12VCE Physics Notes 2014

8/10/2019 12VCE Physics Notes 2014

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MOTION, ELECTRONICS & PHOTONICS,

ELECTRIC POWER, SOUND, LIGHT &

MATTER


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MOTION

HOLIDAYS - Constant Acceleration Equations- Projectile Motion

TERM 1

WEEK 1 - Motion Outcome Begins

WEEK 2 - Motion Outcome

WEEK 3 - Motion Outcome- Draft of Method due (Wed Feb 12)

WEEK 4 - Motion Outcome- Data Analysis due (Mon Feb 17)

WEEK 5 - Motion Outcome due (Mon Feb 24)- Graphical Analysis of Motion- Forces

WEEK 6 - Impulse & Momentum

WEEK 7 - Work, Energy & Power- Collisions,

WEEK 8 - Circular Motion

WEEK 9 - Gravitation

WEEK 10 - Circular Motion & Gravitation Test

ELECTRONICS & PHOTONICS

HOLIDAYS - Electric circuits- Voltage Dividers

TERM 2

WEEK 1 - Diodes

- Voltage Amplifiers- Transducers & Photonics

WEEK 2 - Electronics Test SAC (Mon May 5)


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ELECTRIC POWER,

TERM 2

WEEK 2 - Electronics Test SAC (Mon May 5)- Magnetic Fields- Magnetic Forces

WEEK 3 - Electromagnetic Induction- Induced EMF

WEEK 4 - Motors and Generators

WEEK 5 - Electric power generation

WEEK 6 - Electric Power SAC (Mon June 2)- Electric Power Test

WEEK 7 & 8 - ExamsSOUND

TERM 2

WEEK 9 - Transmission of sound- Waves- Intensity

TERM 3

WEEK 1 - Reflection, transmission & absorption- Diffraction

WEEK 2 - Interference of sound

WEEK 3 - Music- Microphones & loudspeakers

WEEK 4 - Sound SAC (Mon August 4)


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LIGHT & MATTER

TERM 3

WEEK 4 - Light & Waves

WEEK 5 - Photoelectric Effect- Wave/Particle Duality

WEEK 6 - Energy Levels in an Atom- Light Revision

WEEK 7 - Light SAC (Mon August 25)- Exam Revision

ASSESSMENT

School-assessed coursework and an end of year examination will determine the students level ofachievement in Physics.

School-assessed Coursework

Students will be given a score out of 170 that will be a result of the students performance in fiveoutcomes.

Outcome 1 - A plan and report of a student-designed investigation including a conciseevaluation. (Marks allocated = 40)

Outcome 2 - A test on electronics and photonics. This will be done as a 50-minute

classroom activity. (Marks allocated = 30)

Outcome 3 - A written report on Electric Power. This will be done as a 50-minuteclassroom activity. (Marks allocated = 40)

Outcome 4 - A summary report on practical activities related to Sound. This will be doneas a 50-minute classroom activity. (Marks allocated = 30)

Outcome 5 - A test on Light and Matter. This will be done as a 50-minute classroomactivity. (Marks allocated = 30)

The School-assessed coursework contribute 40% to the students study score.

Final examination

This will be a 2 hour external examination and will contribute 60% to the students study score.


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UNIT 3 PHYSICS EXPERIMENTAL INVESTIGATION

CROWN THE COIN

Every year your school runs a carnival. One of the major attractions is the Crown the Coin gamewhere people pay $1 for the chance to release a ball bearing on a metal track in the hope that it will hitand hence win a $2 coin placed near the end of the track. To further frustrate the players chances ofwinning the operator is able to change the height of the end of the track above the ground and theposition of the coin at the start of each game.

For many years, you have been playing this game for the odd win but usually suffer heavy losses.This year is your chance for revenge. Your mother is chairperson of the Carnival OrganisingCommittee and now has the key to the storeroom where this game is kept. You plan to take the keyand experiment on the game so that next month when it makes its reappearance at the carnival youcan win back all that lost money (with interest).

Now you have a real chance to use all that physics you have been studying. It is also a chance to usethe computer for something other than games. You intend to come up with an equation that will allowyou to determine where to put the ball as long as you can estimate where the coin is horizontally fromthe end of the track and can estimate how high the track is above the ground.

In the experimental investigation each student is to perform a practical investigation and write a reportin the range of 1000-1200 words on the investigation.

In order to obtain an excellent result the following points need to be apparent from the report. Thestudent has been able to

Demonstrate advanced understanding of the investigation

Independently design and conduct experimental procedures involving at least two continuousindependent variables.

Systematically collect relevant data. Accurately record insightful and detailed observations.

Estimate uncertainties in data and derived quantities correctly.

Analyse data accurately and fully

Communicate valid conclusions, relating data to hypotheses, taking into account sources oferror and uncertainty and limitations of experimental design


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It is necessary to investigate the following relationships.

1. Relationship between Height of Track & Horizontal Distance (constant Ball Height)

2. Relationship between Height of Ball & Horizontal Distance (constant Height of Track)

Things to consider

How will Height be measured?

What constants should be used?

At least 7 good sets of data are needed for each experiment

REPORT FORMAT

Introduction

Aim of Investigation

The aim of this investigation is to find a relationship between

This will be done in 2 parts

Variables:

Use a table similar to the one below

Variable Symbol Unit of measurement Significance Range

Height of Track

Height of Ball

Mass of Ball

Horizontal Distance

Other variables?

PART HEADING

Aim of Part

The aim of this part of the experiment is to find a relationship between

Apparatus

Use a table similar to the one below

Equipment Purpose Accuracy Reason for choice


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Method

The apparatus was set up as shown in the following diagram.

Diagram

Dot points eliminate the need to use complete sentences

Dont repeat what has been already stated in variables/apparatus

Justify what was done

Discuss how difficulties/potential difficulties can/will be overcome

Consider what you can/should do to gain accurate data. (How many significant figures should beused to measure data? Do measurements need to be repeated?)

Include how data will be analysed.

Results:

The table below displays a summary of the data that was collected.

(The full set of data & graphs should be placed in an appendix at the end of the experiment.)

ExampleConstants

Variable (unit) Variable (unit)

Analysis & Discussion

Graphs of the data were plotted using Microsoft Excel. Linear and power trend lines were insertedand the graph with the greatest R2value was chosen to be the most appropriate curve (only show agraph of the best trend line). An equation was then obtained.

Remember to

Make sure that axes are labelled with a name and unit

Remove background and gridlines unless they are really necessary

Make equation font of a readable size.

Change the equation from x & y to appropriate symbols. Use correct significant figures.

Discuss the graphs and data.

Conclusion

The conclusion is a summary of the experiment. It must link the analysis and discussion to the aim.


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Criterion 3

Analysis and interpretation of experimental data

Inclusion of a clear statement demonstrating an understanding of the relevance of theexperimental data.

Organisation of the data into a convenient form with suitable graphs, charts, tables ordiagrams used to demonstrate the relationship between variables and quantitiesmeasured/observed.

Manipulation of data resulting in the calculation of physical values. Explanation of physicalrelationships.

Accurate measurements and calculations with a correct treatment of significant figures.

Marking scheme

Organisation & presentation of graphs (3 marks)Analysis of data (Parts 1 & 2) (7 marks)

Criterion 4

Evaluation of the practical work

Interpretation of experimental data of significance to the context.

Inclusion of evaluations that can be reasonably inferred from the data.

Discussion of the limitations of the experimental findings in terms of uncertainties.

Explanation of all differences between expected and obtained relationships.

Marking scheme

Discuss & combine results from parts 1 & 2 (5 marks)Theoretical equation & discussion (2 marks)Conclusion (3 marks)

Criterion 5

Knowledge and application of the appropriate physics terms, concepts and relationships

Application of key knowledge and skills about the area of study, throughout the task, withclear definitions and explanations.

Appropriate use of accepted physics symbols, SI units and accepted terminology.

Clear labelling of diagrams and graphs. Correct calculation of physical quantities significant to the context using a level of

mathematics outlined in the study design.

Marking scheme

Symbols & units (5 marks)Labelling (5 marks)Defining terms, clarity (5 marks)


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MEASUREMENT & UNITS

The appropriate S.I. unit must accompany all measured quantities.

The following units should be used when solving a problem unless told otherwise.

QUANTITY UNIT SYMBOL

Length metre mMass kilogram kgTime second s

The following metric prefixes are often used

milli (m) = 10-3

centi (c) = 10-2 kilo = 103

Some units are obtained by combining other units.

All derived units should be made of SI units. If this is not the case, a conversion will be necessaryeg. 72 km/hr = (72 x 1000) / (60 x 60) = 20 m/s.

STANDARD FORM

Any number outside the range 1000 - 0.01 should be expressed in standard form.

Standard form involves expressing a number as a number between one and ten multiplied by a powerof ten.

