MELS 123 ELECTRONIC LOGIC FOR SCIENCE LECTURE NOTES PPT OMOSAE

4
Outline
• Electrical Careers
• Meaning of Repair and Maintenance
• Components and Symbols
• Resistance Colour code
• Electrical Meters
• Electrical Units of Measure
• Engineering Notation
• Hand tools
• Soldiering
•Desoldiering
• cc

Newton’s Contributions
• Calculus
• Light is composed of
rainbow colors
• Reflecting Telescope
• Laws of Motion
• Theory of Gravitation

Newton’s laws of motion
Newton’s laws of motion describe to a high degree of
accuracy how the motion of a body depends on the
resultant force acting on the body.
They define what is known as ‘classical mechanics’.
They cannot be used when dealing with:
(a) speeds close to the speed of light
– requires relativistic mechanics.
(b) very small bodies (atoms and smaller)
– requires quantum mechanics

Newton’s first law of motion
A body will remain at rest or move with a constant
velocity unless it is acted on by a net external
resultant force.
Notes:
1. ‘constant velocity’ means a constant speed
along a straight line.
2. The reluctance of a body to having its velocity
changed is known as its inertia.

Newton’s First Law
(law of inertia)
An object at rest tends to stay at rest and an
object in motion tends to stay in motion
unless acted upon by an unbalanced force.

Examples of Newton’s first law of motion
Box stationary
The box will only move if the push
force is greater than friction.
Box moving
If the push force equals friction
there will be no net force on the box
and it will move with a constant
velocity.
Inertia Trick
When the card is flicked, the coin
drops into the glass because the
force of friction on it due to the
moving card is too small to shift it
sideways.

Balanced Force
Equal forces in opposite
directions produce no motion

Unbalanced Forces
Unequal opposing forces
produce an unbalanced force
causing motion

If objects in motion tend to stay in motion, why don’t
moving objects keep moving forever?
Things don’t keep moving forever because
there’s almost always an unbalanced force
acting upon them.
A book sliding across a table slows
down and stops because of the force
of friction.
If you throw a ball upwards it will
eventually slow down and fall
because of the force of gravity.

(law of inertia)
•MASS is the measure of the amount of
matter in an object.
•It is measured in Kilograms

(law of inertia)
•INERTIA is a property of an object that describes
how ______________________ the motion of
the object
•more _____ means more ____
much it will resist change to
mass inertia

1st Law
•Unless acted upon by
an unbalanced force,
this golf ball would sit
on the tee forever.

• There are four main types of friction:
• Sliding friction: ice skating
• Rolling friction: bowling
• Fluid friction (air or liquid): air or water resistance
• Static friction: initial friction when moving an object
What is this unbalanced force that acts on an
object in motion?

1st Law
•Once airborne,
unless acted on
by an
unbalanced force
(gravity and air –
fluid friction) it
would never
stop!

Newton’s second law of motion

Momentum (p)
momentum = mass x velocity
p = m x v
mass is measured in kilograms (kg)
velocity is measured in metres per second (m/s)
momentum is measured in:
kilogram metres per second (kg m/s)

Momentum has both
magnitude and direction.
Its direction is the same as
the velocity.
The greater the mass of a
rugby player the greater is
his momentum

Question 1
Calculate the momentum of a rugby player, mass 120kg
moving at 3m/s.
p = m x v
= 120kg x 3m/s
momentum = 360 kg m/s

Question 2
Calculate the mass of a car that when moving at 25m/s
has a momentum of 20 000 kg m/s.

Question 2
Calculate the mass of a car that when moving at 25m/s
has a momentum of 20 000 kg m/s.
p = m x v
becomes: m = p ÷ v
= 20000 kg m/s ÷ 25 m/s
mass = 800 kg

Newton’s second law of motion
The acceleration of a body of constant mass is related to the net
external resultant force acting on the body by the equation:
resultant force = mass x acceleration
F = m a

Question 1
Calculate the force required to cause a car of mass 1200 kg to
accelerate at 6 ms -2.
F = m a
F = 1200 kg x 6 ms -2
Force = 7200 N

Question 2
Calculate the acceleration produced by a force of 20 kN on a mass of 40
g.

Question 2
Calculate the acceleration produced by a force of 20 kN on a mass of 40
g.
F = m a
20 000 N = 0.040 kg x a
a = 20 000 / 0.040
acceleration = 5.0 x 105 ms -2

Question 3
Calculate the mass of a body that accelerates from 2
ms -1 to 8 ms -1 when acted on by a force of 400N for 3
seconds.

Question 3
Calculate the mass of a body that accelerates from 2
ms -1 to 8 ms -1 when acted on by a force of 400N for 3
seconds.
acceleration = change in velocity / time
= (8 – 2) ms -1 / 3s
a = 2 ms -2
F = m a
400 N = m x 2 ms -2
m = 400 / 2
mass = 200 kg

Answers
Force Mass Acceleration
4 kg 6 ms -2
200 N 5 ms -2
600 N 30 kg ms -2
5 g 400 ms -2
5 μN 10 mg cms -2
Complete:

Answers
Force Mass Acceleration
24 N 4 kg 6 ms -2
200 N 40 kg 5 ms -2
600 N 30 kg 20 ms -2
2 N 5 g 400 ms -2
5 μN 10 mg 50 cms -2
24 N
40 kg
20
2 N
50
Complete:

Types of force
1. Contact
Two bodies touch when their repulsive molecular forces (due to electrons) equal the force that
is trying to bring them together. The thrust exerted by a rocket is a form of contact force.
2. Friction (also air resistance and drag forces)
When two bodies are in contact their attractive molecular forces (due to electrons and protons)
try to prevent their common surfaces moving relative to each other.
3. Tension
The force exerted by a body when it is stretched. It is due to attractive molecular forces.
4. Compression
The force exerted by a body when it is compressed. It is due to repulsive molecular forces.
5. Fluid Upthrust
The force exerted by a fluid on a body because of the weight of the fluid that has been
displaced by the body. Archimedes’ Principle states that the upthrust force is equal to the
weight of fluid displaced.

6. Electrostatic
Attractive and repulsive forces due to bodies being charged.
7. Magnetic
Attractive and repulsive forces due to moving electric charges.
8. Electromagnetic
Attractive and repulsive forces due to bodies being charged. Contact, friction, tension,
compression, fluid upthrust, electrostatic and magnetic forces are all forms of
electromagnetic force.
9. Weak Nuclear
This is the force responsible for nuclear decay.
10. Electro-Weak
It is now thought that both the electromagnetic and weak nuclear forces are both forms of
this FUNDAMENTAL force.
11. Strong Nuclear
This is the force responsible for holding protons and neutrons together within the nucleus. It
is one of the FUNDAMENTAL forces.

12. Gravitational
The force exerted on a body due to its mass.
It is one of the FUNDAMENTAL forces.
The weight of a body is equal to the gravitational force acting on the body.
Near the Earth’s surface a body of mass 1kg in free fall (insignificant air
resistance) accelerates downwards with an acceleration equal to g = 9.81 ms-2
From Newton’s 2nd law:
ΣF = m a
ΣF = 1 kg x 9.81 ms -2
weight = 9.81 N
In general: weight = mg

Gravitational Field Strength, g
This is equal to the gravitational force acting on 1kg.
g = force / mass
= weight / mass
Near the Earth’s surface:
g = 9.81 Nkg-1
Note: In most cases gravitational field strength is numerically equal to
gravitational acceleration.

Rocket question
Calculate the engine thrust required to
accelerate the space shuttle at 3.0 ms -
2 from its launch pad.
mass of shuttle, m = 2.0 x 10 6 kg
g = 9.8 ms -2

Rocket question
F = m a
where :
F = (thrust – weight)
= T – mg
and so:
T – mg = ma
T = ma + mg

Rocket question
ma:
= 2.0 x 106 kg x 3.0 ms -2
= 6.0 x 106 N
mg:
= 2.0 x 106 kg x 9.8 ms -2
= 19.6 x 106 N
but: T = ma + mg
= (6.0 x 106 N) + (19.6 x 106 N)
Thrust = 25.6 x 106 N

Lift question
A lift of mass 600 kg carries a passenger of mass 100 kg.
Calculate the tension in the cable when the lift is:
(a) stationary
(b) accelerating upwards at 1.0 ms-2
(c) moving upwards but slowing down at 2.5 ms-2
(d) accelerating downwards at 2.0 ms-2
(e) moving downwards but slowing down at 3.0 ms-2.
Take g = 9.8 ms-2

Let the cable tension = T
Mass of the lift = M
Mass of the passenger = m
(a) stationary lift
From Newton’s 1st law of motion:
A stationary lift means that resultant force acting on
the lift is zero.
Hence:
Tension = Weight of lift and the passenger
T = Mg + mg
= (600kg x 9.8ms-2) + (100kg x 9.8ms-2)
= 5880 + 980
Cable tension for case (a) = 6 860 N

(b) accelerating upwards at 1.0 ms-2
Applying Newton’s 2nd law:
ƩF = (M + m) a
with
ƩF = Tension – Total weight
Therefore:
(M + m) a = T – (Mg + mg)
(600kg + 100kg) x 1.0 ms-2
= T – (5880N + 980N)
700 = T – 6860
T = 700 + 6860
Cable tension for case (b) = 7 560 N

(c) moving upwards but slowing down at
2.5 ms-2
Upward accelerations are positive in this
question.
The acceleration, a is now
MINUS 2.5 ms-2
Therefore:
(M + m) a = T – (Mg + mg)
becomes:
(600kg + 100kg) x - 2.5 ms-2
= T – (5880N + 980N)
- 1750 = T – 6860
T = -1750 + 6860
Cable tension for case (c) = 5 110 N

(d) accelerating downwards
at 2.0 ms-2
Upward accelerations are positive in this
question.
MINUS 2.0 ms-2
Therefore:
(M + m) a = T – (Mg + mg)
becomes:
(600kg + 100kg) x - 2.0 ms-2
= T – (5880N + 980N)
- 1400 = T – 6860
T = -1400 + 6860
Cable tension for case (d) = 5 460 N

(e) moving downwards but slowing
down at 3.0 ms-2
This is an UPWARD acceleration
PLUS 3.0 ms-2
Therefore:
(M + m) a = T – (Mg + mg)
becomes:
(600kg + 100kg) x + 3.0 ms-2
= T – (5880N + 980N)
2100 = T – 6860
T = 2100 + 6860
Cable tension for case (e) = 8 960 N

Terminal Velocity
Consider a body falling through a fluid
(e.g. air or water)
When the body is initially released the only significant
force acting on the body is due to its weight, the
downward force of gravity.
The body will fall with an initial acceleration = g
Note: With dense fluids or with a low density body the upthrust
force of the fluid due to it being displaced by the body will also
be significant.
weight

As the body accelerates
downwards the drag force exerted
by the fluid increases.
Therefore the resultant downward
force on the body decreases
causing the acceleration of the
body to decrease.
F = (weight – drag) = ma
Eventually the upward drag force
equals the downward gravity force
acting on the body.

