Download Mechanics, Sound, Heat and Thermodynamics

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UNIT – I
LESSON – 1
CONTENTS
1.0 Aims and Objectives
1.1 Introduction
1.2 Co-efficient of thermal conductivity
1.3 Cylindrical flow of heat
1.4 Determination of thermal conductivity (K), of bad conductor by
Lee’s disc method.
1.5 Convection currents in atmosphere – Change of pressure with height – Lapse rate.
1.6 Calculation of Lapse Rate (For Dry Adiabatic)
1.7 Let us Sum up
1.8 Check your progress
1.9 Lesson end activities
1.10 Points for discussion
1.11 References
1.0 AIMS AND OBJECTIVES
In this lesson you will learn about the processes, namely (i) conduction (ii) convection and
(iii) radiation by which transference of heat takes place. You will also learn to know how to
determine the coefficient of thermal conductivity for a good conductor and bad conductor.
1.1 INTRODUCTION
(i) CONDUCTION: In this process heat is transmitted from one point to the other through the
substance without the actual motion of the particles. When one end of a brass rod is heated in
flame the other end gets heated in course of time. In this case the molecules at the hot end vibrate
with higher amplitude (K.E) and transmit the heat energy from one particle to the next and so on.
Heat is said to be conducted through the rod. However, particles in the body remain in their
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position and so not move. Thus conduction is the transference of heat from the hotter part of a
body to the colder part without the motion of the particles in the body.
Metals are good conductors of heat and wood, glass, brick, cotton, wool, rubber are bad
conductors of heat. For example thick brick walls are used in the construction of a cold storage.
Brick is a bad conductor of heat and does not allow outside heat to flow inside the cold storage.
Also a steel blade appears colder than a wooden handle in winter. Steel is a good conductor of
heat. As soon as a person touches the blade heat flows from the hand (higher temperature) to the
blade to low temperature. Therefore it appears colder. Since wood is a bad conductor of heat,
does not allow heat to flow to the handle.
(ii) CONVECTION: It is the process in which heat is transmitted from one place to another by
the actual movement of the heated particles. It is prominent in the case of liquids and gases. Land
and sea breezes and trade winds are formed due to convection. Suppose water in a container is
heated from below. The layer of water in the bottom gets heated, its density decreases and it
comes up transferring heat. In this case, the transference of heat from the bottom of the vessel to
the top of the vessel is by convection. It is used in ventilation. Rooms are provided with
ventilators near the ceiling. Air in the room gets warmer due to respiration of persons in the
room. Warm air containing more of CO2 (gas) and water vapour has less density and moves
upwards. Fresh air from outside enters the room through the doors and windows. The impure air
moves outside through the ventilators. This phenomenon is continuous.
(iii)RADIATION:
It is the process in which heat in transmitted from one place to the other directly without the
necessity of the intervening medium. We get heat radiations directly from the sun without
affecting the intervening medium. Heat radiation can pass through vacuum. Also it forms a part
of electromagnetic spectrum.
Applications of heat radiations:
(1) White cloths are preferred in summer and dark colored clothes in winter.
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Reason: When heat radiations fall on white clothes, they are reflected back. No heat is
absorbed by the clothes and a person does not get heat from outside in summer. Dark
clothes in winter will absorb the heat radiations falling on them and keep the body
warm.
(2) Polished reflectors are used in electric heaters to reflect maximum heat in the room.
1.2 Co-efficient of thermal conductivity:
T2
T1
x
Fig. 1.1
Suppose there is a
slab of material of area of
cross-section A. Let the opposite faces be maintained at temperatures T1 0C and T2 0C (T1>T2).
Let x be the distance between the faces. It is found that the quantity of Q of heat conducted in a
time t is directly proportional,
to A
to (T1-T2)
to t (time)
and
inversely proportional to X.
Therefore,
Q
AT1  T2  t
x
or
Q
kAT1  T2  t
x
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Where K is called the co-efficient of thermal conductivity of the material of the slab.
Definition: (Thermal Conductivity)
It is defined as the quantity of heat conducted in one second from one face to the opposite
face of a slab of area of cross-section 1sq.Cm, when the distance between the faces is equal to
1cm and the difference in temperature between the faces is equal to 10C.
Temperature gradient:
The quantity (T1-T2)/X represents the rate of fall of temperature with respect to distance.
The quantity (dT/dx) represents the rate of change of temperature with respect to the distance. As
temperature decreases with in crease in distance from the hot end, the quantity (dT/dx) is
negative and is called the temperature gradient.
Therefore,
 dT 
Q   KA t
 dx 
1.3 Cylindrical flow of heat:
Description:
Consider a cylindrical tube of length l, inner radius r1 and outer radius r2. After the steady
state is reached, the temperature on the inner surface is 1 and on the outer surface in 2 in such a
way 1>2. Heat is conducted radially across the wall of the tube. Consider an element of
thickness dr and length l at a distance r from the axis.
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dr
2
+d

r2
r
r1
1
Fig.1.2
Calculation:
The quantity of heat flowing per second across the element.
 d 
Q   KA 
 dr 
Here A=2rl
Therefore
Q  2Krl
d
dr
(1)
Q is a constant after the steady state is reached.
Integrating equation (1) we get,
r2
2
1
1
dr
Q
 2Kl  d
r r


r 
Q log e 2   2Kl  1  
1
1 


1


r 
Q log e 2   2Kl  2   1 
r1 

6
r2
r1
K
2  1   2 
Q log e
Q  2.3026  log 10
K
(2)
r2
r1
2l  1   2 
1.4 Determination of thermal conductivity (K), of a bad conductor (card board) by Lees
disc method.
This apparatus consists of three parts A, B and C.
B
T1
C
Description:
T2
A
Fig.1.3
A is a solid brass plate of
mass m, thickness d, specific heat s and radius
r. It is held horizontal by three threads from a stand attached to three hooks at its sides. The
cardboard C is cut to the same size and is placed above A. B is a hollow brass cylinder placed
above C through which steam is sent. C is of the same cross-section as A or B. Two
thermometers T1 and T2 measure the temperatures of B and A respectively. The thickness t of the
cardboard is first found. Steam is passed through B and its temperature 1 k is measured by the
thermometer T1. Cardboard conducts heat slowly and the temperature gradient along the
cardboard is
d  1   2

dx
t
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Now the cardboard is removed and B is kept directly on A. The temperature of A rises beyond
2. Then B is taken out. A is allowed to cool and its temperature at various intervals of time are
noted. Cooling curve is drawn for about 5 K on either side of 2. The rate of cooling (d/dt) at 2
is found from the graph. Heat lost by lower disc per sec is equal to ms(d/dt). This heat is lost by
the top surface, bottom flat surface and the curved side of total area (2r2 + 2rd), But at the
steady temperature of A, it loses heat through the bottom and sides of area (r2 + 2rd)
Hence the rate of flow of heat is,
d
d
dt

r 2  2rd 
2
dt
(2r  2rd )
ms
 d  r  2d 
  ms


dt  2r  2d 

But we have
d
d
 KA
dt
dx
or
K r 2
1   2
 d   r  2d 
 ms
 2 

t
 dt   2r  2d 
from which K, the thermal conductivity of cardboard is calculated. The unit is watt per metre
per degree Kelvin.
Time (t)
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For various values of , (d/dt) is determined. This is done by drawing tangents to the curve at
various points on the curve.
Note: For experimental determination Ref. Annexure
Determination of thermal conductivity (K) of rubber:
It can be determined in the laboratory applying the principle of cylindrical flow of heat.
T
steam
C
water
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Description:
A known quantity of water is taken in calorimeter C. A rubber tubing whose inner and
outer radii arc r1 and r2 is taken and a known length (50 cm) of it is immersed in water as shown
in the above figure. The initial temperature of water is noted. Let it be 1. Steam is passed
through the rubber tubing for a known time t seconds. Let the final temperature of water be 2
after applying radiation correction.
K of rubber:
Let the temperature of steam is 3. The average temperature on the outer surface of rubber
tubing is,
  2 
4   1

 2 
Calculation:
Let the mass of water = m.
Water equivalent of the calorimeter (ws)= W
Rise in temperature (2-1)
Heat gained by water = (m+w) (2-1)
Quantity of heat flowing per second,
Q
m  w 2   1 
Q  2.3026  log
But
K
t
r2
r
10 1
2l  1   2 
Substituting the values of equation (1) and(2) in (3),
(2)
(3)
10
m  w
K
r2
r
 2.3026  log 2
r1
10 r1




2 
2l  3  1
t
2


2
  1 
Thus K for rubber can be calculated.
1.5 Convection currents in atmosphere – Change of pressure with height – lapse rate:
The pressure of the atmosphere decreases with increase in height from the sea level. The
density of air is not the same at all levels. If the density of air were the same at all levels, the
atmospheric pressure at sea level would correspond to a vertical column of 8km of air only. But
it has been found that air exists even at a height of 20km.
Consider that the pressure of air at a height h = P. For an increase in height dh, the decrease in
pressure in dp.
i.e. dp = -(dh)g
( P  hg )
(1)
where  is the density
[negative sign shows that the pressure decreases with increase in height]
We know,

M
V
[density = (mass/volume)]
Substituting for ,
dp  dh 
M
g
V
But for a perfect dry air,
PV = RT (Gas equation)
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p-dp
dh
p
dp
h
Earth
Fig.1.4
or
 RT 
V 

 P 
dP  (-dh)
Hence
M g
RT
dp
 Mg 
 
dh
p
 RT 
Integrating,
log P  
Where C is a constant.
When
h = 0; P = P0; C = log P0
Substituting,
Mg
hC
RT
(2)
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log P  
log e
i.e;
Mg
h  log P0
RT
P
Mg

h
P0
RT
P  P0 e

(3)
Mg
h
RT
(4)
Equation (2) will be true only under isothermal conditions. But air temperature decreases
uniformly with height. Assuming that the lapse rate is  in the lower portion of the atmosphere
(troposphere).
T = T - αh
Where T in the temperature at the ground.
Substituting in equation (2) we get,
dp
dh
 Mg 
 
 T  h 
p
R

 
log P  
Mg
logT0  R  K
R
log P0 
Mg
logT0  K
R
when h=0; P = P0
K  log P0 
Mg
log T0
R
Substituting this value of K in equation (5),
log P  
Mg
Mg
logT0  h  log P0 
log T0
R
R
(5)
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log
 T  dh 
Mg
P


log 0

P0
R
 T0 
 T  h
Mg
log  0
R
T0

P  P0 e
i.e,




(6)
(7)
1.6 CALCULATION OF LAPSE RATE (For Dry Adiabatic):
The relation between pressure and temperature for an adiabatic process is,
p  1
 constant
T
 1

or P T  constant
Differentiating
(-1) P

Dividing by T– P
-2

dP .T– + P
 -1

(-) T- -1dT = 0
 -1
  1 dp     dT
P
T
0
dp
 dt

p  1 t
(1)
For an increase in height dh, the decrease in pressure in dp,
dP    g dh  
But
PV = RT
dP  
or
MP
g dh
RT
M
g dh
V
 RT 
V 

 P 
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dP
Mg

dh
p
RT
(2)
Equating (1) and (2)
   dT
Mg



dh

1
T
RT


dT
Mg    1 



dh
R   
(3)
The above equation holds good only for perfect dry air. The lapse rate (dT/dR) can be calculated
from equation (3),
Substituting the values,
M=29g; g=981cm/sec2; R=8.31*107erg/grammole/K; =1.4 (ratio of specific heat)
dT
29  981  1.40  1 



dh
8.31 10 7  1.40 
=10-4 Kper cm
or
10-4 C/cm
Thus the lapse rate for a height of,
1Km=10-4105=100C
But it has been found that the average lapse rate is lower than this value. Under average
conditions the lapse rate has been found to be between 50 C and 6.5 0C per km. At night negative
lapse rate is set up because the layers of air nearer the surface may cool.
1.7 LET US SUM UP
In this lesson you have learned about the coefficient of thermal conductivity and determination
of thermal conductivity for a good conductor and a bad conductor. Also in this lesson you have
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learned about the change of pressure with height in the atmosphere with its relation to Lapse rate
and its calculations
1.8 CHECK YOUR PROGRESS
1. Define thermal conductivity
2. What is the reason for the atmospheric changes in convection currents ?
3. Define Lapse rate
1.9 LESSON END ACTIVITIES
1. The opposite faces of a metal plate of 0.2 cm thickness are at different of temperature at
100oC and the area of the plate is 100 sq. m. Find the quantity of heat that will flow through the
plate in one minute if k=0.2 CGS units.
[ Hint: Q = (KA(1-2)t)/d ]
2. A room is maintained at 20oC by a heater of resistance 20 ohms connected to 200 volts mains.
The temperature is uniform throughout the room and the heat is transmitted through a glass
window area 1 m2 and thickness 0.2 cm. Calculate the temperature outside. K=0.2 cal /m-Co-s.
[ Hint: H = (KA(1-2)t)/d where H = (v2) /4.2R ]
1.10 POINTS FOR DISCUSSIONS
(1)
(i) Give the theory of cylindrical flow of heat.
(ii) Describe Lee’s disc method to find the coefficient of thermal conductivity of a
.
(2)
poor conductor.
(i) Explain coefficient of thermal conductivity. What is temperature gradient?
(ii) Describe with a neat diagram explain how you would determine the thermal
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conductivity of rubber.
1.11 REFERENCES
(1) Heat and Thermodynamics by Brij-lal and Subramanyam
(2) Heat and Thermodynamics by Anantha Krishnan
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UNIT-I
LESSON – 2
CONTENTS
2.0 Aims and Objectives
2.1 Stability of the Atmosphere
2.2 Green House Gases
2.3 Newton’s Law of Cooling
2.4 Black Body
2.5 Wein’s Displacement Law
2.6 Let us Sum up
2.7 Check your progress
2.8 Lesson end Activities
2.9 Points for Discussion
2.10 References
2.0 AIMS AND OBJECTIVES
In this lesson you will learn about the stability of atmosphere, troposphere, and
stratosphere. Also you will understand about green house gases, green house effect,
Newton’s law of cooling. You will learn about what is a black body and Wien’s
displacement law.
2.1 STABILITY OF THE ATMOSPHERE:
The atmospheric temperature decreases with altitude upto a height of about 20km. The
rate of fall of temperature is about 50 C per km height and is known as the lapse rate.
The atmosphere is divided into two regions,
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(1) Troposphere or the convective zone:
It is the region in which the temperature falls with increase in height from the ground.
(2) Stratosphere or advective zone:
It is the region in which the decrease in temperature does not take place with increase in
height.
The layer that separates these two regions is known as tropopause. The height of tropopause
is different at different places on the earth. It is about 14 km at the equator and about 10 km at
the poles. The fall in temperature with altitude in the troposphere is due to convection. When the
heat radiations from the sun fall on the fall on the surface of the earth, it gets heated. The
atmospheric air surrounding the earth gets heated by conduction and radiation. The heated air
moves up and convection currents are set up. The heated air that moves up from the regions of
higher to lower pressure is adiabatically cooled (ie it can neither give heat nor take heat from the
surroundings). Similarly when the colder air moves down from lower to higher pressure regions
it is adiabatically heated. Thus a convective equilibrium is set up and there is a gradual fall of
temperature with increase in height.
2.2 GREEN HOUSE GASES:
Sunlight warms the earth’s surface during day time and earth’s surface radiates heat back
to the space. Certain gases in the atmosphere absorb this radiant energy and re-emit the heat back
to the space. Those gases which are capable of absorbing and re-emitting the heat radiation are
called green house gases. The heat from the surface of the earth warms up the green house gases.
These gases in turn emit some heat into space and some back down to the surface. This fraction
of the heat provides global warming in addition to the sun’s direct heat. Without any green house
gases in the atmosphere, the average surface temperature would be very cold ie., around -18 0C.
Today’s greenhouse gases radiate sufficient heat beck to earth to give an average global
temperature of +15 0C. In if future concentration of greenhouse gases increases, there will be
additional global warming. It is therefore necessary to maintain the present level of greenhouse
gases to avoid any drastic change in the climate.
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Greenhouse effect:
The heating up of earth’s atmosphere due to infrared rays which are reflected from the
earth’s surface by the carbon-di-oxide layer in the atmosphere is called greenhouse effect.
If green house gases continue to increase at the present ratio, it is predicted that the
earth’s average surface temperature may go up by 1 to 5 0C, before the end of twenty first
century. This global warming will not be evenly distributed ie., the equatorial region will warm
by 1 0C to 2 0C, the polar regions will warm up most rapidly to an increase of 6 0C to 120C.
Global warming will have two major effects on climate,
I. The polar ice caps will begin to melt and the sea level throughout the world will increase.
This will cause many low-lying coastal regions to submerge. As the polar ice melts, less
sunlight will be reflected back from the snow or more sunlight will be absorbed which will
contribute to the additional global warming.
II. In a warmer world, more water will be returned to the atmosphere by evaporation and
transpiration. There will be an increase in the total rainfall accompanied by stronger wind in
the equatorial region. The present snow regions will have less snow in winter and more heat
in summer. The vegetation may go dry.
2.3 NEWTON’S LAW OF COOLING:
Newton’s law of cooling states that the rate of cooling of a body is directly proportional to the
difference in temperature between the body and the surroundings. By rate of cooling is meant the
amount of heat radiated by the body per second. For any body, the rate of cooling is directly
proportional to the fall in temperature per second. Thus, if a hot body at a temperature T
1
0
C
cools to T2 0C in t seconds, when kept in a surrounding at a temperature T 0C, the fall in
temperature per second.
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 T  T2 
 1

 t 
The mean difference in temperature between the body and the surrounding,
 T  T2 
 1
T
 2 
From Newton’s law it follows that when the body cools from T 1 0C to T2 0C in t seconds,
 T  T2   T1  T2 
 1

 T
 t   2 
Specific Heat of a Liquid by the Method of Cooling:
A calorimeter is weighed empty with a stirrer. It is filled to 2/3 of its capacity with hot water
at about 60 0C. It is suspended inside an enclosure maintained at a constant temperature and a
thermometer is placed in it. The calorimeter is kept stirred and the time taken for the calorimeter
to cool from T
1
0
C to T2 0C (from 55 0C to 50 0C) is found. The calorimeter is cooled to the
room temperature and weighed. The experiment is repeated by filling the calorimeter with the
liquid up to the same level and finding the time taken to cool through the same range of
temperature. The calorimeter is weighed with the liquid. The specific heat of the liquid is
calculated as follows:
Weight of calorimeter and stirrer = w1 gm.
Time taken to cool from T 1 0C to T2 0C with water = t1 sec.
Weight of calorimeter, stirrer and water = w2 gm.
Time taken to cool from T 1 0C to T2 0C with liquid = T2 gm.
Weight of calorimeter, stirrer and water =w3 gm
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Let S1 be the specific heat of the calorimeter and S the specific heat of the liquid.
Rate of cooling of the calorimeter when filled with water,

w1 S1  w2  w1 T1  T2 
t1
Rate of cooling of the calorimeter when filled with liquid,

w S  w
1
1
3
 w1 S T1  T2 
t2
The rate of cooling is the same in both the cases since the mean difference in temperature
between the calorimeter and surrounding is the same in both cases. Therefore

w S
1
1
 w2  w1 T1  T2 
t1
w S
1
1
 w2  w1 
t1


w S  w
1
1
w S  w
1
1
3
3
 w1 S T1  T2 
t2
 w1 S 
t2
from which S is calculated.
2.4 BLACK BODY:
A perfectly blackbody is one which absorbs all the heat radiations (corresponding to all
wavelengths) incident on it. When such a body is placed inside an isothermal enclosure (if a
system is perfectly conducting to the surroundings and the temperature remains constant
throughout the process, it is called isothermal process) it will emit the full radiation of the
enclosure after it is in equilibrium with the enclosure. These radiations are independent of the
nature of the substance. Such radiations in a uniform temperature enclosure are known as
blackbody radiations. Also blackbody completely absorbs heat radiations of all wavelengths.
22
Hence we can say that the blackbody also emits completely the radiations of all wavelengths at
that temperature. It is a good absorber and emitter.
Incident radiations
hole
Black body absorber
Fig.1.9a
Illustrations: Absorber
A hollow copper sphere in taken and is coated with lamp black in its inner surface. A fine hole
is made and a projection is made just in front of the hole. When the radiations enter the hole,
they suffer multiple reflections and are completely absorbed. This body acts as a blackbody
absorber.
Emitter:
Heat
Radiations
Black body emitter
Fig.1.9b
When this body is placed in a bath at a fixed temperature, the heat radiations come out of the
hole. The hole acts as a blackbody radiator or emitter. It should be noted that only the hole and
not the walls of the body acts as the blackbody radiator.
23
Perfectly blackbody:
It is one which absorbs all the heat radiations incident on it. It does not reflect or transmit any
heat radiations. Irrespective of the colour of the incident radiations, the body appears black.
When such a body is heated to a high temperature, it emits the radiations of all wavelengths. The
radiations emitted by a perfectly blackbody depend only upon the temperature of the blackbody
and not on the nature of its material.
Absorptive and Emissive Powers:
Absorptive power:
It is the ratio of the amount of heat absorbed in a given time by the surface to the amount
of heat incident on the surface in the same time.
Emissive power:
It is the ration of the amount of heat radiations emitted by unit area of a surface in one
second to the amount of heat radiated by a perfectly blackbody of unit area in one second under
ideal conditions.
2.5 WEIN’S DISPLACEMENT LAW:
It states that the product of the wavelength corresponding to maximum energy and
absolute temperature is constant.
max T = Constant =0.2892 cm-k
It also shows that with increase in temperature m decreases. Wein has shown that the energy
E max is directly proportional to the fifth power of the absolute temperature.
ie., E max  T5
24
E max = Constant  T5
Rayleigh – Jean’s law:
The energy distribution in the thermal spectrum according to Rayleigh is given by,
E 
8KT
4
Where K is the Boltzmann’s constant
Note: Wein’s law holds good only in the region of shorter wavelengths. It does not holds good at
longer wavelengths. The Rayleigh – Jean’s law holds good in the region of larger wavelengths
but not for shorter wavelengths.
2.6 LET US SUM UP
On going through this lesson you will understand about the stability of the atmosphere, green
house gases that are responsible for green house effect. Also you will come to know about the
Newton’s law of cooling, the sources of black body radiation and the connecting law namely
Wien’s displacement law.
2.7 CHECK YOUR PROGRESS
1. What are green house gases ?
2. State Wien’s displacement law
3. State Newton’s law of cooling
4. What is black body radiation ?
2.8 LESSON END ACTIVITIES
1. Calculate the value of the surface temperature of the moon using Wein’s displacement
law.
[Hint: m  = 2892 x 10-6 m-k, Given m = 14.46 x 10-6 m]
2. Calculate the radiant emittance of a black body at a temperature (i) 400 K (ii) 4000 K.
( = 5.67210-8 M.K.S units)
25
2.9 POINTS FOR DISCUSSION
1. Write short notes on,
(i)
Black body radiation
(ii)
Temperature of the sun.
2. How will you determine the specific heat of liquid by Newton’s law of cooling.
2.10 REFERENCES
1
Heat and Thermodynamics by Brij-lal and Subramanyam
2
Heat and Thermodynamics by Anantha Krishnan
3
Heat and Thermodynamics by R. Murugesan
26
LESSON – 3
CONTENTS
3.0 Aims and Objectives
3.1 Stefan’s Law – Also Called as Stefan – Boltzmann Law
3.2 Determination of Stefan’s Constant (Laboratory Method)
3.3 Solar Constant
3.4 Solar Spectrum
3.5 Temperature of Sun
3.6 Let us Sum up
3.7 Check your progress3
3.8 Lesson end Activities
3.9 Points for Discussion
3.10 References
3.0 AIMS AND OBJECTIVES
In this lesson another important law related to black body radiation namely Stefan’s law will be
discussed. The derivation of it and its determination by experimental method will also be
detailed. Followed by it the Solar constant, solar spectrum and temperature of the sun will be
discussed.
3.1 STEFAN’S LAW – ALSO CALLED AS STEFAN – BOLTZMANN LAW:
According to this law, the rate of emission of radiant energy by unit area of a perfectly
blackbody is directly proportional to the fourth power of its absolute temperature,
R  T4 or R =  T4
(1)
where  is called Stefan’s constant.
If the body is not perfectly black and its emmissivity or relative emittence is e, then
R = e  T4
(2)
27
Here e varies between zero and one, depending on the nature of the surface. For a perfectly
blackbody e = 1. This law is not only true for emission but also for absorption of radiant energy.
When the body has the same temperature as that of the surroundings, the rate of emission and
absorption are equal.
Hence, if a perfectly blackbody at temperature T1 is surrounded by a wall at temperature T2,
the net rate of loss (or gain) of heat energy per unit area of the surface is given by,
R  (T14 - T24)
R =  (T14 - T24)
If the body gas an emissivity e then,
R = e  (T14 - T24)
Mathematical derivation of Stefan’s law:
The fact that blackbody radiations exert pressure similar to a gas, helps in applying
thermodynamics to heat radiations.
Let  be the energy density of radiations inside a uniform temperature enclosure at
temperature T. P is the pressure and V is the volume.
Applying the first law of thermodynamics,
 H = du + P.dV
(1)
Where  H is the quantity of heat supplied to a system, the amount of external work done be
P.dV and the increase in internal energy of the molecules be dV.
Applying thermodynamical relation,
 H 
 P 

 

 V  T  T  V
(2)
28
 U  PV 
 P 

  T

V

T
 T V
 U 
 P 

  T
 P
 V  T
 T V
U  V  P 
(3)

3
 U 

 
 V  T
or
Here  is a function of temperature alone.
Substituting these values in equation (3),
 
