SECTION 2.
STRUCTURE AND MOLECULAR PROPERTIES OF MATTER.


Pressure : Pressure is defined as the force per unit area acting on an object
Pr essure 
Force
Area
F
P
A
Symbol P
Therfore
N
S.I. Units 2
m
called Pascal Pa
Density : Defined as the mass of a substance per unit volume of the substance
mass
Density 
Volume
Symbol 
m

V
kg
S.I. Units 3  kgm3
m
Relative Density : Defined as the density of a substance relative to ( or divided by )
the density of water.
Density of substance
R.D 
Density of water
S.I. Units NONE
And note also that the density of water is
1000 kg / m3
R.D 
Density of substance
Density of water
mass of substan ce
R.D 
mass of water
If Volume of substance
Volume of substance
Voulme of water
= Volume of water
the equation becomes
R.D =

mass of substance
mass of an equal volume of water
So the relative density can be defined as the mass of a substance divided by the mass
of an equal volume of water.
Example 1. The relative density of methanol is 0.8 Calculate (i) the density of
methanol
(ii) the mass of 500 cm3 of methanol
Example 2. The density of iron is 7700 kg/m3
Calculate ( i ) its relative density
(ii)
The mass of a rod of iron 50 cm long and 10 cm in diameter.
Example 3. Calculate the mass of
(i)
50 cm3 of water
(ii)
50 m3 of water
(iii)
500 mm3 of iron
(iv)
500 mm3 of methanol.
Example 4. Calculate the density of gasoline if 51 g occupies 75 cm3
Example 5. A flask has a mass of 25 g when empty 75 g when filled with water and
88 g when filled with glycerine. Calculate the density of glycerine.
Example 6. The mass of a litre of milk is 1.032 kg. The butterfat it contains has a
density of 865 kgm-3And constitutes 4% of the milk by volume. What is the density
of fat free skimmed milk.
1 litre = 1000 cm3
Pressure due to a liquid of height h ( m)
h ( m)
Consider a liquid at height h ( m) in a container where the area of the end of the
container is A m2
Force
Pr essure 
Area
Re call
m

V
and V  A.h
Therefore m  .V  Ah
Ahg

Pr essure 
 gh
A
S.I. Units Pascal Pa
Note:
THE PRESSURE IS THE SAME AT ALL POINTS AT THE SAME DEPTH.

ATMOSPHERIC PRESSURE.
Surrounding the earth is a layer of air consisting of a mixture of gases called the
atmosphere. The earth retains its atmosphere because of the pull of gravity on the air
molecules. The pressure exerted by the atmosphere is called atmospheric pressure.
MEASUREMENT OF ATMOSPHERIC PRESSURE.
Atmospheric pressure is measured using a barometer. A barometer is an instrument
for measuring the pressure of air.
As the atmospheric pressure value is quite large (as we will show ) the value of
atmospheric pressure ( A.P ) is often given in terms of a height of the liquid mercury.
Mercury is used as it has a relative density of 13.6
The simple barometer consists of a tube about 80 cm long which is open at one end.
If the tube is first completely filled with mercury and then inverted into a dish of
mercury it is found that the mercury falls a little way in the tube.
The column of mercury is now supported by the pressure of the air ( Atmospheric
Pressure ) on the free surface of the mercury in the dish.
The height of the mercury in the tube is found to be approximately 76 cm.
Note if water was used the A.P. would support a column 13.6 times this height which
is 10.336 m which is much too large a height to be easily set up.
Atmospheric Pressure = Pressure exerted by 76 cm ( Hg )
As mercury is a liquid
Pressure = 13,600 kgm-3x9.81ms-2 x76x10-2m
Pressure = 1.01396 x 105 Pa
NOTE : An alternative unit of pressure that is used when dealing with Atmospheric
Pressure is called the bar where
1 bar
= 1 x 10 5 Pa
Therefore
A.P.
= 1.014 bar
Or
A.P.
=
1014 m bars
ISOBAR : Line joining points of equal atmospheric pressure.
GAUGE PRESSURE : Pressure above atmospheric pressure.
i.e Pressure due to liquid only
ABSOLUTE PRESSURE : Total Pressure
= Atmospheric Pressure + Gauge Pressure
e.g to calculate the total or absolute pressure at a depth h ( m) in the ocean we would
have to add the atmospheric pressure to the pressure due to the height h (m) of water.
There are two other main types of barometers
(i)
(ii)
Fortin barometer and
Aneroid barometer
Question 1. Calculate the total pressure at a depth of (i) 5m (ii) 25m and (iii) 50 m under sea
water. Given that the density of sea water is 1003 kgm-3 and the atmospheric pressure on the
day is equivalent to 747 mm of Hg. R.D (Hg) =13.6.