Example Speed of light = 300 000 000 m/s = 3 x 108m/s

SIGNIFICANT FIGURES

The number of figures quoted in an answer should indicate the accuracy of a number.

The final answer should only contain as many figures as the least accurate number specified in thequestion.

Example 1.27 kg x 328 m 25 s = 16.6624 kg m s-1

This should be written as 17 kg m s-1because 25 is the least accurate number in the question as it

contained only 2 digits.

Zeros at the beginning of a number are not counted as significant figures.

e.g. 0.02040 has 4 significant figures. The first 2 zeros are not counted.


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VECTORS & SCALARS

Physical quantities may be either vectors or scalars. Vector quantities must have a direction as well asa magnitude. Scalar quantities have magnitude only.

Vector Quantities Scalar Quantities

Displacement DistanceVelocity SpeedAcceleration WorkForce MassImpulse EnergyMomentum Power

Arrows can represent vectors where the length indicates the magnitude and the direction is indicatedby the way the arrow points.

It is often easiest to split vectors into components.

Example 2 m N 60 E = 2 sin 60 E + 2 cos 60 N

Vectors cannot be added or subtracted like numbers.For example 3 m North + 3 m East does not equal 6 m northeast.

Vectors may be added by placing the arrows that head to tail. The resultant vector is drawn from thetail of the first vector to the head of the last vector. The magnitude of the vector and its direction canbe calculated from a scale diagram or geometry.

Subtracting a vector is the same as adding a vector in the opposite direction.

Change in a vector = final vectorinitial vector

The velocity of A relative to B = velocity of Avelocity of B

GRAPHICAL ANALYSIS OF MOTION

A convenient way of displaying information about an object is on a graph. The gradient and/or thearea may give additional information about the object. If dividing the quantities or units of the y andx-axes gives a known quantity or unit, then the gradient is useful. If multiplying the quantities orunits of the y and x-axes gives a known quantity or unit, then the area under the graph is useful.

For motion the following graphs are useful.

Graph Gradient Area

s - t velocityv - t acceleration change in displacementa - t change in velocity

PROBLEMS: Gardiner p16 Set 6 Ques. 3, 4, 7, 8, 11-13, 15, 16


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MOTION EQUATIONS

average speed distance covered

time taken

average velocity change in displacement

change in time

average acceleration change in velocity

change in time

If the acceleration of an object is constant, the following equations can be used

s v u

2 x t

v = u + a t

v2= u2+ 2 a s

s = u t + a t2

s = v t a t2

where s = displacement u = initial velocityv = final velocity a = acceleration

t = time

Note that as s, u, v & a are vectors, they must have a direction. Using positive and negative signs inthe equations can indicate this.

If an object is falling and the only force on it is gravity then its acceleration will be

g = 9.8 m s-2

(often 10 is used)

PROBLEMS: Gardiner p22 Set 7 Ques. 18, 1417, 19, 20Gardiner p32 Set 12 Ques. 19


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FORCES

The motion of an object can be determined if the forces acting on it are known. The way forces affectmotion are summarised by Newton's Laws.

1. A body at rest or in a state of uniform motion will remain in that state unless acted upon byexternal unbalanced forces.

2. A body will accelerate in the direction of the resultant, or net, force that acts on it accordingto the equation

Fnet = m a

3. For every applied force there is an equal and opposite reaction force.

Force is a vector quantity and is measured in Newtons (N). To find the net force, the forces must beadded vectorially.

If the net force on an object is constant, its acceleration will also be constant. This means that theconstant acceleration equations may be used to determine its motion.

When two or more objects are connected it is often best to consider them as a system and examine theexternal forces on the system.

PROBLEMS: Gardiner Set 9 Ques. 14, 911, 3134

MASS & WEIGHT

Mass is a property of an object. It is a measure of how much matter is present. It does not depend onposition.

Weight is a measure of the gravitational force acting on an object and depends on position. This forceis given by

Weight = W = m g

where g is the acceleration due to gravity or the gravitational field strength at a particular point.

PROBLEMS: Gardiner Set 10 Ques. 511

FRICTION

When friction acts on an object it provides a force that acts in the opposite direction to the direction ofmotion. Friction depends on the roughness of the surfaces in contact and how hard they push againsteach other, but not on the areas of the surfaces or the speed. Friction cannot cause motion; it onlyopposes it.

PROBLEMS: Gardiner Set 9 Ques. 1924, 2729Gardiner Set 10 Ques. 1416, 18, 19


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IMPULSE & MOMENTUM

F t is called the impulse on an object.

m v is called the momentum of an object. mv is called the change in momentum.

F t = m v Impulse = Change in momentum

Impulse and momentum are both vector quantities. Their units are N s or kg m s-1

When there are no external forces on a system, momentum is conserved, that is,

momentum before = momentum after

Gardiner Set 9 Ques. 1218Gardiner Set 17 Ques. 1, 2, 4, 69, 12, 13

WORK & ENERGY

Work done F s cos where is the angle between F and s.

Work is a scalar quantity and it is measured in Joules (J).

If F and s are perpendicular then the work done is zero.

When work is done against gravity Work = m g h where h is the height

In this case the work done is equal to the gravitational potential energy gained by the object.

PROBLEMS: Gardiner Set 18 Ques. 15, 7, 8Gardiner Set 19 Ques. 13

KINETIC ENERGY

When an object is moving it has kinetic energy.

KE = m v

Kinetic energy is a scalar quantity and is measured in Joules.

A change in kinetic energy means that work is being done.

Work = change in kinetic energy



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ELASTIC POTENTIAL ENERGY

One means of storing energy is in a spring where it can be used later as kinetic energy, the totalenergy (spring + kinetic) must be constant.

The area under any F - x graph gives the amount of energy stored.

For springs obeying Hookes Law

F = k x

Elastic Potential Energy stored = k x

where k is the spring constant and x is the extension or compression.

PROBLEMS: Gardiner Set 20 Ques. 13

CONSERVATION OF ENERGY

The law of conservation of energy suggests that the total energy of a system remains constant. In thecases where friction can be ignored a change in potential energy involves a corresponding change inkinetic energy.


POWER

Power is a measure of how quickly energy is used.

Power work

time

Power is measured in watts (W) and is a scalar quantity. Another common formula for calculatingpower is

Power = F v

PROBLEMS: Gardiner Set 18 Ques. 2125Gardiner Set 19 Ques. 4


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ELASTIC COLLISIONS

If a collision is elastic the total kinetic energy before collision equals the total kinetic energy aftercollision. Collisions are usually not elastic. Another way of saying that a collision is elastic is to say,the interaction force depends only on the separations of the bodies ". This means that permanentdeformation of the bodies does not occur.

Example: A 1 kg mass moving at 4 m/s makes an elastic collision with a 3kg mass moving inthe same direction at 2 m/s. Find the velocity of the objects after the collision.

Before After

4 m/s 2 m/s x m/s y m/s

[ 1 kg ]--> [ 3 kg ]--> [ 1 kg ]--> [ 3 kg ]-->

momentum: 4 + 6 = x + 3y

so 10 = x + 3yx = 10 - 3y

kinetic energy: 8 + 6 = x + (3)y28 = x + 3 y

(sub. for x) 28 = 100 - 60 y + 9 y + 3 y

so 12y - 60 y + 72 = 0y - 5 y + 6 = 0( y - 3 ) ( y - 2 ) = 0

=> y = 3 or y = 2

If y = 2 ; x = 4, this is not a possible solution as the 1 kg mass cannot get ahead of the 3 kg mass.

so y = 3 ; x = 1 must be the solution.


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INELASTIC COLLISIONS

The majority of collisions are inelastic so kinetic energy is not conserved. For an inelastic collision,only a fraction of the initial kinetic energy is retained.

Example: An object of mass 2 kg travelling at 4 m/s makes a head on collision with anobject of mass 1 kg which is at rest. The system loses 1/4 of its initial kineticenergy. Find the velocity of the objects after the collision.

Before After

4 m/s 0 m/s x m/s y m/s

[ 2 kg ]--> [ 1 kg ]--> [ 2 kg ]--> [ 1 kg ]-->

momentum : 8 = 2x + y

so y = 82x

KE : ( ( 2 ) 4 ) = (2) y + y12 = x + y224 = 2 x + y

(sub. for y) 24 = 2 x + 6432x + 4 x

so 6 x - 32 x + 40 = 03 x - 16 x + 20 = 0( 3x - 10 ) ( x - 2 ) = 0

=> x = 10/3 or x = 2

If x = 10/3 ; y = 4/3, this is not a possible solution as the 2 kg mass cannot get ahead of the 1 kg mass.

so x = 2 ; y = 4 must be the solution.


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COLLISIONS

1. A 2 kg mass moving at 3 m/s makes an elastic, head-on collision with a 1 kg object which isat rest. Calculate the velocity of the objects after the collision.