Therefore there is no longer
any resultant force acting on
the body.
F = 0 = ma
and so: a = 0
The body now falls with a
constant velocity.
This is also known as ‘terminal
speed’
Skydivers falling at their
terminal speed

speed
time from release
terminal speed
initial acceleration = g
resultant force & acceleration

Newton’s third law of motion
When a body exerts a force on another body then the
second body exerts a force back on the first body that:
• has the same magnitude
• is of the same type
• acts along the same straight line
• acts in the opposite direction
as the force exerted by the first body.

Examples of Newton’s third law of motion
1. Earth – Moon System
There are a pair of gravity forces:
A = GRAVITY pull of the EARTH to the LEFT on the MOON
B = GRAVITY pull of the MOON to the RIGHT on the EARTH
A
B
Notes:
Both forces act along the same straight line.
Force A is responsible for the Moon’s orbital motion
Force B causes the ocean tides.

2. Rocket in flight
There are a pair of contact (thrust) forces:
A = THRUST CONTACT push
of the ROCKET ENGINES
DOWN
on the EJECTED GASES
B = CONTACT push
of the EJECTED GASES
UP
on the ROCKET ENGINES
Note: Near the Earth there will also be a pair of
gravity forces. If the rocket is accelerating upwards
then the upward contact force B will be greater than
the downward pull of gravity on the rocket.
A
B

3. Person standing on a floor
There are a pair of gravity forces:
A = GRAVITY pull of the EARTH
DOWN on the PERSON
B = GRAVITY pull of the PERSON
UP on the EARTH
And there are a pair of contact forces:
C = CONTACT push of the FLOOR
UP on the PERSON
D = CONTACT push of the PERSON
DOWN on the FLOOR
Note: Neither forces A & C nor forces D & B are Newton
3rd law force pairs as the are
NOT OF THE SAME TYPE
although all four forces will usually have the same
magnitude.
A
B
C
D
EARTH

Tractor and car question
A tractor is pulling a car out
of a patch of mud using a
tow-rope as shown in the
diagram opposite. Identify
the Newton third law force
pairs in this situation.
G1
G2
T1 T2
C1
C2
F1 F2
1. There are three pairs of GRAVITY forces between the tractor, rope, car and the
Earth - for example forces G1 & G2.
2. There are two pairs of TENSION forces. The tractor exerts a TENSION force to the
LEFT on the rope and the rope exerts an equal magnitude TENSION force to the
RIGHT on the tractor. A similar but DIFFERENT magnitude pair exist between the
rope and the car, T1 & T2.
3. There are eight pairs of CONTACT forces between the eight tyres and the ground -
for example forces C1 & C2.
4. There are eight pairs of FRICTIONAL forces between the eight tyres of the tractor
and car and the ground - for example forces F1 & F2.
For the tractor to succeed the tension force T1 must be greater than the four
frictional forces acting from the ground on the car’s four tyres.

Trailer question
A car of mass 800 kg is towing a trailer of mass
200 kg. If the car is accelerating at 2 ms-2
calculate:
(a) the tension force in the tow-bar
(b) the engine force required
Let the engine force = E
The tension force = T
Car mass = M
Trailer mass = m
Acceleration = a
The forces are as shown in the diagram.
The force acting on the trailer = T
= ma
= 200kg x 2 ms-2
Tension force in the tow-bar = 400 N
The resultant force acting on the car
ΣF = E – T
E – T = Ma
but: T = ma
Hence:
E – ma = Ma
E = Ma + ma
= (M + m) a
= (800kg + 200kg) x 2 ms-2
Engine force = 2000 N
E
T

ACTIVITY
1. State Newton’s first law of
motion and give two
examples of this law.
2. State the equation for
Newton’s second law of
motion.
3. Explain why a heavy
object falls at the same
rate as a heavy one.
5. What does the drag force
acting on a body depend
upon?
6. Describe and explain the
motion of a body falling
because of gravity through
a fluid.
7. What is meant by terminal
speed?

Activity
1. What does the drag force acting on a body depend
upon?
2. Describe and explain the motion of a body falling
because of gravity through a fluid.
3. What is meant by terminal speed?

Newton’s Second Law
Force equals
mass times
acceleration.
F = ma

• Force = Mass x Acceleration
• Force is measured in Newtons
ACCELERATION of GRAVITY(Earth) = 9.8 m/s2
• Weight (force) = mass x gravity (Earth)
Moon’s gravity is 1/6 of the Earth’s
If you weigh 420 Newtons on earth,
what will you weigh on the Moon?
70 Newtons
If your mass is 41.5Kg on Earth
what is your mass on the Moon?

•WEIGHT is a measure of the force of
________ on the mass of an object
•measured in __________
gravity
Newtons

One rock weighs 5 Newtons.
The other rock weighs 0.5
Newtons. How much more
force will be required to
accelerate the first rock
at the same rate as the
second rock?
Ten times as much

Newton’s Third Law
For every action there is an equal and opposite
reaction.

Newton’s 3rd Law
• For every action there is an equal and opposite
reaction.
Book to
earth
Table to
book

Think about it . . .
What happens if you are standing on a
skateboard or a slippery floor and push against
a wall? You slide in the opposite direction
(away from the wall), because you pushed on
the wall but the wall pushed back on you with
equal and opposite force.
Why does it hurt so much when you stub
your toe? When your toe exerts a force on a
rock, the rock exerts an equal force back on
your toe. The harder you hit your toe against
it, the more force the rock exerts back on your
toe (and the more your toe hurts).

• A bug with a mass of 5 grams flies
into the windshield of a moving
1000kg bus.
• Which will have the most force?
• The bug on the bus
• The bus on the bug

• The force would be the same.
• Force (bug)= m x A
• Force (bus)= M x a
Think I look bad?
You should see
the other guy!

Action: earth pulls on you
Reaction: you pull on earth
Action and Reaction on Different Masses
Consider you and the earth

Action: tire pushes on road
Reaction: road pushes on tire

Action: rocket pushes on gases
Reaction: gases push on rocket

Consider hitting a baseball with a bat. If we
call the force applied to the ball by the bat the
action force, identify the reaction force.
(a) the force applied to the bat by the hands
(b) the force applied to the bat by the ball
(c) the force the ball carries with it in flight
(d) the centrifugal force in the swing
(b) the force applied to the bat by the ball

Newton’s 3rd Law
•Suppose you are taking a space walk near the
space shuttle, and your safety line breaks. How
would you get back to the shuttle?

Newton’s 3rd Law
• The thing to do would be to take one of the tools from your tool belt
and throw it is hard as you can directly away from the shuttle. Then,
with the help of Newton's second and third laws, you will accelerate
back towards the shuttle. As you throw the tool, you push against it,
causing it to accelerate. At the same time, by Newton's third law, the
tool is pushing back against you in the opposite direction, which
causes you to accelerate back towards the shuttle, as desired.

Review
Newton’s First Law:
Objects in motion tend to stay in motion
and objects at rest tend to stay at rest
unless acted upon by an unbalanced force.
Newton’s Second Law:
Force equals mass times acceleration
(F = ma).
Newton’s Third Law:
For every action there is an equal and
opposite reaction.

1stlaw: Homer is large and
has much mass, therefore he
has much inertia. Friction
and gravity oppose his
motion.
2nd law: Homer’s mass x
9.8 m/s/s equals his
weight, which is a force.
3rd law: Homer pushes
against the ground and it
pushes back.

WeBWork set # 3.
All about friction.
Similar to problem
4 on set # 2, but now
with friction.
Make sure you
determine the normal
force correctly!
The normal force in
this problem is
directed horizontally.
Determine the net
acceleration of the
blocks in order to
determine their
contact force.
Motion with variable
acceleration!
Make sure you deter-
Mine the correct
Directions of the
friction and normal
forces.

Physics 121.
Quiz Lecture 5.
• The quiz today will have 3 questions.

A quick review:
Newton’s first law of motion.
First Law:
Consider a body on which no net
force acts. If the body is at rest, it
will remain at rest. If the body is
moving with constant velocity, it
will continue to do so.
• Notes:
• Net force: sum of ALL forces
acting on the body.
• An object at rest and an object
moving with constant velocity
both have no acceleration.

A quick review:
Newton’s second law of motion.
Second Law:
The acceleration of an object is directly
proportional to the net force acting on it and it
inversely proportional to its mass. The
direction of the acceleration is in the direction
of the net force acting on the object:
F
  m
r
a

A quick review:
Newton’s third law of motion.
Third law:
Suppose a body A exerts a force
(FBA) on body B. Experiments
show that in that case body B
exerts a force (FAB) on body A.
These two forces are equal in
magnitude and oppositely
directed:
Note: these forces act on
different objects and they do not
cancel each other.
FBA  
r
FAB

Newton’s laws of motion.
Problem solving strategies.
• The first step in solving problems involving forces is to determine all the forces that act on the
object(s) involved.
• The forces acting on the object(s) of interest are drawn into a free-body diagram.
• Apply Newton’s second law to the sum of to forces acting on each object of interest.

Interesting effects.
The rope must always sag!
Why?

Interesting effects.
The force you need
to supply increases
when the height of
your backpack
Increases. Why?

Friction.
• A block on a table may not start
to move when we apply a small
force on it.
• This means that there is no net
force in the horizontal direction,
and that the applied force is
balanced by another force.
• This other force must change its
magnitude and direction based
on the direction and magnitude
applied force.
• If the applied force is large
enough, the block will start to
move and accelerate.

Friction.
• When the applied force exceeds a certain
maximum value, the object will start to move.
• Once the object starts to move, the magnitude
of the force required to keep the object moving
with constant velocity is smaller than the
magnitude of the force required to start the
motion.
• The forces that try to oppose our motion are
the friction forces between the object and
surface on which it is resting.

Friction.
• Based on these observations we
can conclude :
• There are two different friction
forces: the static friction force
(no motion) and the kinetic
friction force (motion).
• The static friction force
increases with the applied
force but has a maximum
value.
• The kinetic friction force is
independent of the applied
force, and has a magnitude
that is less than the maximum
static friction force.

Friction and braking.
• Consider how you stop in your
car:
• The contact force between the
tires and the road is the static
friction force (for most normal
drivers). It is this force that
provides the acceleration
required to reduce the speed
of your car.
• The maximum static friction
force is larger than the kinetic
friction force. As a result, your
are much more effective
stopping your car when you
can use static friction instead of
kinetic friction (e.g. when your
wheels lock up).

Friction and normal forces.
• The maximum static friction force and the
kinetic friction force are proportional to the
normal force.
• Changes in the normal force will thus result
in changes in the friction forces.
• NOTE:
• The normal force will be always
perpendicular to the surface.
• The friction force will be always opposite to
the direction of (potential) motion.

Pushing or pulling:
a big difference.
More Friction Less Friction

Circular motion.
A quick review.
• When we see an object carrying out circular
motion, we know that there must be force
acting on the object, directed towards the
center of the circle.
• When you look at the circular motion of a
ball attached to a string, the force is
provided by the tension in the string.
• When the force responsible for the circular
motion disappears, e.g. by cutting the string,
the motion will become linear.