ARCHIMEDES PRINCIPLE.
Fluid: General term for a liquid or gas
If a body is immersed in a liquid, the pressure of the liquid produces an upthrust on
the body.
When a body is placed in a liquid it will always displace its own volume of fluid.
The upthrust is a force given the symbol U
Archimedes Principle states that
When a body is wholly or partially immersed in a fluid there is an upthrust produced
equal to the weight of the fluid displaced.
This fact can be used to measure the relative density ( R.D.) of a material.
FLOTATION.
When a body is immersed in a fluid there are two forces on the object in the vertical
direction
(i)
weight acting vertically down ( W )
(ii)
Upthrust acting vertically up ( U )
(a)
If W > U object accelerates down
(b)
If W < U object accelerates up
(c)
If W = U the body floats.
The Principle of flotation states : When a body floats its weight is equal to the
upthrust on it.
Archimedes Principle is used to measure relative density of a substance.
We have already shown that the relative density of a substance can be given as
R.D 
mass of substance
mass of an equal volume of water
To use archimedes principle to calculate the relative density of a substance we need to
use a spring balance calibrated in Newtons .
Consider the following set up where a mass m ( kg) is held at the end of a wire. The
mass produces a tension force in the wire and has a weight that acts vertically down.
The mass is stationary in the vertical direction so the total force on the mass in the
vertical direction is zero.
Therefore the Tension force T on the wire acting vertically up is equal to the weight
of the object.
Tension Force
Weight of the object
Spring Balance. Where a pointer on the balance indicates the force acting on the
spring i.e. the weight of the object or the tension force on the wire.
1.
Weigh the substance in air ( W 1 or T 1 ) and to calculate the mass in air divide
the weight by g.
W1
m1 
g
2.
Weigh the substance when it is immersed in water. ( W 2 or T 2)

The mass of the substance in water m 2
W2
m2 
g
There are three forces acting on the substance when it is immersed in the water
1.
It weight acting vertically down
2.
The upthrust acting vertically up
3.
The Tension in the spring acting vertically up T 2

Tension
Weight
Upthrust
Total Force in the vertical direction is equal to zero
Therefore
W - U - T2 = 0
T1 - U - T2 = 0
Therefore
U = T1 - T2
U = Weight of water displaced.
U = Weight of equal volume of water
Therefore
U
 mass of equal volume of water
g
U T1  T2  T1 T2

 
g
g
g g
U
 mass of equal volume of water
g
Therefore Relative Density
RD 
R.D 
= m1  m2
m
m1  m2
mass of substance
apparent loss in mass when subtance is immersed in water

Example 1. The mass of an object in air is 50 g and it appears to have a mass of 35 g
when immersed in water. Calculate (i) relative density of the substance and (ii) the
density of the substance
(iii) If the same object appears to have a mass of 40 g when it is placed in a second
fluid calculate the relative density and density of the fluid.
Given
3
Example 2. A solid cube of material is 0.75 cm on each edge. It floats in oil of
density 800 kg / m3 with one third of the block out of the oil. Calculate ( i )the
Upthrust force on the cube and (ii) the density of the material of the cube.
Example 3. A block of metal weighs 0.096 N . When suspended in water it has an
apparent weight of 0.071 N. Find the density of the metal.
Example 4. A person has a weight of 667 N and a density of 980 kg / m3 can just float
in water with his head above the surface with the aid of a life jacket which is wholly
immersed in the water. Assuming that the volume of its head is 1 / 15 th of his whole
volume and the relative denstiy of the life jacket is 0.25 calculate the volume of the
jacket.
Solutions.
Example 1.
R.D. = mass in air / apparent loss in mass
50 x 10-3 / ( 50 x10-3 – 35 x 10-3)
= 3.33
Density = 3.33 x 1000 kg / m3 = 3,333 kg / m3
R.D. = mass of substance/mass of equal vol of H2O
R.D. = (50x10-3- 40x10-3 ) / (50x10-3-35x10-3)
R.D. = 0.66
Density = 666.67 kg / m3
Example 2. The block is in equilibrium so
weight of the block = Uphrust = weight of
oil displaced
Mass of block x g = mass of oil displaced x g
b
Vb
o
( Volume displaced) x g
b
Vb
o
( 2 Vb / 3 ) x g
Vb = 4.22 x 10-7 m3
o
= 800 kg / m3 then
U = 2.21 x 10-3 N
b
Example 3.
R.D. = weight / apparent loss in weight
= 0.096 / ( 0.096 – 0.071 ) = 3.84
3
Density = R.D. x density of water
= 3.84 x 1000 kg / m3
= 3,840 kg / m3
Example 4.
The volume of the person is 0.07 m3
U = weight of person + weight of jacket
U = 667 N + mass of jacket x g
U = 667 N + density of jacket x Volume x g
U = 667 N + 250 x Volume of jacket x g
U = 667 N + 250 x Vlj x g
Also the value of the Uphrust
U =
water
g [ ( 14/15) Vperson + V lj ]
Equating the two values we get that
V lj = 0.004 m3
VISCOSITY.
Viscosity is a term that describes internal friction in fluids. The more viscous a fluid
the greater the force required to cause one layer of fluid to slide past another.
The viscosity of gases is less than that of liquids, and the viscosity of water and light
oils is less than that of heavy oils and molasses.
Also the viscosity of a fluid increases with decreasing temperature.
Bodies moving through fluids experience a retarding or frictional force. The value of
the force depends on the COEFFICIENT OF VISCOSITY
force.
The value of the coefficient of viscosity of the fluid also determines the rate of flow of
the fluid through a pipe.
Let us consider the following two cases that are related to the viscosity of a fluid.
A.
An object moving through a fluid.
If an object is allowed to fall through a viscous fluid it at first accelerates but then it
reaches a uniform velocity called the terminal velocity
Consider for example a ball bearing dropped into a column of glycerine.
At point A the ball-bearing is accelerating.
However at the point B the ball-bearing is moving at constant velocity and the
acceleration is therefore zero. (v is constant therefore a = 0)
A
B
The retarding force experienced by an object of radius r (m) moving at velocity v (
-2
ms-1
STOKES LAW
v
When the terminal velocity v o ( m s-1 ) is achieved
Then the total force acting on the object in the vertical direction must be zero.
How many forces act on the object in the vertical direction ?
Three
Weight
Vertically down
Upthrust
Vertically up (object immersed in liquid)
Frictional ( Retarding ) Force Vertically up
(viscous liquid)
Consider now therefore a sphere of radius r ( m) moving through a liquid at terminal
velocity vo ( ms-1 )
U
F
Given the following values
W
Density of the sphere material

