PHY 101 Elementary Physics

Purpose: Introduce the students to the general concepts in physics.

Expected Learning Outcomes

At the end of the course the student should be able to:

Explain the theory and behaviour of physical objects.

Observe and describe the properties of heat.

Observe and describe the properties of optics.

Content

Units and Dimensions.

Kinematics. Dynamics. Circular motion. Simple harmonic motion (SHM).

Gravitation of rigid body motion.

Kinetic and static friction. Calorimetry, gases thermal expansion and change

of state.

Transfer of heat, thermometry.

Optics: reflection at plane surfaces, refraction through prism, dispersion

spectra, refraction through lenses, optical instrument, velocity of light,

photometry.

Oscillations and waves, sound and light interference, diffraction, polarisation,

quantum concepts.

Learning and Teaching Methodologies

Teaching is by lecture method consisting of two lecture hours and one hour

of tutorials/two laboratory hours = 1 lecture hour weekly for thirteen weeks.

Assessment:

Type Weighting (%)

Examination 70%

Continuous Assessment 30%

Total 100%

Recommended Texts

1. Hugh D. Young, Roger A. Freedman, Lewis Ford (2006). University

Physics with Modern Physics with Mastering Physics (12th ed.).

Addison Wesley.

2. Hugh D. Young, Roger A. Freedman, T. R. Sandin, A. Lewis Ford

(1999). Sears and Zemansky's University Physics (10th ed.). Addison

Wesley Publishing Company. ISBN: 0201603225.

Contents

PHY 101 Elementary Physics 1

Learning and Teaching Methodologies.............................................................................1Assessment:.......................................................................................................................1Recommended Texts.........................................................................................................2

KINEMATICS 3

Displacement.....................................................................................................................4Speed, Velocity, and Acceleration.....................................................................................5Instantaneous Speed and Velocity.....................................................................................5Acceleration.......................................................................................................................6Kinematics with Graphs....................................................................................................6Position vs. Time Graphs...................................................................................................6Calculating Velocity..........................................................................................................7Average Velocity...............................................................................................................8Vectors...............................................................................................................................8

What’s a Vector?......................................................................................................8Vectors vs. Scalars...................................................................................................8Vector Addition........................................................................................................8Tip-to-Tail Method...................................................................................................8Parallelogram Method..............................................................................................9Adding Vector Magnitudes......................................................................................9Adding Perpendicular Vectors.................................................................................9

Adding Parallel Vectors.........................................................................................10Multiplication by a Scalar......................................................................................10Vector Multiplication.............................................................................................11

Dot Product......................................................................................................................11The Cross Product...........................................................................................................12

One-Dimensional Motion with Uniform Acceleration 12

Two-Dimensional Motion with Uniform Acceleration 14

Unit summary............................................................................................................16

Unit 2 22

Circular motion and rotation 22

Introduction.....................................................................................................................22Centripetal Force.............................................................................................................23

Examples..........................................................................................................23ROTATIONAL KINEMATICS......................................................................................25

Angular Position.....................................................................................................25Angular Displacement............................................................................................26Angular Velocity....................................................................................................26Angular Acceleration.............................................................................................27Angular Frequency.................................................................................................27Angular Period.......................................................................................................27Torque Defined as a Vector Quantity....................................................................28Problems.................................................................................................................30Projectile Motion...................................................................31 Maximum Height...................................................................................................32

Time of Flight..................................................................................................................33Range...............................................................................................................................33Maximum Range with Angle..........................................................................................34

Simple Harmonic Motion (SHM) 35

Introduction.....................................................................................................................35Describing oscillations....................................................................................................35Motion of a simple harmonic oscillator...........................................................................36Graphs of simple harmonic motion.................................................................................37Computing oscillator motion step by step.......................................................................38Elastic potential energy...................................................................................................40Energy flows in an oscillator...........................................................................................41Resonance........................................................................................................................42Period, Amplitude and Frequency...................................................................................42Equations of Simple Harmonic Motion...........................................................................43Displacement, velocity and acceleration vectors of a particle undergoing simple harmonic motion..............................................................................................................................43Examples of SHM...........................................................................................................44

Spring Mass System...............................................................................................44Simple Pendulum...................................................................................................45Torsional Pendulum...............................................................................................46

Summary......................................................................................................................47WORKED EXAMPLES........................................................................................47Problems.................................................................................................................48

Static Friction Vs Kinetic Friction 51

Coefficient of Friction.....................................................................................................53

Optics 54

Introduction 54

Thin Lenses......................................................................................................................54Mirrors.............................................................................................................................57Ray Tracing.....................................................................................................................58Problems on Mirrors and Lenses.....................................................................................59Example...........................................................................................................................59Reflection at plane surfaces.............................................................................................60

The Law of Reflection...........................................................................................60

Reflection.........................................................................................................................61Refraction........................................................................................................................62Dispersion........................................................................................................................63

Optical Instruments 64

General Features of Mirrors and Lenses..........................................................................65Concave Mirrors..............................................................................................................65Convex Mirrors................................................................................................................67The Two Equations for Mirrors and Lenses....................................................................67First Equation: Focal Length...........................................................................................68Second Equation: Magnification.....................................................................................68Convex Lenses.................................................................................................................69Concave Lenses...............................................................................................................71

WAVE OPTICS 72

Young’s Double-Slit Experiment....................................................................................72Diffraction.......................................................................................................................76

Single-Slit Diffraction............................................................................................77Polarization......................................................................................................................77

KINEMATICS

Kinematics derives its name from the Greek word for “motion,” kinema. Kinematics provides us with the language and the mathematical tools to describe motion, whether the motion of a charging pachyderm or a charged particle. Kinematics is the study of how things move. Here, we are interested in the motion of normal objects in our world. A normal object is visible, has edges, and has a location that can be expressed with (x, y, z) coordinates. We will not be discussing the motion of atomic particles or of light.

Kinematics  -  Kinematics is concerned with describing the way in which

objects move.

Displacement -  An objects total change in position. If a man runs around an

oval 400 meter track, stopping at the precise location he began, though he ran

a distance of 400 meters, his total displacement was 0.

Dynamics -  Dynamics focuses on understanding why objects move the way

they do.

Reference frame -  The coordinate system with respect to which motion is

being described.

Speed -  A measure of how fast an object is moving.

Average velocity -  The time-average of the velocity function over a specified

time-interval. (See formula below.)

Instantaneous velocity -  The value of the velocity function at a particular

instant in time. (See formula below.)

Gravitational acceleration  -  The gravitational acceleration of objects near

the earth's surface is the same for all objects regardless of mass and is given by

the number g = 9.8m/s2 .

Scalar-valued function -  A function that outputs scalars (regular numbers).

Most common functions that you are probably familiar with are scalar-valued

functions.

Vector-valued function -  A function that outputs vectors. This means that

while the domain of the function may consist of scalars, the values in the range

are all vectors.

Position function -  A position function can be either scalar-valued (for motion

in one dimension) or vector-valued (for motion in two or three dimensions). At

each point in time its value represents the position of an object at that time.

Velocity function - This function is the time-derivative of the position

function, and gives the velocity of an object at each point in time.

Acceleration function - This function is the time-derivative of the velocity

function, and the second time-derivative of the position function. It gives the

value of the acceleration of an object at each point in time.

Time-derivative - The time-derivative of a function is a new function whose

value at each point represents the rate of change of the original function with

respect to time.

Simple harmonic motion - Periodic motion that can be described by special

types of position functions. Examples of simple harmonic motion include an

object moving in a circle and a ball bouncing up and down on a spring.

Displacement:

An objects total change in position. If a man runs around an oval 400 meter track, stopping at the precise location he began, though he ran a distance of 400 meters, his total displacement was 0.

Displacement

Displacement is a vector quantity, commonly denoted by the vector s that

reflects an object’s change in spatial position. The displacement of an object

that moves from point A to point B is a vector whose tail is at A and whose tip

is at B. The distance that the object travels is equal to the length of path AB.

Speed, Velocity, and Acceleration

Speed is a measure of the distance an object travels in a given length of time:

Velocity is a vector quantity defined as rate of change of the displacement

vector over time:

Average velocity =

Instantaneous Speed and Velocity

The instantaneous velocity or speed at a given moment. We want to know

how fast that object is moving right now. Instantaneous velocity is not a

tricky concept: we simply take the equation above and assume that is very,

very small.

Example

Which of the follow sentences contains an example of instantaneous velocity?

A. “The car covered 500 kilometres in the first 10 hours of its northward

journey.”

B. “Five seconds into the launch, the rocket was shooting upward at 5000

meters per second.” C. “The cheetah can run at 70 miles per hour.”

D. “Moving at five kilometres per hour, it will take us eight hours to get to

the base camp.”

E. “Roger Bannister was the first person to run one mile in less than four

minutes.”

Instantaneous velocity has a magnitude and a direction, and deals with the

velocity at a particular instant in time.

Acceleration

Speed and velocity only deal with movement at a constant rate.

Acceleration is a vector quantity that measures the rate of change of the

velocity vector with time:

Average acceleration =

Kinematics with Graphs

A common way of testing kinematics qualitatively is to present you with a

graph plotting position vs. time, velocity vs. time, or acceleration vs. time

and to ask you questions about the motion of the object represented by the

graph.

Position vs. Time Graphs

Positions vs. time graphs give you an easy and obvious way of determining an

object’s displacement at any given time, and a subtler way of determining

that object’s velocity at any given time. Let’s put these concepts into

practice;

Any point on this graph gives us the position of the ant at a particular

moment in time. For instance, the point at (2,–2) tells us that, two seconds

after it started moving, the ant was two centimetres to the left of its starting

position, and the point at (3,1) tells us that, three seconds after it started

moving, the ant is one centimetre to the right of its starting position.

Let’s read what the graph can tell us about the ant’s movements. For the first

two seconds, the ant is moving to the left. Then, in the next second, it

reverses its direction and moves quickly to y = 1. The ant then stays still at y

= 1 for three seconds before it turns left again and moves back to where it

started. Note how concisely the graph displays all this information.

Calculating Velocity

Velocity measures the rate of change of displacement over time. If

displacement is given here by the vector y, then the velocity of the ant is

Average velocity during this time interval is a simple matter of dividing rise

by run, as we’ve learned in math class.

Average Velocity

How about the average velocity between t = 0 and t = 3? It’s actually easier

to sort this out with a graph in front of us, because it’s easy to see the

displacement at t = 0 and t = 3,

Vectors

What’s a Vector?