T d 

3 dt
3
4 T d

3
3 dt
d
dT
4

T
integrating log = 4logT + constant
  KT 4
or
(4)
Where K is a constant.
Also the total rate of emission per unit area of a blackbody is proportional to the energy
dentsity.
29
R    T4
R =  T4
(5)
Where  is Stefan’s constant.
Note: The value of Stefan’s constant = 5.672 x 10-8 in MKS units
3.2 DETERMINATION OF STEFAN’S CONSTANT (Laboratory Method):
The laboratory apparatus used to determine the Stefan’s constant is shown in Fig 1.12. A
hollow hemispherical metallic vessel A is enclosed in a wooden box W. The inner surface of A is
coated with lamp black and wooden box W is lined with tin plates. The whole apparatus is
placed on a wooden base having a small hole at its centre. The vessel A is heated by passing
steam inside the box and A acts as a black body radiator. The thermometers T , T record the
temperature of A.
T
T
Steam
W
A
C
B
H
G
Rh
E
F
Fig.1.12a
30
A small silver disc B whose upper surface is coated with lamp black is placed at the central hole.
The ebonite covering C is used to cover and uncover the disc B from the radiations of the
enclosure. It can be arranged from outside with the help of the handle H. The disc B is connected
to a thermocouple arrangement. One junction of the thermocouple is immersed in a tube
containing oil. The tube is surrounded by a beaker containing water. A sensitive galvanometer G
is used in the circuit. The leads connected to the terminals of the galvanometer are immersed in
cotton wool in the box F to avoid any distribution effect due to the difference of temperature in
the leads. A rheostat Rh can be used in the circuit to obtain the deflection within the range. The
actual experiment consists of two parts.
1. The thermocouple is first standardized. Before passing steam into the chamber the disc B is
at the room temperature.
A
α
B
C
Deflection
Fig1.12b
The water bath E acts as a hot junction. It is heated and at various temperatures of the hot
junction, the corresponding deflections in the galvanometer are noted. A graph between the
31
difference of temperature of the hot junction and the room temperature along the Y-axis and
galvanometer deflection along X-axis is plotted in fig 1.12b. From the graph,
dT
AB
 tan  
d
BC
(i)
2. The disc is completely covered with C and steam is passed into the chamber. After some
time, the thermometers T. T show constant temperature. The bath E is kept at room temperature.
With the help of the handle H, the cover C is tilted so that the upper surface of the disc B
receives the radiations from the enclosure. The deflections in the galvanometer are observed after
equal intervals of time (say 10 seconds). A graph is plotted between time and deflection
(fig1.12c). A tangent is drawn on the curve at a point D.
dt
EF
 tan  
d
GF
(ii)
Theory:
Let, at any instant, the temperature of the enclosure and the disc be T1 and T2 (degrees
Kelvin) respectively. The disc will absorb more heat from the surroundings and radiate less heat
to the surroundings. Its temperature will rise. From Stefan’s law,
R1 = T14 and R2 = T24
(R1 - R2) = ( T14 - T24)
(iii)
E
Time
(sec)
D
F
G
Deflection
Fig.1.12c

32
Here R1 is the amount of heat radiation absorbed per unit area per second by the disc and R2 is
the amount of heat radiation emitted per unit area per second by the disc.
Let the mass of the disc be m, specific heat S, rate of rise of temperature dT/dt, and area of the
upper surface A.
Then
R
1

 R 2 A
dT
 mS
J
dt

4
 T1  T 4 2 A
dT
 mS
J
dt

JmS

4
A T1  T 4 2
dT
 dt
(iv)
To find (dT/dt), equations (i) and (ii) are used
dT dT d tan 



dt d dt tan 
To find T2, the deflection in the galvanometer corresponding to the point D on graph in fig
1.12c is noted and for this deflection, the temperature difference from the graph fig 1.12bis
noted. To this reading add the room temperature and find T2 in degrees Kelvin.
Substituting these values in equation (iv),
33

JmS tan 

4

A T1  T 4 2 tan 
(v)
Hence  can be calculated.
3.3 SOLAR CONSTANT:
The sun is the source of heat radiations and it emits heat radiations in all directions. The
earth receives only a fraction of the energy emitted by the sun. The atmosphere also absorbs a
part of the heat radiations and air, clouds, dust particles etc. in the atmosphere scatter the heat
and light radiations falling on them. From the quantity of heat radiations received by the earth, it
is possible to estimate the temperature of the sun. Therefore, to determine the value of a constant,
called solar constant, certain ideal conditions are taken into consideration.
Solar constant:
It is the amount of the heat energy (radiation) absorbed per minute by one sq cm of a perfectly
black body surface placed at a mean distance of the earth from the sun, in the absence of the
atmosphere, the surface being held perpendicular to the sun’s rays.
The instruments used to measure the solar constant are called pyrheliometers. The heat energy
absorbed by a known area in a fixed time is found with the help of the pyrheliometers. To
eliminate the effects of absorption by the atmosphere, the value of the solar constant is found at
various altitudes of the sun on the same day under similar sky conditions. If S is the observed
solar constant, S0 the true solar constant and Z the altitude (angular elevation) of the sun, then
S = S0 αsec Z
or
log S = log S0 + sec Zlog α
(i)
(ii)
34
Here α is a constant.
A graph is platted between log S along the y-axis and sec Z along the x-axis. The graph is a
straight line. Produce the graph to meet the y-axis. The intercept on the y-axis gives log S0, the
value of S0 the solar constant can be calculated. The value obtained varies between 1.90 and 2.60
calories per sq cm per minute.
3.4 SOLAR SPECTRUM:
The radiation received from the sun is similar to that of a perfectly black body. In solar
spectrum the wavelength corresponding to maximum energy is about 5000 A for a surface
temperature of 5750 K. The solar spectrum consists of a large number of dark lines called
Franunhofer lines. These lines were first observed by Wollaston in 1802 and later studied by
Fraunhofer in 1814. These lines are observed in the complete range of the spectrum viz. ultraviolet, visible and infrared. Fraunhofer measured the wavelengths of many of these lines
accurately and found that they occupied exactly the same positions as the bright lines emitted by
different gases and vapours.
light source
M3
M2
P
M3
Fig.8.14a
Infra-red spectrometer
Thermopile
35
According to Kirchhoff’s law of radiation, any substance at a lower temperature will absorb the
radiations of those wavelengths which it will emit when excited by electric discharge. This the
solar spectrum is an absorption spectrum and the Fraunhofer lines are the absorption lines of
relatively cooler gases and vapours present in the earth’s atmosphere and the sun’s outer
atmosphere. The central portion of the sun is called the photosphere which is at a very high
temperature. Surrounding the photosphere is the chromosphere which is at a much lower
temperature than the photosphere. The positions of the Fraunhofer lines in the solar spectrum
give the atmosphere. The presence of more than sixty elements in the sun’s atmosphere is found
this way. In fact, helium was first discovered by the study of the Fraunhofer lines in the solar
spectrum before it had been isolated in the laboratory. The most prominent of the Fraunhofer
lines are denoted by the letters of the alphabet. Some of the lines, with their wavelengths and the
elements responsible for their absorption are given below:
Line
Element
Wavelength in Å
A
Atmospheric oxygen
7594
B
Atmospheric oxygen
6867
C
Hydrogen
6563
D1
Sodium
5896
D2
Sodium
5890
F
Hydrogen
4861
G1
Hydrogen
4341
H
Calcium
3969
K
Calcium
3934
Infra-red-Spectrum:
Extending on either side of the visible spectrum, there are invisible radiations which do not
cause the sensation of sight. The radiations beyond the red end of the visible spectrum are called
infra-red radiations and their wavelengths extend up to 400,000 Å. Sun is a powerful source of
36
infra-red radiation. Beyond the violet end of the visible spectrum, the radiations extending up to
a wavelength of 100 Å are called ultraviolet radiations.
The heating effect of the infra-red radiations is used in measuring the wavelength of the
radiation. An infra-red ray spectrometer is shown in Fig 1.14(a). Light from a strong source of
light such as an electric arc is rendered parallel by reflection from a concave stainless steel
mirror M1. This parallel beam is refracted through the rock salt prism P and the emergent
dispersed beam after reflection from the mirrors M1 and M3 is incident on a thermopile or a
bolometer. M2 is a plane mirror and M3 is a concave mirror. Wien’s law of radiation is given by,
mT = 0.2892
where T is the absolute temperature and m is the wavelength of the radiation. The thermopile
readings help in the calculation of temperature and from the temperature,  can be calculated.
As infra-red radiations are not absorbed by air or thick frog, infra-red ray photographs can be
taken over long distances of fog and mist where visible light cannot penetrate. For this, specially
designed photographic plates are used with suitable filters. A solution of iodine in alcohol is a
suitable filter, because this transmits the infra-red radiations and absorbs the visible light. In
World War II, infra-red photography played a very useful part in detecting objects in the dark
through mist, fog and clouds. Infra-red radiations have a wide application in the field of
medicine, industry etc. Infra-red radiations can penetrate deep into the human body and by their
property of heating can dilate the blood vessel at the portion exposed to the radiations. This
enables increased flow of blood.
Ultra-violet Spectrum:
The spectrum that covers the wavelengths from 4000 Å to 100 Å is called the ultra-violet
spectrum. An electric arc of carbon, iron or other materials, mercury vapour lamps, discharge of
electricity through hydrogen contained in quartz tubes are some of the artificial sources that give
37
ultra-violet radiations.
Ordinary glass absorbs the ultra-violet radiations. Hence quartz lenses
and prisms are used. But quartz is double refracting material. If a single prism of quartz is used,
due to the property of double refraction of the prism, two images of the prism, two images of the
slit (ordinary and extraordinary) corresponding to a single wavelength are observed. This will
reduce the sharpness of each image. To compensate for this, the collimating lens is
A
R
L
B
C
Fig1.14
made from right-handed quartz and the telescope objective is made from left-handed quartz.
Similarly the prism used for the dispersion of the incident beam consists of two halves. The two
halves are held together by glycerine. One half is made from right-handed quartz and the other
half is made from left-handed quartz.
In fig 1.14(b) ABD is a right-angled prism of right-handed quartz and ADC is of left-handed
quartz. For recording an ultra-violet spectrum for wavelengths shorter than 1,200 Å, a concave
reflection grating is used. The diffracted beam is photographed. The grating and the
photographic plate are enclosed in a metal chamber which is evacuated.
Ultra-violet radiations have a variety of applications. Sterilisation of rooms in which blood
plasma, drugs, vaccines etc. are prepared and sealed in the containers is done by ultra-violet
radiations. Drugs, poisons, dyes etc. fluoresce under the action of ultra-violet rays. The resolving
38
power of a microscope is increased when ultra-violet light is used for illumination. Fluorescent
tubes depend on the principle of fluorescence effected by ultra-violet radiations.
Electromagnetic Spectrum:
Visible spectrum includes those wavelengths which can stimulate the sense of sight. But there
is no basic difference between light waves and electromagnetic waves produced by electrical
oscillating circuits. The term electromagnetic spectrum is used for the range of wavelengths from
104 meters to 1 Å (10-8cm). There is in fact no limit to the production of electromagnetic waves
of very long wavelengths. The frequency of an alternating current generator can be made as low
as possible by decreasing the speed of the generator. The wavelength of waves transmitted by a
50 cycle transmission line is 5x108 cm. Waves of shorter wavelength can be produced by
electrical oscillators. X-rays and gamma rays represent the waves of very short lengths.
It is interesting to note that the visible range of the spectrum comprises only a small range of
the electromagnetic spectrum extending approximately from 4000 Å in the extreme violet region
to 8000 Å in the extreme red. Beyond the violet region of the visible spectrum is the ultraviolet,
the X-rays and γ-rays.
3.5 TEMPERATURE OF THE SUN:
The sun consists of a central hot portion surrounded by the photosphere. The central portion
has a temperature of the order of 107 K. The photosphere has a temperature of about 6000 K.
This temperature is also called the effective temperature of the sun. Considering the sun as a
perfect black body radiator, the temperature of the sun can be calculated.
Let the mean distance of the sun from the earth be R and S the solar constant. Then, the total
amount of heat energy received by the sphere of radius R in one minute = 4R2S.
39
If r is the radius of the sun, then the amount of the heat energy radiated by 1 sq cm surface of
the sun in one minute,
E = (4R2S / 4r2) = (R / r)2 x S
Taking,
R = 148.48x107 km
r = 6.928x105km
The mean value of
S = 1.94 cals per cm3 per minute.
E = (148.48x107 / 6.928x105)2 x (1.94 / 60) cals per sec -----(i)
Also,
E =  T4
But
 = 5.75x10-5 ergs per cm2 per second
= (5.75x10-5 / 4.2x107) cal per cm2 per second
E = (5.75x10-5 / 4.2x107) . T4 -----(ii)
Equating (i) and (ii),
(5.75x10-5 / 4.2x107) . T4 = (148.48x107 / 6.928x105)2 x (1.94 / 60)
T4 = 5730 K
-----(iii)
This temperature gives the effective temperature of the sun acting as a black body radiator.
The actual temperature of the sun is higher than this value. The temperature of the sun is usually
taken as 6000 K.
Temperature of the sun can also be calculated from Wien’s displacement law,
40
max T = 0.2892
The wavelength of the radiations for the radiations for which the energy is maximum in the
spectrum is 4900x10-8 cm.
Substituting the value of m, the value of T comes out to be 5902 K. This value is in
agreement with the accepted value. Hence, the effective temperature of the sun (photosphere) is
about 6000 K.
3.6 LET US SUM UP
In this lesson an important law viz., Stefan Boltzman law is discussed and its derivation and
determination is explained. Also you learned about solar constant and details of various spectrum
such as solar spectrum, infrared, ultraviolet spectrum. The topic of solar constant, and a method
to obtain temperature of the sun is also described in this lesson.
3.7 CHECK YOUR PROGRESS
1. State Stefan Boltzman law
2. Define Solar Constant
3. Write a note on Solar spectrum
3.8 LESSON END ACTIVITIES
1. Calculate the radiant emittance of a black body at a temperature of i) 400 K and ii) 4000 K
using Stefan’s law.
[Hint: R = T4,  = 5.67 x 10-8 MKS units]
2. If a solar constant at the surface of the earth is 1400 W m-2, compute its value at Jupiter which
is 5.2 AU away from the sun.
[Hint: S2/S1 = (R1/R2)2, Here R1 = 1 AU ]
41
3. The total luminosity of the sun is 3.9 x 1026 watts. The mean distance of the sun from the earth
is 1.496 x 1011 m. Calculate the value of solar constant.
[Hint: S = (E/4R2) ]
3.9 POINTS FOR DISCUSSIONS
(1)
(i) State Stefan’s law of heat radiation and derive the law mathematically.
(ii) Describe an experiment for the verification of Stefan’s constant. What is its
numerical value in M.K.S units?
(2) Define solar constant. How is it experimentally determined? Comment on the source
of energy in the sun.
3.10 REFERENCES
(1) Heat and Thermodynamics by Brijlal and Subramanyam
(2) Treatise on Heat by Srivastava
42
************************************************************************
Annexure-I
************************************************************************
The given cardboard is placed between the disc Lee’s and the steam chamber. Two
thermometers are inserted into the radial holes drilled on the side of the metallic discs. Steam is
then passed through the steam chamber from a boiler until a steady state is reached. The steady
temperatures of the disc 1 and steam chamber 2 recorded by the thermometers are noted.
The cardboard is then removed and the disc heated directly until the temperature rises 5C above
1. Then the steam chamber is removed, and the disc is allowed to cool. A stop clock is started
and the time for every 1C fall in temperature of the disc is noted until the temperature of the disc
falls 5C below 1. The values are tabulated.
A cooling curve is drawn with time along x-axis and the temperature along the y-axis as shown
in figure below,
rom these readings, thermal conductivity of the bad conductor is calculated by usin equation.
Observation:
Mass of the disc
Specific heat of the material of the disc,
M=…kg
s=…Jkg-1K-1
43
Radius of the disc,
r=…m
Thickness of the disc,
l=…m
Thickness of the bad conductor,
d=…m
The steady temperatures of the disc,
1=…m
The steady temperatures of the steam chamber,
2=…m
Rate of cooling at 1 [from the cooling curve],
(d/dt) =Ks-1
Results:
The coefficient of the given thermal conductivity of the bad conductor
K…………Wm-1K-1
44
QUESTIONS
(1)
(i) Give the theory of cylindrical flow of heat.
(ii) Describe Lee’s disc method to find the coefficient of thermal conductivity of a
.
(2)
bad conductor.
(i) Explain coefficient of thermal conductivity. What is temperature gradiant?
(ii) Describe with a neat diagram explain how you would determine the thermal
conductivity of rubber.
(3)
(i) State Stefan’s law of heat radiation and derive the law mathematically.
(ii) Describe an experiment for the verification of Stefan’s constant. What is its
numerical value in M.K.S units?
(4) Define solar constant. How is it experimentally determined? Comment on the source
of energy in the sun.
(5)
Write short notes on,
(iii)
Black body radiation
(iv)
Solar spectrum.
(v)
Temperature of the sun.
45
PROBLEMS
(1) Two sides of a square copper plates of side 15cm and of thickness 0.5cm are kept at 0o C
and 100o C. Calculate the quantity of heat conducted through the plate per hour. The
thermal conductivity of copper is 0.9.
Solution:
 2 
Q  KA 1
t
 d 
2
 15   100   
0.9  
 3600  4186
 
2 
 100   0.5  10 
= 6.1103
(2) Calculate the value of the surface temperature of the moon using Wein’s displacement
law. (Value of m = 14.46 micron).
Wein’s displacement law m
= constant
The value of constant = 289210-6M.K
 = temperature to be calculated
14.4610-6 = 289210-6
 = 200K
(3) Calculate the radiant emittance of a black body at a temperature (i) 400 K (ii) 4000 K.
( = 5.67210-8 M.K.S units)
Solution:
R =  T4
i) R = 5.762  10-8[400]4 = 1452 watts/m2
ii) R = 5.762 10-8[4000]4 = 14520 Kilo.watts/m2
46
(4) Calculate the energy radiated per minute from the filament of an incandescent lamp at
2000 K, if the surface area is 5.010-5 m2
and its relative emittance is 0.85.[  = 5.67210-8]
Solution:
E = A e  t (T4)
Here A = Area; e= relative emittance; t=time
E = 5  10-5  0.85 5.672 10-8 60  (2000)4
E = 2315 Joules
47
UNIT – II
LESSON – 4
CONTENTS
4.0
Aims and Objectives
4.1
First Law of Thermodynamics.
4.1 (a) First Law of Thermodynamics for a Change in State of a Closed Systems
4.2
Isothermal Process
4.2(a) Adiabatic Process.
4.2(b) Isochoric Process
4.3
Gas Equation During an Adiabatic Process
4.3 (a) Slopes of Adiabatic and Isothermals
4.3 (b) Work done During an Isothermal Process
4.3 (c) Work done during and Adiabatic Process.
4.4
Let us Sum up
4.5
Check your progress
4.6
Lesson end activities
4.7
Points for discussion
4.8
References
4.0 AIMS AND OBJECTIVES
In this lesson you will learn the first law of thermo dynamics, adiabatic and isothermal processes
and also will learn the work done during adiabatic and isothermal processes.
4.1 First Law of Thermodynamics.
Joule’s law gives the relation between the work done and the heat produced. It is true
when the whole of the work done is used in producing heat or vice varsa. Here, W= JH where J
is the Joule’s mechanical equivalent of heat. But in practice, when a certain quantity of heat is
supplied to a system the whole of the heat energy may not be converted into work. Part of the
heat may be used in doing external work and the rest of the heat might be used in increasing the
48
internal energy of the molecules. Let the quantity of heat supplied to a system be δH, the amount
of external work done be δW and the increase in internal energy of the molecules be dU. The
term U represents the internal energy of a gas due to molecular agitation as well as due to the
forces of inter-molecular attraction. Mathematically.
δH = dU + δW
….(i)
Equation (i) represents the first law of thermodynamics. All the quantities are measured
in heat units. The first law of thermodynamics states that the amount of heat given to a systems is
equal to the sum of the increase in the internal energy of the system and the external work done.
For a cyclic process, the change in the internal energy of the system is zero because the
system is brought back to the original condition. Therefore for a cyclic process  dU = 0
 δH
and
=  δw
…(ii)
This equation represents Joule’s law. (Both are expressed in heat units).
For a system carried through a cyclic process, its initial and final internal energies are
equal. From the first law of thermodynamics, or a system undergoing any number of complete
cycles.
U2 – U1 = 0

δH
=  δW
H
= W (both are in the heat units)
4.1 (a) First Law of Thermodynamics for a Change in State of a Closed Systems
For a closed system during a complete cycle, the first law of the Thermodynamics in
 δH =
 δW
49
In Practice, however, we are also concerned with a process rather than a cycle. Let the system
undergo a cycle, changing its state from 1 to 2 along the path A and from 2 to 1 along the path B.
This cyclic process is represented in the P-V diagram (Fig. 2.1a).
 δH =
 Δw
Fig. 2.1 (a)
For the complete cyclic process
2A
1B
 δH +  δH =
1A
2B
2A
1B
 δW +  δW
1A
2B
…(i)
Now, consider the second cycle in which the system changes from state 1 to state 2 along the
path A and returns from state 2 to state 1 along the path C. For this cyclic process.
2A
 δH +
1A
1C
2A
 δH =  δW
2C
1A
Subtracting (ii) from (i)
1C
+  δW
2C
…(ii)
50
1B
1C
 δH -  δH =
2B
2C
(or)
1B
 (δH - δW)
2B
2B
1C
 δW +  δW
2B
2C
1C
=  (δH 2C
δW)
…(iii)
Here B and C represent arbitrary process between the states 1 and 2. Therefore, it can be
concluded that the quantity (δH - δW ) is the same for all processes between the states 1 and 2.
The quantity (δH-δW) depends only on the initial and the final states of the system and is
independent of the path followed between the two states.
dE = (δH – δW)
Let
From the above logic, it can be seen that
dE = constant and is independent of the path
This naturally suggests that E is a point function and dE is an exact differential.
The point function E is a property of the system.
Here dE is the derivative of E and it is an exact differential.
or
δH - δW – dE
--- (iv)
δH = dE + δW
--- (v)
Integrating equation (v), from the initial state 1 to the final state 2
1H2
= (E2 – E1) + 1W2
(Note. 1H2 cannot be written as (H2 – H1), because it depends upon the path).
Similarly, 1W2 cannot be written as (W2 – W1), because it also depends upon the path.
Here
1H2
1W2
represents the heat transferred,
represents the work done,
51
E2 represents the total energy of the system in state 2,
E1 represents the total energy of the system in state 1,
At this point, it is worthwhile discussing what this E can possibly mean. With reference
to the system, the energies crossing the boundaries are all taken care of in the form of H and W.
For dimensional stability of Eq. (v), this E must be energy and this must belong to the system.
Therefore, E2 represents the energy of the system in state 2
E1 represents the total energy of the system in state 1
This energy E acquires a value at any given equilibrium condition by virtue of its
thermodynamic state. The working substance for example a gas, has molecules moving in all
random fashion. The molecules have energy associated by virtue of mutual attraction and this
part is similar to the potential energy of a body in macroscopic terms. They also have velocities
and hence kinetic energy. This energy E therefore can be visualized as comprising of molecular
potential and kinetic energies in addition to macroscopic potential and kinetic energies. The first
part, which owes its existence to the thermodynamic nature is often called the internal energy
which is completely dependent on the thermodynamic state and the other two depend on
mechanical or physical state of the system.
E = U + KE + PE + others which depend upon chemical nature etc.
For a closed system (non – chemical) the changes in all others except U are insignificant
and
dE = dU
From equation (v)
δH = dU + δW
Here all the quantities are in consistent units.
52
4.2
Isothermal Process
If a system is perfectly conducting to the surroundings and the temperature remains
constant throughout the process, it is called an isothermal process. Consider a working substance
at a certain pressure and temperature and having a volume represented by the point A (Fig.2.2)
Fig. 2.2
Pressure is decreased and work is done by the working substance at the cost of its internal
energy and there should be fall in temperature. But, the system is perfectly conducting to the
surroundings. It absorbs heat from the surroundings and maintains a constant temperature. Thus
from A to B the temperature remains constant. The curve AB is called the isothermal curve or
isothermal.
Consider the working substance at the point B and let the pressure be increased. External
work is done on the working substance and there should be rise in temperature. But the system is
perfectly conducting to the surroundings. It gives extra heat to the surroundings and its
temperature remains constant from B to A.
53
Thus during the isothermal process, the temperature of the working substance remains
constant. It can absorb heat or give heat to the surroundings. The equation for an isothermal
process is
PV = RT = constant (For one gram molecule of a gas)
For n gram molecules of a gas PV = nRT
4.2(a) Adiabatic Process.
During an adiabatic process, the working substance is perfectly insulated from the
surroundings. It can neither given heat nor take heat from the surroundings. When work is done
on the working substance, there is rise in temperature because the external work done on the
working substance increases its internal energy. When work is done by the working substance, it
is done at the cost of its internal energy. As the system is perfectly insulated from the
surroundings, there is fall in temperature.
Thus, during an adiabatic process, the working substance is perfectly insulated from the
surroundings. All along the process there is change in temperature. A curve between pressure
and volume during the adiabatic process is called an adiabatic curve or an adiabatic.
Examples.
1. The compression of the mixture of oil vapour and air during compression stroke of an internal
combustion is an adiabatic process and there is rise in temperature.
2. The expansion of the combustion products during the working stroke of an engine is an
adiabatic process and there is fall in temperature.
3. The sudden bursting of a cycle tube is an adiabatic process.
Apply the first law of thermodynamics to an adiabatic process, δH = 0,
δH = dU + δW
or
0 = dU + δW
The process that take place suddenly or quickly are adiabatic processes.
54
4.2(b) Isochoric Process
If the working substance is taken in a non-expanding chamber, the heat supplied will
increase the pressure and temperature. The volume of the substance will remain constant. Such a
process is called an Isochoric Process. The work done is zero because there is no change in
volume. The whole of the heat supplied increases the internal energy. Therefore, during the
isochoric process δW = 0.
δH = dU
The heat transferred in such a process
δH = CvdT
C2dT = dU
Hence C2 is the specific heat for one gram-molecule of a gas at constant volume.
4.3 Gas Equation During an Adiabatic Process
Consider 1 gram of the working substance (ideal gas) perfectly insulated from the
surroundings. Let the external work done by the gas be δW.
Applying the first law of thermodynamics
δH = dU + δW
But
and
δH = 0
δW = P.dV
Where P is the pressure of the gas and dV is the change in Volume.
P.dV
0 = dV +
….(1)
J
55
As the external work is done by the gas at the cost of its internal energy, there is fall in
temperature by dT.
dU = 1 x Cv x dT
P.dV
Cv dT +
=0
….(2)
J
For an ideal gas
PV = rT
Differentianting,
P.dV + V.dP = r.dT
Substituting the value of dT in equation (ii),
P.dV + V.dP
CV
P.dV
+
=0
r
J
CV (P.dV + V.dP) + r. P.dV
=0
J
But,
r
= Cp - Cv
J
Cv P.dV + Cv.V.dP + Cp.PdV - Cv.PdV = 0
Cp P.dV + Cv.V.dP = 0
Dividing by Cv.PV
Cp
dV
Cv
V
dP
+
=0
P
Cp
=
But
Cv
dP
dV
+ 
P
=0
V
. . . . (3)
56
Integrating, log
P +  log V = const.
log PV = const..
PV = const.
or
…(4)
This is the equation connecting pressure and volume during an adiabatic process.
Taking
PV = rT
or
P =
rT
V
rT
V = const.
V
But r is const
rTV -1 = const.
TV -1 = const.
Also
rT
V=
P
rT
P