Density of liquid
= 
Volume of sphere
4 r 3
=
3
= 
W = Weight of sphere = mass of sphere x g
U = Upthrust = Weight of liquid displaced
4 3
r of the liquid
3
since a body displaces its own volume of liquid
= Weight of a volume of

F  Retarding (Frictional) Force
Since also the Total Force

= 6rv 0
W U  F  0
F  W U
W  mass of sphere x g
= (Volume of sphere) g
4

W =  r 3 g
3

U  mass of liquid displaced x g
= (Volume of liquid displaced) g
4

U =  r 3 g
3

F  6rv o

F  W U

4 3
4 3
6rv o = r g  r g
3
3
4 3
6rv o = r g(g   )
3
Giving
2g    2
vo 
r
9
Typical Values
Substance
Viscosity coefficient
( N m-2s)
Water
1.005 x 10-3
Sulphuric acid
0.0254
Olive Oil
0.084
Air
1.83 x 10-5
Glycerine
1.6
Note 1. Frictional Force between the glass of the column and the ball-bearing is not taken into account.
Note 2. You must also ensure that the column is vertical and that the ball-bearing
travels down the column in a vertical direction.
Experimental Procedure for calculating the viscosity of glycerine.
A number of different ball bearings with different radii are placed, one at a time, into
the column of glycerine.
Measure the diameter of the ball bearing using a micrometer screw and calculate a
value for its radius.
The terminal velocity of the ball bearing is calculated by marking off two points a
certain distance apart along the tube. Start a stop clock when the ball bearing passes
the first mark and stop the clock when it passes the second mark.
Repeat the process three times for each ball bearing and calculate the average time
taken
o
= distance between marks / time taken
Set up a table of results
r (m)
r2 ( m2)
vo ( ms-1)
Plot a graph of vo(m.s-1) versus r2 (m2)
This gives a straight line graph through the origin

Since
2g    2
vo 
r
9
The slope of the line is
2g   
slope 
9
Or
2g   

9.Slope
Can be looked up and put in as constants in the equation which is then solved to
calculate the value for the coefficient for viscosity for glycerine.
B.
LIQUID FLOWING THROUGH A PIPE.
The rate of flow of a fluid through a pipe depends on
(i)
the coefficient of viscosity of the fluid
(ii)
the radius of the pipe
(iii)
the pressure difference between the two ends of the pipe.
It can be shown that for a liquid flowing through a pipe that the equation for the
quantity of fluid flowing is given by

Pa 4
Q
8l
Where
Q = Volume ( Quantity ) of liquid flowing per
second in m3s-1
P
=
Pressure difference between the two ends of the pipe
a = radius of the pipe in metres
l
=
length of the pipe in metres
To calculate experimentally the viscosity of water we set up the following.
A bell jar with a side connection to a capillary tube has water placed in it.
h
The rate of flow of the liquid is given by
Pa 4
Q
8l
P  Pressure difference between the two ends
P = gh Where h is the height of the water in
the bell jar above the capillary tube
Q=
V
t
Therefore

V
t
gha4

8l
ga4

h
8l
The V / t values for several heights h are found and a graph of V / t versus height h
is drawn.
V / t ( m3s-1)
height h ( m)
And gives a straight line through the origin.
The volume of the water flowing per second is calculated by measuring the mass of
water collected in 100 seconds in grams
1.
Dividing by 103 ( 1000 ) converts grams to
kg for correct S.I. mass units
2.
dividing by 102 calculated the mass of water flowing per second and
as V = mass / density
3.
To convert mass into volume we use the equation volume = mass / density
if the mass of water collected every second is divided by the density of water
(1000 kg /m3) this gives the value for V / t
So the mass collected in grams in 100s = m
Then mass in kg in 100 s = m x 10-3 kg
Mass in kg collected in 1 s = m x 10-3 x 10-2 = m x 10-5 kg.s-1
to convert this volume per unit time
Volume in m3 collected per s = m x10-5x10-3 = m x 10-8 m3.s-1
The viscosity of water can be calculated finding the slope of the graph and using the
equation
Recall the equation is