A vector is a mathematical object possessing, and fully described by, a

magnitude and a direction.

Vectors vs. Scalars

In contrast to a vector quantity, a scalar quantity does not have a direction;

it is fully described by just a magnitude. Examples of scalar quantities

include the number of words in this sentence and the mass of the Hubble

Space Telescope. Vector quantities you’ll likely come across quite frequently

in physics include displacement, s; velocity, v; acceleration, a; force, F;

momentum, p; electric field, E; and magnetic field, B.

Vector Addition

The easiest way to learn how vector addition works is to look at it

graphically. There are two equivalent ways to add vectors graphically: the

tip-to-tail method and the parallelogram method. Both will get you to the

same result, but one or the other is more convenient depending on the

circumstances.

Tip-to-Tail Method

We can add any two vectors, A and B, by placing the tail of B so that it meets

the tip of A. The sum, A + B, is the vector from the tail of A to the tip of B.

Note that you’ll get the same vector if you place the tip of B against the tail

of A. In other words, A + B and B + A are equivalent.

Parallelogram Method

To add A and B using the parallelogram method, place the tail of B so that it

meets the tail of A. Take these two vectors to be the first two adjacent sides

of a parallelogram, and draw in the remaining two sides. The vector sum, A

+ B, extends from the tails of A and B across the diagonal to the opposite

corner of the parallelogram. If the vectors are perpendicular and unequal in

magnitude, the parallelogram will be a rectangle. If the vectors are

perpendicular and equal in magnitude, the parallelogram will be a square.

Adding Vector Magnitudes

Of course, knowing what the sum of two vectors looks like is often not

enough. Sometimes you’ll need to know the magnitude of the resultant

vector. This, of course, depends not only on the magnitude of the two vectors

you’re adding, but also on the angle between the two vectors.

Adding Perpendicular Vectors

Suppose vector A has a magnitude of 8, and vector B is perpendicular to A

with a magnitude of 6. What is the magnitude of A + B? Since vectors A and

B are perpendicular, the triangle formed by A, B, and A + B is a right

triangle. We can use the Pythagorean Theorem to calculate the magnitude of

A + B, which is

Adding Parallel Vectors

If the vectors you want to add are in the same direction, they can be added

using simple arithmetic. For example, if you get in your car and drive eight

miles east, stop for a break, and then drive six miles east, you will be 8 + 6 =

14 miles east of your origin. If you drive eight miles east and then six miles

west, you will end up 8 – 6 = 2 miles east of your origin.

Multiplication by a Scalar

Multiplication is like repeated addition. Multiplying 4 by 3 means adding four

three times: . The multiplication of a vector times a scalar

works in the same way. Multiplying the vector A by the positive scalar c is

equivalent to adding together c copies of the vector A. Thus 3A = A + A + A.

Multiplying a vector by a scalar will get you a vector with the same direction,

but different magnitude, as the original.

Vector Multiplication

There are two forms of vector multiplication: one results in a scalar, and one

results in a vector.

Dot Product

The dot product, also called the scalar product, takes two vectors,

“multiplies” them together, and produces a scalar. The smaller the angle

between the two vectors, the greater their dot product will be.

The dot product of any two vectors, A and B, is expressed by the equation:

Where is the angle made by A and B when they are placed tail to tail

The dot product of A and B is the value you would get by multiplying the

magnitude of A by the magnitude of the component of B that runs parallel to

A. Looking at the figure above, you can get A · B by multiplying the

magnitude of A by the magnitude of , which equals . You would get

the same result if you multiplied the magnitude of B by the magnitude of ,

which equals .

The Cross Product

The cross product, also called the vector product, “multiplies” two vectors

together to produce a third vector, which is perpendicular to both of the

original vectors. The closer the angle between the two vectors is to the

perpendicular, the greater the cross product will be. We encounter the cross

product a great deal in our discussions of magnetic fields. Magnetic force

acts perpendicular both to the magnetic field that produces the force, and to

the charged particles experiencing the force.

The cross product of two vectors, A and B, is defined by the equation:

Where is a unit vector perpendicular to both A and B. The magnitude of the

cross product vector is equal to the area made by a parallelogram of A and

B. In other words, the greater the area of the parallelogram, the longer the

cross product vector.

One-Dimensional Motion with Uniform Acceleration

Most problems will involve objects moving in a straight line whose

acceleration doesn’t change over time. For such problems, there are five

variables that are potentially relevant: the object’s position, x; the object’s

initial velocity ; the object’s final velocity, v; the object’s acceleration, a;

and the elapsed time, t. if you know any three of these variables, you can

solve for a fourth. Here are the five kinematics equations;

The variable represents the object’s position at t = 0. Usually, = 0.

When They Say . . . They

Mean . . .

“. . . starts from rest . . .”

“. . . moves at a constant

velocity . . .”a = 0

“. . . comes to rest . . . ” v = 0

Example

A student throws a ball up in the air with an initial velocity of 12 m/s and

then catches it as it comes back down to him. What is the ball’s velocity

when he catches it? How high does the ball travel? How long does it take the

ball to reach its highest point?

ANSWER

We know the initial velocity, m/s, and the acceleration due to gravity,

m/s2, and we know that the displacement is x = 0 since the ball’s final

position is back in the student’s hand where it started. We need to know the

ball’s final velocity, v, so we should look at the kinematics equation that

leaves out time, t:

Because both x and are zero, the equation comes out to

Solving for x:

How long does it take the ball to reach its highest point?

Let’s choose the one that leaves out x:

Two-Dimensional Motion with Uniform Acceleration

If you’ve got the hang of 1-D motion, you should have no trouble at all with 2-

D motion. The motion of any object moving in two dimensions can be broken

into x- and y-components. Then it’s just a matter of solving two separate 1-D

kinematics equations.

If we break this motion into x- and y-components, the motion becomes easy

to understand. In the y direction, the ball is thrown upward with an initial

velocity of and experiences a constant downward acceleration of g = –9.8

m/s2. This is exactly the kind of motion we examined in the previous section:

if we ignore the x-component, the motion of a projectile is identical to the

motion of an object thrown directly up in the air.

In the x direction, the ball is thrown forward with an initial velocity of and

there is no acceleration acting in the x direction to change this velocity. We

have a very simple situation where and is constant.

We can calculate the x- and y-components separately and then combine them

to find the velocity of the projectile at any given point:

Because is constant, the speed will be greater or lesser depending on the

magnitude of . To determine where the speed is least or greatest, we follow

the same method as we would with the one-dimensional example we had in

the previous section. That means that the speed of the projectile in the figure

above is at its greatest at position F, and at its least at position C.

Unit summary

In this unit you learned Formulae

Key Formulas

Angular Position

Definition of a Radian

Average Angular Velocity

Average Angular

Acceleration

Angular Frequency

Angular Period

Relations between Linear

and Angular Variables

Equations for Rotational

and Angular Kinematics

with Constant Acceleration

Torque As Trigonometric

Function

Component Form of the

Torque Equation

Torque As Cross Product

Newton’s Second Law in

Terms of Rotational Motion

Moment of Inertia

Kinetic Energy of Rotation

Angular Momentum of a

Particle

Component Form of the

Angular Momentum of a

Particle

Angular Momentum of a

Rotating Rigid Body

1An object that experiences 120 revolutions per minute experiences 2

revolutions per second; in other words, it rotates with a frequency of 2 Hz.

We have formulas relating frequency to angular velocity and angular velocity

to linear velocity, so solving this problem is simply a matter of finding an

expression for linear velocity in terms of frequency. Angular and linear

velocity are related by the formula , so we need to plug this formula

into the formula relating frequency and angular velocity:

2.     Frequency and angular velocity are related by the formula , and

angular velocity and angular acceleration are related by the formula .

In order to calculate the washing machine’s acceleration, then, we must

calculate its angular velocity, and divide that number by the amount of time

it takes to reach that velocity:

3 You need to apply the right-hand rule in order to solve this problem.

Extend the fingers of your right hand upward so that they point to the 0-

second point on the clock face, and then curl them around so that they point

downward to the 30-second point on the clock face. In order to do this, you’ll

find that your thumb must be pointing inward toward the clock face. This is

the direction of the angular velocity vector.

4.    The torque on an object is given by the formula , where F is the

applied force and r is the distance of the applied force from the axis of

rotation. In order to maximize this cross product, we need to maximize the

two quantities and insure that they are perpendicular to one another.

Statement I maximizes F and statement III demands that F and r be

perpendicular, but statement II minimizes r rather than maximizes it, so

statement II is false.

5.    The torque acting on the pendulum is the product of the force acting

perpendicular to the radius of the pendulum and the radius, . A free-

body diagram of the pendulum shows us that the force acting perpendicular

to the radius is .

Since torque is the product of and R, the torque is .

6 The seesaw is in equilibrium when the net torque acting on it is zero. Since

both objects are exerting a force perpendicular to the seesaw, the torque is

equal to . The 3 kg mass exerts a torque of N · m in the

clockwise direction. The second mass exerts a torque in the counterclockwise

direction. If we know this torque also has a magnitude of 30g N · m, we can

solve for m:

7.      The rotational equivalent of Newton’s Second Law states that .

We are told that N · m and I = 1/2 MR2, so now we can solve for :

8.     At the top of the incline, the disk has no kinetic energy, and a

gravitational potential energy of mgh. At the bottom of the incline, all this

gravitational potential energy has been converted into kinetic energy.

However, in rolling down the hill, only some of this potential energy becomes

translational kinetic energy, and the rest becomes rotational kinetic energy.

Translational kinetic energy is given by 1 /2 mv2 and rotational kinetic energy

is given by 1 /2 I 2. We can express in terms of v and R with the equation =

v/R, and in the question we were told that I = 1/2 mR2. We now have all the

information we need to solve for v:

9 This is a conservation of momentum question. The angular momentum of

the rock as it is launched is equal to its momentum after it’s been launched.

The momentum of the rock-basket system as it swings around is:

The rock will have the same momentum as it leaves the basket. The angular

momentum of a single particle is given by the formula L = mvr. Since L is

conserved, we can manipulate this formula and solve for v:

Be sure to remember that the initial mass of the basket-rock system is 250

kg, while the final mass of the rock is only 200 kg.

10 Angular momentum, , is a conserved quantity, meaning that the

greater I is, the less will be, and vice versa. In order to maximize angular

velocity, then, it is necessary to minimize the moment of inertia. Since the

moment of inertia is greater the farther the mass of a body is from its axis of

rotation, we can maximize angular velocity by concentrating all the mass

near the axis of rotation.

