Calculate (i) Impedance of the circuit (ii) Current in the circuit (iii) Poten-tial difference across each component (iv) Power factor of the circuit (v) Power consumed.

(Ans. 1000 ohm ; 2 A, 194 V ; 60 V ; 320 Watt)16. A circuit consists of a resistance of 20 Ω in series with an inductance of 95.6 mH and a capacitor

of 318 µF. It is connected to a 500 V, 25 Hz supply. Calculate the current in the circuit and thepower factor. (Ans. 24.3A ; 0.97 lead)

17. A circuit is made up of 10 Ω resistance, 12 µH inductance and 281.5 µF capacitance in series. Thesupply voltage is 10 V (constant). Calculate the value of the current when the supply frequencyis (i) 50 Hz and (ii) 150 Hz. (Ans. (i) 8 A leading (ii) 8 A lagging)

18. An a.c. series circuit has a resistance of 10 Ω, an inductance of 0.2 H and a capacitance of 60 µF.Calculate (i) the resonant frequency (ii) the current (iii) the power at resonance given that ap-plied voltage is 300 V. (Ans. (i) 46 Hz (ii) 20 A (iii) 4 kW)

19. A resistor and a capacitor are connected in series with a variable inductor. When the circuit isconnected to a 240 V, 50 Hz supply, the maximum current given by varying the inductance is0.5 A. At this current the voltage across the capacitor is 250 V. Calculate the value of : (i) theresistance (ii) the capacitance (iii) the inductance. Neglect the resistance of the inductor.

(Ans. (i) 480 Ω (ii) 6.36 µF (iii) 1.50 H)20. A series circuit has the following characteristics : R = 10 Ω ; L = 100/π mH ; C = 500/π µF. Find:

(i) the current flowing when the applied voltage is 100 V at 50 Hz. (ii) the power factor of thecircuit (iii) what value of the supply frequency would produce series resonance.

(Ans. (i) 7.07 A (ii) 0.707 lead (iii) 70.71 Hz)21. An inductor of 0.5 H inductance and 90 Ω resistance is connected in parallel with a 20 µF capaci-

tor. A voltage of 230 V at 50 Hz is maintained across the circuit. Determine the total power takenfrom the source. (Ans. 14.5 Ω)

22. A resistance of 50 Ω, an inductance of 0.15 H and a capacitance of 100 µF are connected inparallel across a 100 V, 50 Hz supply. Calculate (i) the current in each circuit (ii) the resultantcurrent and (iii) the phase angle between the resultant current and supply voltage. Draw thephasor diagram. (Ans. (i) 2A in phases with voltage ; 212 A-lagging V by 90°

3.14 A-leading V by 90° (ii) 2.245A (iii) 12.8°)23. A coil having a resistance of 8 Ω and inductance of 0.0181 H is connected in parallel with a

capacitor having a capacitance of 398 µF and resistance of 5 Ω. If 100 V at 50 Hz are appliedacross the terminals of the above parallel circuit, calculate (i) the total current taken form thesupply and (ii) its phase angle with respect the supply voltage. Draw a complete vector diagramfor the circuit showing the 3 currents and the supply voltage. (Ans. (i) 13.97 A (ii) 12.8°)

24. A coil of inductance 31.8 mH and resistance of 10 Ω is connected in parallel with a capacitoracross a 250 V 50 Hz supply. Determine the value of the capacitance if no reactive current istaken from the supply. (Ans. 159 µF)

25. A 100-V, 80 W lamp is to be operated on 230-V, 50-Hz a.c. supply. Calculate the inductance of thechoke required to be connected in series with the lamp for this operation. The lamp can be takenas equivalent to a non-inductive resistance. If the p.f. of the lamp circuit is to be improved tounity, calculate the value of the capacitor which is to be connected across the circuit.

(Ans. L = 0.823 H ; 10 µF)

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