V
t
ga 4

h
8l
Therefore the slope of a graph of
ga 4
Slope 
8l
Therefore
 
ga 4
8l(slope)
Where
g = 9.81 m s-2
a = radius of the capillary tube
l = length of the capillary tube
-3
V
versus h is
t
PROBLEM SHEET Viscosity
Example 1. How much water will flow in 30 seconds through 200 mm of a capillary
tube
1.5 mm in diameter if the pressure difference between the ends is equivalent to 5 cm
of Hg.
The viscosity of water is given as 1 x 10-3 Nm-2s and the relative density of Hg is
13.6.
Example 2. Find the terminal velocity of a ball bearing of steel of radius 0.5 mm
falling freely in glycerine, given the following values
Density of steel = 7,700 kg m-3
Density of glycerine = 1260 kg m-3
Viscosity of glycerine = 1.6
ELASTICITY IN SOLIDS.
A load or Force is applied to a solid e.g. a piece of wire
If the wire returns to its original position when the load is removed then the load is
within what is called the ELASTIC LIMIT of the material.
e.g.
Apply a vertical Force to a wire A B
A
B
A range of Forces were applied and the extensions produced were measured. The
values are given below
Force Applied (N)
Extension (x10-3m).
10
14
20
.28
30
.42
40
.56
60
.98
70
1.48
Question Plot a graph of Force (N) versus extension ( m).
Force(N)
extension (m)
HOOKE’S LAW
Extension is proportional to the applied force in a wire if the elastic limit is not
exceeded.
F  extension
THEREFORE
F = k (extension)
OR
F = k (ext)
YIELD POINT.
The molecules of the solid begin to slide across each other once the load exceeds the
elastic limit.
The material becomes PLASTIC at this point.
This point is called the YIELD POINT.
This is indicated by a sudden increase in the extension produced
e.g consider the graph that we drew for Force versus extension
Force (N)
A
extension ( m)
The point A is the Yield point.
YOUNGS MODULUS.
Stress is defined as the Force per unit Area
Stress 
Force
Area
S.I.Units Nm2
Strain is defined as the extension produced per unit length

extension
Strain 
original length
No Units
Young’s Modulus is defined as Stress divided by Strain

Youngs Modulus 
S.I.Units

Nm
m 2m

Stress
Strain
N
m2

F
A
ext
l
 Nm2

Fl
A ext
Typical Values
Substance
Young’s Modulus(GN m-2)
Aluminium
70
Brass
90
Copper
110
Steel
192
Tungsten
355
Note 1. G N / m2
2.
= x 109 N / m2
Youngs Modulus is a constant for a given material
Laboratory Experiment to measure Young’s Modulus for a piece of copper wire.
Set up the apparatus as shown in the diagram
wooden block
copper wire
gummed paper to measure extensions
pulley
G clamp
slotted
weight
hanger
To keep the wire taut you will have to apply a weight of say 100 grams to it.
Carefully measure the original length of the wire.
Measure the diameter of the wire using a micrometer screw gauge, measure the value
about 4 to 6 times at different points down the full length of the copper wire and then
calculate an average value. Calculate the cross sectional area of the wire in m2.
Now add extra weights ( loads ) gradually to the wire and note the total extension
produced each time.
Keep adding the weights until the wire breaks. The extension values are going to be
small especially in the elastic region.
One method of measuring the extensions is place a doubled over piece of gummed
paper at some point on the wire. Place a metre stick behind the paper and line up the
meter stick and one edge of the paper as the wire extends the edge will also move
along in front of the meter stick and the extension can be measured.
Draw up a data table as follows
Force ( W = m g )
Extension
Stress = extension/
original length
Strain = Force / area
Plot a graph of Stress versus Strain for the whole test i.e. up until the wire broke.
Redraw the graph of stress versus strain for just the linear portion of the results i.e.
use a different scale for the strain.
Find the slope of this graph and calculate the value for Young’s Modulus for copper.
Question 1. A solid steel column is 2 m long and 9 cm in diameter, calculate its
increase in length when carrying a vertical load of 15 kg, given Y ( steel ) =1.92 x
1011 N / m2
Question 2. A copper wire 2 m long and 2 mm in diameter is stretched by 1 mm what
mass is needed?, given Y(copper )= 1.176 x 1011 N / m2
Question 3.( a) By how much is a cooper wire , length 150 cm and diameter 1.2 mm
stretched when a Force of 1 kN is applied to it
( b) calculate the work done in stretching the wire.
Question 4. A brass wire of length 1 m and diameter 4 mm is stretched by 0.1 mm by
a force of 113 N Calculate (a) the stress (b) strain and (c)the value for young’s
modulus for brass
ELASTICITY IN SOLIDS.
A load or Force is applied to a solid e.g. a piece of wire
If the wire returns to its original position when the load is removed then the load is
within what is called the ELASTIC LIMIT of the material.
e.g.
Apply a vertical Force to a wire A B
A
B
A range of Forces were applied and the extensions produced were measured. The
values are given below
Force Applied (N)
10 20 30 40 60 70
-3
Extension (x10 m).14.28
.42
.56
.98
1.48
Question Plot a graph of Force (N) versus extension ( m).
Force(N)
extension (m)
HOOKE’S LAW
Extension is proportional to the applied force in a wire if the elastic limit is not
exceeded.
F  extension
THEREFORE
F = k (extension)
OR
F = k (ext)
YIELD POINT.
The molecules of the solid begin to slide across each other once the load exceeds the
elastic limit.
The material becomes PLASTIC at this point.
This point is called the YIELD POINT.
This is indicated by a sudden increase in the extension produced
e.g consider the graph that we drew for Force versus extension
Force (N)
A
extension ( m)
The point A is the Yield point.
YOUNGS MODULUS.
Stress is defined as the Force per unit Area
Stress 
Force
Area
S.I.Units Nm2
Strain is defined as the extension produced per unit length