= const.
P
r  T = const.
T -1 = const.
or
P ------
= const
T
or
Thus, during an adiabatic process
(i)
PV = const.
(ii)
TV -1 = const. and
(iii)
P -1
= const.
T
= const.
57
4.3 (a) Slopes of Adiabatics and Isothermal
In an isothermal process
PV = const.
Differentiating,
PdV + VdP = 0
dP
P
= -
or
dV
...(1)
V
In an adiabatic process
PVγ = Const.
Differentiating,
Pγ V -1 dV + VdP = 0
γP
dP
= dV
...(2)
V
Therefore, the slope of an adiabatic is γ times the slope of the isothermal
.
Fig. 4.3 (a)
Hence, the adiabatic curve is steeper than the isothermal curve at a point where the two
curves intersect each other. (Fig:4.3(a).
4.3 (b) Work done During an Isothermal Process
When a gas is allowed to expand isothermally, work is done by it.
58
Let the initial and final volumes be V1 and V2 respectively. In Fig.2.3 (b), the area of the
shaded strip represents the work done for a small change in volume dV. When the volume
changes from V1 to V2.
Work done
=
V2
 P. dV = area ABba
V1
...(i)
Fig. 4.3 (b)
Fig.4.3(b) represents the indicator diagram. Considering one gram molecule of the gas
PV = RT
RT
P=
or
V
W = RT
V2

V1
dV
V
V2
= RT loge
…. (2)
V1
Also
or
P1V1 = P2V2
V2
=
P1
. . . (3)
V1
P2
W = RT x 2.3026 x log 10 P1
…(4)
P2
59
Here, the change in the internal energy of the system is zero (because the temperature
remains constant). So the heat transferred is equal to the work done.
4.3 (c) Work done during and Adiabatic Process.
During an adiabatic process, the gas expands from volume V1 to V2. As shown by the
indicator diagram the work done for an increase in
Fig. 2.3 (c)
Fig:4.3(c) Volume dV = P.dV. Work done when the gas expands from V1 to V2 is given by,
W=
v2

PdV
= Area ABba
V
1
During an adiabatic process,
PV = const = K
K
P=
V
W=
=
K
1
V2

V1
dV
V
1
1
-
1-
V2 -1
Since A and B lie on the same adiabatic
P1V1 =
P2V2 = K
V1 r-1
60
W=
1
K
K
-
V2 -1
1-
1
P2V2
1-
-1
W=
V1 -1
P1V1
-
W=
V2
V1 -1
1
P2V2 - P1V1
…(2)
1-
Taking T1 and T2 as the temperatures at the points A and B respectively and considering
one gram molecule of gas.
P1V1 = RT1
and
P2V2 = RT2
Substituting these values in equation (ii)
W=
1
RT2 - RT1
…(3)
1-r
Here heat transferred is zero because the system is thermally insulated from the surroundings.
4.4 LET US SUM UP
From this lesson you have understood about the first law of thermodynamics and also learned
about the difference between the adiabatic and isothermal processes. The work done during the
adiabatic and isothermal processes is also discussed in this lesson.
4.5 CHECK YOUR PROGRESS
1. State first law of thermodynamics and mention its importance in heat
2. What is isothermal and adiabatic processes ?
3. What are the differences in the slope of isothermal and adiabatic processes ?
61
4.6 LESSON END ACTIVITIES
1. A water fall is 500 meters high. Assuming that the entire kinetic energy gained during fall is
converted into heat. Calculate the rise in temperature of water in the base of the fall.
(g = 9.8 m/s2)
[Hint: i) H = (mgh)/J, ii) H= ms

Using first law of thermodynamics calculate the change in the internal energy of the system if
the latent heat of vaporization is 5.4 x 105 cal/Kg. Given dw=4.027x104 cal.
[Hint: dH = dv+dw]
4.7 POINTS FOR DISCUSSION
1.
2.
(i)
Explain Isothermal and Adiabatic Process.
(ii)
Derive an expression for the work done during an adiabatic process.
(i)
Deduce the equation for the work done during Isothermal Process.
4.8 SOURCES
1. Heat and thermodynamics by R. Murugesan
62
LESSON - 5
CONTENTS
5.0 Aims and Objectives
5.1 Clement and Desormes Method – determination of 
5.2 Irreversible Process
5.3 Reversible Process
5.4 Let us Sum up
5.5 Check your progress
5.6 Lesson end Activities
5.7 Points for Discussion
5.8 References
5.0 AIMS AND OBJECTIVES
In this lesson you will learn the ratio of the specific heat of gat at constant volume and at
constant pressure and its experimental determination. Also you will learn about the
reversible and irreversible processes.
2.9 Clement and Desormes Method – determination of 
Clement and Desormes in 1819 designed an experiment to find , the ratio between the
two specific heats of a gas.
63
Fig. 5.1
The vessel A has a capacity of 20 to 30 litres and is fitted in a box containing cotton and
wool. At the top end, three tubes are fitted as shown in figure.5.1. Through S1, dry air is forced
into the vessel A. The stop cock S1, is closed when the pressure inside A is slightly greater than
the atmospheric pressure. Let the difference in level on the two sides of the manometer M be H
and the atmosphere pressure be P0. The pressure of air inside the vessel is P1.
The stop-cock S is suddenly opened and closed just at the moment when the level of the
liquid on the two sides of the manometer are the same. Some quantity of air escapes to the
atmosphere. The air inside the vessel expands adiabaticaliy. The temperature of air inside the
vessel falls due to adiabatic expansion. The air inside the vessel is allowed to gain heat from the
surroundings and it finally attains the temperature of the surroundings. Let the pressure at the end
be P2 and the difference in levels on the two sides of the manometer be h.
Theory. Consider a fixed mass of air left in the vessel in the end. This mass of air has
expanded from volume V1 (Less than the volume of the vessel) at pressure P1 to volume V2 at
pressure P0. The process is adiabatic as shown by the curve AB (fig 2.4(a) .
64
P2V1 = P0V2
P1

V2
=
P0
V1
… (1)
Finally the point C is reached. The points A and C are at the room temperature. Therefore
AC can be considered as an isothermal.
P1 V1 =
V2
P2V2
P1
=
V1
P2
… (2)
Substituting the value of V2 in equation (i).
V1
P1
P1
r
=
P0
P2
Fig. 5.1(a)
Taking Logarithms,
log P1 – log Po =  (log P1 – log P2)
65
log P1 - log Po
=
log P1 - log P2
But
… (3)
P1 = Po + H and P2 + Po + h
 = log (Po + H)- log Po
log (Po + H) – log (Po + h)
log
(Po + H)
Po
=
log
(Po + H)
Po+h
log
1 +H
Po
log
1+
H-h
Po+h
H
Po
Approximately,  =
H
=
H-h
H-h
H
Hence
=
H-h
...(4)
Similarly,  for any gas can be determined by this method.
Drawbacks. When the stop-cock is opened, a series of oscillations are set up. This is
shown by the up and down movement of the liquid in the manometer. Therefore, the exact
moment when the stopcock should be closed is not known. The pressure any not be equal to the
66
atmospheric pressure when the stop-cock is closed. It may be higher or less then the atmospheric
pressure. Thus the result obtained will not be accurate.
5.2 Ireversible Process
The thermo dynamical state of a system can be defined with the help of the thermo
dynamical coordinates of the system. The state of a system can be changed by altering the
thermo dynamical coordinates. Changing from one state to the other by changing the thermo
dynamical coordinates is called a process.
Consider two states i.e, state A and State B. Changes of state from A to B or vice versa is
a process and the direction of the process will depend upon a new thermo dynamical coordinate
called entropy. All process are not possible in the universe.
Consider the following process:
(1)
Let two blocks A and B at different temperatures T1 and T2 (T1>T2) be
kept
in
contact but the system as a whole is insulated from the surroundings. Conduction
of heat takes place between the blocks, the temperature of A falls and the
temperature of B rises and thermo dynamical equilibrium will be reached.
(2)
Consider a flywheel rotating with an angular velocity its initial kinetic energy is
½ I2. After some time the wheel comes to rest and kinetic energy is utilized in
overcoming friction at the bearings. The temperature of the wheel and the
bearings rises and the increase in their internal energy is equal to the original
kinetic energy of the fly wheel.
(3)
Consider two flasks A and B connected by a glass tube provided with a stop cock.
Let A contain air at high pressure and B is evacuated. The System is isolated from
67
the surroundings. If the stop cock is opened, air rushes from A to B, the pressure
in A decrease and the volume of air increases.
All the above three examples though different, are thermo dynamical processes involving
change in thermo dynamical coordinates. Also, in accordance with the first law of
thermodynamics, the principle of conservations of energy is not violated because the total energy
of the system is conserved. It is also clear that, with the initial conditions described above, the
three processes will takes place.
Let us consider the possibility of the above three processes taking place in the reverse
direction. In the first case, if the reverse process is possible, the block B should transfer heat to A
and initial conditions should be restored. In the second case, if the reverse process is possible, the
heat energy must again change to kinetic energy and the fly wheel start rotating with the initial
angular velocity . In the third case, if the reverse process is possible the air in B must flow back
to A and the initial condition should be obtained.
But, it is a matter of common experience, that none of the above conditions for the
reverse process are reached. It means that the direction of the process cannot be determined by
knowing the thermo dynamical coordinates in the two end states. To determine the direction of
the process a new thermo dynamical coordinate has been devised by Clausius and this is called
the entropy of the system. Similar to internal energy, entropy is also a function of the State of the
system. For any possible process, the entropy of an isolated system should increase or remain
constant. The process in which there is a possibility of decrease in entropy cannot take place.
If the entropy of an isolated system is maximum, any change of state will mean decrease
in entropy and hence that change of state will not take place.
To conclude, process in which the entropy of an isolated system decreases do not take
place or for all processes taking place in an isolated system the entropy of the system should
increase or remain constant. It means a process is irreversible if the entropy decreases when the
68
direction of the process is reversed. A process is said to be irreversible if it cannot be retracted
back exactly in the opposite direction. During an irreversible process heat energy is always used
to overcome friction. Energy is also dissipated in the form of conduction and radiation. This loss
of energy always takes place whether the engine works in one direction or the reverse direction.
Such energy cannot be regained. In actual practice all the engines are irreversible. If electric
current is passed through a wire, heat is produced. If the direction of the current is reversed, heat
is again produced. This is also an example of an irreversible process. All chemical reaction are
irreversible. In general, all natural processes are irreversible.
5.3 Reversible Process
From the thermo dynamical point of view, a reversible process is one in which an
infinitesimally small change in the external conditions will result in all the changes taking place
in the direct process but exactly repeated in the reverse order and in the opposite sense. The
process should take place at an extremely slow rate. In a reversible cycle, there should not be any
loss of heat due to friction or radiation. In this process, the initial conditions of the working
substance can be obtained.
Consider a cylinder, containing a gas at a certain pressure and temperature. The cylinder
is fitted with a frictionless piston. If the pressure is decreased, the gas expands slowly and
maintain a constant temperature (isothermal process). The energy required for this expansion is
continuously drawn from the source (surroundings). If the pressure on the Piston is increased, the
gas contracts slowly and maintains constant temperature (isothermal process). The energy
liberated during compression is given to the sink (surroundings). This is also true for an adiabatic
process provided the process takes place infinitely slowly.
The process will not be reversible if there is any loss of heat due to friction, radiation or
conduction. If the changes take place rapidly, the process will not be reversible. The energy used
in overcoming friction cannot be retraced.
69
5.4 LET US SUM UP
In this lesson you have learned about the relation between the two specific heats of gases namely
Cp and Cv. Also you learned about the reversible and irreversible processes.
5.5
CHECK YOUR PROGRESS
1. State the difference between reversible and irreversible processes.
2. Why gas has got two specific heat ? Give reasons.
5.6 LESSON END ACTIVITIES
1. Explain reversible and irreversible processes with suitable examples
2. 14
POINTS FOR DISCUSSION
1. How do you determine  by Clement and Desormes method
5.8 REFERENCES
1. Heat by Anantha Krishnan
2. Heat and thermodynamics by Brijlal and Subramaniam
70
LESSON – 6
CONTENTS
6.0 Aims and Objectives
6.1 Second law of Thermodynamics
6.2 Carnot’s Reversible Engine.
6.3 Carnot's Engine and Refrigerator
6.4 Carnot's Theorem
6.5 Let us Sum up
6.6 Check your progress
6.7 Lesson end Activities
6.8 Points for Discussion
6.9 References
6.0 AIMS AND OBJECTIVES
In this lesson you will learn the second law of thermo dynamics, and its applications to Carnot
engine also you will derive Carnot’s theorem
6.1
Second law of Thermodynamics
A heat engine is chiefly concerned with the conservation of heat energy into mechanical
work. A refrigerator is a device
to cool a certain space below the temperature of its
surroundings. The first law of thermodynamics is a qualitative statement which does not preclude
the possibility of the existence of either a heat engine or a refrigerator. The first law does not
contradict the existence of a 100% efficient or a self-acting refrigerator.
In practice, these two are not attainable. These phenomena are recognized and this led to
the formulation of a law governing these two devices. It is called second law of thermodynamics.
71
A new term reservoir is used to explain the second law. A reservoir is a device having
infinite thermal capacity and which can absorb, retain or reject unlimited quantity of heat without
any change in its temperature.
Kelvin-Plank statement of the second law is as follows:
“It is impossible to get a continuous supply of work from a body (or engine) which can
transfer heat with a single heat reservoir. This is a negative statement. According to this
statement, a single reservoir at a single temperature cannot continuously transfer heat into work.
It means that there should be two reservoirs for any heat engine. One reservoir (called the
source) is taken at a higher temperature and the other reservoir (called the sink) is taken at a
lower temperature.
According to this statement, zero degree absolute temperature is not attainable because
no heat is rejected to the sink at zero degree Kelvin. If an engine works between any temperature
higher than zero degree Kelvin and zero degree Kelvin, it means it uses a single reservoir which
contradicts Kelvin-Planck’s statement of the second law. Similarly, no engine can be 100%
efficient.
In a heat engine, the engine drawn heat from the source and after doing some external
work, it rejects the remaining heat to the sink. The source and sink are of infinite thermal
capacity and they maintain constant temperature.
First Part: According to Kelvin, the second law can also be stated as follows:
"It is impossible to get a temperature lower than that of its surroundings.
In a heat engine the working substance does some work and rejects the remaining heat to
the sink. The temperature of the source must be higher than the surroundings and the engine will
not work when the temperature of the source and the sink are the same. Take the case of a steam
engine. The steam (working substance) at high pressure is introduced into the cylinder of the
engine. Steam expands, and it does external work. The contents remaining behind after doing
work are rejected to the surroundings. The temperature of the working substance rejected to the
surroundings is higher than the temperature of the surroundings.
72
If this working substance rejected by the first engine is used in another engine, it can do
work and the temperature of the working substance will fall further.
It means that the working substance can do work only if its temperature is higher than
that of the surroundings.
Second Part. According to Clausius:
"It is impossible to make heat flow from a body at a lower temperature to a body at a higher
temperature without doing external work done on the working substance."
This part is applicable in the case of ice plants and refrigerators. Heat itself cannot flow from a
body at a lower temperature, to a body at a higher temperature unless some external works is
done on the working substance. Take the case of ammonia ice plant. Ammonia is the working
substance. Liquid ammonia at low pressure takes heat from the brine solution in the brine tank
and is converted to a low pressure vapour. External work is done to compress the ammonia
vapours to high pressure. This ammonia at high pressure is passed through coils over which
water at room temperature is poured. Ammonia vapour gives heat to water at room temperature
and gets itself converted into liquid again. This high pressure liquid ammonia is throttled to low
pressure liquid ammonia. In the whole process ammonia (the working substance) takes heat from
brine solution (at a lower temperature) and gives heat to water at room temperature (at a higher
temperature). This is possible only due to the external work done on ammonia by the piston in
compressing it. The only work of electricity in the ammonia ice plant is to move the piston to do
external work on ammonia. If the external work is not done, no ice plant or refrigerator will
work. Hence, it is possible to make heat flow from a body at a lower temperature to a body at a
higher temperature by doing external work on the working substance.
Thus, the second law of thermodynamics plays an important part for practical devices
e.g., heat engines and refrigerators. The first law of thermodynamics only gives the relation
73
between the work done and the heat produced. But the second law of thermodynamics gives the
conditions under which heat can be converted into work.
6.2 Carnot’s Reversible Engine.
Heat engines are used to convert heat into mechanical work. Its efficiency is maximum
and it is ideal heat engine.
For any engine, there are three essential requisites:
(1) Source. The source should be at a fixed high temperature T1 from which the heat
engine can draw heat. It has infinite thermal capacity and any amount of heat can be drawn from
it at constant temperature T1.
(2) Sink. The sink should be at a fixed lower temperature T2 to which any amount of heat
can be rejected. It also has infinite thermal capacity and its temperature remains constant at T2.
(3) Working Substance. A Cylinder with non-conducting sides and conducting bottom
contains the perfect gas as the working substance.
Fig. 6.2
A perfect non-conducting and frictionless piston is fitted into the cylinder. The working
substance undergoes a complete cyclic operation. (Fig 6.2)
74
A perfectly non-conducting stand is also provided so that the working substance can
undergo adiabatic operation.
Carnot’s Cycle.
(1) Place the engine containing the working substance over the source at temperature T1.
The working substance is also at a temperature T1. Its pressure is P1 and volume is V1
as shown by the point A in (Fig 6.2a).
Fig. 6.2a
Decrease the pressure, the volume of the working substance increases. Work is done by
the working substance. As the bottom is perfectly conducting to the source at temperature T1 it
absorbs heat. The process is completely isothermal. The temperature remains constant. Let the
amount of heat absorb by the working substance be H1 at the temperature T1. The point B is
obtained.
Consider one gram molecule of the working substances.
Work done from A to B (isothermal process)
75
V2
V2
W1 =

P.dV = RT 1 log
V1
= area ABGE
…(1)
(2) Place the engine on the stand having an insulated top. Decrease the pressure on the
V1
working substance. The volume increase. The process is completely adiabatic. Work is done by
the working substance due to internal energy. The temperature falls. The working substance
undergoes adiabatic change from B to C. At C the temperature is T2 (Fig.6.2a)
Work done from B to C (adiabatic process)
But PV = constant = K
P2V2 = RT1
V3
W2 =
 P.dV
V2
V2
dV
P3V3 = RT2
= 
V
V3
P3V3 = P2V2 = K
KV3 1- – KV2 1-
1-
P3V3 – P2V2
1-
R(T2- T1)
1-
R(T1- T2)
-1
W2 = Area BCHG
(3)
…(2)
Place the engine on the sink at temperature T3. Increase the pressure. The work
done on the working substance. As the base is conducting to the sink, the process is isothermal.
A quantity of heat H2 is rejected to the sink at temperature T2. Finally the point D is reached.
Work done from B to C (isothermal process)
76
V4
W3 = PdV
V3
= RT2 log V4
V3
= - RT2 log V3
V4
W3 = area CHFD
. . . (3)
( The –ve sign indicates that work is done on the working substance.)
(4)
Place the engine on the insulating stand. Increase the pressure. The Volume
decreases. The process is completely adiabatic. The temperature rises finally the point A is
reached.
Work done from D to A (adiabatic process).
V1
W4 =

PdV
V4
=-
R(T1- T2)
-1
W4 = Area DFEA
... (4)
(W2 and W4 are equal and opposite and cancel each other.)
The net work done by the working substance in one complete cycle.
= Area ABEG + Area BDHG – Area CHFD - Area DFEA
= Area ABCD
The net amount of heat absorbed by the working substance
= H1 – H2
Net work
= W1 + W2 + W3 + W4
77
V2
R(T1 - T2)
= RT1 log
+
V3
- RT2 log
-1
V1
V2
+ log
V1
V4
V2
W = RT1 log
R (T1 _ T2)
-1
V3
- RT2 log
V1
V4
...(5)
The points A and D are on the same adiabatic
T1 V1 r-1 = T2V4 r-1
T2
V1
-1
=
T1
V4
...(6)
The points B and C are on the same adiabatic
T1 V2 r-1 = T2V3 r-1
T2
V2
-1
=
T1
V3
...(7i)
From (6) and (7)
-1
-1
V1
V2
=
V4
or
V3
V1
V2
=
V4
or
V3
V2
V3
=
V1
V4
78
From equation (5)
V2
V2
W = RT1 log
+ RT2 log
V1
V1
V2
W= R
log
T1 – T2
V1
W = H1 – H2
Useful output
Efficiency
W
 =
=
Input
H1
Heat is supplied from the source from A to B only.
V2
H1 = RT1 log
V1
W

H1 – H2
=
=
H1
H1
V2
R(T1 – T2) log
V1
=
V2
R(T1 log
V1
H2
=1–
H1
T2
=1–
.
T1
... (8)
79
The Carnot's engine is perfectly reversible. It can be operated in the reverse direction
also. Then it works as a refrigerator. The heat H2 is taken form the sink and external work is
done on the working substance and heat H1 is given to the source at a higher temperature.
The isothermal process will take place only when the piston moves very slowly to give
enough time for the heat transfer to take place. The adiabatic process will take place when the
piston moves extremely fast to avoid heat transfer. Any practical engine cannot satisfy these
conditions.
All practical engines have an efficiency less than the Carnot's engine.
6.3 Carnot's Engine and Refrigerator
Caronot's cycle is perfectly reversible. It can work as a heat engine and also as a
refrigerator. When it works as a heat engine, it absorbs a quantity of heat H1 from the source at a
temperature T1, does an amount of work W and rejects an amount of heat H2 to the sink at
temperature T2. When it works as a refrigerator, it absorbs heat H2 from the sink at temperature
T2. W the amount of work is done on it by some external means and rejects heat H1 to the source
at a temperature T1 (Fig 2.8). In the second case heat flows from a body at a lower temperature to
a body at a higher temperature, with the help of external work done on the working substance
and it works as a refrigerator. This will not be possible if the cycle is not completely reversible.
Coefficient of Performance. The amount of heat absorbed at the lower temperature is
H2. The amount of work done by the external process (input energy) = W and the amount of heat
rejected = H1. Here H2 is the desired refrigerating effect.
80
Fig. 6.3
Coefficient of performance
H2
=
H2
=
W
H1 – H2
Suppose 200 joules of energy is absorbed at the lower temperature and 100 joules of
work is done with external help. Then 200 + 100 = 300 joules are rejected at the higher
temperature.
The coefficient of performance
H2
=
W
H2
=
H1 – H2
200
=
=2
81
300 – 200
Therefore the coefficient of performance of a refrigerator =2.
In the case of a heat engine, the efficiency cannot be more than 100% but in the case of a
refrigerator, the coefficient of performance can be much higher than 100%.
6.4 Carnot's Theorem
The efficiency of a reversible engine does not depend on the nature of the working
substance. It merely depends upon the temperature limits between which the engine works.
"All the reversible engines working between the same temperature limits have the same
efficiency. No engine can be more efficient than a carnot's reversible engine working between
the same two temperatures."
Consider two reversible engines A and B, working between the temperature limits T1 and
T2 (Fig 6.4). A and B are coupled. Suppose A is more efficient than B. The engine A works as a
heat engine and B as a refrigerator. The engine A absorbs an amount of heat H1 from the source
at a temperature T1. It does external work W and transfers it to B. The heat rejected to the sink is
H2 at a temperature T2. The engine B absorbs heat H2 from the sink at temperature T2 and W is
the amount of work done on the working substance. The heat given to the source at temperature
T1 is H1’.
Suppose the engine A is more efficient than B.
Efficiency of the engine A
H1 - H2
 =
W
=
H1
H1
Efficiency of the engine B
H1' - H2'
' =
W
=
H1
H1'
82
Fig. 6.4
Since
 > '
then H1' > H1
Also
W = H1 – H2 = H1' – H2'
H2 ' > H2
Thus, for the two engines, A and B working as a coupled system, (H2' – H2) is the quantity
of heat taken from the sink at a temperature T2 and (H1' – H1) is the quantity of heat taken from
the sink at a temperature T2 and (H1' – H1) is the quantity of heat given to the source at a
temperature T1. Both (H2' – H2) and (H1' – H1) are positive quantities. It means heat flows from
the sink at a temperature T2 (lower temperature) to the source at a temperature T1 (higher
temperature)i.e., heat flows
from a body at a lower temperature to a body at a higher
temperature. But, no external work has been done on the system. This is contrary to the second
law of thermodynamics. Thus  cannot be greater than '. The two engines (reversible) working
between the same two temperature limits have the same efficiency. Moreover, in the case of a
Carnot's engine, there is no loss of heat due to friction, conduction or radiation (irreversible
possesses). Thus, the Carnot's engine has the maximum efficiency. Whatever may be the nature
of the working substance, the efficiency depends only upon the two temperature limits.
In a practical engine there is always loss of energy due to friction, conduction, radiation
etc., and hence its efficiency is always lower than that of a Carnot's engine.
83
6.5 LET US SUM UP
From this lesson you learned the second law of thermodynamics and its application to Carnot’s
reversible engine and refrigerator. Also you learned about the importance of the Carnot’s
theorem and its applications.
6.6 CHECK YOUR PROGRESS
1. State Carnot’s theorem
2. Mention important points adopted in Carnot’s reversible engine
3. State second law of thermodynamics
6.7 LESSON END ACTIVITIES
1. Suppose the efficiency of the heat engine is 20% and heat absorbed by the working substance
from the source is H1 (100 Joule) at temperature T1 (500 K) and heat rejected to the sink H2 at
T2 (300 K). Calculate H2. (Answer: 800 Joules)
[Hint:  = 1- (H2/H1) ]
2. In a Carnot’s engine the temperature of the source and the sink are 500 K and 375 K
respectively. If the engine consume 600 x 103 cal/cycle, find
i) efficiency of the engine (Ans: 25 %)
ii) Work done per cycle (Ans: 6.3 x 105 Joules)
iii) Heat rejected per cycle (Ans: 1.89 x 106 Joules)
6.8 POINTS FOR DISCUSSION
1. .
(i)
State and Prove Carnot's theorem.
(ii)
Write a note on Carnot Refrigerator.
6.9 REFERENCES
1. Heat and thermodynamics by R. Murugesan
2. Heat and thermodynamics by Brijlal and Subramanyam
84
QUESTIONS
1.
State the two laws of Thermodynamics and explain their significance. Derive an
Expression.
2.
3.
4.
(i)
Explain Isothermal and Adiabatic Processes.
(ii)
Derive an expression for the work done during an adiabatic process.
(i)
State and Prove Carnot's theorem.
(ii)
Write a note on Carnot is Refrigerator.
(i)
Deduce the equation for the work done during Isothermal Process.
(ii)
Explain clement and Deforms method for determining .
85
UNIT – III
LESSON – 7
CONTENTS
7.0
Aims and Objectives
7.1
Introduction
7.2
Fundamental principles of impact
7.3
Oblique impact of a smooth sphere on a fixed smooth plane:
7.4
Direct impact of two smooth spheres:
7.5a
7.5
Loss of K.E. due to direct impact of two smooth spheres:
Oblique impact of two smooth spheres:
7.5a
Loss of K. E. Due to oblique impact:
7.6
Let us Sum up
7.7
Check your progress
7.8
Lesson end Activities
7.9
References
7.0 Aims and Objectives
In this lesson you will learn about different conservation laws and also the meaning of impact
and impulse. Also you will learn about the study of the differences in discrete and oblique impact
and the derivation to determine the velocity and loss of Kinetic energy after collision.
7.1 INTRODUCTION
The impulse I of a constant force F acting for a time t is defined as F x t.
ie., I = F x t
By Newton’s second law, F = ma. If u and v are the initial and final velocities of the particle,
a = (v - u)/t
86
v u 
Hence, I = F x t = (m.a)t = m
t