Unit 2

Circular motion and rotation

Introduction

Before discussing the dynamics of uniform circular motion, we must explore its kinematics. Because the direction of a particle moving in a circle changes at a constant rate, it must experience uniform acceleration. But in what direction is the particle accelerated? To find this direction, we need only look at the change in velocity over a short period of time:

Figure %: A particle in Uniform Circular Motion

The diagram above shows the velocity vector of a particle in uniform

circular motion at two instants of time. By vector addition we can see that

the change in velocity, Δv , points toward the centre of the circle. Since

acceleration is the change in velocity over a given period of time, the

consequent acceleration points in the same direction. Thus we define

centripetal acceleration as acceleration towards the centre of a circular

path. All objects in uniform circular motion must experience some form of

uniform centripetal acceleration.

We find the magnitude of this acceleration by comparing ratios of velocity and position around the circle. Since the particle is travelling in a circular path, the ratio of the change in velocity to velocity will be the same as the ratio of the change in position to position. Thus:

= =

Rearranging the equation, =

Thus a =

We now have a definition for both the magnitude and direction of

centripetal acceleration: it always points towards the centre of the circle,

and has a magnitude of v 2/r.

The magnitude of the tension of the string (and therefore the acceleration of the ball) varies according to velocity and radius. If the ball is moving at a high velocity, the equation implies, a large amount of tension is required and the ball will experience a large acceleration. If the radius is very small, the equation shows, the ball will also be accelerated more rapidly.

Centripetal Force

Centripetal force is the force that causes centripetal acceleration. By using Newton's Second Law in conjunction with the equation for centripetal acceleration, we can easily generate an expression for centripetal force.

F c = ma =

Examples

(a)A 2 kg ball on a string is rotated about a circle of radius 10 m. The maximum tension allowed in the string is 50 N. What is the maximum speed of the ball?

Solution

The centripetal force in this case is provided entirely by the tension in the

string. If the maximum value of the tension is 50 N, and the radius is set at

10 m we only need to plug these two values into the equation for centripetal

force:

T = F c = implies that v =

thus

v = = 15.8 m/s

(b)During the course of a turn, an automobile doubles its speed. How much additional frictional force must the tires provide if the car safely makes around the curve?

Solution

Since F c varies with v 2, an increase in velocity by a factor of two must be

accompanied by an increase in centripetal force by a factor of four.

(c) A satellite is said to be in geosynchronous orbit if it rotates around the earth once every day. For the earth, all satellites in geosynchronous orbit must rotate at a distance of 4.23×107 meters from the earth's centre. What is the magnitude of the acceleration felt by a geosynchronous satellite?

Solution

The acceleration felt by any object in uniform circular motion is given by a

= . We are given the radius but must find the velocity of the satellite. We

know that in one day, or 86400 seconds, the satellite travels around the

earth once. Thus:

v = = = = 3076 m/s

Thus

a = = = .224 m/s2

(d)The maximum lift provided by a 500 kg airplane is 10000 N. If the plane travels at 100 m/s, what is its shortest possible turning radius?

Solution

Again, we use the equation F c = . Rearranging, we find that r = .

Plugging in the maximum value for the lift of the plane, we find that

R min = = 500m

ROTATIONAL KINEMATICS

Angular Position

By convention, we measure angles in a circle in a counter clockwise

direction from the positive x-axis. The angular position of a particle is the

angle, made between the line connecting that particle to the origin, O, and

the positive x-axis, measured counter clockwise. Let’s take the example of a

point P on a rotating wheel:

In this figure, point P has an angular position of . Note that every point on

the line has the same angular position: the angular position of a point

does not depend on how far that point is from the origin, O.

We can relate the angular position of P to the length of the arc of the circle

between P and the x-axis by means of an easy equation:

In this equation, l is the length of the arc, and r is the radius of the circle.

Angular Displacement

Now imagine that the wheel is rotated so that every point on line moves

from an initial angular position of to a final angular position of .

The angular displacement, of line is:

For example, if you rotate a wheel counter clockwise such that the angular

position of line changes from = 45º = π/4 to = 135º = 3π/4, as

illustrated below, then the angular displacement of line is 90º or π/2

radians.

For line to move in the way described above, every point along the line

must rotate 90º counter clockwise. By definition, the particles that make up

a rigid body must stay in the same relative position to one another. As a

result, the angular displacement is the same for every point in a rotating

rigid body.

Angular Velocity

Angular velocity, , is defined as the change in the angular displacement

over time. Average angular velocity, is defined by:

Angular velocity is typically given in units of rad/s. As with angular

displacement, the angular velocity of every point on a rotating object is

identical.

Angular Acceleration

Angular acceleration is defined as the rate of change of angular velocity

over time. Average angular acceleration, is defined by:

Angular acceleration is typically given in units of rad/s2

Angular Frequency

Angular frequency, f, is defined as the number of circular revolutions in a

given time interval. It is commonly measured in units of Hertz (Hz), where 1

Hz = 1 s–1. For example, the second hand on a clock completes one

revolution every 60 seconds and therefore has an angular frequency of 1 /60

Hz.

The relationship between frequency and angular velocity is:

For example, the second hand of a clock has an angular velocity of

s. Plugging that value into the equation above, we get

Which we already determined to be the frequency of the second hand of a

clock.

Angular Period

Angular period, T, is defined as the time required to complete one

revolution and is related to frequency by the equation:

Since we know that the frequency of the second hand is 1/60 Hz, we can

quickly see that the period of the second hand is 60 s. It takes 60 seconds

for the second hand to complete a revolution, so the period of the second

hand is 60 seconds. Period and angular velocity are related by the equation

Torque Defined as a Vector Quantity

Torque, like angular velocity and angular acceleration, is a vector quantity.

Most precisely, it is the cross product of the displacement vector, r, from

the axis of rotation to the point where the force is applied, and the vector

for the applied force, F.

To determine the direction of the torque vector, use the right-hand rule,

curling your fingers around from the r vector over to the F vector. In the

example of lifting the lever, the torque would be represented by a vector at

O pointing out of the page.

Example

A student exerts a force of 50 N on a lever at a distance 0.4 m from its axis

of rotation. The student pulls at an angle that is 60º above the lever arm.

What is the torque experienced by the lever arm?

Let’s plug these values into the first equation we saw for torque:

Key Formulas

Angular Position

Definition of a

Radian

Average Angular

Velocity

Average Angular

Acceleration

Angular

Frequency

Angular Period

Relations

between Linear

and Angular

Variables

Equations for

Rotational and

Angular

Kinematics with

Constant

Acceleration

Torque As

Trigonometric

Function

Component Form

of the Torque

Equation

Torque As Cross

Product

Newton’s Second

Law in Terms of

Rotational

Motion

Moment of

Inertia

Kinetic Energy of

Rotation

Angular

Momentum of a

Particle

Component Form

of the Angular

Momentum of a

Particle

Angular

Momentum of a

Rotating Rigid

Body

Problems

Problem

A 2 kg ball on a string is rotated about a circle of radius 10 m. The maximum tension allowed in the string is 50 N. What is the maximum speed of the ball?

Problem :

During the course of a turn, an automobile doubles its speed. How much additional frictional force must the tires provide if the car safely makes around the curve?

Problem :

A satellite is said to be in geosynchronous orbit if it rotates around the earth once every day. For the earth, all satellites in geosynchronous orbit must

rotate at a distance of 4.23×107 meters from the earth's center. What is the magnitude of the acceleration felt by a geosynchronous satellite?

Problem :

The maximum lift provided by a 500 kg airplane is 10000 N. If the plane travels at 100 m/s, what is its shortest possible turning radius?

Problem :

A popular daredevil trick is to complete a vertical loop on a motorcycle. This

trick is dangerous, however, because if the motorcycle does not travel with

enough speed, the rider falls off the track before reaching the top of the

loop. What is the minimum speed necessary for a rider to successfully go

around a vertical loop of 10 meters?

Projectile Motion

Projectile motion occurs when objects are fired at some initial velocity or

dropped and move under the influence of gravity.

One of the most important things to remember about projectile motion is that the effect of gravity is independent on the horizontal motion of the object. The independence of the horizontal and vertical components of the motion of the object is often used in these type of physics problems.

It also accounts for the counterintuitive observation that a bullet fired horizontally from a gun takes the same time to fall to the ground as a bullet that is dropped from the same height.

To be able to solve projectile problems you will often need to resolve the motion into its components.

The horizontal motion is given by:

While the vertical motion is:

Often question about projectiles will ask you to find the maximum height, or the range of the projectile, or even how long it took for the projectile to hit the ground.

Maximum Height

When the projectile reaches its maximum height its vertical component of velocity will be zero. We can use the linear equations of motion to calculate this height, h.

From where v is the final velocity, u is the initial velocity, a is the acceleration in this case provided by gravity and s is the distance covered or in our case the height h.

The final velocity is zero. So, substituting our symbols we have

But vy is v0 sin θ

Time of Flight

We can use the independence of the horizontal and vertical motions of the projectile to determine the time of flight. If we know the initial velocity, we can work out the time. The vertical motion looks like that of a ball being thrown upward. At its highest point the velocity will be zero. We just need to find the time that it takes and multiply by two, since what goes up, must come down.

Recall from the linear equations of motion, v = u + a t

So, rearranging,

Therefore,

Range

Since the trajectory in the absense of air resistance is symmetrical about the point of maximum height, it will be half way through its motion so the range will be twice the distance it takes to travel in this time.

The horizontal distance is just,

Inserting the horizontal component of velocity

and using the argument in the previous section to calculate the time to reach the maximum height, not forgeting to multiply by 2, we obtain

The identity can be used to simplify the expresion further.

If the projectile is fired at some height h0 above the ground that it lands on we must find the additional time that it stays in the air.

To find the time involves solving this quadratic equation in t.

We are only interested in the positive solution since a negative value for time is not physical.

To check that this expression is correct, it should return to the expression for range on level ground when h0 = 0.

Maximum Range with Angle

But what about the greatest range? For a projectile fired on level ground and neglecting air resistance, the range is determined from the equation

Looking at this expression can see it is a constant multiplied by a sine function. It will have a maximum when sine is 1. We know this happens when theta = 90°. So to get 90° theta must be 45°. The maximum range would be

If we didn't know this, we could differentiate the equation for range with respect to θ and set the result equal to zero.