extension
Strain 
original length
No Units
Young’s Modulus is defined as Stress divided by Strain
Stress
Strain
Youngs Modulus 
S.I.Units
Nm
m 2m

N
m2

Typical Values
Aluminium
Young’s Modulus(GN m-2)
70
Brass
90
Copper
110
Steel
192
A
ext
l
 Nm2

Substance
F

Fl
A ext
Tungsten
Note 1. G N / m2
3.
355
= x 109 N / m2
Youngs Modulus is a constant for a given material
Laboratory Experiment to measure Young’s Modulus for a piece of copper wire.
Set up the apparatus as shown in the diagram
wooden block
copper wire
gummed paper to measure extensions
pulley
G clamp
slotted
weight
hanger
To keep the wire taut you will have to apply a weight of say 100 grams to it.
Carefully measure the original length of the wire.
Measure the diameter of the wire using a micrometer screw gauge, measure the value
about 4 to 6 times at different points down the full length of the copper wire and then
calculate an average value. Calculate the cross sectional area of the wire in m2.
Now add extra weights ( loads ) gradually to the wire and note the total extension
produced each time.
Keep adding the weights until the wire breaks. The extension values are going to be
small especially in the elastic region.
One method of measuring the extensions is place a doubled over piece of gummed
paper at some point on the wire. Place a metre stick behind the paper and line up the
meter stick and one edge of the paper as the wire extends the edge will also move
along in front of the meter stick and the extension can be measured.
Draw up a data table as follows
Force ( W = m g )
Extension
Stress = extension/
original length
Strain = Force / area
Plot a graph of Stress versus Strain for the whole test i.e. up until the wire broke.
Redraw the graph of stress versus strain for just the linear portion of the results i.e.
use a different scale for the strain.
Find the slope of this graph and calculate the value for Young’s Modulus for copper.
Question 1. A solid steel column is 2 m long and 9 cm in diameter, calculate its
increase in length when carrying a vertical load of 15 kg, given Y ( steel ) =1.92 x
1011 N / m2
Question 2. A copper wire 2 m long and 2 mm in diameter is stretched by 1 mm what
mass is needed?, given Y(copper )= 1.176 x 1011 N / m2
Question 3.( a) By how much is a cooper wire , length 150 cm and diameter 1.2 mm
stretched when a Force of 1 kN is applied to it
( b) calculate the work done in stretching the wire.
Question 4. A brass wire of length 1 m and diameter 4 mm is stretched by 0.1 mm by
a force of 113 N Calculate (a) the stress (b) strain and (c)the value for young’s
modulus for brass
Wave Motion
A wave is a means of transferring energy through a medium (or a vacuum ) without
any net movement of the medium i.e. only the energy is transferred.
The transfer of energy is said to take place by means of Wave Motion.
There are two types of wave motion:
1. Transverse waves which travel perpendicular of the direction of the vibration
and or the wave velocity is perpendicular to the vibrations.
2.
Longitudinal waves which travel parallel to the direction of the vibration or the
wave velocity is parallel to the direction of the oscillation.
Wave Motion can also be divided into
Mechanical and Electromagnetic Waves
Mechanical Waves
are those that are propagated through some medium which when energy is imparted to
the medium the medium is disturbed and the particles of the medium vibrate.
As the particles of the medium are linked the vibration of each particle affects that of
its neighbours and the added energy propagates by means of the interactions between
the particles of the medium. Mechanical Waves can be either
Transverse ( stretched string) OR
Longitudinal ( sound )
Electromagnetic Waves.
Electromagnetic Waves do not need a medium through which to propagate
( travel ) i.e. they can travel through a vacuum. They are transverse waves.
Electromagnetic wave – A wave of energy having a frequency within the
electromagnetic spectrum and propagated as a periodic disturbance of electric and
magnetic fields when an electric charge accelerates.
In a vacuum electromagnetic waves travel at the speed of light.
The general representation of a wave is a sine wave and we can show the wave in
terms of changing time or changing position and we get the following
Amplitude
wave
displacement
OR
dispalcement
w ave
tim e
Wave Formula
Again as before when we looked at sine waves the maximum displacement is called
the Amplitude.
The distance between two consecutive points in phase is called the wavelength
.
The time taken for one point to undergo one complete cycle or oscillation is called the
periodic time T (s).
The number of complete oscillations made per second is called the frequency f
measured in an S.I. Unit called Hertz = Hz = s-1
1
f 
T
velocity of wave
v = f.
= frequency x wavelength
This wave formula applies to all waves.
The equation for the velocity of the wave comes from the motion relationship
Distance = velocity x time
 =vxT
f=
1
T
Gives the standard wave relationship
v = f. 