2. A 3 kg mass moving at 5 m/s makes an elastic, head-on collision with a 2 kg object which isat rest. Calculate the velocity of the objects after the collision.

3. A 5 kg mass moving at 4 m/s makes an elastic, head-on collision with a 1 kg object moving at2 m/s in the opposite direction. Calculate the velocity of the objects after the collision.

4. A 1 kg object moving at 5 m/s north makes an elastic collision into the back of a 3 kg objectmoving at 1 m/s north. Calculate the velocity of the objects after the collision.

5. A 4 kg object moving at 2 m/s east makes a head-on collision with a 2 kg object which is atrest. If one quarter of the kinetic energy is lost due to the collision, calculate the velocity ofthe objects after the collision.

6. A 1 kg object moving at 4 m/s south makes a head-on collision with a 4 kg object which is atrest. If the total kinetic energy after the collision is only one quarter of the initial kineticenergy calculate the velocity of the objects after the collision.

7. A 2 kg object moving at 4 m/s east makes a head-on collision with a 1 kg object moving westat 2 m/s. If half of the kinetic energy is lost due to the collision, calculate the velocity of theobjects after the collision.

8. A 1 kg object moving at 4 m/s north makes a head-on collision with a 2 kg object moving at 1m/s south. If after the collision the system has only two thirds of the initial kinetic energycalculate the velocity of the objects after the collision.


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PROJECTILE MOTION

When an object is projected at an angle to the earth's surface it does not follow a straight line butinstead it follows a parabolic path. This is because there is acceleration vertically downwards due tothe earth but no acceleration horizontally.

To determine information about an object moving in this way it is best to consider vertical andhorizontal motions separately.

Example 1: An object is projected horizontally from the top of a cliff 20 m high at 30 m/s,calculate

(a) the time taken to reach the ground(b) the horizontal displacement(c) the final velocity

(a) Horizontally Vertically

speed = 30 m s -1 u = 0a = 10 m s -2s = 20 m

Consider vertical motion

u = 0 a = 10 s = 20 m t =?

Use s = u t + 1/2 at so 20 = 5 t

=> t = 2.0 s

(b) Consider horizontal motion speed = 30 time = 2

Use distance = speed x time = 30 x 2 = 60 m

(c) The final velocity has a vertical component and a horizontal component

Horizontal velocity = 30 m/s across,

Vertical velocity found from v= u+ 2as

v = 20 m/s down

The final velocity is the vector sum of these velocities

Using Pythagoras X = 36 m/s

Using trig tan 20/30 so 340

So the final velocity is 36 m/s at an angle of 34 degrees to theground. .


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Example 2: A shell is fired from cannon with an initial velocity of 100 m/s at an angle of 60 tothe ground. Find

(a) the time it takes to reach the ground(b) how far it is from the cannon when it hits the ground

(a) The velocity should firstly be broken up into vertical and horizontal components.

Vertically Horizontally

u = 100 sin 60 = 86.6 speed = 100 cos 60 = 50a = -10 (taking upwards as positive)

Just before it reaches the groundit will be travelling just as fast asit was initially but in the oppositedirection so

v = -100 sin 60

Using v = u + at

t = 17.3 s

(b) Considering horizontal motion

Distance = speed x time = 50 x 17.3 = 866 m

PROBLEMS: Heinemann p. 29 Section 1.4 Ques. 110Gardiner Set 13 Ques. 25, 7, 8, 10, 11


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CIRCULAR MOTION

For an object to move in a circle at a constant speed it is necessary to apply a force perpendicular tothe motion of the object.

For circular motion:

F net m v2

r

where v is the speed and r is the radius of the circle

The direction of this force is towards the centre of the circle.

The acceleration can be found by

a v2

r

During one trip around the circumference of the circle the object moves a distance 2 r.T is the period of the motion (time for one revolution).

v 2 r

T

a 42r

T2

F net m 4 2r

T2

The frequency of the motion is the number of revolutions in one second.

f 1

T

Note thatmv2

r is not a real force but is a net force just like m a.

It is the sum of the forces on the object.


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Example: A car of mass 1000 kg passes over a bridge which is formed by the arc of a circleof radius 10 m.

(i) Find the force exerted by the car on the road at the top of the bridge if the caris travelling at 8.0 m/s.

(ii) What speed would cause the car to be on the point of leaving the bridge at itshighest point?

(i) Fnet = mgN

N is the force of the road on the carthat is equal in magnitude to the forceof the car on the road.

N = m gF net = m g

m v2

r = 3.4 x 10

3

N

(ii) To leave the bridge N = 0 so F net = m g

m v2

r= m g

v = r g= 10 m/s

PROBLEMS: Heinemann p. 61 Section 2.5 Ques. 110Heinemann p. 67 Section 2.6 Ques. 110Heinemann p. 74 Section 2.7 Ques. 110

Gardiner Set 15 Ques. 15, 816


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GRAVITATION

For any two objects an attractive force exists between them given by Newtons Law of Gravitation

F G M m

r2

where G is the gravitational constant 6.67 x 10-11

N m2kg

-2

M is the mass of the body being orbitedm is the mass of the orbiting bodyr is the distance between their centres

From F = m a, the acceleration of a body m, its centripetal acceleration, is

a G M

r2

This is often called the gravitational field strength or the acceleration due to gravity (g).

Equating this expression with the centripetal acceleration formula shows that

r3

T2 k (Keplers Law)

where k is a constant for all bodies orbiting the same body of mass M.

PROBLEMS: Heinemann p. 83 Section 3.1 Ques. 18Heinemann p. 89 Section 3.2 Ques. 19, 11, 12Heinemann p. 98 Section 3.3 Ques. 112Heinemann p. 103 Section 3.4 Ques. 110

MOTION REVISION

PROBLEMS: Heinemann p. 30 Chap.1 Review Ques. 14, 618Heinemann p. 75 Chap.2 Review Ques. 123Heinemann p. 110 Chap.3 Review Ques. 120

Heinemann p. 112 Exam-style Ques. 150


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ELECTRIC CIRCUITS

The current flowing passed a point is given by

I q

t

where q is the charge in coulombs.

The energy possessed by a charge is given by

Energy = V q where V is the voltage.

The current flowing through an ohmic resistor can be related to the potential difference across its endsby the formula

V = I R (Ohm's Law)

The amount of energy used by a resistor is given by

Energy = V I t

The power used in a resistor is given by

Power V I I2 V2

For resistors in series, the current flowing through all resistors is the same. The total voltage given bythe supply is equal to the sum of the potential difference across each resistor and the total resistance isgiven by

RT = R1 + R2 + R3 + ....

For resistors in parallel, each resistor has a potential difference equal to the voltage of the supply. Thetotal current flowing is equal to the sum of the currents flowing through each resistor and the totalresistance is given by the formula

1

T

1

1

1

2

1

3

In most cases the voltage given to a circuit is not equal to the total voltage in the supply. This is due tothe existence of an internal resistance within the supply. The supply can be thought of as a source ofvoltage (EMF) that is in series with a resistor. The actual voltage delivered will be the source voltage

less the voltage taken by the internal resistor.

V I r

Electronic components are also known as transducers. There are 2 types of transducers. Inputtransducers have a non-electrical input and an electrical output. Output transducers have an electricalinput and a non-electrical output.

Two commonly used transducers are Temperature Dependent esistors (TDs or thermistors) andLight Dependent esistors (LDs).

PROBLEMS: Heinemann p. 127 Section 4.1 Ques. 112


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AC ELECTRICITY & THE CATHODE RAY OSCILLOSCOPE

Electricity is usually transmitted as an alternating current (AC). In this case the voltage is notconstant and the current regularly changes direction.

The cathode ray oscilloscope (CRO) is an extremely useful device in electronics. It can be used togive a visual representation of electrical signals in the form of a voltage-time graph.

When suitably calibrated, the grid marked on the screen can be used to make accurate measurements.

The vertical scale measures the voltage.

The horizontal scale measures time.

For a sinusoidal wave pattern, the following measurements can be made.

The peak voltage is given by VPEKVpeak to peak

2

RMS voltage is given by VMVpeak

The frequency of the wave is given by f 1

T

AC meters are designed to give RMS values.

PROBLEMS: Heinemann p. 291 Section 8.1 Ques. 710


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VOLTAGE DIVIDERS

Voltage dividers allow a small, usable voltage to be obtained from a large voltage. A simple voltagedivider consists of two resistors in series.

Consider the following circuit.

The pair of resistors divides the voltage of thebattery into two parts. By choosing the rightpair of resistors you can make the divider giveyou any voltage at I between 0V and 9V

If the resistor R has a low resistancecompared with resistor S, then the potentialdifference across R will be less than thepotential difference across S. The voltage at Iwould be high. If the voltage at I is high

enough then the light would be on.