Circular motion.
A quick review.
• In most cases, the string force
not only has to provide the force
required for circular motion, but
also the force required to
balance the gravitational force.
• Important consequences:
• You can never swing an object
with the string aligned with the
horizontal plane.
• When the speed increases, the
acceleration increases up to
the point that the force
required for circular motion
exceeds the maximum force
that can be provided by the
string.

Circular motion and its connection to friction.
• When you drive your car around a
corner you carry out circular motion.
• In order to be able to carry out this
type of motion, there must be a force
present that provides the required
acceleration towards the center of the
circle.
• This required force is provided by the
friction force between the tires and
the road.
• But remember ….. The friction force
has a maximum value, and there is a
maximum speed with which you can
make the turn.
Required force = Mv2/r.
If v increases, the friction force
must increase and/or the radius
must increase.

Circular motion and its connection to friction.
• Unless a friction force is present you
can not turn a corner …… unless the
curve is banked.
• A curve that is banked changes the
direction of the normal force.
• The normal force, which is
perpendicular to the surface of the
road, can provide the force required
for circular motion.
• In this way, you can round the curve
even when there is no friction …….
but only if you drive with exactly the
right speed (the posted speed).

Air “friction” or drag.
• Objects that move through the air
also experience a “friction” type
force.
• The drag force has the following
properties:
• It is proportional to the cross
sectional area of the object.
• It is proportional to the velocity
of the object.
• It is directed in a direction
opposite to the direction of
motion.
• The drag force is responsible for
the object reaching a terminal
velocity (when the drag force
balances the gravitational force).

Terminal air “friction” or drag.
• The science of falling cats is called
feline pesematology.
• This area of science uses the data
from falling cats in Manhattan to
study the correlation between
injuries and height.
• The data show that the survival rate
is doubling as the height increases
(effects of terminal velocity). E.g.
only 5% of the cats who fell seven to
thirty-two stories died, while 10% of
the cats died who fell from two to
six stories.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.

That’s all!
Next week: gravity keeps us together!

CIRCULAR MOTION
By Omosa Elijah
Lecturer,
Kisii University,
Kenya

Specific Objectives
Lessons Topics
Circular motion
Motion in a circular path at constant speed implies
there is an acceleration and requires a centripetal force.
Angular speed ω = v / r = 2π f
Centripetal acceleration a = v2 / r = ω2 r
Centripetal force F = mv2 / r = mω2 r
The derivation of a = v2/ r
Applications of Uniform Circular Motion

Uniform Circular Motion
Consider an object moving around a circular
path of radius, r with a constant linear speed
, v
The circumference of this circle is 2π r.
The time taken to complete one circle, the
period, is T.
Therefore:
v = 2π r / T
But frequency, f = 1 / T and so also:
v = 2π r f
v
r
v
r
v
r
v
r
Note: The arrows represent the
velocity of the object. As the
direction is continually
changing, so is the velocity.

Question
The tyre of a car, radius 40cm, rotates with a frequency of 20 Hz.
Calculate (a) the period of rotation and (b) the linear speed at
the tyres edge.

Question
The tyre of a car, radius 40cm, rotates with a frequency of 20 Hz.
Calculate (a) the period of rotation and (b) the linear speed at
the tyres edge.
(a) T = 1 / f
= 1 / 20 Hz
period of rotation = 0.050 s
(b) v = 2π r f
= 2 π x 0.40 m x 20 Hz
linear speed = 50 ms-1

Angular displacement, θ
Angular displacement, θ is equal to
the angle swept out at the centre of
the circular path.
An object completing a complete
circle will therefore undergo an
angular displacement of 360°.
½ circle = 180°.
¼ circle = 90°.
θ

Angles in radians
The radian (rad) is defined as the angle swept
out at the centre of a circle when the arc
length, s is equal to the radius, r of the circle.
If s = r
then θ = 1 radian
The circumference of a circle = 2πr
Therefore 1 radian = 360° / 2π = 57.3°
And so:
360° = 2π radian (6.28 rad)
180° = π radian (3.14 rad)
90° = π / 2 radian (1.57 rad)
θ
r
r
s
Also: s = r θ

Angular speed (ω)
angular speed = angular displacement
time
ω = Δθ / Δt
units:
angular displacement (θ ) in radians (rad)
time (t ) in seconds (s)
angular speed (ω) in radians per second (rad s-1)

Angular speed can also be measured in revolutions per second
(rev s-1) or revolutions per minute (r.p.m.)
Question:
Calculate the angular speed in rad s-1 of an old vinyl record player
set at 78 r.p.m.
78 r.p.m.
= 78 / 60 revolutions per second
= 1.3 rev s-1
= 1.3 x 2π rad s-1
78 r.p.m. = 8.2 rad s-1

Angular frequency (ω)
Angular frequency is the same as angular speed.
For an object taking time, T to complete one circle of angular
displacement 2π:
ω = 2π / T
but T = 1 / f
therefore: ω = 2π f
that is: angular frequency = 2π x frequency

Relationship between angular
and linear speed
For an object taking time
period, T to complete a circle
radius r:
ω = 2π / T
rearranging: T = 2π / ω
but: v = 2π r / T
= 2π r / (2π / ω)
Therefore:
v = r ω
and:
ω = v / r

Question
A hard disc drive, radius 50.0 mm, spins at 7200 r.p.m. Calculate
(a) its angular speed in rad s-1; (b) its outer edge linear speed.

Question
A hard disc drive, radius 50.0 mm, spins at 7200 r.p.m. Calculate
(a) its angular speed in rad s-1; (b) its outer edge linear speed.
(a) 7200 r.p.m. = [(7200 x 2 x π) / 60] rad s-1
angular speed = 754 rad s-1
(b) v = r ω
= 0.0500 m x 754 rad s-1
linear speed = 37.7 ms-1

Complete
angular speed linear speed radius
6 ms-1 0.20 m
40 rad s-1 0.50 m
6 rad s-1 18 ms-1
48 cms-1 4.0 m
45 r.p.m. 8.7 cm
Complete

Complete
angular speed linear speed radius
6 ms-1 0.20 m
40 rad s-1 0.50 m
6 rad s-1 18 ms-1
48 cms-1 4.0 m
45 r.p.m. 8.7 cm
Answers
30 rad s-1
20 ms-1
3 m
0.12 rad s-1
0.42 ms-1

Centripetal acceleration (a)
An object moving along a circular
path is continually changing in
direction. This means that even if it is
travelling at a constant speed, v it is
also continually changing its velocity.
It is therefore undergoing an
acceleration, a.
This acceleration is directed towards
the centre (centripetal) of the
circular path and is given by:
a = v2
r
v
r
a
v
r
a

but: v = r ω
combining this with: a = v2 / r
gives:
a = r ω2
and also:
a = v ω

Complete
angular
speed
linear speed radius centripetal
acceleration
8.0 ms-1 2.0 m
2.0 rad s-1 0.50 m
9.0 rad s-1 27 ms-1
6.0 ms-1 9.0 ms-2
33⅓ r.p.m. 1.8 ms-2
Complete

Complete
angular
speed
linear speed radius centripetal
acceleration
8.0 ms-1 2.0 m
2.0 rad s-1 0.50 m
9.0 rad s-1 27 ms-1
6.0 ms-1 9.0 ms-2
33⅓ r.p.m. 1.8 ms-2
Answers
4.0 rad s-1
32 ms-2
1.0 ms-1
2.0 ms-2
3.0 m 243 ms-2
4.0 m
1.5 rad s-1
0.15 m
0.52 ms-1

ISS Question
For the International Space
Station in orbit about the Earth
(ISS) Calculate:
(a) the centripetal acceleration
and
(b) linear speed
Data:
orbital period = 90 minutes
orbital height = 400km
Earth radius = 6400km

(a) ω = 2π / T
= 2 π / (90 x 60 seconds)
= 1.164 x 10-3 rads-1
a = r ω2
= (400km + 6400km) x (1.164 x 10-3 rads-1)2
= (6.8 x 106 m) x (1.164 x 10-3 rads-1)2
centripetal acceleration = 9.21 ms-1
(b) v = r ω
= (6.8 x 106 m) x (1.164 x 10-3 rads-1)
linear speed = 7.91 x 103 ms-1
(7.91 kms-1)

Proof of: a = v2 / r
NOTE: This is not required for A2 AQA Physics
Consider an object moving at
constant speed, v from point A to
point B along a circular path of
radius r.
Over a short time period, δt it covers
arc length, δs and sweeps out angle,
δθ.
As v = δs / δt then δs = v δt.
The velocity of the object changes in
direction by angle δθ as it moves
from A to B.
vA
A B
C
δθ
vB

If δθ is very small then δs
can be considered to be a
straight line and the shape
ABC to be a triangle. Triangle ABC will have the same
shape as the vector diagram
above.
Therefore δv / vA (or B) = δs / r
-vA
vB
δv δθ
The change in velocity, δv
= vB - vA
Which is equivalent to:
δv = vB + (- vA)
A B
C
δθ
vB
δs
r
r

but δs = v δt
and so:
δv / v = v δt / r
δv / δt = v2 / r
As δt approaches zero, δv / δt will become
equal to the instantaneous acceleration, a.
Hence: a = v2 / r
In the same direction as δv, towards the
centre of the circle.
-vA
vB
δv δθ

Centripetal Force
Newton’s first law:
If a body is accelerating it must be
subject to a resultant force.
Newton’s second law:
The direction of the resultant force
and the acceleration must be the
same.
Therefore centripetal acceleration
requires a resultant force directed
towards the centre of the circular
path – this is CENTRIPETAL FORCE.
Tension provides the
CENTRIPETAL FORCE
required by the hammer
thrower.

What happens when centripetal force
is removed
When the centripetal
force is removed the
object will move along a
straight line tangentially
to the circular path.

Other examples of centripetal forces
Situation Centripetal force
Earth orbiting the Sun GRAVITY of the Sun
Car going around a bend. FRICTION on the car’s tyres
Airplane banking (turning) PUSH of air on the airplane’s
wings
Electron orbiting a nucleus ELECTROSTATIC attraction
due to opposite charges

Equations for centripetal force
From Newton’s 2nd law of motion:
ΣF = ma
If a = centripetal acceleration
then ΣF = centripetal force
and so:
ΣF = m v2 / r
and ΣF = m r ω2
and ΣF = m v ω

Question 1
Calculate the centripetal tension
force in a string used to whirl a mass
of 200g around a horizontal circle of
radius 70cm at 4.0ms-1.

Question 1
Calculate the centripetal tension
force in a string used to whirl a mass
of 200g around a horizontal circle of
radius 70cm at 4.0ms-1.
ΣF = m v2 / r
= (0.200kg) x (4.0ms-1)2 / (0.70m)
tension = 4.6 N

Question 2
Calculate the maximum speed that a car of mass 800kg can go
around a curve of radius 40m if the maximum frictional force
available is 8kN.