SOUND
We have already talked about wave motion and to recap there are two types of wave
motion
1.
Transverse are wave perpendicular to the direction of the vibration
e.g Light waves , radio waves, all electromagnetic waves and a stretched string
2.
Longitudinal waves which travel parallel to the direction of the vibration .
e.g. Sound waves
We can also divide wave types into
1. Electromagnetic waves OR
2. Mechanical waves
Electromagnetic waves i.e. all wave motion in the electromagnetic spectrum.
They are transverse waves and are caused by a change in the intensity of an
electromagnetic field. They can travel through a vacuum.
Mechanical Waves These waves are caused by a change in position or density which
is transmitted through a material medium.
They cannot travel through a vacuum
e.g sound
Sound is a longitudinal mechanical wave.
It is the vibrations of the medium itself usually air that transfers the energy from one
place to another.
The air motion is in the same direction as the wave motion.
The velocity of sound varies with the medium through which it is travelling.
Velocity
Medium
330 m. s-1
Air
1450 m . s-1
Water
4000 m . s-1
Timber
5000 m . s-1
Brick
Sound travelling through a material travels by causing a mechanical vibration of the
material itself. The more rigid the material the faster this vibration will travel .
As the density of the material increases so does the velocity of sound through the
material.
Frequency.
The frequency of sound is the number of waves passing a point per second.
Audible sounds are in the range
20 Hz - 20000 Hz
However note
1. The ear is most sensitive between the ranges 2000 Hz - 4000 Hz
And
2.
As a person gets older they tend to start to lose the higher frequencies.
Audible sounds are in the range 20Hz to 20,000Hz and the average human ear is most
sensitive at a frequency of 3000Hz.
The frequency determines the pitch of the sound and in musical terms going up one
octave doubles the pitch
Middle C on the piano
C above this
C below this
f = 256 Hz
f = 512 Hz
f = 128 Hz
Quality of sound.
Any sound consists of one or more frequencies.
Related frequencies are called harmonics where
f = fundamental frequency = 1st Harmonic
2 f = 2 nd Harmonic
3 f = 3rd Harmonic
A musical note contains either
One frequency only or
The simultaneous sounding of related frequencies or harmonics.
Noise is the simultaneous sounding of a number of unrelated frequencies in that no
pattern can be discerned.
Every musical instrument produces a sound with characteristic quality. This arises
from the relative abundances of the various harmonics present.
Intensity of Sound.
Determined by the amplitude of the wave.
The intensity of sound is the energy per second passing 1 m2 perpendicular to the
direction of the sound.
S.I. Units
W / m2
Lowest audible intensity = 10-12 W / m2
Highest tolerable intensity = 100 W / m2
The range of intensity values that we need to consider is therefore from
10-12
up to 102 This is a very large range of values and so we use a log scale
instead
Intensity Level .
This is a log scale
S.I. Unit Decibel
dB.
Compares the intensity of a sound with the standard lowest audible intensity of
10-12 W / m2.
To calculate the intensity level of a sound of intensity I W / m2 use the following
equation
Intensity level
In dB
= 10 log ( I / I o )
Where Io = 1 x 10-12 W / m2 called the Threshold Intensity.
Intensity Level is a log scale measured in dB
Intensity is a linear scale measured in W / m2
Therefore any linear functions such as
Addition , subtraction , multiplication or division must be carried out in the linear
scale of intensity and then converted back to the log scale of intensity level
Convert the following intensity values to intensity level values
1x10-6 W / m2
5 x 10-4 W / m2
20 W / m2
Convert the following intensity level values to intensity values
50 dB,
46 dB,
87 dB
Add the following intensity level values
40 dB + 50 dB,
34 dB + 42 dB
Reduction of sound with distance.
Vibrations set up cause the transfer of mechanical energy away from the source of the
sound.
As this energy travels it is gradually dissipated as heat energy and finally the
vibrations die away completely.
The intensity I d at a distance d metres from a sound of intensity Is is given by
Id =
I s / d2
Example 1 calculate the intensity at a distance of 15 m from a source of intensity
1 x 10-6 W / m2
Example 2 calculate the intensity level at a distance of 5 m from a source of 45 dB
PROBLEM SHEET.
Question 1. Calculate the intensity level of the following intensity values
(i)
6 x 10-6 W/m2 (ii) 2 x 10-3 W/m2
Question 2. Calculate the intensity of a cound which has an intensity level of
(i)
55 dB (ii) 62 dB
Question 3. Calculate the intensity level of two sounds each of intesity 25 dB.
Question 4. Three machines each with an intensity level of 46 dB are switched on in
a room. Calculate the total intensity level in the room.
Question 5. If the machines in question 4 are fitted with sound insulation which
reduces the intensity by 40% calculate the new intensity level value.
Question 6. A sound source produces an intensity level of 75 dB. Calculate the
intensity level at a distance of (I) 3 m (ii) 6 m and (iii) 10 m from the source.
Question 7. A person speaking normally produces an intensity level of 40 dB at a
distance of 1.5 m. If the conditions are such that the intensity level to be audible is 20