If you plot this over a range from 0 to 90°, it is zero at 45°. So the angle that will give the greatest range is 45°

Simple Harmonic Motion (SHM)

Introduction

In addition to linear motion and rotational motion there is another kind of motion that is common in physics. This is the to and fro motion of oscilations or vibrations.

When something oscillates, it moves back and forth with time. It is helpful to trace out the position of an oscillating particle with time so we can define some terminology.

Describing oscillations

Language to describe oscillations

+A

0

–Aperiodic time Tphase changes by 2

Aangle t

Sinusoidal oscillation

time t

amplitude A

f turns persecond

= 2f radian per second

2 radianper turn

Phasor picture

s = A sin t

Periodic time T, frequency f, angular frequency :

f = 1/T unit of frequency Hz = 2f

Equation of sinusoidal oscillation:

s = A sin 2ft s = A sin t

Phase difference /2

s = A s in 2fts = 0 when t = 0

s = A cos 2fts = A when t = 0

t = 0

sand falling from a swinging pendulum leavesa trace of its motion on a moving track

Motion of a simple harmonic oscillator

Motion of harmonic oscillator

large force to left

large displacement to right

zero velocity

mass m

displacementagainst time

velocityagainst time

forceagainst time

right

left

small displacement to right

small velocityto left

mass msmall force to left

right

left

large velocityto left

mass mzero net force

right

left

small displacement to left

small velocityto left

mass m

small force to right

right

left

large displacement to left

mass mlarge force to right

zero velocityright

left

Graphs of simple harmonic motion

Force, acceleration, velocity and displacement

If this is how the displacement varieswith time...

... the velocity is the rate of changeof displacement...

... the acceleration is the rate ofchange of velocity...

...and the acceleration tracks the forceexactly...

... the force is exactly opposite tothe displacement...

Phase differences Time traces varies with time like:

/2 = 90

/2 = 90

= 180

zero

displacement s

force F = –ks

displacement s

cos 2ft

same thing

–sin 2ft

–cos 2ft

–cos 2ft

cos 2ft

acceleration = F/m

velocity v

Computing oscillator motion step by step

These two diagrams show the computational steps in solving the equation for a harmonic oscillator.

Dynamics of harmonic oscillator

How the graph starts

force changesvelocity

0

time

t

zero initialvelocity

velocity would stayzero if no force

How the graph continues

force of springs accelerates mass towardscentre, but less and less as the mass nears thecentre

time

0

trace straighthere because nochange ofvelocity

no force at centre:no change of velocity

trace curvesinwards herebecause ofinwardschange ofvelocity

change of velocitydecreases asforce decreasesnew velocity

= initial velocity+ change ofvelocity

Constructing the graph

if no force, same velocityand same change indisplacementplusextra change indisplacement fromchange of velocity dueto force

= –(k/m) s (t)2

change in displacement = v t

t

t

extra displacement= v t

change of velocity v= acceleration tv = –(k/m) s t

because of springs:force F = –ks

acceleration = F/macceleration = –(k/m) s

Health warning! This simple (Euler) method has a flaw. It always changesthe displacement by too much at each step. This means that the oscillatorseems to gain energy!

extra displacement

Elastic potential energy

The relationship between the force to extend a spring and the extension determines the energy stored.

Energy stored in a stretched spring

no forcework F1 xforce F1

x

larger force

extension x

area below graph= sum of force change in displacement

extra areaF1 x

total area Fx

21

unstretched

F1

Energy supplied

energy stored in stretchedspring = kx2

21

small change xenergy supplied = F x

stretched to extension x by force F:energy supplied = Fx2

1

spring obeysHooke’s law: F = kx

F = 0x = 0

F = kx

Energy stored in stretched spring is kx221

00

x

Energy flows in an oscillator

The energy sloshes back and forth between being stored in a spring and carried by the motion of the mass.

Energy flow in an oscillator

displacement

time

time

s = A sin 2ft

PE = kA2 sin22ft

0

0

potential energy= ks21

2

potential energy

energy in stretched spring

energy carried by moving mass

time

time

0

0

kinetic energy= mv21

2

velocity

mass andspringoscillate

vmax

A

Avmax vmax

The energy stored in an oscillator goes back and forth between stretched spring and movingmass, between potential and kinetic energy

from spring tomoving mass

from movingmass to spring

from movingmass to spring

from spring tomoving massenergy in

stretched spring

energy inmoving mass

kinetic energy

vmax = 2fA

v = vmax cos 2ft

KE = mvmax cos22ft212

12

Resonance

Resonance occurs when driving frequency is equal to natural frequency. The amplitude at resonance, and just away from resonance, is affected by the damping.

Resonant response

10

5

01

Example: ions in oscillating electric field

ions in a crystalresonate andabsorb energy

Oscillator driven by oscillating driver

electricfield

+ – + –

low damping:large maximum responsesharp resonance peak

frequency/natural frequency0 0.5 1 1.5 2.0

10

5

01

frequency/natural frequency0 0.5 1 1.5 2.0

more damping:smaller maximum responsebroader resonance peak

Resonant response is a maximum when frequency of driver is equal to natural frequency of oscillator

narrow rangeat peakresponse

12

wider rangeat peakresponse

12

Simple Harmonic Motion (SHM) of the position of a particle with time

produces a Sinusoidal wave.

Period, Amplitude and Frequency

The time taken for the particle to complete one oscilation, that is, the time taken for the particle to move from its starting position and return to its original position is known as the period. and is generally given the symbol T. The frequency ν is related to the period, it is defined as how many oscillations occur in one second. Since the period is the time taken for one oscillation, the frequency is given by

f = 1/T(1)

The frequency is measured in [s-1]. This unit is known as the Hertz (Hz) in honour of the physicist Heinrich Hertz. The maximum displacement of the particle from its resting position is known as the amplitude. The frequency is also given the symbol f.

Equations of Simple Harmonic Motion

The definition of simple harmonic motion is simply that the acceleration causing the motion a of the particle or object is proportional and in opposition to its displacement x from its equilibrium position.

a(t) ∝ -x(t)

Where k is a constant of proportionality. This remembering that the acceleration is the second derivative of position, also leads us to the differential equation

x''(t) = - k x(t)

Simple Harmonic Motion is closely related to circular motion as can be seen if we take an object that moves in a circular path, like a ball stuck on a turntable. If we consider just the y-component of the motion the path with time we can see that it traces out a wave as shown in Flash 2.

y(t) = A sin(ωt)(2)

We can also see that the period of the motion is equal to the time it takes for one rotation. Therefore, if we know the angular velocity ω = θ/t. For one rotation, ω = 2π/T therefore the period is also equal to

T = 2π/ω(3)

The particle can also at different speeds is conected with the period. The frequency is the number of oscilations per second.

Consider the particle undergoing simple harmonic motion in Flash 3. The displacement with time takes the form of a sinusoidial wave. The velocity of the particle can be calculated by differentiating the displacement. The result is also a wave but the maximum amplitude is delayed, so that when the displacement is at a maximum the velocity is at a minimum and when the displacement is zero the velocity has its greatest

Displacement, velocity and acceleration vectors of a particle undergoing simple

harmonic motion

We set this out mathematically, using a differential equation as in equation (4). We specify the equation in terms of the forces acting on the object. The acceleration is the second derivative of the position with respect to time and this is proportional to the position with respect to time. The minus sign indicates that the position is in the opposite direction to the acceleration.

m y''(t) = - k y(t)(4)

The derivation of the solution can be found here

For which the general solution is a wave like solution.

y(t) = c1 cos(ωt) + c2 sin(ωt)

Where, ω is the angular frequency. (ω=2πf) The values of c1 and c2 are determined by the initial conditions. Specifically, c1 = y0 and c2 = v0/ω These two initial conditions specify the starting position and the initial velocity.

The general solution can also be written more compactly as

y(t)= A cos(ωt - φ)(5)

Where φ = tan-1(ωy0/v0), A = (y02 + (v0/ω)2)1/2

Differentiating once with respect to time, we obtain the velocity. (The derivative of cos x = - sin x)

v = y'(t)= - ωA sin(ωt - φ)(6)

Finally, the acceleration is the derivative of the velocity with respect to time. (The derivitive of -sin x = - cos x)

a = y''(t) = - ω2A cos(ωt - φ)(7)

Substituting equations (5) and (7) into equation (4) we verify that this does indeed satisfy the equation for simple harmonic motion. With the constant of proportionality k = ω2

Thus

a(t) = - ω2y(t)

The time for the maximum velocity and acceleration can be determined from these equations. From equation (6) the maximum magnitude of the velocity occurs when sin(ωt - φ) is 1 or -1. Therefore the maximum velocity is ±ωA. Intuitively, we can imagine that this velocity occurs when the oscillating system has reached the equilibrium position and is about to overshoot. The minus sign indicates the direction of travel is in the opposite direction.

The maximum acceleration occurs where the argument of cosine in eqn (7) is also -1 or 1. Thus the maximum acceleration is ±ω2A which occurs at the ends of the oscillations, as this is where the direction changes.

Examples of SHM

Spring Mass System

Consider a spring of spring constant k conected to a mass m. If the mass is displaced from its equilibrium position by a distance x a force F will act in the opposite direction to the displacement. From Hooke's Law the magnitude of the force is given by

F= - k x

When the mass is released it the force will act on the mass to bring it back to its equilibrium position. However, if there is no friction the inertia of the mass will cause it to overshoot the equilibrium position and the force will act in the opposite direction slowing it down and pulling it back.

The action of this force on the mass keeps it oscillating backwards and forwards.

The spring mass system consists of a spring with a spring constant of k attached to a mass, m. The mass is displaced a distance x from its equilibrium position work is done and potential energy is stored in the spring. If the mass is displaced by a small distance dx, the work done in stretching the spring is given by dW = F dx.

The force on the spring is assumed to obey Hooke's law, therefore, the restoring force is proportional to the extension. The work done is then dW = - k x dx

Making the total work W = - ∫k x dx = -1/2 k x2 + C

At the time of release, the energy of the system will consist totally of potential energy. PE= -1/2 kx2 and the potential energy as a function of time is

PE(t) = 1/2k A cos2(ωt - φ)

As the spring pulls the mass toward the equilibrium position, the potential energy is transformed into kinetic energy until at the equilibrium position the kinetic energy will be maximised. The KE is 1/2 mv2 and the maximum velocity occurs at x=0. From equation (6), the velocity will be ω A

Therefore the kinetic energy is 1/2 mω2A2 sin2(ωt - φ)

From the conservation of mechanical energy, the sum of energy between kinetic energy and potential energy will always be the same.