Question 2
Calculate the maximum speed that a car of mass 800kg can go
around a curve of radius 40m if the maximum frictional force
available is 8kN.
The car will skid if the centripetal force required is greater
than 8kN
ΣF = m v2 / r
becomes: v2 = (ΣF x r ) / m
= (8000N x 40m) / (800kg)
v2 = 400
maximum speed = 20ms-1

tan θ = v2/rg
mg
R
q
Rsinq
Figure 2
Car on banked track (unpowered)
In this case it is the component of the weight of the car towards the
centre of the curve that provides the centripetal force. Notice that there is
no additional force mv2/r - the component of weight is the centripetal
force. (See Figure 2)
If the track is banked at and angle θ to the
horizontal then:
Resolving horizontally and vertically:
Rsinθ = mv2/r and Rcosθ = mg
where v is the maximum speed that the
car can take the corner.
so :
Notice that it can take the curve even if there is no friction
between the tyres and the road.

Example problems
1. Calculate the maximum speed at which a car can
travel on a frictionless banked track of radius 75 m if
the angle of banking with the horizontal is 25o. (g =
9.8 ms-2)
Using: tan θ = v2/rg; so v2 = rgtanθ = 75x9.8xtan25
= 343; Giving maximum speed (v) = 18.5 ms-1.
2. Calculate the angle of banking needed on a test
track of radius 200m if the car is to travel round at 120
mph (54.54 ms-1) without coming off the track. (g =
9.8 ms-2)
Using: tanθ = v2/rg tan θ = 54.542/[200x9.8] = 1.518
Therefore: Angle of banking (θ) = 56.6O

Motion in a vertical circle
If an object is being swung round on a string in a
vertical circle at a constant speed the centripetal
force must be constant but because its weight
(mg) provides part of the centripetal force as it
goes round the tension in the string will vary.
(See Figure 1)
Figure 1
mg
mg
mg
mg T3
T1
T2
T2

Let the tension in the string be T1 at the bottom of the
circle, T2 at the sides and T3 at the top.
At the bottom of the circle :
T1 - mg = mv2/r so T1 = mv2/r + mg
At the sides of the circle:
T2 = mv2/r
At the top of the circle:
T3+ mg = mv2/r so T3 = mv2/r - mg
So, as the object goes round the circle the tension in
the string varies being greatest at the bottom of the
circle and least at the top. Therefore if the string is to
break it will be at the bottom of the path where it has
to not only support the object but also pull it up out of
it straight-line path.

Question 3
A mass of 300g is whirled around a
vertical circle using a piece of string of
length 20cm at 3.0 revolutions per
second.
Calculate the tension in the string at
positions:
(a) A – top
(b) B – bottom and
(c) C – string horizontal
The angular speed, ω = 3.0 rev s-1
= 6 π rad s-1
C
B
A

(a) A – top
Both the weight of the mass and the tension
in the string are pulling the mass towards the
Therefore: ΣF = mg + T
and so: m r ω2 = mg + T
giving: T = m r ω2 – mg
= [0.300kg x 0.20m x (6π rads-1)2]
– [0.300kg x 9.8 Nkg-1]
= [21.32N] – [2.94N]
tension at A = 18.4N
mg
T

(b) B – bottom
The weight is now acting away from the
Therefore: ΣF = T – mg
and so: m r ω2 = T – mg
giving: T = m r ω2 + mg
= [0.300kg x 0.20m x (6π rads-1)2]
+ [0.300kg x 9.8 Nkg-1]
= [21.32N] + [2.94N]
tension at B = 24.3N
mg
T

(c) C – horizontal string
The weight is acting perpendicular to
the direction of the centre of the circle.
It therefore has no affect on the
centripetal force.
Therefore: ΣF = T
and so: m r ω2 = T
giving: T = m r ω2
= [0.300kg x 0.20m x (6π rads-1)2]
tension at C = 21.3N
mg
T

Example problem
A stunt pilot of mass 80 kg flies in a vertical circle of
radius 350 m at a constant speed of 70 ms-1. (Take g =
9.8 ms-2)
Calculate the force of the seat on him at:
(a) the top of the circle
(b) the sides of the circle when he is moving vertically
(c) the bottom of the circle
(a) T = mv2/r – mg = 80x702/350 – 80x9.8 =
1120 – 784 = 336 N
(b) T = mv2/r – mg = 80x702/350 = 1120 N
(c) T = mv2/r – mg = 80x702/350 + 80x9.8 =
1120 + 784 = 1904 N

Question 4 meow!
Calculate the maximum speed that Pat can drive over
the bridge for Jess to stay in contact with the van’s roof
if the distance that Jess is from the centre of curvature is
8.0m.

Jess will remain in contact with the van’s roof as
long as the reaction force, R is greater than zero.
The resultant force, ΣF downwards on Jess, while
the van passes over the bridge, is centripetal and
is given by:
ΣF = mg - R
and so: m v2 / r = mg - R
The maximum speed is when R = 0
and so: m v2 / r = mg
v2 / r = g
v2 = g r
maximum speed, v = √ (g r)
= √ (9.8 x 8.0)
= √ (78.4)
maximum speed = 8.9 ms-1
mg
R
Forces on
Jess

From Particles to Rigid Bodies
• Particles
• No rotations
• Linear velocity v only
• Rigid bodies
• Body rotations
• Linear velocity v
• Angular velocity ω

Coordinate Systems
• Body Space (Local Coordinate System)
• bodies are specified relative to this system
• center of mass is the origin (for convenience)
• We will specify body-related physical properties
(inertia, …) in this frame

Coordinate Systems
• World Space
• bodies are transformed to this common system
p(t) = R(t) p0 + x(t)
• x(t) represents the position of the body center
• R(t) represents the orientation
• Alternatively, use quaternion representation

Coordinate Systems
Meaning of R(t): columns represent the coordinates of the body space base
vectors (1,0,0), (0,1,0), (0,0,1) in world space.

Kinematics: Velocities
• How do x(t) and R(t) change over time?
• Linear velocity v(t) = dx(t)/dt is the same:
• Describes the velocity of the center of mass (m/s)
• Angular velocity (t) is new!
• Direction is the axis of rotation
• Magnitude is the angular
velocity about the axis
(degrees/time)
• There is a simple
relationship between
R(t) and (t)

Dynamics: Accelerations
• How do v(t) and dR(t)/dt change over time?
• First we need some more machinery
• Forces and Torques
• Momentums
• Inertia Tensor
• Simplify equations by formulating accelerations terms
of momentum derivatives instead of velocity
derivatives

Forces and Torques
• External forces Fi(t) act on particles
• Total external force F= Fi(t)
• Torques depend on distance from the center of mass:
i (t) = (ri(t) – x(t)) * Fi(t)
• Total external torque
 =  ((ri(t)-x(t)) * Fi(t)
• F(t) doesn’t convey any information about where the various forces
act
• (t) does tell us about the distribution of forces

Linear Momentum
• Linear momentum P(t) lets us express the effect of total force F(t) on body
(simple, because of conservation of energy):
F(t) = dP(t)/dt
• Linear momentum is the product of mass and linear velocity
• P(t) = midri(t)/dt
= miv(t) + (t) £ mi(ri(t)-x(t))
• But, we work in body space:
• P(t)= miv(t)= Mv(t) (linear relationship)
• Just as if body were a particle with mass M and velocity v(t)
• Derive v(t) to express acceleration:
dv(t)/dt = M-1 dP(t)/dt
• Use P(t) instead of v(t) in state vectors

• Same thing, angular momentum L(t) allows us to express the effect of
total torque (t) on the body:
(t) = dL(t)/dt
• Similarily, there is a linear relationship between momentum and
velocity:
I(t)(t)=r(t)xP(t)=L(t)
• I(t) is inertia tensor, plays the role of mass
• Use L(t) instead of (t) in state vectors
Angular momentum

HEAT ENERGY
• SPECIFIC HEAT CAPACITY
• SPECIFIC LATENT HEAT

Thermal energy
• Thermal energy is the energy of an object due to its temperature.
• It is also known as internal energy.
• It is equal to the sum of the random distribution of the kinetic and
potential energies of the object’s molecules. Molecular kinetic energy
increases with temperature. Potential energy increases if an object
changes state from solid to liquid or liquid to gas.

Temperature
Temperature is a measure of the degree of hotness of
a substance.
Heat energy normally moves from regions of higher to
lower temperature.
Two objects are said to be in thermal equilibrium with
each other if there is not net transfer of heat energy
between them. This will only occur if both objects are
at the same temperature.

Absolute zero
Absolute zero is the lowest
possible temperature.
An object at absolute zero has
minimum internal energy.
The graph opposite shows that the
pressure of all gases will fall to zero
at absolute zero which is
approximately - 273°C.

Temperature Scales
A temperature scale is defined by two fixed points
which are standard degrees of hotness that can be
accurately reproduced.

Celsius scale
symbol: θ
unit: oC
Fixed points:
ice point:
0oC: the temperature of pure melting ice
steam point:
100oC: the temperature at which pure water boils at
standard atmospheric pressure

Absolute scale
symbol: T
unit: kelvin (K)
Fixed points:
absolute zero:
0K: the lowest possible temperature.
This is equal to – 273.15oC
triple point of water:
273.16K: the temperature at which pure water exists in thermal
equilibrium with ice and water vapour.
This is equal to 0.01oC.

Converting between the scales
A change of one degree celsius is the same as a change of one
kelvin.
Therefore:
oC = K - 273.15
OR K = oC + 273.15
Note: usually the converting number, ‘273.15’ is approximated
to ‘273’.

Complete (use ‘273’):
Situation Celsius (oC) Absolute (K)
Boiling water 100 373
Vostok Antarctica 1983 - 89 184
Average Earth surface 15 288
Gas flame 1500 1773
Sun surface 5727 6000

Specific heat capacity, c
The specific heat capacity, c of a substance is the
energy required to raise the temperature of a unit
mass of the substance by one kelvin without change of
state.
ΔQ = m c ΔT
where:
ΔQ = heat energy required in joules
m = mass of substance in kilograms
c = specific heat capacity (shc) in J kg -1 K -1
ΔT = temperature change in K

If the temperature is measured in celsius:
ΔQ = m c Δθ
where:
c = specific heat capacity (shc) in J kg -1 °C -1
Δθ = temperature change in °C
Note:
As a change one degree celsius is the same as a change
of one kelvin the numerical value of shc is the same in
either case.

Examples of SHC
Substance SHC (Jkg-1K-1) Substance SHC (Jkg-1K-1)
water 4 200 helium 5240
ice or steam 2 100 glass 700
air 1 000 brick 840
hydrogen 14 300 wood 420
gold 129 concrete 880
copper 385 rubber 1600
aluminium 900 brass 370
mercury 140 paraffin 2130

Answers
Substance Mass SHC
(Jkg-1K-1)
Temperature
change
Energy (J)
water 4 kg 4 200 50 oC 840 000
gold 4 kg 129 50 oC 25 800
air 4 kg 1 000 50 K 200 000
glass 3 kg 700 40 oC 84 000
hydrogen 5 mg 14 300 400 K 28.6
brass 400 g 370 50oC to 423 K 14 800
Complete:

Question
Calculate the heat energy required to raise the
temperature of a copper can (mass 50g) containing
200cm3 of water from 20 to 100oC.