t

= m (v - u)
Thus, the impulse of a force is equal to the change in momentum produced.
Impulsive force:
An impulsive force in an infinitely great force acting for a very short interval of time, such
that their product is finite.
The force and the time cannot be measured because one is too great and the other is too
small. Their product which is definite is capable of measurement. We can say that the impulse of
the impulsive force is always measured by the change in momentum (product of mass and
velocity) produced.
Examples of impulsive force are,
(a) the blow of a hammer on a pile,
(b) the force exerted by the bat on a cricket ball.
Collisions:
In a collision, a large force acts on each colliding particle for a short time. The force is called
impulse force. For example when a bat hits a ball, a bat exerts a large force on the ball. Both the
ball and the bat are deformed during the collision. The force of interaction may be due to
different causes in different cases. Thus in the collision between two billiard balls, the force of
interaction is due to elasticity. It arises in existence only when the two billiard balls come into
physical contact. In the case of scattering of -particles by the nuclei, it is the electrostatic force
that causes interaction.
87
Elastic and In Elastic Collisions:
There are two types of collisions,
(i)
(i)
Elastic
(ii)
In Elastic
Elastic collisions are those in which the total kinetic energy before and after the collision
remains unchanged. Collisions between atomic, nuclear and fundamental particles are the
true elastic collisions. Collisions between ivory or glass balls can be treated as
approximately elastic collisions. In such a collision between two particles, we have,
m1u1 + m2u2 = m1 v1 + m2v2 and
1
1
1
1
2
2
2
2
m1u1  m 2 u 2  m1 v1  m 2 v 2
2
2
2
2
where m1 and m2 are the respective masses of the two particles and u1, u2 and v1, v2 their
velocities before and after the collision.
(ii) If the K.E. is not conserved, the collision is said to be inelastic. When two bodies stick
together after collision, the collision is said to be completely inelastic. For example, the
collision between a bullet and its target is completely inelastic when the bullet remains
embedded in the target.
Completely Inelastic Collision: Suppose a body of mass m1 moving with a velocity u1 collides
with a body of mass m2 moving with velocity u2 in the same direction. The two bodies stick
together after collision and they move with a final common velocity V in the same direction as
the original. It is not necessary to restrict the discussion to one dimensional motion. Using the
conservation of momentum principle,
m1u1 + m2u2 = (m1 + m2) V.
From this the value of V can be determined if u1 and u2 are known.
88
7.2 FUNDAMENTAL PRINCIPLES OF IMPACT
1. Newton’s law of impact-coefficient of restitution: When two bodies impinge directly, their
relative velocity after impact is in a constant ratio to their relative velocity before impact and
is in the opposite direction. This constant ratio depends only on the material of the bodies and
not on their masses or velocities. It is called the coefficient of restitution and is denoted by
the letter e. If u1, u2 be the velocities of two bodies before the impact and v1, v2 the velocities
after impact,
(v1- v2) / (u1- u2) = -e or v1- v2 = -e(u1- u2)
where (u1- u2) and (v1- v2) are their relative velocities, before and after the impact. e lies between
0 and 1. If e = 0, the bodies are called perfectly plastic bodies. If e = 1, the bodies are called
perfectly elastic bodies. For two glass balls, e = 0.94; For two lead balls, e = 0.2
The ratio, with a negative sign, of the relative velocity of two bodies after impact to their relative
velocity before impact is called the coefficient of restitution.
2. Motion of two smooth bodies perpendicular to the line of impact: When two smooth bodies
impinge, there is no tangential action between them. Hence there is no change of momentum
along the tangent. Hence, there is no change of velocity for either body along the tangent. In
other words, there is no change in the velocity of a body in direction perpendicular to the
common normal due to impact.
3. Principle of conservation of momentum: The total momentum of two bodies after impact
along the common normal should be equal to the total momentum before the impact along
the same direction.
The above three principles are sufficient to determine the change in motion of two impinging
smooth bodies.
89
7.3 OBLIQUE IMPACT OF A SMOOTH SPHERE ON A FIXED SMOOTH PLANE:
V
U
α

C
X
P
Fig.7.3
Let XY be the fixed plane. Let the sphere strike the fixed plane at point P. Then if C is the
centre of the sphere, CP is the common normal at the point of contact of the plane and the
sphere. Let u and v be the velocities of the sphere before and after impact making angles  and 
respectively with the common normal CP (fig 3.3). By Newton’s experimental law, the relative
velocity of the sphere along the common normal after impact is –e times its relative velocity
along the common normal before impact.
v cos  - 0 = -e(-u cos  - 0)
(or)
v cos  = eu cos 
---(1)
Since both the sphere and the plane are smooth, there is no force in a direction parallel to the
plane. Hence the velocity of the sphere resolved parallel to the plane is unaltered by the impact.
Dividing (1) by (2),
v sin  = u sin 
---(2)
cot  = e cot 
---(3)
Squaring and adding (1) and (2),
v2 = u2 (sin2  + e2 cos2 )
Equations (4) and (3) give the velocity and direction of the sphere after impact.
---(4)
90
POINTS:
1. The impulse of the plane of the sphere is measured by the change of momentum of the
sphere measured along the normal.
I = m [ v cos  - (-u cos )] = m [ v cos  + u cos ]
= m [eu cos  + u cos ]
I = mu (1+e) cos 
2. If e =1, then  =  and v = u i. e., when a perfectly elastic sphere impinges on a fixed
smooth plane its velocity is unaltered in magnitude by the impact and the angle of reflection is
equal to the angle of incidence.
3. If e = 0, then cot  = 0 or  = 90o [from (3)] and v = u sin  [from (4)]. Thus if the sphere
and the plane are both inelastic, the sphere moves along the plane with velocity u sin .
4. If  = 0, then cot  = 0 and v = eu from (3) and (4) i. e., if a sphere impinges normally on a
horizontal plane, it rebounds vertically with velocity eu.
5. The change in K.E. of the sphere due to impact on the plane is given by,
1
1
1
mv 2  u 2  mv  u v  u   I v  u 
2
2
2
Here, m (v – u) = I = Impulse of the force of the sphere on the plane.
91
7.4 DIRECT IMPACT OF TWO SMOOTH SPHERES:
m1
U1
m2
V1
U2
V2
Fig.7.4
A smooth sphere of mass m1 moving with a velocity u1 impinges on another smooth sphere of
mass m2 moving in the same direction with velocity u2. If e is the coefficient of restitution
between them, find the velocities of the spheres after impact.
Since the spheres are smooth, there is no impulsive force on either along the common tangent.
Hence in this direction their velocities after impact are the same as their original velocities i. e.,
zeroes. Let v1 and v2 be the velocities of the two spheres along the common normal after impact
[fig 7.4].
By the principle of conservation of momentum,
m1 v1 + m2 v2 = m1u1 + m2u2
---(1)
By the Newton’s experimental law,
v1 - v2 = -e (u1 -u2)_
Multiplying (2) by m2 and adding to (1),
---(2)
92
v1 (m1 + m2) = m2u2 (1 +e) + u1 (m1 - em2)
v1 
m2 u 2 1  e u1 m1  em2 
m1  m2 
---(3)
Multiplying (2) by m1 and subtracting from (1),
v2 (m1 + m2) = m1u1 (1 +e) + u2 (m2 – em1)
v2 
m1u1 1  e  u 2 m2  em1 
m1  m2 
---(4)
Equations (3) and (4) give the velocities of the two spheres after impact.
POINTS:
1. The impulse of the blow on the sphere of mass m1 = change of momentum
produced in it  m1 v1  u1  
m1m2 1  e  u 2 u2  u1 
m1  m2 
This is equal and opposite to the
impulse of the blow on the sphere of mass m2.
2. If e = 1 and m1 = m2 then, v1 = u2 and v2 = u1. Thus, if two equal perfectly, elastic spheres
impinge directly, they interchange their velocities.
7.4(a) LOSS OF K.E. DUE TO DIRECT IMPACT OF TWO SMOOTH SPHERES:
Let m1, m2 be the masses, u1 and u2, v1 and v2 their velocities before and after impact and e
the coefficient of restitution. Then, by the principle of conservation of linear momentum,
m1 v1 + m2v2 = m1u1 + m2u2
By Newton’s experimental law,
---(1)
93
v1 – v2 = -e (u1 -u2)
---(2)
Square both equations, multiply the square of the second by m1m2 and add the results. Then,
Note:
When e = 1, the loss of K.E. is zero. In general e < 1 so that (1 – e2) is positive.
(u1 -u2)2 is
always positive. Hence, there is always a loss of K. E. due to impact. The K. E. lost during
impact is converted into (i) sound, (ii) heat or (iii) vibration or rotation of the colliding bodies.
1 m1 m 2 u1  u 2 
where = 0, the loss in K.E.
2
m1  m 2
2
94
7.5 OBLIQUE IMPACT OF TWO SMOOTH SPHERES:
Fig.7.5
A smooth sphere of mass m1 moving with velocity u1 impinges obliquely on a smooth sphere of
mass m2 moving with velocity u2. If the directions of motion before impact make angles  and 
with the common normal, the velocities and direction of the spheres after impact can be
determined.
Let AB be the common normal [fig 7.5]. Let v1 and v2 be the velocities of the two spheres after
impact making angles  and  with the common normal AB. Before impact velocities along the
common normal AB are u1 cos  and u2 cos  and velocities perpendicular to AB are u1 cos 
and u2 sin . After impact velocities along AB are
v1 cos  and v2 cos  and perpendicular to
AB are v1 sin  and v2 sin .
By the principle of conservation of momentum, the total momentum of the two spheres along
the common normal is unaltered by the impact.
m1 v1 cos  + m2v2 cos  = m1u1 cos  + m2 u2 cos 
By Newton’s experimental law (on relative velocities along the common normal).
---(1)
95
v1 cos  - v2 cos  = -e (u1 cos  - u2 cos )
---(2)
Since there is no force perpendicular to the common normal AB, the velocities of the spheres
perpendicular to the common normal AB remain unaltered due to impact. Hence,
and
v1 sin  = u1 sin 
---(3)
v2 sin  = u2 sin 
---(4)
Multiplying (2) by m2 and adding to (1),
v1 cos 
m
1
 em2 u1 cos   m2 1  e u2 cos 
m1  m2
---(5)
Multiplying (2) by m1 and subtracting from (1),
v2 cos  
m
2
 em1 u 2 cos   m1 1  e u1 cos 
m1  m2
-----(6)
Squaring (3) and (5) and adding we get v12 and hence we can find v1. Dividing (3) and (5) we
get tan . Similarly, from (4) and (6) we can get v2 and tan . Therefore, v1, v2,  and  are
determined uniquely.
7.5(a) LOSS OF K. E. DUE TO OBLIQUE IMPACT OF SMOOTH SPHARES:
The velocities of the spheres perpendicular to the common normal are unaltered. Therefore, the
loss of K. E. is the same as in the case of direct impact if we substitute
for u1 and u2 respectively.
u1 cos  and u2 cos 
96
The loss in K.E. =
m1 m 2 1  e 2 
u1 cos   u 2 cos  2
2m1  m 2 
7.6 Let us Sum Up
From this lesson you learned about different conservation laws and also the meaning of impact
and impulse. Also in this lesson it is discussed about the differences in direct and oblique impact
and the derivation to determine the velocity and loss of kinetic energy after collision.
7.7 Check Your Progress
1. Define impulse and impact
2. State different conservation laws
3. State the differences between Direct and Oblique impact
7.8 Lesson end Activities
1) (a) Explain (i)Direct and (ii) Oblique impact
(b) Derive an expression for the final velocity and loss of K.E in case of Oblique
impact
7.9 References
1. Dynamics by D.S. Mathur
97
LESSON – 8
CONTENTS
8.0
Aims and Objectives
8.1
Banking of tracks.
8.2
Motion in a vertical circle
8.3
Friction
8.3a Laws of friction
8.3b Angle of friction and coefficient of friction
8.3c Static and Dynamic friction
8.3d Methods of reducing friction
8.4
(a) Motion down a rough inclined plan
(b) Motion up a rough inclined plan
8.5
Let us Sum up
8.6
Check your progress
8.7
Lesson end activities
8.8
Points for discussion
8.9
References
8.0 Aims and Objectives
In this lesson you will come to know about the reason for banking of curves. Also you will learn
about the laws of friction, angle of friction and the limitations of friction. You will also
understand its advantages and disadvantages. Also you will come to know about the equivalence
of a body in a rough inclined plane and its related derivations.
98
8.1 BANKING OF TRACKS:
Let O be the centre of a circular path of radius of
curvature r. A person moving on this circular path bends
away from the normal by an angle . The weight of the
person acts vertically downwards through his centre of
gravity.
Let R be the reaction of the ground. This
force R can be resolved into two rectangular
components R cos  vertical to the ground and R sin 
towards the centre of the circular path. The component R
cos  balances the weight mg and the component R sin 
fig.8.1
provides the necessary centripetal force for the person to move on the circular path.
mv 2
R sin  
r
---(1)
R cos  = mg
Dividing (1) by (2),
tan  
v2
rg
-----(2)
Here  gives the angle through which the cyclist bends from the vertical.
Similarly in the case of a railway track (fig 8.1),
tan  
v2
rg
When a train moves along a curve, the outer rail is raised by a height h over the inner rail such
that sin  = h / d where d is the distance between the two rails and
tan  = v2 / rg.
Since the angle  is small.
sin  = tan 
h v2

d rg
or
h
v2d
rg
. . . . (3)
While constructing curved rail tracks the value of h is obtained for fixed values of v, d and r.
Therefore, on a
99
Fig 8.1a
curved rail track, every train has to move with a particular speed specified for the track.
8.1 a) MOTION IN A VERTICAL CIRCLE:
When a stone or a body tied to the end of a string is whirled in a vertical circle, the motion is
not uniform.
Let v1 be the velocity of the stone while it posses the lowest point. The tension in the string in
the upward direction = T1. The weight mg acts in the downward direction
2
T1 - mg = mv1 / R
or
T1 = (mv12 / R) + mg
[fig 8.1(b)].
V2
---(1)
T1
At the highest point, the velocity is v2 and the tension in the string is T2.
The weight mg acts vertically downward.
O
T1
R
2
T2 + mg = mv2 / R
2
or
T2 
mV 2
 mg
R
-----(2)
V
Fig.8.1b
When the tension in the string T2 = 0,
2
mv 2
 mg
R
or v 2 
gR
----(3)
The velocity v2 in equation (3) represents the critical velocity. If the velocity v2 is below this
critical velocity, the string becomes slack at the highest point and the stone falls downward
instead of moving along the circular path
100
In order that the velocity v2 at the top is not below the critical velocity R,the minimum
velocity v1 at the bottom should be such that v2 becomes equal to
gR
at the top.
When a stone rises through a height 2R, its potential energy increases by an amount equal to the
decrease in the kinetic energy,
½ mv12 = mg (2R) + ½ mv22
v12 = 4 g R + v22
v22 = gR
But,
v12 = 4 g R + g R = 5 g R
v1  5 gR
or
known as centrifugal force. Its magnitude is (mv2 / r). It is not force of reaction. Centrifugal
force is a fictitious force and holds good in a rotating frame of reference.
When a car is turning round a corner, the persons sitting inside the car experience an outward
force. This is due to the fact that no centripetal force is provided by the passengers. Therefore to
avoid this outward force, the passengers are to exert an inward force.
CENTRIFUGE:
It is an appliance to separate heavier particles from the lighter particles in a liquid. The liquid
is rotated in cylindrical vessel at a high speed with the help of an electric motor. The heavier
particles move away from the axis of rotation and the lighter particles move near to the axis of
rotation.
Application:
(1) Sugar crystals are separated from molasses with the help of a centrifuge.
(2) In cream separators, when the vessel containing milk is rotated at high speed, the lighter
cream particles collect near the axle while the skimmed milk moves away from the axle.
101
(3) In drying machines, the wet clothes are rotated at high speed. The water particles fly off
tangentially through the holes in the wall of the outer vessel.
(4) Honey is also separated from bees wax with the help of a centrifuge.
8.2 FRICTION:
In the universe, no body is perfectly smooth. When a body slides over the surface of another
body, there is an opposing force acting opposite to the direction of motion. This opposing force
is called fiction. When a ball is rolled on a rough surface, it comes to rest in a short time due to
friction. If the surface is smooth, friction will be less and the ball rolls for a longer time. Friction
is not only an evil but it is also a necessity.
LIMITING FRICTION:
F
R
A
T
mg
Fig.3.4
Let T be a table and A a block of wood. A string is tied to the block and it passes over a pulley
and carries a scale pan at the other end. Let mg be the weight of the block and R the normal
reaction. The force mg balances the normal reaction R (fig 3.4).
Initially, when there is no mass in the scale pan the block does not move. The force of pull =
Sg (S is the mass of the scale pan). This is due to the fact that a force of friction F acts in the
opposite direction and these two forces balance one another. Adding weights in the scale pan, the
force of pull increases and accordingly the force friction also increases and the block does not
move. The force of friction increases to a certain limiting value called the force of limiting
102
friction, and if the force of pull is increased beyond this, the body begins to move in the direction
of pull.
Thus, when one body is placed over the other, a force acts between the surfaces in contact and
this force is called friction and it is dependent on the nature of the surfaces in contact and the
weight of the body A.
As the value of P is gradually increased, the force of pull (S+P) gradually increases, the force
of friction also increases and when the force of friction is equal to the force of limiting friction,
the body A just begins to move.
Laws of limiting friction:
Fig.3.4.1
(i)
The force of limiting friction is directly proportional to the normal reaction for the two
surfaces on contact and acts in a direction opposite to the direction of the force of pull
(fig 3.4.1).
F  R (when the body A just begins to move).
F = R
where  is the coefficient of friction,

F
R
103
The coefficient of friction is defined as the ratio of the force of limiting friction to the normal
reaction.
(ii)
The force of limiting friction is independent of the size and shape of the bodies in
contact as long as the normal reaction remains the same.
8.2(a) Verification of the first law:
Fig. 8.2a
Keep a mass m1 on the block of mass M. The normal reaction R1 = (M + m1) g. Gradually
increase the weights in the scale pan P1 so that the block just begins to move (fig 82(a)). Then the
force of limiting friction,
F1 = (S + p1) g and R1 = (M + m1)g
Similarly keep different masses m2, m3 etc., on the block A and find the
corresponding
values p2 p3 etc., when the block just begins to move,
R1 = (M + m1) g ; R2 = (M + m2) g ; R3 = (M + m3);
F1 = (S + p1)g ; F2 = (S + p2)g ; F3 = (S + p3)g
Calculate the ratio F1 / R1 ; F2 / R2 ; F3 / R3 . It will be found that these ratios are equal.
104
F
R
is a constant or F  R.
Further, the mean value of F
R
gives the coefficient of friction between the surfaces.
Second law:
R
R
B
A
mg
R
mg
C
mg
Fig.8.2(a)
A, B, C are three different bodies of the same mass but of different shapes and sizes. (fig 8.2a). It
will be found by experiment that the forces of limiting friction are the same in the three cases.
That is force of limiting friction is independent of the size and shape of the bodies in contact as
long as the normal reaction remains constant.
105
8.2(b) ANGLE OF FRICTION AND COEFFICIENT OF FRICTION:
Let ABC be an inclined plane on which a body of mass m is placed. Let the angle of
inclination of the plane be . The weight mg can be resolved into two rectangular components (i)
mg cos  perpendicular to the plane AB and (ii) mg sin  parallel to the plane.
R
R
B
R

R
F
mg sin
mg cos
mg

C
A
Fig.8.2(b)
The force mg cos  balances the normal reaction R and the force mg sin  gives the necessary
force for the body to move down the plane. The force of friction F acts in the opposite direction
and these two forces balance one another if F is less than the limiting friction (fig 8.2(b)).
Now suppose that the angle of inclination of the plane is increased. Then the component mg
sin  also increases (because the value of sin  increases as  increases).
At a particular value of the , body just begins to slide down the inclined plane. At this stage
the value of mg sin  is equal to the force of limiting friction.
R = mg cos 
and
F = mg sin 
---(1)
---(2)
106
F mg sin 

 tan 
R mg cos 
 = tan 
. . . (3)
 is called the angle of friction.
The tangent off the angle of friction is equal to the coefficient of friction between the two
surfaces in contact.
Angle of friction is defined as the tangent of the angle of inclination of the plane when the
body just begins to slide down.
 = tan 
Angle of friction is also defined as the angle made by the resultant of normal reaction and the
force of limiting friction with the normal. Here R is the resultant of F and R. The angle between
R and R is angle of friction.
8.2(c) STATIC AND DYNAMIC FRICTION:
(i) Laws of static friction:
(1) The force of limiting friction is directly proportional to the normal reaction for the same two
surfaces in contact and acts in a direction opposite to the direction of the force of pull.
(2) The force of limiting friction is independent of the size and shape of the bodies in contact
if the normal reaction remains the same.
(ii) Laws of dynamic friction:
(1) The dynamic friction is proportional to the normal reaction and it is less than the static
friction.
(2) The dynamic friction does not depend upon the velocity, provided the velocity is neither
too large nor too small.
107
(iii) Distinction between static and dynamic friction:
Friction is a self-adjusting force which increases with the applied force so long as the body
remains at rest. Static friction is defined as the force required to bring about just motion of one
body over the surface of another. Once the body starts moving, the force
B
Rest
Motion
D
C
E
Applied Force
A
Fig.3.4.5
of friction decreases by a small amount and remains constant so long as the motion of the body is
maintained. The force of friction does not increase further, even if the applied force is increased
(fig 3.4.5). From A to B, the force of friction increases with the applied force and the body is at
rest. At B, the force of friction is maximum and BE measures the static or limiting friction. With
increase in the applied force beyond E, the force of friction decreases by a small amount (BC)
and thereafter the force of friction remains constant. The dynamic friction is equal to PQ in the
graph. Here PQ is less than BE. Therefore dynamic friction is less than the static friction.
When a cart is at rest, the irregularities between the surfaces of the cart wheel and the ground
get interlocked and the projections of one closely fit into the depressions of the other. To
separate these inter locking, force is required. When the cart is in motion, perfect interlocking
does not take place and no force is required to separate them. Further, at the start, the horse has
to apply (i) force F1 to overcome the force of friction (static) and (ii) force F3 to bring the cart
into motion. On the other hand, when the cart is in motion, the horse has to apply a force F2 only
108
to overcome the dynamic friction. Since F1 > F3, F1 + F2 must be greater than F3. Hence a horse
has to apply a greater force to start the cart than to keep the cart in motion.
(iv) Friction, a Necessity:
Friction is a necessity: Without friction between the feet and the ground, it will not be
possible to walk.
(2) The tyres of a motor car and bicycle are made rough to increase friction.
(3) In the absence of friction, the brakes of a motor car cannot work.
(4) It is the friction between the belt and the pulley that helps in the rotation of the various
parts of a machine.
(5) When the ground becomes slippery after rain, it is made rough by spreading sand etc.
8.2(d) Methods of Reducing Friction:
(1) Polishing:
Friction between two surfaces can be reduced by polishing them. The interlockings and
projections are minimized by this method. Jewel bearings in watches and highly polished
agate knife edges in balances are used to minimize friction considerably.
(2) Ball bearings:
Rolling friction is less than sliding friction. Consequently in rotating machinery, the shafts are
fixed on ball bearings so that the friction is reduced considerably. The free wheel of a cycle,
the axle of a motor car, the shafts of motors, dynamos etc., are provided with ball bearings
(8.2(d)).
Fig.8.2(d)
109
(3) Lubricants:
A lubricant is a substance (a solid or a liquid) which forms a thin layer between the two
surfaces in contact. It also fills the depressions present in the surfaces of contact and reduces
friction considerably. In light machinery thin oils with low viscosity are used. In heavy and
fast moving machinery thick oils and solids (grease) are used. Friction between two surfaces
is decreased by using lubricants. Thin oil is used in watches. In heavy machinery grease (thick
oil) is used so that it is not squeezed out by the weight of the rotating parts.
(iv) Cone friction:
A cone with vertex at the point of contact of the two bodies, with normal reaction as axis and the
semi-vertical angle equal to the angle of friction is known as cone of friction.
8.4(a) MOTION DOWN A ROUGH INCLINED PLANE:
R
R
w sin
w cos
w