This circuit allows the globe to be providedwith the voltage it needs in order to operate normally provided appropriate values of R and S are used.

If the addition of the globe is not to change the properties of the voltage divider circuit its resistanceshould be greater than or equal to 10 times the resistance of S. (Rload> 10 RS)

The disadvantage of this voltage divider circuit is that the energy across resistor R is wasted.

PROBLEMS:

Question 1

Consider the simple circuit as shown below.

(a) What is the potential differencebetween terminals X and Y above?

(b) A very large resistance of size 10 mega ohms is connected across the terminals X and Y.What effect would this resistance then have on the potential difference across X and Y?

The potential difference across X and Y with the 10Mresistance in place would be:

A. significantly less than for the first circuitB. significantly greater than for the first circuitC. approximately the same as the first circuit


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Question 2

A student designs a variable heatingelement for heating liquids. The circuit isshown.

The heating element is ohmic and has aresistance of 100 ohms. The variableresistor Rvhas a range of 0 to 100 ohms.

(a) Discuss and explain the behaviour of the heating element when the variable resistor is set tozero. Determine the power used by the heating element.

(b) The variable resistor is now set to its maximum value of 100 ohms. Use calculations todetermine the flowing through the heating element, and calculate its power consumption.

(c) The student wants to increase the maximum power output of the heating element andmodifies the circuit by removing the 10 ohm resistor. Comment on the safety of this circuitdesign.


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Question 3

The following diagram shows two components, R and X, connected in series to a 9.0 V supply.

R is a variable resistor which follows Ohm's law. X is a component which does not follow Ohm's law.

The current-voltage graph below shows how the current in X varies with the potential differenceacross it.

R is set so that the current in X is 0.10 A.(a) What is the potential difference across R?

(b) Determine the resistance of R at this setting.

(c) Calculate the power loss in R at this setting.


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Question 4

The cold storage room at a supermarket warehouse is equipped with a temperature sensor that triggersa light to flash if the temperature rises over a certain level. The sensor contains a thermistor (T) whichhas resistance characteristics shown below. The alarm is controlled by the voltage divider circuitbelow and is triggered to turn on when VOUT rises past 8.0 V.

(a) What is the resistance of the thermistor when the temperature is 0 C?

(b) Determine the temperature at which the alarm is triggered.

(c) The battery for the sensor unit runs flat and is replaced; however the replacement battery isdefective and provides a voltage of only 10V. How will this affect the way the circuit triggersthe alarm? Use calculations in your answer.


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Question 5

The lighthouse at Point Lonsdale switches on automatically when the light intensity drops due to fogor nightfall. The switching system contains a light-dependent resistor (LDR) that has thecharacteristics shown below.

The LDR is part of a voltage divider circuit shown.

(a) What is the resistance of the LDR when the light intensity is 30 W/m2?

(b) The switching unit is activated when the voltage across it rises past 8.0V.Where should the switching unit be placed in the voltage divider circuit; X or Y?Explain your answer and determine the light intensity at which the lighthouse turns on.


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THE DIODE

A diode is a device that has a large resistance (> 10M) in one direction and a small resistance(< 5 ) in the other direction

A diode usually has a band around one of its ends.

When the banded end of the diode directed towards the negative terminal the diode is said to beforward biased. In this situation the diode has a low resistance so a large current can flow.

When the banded end of the diode directed towards the positive terminal the diode is said to bereverse biased. In this situation the diode has a very high resistance so almost no current can flow.

Diodes do not obey Ohms Law. graph of current against voltage has the following shape.

The maximum forward biased voltage is often called the switch-on voltage.

Diodes are often used to protect components that should only have current flow through them in acertain direction.

When a diode is connected to AC only half of the current will be allowed to flow.

Another group of diodes called light emitting diodes (LEDs) behave the same as normal diodes butalso give off light.

A diode is represented by the following symbol.

PROBLEMS: Heinemann p. 135 Section 4.2 Ques. 15, 6 (b), 710


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Question 3

A diode and a light globe have I-V characteristics shown below.

DIODE LIGHT GLOBE

(a) The diode and light globe are connected as shown below.Discuss whether the globe lights up or not, and determinethe current that flows through the globe.

(b) The diode and light globe are now connected as shown below.Again discuss whether the light globe glows, and determine the

current that flows through it.

Question 4

Two non-ohmic resistors X and Y are connected in parallel. These non-ohmic resistors show voltage-current curves as follows.

What is the total current flowing in this circuit?


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AMPLIFICATION

Transistors are able to act as amplifiers. A small change in the base current causes a large change inthe current between the collector and the emitter. A practical example of this is when a microphoneinput provides a small variation at the base of a transistor which leads to larger variations between thecollector and the emitter. This enables sounds to be amplified electronically.

A base voltage of about 0.6 V is needed for the transistor to begin to conduct electricity. If thevoltage is below this it is in cut off and no current flows. bove about 0.7 V it is in saturation andmaximum current flows. Between 0.6 V and 0.7 V the relationship between input and output currentis linear.

The performance of voltage amplifiers is often displayed in the form of a Vout versus Vin graphwhere the gradient of the graph gives the gain.

Voltage gain Vout Vin

If the gradient of the graph is negative then the amplifier is an inverter and the output signal will bethe negative of the input signal.

Amplification can only occur within a small range of values for Vin. Attempting to amplify voltagesoutside the amplification area can lead to distortion and clipping.

PROBLEMS:

Heinemann p. 142 Section 4.3 Ques. 17

Vin

Vout


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Question 1

A voltage amplifier has a gain of 50. Explain what is meant by the term "gain".

Question 2

Brian is investigating the performance of an amplifier. He feeds the alternating signal shown belowinto the amplifier which has the following VOUT- VINcharacteristics.

(a) Calculate the magnitude of the gain of this amplifier.

(b) The input signal is fed into the amplifier. Determine the peak value of the output voltage

produced.

(c) Brian increases the amplitude of the input signal to a peak voltage of 0.10V.On the graph below, sketch and label the output signal produced by the amplifier.


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Question 4

Sue and Alan construct the following circuit in order to measure the gain of a voltage amplifier.

The diagrams below are the traces from a cathode ray oscilloscope (CRO) screen that Sue and Alanobserved during experiments.

Connected between points A and CCalibrated control settings:Y Amplifier = 0.2 V per square

(a) Calculate the peak to peak voltage between A and C

Connected between points D and ECalibrated control settings:Y Amplifier =1 V per square

(b) Calculate the peak to peak voltage between D and E

(c) Calculate the peak to peak voltage between B and C

(d) What is the 'gain' of this voltage amplifier?

(e) State whether this voltage amplifier inverts or does not invert the input signal


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MAGNETIC FIELDS

A magnetic force can be seen to act on current carrying wires and charged particles as well as somemetallic objects.

As with gravitation and electrostatics we can imagine a field being associated with magnetism.Magnetic field strength is measured in Teslas (T). Magnetic field lines go from a north pole to a southpole. Magnetic field is a vector quantity.

When compasses are in the vicinity of a current carrying wire they are deflected. They indicate that amagnetic field is surrounding the wire. The direction of this field can be found by using the righthand rule.

If the thumb of the right hand is pointing in the direction of the conventional current flow, theaccompanying magnetic field will circle the wire in the direction indicated by the fingers when theyare brought together to form a fist.

Magnetic field lines will also pass through and around a loop. If the fingers are curled in the directionof the current flow the thumb will indicate the direction in which the field will pass through the loop.

On a diagram a magnetic field can be drawn as either a set of parallel lines pointing in the direction ofthe magnetic field, a set of crosses if the magnetic field is inwards or a set of dots if the field isoutwards.

The current in a wire is drawn as either a single line pointing in the direction of the current, a crossinside a circle if the current field is inwards or a dot inside a circle if the current is outwards.

PROBLEMS:

1. For each of the cases shown in figure below, 2. The figure below gives the directionwhat is the direction of the magnetic field at of the magnetic field at a point Ppoint X due to the currentI in a wire? associated with a current in a wire.(X is in the plane of the page.) For each case use the key to give the

direction of the current in the wire.


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MAGNETIC FORCE ON A WIRE

If a current flows through a wire that is perpendicular to a magnetic field it experiences a force thatcan be found using the formula

F I L B sin

I is the current flowing and L is the length of the wire within the magnetic field B. Note that if B andI are perpendicular the force is a maximum but if they are in the same or opposite directions the forceis zero. is the angle between B and I.

When the thumb of the right hand is pointed in the direction of the conventional current and thefingers are pointed in the direction of B, the palm of the right hand will then point in the direction ofthe force.

PROBLEMS:

1. What is the direction of the force on the current carrying conductor in the magnetic field, foreach case in figure below?


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2. In Figure below, PQRST is a loop of wire carrying a current, placed in a uniform magneticfield that is directed to the right.