• Metal has known mass, m.
• Initial temperature θ1 measured.
• Heater switched on for a known time, t
• During heating which the average p.d., V and electric
current I are noted.
• Final maximum temperature θ2 measured.
• Energy supplied = VIt = mc(θ2 - θ1 )
• Hence: c = VIt / m(θ2 - θ1 )

Example calculation
Metal mass, m. = 500g = 0.5kg
Initial temperature θ1 = 20oC
Heater switched on for time, t = 5 minutes = 300s.
p.d., V = 12V; electric current I = 2.0A
Final maximum temperature θ2 = 50oC
Energy supplied = VIt = 12 x 2 x 300 = 7 200J
= mc(θ2 - θ1 ) = 0.5 x c x (50 – 30) = 10c
Hence: c = 7 200 / 10
= 720 J kg -1 oC -1

Measuring SHC (liquid)
Similar method to metallic solid.
However, the heat absorbed by
the liquid’s container (called a
calorimeter) must also be
allowed for in the calculation.

Electrical heater question
What are the advantages and
disadvantages of using paraffin rather
than water in some forms of portable
electric heaters?

Climate question
Why are coastal regions cooler in summer but milder in
winter compared with inland regions?

Latent heat
This is the energy required to change the
state of a substance. e.g. melting or
boiling.
With a pure substance the temperature
does not change. The average potential
energy of the substance’s molecules is
changed during the change of state.
‘latent’ means ‘hidden’ because the heat
energy supplied during a change of state
process does not cause any temperature
change.

Specific latent heat, l
The specific latent heat, l of a substance is the energy
required to change the state of unit mass of the
substance without change of temperature.
ΔQ = m l
where:
ΔQ = heat energy required in joules
m = mass of substance in kilograms
l = specific latent heat in J kg -1

Examples of SLH
Substance State change SLH (Jkg-1)
ice → water solid → liquid
specific latent heat of fusion
336 000
water → steam liquid → gas / vapour
specific latent heat of vaporisation
2 250 000
carbon dioxide solid → gas / vapour
specific latent heat of sublimation
570 000
lead solid → liquid 26 000
solder solid → liquid 1 900 000
petrol liquid → gas / vapour 400 000
mercury liquid → gas / vapour 290 000

Complete:
Substance Change SLH
(Jkg-1)
Mass Energy (J)
water melting 336 000 4 kg 1.344 M
water freezing 336 000 200 g 67.2 k
water boiling 2.25 M 4 kg 9 M
water condensing 2.25 M 600 mg 1 350
CO2 subliming 570 k 8 g 4 560
CO2 depositing 570 k 40 000 μg 22.8

Question 1
Calculate (a) the heat energy required to change 100g of
ice at – 5oC to steam at 100oC.
(b) the time taken to do this if heat is supplied by a
500W immersion heater.
Sketch a temperature-time graph of the whole process.

Question 2
A glass contains 300g of water at 30ºC. Calculate the
water’s final temperature when cooled by adding (a)
50g of water at 0ºC; (b) 50g of ice at 0ºC. Assume no
heat energy is transferred to the glass or the
surroundings.

Work done by an ideal gas during expansion
Consider an ideal gas at a pressure P enclosed in a cylinder
of cross sectional area A.
The gas is then compressed by pushing the piston in a
distance dx, the volume of the gas decreasing by dV. (We
assume that the change in volume is small so that the
pressure remains almost constant – at P).
Work done on the gas during this compression = dW
Force on piston = PA
So the work done during compression = dW = PAdx
= PdV

The first law of thermodynamics can then be written
as:
dU = dQ + dW = dQ + PdV
In the PV diagram net work is done by the gas if it
expands at the higher temperature and net work is
done on the gas if is compressed at the higher
temperature.
dV
dx
P,V
F
A
dU = dQ + dW = dQ + PdV

It is impossible for there to be a net transfer
of heat from a cold body to a hot body in an
isolated system
Things wear out
Things get worse
The second law of thermodynamics
The first law of thermodynamics relates the input of heat
energy to the mechanical work that may be obtained from it.
It says nothing about the way that this conversion may take
place, however, nor does it put any restrictions on it.
The second law of thermodynamics states the way in which
these changes of energy may take place. It is considered by
many eminent scientists to be one of the most fundamental
laws of Physics and yet in one of its forms it may be stated in
the following very simple manner:
Some alternative ways of stating the second law are:

241
Waves are everywhere in nature
• Sound waves,
• visible light
waves,
• radio waves,
• microwaves,
• water waves,
• sine waves,
• telephone chord
waves,
• stadium waves,
• earthquake
waves,
• waves on a
string,
• slinky waves

242
What is a wave?
• a wave is a disturbance that travels through a medium from one
location to another.
• a wave is the motion of a disturbance

A. Waves
1. The nature of waves
a. A wave is a rhythmic
disturbance that transfers energy.
b. All waves are made by
something that vibrates.

b. Compressional
(longitudinal)
2. Mechanical waves need a matter medium to travel
through.
(sound, water, seismic)
3. Two basic types of waves:
a. Transverse

4. Wave properties:
a. Wavelength - distance from a point on a
wave to the same corresponding point on the
next wave.
b. Frequency - number of waves that pass a
point in one second (expressed in Hz).

246
Transverse Waves
• The differences between the two can be seen

247
Anatomy of a Wave
• Now we can begin to describe the anatomy of our waves.
• We will use a transverse wave to describe this since it is easier to see
the pieces.

248
Anatomy of a Wave
• In our wave here the dashed line represents the
equilibrium position.
• Once the medium is disturbed, it moves away from
this position and then returns to it

249
Anatomy of a Wave
• The points A and F are called the CRESTS of the
wave.
• This is the point where the wave exhibits the
maximum amount of positive or upwards
displacement
crest

250
Anatomy of a Wave
• The points D and I are called the TROUGHS of the
wave.
• These are the points where the wave exhibits its
maximum negative or downward displacement.
trough

251
Anatomy of a Wave
• The distance between the dashed line and point A
is called the Amplitude of the wave.
• This is the maximum displacement that the wave
moves away from its equilibrium.
Amplitude

252
Anatomy of a Wave
• The distance between two consecutive similar
points (in this case two crests) is called the
wavelength.
• This is the length of the wave pulse.
• Between what other points is can a wavelength be
measured?
wavelength

253
Anatomy of a Wave
• What else can we determine?
• We know that things that repeat have a frequency
and a period. How could we find a frequency and
a period of a wave?

c. Wavelength has an inverse relationship to
wave frequency.
d. Wave velocity depends on the type of
wave and medium.
1) Sound is faster in more dense media
and in higher temps.
2) Light is slower in more dense media,
but faster in a vacuum.

255
Basic definitions:
Wavelength: the distance between any two successive
corresponding points on the wave, that is, between two
maxima or two minima (l)
Displacement: the distance from the mean, central,
undisturbed position at any point on the wave (y)
Amplitude: the maximum displacement (a) from zero to
a crest or a trough
Frequency: the number of vibrations per second made
by the wave (f)
Period: the time taken for one complete oscillation (T=
1/f)

256
Wave frequency
• We know that frequency measure how often something happens over
a certain amount of time.
• We can measure how many times a pulse passes a fixed point over a
given amount of time, and this will give us the frequency.

257
Wave frequency
• Suppose I wiggle a slinky back and forth, and count that 6 waves pass
a point in 2 seconds. What would the frequency be?
• 3 cycles / second
• 3 Hz
• we use the term Hertz (Hz) to stand for cycles per second.

258
Wave Period
• The period describes the same thing as it did with a pendulum.
• It is the time it takes for one cycle to complete.
• It also is the reciprocal of the frequency.
• T = 1 / f
• f = 1 / T
• let’s see if you get it.

259
Wave Speed
• We can use what we know to determine how fast a wave is moving.
• What is the formula for velocity?
• velocity = distance / time
• What distance do we know about a wave
• wavelength
• and what time do we know
• period

260
Wave Speed
• so if we plug these in we get
• velocity =
length of pulse /
time for pulse to move pass a fixed point
• v = l / T
• we will use the symbol l to represent wavelength

261
Wave Speed
• v = l / T
• but what does T equal
• T = 1 / f
• so we can also write
• v = f l
• velocity = frequency * wavelength
• This is known as the wave equation.
• examples

3)
e. Amplitude - size related to the energy
carried by the wave.
1) Transverse - how high above or how
low below the nodal line.
2) Compressional - how dense the
medium is at the compressions & rarefactions.

263
Wave Behavior
• Now we know all about waves.
• How to describe them, measure them and analyze
them.
• But how do they interact?
• We know that waves travel through mediums.
• But what happens when that medium runs out?

264
Positive x direction : y = a sin (t – kx)
Negative x direction : y = a sin (t + kx)
y = a sin (t + kx)
x
Figure 2
We will consider here the motion of a sine wave (Figure 2), since this type
is the most fundamental. However it can be shown that any other wave
may be built up from a series of sine waves of differing frequency.
Phase: a term related to the displacement at zero time (e) (see below)
We can express a wave travelling in
the positive x direction by the
equation:
and for one travelling in the opposite
direction:
where k is a constant and  = 2pf.

265
The sign gives the direction of the motion. We can separate each equation into two
terms:
(a) a term showing the variation of displacement with time at a particular place - for
example, when x = 0 y = a sin (t), that is, the variation of displacement with time at the
particular place x = 0.
(b) a term showing the variation of displacement with distance at a particular time - for
example, when t = 0 y = a sin (kx), that is, the variation of displacement with distance at
a particular time t = 0.
An alternative form of the equation can be proved as follows.
Since the period T = 1/f where f is the frequency and  = 2pf we have  = 2p/T.
Also when t = 0 y = 0 at x = 0,l/2, l...and so on, and so k = 2p/l. The equation may
therefore be written:
y = a sin 2p(t/T + x/l)

266
Example problem
A certain travelling wave has frequency (f) of 200 Hz, a wavelength (l) of 2m and
an amplitude (a) of 0.02 m. Calculate the displacement (y) at a point 0.3m from the
origin at a time 0.01s after zero displacement at that point.
The period of the wave = 1/f = 1/200 = 0.005 s-1
y = a sin 2p(t/T + x/l) = 0.02 sin [2p(0.01/0.005 + 0.3/2)]
= 0.02 sin[2p(2 + 0.15)] = 0.02 sin[13.5] = 0.02x0.81 = 0.016 m

5. Wave behavior:
a. Reflection - the bouncing back of a wave.
1) Sound echoes
2) Light images in mirrors
3) Law of reflection
i = r

268
Boundary Behavior
• The behavior of a wave when it reaches the end of its medium is
called the wave’s BOUNDARY BEHAVIOR.
• When one medium ends and another begins, that is called a
boundary.