Fig.8.3a
Suppose a body of mass M is placed on a rough inclined plane, inclined at an angle . The
component of the weight of the body parallel to the plane causes the body to slide down the
plane. The force of friction called into action opposes the sliding and for small values of , the
body will be in equilibrium under the actions of,
(1) the weight of the body Mg = W;
110
(2) the frictional force F which opposes the motion; and
(3) the force of reaction R which acts in a direction perpendicular to the plane.
As increases, the tendency to slide down increases and consequently the force of friction
goes on increasing till it reaches its maximum value R, where  is the coefficient of friction.
Let us consider the case when the body is just at the point component of the weight of the body
parallel to the inclined plane.
Resolving W into its components.
W cos  = R
---(1)
W sin  = R
---(2)
Dividing equation (2) by equation (1),
sin  R


cos 
R
tan  = .
But  = tan, where  is called the angle of friction.
Therefore  = .
8.4(b) MOTION UP THE ROUGH INCLINED PLANE:
Suppose the angle of inclination () of the plane is greater than the angle of friction . Then an
external force is necessary to keep the body at rest or to make it to slide upwards. Let Q be the
force when the body is just on the point of sliding up. Now the frictional force acts in the
downward direction opposing the motion of the body.
Q  W sin   R
Q  W sin  Mg cos . (since R= Mg cos)
Q  W sin    cos 
Q  W sin  
sin 
cos 
cos 
111
Q W
sin  . cos   sin  . cos 
cos 
Q W
sin    
cos 
8.5 Let us Sum Up
In this lesson you learned about the reason for banking of tracks or curves and about the laws of
friction, friction angle and the friction limitations. You also learned about the advantages and
disadvantages of it. Also it is discussed in this lesson about the equivalence of a body in a rough
inclined plane and its related derivation.
8.6 Check your progress
1. State the reason why a man going in a bicycle lean sideways while negotiating a curve.
2. What is friction and state the laws of friction ?
3. Define Angle of friction and cone of friction
8.7 Lesson end Activities
1. A cyclist turns round a curve at 15 km/ hour. If he turns at double the speed, the tendency to
over takes is
a) quadrupled
b) doubled
c) halved
d) unchanged
Choose the correct answer
8.8 Points for discussion
1) (a) write short notes on
(i)
Laws of friction
(ii)
Angle of friction
(iii)
Cone of friction
112
2) Explain different methods of reducing friction.
3) With necessary diagrams, explain the equilibrium of a body on a rough inclined
plane to the horizontal during its up and down motion.
8.9 References
1. Dynamics by D.S. Mathur
113
UNIT – IV
LESSON – 9
CONTENTS
9.0
Aims and Objectives
9.1
Kepler’s laws of planetary motion.
9.2
Newton’s universal law of gravitation.
9.3
Newton law from Kepler’s laws.
9.4
Boy’s experiment.
9.5
Gravitational mass – Gravitational potential.
9.5 a Gravitational potential and field at a point due to spherical shell.
9.6
Variation of g at the poles and at the equator.
9.6a Variation of g with altitude
9.6b Variation of g with depth
9.6c Variation of g with rotation of earth
9.7
Satellites
9.8
Let us Sum up
9.9
Check your progress
9.10
Lesson end Activities
9.11
Points for Discussion
9.12
References
9.0 AIMS AND OBJECTIVES
From this lesson you will understand about the Kepler’s laws of planetary motion and its
importance in the present day life. The meaning of Newton’s law of gravitation and the
derivation of Newton’s law from Kepler’s law is also discussed in detail in this lesson. Moreover
you will come to know about the variation of acceleration of gravity at poles and equator. Also
the principle of satellites is discussed.
114
9.1 KEPLER’S LAWS OF PLANETARY MOTION
The solar system consists of the sun and the various planets revolving around the sun in
different elliptical orbits. The detailed work concerning their motion was done by Copernicus,
Tycho Brahe, and Kepler. Kepler formulated three basic laws of planetary motion using the
comprehensive observation and data collected by Tycho Brahe.
The three laws of planetary motion are:
(1) Shape of the orbit: Every planet moves in an elliptical orbit with the sun being at one of
its foci.
(2) Velocity in the orbit: The radius vector drawn from the sun to the planet sweeps out equal
areas in equal intervals of time.
Consider a planet moving in an elliptical orbit with the sun at the focus S (fig 9.1). Let the
radius vector of the planet describe a small angle d in time dt while moving from the position A
to B. Arc AB = R d.
Area swept in time dt
= Area ABS
= (½)R  R d = ½ R2d
Areal velocity
= (½R2d) / dt
= ½R2(d/dt) = ½R2 = constant
115
Fig 9.1
In this case ½R2  represents the areal velocity of the radius vector. When R decreases, 
increases but ½R2  is always constant, i.e., the areal velocity is constant.
Area of the ellipse = ab
Here a and b are semi-major and semi-minor axes of the ellipse respectively. Time period of
revolution of the planet around the sun,
T = Area / Areal velocity
T = ab / ½R2  = 2ab / R2 
(3) Time periods of planets: The square of the time period of revolution of a planet around the
sun is proportional to the cube of the semi-major axis of the orbit. Thus, if T1 and T2 are
the time periods of revolution and a1 and a2 are the semi-major axes of the planets
respectively,
T12 / a13 = -( T22 / a23) = constant
or
T2 / a3 = constant
116
T2  a3
9.2 NEWTON’S UNIVERSAL LAW OF GRAVITATION:
Newton derived the universal law of gravitation from Kepler’s Laws. Assuming the orbit of
the planet around the sun to be a circle of radius R, the centripetal force acting on the planet is
given by,
F = mv2 / R
But,
v = R
F = mR2

 = 2 / T
Also,

F = mR(2 / T)2
F = 42mR / T2
---(1)
According to Kepler’s third law,
T2  R3
T2 = kR3
---(2)
where k is a constant.
From equations (1) and (2),
F = 42mR / kR3 = (42 / k)m / R2
F = Km / R2
[Here 42 / k = K]
117
F  m / R2
or
---(3)
Thus, the force of attraction acting on the planet provides the necessary centripetal force for
the planet to move in its orbit and this force is,
(i)
Directly proportional to its mass
(ii)
Inversely proportional to the square of its distance from the sun.
However, the force of attraction in mutual and directed along the line joining the two
bodies therefore the force is also proportional to the mass of the body i.e. Taking M as the
mass of sun,
F  Mm / R2
---(3)
F = GMm / R2
---(4)
Here G is the constant of gravitation.
The motion of the planets round the sun led Newton, to formulate the Universal law of
gravitation. According to Newton, this law is true not only for the heavenly bodies but is also
true for any two bodies in this universe. The law holds good for the force of attraction between
the sun and the earth, for the attraction between the moon and the earth, for the attraction of an
electron in the orbit and the nucleus in the atom etc.
According to this law:
Every body in this universe attracts every other body with a force which is directly
proportional to the product of their masses and inversely proportional to the square of the
distance between their centres.
118
Consider two bodies A and B masses M and m respectively. The distance between their
centres,
Fig 5.3
Fig 9.3
= R (fig 5.3)
F  Mm
---(1)
F  1 / R2 ---(2)
Combining (1) and (2),
F  Mm / R2 or F = G [Mm / R2]
Here G is the universal constant of gravitation.
If
M = m = 1,
and
R = 1,
F=G
The constant of gravitation is numerically equal to the force of attraction between two unit
masses at a unit distance apart.
In SI units
G = 6.670  10-11 newton-m2 / kg2
119
9.3 NEWTON’S LAW FROM KEPLER’S LAWS:
Let M be the mass of sun, m1 and m2 be the masses of two planets and distances d1 and d2
from the sun. Then the forces of attraction between the sun and each planet is,
F1 = G . Mm1 / d12
F2 = G . Mm2 / d22
and
F1 / F2 = (m1 / m2) . (d22 / d12)
Let T1 and T2 be the periods of revolution of the two planets around the sun.
Then the central forces are,
f1  m1d1 (2 / T1)2
f2  m2d2 (2 / T2)2
and
From Kepler’s III law, T12  d13 and T2  d23.
Hence f1  m1d1 / d13

or
f1  m1 / d12
and
f2  m2 / d22.
f1 / f2 = (m1 / m2) . (d22 / d12) which is Newton’s law of gravitation.
9.4 BOY’S EXPERIMENT:
Vernon Boye’s devised a method in 1895 to find the constant of gravitation eliminating many
of the errors in the previous methods.
120
Fig 9.4
The apparatus consists of a plane mirror strip AB about 6 long suspended by a phosphor bronze
suspension about 18 long. The attracted masses (m, m) are gold spheres each 0.25 in diameter
and are suspended at two different lovels from the ends of the mirror. The attracting spheres M,
M are suspended separately to be in level with m, m and are made of lead each of diameter 4.5.
The entire suspended system is enclosed in a double walled glass vessel. The deflections are
observed by using a scale and telescope in front of the mirror(not shown in fig). The arrangement
of the heavy spheres at different levels helps in keeping the attraction by the masses to be
minimum.
To start with, one of the smaller spheres is kept in front of a bigger sphere and the other
behind the second larger sphere at the same distance d between them. Due to gravitational
couple, the mirror turns through an angle. Using a telescope and scale, the angle of rotation can
be found. Now the lid from which the larger masses are suspended is rotated by 180 so that the
couple is reversed. Then the mirror deflects by 2 between the two positions. The corresponding
shift s in the scale reading is noted.
Then
2 = s / 2D
or
 = s / 4D
121
and  can be found. D is the distance between the mirror and scale. If C is the couple required
per unit twist, the elastic couple = C. Gravitational force between each set of spheres = G . (Mm
/ d2) . a.
If a is the length of the plane mirror, the moment of the gravitational couple becomes G .
(Mm / d2) . a.
Hence
C. (s / 4D) = G . (Mm / d2) . a.
The spheres are allowed to oscillate freely and the period of oscillation T is found. Then T =
2 (I/C), where I is the moment of inertia about the axis of suspension.

C = (42I / T2)
or (42I / T2) . (s / 4D) = G (Mm / d2) . a
from which G can be calculated. Boy’s value for G was 6.6576  10-11 newton-meter2 / kg2.
Mass and density of the earth:
A knowledge of the absolute value of G helps in computing the mass and mean density of the
earth.
Let M be the mass of the earth.
R be the radius of the earth.
d be the mean density of the earth.
g be the acceleration due to gravity at the earth’s surface.
Consider a particle of mass m on the earth’s surface. The force on the mass,
mg = G . (M . m / R2), from Newton’s law.
Then M = gR2 / G.
Also M = 4/3 R3d.

d = (3g / 4RG).
9.5 GRAVITATIONAL MASS:
According to the law of gravitation, the gravitational force of attraction of a body towards the
centre of the earth is equal to the weight of the body.
122
Let the weight of the body be W,
W = mg
m=W/g
This mass is called the gravitational mass of the body. This is determined with the help of a
beam balance. It will be the same even on the surface of the moon. If the value of g is less on the
moon, the weight of the body will also be less but the gravitational mass is the same.
Note. There is no relation between the4 inertial mass and the gravitational mass of body. But,
for all practical purposes, both are taken to be identical and their units are the same.
Gravitational Potential:
Consider a particle A of mass m. P is a point at a distance r from A. The gravitational
intensity at P is given by,
F = GM / r2
---(1)
Gravitational potential at a point is defined as the amount of work done in moving a unit mass
from the point to infinity against the gravitational force of attraction (fig 5.19)
A
P
M
r
Fig 9.5
F = - dV/dr
or
---(2)
dV = -F  dr = (-GM / r2) dr
Integrating between the limits r and infinity,
 dV = r (- GM / r2) dr = [GM / r]r = [(GM / ) – (GM / r)]
V = - (GM / r)
---(3)
The gravitational potential at a point due to a point mass
= - (GM / r)
It is maximum & zero at infinity. At all other points it is less than zero i.e. negative value.
123
Gravitational potential at a point is a scalar quantity.
9.5(a) GRAVITATIONAL POTENTIAL AND FIELD AT A POINT DUE TO A
SPHERICAL SHELL:
(i) At a point outside the shell
Consider a uniform spherical shell of mass M and radius (fig. 9.5(a)). Let  be the density per
unit area of the shell. The planes AC and BD cut the shell vertically and the element between the
two planes is a ring of radius AE = a sin . The thickness of the element AB = a.d.
Surface area of the element
= 2 (a sin ) a.d = 2a2 sin  d
m = (2a2 sin  d) 
Mass of the element,
AP = x
Fig 9.5(a)
Potential at P due to the element,
dV = - (Gm / x)
dV = - [G (2a2)  sin  d / x]
In the  OAP,
x2 = a2 + R2 – 2a R cos 
Differentiating,
2x dx = 2a R sin  d
---(1)
124
x = (aR sin  d / dx)
Substituting this value of x in equation (i),
dV  
G 2a 2  sin  ddx
aR sin  d
 2Ga 
dV  
dx
 R 
---(2)
Integrating for the whole shell,
R a
 2Ga 
V 
dx
R 
R a 
4a 2 
V 
R
But,
M = 4a2

V = - (GM / R)
---(3)
Thus, for a point outside the shell, the shell behaves as if the whole of its mass is concentrated
at the centre of the shell.
V1/R
Further
---(4)
Gravitational field
E = - (dV / dR)
E = - (d / dR) [-(GM / R)]
^
GM R
E
R2

The negative sign indicated that the field is towards the centre of the shell.
(i)
At a point inside the shell
Potential at a point P inside the shell, due to the element (fig 9.5(b))
 2Ga 
dV  
 dx
 R 
For the whole shell, integrate between the limits (a-R) and (a+R)
125
aR
V 
 2Ga 

dx
R


aR

V=-G(4a)
M=4a2
Fig 9.5(b)
 GM 
V  

 a 
….(5)
Expression (v) shows that 5 is independent of R. It is the same at all points inside the shell. This
potential is equal to the potential is equal to the potential on the surface of the shell and is
constant.
E
E
dV
dR
d   GM 

 
dR   a 
As (-GM/a) is constant E=0.
Therefore, the gravitational field at any point inside a spherical shell is zero.
126
9.6 VARIATION OF g AT THE POLES AND AT THE EQUATOR:
The shape of the earth is slightly ellipsoidal. It is bulging at the equator and flattened at
the poles. Its equatorial radius is more than the polar radius.
g
As
g
GM
R2
1
R2
The value of g at the poles is greater as compared to its value at the equator.
4.6(a)Variation of g with altitude
Let P be a point on the surface of the earth and Q another point at an altitude h . R is the radius
on the earth and M is the mass of the earth. A body of mass m when placed at the point P
experiences a force mg towards the centre of the earth.
Fig 9.6(a)

mg 
GMm
R2
. . . (1)
when the body is at Q, let the acceleration due to gravity be g’
mg ' 
GMm
2
 h
R
(2)
127
Dividing (2) by (1)
g'
R2

2
g R  h 
 2h 
g '  g 1  
R

(3)
Therefore the acceleration due to gravity decreases with the altitude
9.6(b)Variation of g with depth:
Let g and g’ be the acceleration due to the gravity at point P and Q respectively. At point
P the whole mass of the earth attracts the body and at Q it attracted by the mass of the earth of
radius (R-h)

mg 
GMm
R2
(1)
mg 
GM ' m
R  h 2
(2)
Fig 9.6(b)
4
8
Here M  R 3 
128
M '
4
3
 R  h 
8
where  is the mean density of the earth.
Dividing (ii) by(i) we get
4
R  h 2 R 2
g m R2
 .
 3
g m R  h 4
R 2 R 2 R  h 
3
 h
g   g 1  
 R
Therefore g decreases with depth
9.6(c)Variation of g with rotation of earth
Consider a body of mass m on the surface of the earth at a latitude .
Here

OP = R
BP = R cos 
Let  be the angular velocity of the earth.
Force mg acts along PO. Resolve mg in to two rectangular components (i) mg sin  along
PA and (ii) mg cos  along PB (fig 9.6(b).
Due to the rotation of the earth, a body at P experiences an outward force
Fig 9.6(c)
129
= mv2 / r = mr2
r = BP = cos 
The net force along PC = mg cos  - m R2 cos .
The resultant force experienced by P is along PQ which is the resultant of the forces
along PA and PC (represented vectorially).
 F
mg sin    mg cos   mR
 mg 1 
R 2  4 cos 2  2 R 2 cos 2 

g2
g
Here
2
2
2
cos  
R 2 4 cos 2 
is negligible
g2

F  mg 1 
2R 2 cos 2 
g
 1  R 2 cos 2  

mg '  mg 
g


 R 2 cos 2  

g '  g 1 
g


(1)
Special cases
(i)At the equater,
  0, cos   1

R 2 

g '  g 1 
g 

But
R 2
1

g
289
1  288

g '  g 1 

 289  289
(2)At the pole  = 90 , cos = 0

g’ = g
Therefore the acceleration due to gravity at the poles is greater than that at the equator.
130
9.7 SATELLITES:
In the solar system, different planets revolve round the sun. The radii of the orbits and
their time periods of revolution are different for different planets. In these cases, the force of
gravitation between the sun and the planet provides the necessary centripetal force. Similarly the
moon revolves around the earth and the moon provides the necessary centripetal force for the
moon to be in its orbit. Here moon is the satellite of the earth.
From 1957, many artificial satellites around the earth have been launched. The satellites
are put into orbits with help of multi-stage rockets.
Orbital velocity. Suppose a satellite of mass m is moving in a orbit of radius r around the earth.
If v is the velocity of the satellite in its orbit, the necessary centripetal forcethe force of attraction
between the earth and (mv2 / r) is provided by the gravitational force of attraction between the
earth and the satellite.
Gravitational force = GMm / r2
Here, M is the mass of the earth
 mv2 / r = GMm / r2 or v = [GM / r]1/2
--------------(1)
If g is the acceleration due to gravity at the position of the satellite
mg 
GMm
r2
or
GM
 gr
r
………………………(2)
From equations (1) and (2)
v  gr
For a satellite, revolving very near the earth’s surface,
v  gr
v  gR
Taking the velue of
g = 9.8 m/s2
…………………………(3)
131
R= 64 x 105 m.
v = [9.8 x 64 x 105]1/2 = 7920 m/s
= 7.92 km/s.
9.8 Let us sum up
In this lesson you learned about the Kepler’s laws of planetary motion , Universal law of
gravitation due to Newton and derivation of it from Kepler’s laws. The concept of acceleration
due to gravity is also discussed in particular about its variation at poles and equators. Also the
principle of satellite is discussed.
9.9 Check your progress
1. State Kepler’s laws of planetary motion
2. Define Newton’s Universal law of gravitation and mention its merits
3. Derive Newton’s law from Kepler’s laws
4. Write a note on Satellites
9.10 Lesson end Activities
Choose the correct answer
1. Weightlessness experienced while orbiting the earth in spaceship is a result of
a) inertia
b) zero gravity
c) centre of gravity
d) acceleration
2. Show that the objects lying at the equator will fly off the surface of earth, if the speed of
rotation of the earth increases seventeen times its present rate.
9.11 Points for Discussion
1) (i) State and derive Kepler’s laws of planetary motion
(ii) Determine G by Boy’s method
132
2) (i) Discuss the variation of g with (a) altitude (b) depth (c) lattitude
(ii) What is gravitational potential ? Discuss.
3) Determine the gravitational potential and field due spherical shell. Discuss different
cases.
9.12 References
1) Ancillary Physics by Narayanamurthy
2) Modern Physics by Brijlal and Subramanyam
133
LESSON – 10
10.0
Aims and Objectives
10.1
Young’s modulus of a material
10.1(a) Relation between Y,K,N
10.1(b) Poisson ratio
10.1(c) Relation Between Three Modui Of Elasticity
10.1(d) Poisson Ratio
10.2
Energy stored in a stretching wire
10.2 (a) Work done in a stretching wire
10.3
Work done in twisting
10.4
Bending of beams
10.5
Cantilever
10.6
Young’s modulus by non-uniform bending
10.7
Young’s modulus by uniform bending
10.8
I form of girders
10.9
Let us Sum up
10.10 Check your progress
10.11 Lesson end activities
10.12 Points for discussion
10.13 References
10.0 AIMS AND OBJECTIVES
In this lesson you will learn about the different categories of modulus of elasticity and its
inter relation. The importance of I – form girders will also be discussed in this lesson . Also
you will learn about the importance of uniform and non uniform bending of beams and the
determination of bending moment.
134
10.1 YOUNG’S MODULUS OF A MATERIAL:
Suppose that a cylindrical wire of a length L and radius r is suspended from a rigid
support. If a mass M is attached to its free end, a force Mg will act on the wire. This force will
increase the length of the wire. Let the increase in length be l.
Now longitudinal strain is the ratio of increase in length to the original length.
Fig 10.1
 longitudanal strain 
increasein length l

orginal lenth
L
Longitudinal stress is the ratio of the applied force to the area on which it acts. In this
case the applied force Mg isacting on the area of cross-section of the wire.
 longitudanal stress 
Youngs mod ulus 
applied force
Mg

area of cross sec tion r 2
longitudanal stress
longitudanal strain
Y
Mg / r 2
l/L
135
Y
MgL
r 2 l
Three different modulus of elasticity
Just as there are three kinds of stress and strain, there are three kinds of modulus of
elasticity.
The modulus of elasticity related to the changes in length, is called the Young’s modulus.
Young’s modulus is defined as the ratio of the longitudinal (tensile) stress to the longitudinal
(tensile) strain.
Youngs mod ulus 
longitudanal stress
longitudanal strain
The modulus of elasticity related to the changes in volume is called the volume elasticity
or bulk modulus. Bulk modulus Id defined as the ratio of the volume stress to the volume strain.
Bulk mod ulus 
Volume stress
volume strain
The modulus of elasticity related to the changes in shape is called the modulus of rigidity.
Modulus of rigidity is defined as the ratio of the shearing stress to the shear.
 Modulus of rigidity 
shearing stress
shear strain
These three module of elasticity are also called the elastic constant.
136
10.1 BULK MODULUS OF A MATERIAL
Consider a body in form of a sphere of volume V under pressure P. Suppose that an
additional pressure dP is applied to the sphere, from all sides, at right angles to its surface (fig
4.8). Due to the pressure, the sphere is compressed and its volume decreases. Let the change in
volume be dV.
Volume strain is the ratio of the change in volume to the original volume.
Volume strain 
change in volume dV

orginal volume
V
Volume stress is equal to the applied force per unit area.
But by definition, the applied force per unit area is equal
to the pressure.
Volume stress 
applied force
 applied pressure  dP
area
K 
dP
dV
V
V
Fig 10.1(a)
dP
dV
10.1(b)Modulus of rigidity of a material.
Consider a solid in the form of a cube. The vertical section ABCD of the cube is shown
in the figure. Let a tangential force F be applied to the upper surface of the cube, while the lower
surface is kept fixed. The force will cause a change in the shape of the cube.
137
Fig 10.1(b)
Due to the force, each layer of the cube, except the fixed layer, gets displaced in the direction of
the force. The shearing strain is defined as the ratio of the lateral displacement of any layer to its
distance from the fixed layer.
 Shearing strain or Shear 

lateral displacement of a layer
its dis tan ce from the fixed layer
AA '
 tan 
AD
The angle  is usually very small and hence we can write tan  = 
Therefore,
Modulus of rigidity 
F
n 

A

F
A
shearing stress 
shearing stress
shear
tan gential force F

area
A
138
The SI and CGS unitsof (i)Stress (ii)Strain and (iii)Modulus of elasticity. Also derive their
dimension.
Ans. (i) Stress is the ratio of force to the area.
 Unit of stress 
unit of force
unit of area
Thus the SI unit of stress is Newton / (metre)2, i.e., (N/m2) and its CGS unit is dyne/cm2.
(ii) Strain is a ratio of two similar quantities; therefore it has no units. Strain is a pure number.
(iii) Modulus of elasticity 
stress
strain
As strain has no units, the unit of modulus of elasticity is the same as that the stress, i.e., N/m2
in the SI system and dyne/cm2 in the CGS system.
stress  force 
area 

M
L1T 2 
L2
1
 M 1 L1T 2 
As strain is a pure number, its dimensions are zero in mass as well as in length and time, so
that dimensions of strain can be written as [M0 L0 T0].
Therefore the dimensions of modulus of elasticity are given by,
Modulus of elasticity 
stress M 1 L1T 2 
 1 1 2 

strain
L2   M L T
Thus the units and dimensions of modulus of elasticity and those of stress.
139
10.1(c)Relation Between Three Modui Of Elasticity
suppose three stresses P,Q and R acted perperdicular to three faces ABCD, ADHE and ABFE of
the cube of unit volume(fig 10.1(c))each of these stresses will produce an extension in its own
direction and compression along the other two perpendicular directions.
Fig. 10.1(c)
If  is the extension per unit length, then the elongation along the direction of P will be P .if 
is the contraction per unit stress then the elongation along the direction of P due to other two
stresses will be Q and R
 the net change in the dimension along the direction of P due to all the stresses is
e= P-Q- R
similarly the net change in the dimension along the direction of Q f=Q-P- R and the net
change in the dimension along the direction of R g = =R-P- Q
Case (i):
If only P acts and Q=R=0 then it is a case of longitudinal stress
Linear strain = P
Young’s modulus Q=
(i.e.) q=1/
linear stress
p

linear strain P
or
=1/q
140
Case (ii):
if R=0 and P=-Q then the change in the dimension along the direction of P is e=P-(-P)
(i.e.) e= (+) P
Angle of shear =2e*=2(+) P
Rigidity modulus
n
P


P
2   P
or 2   
1
n
If P=Q=R the in the volume is =e+f+g
=3e=3(-2) P
(since e = f =g)
Bulk strain = 3(-2) P
Bula modulus k =P/(-2)P
From (2) 2    
2  2  
or (-2)) = 1/3k
…………(3)
1
n
1
n
From (3), (-2)) = 1/3k
Adding (4) and (5)
3 