What is the direction of the magnetic force acting on section:

(a) PQ of the wire?

(b) QR of the wire?

(c) ST of the wire?

3. What is the magnitude of the force per metre on a wire carrying a current of 0.80 A in auniform magnetic field of flux density 2.0 102 T, if the wire is perpendicular to themagnetic field?

4. A wire 1.2 m long carrying a current of 0.65 A has a force of 3.9 102N exerted on it by auniform magnetic field at right angles to the wire. What is the magnitude of the magnetic fluxdensity?

5. A wire 80.0 cm long is at right angles to a magnetic field of flux density 4.0 10

2

T. If thewire experiences a force of 4.8 102N, what is the current in the wire?


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6. A conductor carrying a current of 4.0 A is placed between the poles of a strong pair ofmagnets of strength 1.0 T, as shown below. The current direction is out of the plane of thepage. What are the magnitude and the direction of the force on a 12 cm length of theconductor?

7. A stream of electrons is directed to the right, through a magnetic field that is directed into theplane of the page. In which direction will the stream of electrons be deflected?

Gardiner p. 178 Ques. 14


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ELECTROMAGNETIC INDUCTION

When a wire is moved through a magnetic field, charges within the wire separate so that positivecharges move to one end of the wire and negative charges move to the other. This separation ofcharges causes a potential difference or EMF (electromotive force) to exist between the ends of thewire. The value of the EMF is found by

L B v sin

is the potential difference between the ends of the wireL is the length of the wireB is the magnetic field strengthv is the velocity of the wire. is the anglebetween v and B.

PROBLEMS:

1. A straight wire of length l is moved horizontally to the left, with a constant speed v, through avertical magnetic field of flux densityB, as shown in the figure below.

(a) What is the magnitude of the emf induced in the wire when:

(i) the wire is moving with a speed v to the left, as shown?

(ii) the wire is stationary?

(iii) the wire is moving to the right with a speed of 2v?

(iv) the wire is moved vertically upwards in the direction of the field with a speed

of v?

(b) What would be the magnitude of the induced emf if, instead, the wire was heldstationary and the magnetic field was moved with constant speedv:

(i) to the left?

(ii) vertically upwards?


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2. A straight wire is moved at a constant speed of 4.0 m s1 at right angles to a magnetic field ofmagnitude 0.25 T. If the emf induced in the wire is 0.50 V, what is the length of the wire?

3. A metal rod PQ of length 15 cm is moved vertically downwards with a constant speed of0.80 m s1through a magnetic field of 3.0 103T, as shown in the figure below.

(a) What is the emf induced in the rod PQ?

(b) Which end of the rod gains a net negative charge?

4. A conducting rod AB slides along metal conducting rails ED and FC in a uniform magneticfield of flux density 0.36 T, as shown in the figure below.

AB is moved to the right with a speed of 2.5 m s1. There is a resistance of 8.0 ohms insection EF and this is the only resistance of the circuit.

(a) What is the emf induced in AB?

(b) What is the induced current in AB?

(c) What is the direction of the current in AB?

(d) What is the magnitude and the direction of the force needed to keep AB in motion?

(e) What is the net force on AB?


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INDUCED EMF

When a loop is moved through a magnetic field, a current will flow around the loop only when thereis a change in either the area of the loop or the size of the magnetic field.

We define the magnetic flux () through the loop as being

B (measured in Webers).

A change in the magnetic flux will produce an EMF () that is given by

N

t

where N is the number of turns of wire making up the loop.

Note that the negative sign indicates that the EMF and hence the current produced will oppose anychange in flux. Also as the EMF depends on time, as time increases the EMF will decrease if the fluxdoes not continually change.

Rotating the loop can also change the flux. A rotation can either change the area through which thefield passes or cause the field to enter the loop from the opposite direction.

PROBLEMS:

1. The magnetic flux through a loop of wire changes from 5.0 103Wb to zero in 0.020 s.What is the emf induced in the loop of wire?

2. The magnetic flux through a loop of wire changes from 4.5 104Wb to 2.0 104Wb in0.0010 s. What is the emf induced in the loop?

3. An emf of 1.5 millivolts is induced in a coil of wire when the magnetic flux threading the coilincreases from zero to a value,x, in 1.5 milliseconds.

(a) What is the rate of change of flux threading the coil?

(b) What is the magnitude ofx?

4. A square loop of wire of side length 5.0 cm lies between the poles of a large electromagnet.When the electromagnet is turned on, the magnetic field increases from zero to 0.80 T in0.50 s. What is the average magnitude of the emf induced in the loop?


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5. A coiled spring loop of wire of radius 12 cm is heldbetween the poles of a strong magnet as shown in thediagram.

(a) The magnetic field strength between the poles of

the magnet is 1.6 NA

1

m

1.

What is the magneticflux through the loop?

(b) The coiled spring loop is released and it contracts to a circle of radius 4.00 cm in0.60 s. What is the average emf induced between the ends of the loop?

6. A coil containing 500 turns of insulated copper wire is threaded by a magnetic flux of4.8 104Wb. This magnetic flux is reduced to zero in 0.012 s. What is the magnitude of theaverage emf induced in the coil?

7. A magnetic field of flux density 0.010 T is perpendicular to a coil of 50 turns of wire, each ofarea 15 cm2. What is the magnitude of the emf induced in the coil if the direction of the fieldis reversed in 0.012 s?

8. A circular loop of wire of diameter 12 cm has an electrical resistance of 4.0 . It is placed ina magnetic field of flux density 0.010 T so that the plane of the loop is perpendicular to the

field. The loop is removed from the field in 0.08 s.

(a) What is the magnitude of the emf induced in the loop?

(b) What is the current flowing through the loop?

(c) How much electrical energy is dissipated when the loop is removed from the field?


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9. A circular coil consisting of 12 turns and of areaA is held withthe plane of the coil perpendicular to a uniform magnetic field ofmagnitude B. The entire coil is within the magnetic field (seediagram).

(a) The magnitude of the magnetic field is halved in time t. What is the magnitude of theinduced emf in the coil in terms ofA,B and t?

(b) Which one or more of the following changes would cause a clockwise current to flowthrough the coil?

A. Decreasing the strength of the magnetic field.B. Increasing the strength of the magnetic field.C. Reversing the direction of the magnetic field.D. Distorting the shape of the coil to reduce its area.E. Pulling the coil to the right so that it moves out of the magnetic field.

10. The figure below shows the north pole of a bar magnetbeing pushed into a solenoid. For each of the followingsituations state whether the current flows through themeter from X to Y or from Y to X. If no current flows,write N.

(a) The North Pole is moved towards the solenoid.

(b) The magnet is held stationary in the solenoid.

(c) The north pole of the magnet is withdrawn from the solenoid.

(d) The south pole of the magnet is moved towards the solenoid.

(e) The south pole of the magnet is removed from the solenoid.

(f) The North Pole is held stationary and the solenoid is moved to the right.

(g) The North Pole is held stationary and the solenoid is moved to the left.

(h) The magnet is rotated clockwise about its midpoint.

(i) Both the magnet and the solenoid are moved to the right with the same speed.


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12. The figure below shows two coils, AB and CD. Thecurrent flowing through AB produces a magneticfield that threads the coil CD. For the followingsituations state whether the current flows through thegalvanometer connected to coil CD from X to Y orfrom Y to X. If no current flows through thegalvanometer, write N.

(a) The current in AB is increased from zero when the switch is closed.

(b) The current in AB is reduced to zero when the switch is opened.

(c) The sliding contact on the variable resistor is moved to the left with the switch closed.

(d) The sliding contact on the variable resistor is moved to the right with the switchclosed.

(e) A steady current flows through AB.

13. A coil of wire has an area of 0.050 m2. The coilis placed between the poles of a strongelectromagnet of uniform flux density 0.80 T asshown in the figure below. When the currentthrough the electromagnet is turned off, themagnetic field drops to zero.

If the magnetic field falls to zero in a time of1.2 s, what is the magnitude of the average emfproduced in the loop?

14. Two flat horizontal coils are placed as shown inthe figure below.

Use the following key to answer parts a)d):

A. The current in coil 2 is clockwise, lookingfrom above.

B. The current in coil 2 is anticlockwise,looking from above.

C. No current flows in coil 2.

(a) The sliding contact of the rheostat is moved from X to Y.

(b) The sliding contact of the rheostat is held at Y.

(c) The sliding contact is removed from the rheostat at Y.

(d) The sliding contact is held midway between X and Y and coil 2 is moved towardscoil


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MOTORS & GENERATORS

For a rectangular coil of N turns (2 sides with length L and width W) one side rotating in a circle

max 2 N L B v sin 90 L B v 2 N L B (2W/2 f) 2 NBf

Instantaneous EMF = max sin (2f t)

PROBLEMS:

1. An electric motor is set up as shown.

. In which direction must the current flow for thecoil to rotate in a clockwise direction as seen from

the end of the coil AB?