269
Fixed End
• One type of boundary that a wave may encounter is that it may be
attached to a fixed end.
• In this case, the end of the medium will not be able to move.
• What is going to happen if a wave pulse goes down this string and
encounters the fixed end?

270
Fixed End
• Here the incident pulse is an upward pulse.
• The reflected pulse is upside-down. It is inverted.
• The reflected pulse has the same speed, wavelength, and amplitude
as the incident pulse.

272
Free End
• Another boundary type is when a wave’s medium is attached to a
stationary object as a free end.
• In this situation, the end of the medium is allowed to slide up and
down.
• What would happen in this case?

273
Free End
• Here the reflected pulse is not inverted.
• It is identical to the incident pulse, except it is moving in the opposite
direction.
• The speed, wavelength, and amplitude are the same as the incident
pulse.

b. Refraction - the bending of a
wave caused by a change in speed as
the wave moves from one medium to
another.

The girl sees the boy’s foot closer
to the surface than it actually is.
If the boy looks down at his feet, will they seem closer to him than they really are?
No! He is looking straight down and not at an angle. There is no refraction for him.

277
Change in Medium
• Our third boundary condition is when the medium of a wave changes.
• Think of a thin rope attached to a thin rope. The point where the two
ropes are attached is the boundary.
• At this point, a wave pulse will transfer from one medium to another.
• What will happen here?

278
Change in Medium
• In this situation part of the wave is reflected,
and part of the wave is transmitted.
• Part of the wave energy is transferred to the
more dense medium, and part is reflected.
• The transmitted pulse is upright, while the
reflected pulse is inverted.

279
Change in Medium
• The speed and wavelength of the reflected
wave remain the same, but the amplitude
decreases.
• The speed, wavelength, and amplitude of the
transmitted pulse are all smaller than in the
incident pulse.

280
Change in Medium Animation
Test your understanding

c. Diffraction - the
bending of a wave
around the edge of
an object.
1) Water waves
bending around islands
2) Water waves passing
through a slit and spreading
out

Less occurs if wavelength is
smaller than the object.
More occurs if wavelength is larger
than the object.
3) Diffraction depends on the size of the
obstacle or opening compared to the
wavelength of the wave.

4) AM radio waves are longer and can
diffract around large buildings and
mountains; FM can’t.

d. Interference - two or more
waves overlapping to form a new
wave.
All we have left to discover is how
waves interact with each other.
When two waves meet while
traveling along the same medium it
is called INTERFERENCE.

1) Constructive (in phase)
Sound waves that constructively interfere are
louder

2) Destructive (out of phase)
Sound waves that destructively interfere are
not as loud

e. Standing wave - a wave pattern
that occurs when two waves equal in
wavelength and frequency meet from
opposite directions and continuously
interfere with each other.
node antinode

288
Constructive Interference
• Let’s consider two waves moving towards each other, both having a
positive upward amplitude.
• What will happen when they meet?

289
• They will ADD together to produce a greater
amplitude.
• This is known as CONSTRUCTIVE INTERFERENCE.

290
Destructive Interference
• Now let’s consider the opposite, two waves moving towards each
other, one having a positive (upward) and one a negative (downward)
amplitude.
• What will happen when they meet?

291
• This time when they add together they will
produce a smaller amplitude.
• This is know as DESTRUCTIVE INTERFERENCE.

292
Check Your Understanding
• Which points will produce constructive interference and
which will produce destructive interference?
 Constructive
G, J, M, N
 Destructive
H, I, K, L, O

f. Resonance - the ability of an object to
vibrate by absorbing energy at its natural
frequency.

B. Sound
1. Energy is transferred from particle to
particle through matter.
2. How we hear
a. Outer ear collects sound.
b. Middle ear amplifies sound.
c. Inner ear converts sound.

3. Properties of sound
a. Intensity and loudness
1) Intensity depends on the
energy in a sound wave.
2) Loudness is human perception
of intensity.
3) Loudness is measured on the
decibel scale.

a) Threshold of hearing (0 db)
b) Threshold of pain (120 db)

5) Ultrasonic sound has a frequency greater than 20,000
Hz.
a) Dogs (up to 35,000 Hz)
b) Bats (over 100,000 Hz)
c) Medical diagnosis
6) Infrasonic sound has a frequency below 20 Hz; they
are felt rather than heard
(earthquakes, heavy machinery).

c. Speed of sound
1) 332 m/s in air at 0 C.
2) Changes by 0.6 m/s for every Celsius degree
from 0 C.
3) Subsonic – slower
4) Supersonic – faster than sound (Mach 1 =
speed of sound)
5) Sonic boom (pressure cone)

d. The Doppler effect – the change in pitch
due to a moving wave source.
1) Objects moving toward you cause a
higher pitched sound.
2) Objects moving away cause sound of
lower pitch.
3) Used in radar by police and
meteorologists and in astronomy.

4. Musical sound
a. Noise has no pattern.
b. Music has a pattern and deliberate
pitches.
c. Sound quality describes differences
of sounds that have the same pitch and
loudness.
d. Every instrument has its own set of
overtones.

e) Beats are pulsing variations of loudness caused by interference of sounds of slightly
different frequencies.

5. Uses of sound
a. Acoustics – the study of sound.
Soft materials dampen sound; hard
materials reflect it (echoes and
reverberations).
b. SONAR – Sound Navigation and Ranging
(echolocation).
c. Ultrasound imaging
d. Kidney stones & gallstones.

 Diffraction sometimes seems a ‘mysterious’ phenomenon, which is difficult to understand.
 We have noted that electromagnetic waves (light, X-rays etc.), water waves (i.e. elastic
waves in a solid or fluid), matter waves (electrons, neutrons) etc. can be diffracted.
 In fact it is best to start understading diffraction using water waves with a single slit.
 Diffraction can be thought of as a special case of constructive and destructive interference
(a case where there is a large number of scatterers*).
 What are these scatterers?
A: Any entity which impedes (partially and ‘redirects’) the path of wave can be conceived as a
scatterer. Scatterers has to be understood in conjunction with the wave being considered
[i.e. an entity may be a scatter for one kind of waves, but not for another (e.g. an array of atoms is a scatterer for X-
rays, but it is not a scatterer for water waves)].
 In a periodic array they can be entities of the motif [i.e. a geometrical entity (atoms, ions, blocks of
wood), physical property (e.g. aligned spins) or a combination of both].
 Experiments have been conducted where ‘matter waves’ have been diffracted from a
crystal made of electromagnetic radiation (waves)! (Atoms diffracted from a Laser lattice).
 We will use some ‘crude’ analogies and some ‘schematic cartoons’ to get a hang of this
phenomenon → these should not be taken literally.
Diffraction
* Usually in a periodic array.
** Though other simple configurations may be envisaged.

 The bare minimum is one edge*. Two edges forming a single slit is better to get a better
picture.
What is the minimum I need to see diffraction/interference?
* Which blocks part of the wave.

Let us start by throwing some balls on a wide slit.
The balls in the gap pass ‘right through in a straight line’ (well most of the ones!), while the ones
blocked by the obstruction reflect back (reflection not shown).
Warning: these cartoons do not depict diffraction- they are a way to start visualizing the issues!
0
1
Screen
‘Intensity*’
on
‘screen’
Obstacle
* Intensity ~ no. of balls/area/time
Geometrical shadow region
of zero intensity
If we shine ‘incoherent light’
we will get a similar ‘intensity’
distribution.
Near the edges the intensity
will be different (but we will
ignore this for now)

What about the ones hitting the edge?
This is not what happens in diffraction. This is to tell you that ‘watch out for sharp corners’!!
More cartoons on network
0
1
Screen
Intensity
on
screen
Obstacle
Altered ‘intensity’ pattern
(this is not one peak but a broad
diffuse one as the way the balls hit
the barrier edge will send them off
in different angles)
Centre of mass near edge.
Glancing angle collision.

What if the slit width is of the order of the ‘ball size’?
0
Region
of
geometrical
shadow
There is ‘intensity’in the
‘geometrical shadow’
region as well!
 So we have seen that even with macroscopic balls it is possible to get ‘intensity’ in the region
of the geometrical shadow.
 For this effect to be prominent we have noticed that the slit width has to be of the order of the
‘size’ of the ball.

 Consider a series of speed breakers (bumps) on the road. Let a vehicle arrive at a velocity ‘v’.
Another ‘crude’ analogy to understand diffraction

N
B
a
q
A
P
Figure 2
wide slit narrow slit
telescope
adjustable slit
spectrometer
Figure 1 (b)
Figure 1 (a)
Fraunhofer diffraction - Single slit
The diffraction at a single slit of width a is shown in Figure 2. Diffraction occurs in all
directions to the right of the slit but we will just concentrate on one direction towards a
point P in a direction q to the original direction of the waves. Plane waves arrive at P
due to diffraction at the slit AB. Waves coming from the two sides of the slit have a path
difference BN and therefore interference results.
But BN = a sin q, and if this is equal to the wavelength of the light (l) the light from the
top of the slit and the bottom of the slit a will cancel out.and a minimum is observed at P.

Fraunhofer diffraction - Single slit
This is because if the path difference between the two extremes of the slit is exactly one
wavelength there will be points in the upper and lower halves of the slit that will be half a
wavelength out of phase.
Therefore the general condition for a minimum for a single slit is:
The Fraunhofer diffraction due to a single slit is very easy to observe. An adjustable slit is
placed on the table of a spectroscope and a monochromatic light source is viewed
through it using the spectroscope telescope (see Figure 1(a)). An image of the slit is seen
as shown in Figure 1(b). As the slit is narrowed a broad diffraction pattern spreads out
either side of the slit, only disappearing when the width of the slit is equal to or less than
one wavelength of the light used.
The path difference between light from the top and bottom of the slit is written ml where
m is the number of wavelengths ‘fitting into’ BN. m is also known as the ‘order’ of the
diffraction image.
If the intensity distribution for a single slit is plotted against distance from the slit, a graph
similar to that shown that shown in Figure 3 will be obtained. The effect on the pattern of a
change of wavelength is shown in Figure 4
ml = a sin q
where m = 1,2,3,4 and so on.

Single slit diffraction
-q +q
Intensity
Angle of diffraction (q)
Intensity
Angle of diffraction (q)
-q +q
Figure 4
Blue light – short
wavelength giving a narrow
diffraction pattern
Red light – long
wavelength giving a
broad diffraction pattern
These two diagrams show the effect of a change of wavelength on the
single slit diffraction pattern. The pattern for red light is broader than that for
blue because of the longer wavelength of red light.