1 1

n 3k
1
1

3n 9k
From (1)
1 1
1


q 3n 9k
9 3 1
 
q n k
This is the relation between the three moduli of elasticity
10.1(d) Poisson Ratio:
The ratio of the lateral contraction per unit stress to the longitudinal elongation per unit
stress in called Poissson’s ratio. Consider a wire being stretched by a load at one end, and the
other end being fixed. If  in the longitudinal elongation produced by unit stress, there will be a
corresponding contraction  in the perpendicular plane. Then Poisson’s ratio
141
=/
It is also related to the moduli of elasticity as
 = (3k-2n/6k+2n)
 = (q/2n)-1
and
Note: Its value ranges between -1 and +1/2 theoretically and in all practical cases between 0 and
+1/2.
10.2 ENERGY STORED IN A STRESSED WIRE:
Let us consider a wire of length l, are of cross-section a of a material of Young’s modulus q.
Let a force F stretch the wire by a small distance dx. Then the work done is F. dx. The total work
done in stretching the wire by a distance x is then given by,
x
W   F .dx
0
The Young’s modulus q 
Stretching force F 
x
Then,
W 

0
F
stress
Fl
 a
strain x
ax
l
qax
l
qa
1 qax 2 1 qax
xdx 
 .
.x
l
2 l
2 l

1 qax l x
.
.   a..l
2 l a l

1 streaching force elongation


 volume
2
area
length

1
 stress  strain  volume
2
Hence the potential energy stored in a stretched wire = ½  stress  strain  volume.
142
10.2(a) WORK DONE IN A STRETCHING WIRE:
If a stretching force F applied to a wire of length l and area of cross-section a produces an
extension dx, the work done is given by,
………………(1)
W  F .dx
The work done in stretching a wire from its original length l to a length (l + x) will be,
x


W  dW  F .dx
………………..(2)
0
But
q
F
stress
Fl
 a
strain x
ax
l
Therefore
F
qax
l
………………(3)
………………(4)
Substituting the value of F in equ (2) we get
x
W 

0
qax
qa
dx 
l
l
x
 xdx
0

qa x 2
l 2

1 qax
.
.x
2 l

1
F.x
2

1
streaching force  extension
2
This work is stored in the stretched wire in the form of elastic potential energy.
1 qax
.x / al
2 l
Work done per unit volume of the material of the wire  .
(since volume =al)

1 qx x
.
2 l l

1
 stress  strain
2
= average stressstrain
143

It is also
1
2
q  strain 
2
10.3 WORK DONE IN TWISTING:
A cylinder of length l and radius r is twisted through an angle  when the couple acting on it
is
nr 4
C

2l
………………….(1)
Where n is the rigidity modulus of the material of the cylinder. C is also the restoring couple that
comes into play in the cylinder due to the twisting couple.
In the case of twisting couple, the magnitude of the couple is not constant; it increases as 
increases.
For increasing the angle of twist by a small amount d during which the twisting couple is
assumed to be practically constant, the work done against restoring couple is
dw = Cd
…………….(2)
On substituting for C,
nr 4
dW 
 d
2l
The total work done in twisting the cylinder from its original position through an angle  is

W 
nr 4
0 2l d
nr 4  2

2l 2
=

1  nr 4 

 
2  2l

1
C   average couple  twist
2
144
Where C is the final couple, i.e., to produce a twist equal to .
This is the elastic potential energy in the twisted state.
10.4 BENDING OF BEAMS:
A beam is a homogeneous material of uniform cross-section whose breadth and thickness are
negligibly small in comparison with its length. It can be bent by applying an external couple.
Inner layers below the central layers of the bent beam along its length are subjected to a
compression strain while the layers above it are elongated. The length of the middle layer along
the length of the beam does not change. This is called the neutral surface. As the beam is bent by
the external couple, elastic forces come into play and oppose this. In the bent position, the couple
due to elastic forces balance the external couple and the moment of this couple is called the
bending moment.
Fig 10.4
Let us now derive an expression for the bending moment of a beam. ABCD represents the
cross-section for the bent beam along its length. PQ is the neutral axis. PQ is another filament at
a distance y from PQ. Let the beam be bent in the form of an arc of a circle of radius R at PQ
with centre at O. Let  be the angle POQ.
Then
PQ = R
PQ = (R + y) 
Extension = PQ - PQ = y
145
Extentional strain 
increasein length y
y


roginal length
R R
Let q be the Young’s modulus of the material of the beam. Then,
Longitudinal stress on PQ  q  strain 
q. y
R
y
R
If a is the area of cross-section of the filament, longitudinal force  q . a
y
R
Moment of this force about PQ  q. .a. y 
qay 2
.
R
Hence the sum of the moments of all the forces on the entire beam is,

qay 2
R
 ay
2
Or
q
R
 ay
2
.
is called the geometric or second moment of inertia and can be denoted by i.
Therefore, internal bending moment 
qi
.
R
For rectangular cross-sections of breadth b and thickness d, i 

ay 2  bd .
d bd 2

12 12
For circular cross-sections of radius r,
i

ay 2  r 2 .
r 2 r 4

4
4
10.5 Cantilever
Consider a light horizontal beam clamped at one end and suspended by weight Mg at the
other. This beam is called a cantilever. The bending is non-uniform. Let l be the length of the
cantilever. The free end of the beam is depressed by a distance . Let q be the Young’s modulus
of the material of the beam. The cantilever is bent in the form of an arc of radius R and the
elastic restoring couple =
q.i
R
where I is the moment of inertia
146
Let y be the depression of point at a distance x from the clamped end. Then the bending
moment = Mg (l - x).
  dy  2 
1    
  dx  
It is known from geometry that R  
d2y
dx 2
 dy 
 
 dx 
is small in practical cases and R 
1
d2y
dx 2
3
2
.
.
At equilibrium,
q.i
 Mg l  x 
R
or qi
d2y
 Mg l  x 
dx 2
Integrating,
qi

dy
x2 
 Mg  lx    C1
dx
2 

where C1 = constant of integration.
We know, at x = 0,
Hence,
dy
 0 . Then C1 = 0.
dx
qi
dy

x2 
 Mg lx  
dx
2 

Integrating again,
 x2 x3 
qiy  Mg  l
   C 2
6 
 2
147
when x = 0; y = 0 and C2 = 0
 x 2 x3 
 l
qi
y

Mg
 
Hence
6
 2
The depression at the free end is .
Therefore, when x = l, y = .
l3 l3 
qi   Mg   
6 
 2
Then,
Mgl 3
q
3i
Or
10.5(a) Young’s Modulus by Non-Uniform Bending
Consider a beam of uniform cross-section supported symmetrically on two knife-edges
separated by l and loaded (Mg) at the middle. The arrangement is called non-uniform bending.
This is equal to two cantilevers with the points at the knife-edges as points of clamping and
loaded at the centre of the beam. Beyond the knife-edge is Mg / 2 and the length of each
cantilever is (l/2).
For a rectangular beam i 
bd 3
. Then, the equation of the cantilever becomes
12
3
Mg  l 
. 
Mgl 3
2 2
q

bd 3
4bd 3 .
3. .
12
148
Fig. 10.5a
The experiment consists in fixing a vertical pin at the centre of the beam and viewing its tip by
a traveling microscope. The beam is loaded and unloaded in steps and the depressions are
observed when the horizontal crosswire of the microscope coincides with the tip of the pin. If 
is the depression for a mass M, we have q 
gl 3  M   2
. nm and can be calculated. The
4bd 3   
experiment is repeated for different distances between the knife-edges.
10.5(b) Young’s Modulus by Uniform Bending:
Consider a light metre scale CD placed symmetrically on two knife-edges A and B and loaded
with two equal weights Mg at each end. From symmetry the reactions vertically upwards at A
and B are each Mg. P is the centre of the beam at which a vertical pin is attached. The point is
raised due to the load and the beam forms an arc of a circle.
Fig. 10.5(b)
149
The forces Mg at A and C form a couple of moment Mg . a, where a = AC or BD.
qi
R
The bending moment =

q.i
 Mg.a
R
Since this moment is constant, the bending is called uniform.
If h is the elevation of the mid-point of the beam and R the radius of curvature, we have
l2
h2 R  h   , where l is the distance between knife-edges. When h is small compared to R,
4
2 Rh 
l2
4
l2
R
8h
q.i
 Mga
l2
8h
Hence
or
and
i 
q
Mgal 2
8hi
i
bd 3
for rectangular cross-section and
12
r 2
for circular cross-section.
4
Experiment
The beam is supported symmetrically on two knife-edges. Equal weights are suspended from
the two ends by weight hangers. A pin is fixed vertically at the mid-point and its tip is viewed by
a traveling microscope. Weights are added in steps and in each case the vertical scale reading
corresponding to the coincidence of the horizontal crosswire at the tip of the ping is noted. The
experiment is repeated while unloading the weights. The average elevation for a convenient load
150
M is calculated. For ordinary metre scale of breadth b and thickness d, we have the Young’s
modulus q 
3gal 2  M 
 
2bd 3  h 
and is calculated.
10.6 I FORM OF GIRDERS:
When a beam is supported at the edges and loaded at the middle, the centre of the beam is
lowered. This depression at the mid-point of the beam varies.
(i) directly as the load applied;
(ii) as the cube of the distance between the knife-edges;
(iii) inversely proportional as the Young’s modulus of the material
(iv) inversely as the breadth and cube of thickness of the beam.
Thus, when girders are erected, they are subjected to this type of “non-uniform” bending due
to their own weight and the centre sags. To reduce the bending to the minimum, d has to be large
and the mass of beam small. These go contrary to each other. The problem can be got over by
having the cross-section of the girder in the form I which satisfies both the conditions of larger
depth and smaller mass. The top and bottom surfaces of the girder experience maximum stress
and hence more material should be at these positions. Thus the I form for the girder is the most
suitable one.
10.7 Let us sum up
In this lesson you learned about the different types of modulus of elasticity and its relation
between them . Also the importance of I – form girders in the construction of big buildings is
discussed. You also learned about the uniform and non uniform bending and the determination of
bending moment.
10.8 Check your progress
1. Define Poisson’s ratio
2. Derive the workdone in stretching a wire
3. What is the importance of I – form girders
151
10.9 Lesson end Activities
1. An elongation of 0.1 % in a wire of cross section area 10-6 m2 causes tension of 100 N.
Determine the youngs modulus.
10.10 Points for Discussion
1) (i) Define and derive three moduli of elasticity.
(ii) find the relation between them
2) (i) Derive an expression for the bending moment
(ii) Determine the young’s modulus of a material by (a) uniform bending (b) non uniform
bending
10.11 References
1) Mechanics by D.S. Mathur
2) Ancillary Physics by Narayanamurthy
152
LESSON – 11
CONTENTS
11.0 Aims and Objectives
11.1
Torsion in a wire
11.2
Static torsion
11.3
Torsion pendulum
11.3a Searle’s method for the determination of the young’s modulus of the
material of the wire
11.4
Application of modules of elasticity
11.5
Let us Sum up
11.6
Check your progress
11.7
Lesson end Activities
11.8
Points for Discussion
11.9
References
11.0 AIMS AND OBJECTIVES
To aim of this lesson is to introduced the concepts of reaction taken place in wire
twisting and application of modulus electricity.
11.1 TORSION IN A WIRE:
Let us consider a cylindrical wire rigidly clamped at the top end and twisted at the bottom. It
is said to be under torsion. An elastic couple called torsional couple comes into play and the wire
will be in equilibrium in the deformed state.
Let l be the length of the wire, a its radius and n the rigidity modulus of its material. Let the
wire be fixed at the top and twisted at the bottom by a couple of moment C. Consider an annular
tube of radius r and thickness dr. When the wire is twisted through an angle , the radius OD
rotates by an angle  to OD1. The longitudinal line rotates by an angle DAD1 =  which is the
angle of shear.
153
Fig. 11.1
From the figure DD1 = l = r or  
Then the rigidity modulus n 
Hence stress produced n  n.

r
.
l
stress
stress

angle of shear

r
and the corresponding force = stress area
l
nr 
. 2  r . r
l
Moment of this force about the axis
nr 
2n 
. 2  r . r . r 
. r 3 . r
l
l

Moment of all the forces acting on the entire wire
a
=

0
2n 3
na 4
r r 
.
l
2l
This, when in equilibrium is equal to the elastic couple.
If c is the couple required to produce unit twist, c 
c
na 4
.
2l
na 4
. or couple per unit twist
2l
154
11.2 STATIC TORSION:
The Searle’s static torsion apparatus consists of a thin horizontal metal rod clamped at one
end and attached to the centre of a vertical wheel at the other. There is a groove along the
circumference of the wheel over which a rope is fixed to one point and passes over it. A weight
hanger is attached to one point and passes over it. A weight hanger is attached at the free end of
the rope. Two circular arcs divided into degrees are clamped to the base of the instrument. Two
pointers, tightly screwed to the rod in front of the scales measure the angles of twist of the rod.
(i)Direct method:
At first without any load in the weight hanger, the screws are adjusted so that the scales read
zero. Now weights are added in steps in the weight hanger and the scale readings are noted. The
readings are noted while unloading. Then the couple is acted in the opposite direction and the
average value of the angle of twist  for a load M is calculated for each scale.
Fig. 11.2
Let l1 and l2 be the distances of the pointers from the clamped end of the rod. Let 1 and 2 be
the corresponding deflections.
Then l = l1 – l2 and  = 1 - 2
External couple = MgR.
Elastic couple =
Then
na 4
.
2l
2 lg R M
na 4
. .nm  2 and is calculated.
.  MgR and n 
4
a

2l
155
(ii)Telescope method:
A small plane mirror is stuck to the rod at a distance from the clamped end. A scale and
telescope are arranged in front of the mirror to measure the twist of the mirror.
At first, without any load in the weight hanger, the reading zero at the centre of the 25-0-25
scale is made to coincide with the horizontal crosswire. Weights are added in steps to the weight
hanger to give clockwise couple and the corresponding scale readings are noted. The readings
are again noted while unloading. Now the rod is allowed to rotate in anticlockwise direction and
the scale readings are noted in the same manner. The shift in the scale reading(s) for a definite
mass M in the hanger is calculated. The distance D between the scale and mirror is measured.
Then the twist  
s
.
2D
The rigidity modulus of the material of the rod n 
2 lg R M
. 2 D Newtons per square metre and is
a 4 s
calculated.
11.3 TORSION PENDULUM:
The torsion pendulum consists of a circular or rectangular metal disc being suspended by a
thin uniform wire by an adjustable chuck. The other end of the wire is clamped from a rigid
support. The length of the wire from the bottom of the support to the disc is adjusted to a
convenient value. Two equal masses, m each, are placed symmetrically on the disc at a distance
d1 from the centre. The disc is rotated so that the wire is given a twist. The period of torsional
oscillations T1 is found. The experiment is repeated for a larger distance d2 between the centre of
disc and the centre of the mass and the corresponding period T2 is found. If I0 is the moment of
156
inertia of the suspension without the masses about the wire and i0 the moment of inertia of each
mass about a parallel axis through its C.G. we have
T1
and
2
T2
2
4 2
I 0  2 i 0  2 md

c
4 2

I 0  2 i0  2 md
c
2
1
2
2


Then
4 2
T2  T1 
.2md 2 2  d 2 1 .
c
2
2
Substituting c 
na 4
16  lm d 2 2  d 2 1
we have n 
2l
T 2 2  T 1 2 . a 4

and is found.
The experiment can be repeated for different lengths of suspension.
Theory
Consider a wire of length l and radius a under torsional oscillations. Let c be the couple per
unit twist 
na 4
.
2l
For a twist , the potential energy or work done in twisting  1 c  2
2
If I is the moment of inertia of the system about the axis of suspension and  the angular
velocity, the kinetic energy is given by
1 2 1  d 
I  I .

2
2  dt 
From the law of conservation of energy,
2
1 2 1  d 
c  I .
  cons tan t
2
2  dt 
Differentiating with respect to time,
1
d 1
d d 2
.c.2 .
 .I .2 .
.
0
2
dt 2
dt dt 2
or
d 2 c
 .  0
dt 2 I
2
.
157
d 2
c
  .
2
dt
I
This is a case of S.H.M.
Thus the period of torsional oscillations is given by T  2
I
c
and T  2
2 Il
na 4
11.3(a) Searle’s Method for the Determination of Young’s Modulus of the Material Wire:
By using Searle’s method, we can accurately determine the Young’s modulus of a material
available in the form of a wire.
Principle: In this method, a long wire is strectched by the application of a force and the
elongation in the wire is measured. The longitudinal stress and strain are then calculated and
their ratio gives the Young’s modulus of the material.
Fig. 11.3(a)
Apparatus: Even when a large force is applied to a wire, the increase in its length is very small.
In order to measure this elongation accurately, Searle’s apparatus is used. It consists of two
rectangular metal frames P and Q, tied loosely together by a third frame R (fig. 11.3(a)). Due to
the loose coupling, relative displacement can take place between frames P and Q. One end of a
158
spirit level L is fixed to the frame P, while its other end is supported by the tip of the micrometer
screw S. The screw S can move in a nut fixed to the frame Q. Instead of using only one wire, two
long and identical wires A and B of the material are suspended from the same rigid support and
to the free ends of these wires, the frames P and Q are fixed. A sufficie4ntly heavy load is
attached at the lower end of the frame P, sothat the wire A remains straight without any kinks. A
hanger H is suspended from the frame Q. The hanger hepls to stretch the wire B without any
kinks. The weight of the hanger, the wire B can be elongated. Thus the wire B is the
experimental wire, while the wire A is the reference wire used for comparison.
Procedure: With the hanger suspended from the frame Q, the original length (L) of the
experimental wire B is measured. Using a micrometer screw gauge, thje diameter of the wire is
measured at difffernt points along its length and its average radius (r) is accurately determined.
By adjusting the screw of the Searle’s apparatus, the sprit level is made horizontal and the
micrometer scale reading is noted. This is the zero reading. The load (M) applied to the wire is
now increased in steps of half kilogram. During each step the sprit level is made horizontal by
adjusting the screw and the scale reading is taken. The difference between this reading and the
zero gives the extension (l) of the wire corresponding to the applied load. In this way, about five
or six readings are taken.
The load is then decreased, again in steps of half the kilogram, and the procedure is
repeated. This enables us to determine the average extension of the wire corresponding to each
load.
Young’s modulus can be calculated by using the formula
Y
MgL
r 2 l
A graph of extension against load (M) in kg can be plotted and from the slope, the average
values of M/l can be found. Substituting this value in the formula, average value of Y can be
calculated.
159
Elimination of errors: Use of searle’s apparatus enables us to eliminate the following sources
of error in the experiment (i) due to the applied heavy load, the support bends by a small amount.
This is called yielding of support. If only a single wire is used to determine the Young’s
modulus, the yielding of support introduces an error in the measurement of extension in the wire.
(ii) During the experiment, the room temperature may change, causing a change in the length of
the wire. Due to this, an error is introduced; if only one wire is used.
Both these errors are eliminated in searle’s method, where two wires are used. Due to
yielding of support, both the wires are lowered equally. Also the effect of changes in temperature
is same for both wires. As we measure the elongation in the experimental wire by comparing its
length with the reference wire, both the above errors are eliminated. Therefore the young’s
modulus can be accurately determined by searle’s method.
11.4 Applications of modulus of elasticity
1) Most of us would have seen a crane used for lifting and moving heavy loads. The crane has a
thick metallic rope. The maximum load that can be lifted by the rope must be specified. This
maximum load under any circumstances should not exceed the elastic limit of the material of the
rope. By knowing this elastic limit and the extension per unit length of the material, the area of
cross section of the wire can be evaluated. From this the radius of the wire can be calculated.
2) While designing a bridge one has to keep in mind the following factors.
(i) Traffic load
(ii) Weight of the bridge
(iii) Force of winds.
The bridge is so designed that it should not neither bend too much nor break.
11.5 Let us sum up
From this lesson you learned about the reaction taking place in wire in twisting. Also its
application in Torsion pendulum. Finally about the application of modulus of elasticity is
discussed.
160
11.6 Check your progress
1. What is meant by static torsion
2. State the principle of Torsion pendulum
3. Mention any two applications of modulus of elasticity
11.7 Lesson end Activities
1) Explain the Searls method for the determination of Youngs modulus
11.8 Points for Discussion
1) (i) Derive an expression for the couple per unit twist
(ii) How you determine the rigidity modulus of a material by torsion pendulum
11.9 References
1) Modern Physics by Brij Lal and Subramanyam
2) Mechanics by D.S. Mathur
3) Ancillary Physics by Narayanamurthy
161
UNIT – V
LESSON - 12
CONTENTS
12.0
Aims and Object ives
12.1
S imple Har mo nic Mot ion
12.1(a) Represent at io n o f SHM by a wave
12.1(b) Different ial Equat ion o f SHM
12.2
Wave Mot ion
12.2(a) Charact er ist ics o f Wave Mo t io n
12.2(b) Transverse Wave Mot ion
12.2(c) Longit udinal Wave Mot io n
12.2(d) Propert ies o f Lo ng it ud ina l Progressive Waves
12.3
Defin it io ns
12.3a Relat io n bet ween Frequency and Wavelengt h
12.4
The resultant of two S.H. motions in the same directions when they have the same
periodic time.
12.5
Let us Sum up
12.6
Check your progress
12.7
Lesson end Activities
12.9
Points for discussion
12.10 References
12.0 Aims and Objectives
To int ro duce t he simp le Har mo nic Mot ion were discu ssed in t his lesso n
and Wave Mo t ion is also deeply discussed belo w.
162
12.1 Simp le Harmoni c Motion
The propagat io n o f a simple har mo nic wave t hrough a mediu m can be
t ransverse or lo ngit ud inal.
In a t ransver se wave, t he part icles of t he mediu m
vibr at e perpendicular to t he d irect ion o f propagat ion and in a lo ngit udina l wave,
t he part icles o f t he med iu m vibrat e par alle l t o t he direct io n o f propagat ion.
When a st o ne is dropped on t he sur face o f wat er, t ransver se waves are produced.
Pro pagat io n o f sou nd t hrough at mospher ic air is in t he fo r m o f lo ng it udina l
waves.
When a wave is propagat ed t hrough a medium, t he p art icles o f t he
med iu m are displaced fro m t heir mean posit io ns o f r est and t he rest o r ing fo rces
co me int o p lay. These rest oring fo rces are due t o t he elast icit y o f t he mediu m,
grav it y and sur face t ensio n. Due t o t he per iodic mot io n o f t he part icles o f t he
med iu m, a wave mo t io n is produced.
At any inst ant , t he cont our o f all t he
part icles o f t he med ium co nst it ut es a wave.
Let P be a part icle mo ving on t he cir cumference o f a circle o f radiu s 'a'
wit h a u nifo r m velo cit y  (Fig 12.1). Let  be t he u nifor m angular velocit y o f
t he part icle ( = a ). The circle a lo ng which P mo ves is called t he ci rcle of
reference. As t he part icle P mo ves round t he circle cont inuously wit h unifor m
velo cit y, t he fo ot o f t he perpend icular M, vibr at es alo ng t he dia met er YY' (or
XX'). If t he mot ion o f M is p er iod ic i.e. it t akes t he same t ime t o vibr at e once
bet ween t he po int s Y and Y'. At any inst ant , t he dist ance of M fro m t he cent re
O o f t he circle is called t he d isplacement . I f t he part icle mo ves fro m X t o P in
t ime t , t hen POX = MPO =  = t
163
Fig
12.1
Fro m t he Δ MP O
S in = sin t
or
OM
= OM/a
= y = a sin t
…(1)
OM is called the displacement of the vibrating particle. The displacement of a vibrating particle
at any instant can be defined as its distance from the mean position of rest. The maximum
displacement of a vibrating particle is called its amplitude.
Displacement = y = a sin t
@
The r at e o f change o f displacement is called t he velo cit y o f t he vibr at ing
part icle.
Velocity
= dy/dt = + a cos t
…(2)
The r at e of change o f velocit y o f a vibr at ing part icle is called it s
accelerat io n.
Accelerat ion
= Rat e of change of velo cit y
= d/dt (dy/dt )
= d 2 y/dt 2 = -a 2 sin t
…(3)
164
The changes in the displacement, velocity and acceleration of a vibrating particle in one
complete vibration are given in the following table.
Velo cit y
dy / dt = a
cos t
Ang le
t
Po sit io n
of t he
vibr at ing
part icle M
Disp lace
ment
Y = a sin t
O
O
Zero
+a 
Zero
 / 2
Y
+a
Zero
--a 2

O
Zero
--a
Zero
3 / 2
Y'
-a
Zero
+a  2
O
Zero
+a 
Zero
2
Accelerat ion
d 2 y / dt 2 a 2
sin t
Thu s t he velo cit y o f t he vibrat ing part icle is maximum ( in t he direct io n
OY o r OY') at t he mean po sit io n o f r est and zero at t he maximum po sit io n o f
vibr at io n. The accelerat io n o f t he vibrat ing part icle is zero at t he mean po sit io n
o f rest and maximu m at t he maximum posit io n o f vibrat io n. The accelerat io n is
always direct ed towards t he mean posit io n o f rest and is dir ect ly proport io nal t o
t he d isp lacement o f t he vibrat ing part icle.
This t yp e of mot io n where t he
accelerat io n is dir ect ed t owards a fixed p oint (t he mean posit io n o f rest ) and is
pro port io nal t o t he d isplacement of t he vibr at ing part icle is called simp le
har mo nic mo t ion.
12.1(a) Rep resentation of SHM by a wave
Let P be a p art icle mo ving o n t he circumference o f a circle o f rad ius 'a'.
The fo ot of t he perpend icu lar vibr at es o n t he diamet er YY'
Y = a sin t = a sin 2 (t /T)
The
d isp lacement
graph
is
a
sine
curve
represent ed
by
ABCDE
[Fig.5.1(a)] The mot ion o f t he part icle M is simp le har mo nic. T his is t he t yp e
165
o f mot io n t hat can be exp ect ed in t he case o f elast ic media, where t he defo r ming
o bey Ho oke's law.
T he d ist ance AE, aft er which t he cur ve repeat s it self, is
called t he wavelengt h and it is denot ed by .
Fig
12.1(a)
12.1(b)
Differentia l Equation o f SHM
For a part icle vibr at ing simple har mo nically, t he general equat io n o f
d isp lacement is,
y = a s in ( t + )
…. (1)
Here 'y' is t he displacement and 'a' is t he amplit ude and  is epo ch o f t he
vibr at ing part icle.
Different iat ing equat io n (1) wit h respect to t ime
dy/ dt = a co s( t + )
Here
dy/dt
represent s
t he
velo cit y
Different iat ing equat io n (2) wit h respect to t ime
d 2 y/dt 2 = a  2 sin( t + )
But a sin ( t + ) = y
… (2)
of
t he
vibr at ing
part icle.
166
d 2 y/dt 2 =  2 y
d 2 y/dt 2 +  2 y =
or
O
…(3)
Here d 2 y/dt 2 represent s t he acceler at io n of t he part icle.
Equat io n
(3) represent s t he d ifferent ial equat io n of simple har mo nic mot ion.
It also shows t hat in any p heno menon where an equat ion similar t o
equat io n (3) is o bt ained, t he bod y execut es simple har mo nic mo t io n.
The
general so lut io n o f equat ion (3) is
y = a sin ( t + )
Also the time period of a vibrating particle can be calculated from equation (3).
We know
 = d 2 y/dt 2
y
 = Accelerat ion
Disp lacement
T =
2