dB calculate the maximum distance at which the person can be clearly heard.
Question 8. If the intensity level of a sound increases from 30 dB to 40 dB calculate
the corresponding increase in intensity.
Doppler Effect
This is the effect produced if the observer and the source are moving relative to one
another.
We can have
1.
A moving source and a stationary observer
2.
A stationary source and a moving observer
3.
Both source and observer moving relative to one another.
We will look at the case when a source of sound is moving relative to a stationary
observer, a change is frequency is observed known as the Doppler Effect.
NOTE : A sine wave can be represented as a wave which varies periodically between
a maximum and minimum value.
The sine wave can also be represented by a series of parallel lines where each line
represents the crest of a wave
Consider a source of sound stationary and a stationary observer.
Let
f = The frequency of sound waves
v = the velocity of sound in air
If the source is stationary then in a time of one second is will have emitted f waves.
The f waves will occupy a distance of v (m) from the source.
So in this case we can represent the sound waves as follows
v/f
Source
observer
V(m)
v/f is the distance between two consecutive waves in phase which is the wavelength
of the wave
Therefore
Consider now if the source is moving towards the stationary observer with a velocity
us
Then the source will in one second produce f waves however they will only occupy a
distance of v – u s
s at t=0
s at t =1
observer
us
v – us
v(m)
Now the distance between two consecutive points in phase is
v - us
f
The apparent frequency therefore is
f1
– us / f )
f1 = ( v / v – us ) f
Where f1 is the apparent frequency at the observer
f is the actual frequency of the source
v = the velocity of sound in air and
us = the velocity at which the source is moving towards the stationary observer.
Consider now source moving away from a stationary observer with a velocity us.
We can represent the waves as follows
S at t=1
s at t=0
us
observer
v
v + us
Now the distance between two consecutive points in phase is
2
= v + us
f
and therefore the apparent frequency
f 2 = v / ( v + us / f )
f 2 = ( v / v + us ) f
Where
f2 is the apparent frequency at the observer
f is the actual frequency
v is the velocity of sound in air and
us is the velocity of the source away from the observer.
Example 1. A train whistle emits a sound of frequency 500 Hz, and approaches a
station at a steady speed of 40 m.s-1. Calculate the apparent change is frequency
noticed by a stationary observer at the station as the train passes through.
Take v the velocity of sound in air = 330 m.s-1
Example 2. A train approaches a station at a steady speed. If the apparent frequency
of its whistle is 569 Hz as it approaches and 496 Hz as it moves away from the station
calculate
(1)
the speed of the train and
(2)
actual frequency of the whistle.
Example 3. The frequency of the horn of a car is 600 Hz.(a) How fast must a car be
approaching a stationary listener if the sound of its horn is to appear to be at its
loudest.?
(b) How fast must the car recede from the observer if the horn is inaudible because its
frequency is too low ?
Standing or stationary waves.
A stationary wave is produced when two waves of the same frequency, speed and
amplitude but moving in opposite directions meet.
Waves 1 and 2 have the same speed amplitude and frequency but are moving in
opposite directions in the medium.
In a lot of case wave 2 is produced by the reflection of wave 1.
The points N1 , N2, N3 are called NODES and they are positions where there are
no vibrations of the medium, or the points where the amplitude is zero.
The points A1, A2, A3 are called ANTINODES and they are the positions where the
vibrations are at a maximum, or the points where the amplitude is maximum.
The distance between two consecutive nodes or two consecutive antinodes =
Waves on a string.
The distance between two consecutive nodes is half a wavelength.
This has implications in several fields. For instance, in a guitar string, the ends of the
string are nodes.
By changing the position of one of the nodes through frets, the guitarist changes the
effective length of the vibrating string and thereby the note played.
The lowest frequency produced by any particular instrument is known as the
fundamental frequency. The fundamental frequency is also called the first harmonic
of the instrument.
This is the case for the first harmonic or fundamental frequency of a guitar string. The
diagram below depicts this length-wavelength relationship for the fundamental
frequency of a guitar string.
The second harmonic of a guitar string is produced by adding one more node between
the ends of the guitar string. And of course, if a node is added to the pattern, then an
antinode must be added as well in order to maintain an alternating pattern of nodes
and antinodes..
The standing wave pattern for the second harmonic is
This is also
the case for other stringed instruments.
When the
same musical note is played on different
instruments the sounds can appear different?
Stringed instruments sound different for many reasons.
When a string vibrates, its wavelength and frequency of vibration strongly affect the
sound it makes.