2. Why is it necessary for the current direction in the coil of a DC motor to be reversed everyhalf turn?

3. DC motor has a part called a commutator. What is the purpose of this part?


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4. The figure shows a rectangular coil placed between thepoles of a strong magnet as part of a DC motor.

The coil has 40 turns of wire and each turn is rectangularwith length 0.060 m and width of 0.040 m. The magneticfield is uniform and has a magnitude of 0.050 T. The coilis free to rotate when placed between the poles of themagnet. A current of 1.5 A flows through the coil.

(a) At the instant represented in the diagram, the plane of the coil is parallel to themagnetic field. Which one of the following correctly gives the direction of the forceon the side AD?

A. upwardsB. downwardsC. to the rightto the left

(b) What is the magnitude of the force on the side (i) AD and (ii) CD when the plane ofthe coil is parallel to the field?

(c) In what ways could the magnitude of the force on the coil be increased?

5. The figure below shows the basic features of a small DC electric motor. WXYZ is the rotatingcoil, connected to a DC battery. The direction of the magnetic field is parallel to the plane ofthe coil.

(a) With the coil in the position shown and looking from point P, will the coil rotateclockwise or anticlockwise?

(b) The magnetic field has magnitude of 0.24 T and the length of the coil within themagnetic field is 0.060 m. What is the maximum force on the side ZY when a currentof 4.0 A flows through the coil?


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7. A 50-turn coil of wire has dimensions 0.16 m by 0.10 m as shown. A current of 2.0 A flowsthrough the coil. The coil is vertical and there is a magnetic field of 0.12 T through the coil tothe right.

(a) What is the magnitude and direction of the force on the following sides?

(i) AB

(ii) BC

(b) The coil forms part of an electric motor. In which direction (A, B, C or D) will thecoil start to rotate?

A. clockwiseB. anticlockwiseC. not possible to tellD. it will remain at rest

8. The armature of an AC generator is rotating at a constant speed of 30 revolutions per secondin a horizontal field of flux density 1.0 Wb m2. The diameter of the cylindrical armature is 24cm and its length is 40 cm. What is the maximum emf induced in the armature if it has 30turns?

9. A flat rectangular coil 15 cm by 20 cm has 300 turns. An alternating emf of peak value 340 Vis produced when the coil rotates at 3000 revolutions per minute in a uniform magnetic field.What is the value of the magnetic field strength?


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10. A rectangular coil of 30 turns and area 100 cm2rotates at 1200 revolutions per minute in auniform magnetic field of flux density 0.50 T. Find:

(a) the frequency of the generated emf

(b) the maximum emf

(c) the RMS emf

(d) the equation that gives the emf at any instant.

11. A rectangular coil of length 15 cm and width 12 cm, having 60 turns, rotates in air about itsaxis at 1200 revolutions per minute in a uniform magnetic field of flux density 0.40 T.

(a) What is the emf when:

(i) its plane is perpendicular to the magnetic field?

(ii) its plane is parallel to the magnetic field?

(b) Draw a graph to show the time variation of the emf over an interval of 1/20 of a

second.

(c) What is the magnitude of the peak emf?

(d) What is the RMS voltage?

12. The armature of a 50 Hz AC generator rotates in a magnetic field of strength 0.15 T. If thearea of the coil is 2.5 102m2, how many turns must the coil contain if the maximum emfproduced is 150 V?

13. How does a generator differ from a motor?


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14. A student sets up a coil and a magnetic field to demonstrate an electric generator, as shownbelow.

The ends of the coil are connected to slip rings A and B. When the coil is turned at a slowspeed a small current is observed on the meter.

(a) The coil is turned in a clockwise direction, when looking from position C.When is the magnetic flux through the coil:

(i) maximum?

(ii) minimum?

(b) Describe the current through the meter.

(c) To obtain greater accuracy, the student replaces the slip ring with a split ring

commutator and the galvanometer with a CRO. The student observes the trace on theCRO as the voltage varies with time and the ends of the coil make contact with thesplit ring commutator. The coil is rotated as before. Which one of the followinggraphs best represents the variation of voltage with time as seen on the CRO?


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HOUSEHOLD ELECTRICITY SUPPLY

Household electricity is produced by electromagnetic induction caused by the spinning ofelectromagnets between wire coils. In Victoria, this spinning is caused by steam that is obtained byusually using coal to heat water.

When electricity travels through the transmission wires, an amount of energy is lost. Using smallcurrents can reduce this energy loss. If however, a large amount of power is to be transmitted then thetransmission lines must carry a very high voltage.

TRANSFORMERS

Voltages can be changed to different values with a minimal loss of power using transformers, whichwork by electromagnetic induction. A simple transformer consists of two coils of wire of differentnumbers of turns wound onto a soft metal core.

For a transformerV1

V2

N1

N2

The input voltage must be AC. The voltage produced is an AC voltage.

Virtually no power is lost so it can be said that

V1I1= V2I2

The effective voltage is often called the RMS voltage.

In Australia electricity for domestic use is provided at 240 V RMS.


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PROBLEMS:

1. Does a transformer transform voltage, current or energy?

2. Why is alternating voltage necessary for a transformer?

3. For a transformer with a 1000-turn primary, what voltage is available at the 200-turnsecondary if the primary coil is supplied with 240 V AC?

4. For a transformer with 200 turns on the primary coil, supplied with 100 V AC, how manyturns are required to give 600 V AC at the secondary?

5. A transformer with 1500 turns in the primary coil is supplied with 240 V AC. How manyturns are required in the secondary to give 60 V AC?


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6. A transformer is connected to an AC source that can deliver 30 A. The secondary coil of thetransformer can deliver a maximum current of 10 A.

(a) What type of transformer is used?

(b) Calculate the ratio between the number of turns in the primary and the number ofturns in the secondary.

(c) Explain why an alternating current, and not a direct current, is used in a transformer.

7. A step-down transformer with a primary coil of 500 turns and a secondary coil of 250 turns.The primary coil is connected to a 240 V AC mains supply.

(a) What is the secondary voltage?

(b) Explain how the transformer operates.

(c) If the primary current is 2.4 A, what would the secondary be if the transformer is 100per cent efficient?

(d) Why would you expect the efficiency of the transformer to be less than 100 per cent?


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9. What material is normally used for the core of a transformer? The material of the core islaminated. Why is this?

10. A step-up transformer is connected to an AC generator that delivers 120 V and 80 A. Theratio of the number of turns in the secondary coil to the number of turns in the primary is 500.

(a) What is the voltage in the secondary coil?

(b) What is the power input?

(c) What is the maximum power output?

(d) What is the maximum current in the secondary?

11. An ideal transformer is supplied with an AC voltage of 24 V RMS and a current of 1.5 A, asshown below.

(a) What is the peak voltage of the primary of the transformer?

(b) What is the RMS secondary voltage?

(c) What is the RMS secondary current?

(d) What is the power supplied to the transformer?

(e) What is the power supplied to the load?


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12. There are 100 turns in the primary of a step-up transformer and 1000 turns in the secondary.Is there a power gain in the secondary circuit? Explain your answer.

13. A large, steady DC voltage is supplied to the primary coil of a transformer, but no secondaryvoltage or current is obtained. Why not?

14. Below is the graph of voltage against time for an AC supply.

What is the:

(a) peak voltage?

(b) RMS voltage?

(c) peak to peak voltage?

(d) average voltage?

(e) period?

(f) frequency?

15. To generate electricity in a power station, an electromagnet is rotated close to some coils ofwire. To produce the AC voltage the electromagnet must rotate at exactly 50 Hz.

(a) If the generator at a power station produces an output emf of 20 kV and a current of17 500 A, what is the power output?

(b) If the electromagnet rotated at 25 Hz what must be done to the magnetic field to

produce the same output voltage?

(c) Why must the coil be rotated to produce an emf?


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16. A 60 W globe is operated from a 240 V RMS AC supply.

(a) What is the RMS current flowing through the globe?

(b) What is the resistance of the globe when it is lit?

(c) What is the peak value of:

(i) the supply voltage?

(ii) the current through the globe?

17. What is the average value of the current from a 240 V AC supply to a 100 W lamp?

18. What advantage does AC have over DC for long distance power transmission?

19. The current through a car lamp is found to be the same as the effective current through a

household lamp, yet the household lamp is much brighter. Explain why this is so.

20. A person accidentally makes good electrical contact with a 6 V car battery to the earth. Thecurrent path is from the left arm to the right hand with a resistance of approximately 1000ohms.

(a) What is the current flow?

(b) Is this dangerous?

21. A person accidentally establishes good electrical contact with a 240 V RMS AC supply. Thecurrent path is from the left hand to the left forearm with a resistance of approximately 500ohms.