= 2 Disp lacement
Acceler at ion
12.2 Wave Motion
It is a for m o f dist ur bance, which t ravels t hrough t he mediu m due t o
t he repeat ed per io d ic mot io n o f t he part icles o f t he mediu m abo ut t heir mean
posit ions, t he dist ur bance being handed over fro m one part icle t o t he next .
When a st one is dro pped int o a pond, waves are pro duced at t he po int wher e t he
st o ne st r ikes t he wat er in t he pond. The waves t ravel out ward, t he part icles o f
wat er vibr at e o nly up and down about t heir mean po sit io ns. Wat er part icles do
no t travel alo ng wit h t he wave.
S imilar ly, when a t uning fork is set int o
167
vibr at io n, it pro duces waves in air. T he wave t ravels fro m one part icle t o t he
next , but t he part icles o f air vibrat e abo ut t heir mean posit io ns.
12.2(a) Characteri stics of Wave Motion
1. It is t he d ist urbance produced in t he mediu m by t he repeat ed per io d ic
mo t io n of t he part icles o f t he medium.
2. Only t he wave t ravels fo rward whereas t he part icles o f t he medium vibr at e
abo ut t heir mean po sit io ns.
3. There is regular p hase change bet ween t he var iou s part icles o f t he
medium. T he part icle ahead st art s vibrat ing a lit t le lat er t han part icle just
preced ing it .
4. The velo cit y o f t he wave is different fr om t he velo cit y wit h which t he
part icles o f t he mediu m are vibrat ing about t heir mean posit io ns.
The
wave t ravels wit h a unifor m ve locit y whereas t he velo cit y o f t he part icles
is d ifferent at differ ent posit ions.
It is maximum at t he mean posit io ns
and zero at t he ext reme posit io n o f t he par t ic les.
12.2(b)Transverse Wav e Motion
In t his t ype o f wave mot ion, t he part icle s of t he medium vibrat e at r ig ht
ang les t o t he direct io n o f propagat io n of t he wave.
To underst and t he pro pagat io n o f t ransverse waves in a mediu m consider
nine part icles o f t he med iu m and t he circle o f refer ence ( Fig 12.2b).
The
part icles are vibrat ing abo ut t heir mean posit ions, up and down and t he wave is
t raveling fro m left t o right . The dist urbance t akes T / 8 seconds, t o t ravel fro m
o ne part icle t o t he next .
(1)
At t = 0, all t he part icles ar e at t heir mean posit ion.
168
(2)
Aft er T/ 8 second s, t he part ic les 1 t ravels a cert ain dist ance upward
and t he d ist ur bance reaches part icle 2.
(3)
Aft er 2T / 8 seco nd s, part icle 1 has reached it s ext reme po sit io ns
and t he d ist ur bance has reached part icle 3 .
Fig
12.2(b)
169
(4)
Aft er 3T / 8 seconds, part icle 1 has co mplet ed 3/ 8 o f it s vibrat io n
and t he d ist ur bance has reached part icle 4. The posit ion o f part icles 2 and 3 ar e
also sho wn in t he diagram.
(5)
In t his way aft er T / 2 (or) 4T / 8 seconds, part icle 1 has co me back
to it s mean po sit io n and t he d isplace ment o f t he p art icles 2, 3, and 4 are as
shown in t he diagram. The dist ur bance has reached part icle 5.
In t his wa y t he pro cess co nt inu es and t he posit io n o f t he part icles aft er
5T / 8, 6T / 8, 7T / 8 and T (o r) 8T / 8 seconds are sho wn in t he d iagram.
Aft er T seconds, t he part icles one, five and nine are at t heir mean
posit ions. T he wave has reached part icles 9. Part icles 1 and 9 ar e in t he same
phase. The wave has t raveled a dist ance bet ween part icles 1 and 9 in t he t ime
t he part icle 1 has co mp let ed one vibrat io n.
The t op point on t he wave at t he maximu m dist ance fro m t he mean
posit ion is t he crest, whereas t he po int at t he maximum d ist ance belo w t he mean
posit ion is t he t rough.
alt er nat ely fo r med.
represent s t he wave.
T hus in a t ransver se wave, cr est s and t roughs ar e
The co nt our o f t he displaced part icles o f t he mediu m
In t he case o f t ransverse (or lo ng it ud ina l) progressiv e
waves, t his cont our cont inuo usly changes posit io n in space and t he wave seems
to advance in t he direct io n o f propagat io n.
170
12.2(c) Longitudinal Wave Motion
In t his t ype o f wave mo t io n, part icles of t he med ium vibrat e alo ng t he
d irect o r of pro pagat io n of t he wave.
Fig
12.2(c)
Co ns ider nine part icles of t he medium and t he circ le o f reference
(Fig. 12.2(c).
The wave t ravels fro m left t o right and t he part icles vibrat e about t heir
mean po sit io n. Aft er T/8 seconds, t he part ic le 1 goes to t he r ight and co mplet es
1/8 o f it s vibrat io n. The dist ur bance reaches t he part icle 2. Aft er T / 4 second s
t he part icle 1 has reached it s ext reme r ight posit ion and co mp let es 1/ 4 of it s
vibr at io n and t he part ic le 2 co mplet es 1/8 of it s vibrat ion.
T he dist ur bance
reaches t he part icle 3. The process cont inues.
Aft er one co mplet e t ime per io d, t he po sit io ns o f t he var ious part icles are
as sho wn in t he diagr am. The wave has r eached part icle 9. Here 1 and 9 are in
t he same phase.
Here part icles 1, 5 and 9 are at t heir mean posit io ns.
part icles 1 and 3 are clo sed t o t he part icle 2.
The
T his is t he posit io n o f
condensation. S imilar ly part icles 9 and 8 are clo se t o t he p art icle 7. T his is
also t he posit io n o f co ndensat ion or co mp ressio n. On t he ot her hand, part icles 4
and 6 are far away fro m t he part icle 5.
Hence, in a lo ng it udinal wave mot ion,
alt er nat ely for med.
This is t he po sit io n o f rarefaction.
condensat io n and rarefact io n are
171
12.2(d)
Prop erties of Longitudina l Prog ressive Waves
1. The part icles o f t he mediu m vibrat e simple har mo nically alo ng t he
direct ion o f propagat ion o f t he wave.
2. All t he part icles have t he same amp lit ude, fr equency and t ime p er iod
3. There is a gradual phase d iffer ence bet ween t he successive part icles.
4. All t he part icles vibr at ing in p hase will be at a d ist ance equal t o n.
Here n=1, 2, 3 et c.
It means t he minimu m d ist ance bet ween t wo
part icles vibrat ing in p hase is equal t o t he wavelengt h.
5. The velo cit y o f t he p art icle is maximu m at t heir mean po sit io ns and it
is zero at t heir ext reme posit io ns.
6. When t he part ic le mo ves in t he same direct ion as t he propagat io n to
t he wave, it is a regio n o f co mpressio n.
7. When t he part icle mo ves in a d irect ion opposit e to t he direct io n o f
propagat io n o f t he wave, it is in a regio n of rarefact ion.
8. When t he part icle is at t he mean posit io n, it is a regio n o f maximu m
co mpressio n or rarefact io n.
9. When t he part icle is at t he ext reme posit ion, t he medium around t he
part icles has it s nor mal densit y, wit h co mpr essio n o n one side and
rarefact io n o n t he ot her.
10.Due t o t he repeat ed per iodic mot io n o f t he part icles, co mpressio ns and
rarefact io ns
are pro duced
cont inuously.
T he co mpressio ns
and
rarefact io ns t ravel fo rward alo ng t he wave and t ransfer energy in t he
direct ion o f propagat ion o f t he wave.
12.3
Defin itions
( i) Wavelength: It is t he dist ance t raveled by t he wave in t he t ime in
which t he part icle o f t he mediu m co mplet es o ne vibrat ion. It is a lso defined as
t he d ist ance bet ween t wo nearest part icles in t he same phase.
172
The dist ance AB ( Fig 12.3) is equal t o t he wave lengt h .
Fig
12.3
(ii) Frequency : It is t he number of vibrat io ns made by a part icle in one
seco nd (n).
(iii) Amp litude : It is t he maximu m disp lacement of t he p art icle fro m it s
mean po sit io n o f rest . In t he d iagram CD is t he amplit ude.
(iv) Time period : It is t he t ime t aken by a part icle t o co mplet e o ne
vibr at io n (T).
Suppose freq uenc y
= n
T ime t aken t o comp let e n vibrat ions
= 1 seco nd
T ime t aken t o comp let e 1 vibrat ion
= 1/n seco nd
Fro m t he defin it io n o f t ime per io d, t ime t aken t o comp let e one vibrat io n
is t he t ime per io d (T)
T =
1/n
or nT = 1
Frequ ency x T ime per io d = 1
173
(v) Vibration : It is t he to and fro mot io n o f a part icle fro m one ext reme
posit ion t o t he ot her and back again. It is also equal t o t he mot ion o f a part icle
fro m t he mean po sit io n t o o ne ext reme posit io n, t hen t o t he o t her ext reme
posit ion and fina lly back t o t he mean posit ion.
Phase :
It is defined as t he rat io of t he displacement of t he vibr at ing
part icle at any inst ant to t he amplit ude of t he vibr at ing part icle o r it is defined
as t he fr act io n o f t he t ime int er val t hat has elapsed since t he part icle crossed t he
mean posit io n o f r est in t he posit ive dir ect ion o r it is also equal t o t he angle
swept by t he rad ius vect o r since t he vibrat ing part ic le last cro ssed it s mea n
posit ion o f rest .
12.3(a) Relation bet ween Frequ ency and Wavelength
Velocit y o f a wave is t he dist ance t raveled by t he wave in one seco nd.
Velocit y =
dist ance /t ime
Wavelengt h () is t he dist ance t ravelled by t he wave in o ne t ime per io d
(T).