U = KE(t) + PE(t) = 1/2 mω2A2 sin2(ωt - φ) + 1/2k A cos2(ωt - φ)

1/2mω2x2 = 1/2k x2

is transferred into kinetic energy by moving the mass. For a particle undergoing simple harmonic motion, the displacement x is given by equation 1. The potential energy is gradually transferred to kinetic energy. Kinetic energy is given by 1/2 mv2. The velocity is given by equation (6). Summing the kinetic energy and potential energy we obtain,

U= (1/2)A2( k + m)

= 1/2 k A2(8)

Since k = mω2 and cos2x + sin2x = 1

Flash 5 shows the change in energy betweeen kinetic energy and potential

energy with time. The blue line shows the potential energy. It is highest at

the positions of maximum displacement. The red line shows the kinetic

energy. It has a maximum when the velocity is greatest, ie. as it passes the

equilibrium position. The green line shows the total mechanical energy of

the system, ie. the sum of the potential energy and kinetic energy. The total

energy remains constant because there are no losses to friction, heat or air

resistance.

Simple Pendulum

Another common example used to illustrate simple harmonic motion is the simple pendulum. This idealised system has a one end massless string suspended a mass m and the other end fixed to a stationary point. If the mass is displaced by a small distance, the angle moved is small.

The torque on the fixed point P is τ = Iα

- mg sin θ(t)L = mL2 θ''(t)

θ''(t) +g/L θ(t) = 0

This has the same form as simple harmonic motion equation, x''(t) - ω2 x(t), and so the solution is θ(t) = θ0 cos(ωt - φ)

the angular frequency is ω = (g/L)1/2.

It is interesting to note that the mass does not appear in this equation. This means that the frequency of the period only depends on the length of the string and the force of gravity. Pendulums with shorter strings will oscillate faster than pendulums with longer strings. And the same pendulum on the moon, where the force of gravity is 1/6th that of the gravity on the Earth, will also take longer to oscillate.

We have glossed over one important aspect, in that this analysis is true only for small angles of theta. We had to make the approximation that sin θ is aproximately the same as θ which is true only for small angles. The real differential equation

θ''(t) +g/L sin θ(t) = 0

is non-linear and cannot be solved analytically.

Torsional Pendulum

A mass suspended to a fixed support by a thin wire can be made to twist

about its axis. This is known as a torsional pendulum. The mass attached to

the wire rotates in the horizontal plane. In this case θ is the angle of

rotation. When the wire is untwisted and in equilibrium, the angle θ is 0

degrees. It is the twisting of the wire that creates a restoring torque due to

the resistance of the wire to the deformation. For small angles of θ the

magnitude of the torque is proportional to the angle θ

τ = - k θ

Where k is the torque constant of the wire. As with the simple pendulum the equation of motion is

τ = Iα

Where I is the moment of inertia

kθ(t) = - Iθ''(t)

Once again we have formed the equation for simple harmonic motion and can write the solution as θ(t) = A cos(ωt - φ)

The angular frequency, ω is given by

ω = (k/I)1/2

Summary

An oscillation follows simple harmonic motion if it fulfils the following two rules:

Acceleration is always in the opposite direction to the displacement from the equilibrium position

Acceleration is proportional to the displacement from the equilibrium position

The acceleration and displacement are linked by the following equation:

a(t) = - ω2 x(t)

Here ω is called the angular frequency of oscillation, and is given by 2π /T or 2π f

T is the period of oscillation (s), f=1/T = frequency of oscillation (Hz) and x is the displacement (m).

Using Newton's 2nd Law (F=ma) we can show that the Force on the object due to inertia will be:

F= -mω2 x

Using Hooke's Law for springs (F=-kx), we know that the force on the oscillating mass due to the springs is simplyF=-kx

These two forces are always in balance, so m ω2 x - kx = 0

From this we can find the resonant frequency:

ω2=(kx) / (mx) = k/m,

ω = (k/m)1/2

f = 1/ 2π * [(k/m)1/2]

WORKED EXAMPLES

1. A particle that hangs from a spring oscillates with an angular frequency of 2 rad/s.  The spring is suspended from the ceiling of an elevator car and hangs motionless (relative to the car) as the car descends at a constant speed of 1.5 m/s.  The car then suddenly stops.  Neglect the mass of the spring.(a) With what amplitude does the particle oscillate?(b) What is the equation of motion for the particle?  (Choose the upward direction to be positive.)

Solution: (a) When traveling in the elevator at constant speed, the total force on the mass is zero.  The force exerted by the spring is equal in magnitude to the gravitational force on the mass, the spring has the equilibrium length of a vertical spring.  When the elevator suddenly stops, the end of the spring attached to the ceiling stops. The mass, however has momentum, p = mv, and therefore starts stretching the spring.  It moves through the equilibrium position of the vertical spring with its maximum velocity vmax = 1.5 m/s.Its velocity as a function of time is v(t) = -ωAsin(ωt + φ). Since vmax = ωA and ω = 2/s, the amplitude of the amplitude of the oscillations is A = 0.75 m.(b) The equation of motion for the particle is d2x/dt2 = -(k/m)x = -ω2x. Its solution is x(t) = Acos(ωt + φ) = 0.75 mcos((2/s)t + φ). If we choose the t = 0 to be the time the elevator stops and let the upward direction be positive, then x(0) = 0, and v(0) = -1.5 m/s.  We therefore need φ to be π/2.

2. A mass-spring system oscillates with an amplitude of 3.5 cm.  If the force constant of the spring of 250 N/m and the mass is 0.5 kg, determine(a) the mechanical energy of the system,

(b) the maximum speed of the mass, and(c) the maximum acceleration .

Solution: (a) We have m = 0.5 kg, A = 0.035 m, k = 250 N/m, ω2 = k/m = 500/s2, ω = 22.36/s.The mechanical energy of the system is E = (1/2)kA2 = 0.153 J.(b) The maximum speed of the mass is vmax = ωA = 0.78 m/s.(c) The maximum acceleration is amax = ω2A = 17.5 m/s2.

Problems

A particle oscillates with simple harmonic motion, so that its displacement

varies according to the expression x = (5 cm)cos(2t + π/6) where x is in

centimeters and t is in seconds.  At t = 0 find

(a) the displacement of the particle,

(b) its velocity, and

(c) its acceleration.

(d) Find the period and amplitude of the motion.

Solution:

(a)  The displacement as a function of time is x(t) = Acos(ωt + φ).  Here

ω = 2/s, φ = π/6, and A = 5 cm.  The displacement at t = 0 is x(0) = (5

cm)cos(π/6) = 4.33 cm.

(b)  The velocity at t = 0 is v(0) = -ω(5 cm)sin(π/6) = -5 cm/s.

(c)  The acceleration at t = 0 is a(0) = -ω2(5 cm)cos(π/6) = -17.3 cm/s2.

(d)  The period of the motion is T = πs, and the amplitude is 5 cm.

A 20 g particle moves in simple harmonic motion with a frequency of 3

oscillations per second and an amplitude of 5cm.

(a) Through what total distance does the particle move during one cycle of

its motion?

(b) What is its maximum speed?  Where does that occur?

(c) Find the maximum acceleration of the particle.  Where in the motion

does the maximum acceleration occur? 

Solution:

(a)  The total distance d the particle moves during one cycle is from x =

-A to x = +A and back to x = -A, so d = 4A = 20 cm.

(b)  The maximum speed of the particle is

vmax = ωA = 2πfA = 2π 15 cm/s = 0.94 m/s.

The particle has maximum speed when it passes through the

equilibrium position.

(c) The maximum acceleration of the particle is

amax = ω2A = (2πf)2A = 17.8 m/s2.

The particle has maximum acceleration at the turning points, where it

has maximum displacement.

A 1kg mass attached to a spring of force constant 25N/m oscillates on a

horizontal frictionless track.  At t = 0 the mass is released from rest at x =

-3cm, that is the spring is compressed by 3cm.  Neglect the mass of the

spring.  Find

(a) The period of its motion,

(b) the maximum value of its speed and acceleration, and

(c) the displacement, velocity and acceleration as a function of time.

Solution:

(a) The period is T = 2πSQRT(m/k) = 2πSQRT(1 s2/25) = 1.26 s.

(b) The angular acceleration is ω = SQRT(k/m) = 5/s.

The maximum speed is vmax = ωA = 15 cm/s.

The maximum acceleration of the particle is amax = ω2A = 0.75 m/s2.

x(t) = Acos(ωt + φ) = (3 cm)cos((5/s)t + π) = -(3 cm)cos((5/s)t),

v(t) = -ωAsin(ωt + φ) = (15 cm/s)sin((5/s)t),

a(t) = -ω2Acos(ωt + φ) = (0.75 m/s2)cos((5/s)t).

Assume a mass suspended from a vertical spring of spring constant k.  In equilibrium the spring is stretched a distance x0 = mg/k.  If the mass is displaced from equilibrium position downward and the spring is stretched an additional distance x, then the total force on the mass is mg - k(x0 + x) =

-kx directed towards the equilibrium position.  If the mass is displaced upward by a distance x, then the total force on the mass is mg - k(x0 - x) = kx, directed towards the equilibrium position.  The mass will execute simple harmonic motion.  The angular frequency ω = SQRT(k/m) is the same for the mass oscillating on the spring in a vertical or horizontal position.  The equilibrium length of the spring about which it oscillates is different for the vertical position and the horizontal position.

Assume an object attached to a spring exhibits simple harmonic motion.  Let one end of the spring be attached to a wall and let the object move horizontally on a frictionless table.

What is the total energy of the object?

The object's kinetic energy is

K = (1/2)mv2 = (1/2)mω2A2sin2(ωt + φ),

Its potential energy is elastic potential energy.  The elastic potential energy stored in a spring displaced a distance x from its equilibrium position is U=(1/2)kx2.  The object's potential energy therefore is

U = (1/2)kx2 = (1/2)mω2x2 = (1/2)mω2A2cos2(ωt + φ).

The total mechanical energy of the object is

E = K+U = (1/2)mω2A2(sin2(ωt + φ)+cos2(ωt + φ)) = (1/2)mω2A2.

The energy E in the system is proportional to the square of the amplitude.

E = (1/2)kA2.

It is a continuously changing mixture of kinetic energy and potential energy.