 In these set of slides we will try to visualize how constructive and destructive interference
take place (using the Bragg’s view of diffraction as ‘reflection’ from a set of planes).
 It is easy to ‘see’ as to how constructive interference takes place; however, it is not that
easy to see how ‘rays’ of the Bragg angle ‘go missing’.
Understanding constructive and destructive interference

ml = d sin q
Figure 1
A
q
B
d
Fraunhofer diffraction - double slit
For the double slit we simply have light from two adjacent slits meeting at the eyepiece. In this case the
formula for a maximum (a place where the light waves ‘add up’) is:
where d is the distance between the centres of the two
slits (Figure 1).
The intensity of the interference pattern produced by
two sources is simply varied by the diffraction effects.
We will have cos2 fringes modulated by the diffraction
pattern for a single slit. The intensity distribution is
shown in Figure 2.

Example problems
1. Calculate the wavelength of the monochromatic light where the second
order image is diffracted through an angle of 25o using a diffraction
grating with 300 lines per millimetre.
Grating spacing (e) = 10-3/300 m = 3.3x10-6 m
Wavelength (l) = esin25/2 = [3.3x10-6 x 0.42]/2 = 6.97 x 10-7 m = 697 nm
2. Calculate the maximum number of orders visible with a diffraction grating of
500 lines per millimetre, using light of wavelength 600 nm.
Maximum angle of diffraction = 90o e = 10-3/500 = 2x10-6 m
Therefore m = esinq/l = 2x10-6/600x10-9 = 3.33
Therefore maximum number of orders = 3, and a total of seven images of
the source can be seen (three on each side of a central image).

Young's double slit experiment
Light from a monochromatic line source passes through a lens and is focused on to a single slit S. It then
falls on a double slit (S1 and S2) and this produces two wave trains that interfere with each other in the
region on the right of the diagram.
S
d
D
xm
S1
S2
P
R
O
T
Figure 2
The formula relating the dimensions of the apparatus and the wavelength of light may be proved as
follows.
Consider the effects at a point P a distance xm from the axis of the apparatus. The path difference at P is S2P - S1P.
For a bright fringe (constructive interference) the path difference must be a whole number of wavelengths and for a dark
fringe it must be an odd number of half-wavelengths (Figure 2).
Consider the triangles S1PR and S2PT.
S1P2= (xm – d/2)2 + D2
S2P2 = (xm
2 + d/2)2 + D2
S2P – S1P = xmd/D
within the limits of experimental accuracy for D would be
at least 50 cm while d would be less than 1 mm making
the triangle S1S2P very thin.
Therefore
For a bright fringe: ml = xmd/D For a dark fringe: (2m + 1)l/2 = xmd/D

Where m = 0,1,2,3 etc. and so the m th bright fringe for m = 3 is 3lD/d from
the centre of the pattern
The distance between adjacent bright fringes is called the fringe width (x)
and this can be used in the equation as
Wavelength (l) = xd/D and Fringe width (x) = lD/d
Figure 3
Note that the fringe width is directly proportional to the wavelength, and so light with a
longer wavelength will give wider fringes. Although the diagram shows distinct light and
dark fringes, the intensity actually varies as the cos2 of angle from the centre.
If white light is used a white centre fringe is observed, but all the other fringes have
coloured edges, the blue edge being nearer the centre. Eventually the fringes overlap
and a uniform white light is produced.
The separation of the two slits should be of the same order of magnitude as the
wavelength of the radiation used.

Example
Calculate the fringe width for light of wavelength 550 nm in a
Young's slit experiment where the double slits are separated by
0.75 mm and the screen is placed 0.80 m from them.
Fringe width (x) = lD/d = 550 x 10-9 x 0.80/0.75 x 10-3 = 5.9 x 10-4
m = 0.59 mm

Here we see waves scattered from two successive planes interfering constructively.
(press page down button to see the successive graphics)
Note the phase difference of p introduced during the scattering by the atom.

Assuming that path difference of l gives constructive interference:
Similar to the path difference of l, path difference of 2l, 3l… nl also constructively interfere.
All Constructively
interfere
Also to be noted is the fact that if the path difference between Ray-1 and Ray-2 is l then the path
difference between Ray-1 and Ray-3 is 2l and Ray-1 and Ray-4 is 3l etc.
Going across
planes

Exact destructive interference (between two planes, with path difference of l/2) is easy to
visualize. The angle is not Bragg’s angle (let us call it qd).

At a different angle q’ the waves scattered from two successive planes interfere (nearly)
destructively
Warning:
this is a schematic

 In the previous example considered q’ was ‘far away’ (at a larger angular separation) from q
(qBragg) and it was easy to see the (partial) destructive interference.
 In other words for incidence angle of qd (couple of examples before) the phase difference of
p is accrued just by traversing one ‘d’.
 If the angle is just away from the Bragg angle (qBragg), then one will have to go deep into the
crystal (many ‘d’) to find a plane (belonging to the same parallel set) which will scatter out
of phase with this ray (phase difference of p) and hence cause destructive interference.
 In the example below we consider a path difference of l/10 between the first and the second
plane (hence, we will have to travel 5 planes into the crystal to get a path difference of l/2).

 If such a plane (as mentioned in the page before) which scatters out of phase with a off
Bragg angle ray is absent (due to finiteness of the crystal) then the ray will not be cancelled
and diffraction would be observed just off Bragg angles too  line broadening!
(i.e. the diffraction peak is not sharp like a -peak in the intensity versus angle plot)
 Line broadening can be used to calculate crystallite size (grain size).
 This is one source of line broadening. Other sources include: residual strain, instrumental
effects, stacking faults etc.

Standing Wave & Resonance
pattern that results when 2 waves, of same f, & A travel in
opposite directions.
Often formed from pulses reflect off a boundary. Waves
interfere constructively (antinodes) & destructively (nodes)
at fixed points.

Standing waves have no net
transfer of energy – no direction
propagation of energy.

Standing Waves form at natural frequencies of the
material. Occur when material resonates.
When system is disturbed it vibrates at many frequencies.
Standing wave patterns continue. Other frequencies to die
out.
Since there is resonance, the amplitude of particular
wavelengths/frequencies will be amplified.

Relation of Wavelength to String Length
for Standing Waves L = 1/2l.

General expression relating wavelength to
string length for standing waves:
• n ( ½ l) = L
• n is a whole number
• A whole number of half l’s
must fit.

Although we would perceive a string vibrating
as a whole,
it vibrates in a pattern that appears erratic producing
many different overtone pitches. What results are
particular tone colors or timbres of instruments and
voices.

Each standing wave pattern= harmonic.
Harmonics
The lowest f (longest l) at which a string can form
standing wave pattern is the fundamental f or the first
harmonic.

String Length L, l & Harmonics
Standing waves can form on a string of length L,
when the l can = ½ L, or 2/2 L, or 3/2L etc.
Standing waves are the overtones or harmonics.
L = nln. n = 1, 2, 3, 4 whole number harmonics.
2

Frequencies
Standing waves form where ½ l fits the string exactly, calculate
f:
L
nv
f
2

f
v

l
l
f
v 
2
l
n
L 
Substitute v/f for l.
f
nv
L
2

Must know speed in material.
n = harmonic

1st standing wave forms when l = 2L
First harmonic frequency is when n = 1 as below.
L
nv
f
2
1 
When n = 1, f = v/l . This is fundamental
frequency or 1st harmonic.
First harmonic has largest amplitude.

L
nv
f
2
2 
For second
harmonic n = 2.
f2 = v/L
Other standing waves with smaller
wavelengths form other frequencies that
ring out along with the fundamental.

In general,
The harmonic frequencies can be found
where n = 1,2,3… and n corresponds to the
harmonic. v is the velocity of the wave on
the string. L is the string length.
L
nv
fn
2


A resonant air column is
simply a standing longitudinal
wave system, much like
standing waves on a string.
closed-pipe resonator
tube in which one end is open
and the other end is closed
open-pipe resonator
tube in which both ends
are open

Pipes –
the open end has antinode (low P).

Standing Waves in Open Pipe
Both ends must be antinodes.
How much of the wavelength is the fundamental?

The 1st harmonic or fundamental can fit ½ l into the tube.
Just like the string L = nl
2
fn = nv
2L
Where n, the harmonic is an integer.

Closed pipes must have a node at closed end and
an antinode at the open end.
How many wavelengths? L = l
4

Here is the next harmonic.
How many l’s?
L = 3l
4

There are only odd harmonics possible – n = odd
number only.
L = 1/4l.
L = 3/4l.
L = 5/4l.
fn = nv where n = 1,3,5 …
4L

Application: When waves propagate through a
tall building, the building resonates like a tube
open at two ends.
• What is the equation that relates
frequency to wave velocity and
building height?
L
nv
f
2
1 

The building is 360 m tall and allows waves to travel
through it at 2400 m/s, what frequency wave will cause
the most damage to it? Explain why.
(Hint: What is the resonant frequency)?
• 3.3 Hz

• Hwk Read Homer section 4.5.
• Do Formative Assessment 4.5.

Do Now
1. In an open tube resonator, there is a pressure ________ at both
ends.
2. In a close tube resonator, there is a pressure _________ at the
closed end and a pressure ________ at the open end.
node
node
anti-node
***Note that this is the opposite of standing wave formations
which represent air displacement.

DO NOW
1. The diagram above depicts the standing wave pattern of the
fundamental frequency or __________ harmonic of a
_________ tube resonator.
2. The wavelength is equal to _________ times the length of the
tube.
L
first
closed
four

Drawing Standing waves
• The basis for drawing the standing wave patterns for air
columns is that vibrational anti-nodes will be present at any
open end and vibrational nodes will be present at any
closed end. For a tube closed at one end…
= nodes
= anti-nodes
l1 = 4L
l3 =
4L
3
l5 =
4L
5
In general, for the nth harmonic
of a closed tube…
Where n = 1,3,5…(odd)
ln =
4L
n

Drawing Standing waves
• The basis for drawing the standing wave patterns for air
columns is that vibrational anti-nodes will be present at any
open end and vibrational nodes will be present at any
closed end. For a tube open at both ends…
= nodes
= anti-nodes
l1 = 2L
l2 = L
l3 =
L
3
In general, for an nth harmonic
of an open tube…
Where n = 1,2,3,4… (all integers)
ln =
2L
n

Do NOW
• Take out your homework…
• For a closed pipe…
1. The first harmonic has __________ of a wavelength
2. The third harmonic has __________ of a wavelength
3. The fifth harmonic has _________ of a wavelength
4. The seventh harmonic has _________ of a wavelength
5. The nineth harmonic has __________ of a wavelength.
1/4
3/4
5/4
7/4
9/4

REview
• One person reads the question on the green side of the card ("who
has: material a wave travels through"?),
• The answer is on the purple side of someone else's card and they
respond ("I have: medium")

Do Now
• Take out a writing utensil, your notecard (if you have it), and a
calculator. Answer the following question on your DO NOW…
1. What grade do you expect to get on the practice test today?
2. What grade to you want to get on the test on Monday?