(a) What is the current flow?

(b) Is this dangerous?


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22. A set of 12 Christmas tree lights operates from a 240 V RMS supply. The 12 globes areconnected in series. Each light dissipates 4.0 W of power.

(a) What is the RMS voltage across each globe?

(b) What is the RMS current through each globe?

(c) One of the globes is removed and a voltmeter with a resistance of 2.0 104ohms isconnected across the terminals of the globe socket. What would be the RMS voltagereading on the voltmeter?

(d) Another set of 12 different globes, in parallel with each other, are connected to the240 V RMS supply. Each globe dissipates 4.0 W of power. What is:

(i) the current through each of these globes?

(ii) the potential drop across each globe in the parallel setup?

(e) Explain the difference in the function of the two systems when one globeblows.

23. How many joules of electric energy are there in 1 kilowatt-hour?

24. 33 MW of power with a voltage of 66 kV arrives at a town substation, from a generator, overtransmission lines that have a total resistance of 6.0 .

(a) What is the current in the transmission lines?

(b) What is the power loss over the transmission lines?

(c) What is the emf at the generator end of the transmission lines?

(d) What is the power output of the generator?

(e) What fraction of the power generated is lost in the transmission lines?


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25. Two transmission lines each have a resistance of 2.5 and carry a current of 500 A. The emfof the generator connected to the transmission lines is 12 kV.

(a) What is the power loss over the transmission lines?

(b) What is the power input from the generator?

(c) What is the voltage at the end of the transmission lines?

(d) What is the power available at the end of the transmission lines?

26. 20 kW of power is available from a generator, at 250 V, for transmission to consumers in a

town some distance from the generator. The transmission lines over which the power istransmitted have a resistance of 1.2 .

(a) How much power is lost if the power is transmitted at 250 V?

(b) What would be the voltage at the end of the transmission lines?

(c) Would the townspeople be able to use normal 240 V electrical appliances? Why?

(d) How much power would be lost if, instead, the voltage was stepped up by atransformer at the generator to 6.0 kV?

(e) What would be the voltage at the town if the power was transmitted at 6.0 kV?

27. When electrical energy is transmitted over long distances, this is done through thick wires oflow resistance and at high voltage. Why is high voltage used?


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28. The transmission system designed to deliver 200 MW of electrical power to a town from adistant generating plant has a power loss of 20 MW in the transmission wires. The voltageinput to the transmission system at the power plant is 500 kV.

(a) What is the current in the transmission wires?

(b) What is the voltage when the power reaches the town?

(c) What is the total electrical resistance of the transmission wires?

(d) Why is the electrical power transmitted at such high voltage?

29. A farmer living near a waterfall has a small water-driven power generator capable of anoutput of 1000 V AC RMS voltage. The electric power is then transmitted by transmissionlines to his farmhouse several kilometres away. A step-down transformer then provides 240 VAC RMS for the house and farm machinery.

(a) What is the peak voltage across the terminals of the generator?

(b) There are 1600 turns on the primary winding of the transformer. How many turns arethere on the secondary winding?

(c) The power generator is capable of producing 50 kW of electric power when all thefarm machinery and household appliances are operating. The total resistance of thetransmission wires is 0.25 ohm.

(i) What is the RMS current in the transmission wires when the generator isproducing maximum power?

(ii) What is the power loss in the transmission wires?

(iii) What steps could be taken to reduce this power loss?


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30. Electric power is supplied to a factory by a power station some distance away. The powerstation is capable of producing 0.90 MW of power and the voltage across the terminals of thegenerator is 9.0 kV. The wires connecting the generator of the power station to thetransformer at the factory have a total resistance of 0.25 ohm.

(a) What is the current (RMS) flowing through the wire to the factory?

(b) What is the power loss in the connecting wires?

(c) The factory has a step-down transformer to change the voltage to 1000 V. What is theratio of the number of turns in the primary coil to that in the secondary coil of thistransformer?

(d) If the transmission voltage had been 1.0 kV, and if the generator was supplying thesame power, would the power losses in the transmission wires have been more orless? Explain.

31. An electricity supply commission in a country area supplies power from a 12 kV power lineto a farm some distance from the supply line. A transformer assumed to be 100 per centefficient is used and, after connection to the house, the secondary voltage was measured to be

240 V. The ratio of primary turns to secondary turns in the transformer is 48:1.

(a) What is the RMS voltage across the primary coil?

(b) During a load test after installation, the RMS voltage across the secondary coil ismeasured as 230 V and the RMS current as 50 A when all the electrical appliancesare turned on. What is the power use in the house?

(c) What is the RMS primary voltage during the load test?

(d) During the load test what would be the RMS current in the primary coil?

(e) During the load test, what is the power loss in the transmission wires from the supplyline to the primary of the transformer?


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32. To supply electrical energy to a house in a country area, a transformer is used to step downthe voltage from 6.0 kV RMS to 240 V RMS. The wires connecting the step-downtransformer to the house have a total resistance of 0.030 ohm.

(a) What is the peak voltage across the output terminals of the transformer?

(b) If there are 10 000 turns in the primary winding of the transformer, how many turnsare there in the secondary winding?

(c) At one particular time, 20 kW of electric power is being drawn from the outputterminals of the transformer.

(i) What is the RMS current flowing in the transmission wires from the

transformer to the house?

(ii) Calculate the power loss in these transmission wires.


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SOUND

Vibrations cause sounds. These vibrations need a medium to travel through. A particle vibratingwithin a medium causes nearby particles to vibrate. In this way sound is transferred from one place toanother. Sound cannot travel through a vacuum. Sounds travel more quickly through a densermedium. Sound travels through air at a speed of about 330 metres per second.

TRANSMISSION OF SOUND

The way sound is transmitted can be understood by considering the effect that a moving loudspeakerhas on the air in front of it. As the speaker moves outwards the air in front of it is compressed. Thiscompression in turn causes further compressions in front of it. When a speaker moves backwards itcreates a space, a rarefaction, that air moves back into. The air is be compressed again as the speakermoves forwards. This causes the air molecules to vibrate backwards and forwards as longitudinalwaves.

When the variation from normal air pressure is a maximum (at a compression or rarefaction) a

pressure antinode exists.

When the variation from normal air pressure is a minimum a pressure node exists.


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WAVES

Waves are disturbances that travel through space. Longitudinal waves travel in the direction of thewave and transverse waves travel at right angles to the direction of the wave. Both types of waveshave similar behaviour except those longitudinal waves will not travel through a vacuum. Sound is alongitudinal wave.

Waves can be displayed by using two graphs:* a displacement - distance graph and* a displacement - time graph

The displacement - distance graph shows the shape of the wave and the position of each particle forone particular value of time. The distance between two adjacent and corresponding displacements iscalled the wavelength ().The maximum displacement of the graph is the Amplitude of the wave.

Wavelength ()

Amplitude

Distance

Displacement


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INTENSITY

The intensity of a sound is dependent upon the amplitude and is a measure of the power delivered bythe sound waves to a given area. With very intense sounds it is possible to feel the air compressions

of the sound waves. Intensity is measured in watts per square metre (W m-2

).

Intensity Power

area

The greater the amplitude, the louder the sound.

Intensity (Amplitude)2

Intensity also decreases with distance.

I 1

d2

Sound meters can be used to measure how intense a sound is. The intensity of sound is oftenmeasured in decibels. The decibel scale gives a good idea of the way in which the ear hearsintensities.

L(in dB)10log10

I

Io

Another way of indicating intensity levels is using the dB(A) scale. This scale is used by sound meterswhich take into consideration the frequency of the sound being measured, in a manner similar to thehuman ear, with the meter being more sensitive to sound in the middle range of human hearing.

Typical sound intensities are: dB W / m

Threshold of hearing (1000 Hz) 0 10-12

Normal conversation 60 10-6

Noisy factory 90 10-3

A loud clap of thunder 110 10-1

Pain threshold 120 100

Doubling the intensity in W m-2

corresponds to a 3-dB increase.

Loudness is similar to intensity however because the human ear is more sensitive to certainfrequencies, sounds of the same intensity may not have the same loudness.

A 10-dB increase is heard as a doubling of the loudness.


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VIEWING SOUNDS

The type of wave produced can be displayed using a device called a Cathode Ray Oscilloscope(CRO). Sounds can be changed into electrical signals and displayed on the screen of the CRO. Theappearance of the display on the screen is called the waveform of the sound.

It is possible to produce sounds of particular frequencies using a device called a Signal Generation.The waveform of these sounds can be displayed on the CRO and the sounds can be heard using aloudspeaker.

The figure below shows the CRO displays obtained with notes of different frequency and volume orloudness.

Different musical instruments have different waveforms even when they are playing notes of the samepitch and volume. This difference in shape relates to the tone of the note.