Wavelengt h
Velocit y =
=
Time per iod
T
But , frequency x t ime per io d = 1
n x T = 1
T = 1/n
V =
/T
=
/ 1 / n
V = n
12.4
The Resultant of two S.H. motions in the same directions when they have the same
periodic time.
174
The two S.H. motions having the same periodic time can be represented by the equations
and
y1 = a1 sin (t - 1)
y2 = a2 sin (t - 2)
Since the displacements y1 and y2 are in the same direction, the resultant displacement y
at any time t is obtained by their algebraic addition, ie.
y =y1 +y2
=a1 sin (t - 1)+a2 sin (t _2)
= a1 (sin t cos 1 – cos t sin 1) + a2 (sin t cos 2 – cos t sin 2)
= sin t (a1 cos 1 +a2 cos 2) (a1 sin 1 + a2 2)
Now take angle  (Fig. 5.5) such that a cos  = a1 cos 1+a2 cos 2 and a sin  = a1 sin
1+a2 sin 2.
Then, y=a sin t cos  - a cos t sin 
= a sin (t - ).
This represents a S.H, motion of the same periodic time as the component motions and
having an amplitude = a and epoch angle = . From Fig.12.4, a is given by the relation
Fig 12.4
a2 = (a1 sin 1+ a2 sin 2)2 + (a1 cos 1 + a2 cos 2)2
= a12 (sin2 1 + cos 2 1) + a22 (sin2 2 + cos2 2)2
175
+ 2a1a2 (sin 1 sin 2 + cos 1 cos 2)
= a12 + a22+ 2a1a2 cos (1-2)
and  is given by the relation
tan  =
a1 sin 1 + a2 sin 2
a1 cos 1 + a2 cos 2
Proceeding in a similar manner as above, it can be shown that, for a number of simple
harmonic motions acting on a point in the same direction when their periodic times are equal,
amplitude of the resultant motion is given by
a2= ( a sin  )2 + ( a cos )2
Where and  a cos  are the sums of the y and x components of amplitudes respectively.
The epoch of the resultant is given by
tan  =
12.5
a sin 
a cos 
Let us Sum up
From this lesson you have learned the meaning of the simple harmonic motion (SHM) of a wave
and its differential equation. Also you have understood about the different types of waves
traveling in a medium namely transverse and longitudinal wave motion and its properties.
12.6
Check your progress
1) State the differences between transverse and longitudinal waves
2) Give any two properties of longitudinal progressive waves
3) Show c = 
12.7
Lesson end Activities
1) A particle executes simple harmonic motion with a period of 6 seconds and amplitude 3 cm.
Find its maximum velocity.
176
2) A transverse wave is described by the equation y = y0 sin 2
(ft – (x/ )). Find the maximum
particle velocity which is equal to four times of the wave velocity.
12.8
Points for Discussion
1) Give the characteristics of transverse and longitudinal wave motion. Also represent simple
harmonic motion by a wave.
2) Derive the equation for the resultant of two simple harmonic motion in the same direction
having same period of time.
12.9 References
1) Text book of Sound by Singhal
2) Ancillary Physics by Narayan murthy
LESSON - 13
CONTENTS
13.0
Aims and Objectives
177
13.1
Composition of two simple harmonic motions about a point in the same direction when
their periodic times are somewhat different. Beats.
13.2
Composition of two Simple Harmonic Waves having slightly different frequencies:
Beats.
13.3
Beats
13.4
Stationary or Standing Waves
13.4(a) Composition of two S.H. Waves of equal amplitudes and wavelength traveling in
opposite directions.
13.4(b)
Changes wit h dist ance :
13.4(c)
Changes wit h t ime :
13.5
Pro pert ies o f St at ionar y Lo ngit udina l Waves.
13.6
Melde's E xper iment
13.7
Acoustics
13.8
Ultrasonic
13.9
Doppler's Effect
13.10 Let us Sum up
13.11 Check your progress
13.12 Lesson end Activities
13.13 Points for Discussion
13.14 References
13.0 Aims and Objectives
To introduced the composition of two SHMs and ultrasonic and Doppler effect.
178
13.1
Composition of two simple harmonic motions about a point in the same direction
when their periodic times are somewhat different. Beats.
If two simple harmonic motions having some what different periodic times start at some
moment in step, or in same phase, the amplitude of the resultant is equal to the sum of the
amplitudes of the component motions in the beginning. As time passes, the two motions continue
to get more and more out of step or their phase difference increases, with the result that the
amplitude of the resultant motion goes on becoming less and less. After a certain number of
vibrations when the phase difference between the component vibrations becomes . i.e., they are
in opposite phase, the amplitude of the resultant vibration equals the difference between the
amplitudes of the component vibrations. After this the amplitude of the resultants motion again
goes on increasing and when the phase difference between the component vibration becomes 2
, the two motion are in step again or coincidence occurs and the amplitude of the resultant
vibration becomes equal to the sum of the amplitudes of the component vibrations once again.
After this, the two motions get out of step gradually again resulting in gradual fall in amplitude
of the resultant motion. The result is illustrated by displacement curve (3) in Fig.13.1 where the
component vibrations are represented by the displacement curves (1) and (2) Curve 3 is obtained
by the algebraic addition of the displacements of curves 1 and 2.
Fig.13.1
Analytical Treatment. Let us now treat the above case analytically.
Let the periodic time of the two S.H. motions be T1 and T2. Then their frequencies N1
and N2 are = 1/T1 and 1/T2 respectively.
179
Suppose that the two S.H. motions are represented by the equations.
y1 =a1 sin wt
y2 =a2 sin (w+) t = a2 sin (wt + t)
and
Since frequency of vibration = Angle Traveled in / second
Angle Traveled in / revolution
N1 = w/2 and N2 w+/2
Hence, N2 – N1 = /2
..(1)
The phase difference between the two S.H. motion at any time t equals
t. It, varies with the time.
The resultant displacement y at time t is given by
y
= y1 + y2
= a1 sin t + a2 sin (t - t)
= a1 sin t + 2 sin t cos t + a2 cos t sin t
= sin t (a1 + a2 cos t) + cos t a2 sin t
Now take an angle  [Fig.13.1(a)] such that
a cos  = a1 + a2 cos t
and
a sin  = a2 sin t
They, y = a sin t cos  + a cos t sin 
= a sin (t + )
180
Fig.13.1(a)
This equation represents almost a S.H. motion of amplitude 'a' given by
a2 = (a1 + a2 cos t)2 + (a2 sin t)2
= a12 + a22 (sin2 t + cos3 t)2 + 2a1a2 cos t
= a12 + a22 + 2a1a2 cos t
…(2)
While  is given by
tan  =
a2 sin t
a1 + a2 cos t
…(3)
It is clear from the study of equation (2) that the amplitude (a) of the resultant motion
changes with time. Also, from equation (3) it follow that the phase difference between the
resultant motion and the S.H. motion represented by the equation y1 = a1 sin wt, i.e, , also
changes with time.
When t
and
=
0 or 2n, where n is an integer, cos t = 1
a2 = a12 + a22 + 2a1a2
= (a1 + a2)2
or
a
= a1 + a2
When t = (2n + 1 ) , cos t = - 1
and
a2 = a12 + a22 + 2a1a2
= (a1 - a2)2
or
a
= a1 - a2
181
Thus with the passage of time, the amplitude of the resultant various between the
maximum value a1 + a2 and the minimum value a1 –a2. The first maximum value of amplitude
occurs when t = 0 or t =0, the second when t = 2 x 1 x  = 2 – or t =2/, when at = 2 x 2 =
4 or t = 4/ and so on. The interval between the two successive maximum values of amplitude
is hence = 2/. The first minimum value of amplitude occurs when t =  or
t = /, the
second when t=3 or t= 3/, the third when at = 5 or t = 5/ and so on. Again, the interval
between two minimum values of amplitude is = 2/. Thus the interval between two successive
maximum or two successive minimum values of amplitudes = 2/ , or the maximum or
minimum values of amplitudes amplitude occur  /2
times in one second, which is equal
to the difference in frequencies N2 – N1 (equation (i) of the component motions.
It should be noted that the interval between two successive maximum values of
amplitudes = 2 /  and that these will be distinguishable only if this interval is large as
compared with T1 = 2/ or T2 = 2/+, i.e.,
 / 2 is small as compared with
/2 or  + /2. Thus the difference in frequencies N2 – N1 should be small as compared with
N1 or N2 or the two S.H. motions should differ only slightly in periodic time of the changes are
to be noticed.
The phenomenon of beats, heard when two sound notes of nearly equal frequency are
sounded together, is explained in this way. The resultant note or beat note rises and falls in
intensity at a time rate equal to the difference of the frequencies of the two superposed notes.
13.2
Composition of two Simple Harmonic Waves having slightly
different frequencies: Beats.
If two simple harmonic waves of slightly different frequencies affect a particle
simultaneously, its amplitude of vibration changes between maximum and minimum. If the two
are in step at any moment, the resultant displacement of the particle is equal to the sum of
displacement due to the component waves. The amplitude of the resultant vibration is maximum
at the moment. As the time passes, the two get more and more out of step or their phase
difference increases with the result that the amplitude of the resultant goes on becoming less and
182
less. After a certain number of vibrations when the phase difference is = , the resultant
amplitude equals the difference between amplitudes of the component waves and is minimum.
The amplitude of the resultant goes on increasing with further increase in phase difference and is
maximum again when this becomes = 2 and the two motions are in step again.
Analytical Treatment
We shall now deal with the above case analytically. Let two S.H. waves of amplitudes a1
and a2 and frequencies N1 and N2 affect a particle simultaneously and let the time be reckoned
from this moment. The displacement y1 and y2 due to the two waves individually at time t will be
y1= a sin 2  N1 t
given by
and y2= b sin 2  N2 t
The resultant displacement y at the time is given by
y= y1 + y2
= a sin 2  N1t + b sin 2 N2t
Substituting N1 + v for N2, where v =N2 – N1, the difference in frequencies of the two
waves, we have
y
= a sin 2  N1t + b sin 2 (N1 + v) t
= a sin 2  N1t + b sin 2 N1t cos 2  vt + b cos 2 N1t
sin 2 vt
= sin 2  N1t (a+ b cos 2 vt)
+ cos 2 Nt+b sin 2 vt
Take  (Fig .5.7) such that R sin  = b sin 2 vt
R cos  = a+b cos 2 vt
Then, y= R (sin 2N1t cos  + cos 2 N1t sin )
= R (sin 2N1t + )
183
Fig 13.2
Evidently, the phase difference  between the resultant vibration of the particle and that
due to the wave of frequency N1 is a function of t and varies with time.
The amplitude R of the resultant vibration is given by
R2 = (b sin 2vt)2 + (a+b cos 2 vt)2
= (a2 + b2 + 2ab cos 2 vt
This resultant amplitude changes with time. It is maximum when
cos 2 vt = + 1, then,
R2 = a2 +b2 2ab = (a+b)2
or
R = a+b,
i.e, the resultant amplitude is equal to the sum of amplitudes of the component waves. This
happens when 2 vt = n x 2, where n is an integer, or when t = n/v, i.e, 1/v, 2/v, 3/v, 4/v, etc.
the interval between two maxima = 1/ v.
Thus the maximum amplitude of the resultant vibration occurs v times per second, which
is equal to N2 – N1 the difference in frequencies of the component waves.
Again, the resultant amplitude is maximum when
cos 2 vt = -1.
then – R2 = a2 + b2 – 2ab = (a-b)2
or
R = a-b,
the resultant amplitude is equal to the difference between the amplitudes of the component
waves. This happens when
2 vt = (2n+1) 
or
t = 2n+1 / 2v, i.e., 1/ 2v, 3 / 2v, 5 / 2v, 7 / 2v, etc.
184
The
interval between two successive maxima is 2 /2v or 1/v. Thus the minimum
amplitude also occurs v times per second.
The maximum and minimum amplitudes occurs alternately, the maximum occurring at t
= 0, the minimum at t = 1/2v, the next maximum at
t = 2 /2v or 1 / v, the next minimum at t =
2 /2v, then maximum at 4 / 2v or 2 / v and the minimum at t = 5 / 2v and so on.
It should be noted that the difference in frequencies of the two waves N2 – N1 = v should
be small as compared with N1 or N2 in order that the changes in amplitude may be
distinguishable.
13.3
Beats
It has been shown mathematically that when two S.H. sound waves of slightly different
frequencies set a particle simultaneously into vibrations, the particle, under the joint influence of
the two waves, vibrates with varying amplitude. If the two waves enter the ear, the membrane of
the ear vibrates alternately with maximum and minimum amplitude. The intensity of sound,
which is proportional to the square of the amplitudes then changes alternately between certain
maximum and minimum values, the number of maxima or minima of sound heard per second
being equal to the difference in frequencies of the two waves starting from the two sources of
sound. This phenomenon of waxing and waning of sound is called beats. Beats are thus heard
when two notes of slightly different frequencies are sounded together and their frequency is
equal to the difference in frequencies of the two notes.
We know that the vibration in amplitude will become perceptible when N2 – N1 is small
as compared with N1 or N2. Beats are thus heard when two notes of nearly but not quite the same
frequency are sounded together. When the difference in frequencies of the two notes is more than
10 per sound, is becomes difficult to hear the beats separately.
185
Demonstration of Beats.
Place two tuning forks of the same frequency fitted on suitable resonance boxes on a
table, with the open ends of the resonance boxes facing each other. Strike the two tuning forks. A
continuous loud sound is heard. It does not rise and fall. Stick a little wax to a prong of one
tuning fork so as to reduce its frequency and thereby create difference in the frequencies of the
two tuning forks. Sound the two tuning forks together once again. Beats will be heard.
Fig 13.3
Beats can also be demonstrated by means of a Knipp's Tube, Beats are heard when it is
blown at A. The tube AN and A'N' act as closed end organ pipes with nodes at the closed ends
and antinodes at the open ends. They produce notes of slightly different frequencies, which give
rise to beats. By proper adjustments, the intensity of beats may be made so great as to become
painful to the ear.
Practical Application of Phenomenon Beats
The phenomenon of beats is used for tuning a note to any particular frequency. The note
of the desired frequency is sounded together with the note to be tuned. If there is a slight
difference in frequencies of the two, beats are produced. When they are exactly in union, i.e.,
186
have exactly the same frequency, they do not give any beats when sounded together, while they
give the same number of beats with a third note of slightly different frequency.
When the shank of a vibrating tuning fork is placed on the sound board of a monochord,
with a wire stretched over two bridges, beats are heard if the frequency of vibration of the wire is
almost equal to the frequency of the tuning fork. The beats become distinct when sound is dying
away. For finding the exact length for unison, two positions of one of the bridges are found at
which one beat per second is heard. The position of the bridge for unison exists midway between
these two positions. This can be shown as follows.
N+1=1/2l1
N-1= 1/2l2
and
F/m
… (1)
F/m
… (2)
N = 1/2l F/m
… (3)
Adding (i) and (ii), we have
2N
Hence
l
=
(1/2) F/m [ (1 / l1)+ (1 / l2)]
=
(l/2)
=
=
(2 / l) = (1/ l1) +
(2 l1 l2) / (l1+l2)
F/m
(2 / l)
(From (3)
(1/ l2) = (l1+l2) / (l1 l2)
2l1 (l1+) / (2l1+) if l2 – l1 = 
=
l1 (2l1+) / (2l1+
)
=
l1+ ( / 2)if  is small as compared with l1
Which will actually be found to be the case.
13.4
Stationary or Standing Waves
We have learnt that the superposition of two S.H. waves of the same wavelength or
frequency traveling in the same direction gives rise to a progressive S.H. wave whose amplitude
depends upon the phase difference between the component waves If, however, two such waves
be travelling in opposite directions, they mutually interfere with the result that the progressive
nature of the wave is lost. At certain points in the path of the two component waves, called
187
nodes, the medium remains perpetually at rest, while at some others, called antinodes, the
displacement of particles is always maximum, and at other points in the path of the two waves,
the different particles vibrate about their mean positions with different amplitudes. The resultant
wave, so formed, is called a Stationary or a Standing waves.
13.4(a) Composition of two S.H. Waves of equal amplitudes and wavelength traveling in
opposite directions.
Let the two S.H. waves of equal wavelength  and amplitudes a be traveling along the
X-axis with velocity v, the first one travel ling towards the right and the second towards the left.
The treatment given here applies to the transverse waves with displacements in the same plane
also to the longitudinal waves. For representing the waves graphically, we shall show the
displacements of particles above the line of transmission in transverse waves or towards the
right in longitudinal waves above the X-axis, taking such a displacement as positive, and those
below the line of transmission or towards the left below the X-axis, taking such a displacement
as negative.
Fig 13.4(a)
The two oppositely moving waves arriving in zero phase at A are represented by wave
curves 1 and 2 in Fig.13.4(a). Point A has zero displacement in each case and lies on the
descending part of each curve looked at in the direction of propagation of the wave. The
188
displacement at A will be found to increase to some positive value for each after a small interval
of time, when the first wave has advanced a little towards the right or the second wave has
advanced a little towards the left. The subsequent wave curves after a small interval of time are
indicated by dotted lines in the figure.
Next is to be noticed that in the case of longitudinal waves, as the first wave advances
towards the right, the displacement as A increases towards the right, i.e, in the direction of
motion of the wave. As the second moves forward towards the left, the displacement at A
increases in this case also towards the right which direction is opposite to the direction of motion
of this wave. The particle at A will thus have motion in the direction of motion of the first wave
and in the opposite direction to the direction of motion of the second wave i.e, we have
compression at A due to the first wave and rarefaction at this point due to the second wave.
(Keeping in mind that in an associated wave curve compression lies on the descending part of
the curve and rarefaction on its ascending part, if we denote the displacement in the direction of
motion of the wave above. The X-axis the wave curve I may be looked at with the book as such
and the wave curve 2 with the look upside down. A will be seen to lie on the descending part of
wave curve 1 and on the ascending part of wave curve 2 when the two curves are looked at in
this manner).
Let t he waves cross A in zero phase at t = 0. Co nsider t he displacement s
due t o t he t wo waves at t ime t at a po int B d ist ant x fro m A t owards t he left .
Due t o t he fir st wave, t he vibrat ing part icle at B is ahead o f t he vibr at ing
part icle at A in p hase while due t o t he second wave t he vibrat ing part icle at B is
lagg ing in phase behind t he part icle at A.
The disp lacement y 1 at B at t ime t
due t o t he fir st wave is given by
y 1 = (a sin 2 /  ) (vt + x)
The d isp lacement y 2 at B at t ime t due to t he second wave will be g ive n
y 2 = (a sin 2 / ) (vt - x).
by
The resultant displacement at B at time t is obtained by the principle of superposition, viz.,
y
= y1 + y 2
189
= (a sin 2 / ) (vt + x) + (a sin 2 / ) (vt - x)
= (2a sin 2 / ) vt
(cos 2 / ) x
…..(1)
This is t he equat ion o f t he result ant wave. From t his equat ion,
S lo pe o f t he wave cur ve or st rain at any p oint is given by
dy / dx
= -(4a / ) sin (2 / ) vt sin (2 / ) x
while
….(2)
Velocit y o f t he vibr at ing part ic le at any point is given by
dy / dx
= (4av / ) cos (2 / ) vt cos (2 / ) x while
….(3)
13.4(b)Changes with distan ce :
(a) From eq uat ions ( i), ( ii) and ( iii), we find t hat at places where
(2 / )
cos
x = 0 or sin (2 / ) x = + 1, t here will be zero displacement ( y),
max imu m slope o f wave cur ve or maximum st rain (co mpr essio n or rarefact io n
d y/ d x ) and zero velocit y o f t he vibrat ing part icles (d y /dt ) at all t imes. T hese
posit ions are t he nodes. They occur at place where
co s
(2 /) x = 0
or
(2 / ) x = (2n + 1)  / 2, where n is an int eger,
or
(b)
x = (2 n + 1)  / 4 = (n / 2) +  / 4
Ag ain fro m equat io n ( i), ( ii), and ( iii) we find t hat at places where
(2 / ) x = +1 or sin (2 / ) x = 0, t here will be maximum
Fig. 13.4(b)
190
d isp lacement , zero slope o f wave cur ve or st rain (co mpressio n or rarefact io n)
and max imu m velocit y o f vibr at ing part icles at all t imes. T heses posit io ns are
ant ino des. They occur at places where
co s (2 / ) x = +1
or
co s (2 / ) x = n, where n is an int eger
or
x = n / 2
The posit io ns o f no des and ant inodes in a st at ionar y wave case are sho wn
in F ig. 5.9( b) N 0 , N 1 , N 2 , N 3 and N 4 are nodes while A 0 , A 1 , A 2 , A 3 and A 4 ar e
ant ino des. The dist ance bet ween
t wo co nsecut ive nodes, or t wo consecut ive
ant ino des, is half a wavelengt h (/2), while t he dist ance bet ween a node and t he
next ant inode is (/4).
13.4(c)
Changes with time :
(a) Fro m equat io ns ( i), ( ii) and ( iii), we find t hat at t imes given by
sin (2 / )vt = 0 or co s (2 / )vt = + 1, t he displacement and slope o f t he
slo pe o f t he wave cur ve o r st rain (compr essio n or rarefact io n) are zero
ever ywhere and t he velo cit y o f each vibrat ing part icle is great er t hat at an y
ot her t ime.
Fig 13.4(c)
No w sin (2 / )vt = 0 when
(2 / )vt = n where n is an int eg er
191
And if T is t he per io dic t ime, T = ( / t ).
Therefor e
(2/T)t = n
when t = nT / 2
The stage is shown in Fig. 13.4(c) and it recurs or repeats itself after intervals of half the
periodic time or T/2.
(b)
by
Again, fro m equat io ns (1), (2) and (3), we find t hat at t imes g iven
sin (2/) vt = +1
o r co s (2/)vt = 0, t he displacement and slop e o f
curve o r st rain (co mpressio n o r rarefact ion) are maximu m ever y wher e while t he
velo cit y o f each vibrat ing part icle is zero.
Ever y vibr at ing part icle is at it s
ext reme posit io n at t his mo ment and st at ionar y for an inst ant .
Fig 13.4 (d)
or
or
sin (2 / )vt = +1 when
(2 / )vt = (2n + 1)  /2, where n is an int eger
(2 / T) t = (2n + 1)  /2, where T is t he per iodic t ime,
t = (nT / 2) + (T / 4)
The st age is shown in Fig 13.4(d) and occurs a quart er per iod (T/4) aft er
t he st age sho wn at Fig 13.4(c).
The wave cur ves for t he result ant st at io nar y wave, as t he t ime progresses,
are sho wn in Fig. 13.4(e). Wave cur ve No.1 represent s t he S.H. wave t raveling
to wards t he r ight , No.2 t he S.H. wave t raveling t owards t he left and No .3 t he
resu lt ant wave.
192
It is clear t hat N 0 , N 1 and N 2 are nodes where displacement is zero while
A 0 , A 1 , A 2 and A 3 are ant ino des having great er displacement s.
The arrows in t he vert ical d ir ect ion indicat e t he direct io n o f mot ion o f t he
vibr at ing part icles at t he point s aft er t he inst ant s. Curves ( i) and ( iv) indicat e
st at io nar y inst ant s.
Aft er an int er val equal t o t he p er iodic t ime T t he wave
curve repeat s it self.
Fig 13.4(e)
13.5
Prop erties of Stationa ry Longitudina l Waves.
We can no w su mmar ise t he propert ies o f st at ionar y waves produced by t he
superpo sit io n o f t wo set s o f lo ngit udinal waves o f equal amplit ude and p er io dic
t ime, t raveling in o ppo sit e direct io ns.
1.
At t he nodes t here is no di sp lacement of part icles.
2.
At any inst ant t he displacement at t he ant inode is great er t han at any o t her
point .
3.
At any inst ant , t he velocit y o f t he medium at t he ant inode is great er t ha n
t he velocit y o f t he mediu m at any ot her point s.
193
4.
The med ium at each po int mo ves o r vibr at es har mo nically.
5.
At any inst ant , t he who le o f t he medium bet ween t wo consecut ive nodes is
mo ving in t he same direct io n and also t he who le o f t he medium bet ween
t wo co nsecut ive ant ino des is in t he same condit io n – all co mpressed o r all
rarefied.
The vibrat ions occur alt er nat ely in t he t wo direct ions in
co nsecut ive seg ment s bet ween nodes and co mpressio n and rarefact io n
o ccur alt ernat ely in consecut ive segment s bet ween t wo consecut ive
ant inodes.
6.
The d ist ance bet ween t wo consecut ive nodes or t wo consecut ive ant inodes
is half the wave- length o f t he progressive waves.
7.
The t ime in which t he st at ionar y wave go es t hrough all it s changes at any
point is t he same as t he per io dic t ime o f each progressive wave, i. e. t he
t ime in wh ich each progressive wave will advance one wavelengt h.
8.
Co mpressio n and rarefact io n in a st at ioner y wave do not mo ve alo ng as in
a progressive wave; t hey simply appear and d isappear at any place t o be
succeed ed by t he o pposit e condit io n.
9.
A node is t he cent re o f maximu m co mpr essio n and maximum rarefact io n
alt er nat ely.
Co mpressio n at one node is fo llowed by r arefact io n at t he
next node.
10.
At an ant ino de, t he med iu m has it s nor mal densit y. Each node is a po int
o f max imu m and minimum densit y alt er nat ely, t he average densit y at t he
no de being t he same as at t he ant inode or elsewhere.
13.6
Melde' s Experi ment
One end o f a st r ing is t ied t o t he prong o f a t uning fork while
194
Fig 5.11(i)
Fig 13.6(i)
its other end passes over a pulley and is tied to a hanger or a scale pan, hanging vertically, in
which weights are placed. When the tuning fork is fixed horizontally and struck, the string is
split up into a number of vibrating loops (p) (Fig 13.6(i). Any exact number can be adjusted by
changing the weights. The frequency of the tuning fork (n) in this case is the same as that of the
vibrating string. If L is the length of the string between the tuning fork and the pulley, then the
distance between two successive nodes
= L / P and n = 1/ 2(L/P) T/m
where T is the tension or weight of the hanger, or scale pan, plus the weight in it, and m is the
mass per unit length of the string.
Hence, in this case n = (P / 2L) T/m
195
When the tuning fork is fixed vertically and struck, the string ia again spilt up into a number of
loops [Fig 13.6(ii)].
In this case, the string is tightened straight when the prong is at the end of the swing away from
the pulley and is most loose when the prong is nearest to the pulley. The string thus completes
quarter vibration during the time that the tuning fork completes half vibration, i.e., the frequency
of the tuning fork. In this case, therefore,
n / 2 = (P / 2L) T/m
Its vibration are illustrated in Fig: 13.6(iii)
Fig: 13.6(iii)
Note. The effective length of the string is slightly greater than the actual length since the
ends of the string have a slight transverse vibration.
196
13.7 Acoustics
It is sometimes observed that a speech made in certain hall or building or sound in a
talkie house is not audible at certain places in the hall or there is so much of interferences that it
is difficult to understand what is being spoken. It is, in fact, necessary to keep certain important
points in mind while designing and constructing a place for public speaking or entertainment so
that one can follow the utterance from every point, and there are no acoustical defects.
From a series of experiments, it has been found that sound, which can be heard directly
up to a distance of 100 feet in front, can be heard up to 75 feet on each side and 30 feet at the
back. This gives an idea about the position of the stage when a speech is meant to be heard on all
sides.
In a closed hall, however, some other factors also play part – particularly the reflection
from walls. Two requirements are desirable in this case from the acoustical point of view, i.e,
sufficient loudness and clarity of sound at all places. The following factors have to be considered
for this purpose.
Loudness.
A large wall reflects back the sound and thus increases its loudness. Therefore, a wall at
the back of the speaker or a soundboard placed near him and directed towards the audience
enables a speaker to be heard at a greater distance. Bronze vases and similar devices serve as
resonators and reinforce sound. Loud-speakers, fitted above the heads of the audience, serve well
for electrical amplification of sound. The quality of sound, however, suffers by their use, the
tones of lower frequency being usually amplified better than those of higher frequency.
13.8
Ultrasonic
197
A simple valve type oscillating circuit for producing continuous ultrasonic waves is
shown in Fig. 13.8.
Fig 13.8
The vibration of the quartz crystal control the frequency of the tuned oscillatory circuits in the
grid, and the valve amplifies the electric oscillations and feeds them back into the crystal circuit.
Ultrasonic waves of high frequency can be obtained by this method.
Piezo-electric effect can be induced in barium titanate. This can be given the shape of a
tube with a jet to emit intensive ultrasonics.
For powerful ultrasonic waves at a point, a quartz plate of the shape of a concave mirror
is used. Such a projector concentrates the waves.
The piezo-electric quartz crystal may be used for detecting ultrasonic waves. These
waves produce varying electric charges on faces perpendicular to those which receive them.
These are amplified and detected. The principle is used in ultrasonic Inter-ferometer.
APPLICATIONS
198
1.
Signalling: Ultrasonic waves have an advantage over audible sound waves for signaling
in any particular direction because the former, on account of their high frequency, can be sent out
in the form of a sharp beam. If a vibrating quartz crystal in the form of a disc of radius r be used
for transmission, deal of energy is radiated in this form without the amplitude of the vibrating
being large while, for radiating energy at the same rate at ordinary sound frequencies, the
amplitude must be much larger. Thus, ultrasonic found an important in directional signaling.
Supersonic microscope, using supersonics of high frequency, with wave-length
approaching that of light, has been invented. It is used for examining objects concealed from the
human eyes.
2.
Detection of aircraft, submarines: (SONAR – Sound Navigation and ranging). A piezo-
electric quartz crystal oscillator is used for sending out a beam of ultrasonic waves. This is
reflected back from an aircraft or submarine, if one comes in its way, and the reflected beam can
be detected by means of a quartz receiver. Submerged submarines may thus be located by this
method.
A new devices, called sea-scanner, sends out ultrasonic waves ahead of a ship and on its
two sides in the water as well as down wards and receives the enhoes from fish, reefs, etc., on a
radar-type screen as well as in a loud-speaker through a hydrophone. The sound heard by
reflection from different types of objects is different. From the steady or changing pitch of
sound, it can be ascertained whether the object is stationary or moving towards or away from the
listener. The device, as expected, has been found very useful for navigation.
3.
Depth of Sea. The depth of the sea can be found by having an arrangement for recording
the time that elapses between the emission of the ultrasonic waves and their reception after
reflection from the bottom of the sea. Equipments, based on this and giving directly the depth,
have been manufactured. The device is called a fathometer. The transmitter and receiver are
both placed in water. The indicator consists of a metal stylus at the end of an arm, which touches
a paper impregnated with solution of potassium iodide and starch and rotating at a constant speed
199
of 90 revolutions per minute on a back metal plate. As soon as the stylus passes zero on the paper
scale, a pulse of ultrasonic waves is sent by the transmitter. The echo passes after amplification
through the stylus and the sensitized paper to the blackmetal plate, liberating iodine. A brown
mark is thus made on the paper, indicating the depth of the sea.
4.
Heating the Effect: Intense heat is produced when ultrasonic pass through a substance.
Thus, by passing an intense beam of ultrasonics through water at zero degree centigrade with ice
floating on its surface, the water may be made to boil without melting the ice. Joint of
therplastics materials weld with the heat produced by the concentration of ultrasonics in a
wonderful manner.
5.
Mechanical Effect: A glass rod oscillating with ultrasonic frequency combined with its
rotary motion will bore a hole in a piece of steel or glass plate. It passes through steel or glass
like a hot knife through butter. Squares, triangular or zig-zag holes may be bored by means of an
ultrasonics drill. It will bore through the hardest metal or alloy and even through a diamond,
moving to end fro about a thousandth of an inch with each stroke.
Ultrasonic waves of high frequency are capable of shaking off the oily sludge from
metals and are used for removing grease, dust and metal filling by manufacturers of cars,
cameras and delicate instruments.
6.
Experiments have been successfully tried for cleaning clothes with untrasonics. The
ultrasinics shake the dust particles out of the clothes into the detergent without the lease damage
to the clothes. An ultrasonics washing machine of average size can do the washing in a few
minutes and needs low power
7.
An ultrasonics cleaning machines has recently been developed for cleaning very small
objects such as needles, minute parts of watches, ball bearings, electronic assemblies, filters,
gauges etc.
200
8.
American doctors are using the waves for relieving pain in arthritis. A smooth metal head
vibrating 800,000times per second is passed over the skin. The vibrations penetrate the tissues
and a deep massage relieves the pain.
13.9 Doppler's Effect
It is of common observation that the pitch or the shrillness of the sound heard by a
listener when a railway locomotive engine, blowing its whistle, approaches him is higher than
the pitch of the same sound heard by him when the engine is receding. The former is fairly
sharper than the latter. In fact it is found that whenever there is relative motion between a source
emitting waves and an observer, the frequency of the waves received differs from the frequency
of the waves emitted by the source. The phenomenon is known as Doppler's Effect.
The apparent or observed frequency is higher than the actual frequency of the waves
emitted by the source when the distance between the source and the observer is continuously
decreasing while it is lower when the distance is increasing. Thus when a source of sound
approaches a listener or when a listener approaches the source or both approach each other, the
pitch of the note heard is found to be higher then when the source of sound and listener are
stationary; and the pitch is lower when the source recedes from the listener or when the listener
moves away from the source or both move away from each other. The effect in different cases is
discussed below.
Case 1:
When the listener is at rest and the source is in motion.
Imagine a listener to be at rest at L. Suppose a source of sound at S moves towards the
listener through a distance
sound
------------ v -------------------------
--- vs-------------
o--------------------o---------------------o--------------------------o
S
S'
O
SOURCE
Fig. 13.9(a)
L
201
SS' in 1 second. Then SS' is the velocity of the source (vs) per second. Let S=v be the velocity
of sound emitted by the source Fig.13.9(a). Let the frequency of the note emitted by the source
be n per second. The wavelength
of the waves  will be = v / n
The sound waves emitted by the source S will reach  after 1 second while the source
will reach S' in the meanwhile after emitting n waves. The n waves will thus be concentrated
within distance S' = v - vs.
The wavelength of these waves becomes equal to 1 = (v-vs) / n
This is smaller than the wavelength of the original waves
1 = v/n given off by the source.
The effect is illustrated in 13.9(b)
(a) When source is stationary;
(b) When source is moving towards the listener.
Fig 13.9(b)
(Change in wavelength when a source of sound moves towards the listener)
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The wave-fronts n,n-1, n-2,…………..2, 1 in Fig. 13.9(b) are sections of spheres with
centres at S,k,l,……….q and r, where Sk=kl=…qr=rS', each being equal to the distance through
which the source moves forward during
the time 1/n
These condensed waves travel on with the original velocity of the waves and cross the
listener with the same velocity (v). The number of waves which cross the listener in 1 second is,
therefore, the number of condensed waves contained within the distance . The number of these
contained in the distance -s, being , the number contained in the distance  shall be
[n/(v/v-vs)]v=[v/v-vs]n
…(1)
The frequency of the note heard is thus higher than the frequency of the note emitted by the
source.
The apparent frequency n1 may also be obtained by dividing the velocity v of the wave as
it crosses the listener by its wavelength 1. Thus
n1 = v / 1 = [v/( v-vs)]n
Distances measured in the direction in which the sound travels in order to reach the
listener are taken as positive while those measured in the opposite direction are taken as
negative.
If the source recedes from the listener, it moves in the opposite direction to the direction
in which sound travels in order to reach the listener and v8 is thus negative in this case. The
pitch of the note heard will, therefore, be lower low than the pitch of the note n emitted by the
source are contained in a distance  + 5  equals. Then  becomes
(v + vs) / n and n1 = [v / (v + vs)]n
Case 2.
When the source is at rest and the listener is in motion.
Imagine the source at S and the listener at L. Suppose that the listener moves through the
distance LL' away from the source in 1 second. Then LL' = vL is the velocity of the listener or
distance moved by him per second. Let L = v be the velocity of sound or the distance moved by
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sound per second [Fig.13.9(c)]. All the waves between L and  would have crossed the listener
in 1 second if the listener had been at rest. As however, he himself moves from L to L' in the
meanwhile, the waves that actually cross him in 1 second are those which lie between L' and .
If the frequency of the waves emitted by the source be n per second, the waves which lie
between L and  or within the distance L =v are n and, therefore those which lie between L'
and  or within the distance L' = v-vL are
= [(v-vL)/ v]x n
Fig. 13.9(c)
The frequency of the note heard by the listener is thus equal to
n' = [ (v-vL)/ v]x n
. . . (2)
The pitch of the note is thus reduced when the listener moves away from the source, i.e.,
in the same direction along which the sound travels in order to reach the listener or when vL is
positive.
As the source does not move, all the spherical wave-fronts of the waves starting from the
source continue to have their centre at S in this case.
If the Listener moves towards the source, i.e., in opposite direction to that in which the
sound travels in order to reach the listener, vL will be negative and the frequency of the sound
heard will be higher than the frequency of the sound emitted by the source. In this case the
number of waves which cross the listener per second are those contained in a distance
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v+vL. The frequency n' of the note heard is [(v+vL)/v]n
Case 3: When the source and the listener are both in motion
Suppose that the source travels with a velocity vs while at the same time the listener
travels with a velocity vL. Let v be the velocity of sound and n the frequency of the note emitted
by the source.
First imagine the listener to be at rest. The frequency of the note heard by him will be
equal to
n1 = (n/v-vs)
…(From(1) above)
Now if the listener travels with a velocity vL the frequency of the note heard will be
n' = [v-vL/v]x n
…(From(2) above)
= (v-vL/v) x (v/v-vs) x n
= ( v-vL/v-vs ) x n
…(3)
Limitation of Doppler's Principle
Doppler's principle is applicable as long as the relative velocity between the source o
sound and listener is less than the velocity of sound itself. The principle is not a applicable if the
source moves towards the listerner with supersonic velocity or velocity greater than that of
sound.
13.10 Let us Sum up
From this lesson you learned about the composition of two SHMs about a point in the same
direction and how it results in beats. Also you studied the properties of stationary waves which is
the composition of two SHMs of equal amplitude and wavelength traveling in opposite
directions. Finally you learned about ultrasonics and Doppler effect.
13.11 Check your progress
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1) What is Doppler effect ?
2) What are the factors to be considered for an auditorium with regard to acoustics
3) What are ultrasonic waves
4) What are beats
13.12 Lesson end Activities
1) The distance between the source and a cliff is d. Find the time take to hear the first echo.
2) With what velocity a moving source come towards a stationary listener so that the apparaent
frequency noted by the listener is twice the real frequency.
13.13 Points for Discussion
1) Explain how you would determine the frequency of electrically maintained tuning
fork by Melde’s experiment.
2) A) What is Doppler effect and discuss different cases?
B ) Mention few properties and application of ultrasonic.
13.14 References
1) A text book of sound by Singhal
2) Ancillary Physics by Narayanamurthy
3) Text Book of Physics by Brijlal and Subramanyam
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Model Question Paper – I
Core Paper (B.Sc., Physics D.S.E.)
General Physics – I
Heat, Thermodynamics, Mechanics, Properties of Matter and Sound
Answer any five questions
All question carry equal marks
1) (a) Define Co-efficient of Thermal Conductivity
(b) Explain how you would determine the thermal conductivity of a bad conductor by
Lee’s disc method.
(c) Explain stability of atmosphere.
2) (a) Explain First Law of Thermodynamics.
(b) With a neat diagram explain how you would determine  by
Clement and Desormer method.
(c) State Carnot’s theorem.
3) (a) Explain Bauking tracks motion of vertical circle.
(b) Derive the final velocity and loss of kinetic energy with case of (i) Direct
impact and (ii) Oblique impact.
4) (a) State Kepler’s law of planetary motion.
(b) Explain the three types of Elastic Module. Find the relation between them.
(c) What is I - form girders.
5) (a) What is S.H.M?
(b) Derive an expression for the composition of two S.H.M’s at right angles to each
other.
(c) Mention few properties and application of ultrasonic.
6) (a) What is Poisson’s ratio?
(b) Determine Young’s modulus of a material by (i) Uniform bending and (ii) Non –
Uniform bending.
7) (a) What are stationary waves and give its properties?
(b) Explain how you would determine G by Boy’s method.
8) (a) What are Impulse Impact?
(b) Explain the working of Carnot’s engine and find its efficiency.
(c) Write a note on Carnot’s refrigerator.
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Core Paper (B.Sc., Physics D.S.E.)
General Physics – I
Heat, Thermodynamics, Mechanics, Properties of Matter and Sound
Model question paper – II
1) (a) Define (i) Wein’s law (ii) Rayleigh – Jeans Law.
(b) Explain how you would determine the Stefan’s constant along with the
mathematical derivation of Stefan’s law.
(c) Write a note on solar spectrum.
2) (a) What is adiabatic and isothermal process.
(b) Derive an expression for the work done in an adiabatic expansion of gas.
(c) State second law of Thermodynamics.
3) (a) State Laws of friction.
(b) Derive an derivation for the equilibrium of a body on a rough inclined plane,
when the inclination is greater than the angle of friction.
(c) What is an impulse?
4) (a) Explain and derive an expression for the variation of ‘g’ with (i) latitude (ii)
altitude and (iii) depth.
(b) Determine the three Modulii by Searls’s method.
5) (a) What are progressive wave and give its properties?
(b) Explain how you would determine the frequency of electrically maintained tuning
fork by Melde’s experiment.
6) (a) What is Lapse rate?
(b) Explain how are would determine experimentally the specific heat of liquid?
(c) What is Greenhouse effect?
7) (a) What is Doppler effect and discuss different cases?
(b) Derive and expression for the couple per unit twist.
8) (a) What is gravitational potential?
(b) Derive an expression for the Gravitational field at a point due to spherical
shell.
(c) Define angle of friction and cone of friction.