For any object executing simple harmonic motion with angular frequency w, the restoring force F = -mω2x obeys Hooke's law, and therefore is a conservative force.  We can define a potential energy U = (1/2)mω2x 2, and the total energy of the object is given by E = (1/2)mω2A2.

Problems:

A particle that hangs from a spring oscillates with an angular frequency

of 2 rad/s.  The spring is suspended from the ceiling of an elevator car

and hangs motionless (relative to the car) as the car descends at a

constant speed of 1.5 m/s.  The car then suddenly stops.  Neglect the

mass of the spring.

(a) With what amplitude does the particle oscillate?

(b) What is the equation of motion for the particle?  (Choose the upward

direction to be positive.)

Solution:

(a) When traveling in the elevator at constant speed, the total force on the

mass is zero.  The force exerted by the spring is equal in magnitude to the

gravitational force on the mass, the spring has the equilibrium length of a

vertical spring.  When the elevator suddenly stops, the end of the spring

attached to the ceiling stops. The mass, however has momentum, p = mv,

and therefore starts stretching the spring.  It moves through the

equilibrium position of the vertical spring with its maximum velocity vmax

= 1.5 m/s.

Its velocity as a function of time is v(t) = -ωAsin(ωt + φ).

Since vmax = ωA and ω = 2/s, the amplitude of the amplitude of the

oscillations is A = 0.75 m.

(b) The equation of motion for the particle is d2x/dt2 = -(k/m)x = -ω2x. Its

solution is

x(t) = Acos(ωt + φ) = 0.75 mcos((2/s)t + φ).

If we choose the t = 0 to be the time the elevator stops and let the upward

direction be positive, then x(0) = 0, and v(0) = -1.5 m/s.  We therefore

need φ to be π/2.

A mass-spring system oscillates with an amplitude of 3.5 cm.  If the force

constant of the spring of 250 N/m and the mass is 0.5 kg, determine

(a) the mechanical energy of the system,

(b) the maximum speed of the mass, and

(c) the maximum acceleration .

Solution:

(a) We have m = 0.5 kg, A = 0.035 m, k = 250 N/m, ω2 = k/m = 500/s2, ω

= 22.36/s.

The mechanical energy of the system is E = (1/2)kA2 = 0.153 J.

(b) The maximum speed of the mass is vmax = ωA = 0.78 m/s.

(c) The maximum acceleration is amax = ω2A = 17.5 m/s2.

Static Friction Vs Kinetic Friction

Friction is the force resisting the relative motion of two surfaces in contact

or a surface in contact with a fluid. If you try to slide a heavy box resting on

the floor, you may find it difficult to get the box moving. Static friction is the

force that is acting against the box. If you apply a light horizontal push that

does not move the box, the static friction force is also small and directly

opposite to your push. If you push harder, the friction force increases to

match the magnitude of your push. There is a limit to the magnitude of

static friction, so eventually you may be able to apply a force larger than the

maximum static force, and the box will move. The maximum static friction

force is sometimes referred to as starting friction.

The normal force is defined as the perpendicular component of the force

exerted by the surface. In this case, the normal force is equal to the weight

of the object.

Once the box starts to slide, you must continue to exert a force to keep the

object moving, or friction will slow it to a stop. The friction acting on the

box while it is moving is called kinetic friction. In order to slide the box with

a constant velocity, a force equivalent to the force of kinetic friction must be

applied. Kinetic friction is sometimes referred to as sliding friction. Sliding

friction is caused not by surface roughness but by chemical bonding

between the surfaces. Surface roughness and contact area, however, do

affect sliding friction for micro- and nano-scale objects where surface area

forces dominate inertial forces.

Both static and kinetic friction depend on the surfaces of the box and the

floor, and on how hard the box and floor are pressed together. To

distinguish the two frictions, let us consider the graph below.

In making a distinction between static and kinetic coefficients of friction, we

are dealing with an aspect of "real world" common experience with a

phenomenon which cannot be simply characterized. The difference between

static and kinetic coefficients obtained in simple experiments like wooden

blocks sliding on wooden inclines roughly follows the model depicted in the

friction plot from which the illustration above is taken. This difference may

arise from irregularities, surface contaminants, etc. which defy precise

description. When such experiments are carried out with smooth metal

blocks which are carefully cleaned, the difference between static and

kinetic coefficients tends to disappear. When coefficients of friction are

quoted for specific surface combinations are quoted, it is the kinetic

coefficient which is generally quoted since it is the more reliable number.

Coefficient of Friction

The coefficient of friction is a dimensionless quantity symbolized by the

Greek letter μ and is used to approximate the force of friction (static or

kinetic).

The coefficient of static friction is defined as the ratio of the maximum static

friction force (F) between the surfaces in contact to the normal (N) force.

The coefficient of kinetic friction is defined as the ratio of the kinetic friction

force (F) between the surfaces in contact to the normal force:

The friction force is directed in the opposite direction of the resultant force

acting on a body. In the case of kinetic friction, the direction of the friction

force may or may not match the direction of motion: a block sliding atop a

table with rectilinear motion is subject to friction directed along the line of

motion; an automobile making a turn is subject to friction acting

perpendicular to the line of motion (in which case it is said to be 'normal' to

it). A motionless body is subject to static friction. The direction of the static

friction force can be visualized as directly opposed to the force that would

otherwise cause motion, were it not for the static friction preventing

motion. In this case, the friction force exactly cancels the applied force, so

the net force given by the Vector sum, equals zero. It is important to note

that in all cases, Newton's first law of motion holds. The coefficient of

friction is an empirical measurement – it has to be measured

experimentally, and cannot be found through calculations. Rougher surfaces

tend to have higher effective values. Most dry materials in combination

have friction coefficient values between 0.3 and 0.6. Values outside this

range are rarer, but Teflon, for example, can have a coefficient as low as

0.04. A value of zero would mean no friction at all, an elusive property –

even Magnetic levitation vehicles have drag. Rubber in contact with other

surfaces can yield friction coefficients from 1.0 to 2.

Optics

Introduction

Thin Lenses

When the size of the physical and optical objects of a system are much

larger than the wavelength of the light (or as λ→ 0), we are in the realm of

geometrical optics. Optical systems in which the wave nature of light must

be taken into account (interference, diffraction) are called physical optics.

Of course, every real system experiences diffraction effects, so geometric

optics is necessarily an approximation. However, the simplicity arising from

treating only rays which move in straight lines affords many uses.

A lens is a refracting device (a discontinuity in the medium) that

redistributes the energy being propagated by electromagnetic radiation.

This is usually achieved by re-shaping the wave front, most usefully by

turning spherical waves into plane waves and vice-versa. Lenses that cause

an incoming plane wave to bend towards the axis through its middle are

called converging or convex lenses. They are thicker at their midpoint than

at their edges. Concave lenses, on the other hand, are thicker at their edges

than in the middle; they cause an incoming plane wave to bend away from

its central axis and are hence also known as a diverging lenses. Both of

these are illustrated in.

Figure %: Concave and convex lenses.

For a converging lens, the point to which a plane wave converges is called

the focal point or focus. For a diverging lens, it is the point from which

incoming spherical waves must emerge in order to produce plane waves

upon passing through the lens.

Lenses that have only two refracting surfaces are called simple. Also, lenses

that have a thickness that is negligible compared to the overall path length

of the light traversing them are called thin. Here we will only consider thin, simple

lenses. To first-order, the focal length of such a lens is given by:

= (n l -1) -    

where n l is the index of refraction of the lens, R 2 is radius of curvature of

the left surface (from which the light approaches), and R 1 is the radius of

curvature of the right surface (through which the light leaves the lens). This

is known as the lens-maker's equation. We can derive it by considering a

spherical wave emanating from the centre of the sphere with the same

radius R 1 as one side of the lens. From it is clear that tanθ' = y/R 1.

Figure %: Derivation of the lens-makers equation.

But since the angle θ' is small in the thin lens approximation, we can say θ'

= y/R 1 . Using a small angle approximation to Snell's Law we can write n l

θ' = θ , and thus the downwards deflection of the ray is θ - θ' = (n l -1)θ' = (n

l -1)y/R 1 . The distance at which this ray intersects the axial line must be the

focal length and is given by: f = y/ (θ - θ') = R 1/ (n 1 - 1). If we consider a

convex lens, a system of two Plano-convex (planar on one side) lenses, we

can use the formula that 1/f = 1/f 1 +1/f 2 to arrive at the lens-makers

equation.

By far the most important formula in geometrical optics, however, relates

the position of an object placed in front of a lens to the position of its image,

formed by the lens. In the distance between the object and the lens is s o

and the distance between the lens and the image is s i.

Figure %: The Gaussian Lens Formula

Then

+ =    

There are certain sign conventions to be applied with this formula, and with

those to follow. s o > 0 if the object is on the same side of the lens as the

direction from which the light is coming, s o < 0 , otherwise. f > 0 if the

focal point is on the opposite side of the lens to that from which the light is

coming. S i < 0 if the image is on the opposite side of the lens to that from

which the light is coming. R > 0 if the centre of the sphere is on the

opposite side of the lens to that from which the light is coming. The height

of an object, y o, or its image, y i, is considered positive if it lies above the optical axis

(the central axis or axis of symmetry of the lens). Note that a planar interface has a focal length

of infinity. The "transverse magnification" of a thin lens is given by:

M T = = -    

From the sign conventions, M T > 0 implies that the image is upright, while

M T < 0 implies that it is inverted.

Mirrors

There are also two basic types of spherical mirrors. Concave mirrors reflect

incoming plane waves to a focal point directly in front of the mirror (they

are converging mirrors). Convex mirrors reflect incoming plane waves into

outward-moving spherical waves with the centre of the sphere appearing to

be behind the mirror (they are diverging mirrors).

Figure %: Concave and Convex mirrors.

The focal length of a mirror is f = - , where R is the radius of curvature of the

mirror. Also the very same relation between the image and object distances applies:

= +    

Applying the sign conventions that f , s o , and s I are positive in front of the

mirror, f > 0 for concave mirrors and f < 0 for convex mirrors. Note that

images for which s I is positive are called real images, and are those for

which a screen can be placed at the position of the image in order to

observe it; images for which s i is negative are called virtual. No virtual

image can be formed on a screen--any image seen in a mirror is an example

of a virtual image. An alternative formulation of these definitions is to say

that for real images light rays really do pass through where the image

forms; for virtual images light rays only appear to be coming from the

position of the image.