Refraction
A PowerPoint Presentation by
Omosa Elijah,
Kisii University
@ 2017

Objectives: After completing this
module, you should be able to:
• Define and apply the concept of the index of refraction
and discuss its effect on the velocity and wavelength of
light.
• Determine the changes in velocity and/or
wavelength of light after refraction.
• Apply Snell’s law to the solution of problems
involving the refraction of light.
• Define and apply the concepts of total internal
reflection and the critical angle of incidence.

Refraction Distorts Vision
Water
Air
Water
Air
The eye, believing that light travels in
straight lines, sees objects closer to the
surface due to refraction. Such distortions
are common.

Refraction
Water
Air
Refraction is the
bending of light as it
passes from one
medium into another.
refraction
N
qw
qA
Note: the angle of
incidence qA in air and
the angle of refraction
qA in water are each
measured with the
normal N.
The incident and
refracted rays lie in the
same plane and are
reversible.

The Index of Refraction
The index of refraction for a material is the ratio
of the velocity of light in a vacuum (3 x 108 m/s)
to the velocity through the material.
c
v
c
n
v

Index of refraction
c
n
v

Examples: Air n= 1; glass n = 1.5; Water n =
1.33

Example 1. Light travels from air (n = 1) into
glass, where its velocity reduces to only 2 x 108
m/s.
What is the index of refraction for glass?
8
8
3 x 10 m/s
2 x 10 m/s
c
n
v
 
vair = c
vG = 2 x 108 m/s
Glass
Air
For glass: n = 1.50
If the medium were water: nW = 1.33. Then you should
show that the velocity in water would be reduced from c
to 2.26 x 108 m/s.

Analogy for Refraction
Sand
Pavement
Air
Glass
Light bends into glass then returns along original
path much as a rolling axle would when
encountering a strip of mud.
3 x 108 m/s
3 x 108 m/s
2 x 108 m/s vs < vp

Deriving Snell’s Law
Medium 1
Medium 2
q1
q1
q2 q2
q2
Consider two light
rays. Velocities are
v1 in medium 1 and
v2 in med. 2.
Segment R is common
hypotenuse to two rgt.
triangles. Verify shown
angles from geometry.
v1
v1t
v2
v2t
q1
R
1 2
1 2
sin ; sin
v t v t
R R
q q
 
1
1 1
2
2 2
sin
sin
v t
v
R
v t v
R
q
q
 

Snell’s Law
q1
q2
Medium 1
Medium 2
The ratio of the sine of the angle
of incidence q1 to the sine of the
angle of refraction q2 is equal to
the ratio of the incident velocity
v1 to the refracted velocity v2 .
Snell’s
Law:
1 1
2 2
sin
sin
v
v
q
q

v1
v2

Example 2: A laser beam in a darkened room
strikes the surface of water at an angle of 300.
The velocity in water is 2.26 x 108 m/s. What is
the angle of refraction?
The incident angle is:
qA = 900 – 300 = 600
sin
sin
A A
W W
v
v
q
q

8 0
8
sin (2 x 10 m/s)sin60
sin
3 x 10 m/s
W A
W
A
v
v
q
q  
qW = 35.30
Air
H2O
300
qW
qA

Snell’s Law and Refractive Index
Another form of Snell’s law can be derived from the
definition of the index of refraction:
from which
c c
n v
v n
 
1 1 2
1
2 2 1
2
;
c
v v n
n
c
v v n
n
 
1 1 2
2 2 1
sin
sin
v n
v n
q
q
 
Snell’s law for velocities and indices:
Medium 1
q1
q2
Medium 2

A Simplified Form of the Law
1 1 2
2 2 1
sin
sin
v n
v n
q
q
 
Since the indices of refraction for many common
substances are usually available, Snell’s law is often
written in the following manner:
1 1 2 2
sin sin
n n
q q

The product of the index of refraction and the
sine of the angle is the same in the refracted
medium as for the incident medium.

Example 3. Light travels through a block of glass,
then remerges into air. Find angle of emergence
for given information.
Glass
Air
Air
n=1.5
First find qG inside glass:
sin sin
A A G G
n n
q q

500
qG
q
0
sin (1.0)sin50
sin
1.50
A A
G
G
n
n
q
q  
qG = 30.70
From geometry, note angle qG same
for next interface.
qG
sin sin sin
A G G A
A A
n n n
q
q q
 
Apply to each interface:
qe = 500
Same as entrance angle!

Wavelength and Refraction
The energy of light is determined by the frequency of
the EM waves, which remains constant as light passes
into and out of a medium. (Recall v = fl.)
Glass
Air n=1
n=1.5
lA
lG
fA= fG
lG < lA
;
A A A G G G
v f v f
l l
 
; ;
A A A A
G G G G
v f v
v f v
l l
l l
 
1 1 1
2 2 2
sin
sin
v
v
q l
q l
 

The Many Forms of Snell’s Law:
Refraction is affected by the index of refraction,
the velocity, and the wavelength. In general:
1 2 1 1
2 1 2 2
sin
sin
n v
n v
q l
q l
  
All the ratios are equal. It is helpful to recognize
that only the index n differs in the ratio order.
Snell’s Law:

Example 4: A helium neon laser emits a beam of
wavelength 632 nm in air (nA = 1). What is the
wavelength inside a slab of glass (nG = 1.5)?
nG = 1.5; lA = 632 nm
;
G
A A A
G
G A G
n n
n n
l l
l
l
 
(1.0)(632 nm)
1.5
421 nm
G
l  
Note that the light, if seen inside the glass, would
be blue. Of course it still appears red because it
returns to air before striking the eye.
Glass
Air
Air
n=1.5
q
qG
q
qG

Total Internal Reflection
Water
Air
light
The critical angle qc is the
limiting angle of
incidence in a denser
medium that results in an
angle of refraction equal
to 900.
When light passes at an angle from a medium of
higher index to one of lower index, the emerging
ray bends away from the normal.
When the angle reaches a certain
maximum, it will be reflected
internally.
i = r
Critical angle
qc
900

Example 5. Find the critical angle of
incidence from water to air.
For critical angle, qA = 900
nA = 1.0; nW = 1.33
sin sin
W C A A
n n
q q

0
sin90 (1)(1)
sin
1.33
A
C
w
n
n
q  
Critical angle: qc = 48.80 Water
Air
qc
900
Critical angle
In general, for media where
n1 > n2 we find that:
1
2
sin C
n
n
q 

Total refraction in everyday life
• Atmospheric refraction
- the atmosphere made up of layer with different density
and temperature air -->these layers different index of
refraction --> light refracted
- distortion of the shape of Moon or Sun at horizon
- apparent position of stars different from actual one
- if light goes from layers with higher n to layers with lower
--> total refraction: -mirages, looming
• Light guides: optical fibers: used in communication,
medicine, science, decorative room lighting, photography
etc…..

Dispersion by a Prism
Red
Orange
Yellow
Green
Blue
Indigo
Violet
Dispersion is the separation of white light
into its various spectral components. The
colors are refracted at different angles due
to the different indexes of refraction.

Light going through a prism bends toward the base
n1
n2
n =1
1
Bending angle depends on value of n2
n1

Dispersion
• The index of refraction of a medium
depends in a slightly manner on the
frequency of the light-beam
• Different color rays deflect in different
manner during refraction: violet light is
deflected more than red…..
• By refraction we can decompose the white
color in its constituents--> A prism separates
white light into the colors of the rainbow:
ROY G. BIV
• We can do the opposite effect
too…..recombining the rainbow colors in
white light
• Atmospheric dispersion of light: rainbow
(dispersion on tinny water drops) or halos
(dispersion on tiny ice crystals)

Summary
1 2 1 1
2 1 2 2
sin
sin
n v
n v
q l
q l
  
Snell’s Law:
Refraction is affected by the index of refraction,
the velocity, and the wavelength. In general:
c = 3 x 108 m/s
v
Index of refraction
c
n
v

Medium n

Summary (Cont.)
The critical angle qc is the
limiting angle of incidence in
a denser medium that results
in an angle of refraction
equal to 900.
In general, for media where n1 > n2
we find that:
1
2
sin C
n
n
q 
n1 > n2
qc
900
Critical angle
n1
n2

Types of Lenses
• If you have ever used a microscope, telescope, binoculars, or a
camera, you have worked with one or more lenses.
• A lens is a curved transparent material that is smooth and regularly
shaped so that when light strikes it, the light refracts in a predictable
and useful way.
• Most lenses are made of transparent glass or very hard plastic.

Types of Lenses
• By shaping both sides of the lens, it is possible to
make light rays diverge or converge as they pass
through the lens.
• The most important aspect of lenses is that the light
rays that refract through them can be used to magnify
images or to project images onto a screen.

Types of Lenses
• Relative to the object, the image produced by a thin lens can be real
or virtual, inverted or upright, larger or smaller.

Lenses
• For materials that have the entrance
and exit surfaces non-parallel: the
direction of light beam changes
• The best results obtained by lenses:
piece of glass with spherical surfaces
• Two main groups:
- those that converge light rays (like
concave mirrors)
- those that diverge the light rays
(like convex mirrors)
• Converging and Diverging lens
• Characteristic points and lines:
- center of lens
- optical axis
- focal point (on both sides)
- focal length (equal on both sides)

Lens Terminology
• The principal axis is an
imaginary line drawn
through the optical centre
perpendicular to both
surfaces.
• The axis of symmetry is
an imaginary vertical line
drawn through the optical
centre of a lens.

Lens Terminology
• Both kinds of lenses have
two principal foci.
• The focal point where the
light either comes to a focus
or appears to diverge from
a focus is given the symbol
F, while that on the
opposite side of the lens is
represented by Fʹ.

Lens Terminology
• The focal length, f, is the
distance from the axis of
symmetry to the principal
focus measured along the
principal axis.
• Since light behaves the
same way travelling in
either direction through a
lens, both types of thin
lenses have two equal focal
lengths.

Drawing a Ray Diagram for a Lens
A ray diagram is a useful tool for predicting and understanding how
images form as a result of light rays emerging from a lens.
• The index of refraction of a lens is greater than the index of
refraction of air

• The light rays will then bend, or refract, away from the
lens surface and toward the normal.
• When the light passes out of the lens at an angle, the
light rays refract again, this time bending away from
the normal.
• The light rays undergo two refractions, the first on entering
the lens and the second on leaving the lens

Constructing the images produced by lenses
• We can construct the images by the same principles that we used in curved mirrors
• If the lens are ideal ones for each object point we have one image point
• Following two special rays are enough to get the picture
• special rays: - 1. going through the center of the lens (no refraction);
• -2. through the focal point (parallel to the optical axis);
• - 3. parallel with the optical axis (through the focal point)

• A thin lens is a lens that has a thickness that is slight
compared to its focal length. An example of a thin lens
is an eyeglass lens. You can simplify drawing a ray
diagram of a thin lens without affecting its accuracy by
assuming that all the refraction takes place at the axis
of symmetry.

MELS 123 ELECTRONIC LOGIC FOR SCIENCE LECTURE NOTES PPT OMOSAE

MELS 123 ELECTRONIC LOGIC FOR SCIENCE LECTURE NOTES PPT OMOSAE

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