However, there can be additional vibrations in the string at the same time called
"harmonics" that "stack on top" of the main vibration.
In other words, stringed instruments are actually making multiple different vibrations,
and therefore multiple sounds, at once.
The combination of these different vibrations, or harmonics, give different
instruments the different sound textures that you can hear.
Demonstrations. Go to the following web site and follow the instructions to see a
demonstration of the harmonics on a violin.
http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/Standi
ngWaves1.html
and to create your own harmonics go to
http://library.thinkquest.org/19537/cgi-bin/showharm.cgi
BEATS.
When two notes of slightly different frequencies but of similar amplitudes are
sounded together , the loudness ( intensity ) increases and decreases periodically and
beats are said to be heard.
The beat frequency is equal to the difference between the two almost equal
frequencies
Beat frequency = f1 - f 2 ( Hz )
Resonance
All bodies which are capable of vibrating have a natural frequency with which they
vibrate when displaced.
We can also apply a frequency to the body to causes it to vibrate at a frequency other
than its natural frequency which is called the forced vibration.
When the forced vibration is removed the body will return to its natural frequency.
If the forced vibration and the natural frequency are equal then the amplitude of the
vibration of the body increases and this is called resonance.
Therefore
Resonance occurs when the forced frequency applied to the body is equal to the
natural frequency of the body.
Resonance tube.
A column of air in a closed pipe will resonate with a driving frequency to produce
harmonics such as those illustrated in Fig.1.
The column of air vibrates with a node at the closed end and an antinode at the open
end. Therefore the length of the tube is
The velocity of sound in air can be calculated using the equation
Investigation of Standing Waves on Strings using the Sonometer
The relationship between frequency, length and tension for the resonance of a
stretched wire is investigated. This allows the linear density of the wire to be
determined.
When a string is set in vibration the fundamental mode of vibration - or first harmonic
– is as shown in Fig.1.
Fig.1
Fundamental
L=
mode
frequency,
f
For a wire of linear density (mass per unit length), , length l and under a tension T,
the frequency of vibration, f, is given by the equation
F = 1 / 2 L  T / mL
Experimental Procedure to use the sonometer to calculate the linear density of a
material
Set up the apparatus as shown in Fig.2. Connect leads from the variable frequency
function generator to the points X and Y using crocodile clips.
Fig.2
l
X
Y
T
Hollow sounding box
1. Straddle the wire (without touching it) with the large horse-shoe magnet.
2. Place a load to produce a tension T on the wire as shown.
3. Tune the signal generator by varying the frequency until a standing wave pattern
similar to that shown in Fig.1. is obtained. You may also need to vary the position
of the moveable bridge and hence, change the length l. Note the frequency and
length at which the standing wave pattern occurs.
4. Repeat step 4 for a number of different tensions.
5. Measure the length of wire l.
6. Plot a graph of f2 vs T/4l2 and from the slope of the graph find a value for the
linear density of the wire.
Ultrasound
Produced at frequencies above 20,000 Hz.
Frequencies up to several million hertz can be produced inaudible to the human
ear.
Very small wavelengths means that the waves are very penetrating and can be
beamed into a very small area with great accuracy.
Production.
Ultrasound is produced by vibrating crystals usually quartz , using piezoelectric
effect.
When the crystal is stretched electric charges are produced on its surface
If a high frequency a.c current is applied to the crystal the faces of the crystal with
expand / contract causing the crystal to vibrate at a frequency equal to the a.c
frequency.
Conversely when the crystal is compressed the electric charges are reversed
The vibrations produce a wave motion where the frequency of the wave is equal to the
frequency of the a.c source.
In this case the crystal acts as a transducer.
A transducer is a device that converts one type of vibration into another in this case
Electrical to mechanical producing ultrasound.
Advantages.
1.
Not confused with audible sounds.
2.
High frequency implies very penetrating and can be confined to very narrow
beams..
Applications.
Echo sounding
(a) An electrical device fitted to the bottom of a ship sends out regular ultrasound
impulses which are reflected back.
The time interval between the emission and the arrival back of the sound wave
depends directly on the depth of the sea.
(b)
Echo sounding in medicine.
When the wave is emitted it will be reflected back when it hits an obstacle and again
the time interval between the emission and the arrival back of the wave depends on
the position of the obstacle.
Crystal
s (m)
obstacle that produces the reflection
t1 = time zero ultrasound wave sent by the crystal
t2 = reflection pick up at the crystal
2 . s = the distance travelled by the wave
v = velocity of sound in medium through which the wave is travelling
distance
2s
v

time
t 2  t1
Therefore
v(t 2  t1 )
s
2