Mirrors have an advantage over lenses in that they do not suffer chromatic

aberration. This phenomenon arises due to dispersion, causing the lens to

have not just one focal length but a small band of focal lengths

corresponding to the different amounts by which it refracts the different

colours. This means that it is impossible to focus colored images precisely

with a lens. Mirrors, because they do not rely on refraction, do not suffer

this problem. Moreover, it is important to remember that all the formulas

we encountered here were derived using the first order approximation to

the sine function appearing in Snell's Law: sinθ θ. Of course this ignores

higher order terms in θ 3, etc. Corrections arising from this and other

considerations cause aberrations (or deviations) from the simple equations

developed here for spherical lens and mirror systems. In fact, there are five

primary, monochromatic aberrations called spherical aberration, coma,

astigmatism, field curvature, and distortion. They are collectively known as

the Seidel aberrations.

Ray Tracing

Often it will be useful to determine the approximate position of an image,

given the position of the object and the focal length in a lens or mirror

system without resorting to the lens equation. We can do this by drawing

diagrams and mapping out the path of the light rays. This process is known

as analytical ray tracing. The basic strategy is to select a significant point

on the object (like the top) and to draw several principal rays from that

point. For a mirror, the three principle rays, shown in , are: i) parallel to the

axis, returning through the focal point; ii) to the midpoint of the mirror,

reflecting at an equal angle on the opposite side to the central axis; and iii)

through the centre of the sphere of which the mirror is a part, returning

along the same path.

Figure %: Principal rays for a mirror.

The principal rays for a lens are similar: i) parallel to the central axis,

refracting through the focal point; ii) a straight ray through the centre of

the lens; and iii) through the focal point on the near side, refracting parallel

to the axis.

Figure %: Principal rays for a lens.

At least two principal rays must be drawn from the object; the point where

the principal rays cross (or appear to cross, for a virtual image) is the

location of the image. To determine whether an object is magnified or

diminished, it would be necessary to choose one other point (such as the

base) and compare this distance between the two points in the image to

their positions in the object. To find the locations of virtual images it is

necessary to trace the light rays backwards behind the mirror or lens.

Ray tracing is especially useful when complicated systems of mirrors and/or

lenses need to be analyzed. Ray tracing can give a rough but quick idea of

how the system will behave. For example, it can fairly quickly be

determined that a single concave lens will always produce virtual,

diminished, upright images, irrespective of the position of the object.

However, for a convex lens, the location of the image depends on the

location of the object. Recall that real objects and real images have s o > 0 ,

or s I > 0 , while these distances are negative for virtual objects and images

(virtual objects can arise when the image for one lens becomes the object

for another in a lens system). F > 0 corresponds to converging lenses or

mirrors and f < 0 corresponds to diverging lenses or mirrors. Positive y o or

y I correspond to upright objects and images respectively. A negative

magnification corresponds to an inverted image.

Problems on Mirrors and Lenses

Example

1. Where is the image of an object placed 7 centimetres away from a 5

centimetre focal length convex lens? Concave mirror? Are the images real

or virtual in each case?

Solution

A convex lens has a positive focal length. We can apply the lens equation:

1/f = 1/s o +1/s I. This gives: 1/5 = 1/7 + 1/s i. Solving for s I we find s I =

17.5 centimetres. The image is real since s I is positive. A concave mirror

also has a positive focal length so we get the same result (the image is real

and 17.5 centimetres in front of the mirror).

2. Determine whether the images of the objects placed at the following

positions will be upright or inverted. Concave mirror, s o = 2f ; convex

mirror, f < s o < 2f ; concave lens, f > s o .

Solution

The orientation of the image is determined by the sign of the transverse

magnification M T = . For a concave mirror the focal length is positive.

Thus 1/f = 1/2f + 1/s i, and since 2f > f we have 1/f > 1/2f and s i must be

positive, so the magnification must be negative (the object and image

distances are both positive). Hence the first image is inverted. For a convex

mirror the focal length is negative, thus -1/f = 1/s o +1/s iâá’1/s i = - (1/f +

1/s o), where f and s o are now considered positive quantities. Thus s I must

be negative, the magnification positive and the image upright. In the third

case, concave lenses have a negative focal length, thus -1/f = 1/s o +1/s I,

and we have the same situation as before, so the image is upright.

Reflection at plane surfaces

The Law of Reflection

Consider the diagram. The direction of the reflected wave is determined by the phase difference between the scatterers on the surface. This, in turn, is caused by the angle made by the incident wave and the surface (the angle of incidence, θ i). If AB is an incoming wave front and CD is an outgoing wave front such that the spherical wave emitted from A will be in- phase with the wave just emitted from D (this is true is AB = CD ).

Figure %: Wave fronts reflecting from a surface.

This is the condition for all the surface waves to be in-phase. From the

triangles ABD and ACD, which have a common hypotenuse, we can

conclude = , where θ r is the angle of the reflected wave. But

clearly, BD = AC, so:

sinθ i = sinθ râá’θ i = θ r    

This is called the "law of reflection." To state the law in its full generality, it

is also necessary to say that the incident ray and the reflected ray, as well

as the ray perpendicular to the surface, all lie in the same plane. Recall

also, that the angles of incidence and reflection correspond to the angles

between the rays and the normal to the surface (also, to the angles between

the wave fronts and the surface, as in the diagram).

Reflection from a smooth surface (such as a mirror) is called specular reflection (any irregularities in the surface are small compared to λ). When the surface is rough in comparison to λ, diffuse reflection results.

Reflection

We call the ray of light that strikes a reflective surface an incident ray, and

the ray that bounces back a reflected ray. The angle of incidence, is the

angle between the normal—the line perpendicular to the reflective surface

—and the incident ray. Similarly, the angle of reflection is the angle

between the normal and the reflected ray.

The law of reflection tells us that angle of incidence and angle of reflection

are equal:

The reflection of a ray of light works in just the same way as a ball bouncing

off a wall, except gravity has no noticeable effect on light rays.

Refraction

In addition to reflecting light, many surfaces also refract light: rather than

bouncing off the surface, some of the incident ray travels through the

surface, but at a new angle. We are able to see through glass and water

because much of the light striking these substances is refracted and passes

right through them.

Given a ray travelling from a medium with index of refraction into a

medium with index of refraction , Snell’s Law governs the relationship

between the angle of incidence and the angle of refraction:

Example

A ray of light passes from a liquid medium into a gas medium. The incident

ray has an angle of 30º with the normal, and the refracted ray has an

angle of 60º with the normal. If light travels through the gas at a speed of

m/s, what is the speed of light through the liquid medium? Sin 30º =

0.500 and sin 60º = 0.866.

SOLUTION

We know that the index of refraction for a substance, n, gives the ratio of

the speed of light in a vacuum to the speed of light in that substance.

Therefore, the index of refraction, in the liquid medium is related to the

speed of light, in that medium by the equation = c/ ; similarly, the

index of refraction, in the gas medium is related to the speed of light,

in that medium by the equation = c/ . The ratio between and is:

We can calculate the ratio between and using Snell’s Law:

Since we know that the ratio of / is equal to the ration of / , and since

we know the value for , we can now calculate the value for :

Given m/s, we can also calculate that the index of refraction for

the liquid substance is 2.1, while the index of refraction for the gas

substance is 1.2.

Dispersion

The phenomenon of dispersion explains why we see a rainbow when

sunlight refracts off water droplets in the air. The white light of the sun is

actually a mixture of a multitude of different wavelengths.

When this white light passes through water droplets in the air, the different

wavelengths of light are refracted differently. The violet light is refracted at

a steeper angle than the red light, so the violet light that reaches our eyes

appears to be coming from higher in the sky than the red light, even though

they both come from the same ray of sunlight.

Optical Instruments

The reflection and refraction we’ve dealt with so far have focused only on

light interacting with flat surfaces. Lenses and curved mirrors are optical

instruments designed to focus light in predictable ways. While light striking

a curved surface is more complicated than the flat surfaces we’ve looked at

already, the principle is the same

.

The four basic kinds of optical instruments are concave mirrors, convex

mirrors, convex (or converging) lenses, and concave (or diverging) lenses.

General Features of Mirrors and Lenses

The principal axis of a mirror or lens is a normal that typically runs

through the centre of the mirror or lens.

The vertex, represented by V in the diagram, is the point where the

principal axis intersects the mirror or lens.

Spherical mirrors have a centre of curvature, represented by C in the

diagram, which is the centre of the sphere of which they are a slice.

The radius of that sphere is called the radius of curvature, R.

The focal length, f, is defined as the distance between the vertex and the

focal point. For spherical mirrors, the focal length is half the radius of

curvature, f = R/2.

Concave Mirrors

Suppose a boy of height h stands at a distance d in front of a concave

mirror. By tracing the light rays that come from the top of his head, we can

see that his reflection would be at a distance from the mirror and it would

have a height . As anyone who has looked into a spoon will have guessed,

the image appears upside down.

The image at is a real image: as we can see from the ray diagram, the

image is formed by actual rays of light. That means that, if you were to hold

up a screen at position , the image of the boy would be projected onto it.

You may have noticed the way that the concave side of a spoon can cast

light as you turn it at certain angles. That’s because concave mirrors project

real images.

You’ll notice, though, that we were able to create a real image only by

placing the boy behind the focal point of the mirror. What happens if he

stands in front of the focal point?

The lines of the ray diagram do not converge at any point in front of the

mirror, which means that no real image is formed: a concave mirror can

only project real images of objects that are behind its focal point. However,

we can trace the diverging lines back behind the mirror to determine the

position and size of a virtual image. Like an ordinary flat mirror, the image

appears to be standing behind the mirror, but no light is focused on that

point behind the mirror. With mirrors generally, an image is real if it is in

front of the mirror and virtual if it is behind the mirror. The virtual image is

right side up, at a distance from the vertex, and stands at a height .

You can test all this yourself with the right kind of spoon. As you hold it at a

distance from your face, you see your reflection upside down. As you slowly

bring it closer, the upside-down reflection becomes blurred and a much

larger reflection of yourself emerges, this time right side up. The image

changes from upside down to right side up as your face crosses the spoon’s

focal point.

Convex Mirrors

The focal point of a convex mirror is behind the mirror, so light parallel to

the principal axis is reflected away from the focal point. Similarly, light

moving toward the focal point is reflected parallel to the principal axis. The

result is a virtual, upright image, between the mirror and the focal point.