PROBLEMS: (Assume speed of sound in air = 340 m s-1)

1. Below is the displacementdistance graph for a periodic wave.

(a) What is the wavelength of the periodic wave?

(b) What is the amplitude of the wave?

Loud

Quiet

Low pitch & frequency

High pitch & frequency


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2. The range of frequencies of the audible spectrum is from 20 Hz to about 20 kHz. To whatrange of wavelengths in air do these frequencies correspond?

3. A sound wave of wavelength 34 cm travels at 340 m s1in air. What is the frequency of the

sound?

4. Students are listening to an audio frequency oscillator that is producing a sound of frequency5000 Hz. What is the wavelength of this sound?

5. When a trumpet is played, the sound waves that are produced spread out in all directions fromthe source.

(a) Which one of the following best describes the motion of the air particles at a distanceof 12.0 m from the trumpet?

A. The air particles are vibrating in the direction of motion of the sound waves.

B. The air particles are vibrating at right angles to the direction of motion of thesound waves.

C. The air particles are moving away from the trumpet at 340 m s1.

D. The air particles are moving away from the trumpet with a speed greater than340 m s1.

E. The air particles are moving away from the trumpet with a speed less than340 m s1.

(b) What is the sound level, in dB, corresponding to an intensity of 2.0 x 107W m2?

(c) What is the intensity of sound in, W m2produced by a trumpet with a loudnesslevel of 30 dB?


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6. The figure below shows the trace produced on a cathode ray oscilloscope when a microphonepicks up a single note sung by an opera singer.

Draw on the above graph, the oscilloscope trace that would result from:

(a) a louder sound of the same frequency

(b) a note of the same amplitude but with half the frequency

(c) a sound from an audio frequency oscillator having the same pitch as the singer.

7. A sound wave consists of regions of air pressure that alternate from slightly higher to slightlylower than normal. At one particular instant, the variation in the air pressure of a sound wavea long way from its source is as shown below.

(a) What is the frequency of the sound wave?

(b) Which of the points A to H in the diagram correspond to point(s) where the soundwave is causing zero displacement of the air particles?

(c) Show the pressure variations of the wave:

(i) a quarter of a period later

(ii) half a period later.

(d) The intensity of a sound decreases markedly with distance from a source. Why?


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REFLECTION OF SOUNDS

Sound waves are able to reflect, or bounce off, objects. This property is most obvious when echoesare heard. Echoes can be used to determine how far away objects are. The time it takes for the soundwave to bounce back to where it came from determines the distance with a shorter time meaning ashorter distance

Example: A person shouts "Hello" into a cave and hears an echo after one second. How deep is thecave if the speed of sound is 300 m/s?

If the total time of travel is one second then it must have taken the sound 0.5 seconds to reach the cavewall and 0.5 seconds to come back. If sound travels at 300 m/s then in 0.5 seconds it would travel

0.5 x 300 = 150 metres

Water waves can be used to demonstrate the law of reflection.

The angle of incidence and the angle of reflection are measured between the normal and the direction

of travel of the waves.

TRANSMISSION & ABSORPTION

When sound waves strike a material the energy can be transmitted, absorbed and reflected. Theamount that is reflected depends on the relative acoustic impedance of the materials. If these valuesare similar there is little reflection. Air is a substance with little acoustic impedance. Some of thesound that is not reflected will be absorbed by the material and changed into other forms of energyand some will be transmitted through the material. The amount of absorption depends on theabsorption coefficient of the material. Soft materials are usually the best absorbers.

In many theatres curtains cover the brick walls to prevent the reflection of sound by absorption. Thebrick walls reflect outside sounds, which prevents noise from entering the theatre.


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DIFFRACTION OF WAVES

Aim:

The aim of this experiment is to observe the way in which water waves diffract.

Apparatus:

Ripple tank, light source, straight barriers, paper, and wave generator

Method:

The light source is placed above the ripple tank that is filled with water. The sheet of paper is placedunder the ripple tank. The behaviour of the waves is noticed by looking at the paper.

Part 1 - Waves passing an obstacle

Use the wave generator to produce straight pulses.

(i) Direct the waves towardsa straight barrier. Thepattern seen should besimilar to that shown.

(ii) Do the waves move into the region of water directly behind the barrier?


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(iii) How does the bending of the waves changea with a change in the wavelength of the waves?Label the following diagrams appropriately.

Part 2 - Waves passing through a gap

Use the wave generator to produce straight pulses. Place two barriers in the water so that there is agap of about 2-cm between them.

(i) Direct the waves directly towards thegap and use a diagram to show howthe waves appeared after they passed thegap. The pattern seen should be similarto that shown.

(ii) Do the waves move into the region of water directly behind the barriers?


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(iii) How does the bending of the wave change with a change in the wavelength of the waves?Label the following diagrams appropriately.


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(iv) How does the bending of the wave change with a change in the size of the gap? Label thefollowing diagrams appropriately.

Medium gap


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DIFFRACTION OF SOUND

Diffraction is the way waves travel around corners. When waves are forced through a narrow gap theyemerge as circular pulses. This is more obvious when the gap is very small compared with thewavelength. Diffraction is seen for water waves.

Sound waves have the ability to bend around corners or diffract like water waves. Low pitch soundsbend more than high pitch sounds, and the bending is greatest when the gap through which the soundwaves travel is smallest. This is why when music is heard from a stereo from outside an open windowthe high-pitched sounds are only clearly heard when the listener is opposite the window.

PROBLEMS:

1. Parallel water waves of wavelength 2.0 cm are incident on a gap between two barriers. Thewidth of the gap is about 5 cm.

(a) Draw the shape of the wave pattern after the waves have passed through the aperture.

(b) The wavelength is now increased until the wavelength is approximately 5 cm. Whateffect will this have on the wave pattern behind the aperture?

(c) What phenomenon is illustrated by this example?

2. What effect do you expect on the bending of the waves (i.e. the amount of diffraction), when:

(a) the slit is made narrower without changing the wavelength?

(b) the slit is made wider without changing the wavelength?

(c) the wavelength is increased without changing the width of the slit?

(d) the wavelength is decreased without changing the width of the slit?


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3. Waves diffract around the edge of a barrier or obstacle. What effect does decreasing thewavelength have on the amount of diffraction?

4. If the slit width is less than the wavelength then which of the following statements are true forwater waves in a ripple tank? (One or more answers)

A. No diffraction occurs.B. The slit acts as a point source.C. The sides of the slit cast a sharp shadow.D. The water waves are strongly diffracted.

5. Waves in a ripple tank with a constant depth of water approach (a) a narrow opening and (b) awide opening in a barrier. Sketch the shape of the waves after they have passed through thebarrier, for each of the two cases.


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INTERFERENCE OF WAVES

The aim of this exercise is to relate the position of nodal lines produced by interference to the pathdifference as measured by the distance from points on the nodal lines to the sources of the waves.

The diagram below is a photograph of waves produced in a ripple tank.

X Y

The first 3 nodal lines to the left of the central maximum are to be drawn on the sheet of paper as wellas the positions of the sources X & Y.

Five points are to be marked on the diagram (A, B, C, D & E) which indicate five different points onthe first nodal line to the left of the central maximum.

A point F is to be marked such that it is on the second nodal line to the left of the central maximumand as far as possible from the two sources.

A point G is to be marked such that it is on the third nodal line to the left of the central maximum andas far as possible from the two sources.

The distances from X & Y to each of the points A - G are then to be measured and the pathdifferences calculated.


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Results:

Wavelength of waves = cm

TABLE 1

Line Distance (cm) Line Distance (cm) Path difference(cm)

AX AY

BX BY

CX CY

DX DY

EX EY

TABLE 2

Line Distance (cm) Line Distance (cm) Path difference

(cm)EX EY

FX FY

GX GY

Analysis of results:

1. What do you notice about the values for the path difference in Table 1?

2. Calculate the average path difference for Table 1 and then relate this value to the wavelengthof the waves.

3. What do the results obtained in Q1 & 2 suggest about the path difference for points on thefirst nodal line?

4. For Table 2, relate the path differences obtained to the wavelength of the waves. What doesthis suggest about points on nodal lines and their path differences?

5. Given that maximums occur between nodal lines, suggest how the path differences of pointson maximums relate in general to wavelength of the waves.


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INTERFERENCE

When two or more sets of waves meet, they will mix with each other. This phenomenon is calledinterference. The resulting interference pattern is the sum of these waves. This is superposition. If twospeakers produce the same note there will be points where the sound is loud and other points where nosound is heard.

Consider water waves with crests and troughs being emitted at the same time. When a crest meets acrest or a trough meets a trough a double crest or trough will be produced if the amplitudes are equal.If a trough meets a crest they will cancel each other out resulting

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12VCE Physics Notes 2014

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