You’ve experienced the virtual image projected by a convex mirror if you’ve

ever looked into a polished doorknob. Put your face close to the knob and

the image is grotesquely enlarged, but as you draw your face away, the size

of the image diminishes rapidly.

The Two Equations for Mirrors and Lenses

So far we’ve talked about whether images are real or virtual, upright or

upside down. We’ve also talked about images in terms of a focal length f,

distances d and , and heights h and . There are two formulas that relate

these variables to one another, and that, when used properly, can tell

whether an image is real or virtual, upright or upside down, without our

having to draw any ray diagrams. These two formulas are all the math you’ll

need to know for problems dealing with mirrors and lenses.

First Equation: Focal Length

The first equation relates focal length, distance of an object, and distance of

an image:

Values of d and f are positive if they are in front of the mirror and

negative if they are behind the mirror. An object can’t be reflected unless

it’s in front of a mirror, so d will always be positive. However, as we’ve

seen, f is negative with convex mirrors, and is negative with convex

mirrors and with concave mirrors where the object is closer to the mirror

than the focal point. A negative value of signifies a virtual image, while a

positive value of signifies a real image.

Note that a normal, flat mirror is effectively a convex mirror whose focal

point is an infinite distance from the mirror, since the light rays never

converge. Setting 1/f = 0, we get the expected result that the virtual image

is the same distance behind the mirror as the real image is in front.

Second Equation: Magnification

The second equation tells us about the magnification, m, of an image:

Values of are positive if the image is upright and negative if the image is

upside down. The value of m will always be positive because the object itself

is always upright.

The magnification tells us how large the image is with respect to the object:

if , then the image is larger; if , the image is smaller; and if m = 1,

as is the case in an ordinary flat mirror, the image is the same size as the

object.

Because rays move in straight lines, the closer an image is to the mirror, the

larger that image will appear. Note that will have a positive value with

virtual images and a negative value with real images. Accordingly, the

image appears upright with virtual images where m is positive, and the

image appears upside down with real images where m is negative.

Example

A woman stands 40 cm from a concave mirror with a focal length of 30 cm.

How far from the mirror should she set up a screen in order for her image

to be projected onto it? If the woman is 1.5 m tall, how tall will her image be

on the screen?

How far from the mirror should she set up a screen in order for her image

to be projected onto it?

Solution.

The question tells us that d = 40 cm and f = 30 cm. We can simply plug

these numbers into the first of the two equations and solve for , the

distance of the image from the mirror:

Because is a positive number, we know that the image will be real. Of

course, we could also have inferred this from the fact that the woman sets

up a screen onto which to project the image.

How tall will her image be on the screen?

We know that d = 40 cm and we now know that = 120 cm, so we can plug

these two values into the magnification equation and solve for m:

The image will be three times the height of the woman, or m tall.

Because the value of m is negative, we know that the image will be real, and

projected upside down.

Convex Lenses

Lenses behave much like mirrors, except they use the principle of

refraction, not reflection, to manipulate light. You can still apply the two

equations above, but this difference between mirrors and lenses means that

the values of and f for lenses are positive for distances behind the lens

and negative for distances in front of the lens. As you might expect, d is still

always positive.

Because lenses rely on refraction to focus light, the principle of dispersion

tells us that there is a natural limit to how accurately the lens can focus

light. For example, if you design the curvature of a convex lens so that red

light is focused perfectly into the focal point, then violet light won’t be as

accurately focused, since it refracts differently.

A convex lens is typically made of transparent material with a bulge in the

centre. Convex lenses are designed to focus light into the focal point.

Because they focus light into a single point, they are sometimes called

“converging” lenses. All the terminology regarding lenses is the same as the

terminology we discussed with regard to mirrors—the lens has a vertex, a

principal axis, a focal point, and so on.

Convex lenses differ from concave mirrors in that their focal point lies on

the opposite side of the lens from the object. However, for a lens, this

means that f > 0, so the two equations discussed earlier apply to both

mirrors and lenses. Note also that a ray of light that passes through the

vertex of a lens passes straight through without being refracted at an angle.

In this diagram, the boy is standing far enough from the lens that d > f. As

we can see, the image is real and on the opposite side of the lens, meaning

that is positive. Consequently, the image appears upside down, so and m

are negative. If the boy were now to step forward so that d < f, the image

would change dramatically:

Now the image is virtual and behind the boy on the same side of the lens,

meaning that is negative. Consequently, the image appears upright, so

and m are positive.

Concave Lenses

A concave lens is designed to divert light away from the focal point, as in

the diagram. For this reason, it is often called a “diverging” lens. As with

the convex lens, light passing through the vertex does not bend. Note that

since the focal point F is on the same side of the lens as the object, we say

the focal length f is negative.

As the diagram shows us, and as the two equations for lenses and mirrors

will confirm, the image is virtual, appears on the same side of the lens as

the boy does, and stands upright. This means that is negative and that

and m are positive. Note that h > , so m < 1.

Summary

There’s a lot of information to absorb about mirrors and lenses, and

remembering which rules apply to which kinds of mirrors and lenses can be

quite difficult. However, this information is all very systematic, so once you

grasp the big picture; it’s quite easy to sort out the details. In summary,

we’ll list three things that may help you grasp the big picture:

1. Learn to draw ray diagrams: Look over the diagrams of the four kinds of optical instruments and practice drawing them yourself. Remember that light refracts through lenses and reflects off mirrors. And remember that convex lenses and concave mirrors focus light to a

point, while concave lenses and convex mirrors cause light to diverge away from a point.

2. Memorize the two fundamental equations: You can walk into SAT II Physics knowing only the two equations for lenses and mirrors and still get a perfect score on the optical instruments questions, so long as you know how to apply these equations. Remember that f is positive for concave mirrors and convex lenses, and negative for convex mirrors and concave lenses.

3. Memorize this table: Because we love you, we’ve put together a handy table that summarizes everything we’ve covered in this section of the text.

Optical InstrumentValue

of d ´

Real or

virtual?

Value

of f

Upright or

upside down?

Mirrors ( and f are

positive in front of

mirror)

Concave

d > f + Real + Upside down

Concave

d < f – Virtual + Upright

Convex – Virtual – Upright

Lenses ( and f are

positive on far side of

lens)

Convex d

> f + Real + Upside down

Convex d

< f – Virtual + Upright

Concave – Virtual – Upright

Note that when is positive, the image is always real and upside down, and

when is negative, the image is always virtual and upright.

WAVE OPTICS

Young’s Double-Slit Experiment

The wave theory of light came to prominence with Thomas Young’s double-

slit experiment, performed in 1801. We mention this because it is often

called “Young’s double-slit experiment,” The double-slit experiment proves

that light has wave properties because it relies on the principles of

constructive interference and destructive interference, which are

unique to waves.

The double-slit experiment involves light being shone on a screen with two

very narrow slits in it, separated by a distance d. A second screen is set up a

distance L from the first screen, upon which the light passing through the

two slits shines.

Suppose we have coherent light—that is, light of a single wavelength ,

which is all travelling in phase. This light hits the first screen with the two

parallel narrow slits, both of which are narrower than . Since the slits are

narrower than the wavelength, the light spreads out and distributes itself

across the far screen.

At any point P on the back screen, there is light from two different sources:

the two slits. The line joining P to the point exactly between the two slits

intersects the perpendicular to the front screen at an angle .

We will assume that the two screens are very far apart—somewhat more

precisely, that L is much bigger than d. For this reason, this analysis is

often referred to as the “far-field approximation.” This approximation allows

us to assume that angles and , formed by the lines connecting each of the

slits to P, are both roughly equal to . The light from the right slit—the

bottom slit in our diagram—travels a distance of l = d sin more than the

light from the other slit before it reaches the screen at the point P.

As a result, the two beams of light arrive at P out of phase by d sin . If d sin

= (n + 1/2) , where n is an integer, then the two waves are half a

wavelength out of phase and will destructively interfere. In other words, the

two waves cancel each other out, so no light hits the screen at P. These

points are called the minima of the pattern.

On the other hand, if d sin = n , then the two waves are in phase and

constructively interfere, so the most light hits the screen at these points.

Accordingly, these points are called the maxima of the pattern.

Because the far screen alternates between patches of constructive and

destructive interference, the light shining through the two slits will look

something like this:

Note that the pattern is brightest in the middle, where = 0. This point is

called the central maximum. If you encounter a question regarding double-

slit refraction on the test, you’ll most likely be asked to calculate the

distance x between the central maximum and the next band of light on the

screen. This distance, for reasons too involved to address here, is a function

of the light’s wavelength ( ), the distance between the two slits (d), and the

distance between the two screens (L):

Diffraction

Diffraction is the bending of light around obstacles: it causes interference

patterns such as the one we saw in Young’s double-slit experiment. A

diffraction grating is a screen with a bunch of parallel slits, each spaced a

distance d apart. The analysis is exactly the same as in the double-slit case:

there are still maxima at d sin = n and minima at d sin = (n + 1/2) .

The only difference is that the pattern doesn’t fade out as quickly on the

sides.

Single-Slit Diffraction

The setup is the same as with the double-slit experiment, only with just one

slit. This time, we define d as the width of the slit and as the angle

between the middle of the slit and a point P.

Actually, there are a lot of different paths that light can take to P—there is a

path from any point in the slit. So really, the diffraction pattern is caused by

the superposition of an infinite number of waves. However, paths coming

from the two edges of the slit, since they are the farthest apart, have the

biggest difference in phase, so we only have to consider these points to find

the maxima and the minima.

Single-slit diffraction is nowhere near as noticeable as double-slit

interference. The maximum at n = 0 is very bright, but all of the other

maxima are barely noticeable. For this reason, we didn’t have to worry

about the diffraction caused by both slits individually when considering

Young’s experiment.

Polarization

Light is a transverse wave, meaning that it oscillates in a direction

perpendicular to the direction in which it is travelling. However, a wave is

free to oscillate right and left or up and down or at any angle between the

vertical and horizontal.

Some kinds of crystals have a special property of polarizing light, meaning

that they force light to oscillate only in the direction in which the crystals

are aligned. We find this property in the crystals in Polaroid disks.

The human eye can’t tell the difference between a polarized beam of light

and one that has not been polarized. However, if polarized light passes

through a second Polaroid disk, the light will be dimmed the more that

second disk is out of alignment with the first. For instance, if the first disk is

aligned vertically and the second disk is aligned horizontally, no light will

pass through. If the second disk is aligned at a 45º angle to the vertical, half

the light will pass through. If the second disk is also aligned vertically, all

the light will pass through.