INTRODUCTORY PHYSICS A (P14A/PHYS1410)

WORKBOOK

Includes

MECHANICS

WAVES & OPTICS

HEAT & THERMODYNAMICS

Department of Physics The University of the West Indies, Mona

TABLE OF CONTENTS

MECHANICS

Vectors – A Review M-1

2-D Motion M-7

Projectile Motion M-10

Work and Energy M-15

Conservation of Energy,

Mechanical Energy & Potential Energy M-21

Center of Mass M-27

Collisions M-32

Rotational Motion M-37

Angular Momentum M-45

Rolling M-48

Simple Harmonic Motion M-51

Additional Material

Rotational Inertias M-I

P14 Mathematics Notes M-II

WAVES & OPTICS

Waves Part 1 W-1

Waves Part 2 W-7

Waves Part 3 W-12

Wavefronts, Rays and

the Doppler Effect W-21

Optics 1 W-26

Young’s Double Slit Experiment W-31

Thin Film Interference W-41

Newtons’ Rings W-45

Diffraction W-48

HEAT & THERMODYNAMICS

Temperature H-1

Ideal Gases H-2

Heat H-3

Thermodynamics H-4

Kinetic Theory H-11

Adiabatic Expansion of an Ideal Gas H-14

Heat Engines and Efficiencies H-15

Refrigerator H-19

Entropy H-21

MECHANICS

M-1

VECTORS – A REVIEW

Introduction There are 2 types of physical quantities:

Description

Examples

Scalars

Vectors

We have had lots of experience with scalars. Let us quickly review how we handle vectors. Representing Vectors

a) Graphically By an arrow:

Where, length = Arrowhead (orientation) =

Note: (i) We need an appropriate scale. (ii) Irrespective of where drawn, if vectors are // & of same

length then they are equal. (iii) We use an arrow over ‘a’ to indicate a vector. If no

arrow => magnitude. (iv) A –ve vector = vector of same magnitude but opposite

direction.

b) By Components

A graphical representation is not always convenient (plus you are doing P14A now!). Therefore another strategy is to resolve the vector into perpendicular components. For our purposes we will resolve into 2 components: one along the x axis and one along the y axis.

Consider the vector a below.

Example 1 (i) Draw vector a representing a velocity of 6 ms-1 in a north easterly direction. (ii) Also draw -a .

ax =

ay =

And to get back the magnitude and direction from the components: a =

tanθ =

θ

a

Note: (i) In equations ‘a’

is magnitude. (ii) θ is angle to

horizontal. What if it is not?

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c) By Unit Vectors

If we are given (or can figure out) the x and y components of a vector i.e. ax, ay then it is much easier (and makes life much simpler) to represent the vector in terms of unit vectors.

That is, if we represent the x, y and z directions by a unit vector (magnitude one), and call these unit vector i, j, and k, then we can represent the vector a as: a = axi + ayj. Example 3 Vector a above can be written as: Note: (i) With unit vector representation, we can easily visualize the vector.

(ii) If given vector in terms of unit vector e.g. a = ax i + ay j we can get back original vector That is, magnitude of a =

tanθ =

Adding Vectors Consider the following two displacement vectors. How would we add them?

Note: We can only add similar vectors i.e. displacement and displacement!

a) Graphically

(i) Draw 1st vector to scale (ii) Draw 2nd vector to scale with tail to head (iii) Connect tail of 1st to head of 2nd. (iv) Measure to get angle and length.

Note arrow direction of resultant!!

Example 2 A displacement vector a has magnitude 125 km and is directed 680 above the horizontal. What are the values of ax and ay?

53.13

a

5 km

36.87

b 10 km

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b) By Components

(i) Draw both vectors on same axis (ii) Make sure tails together (iii) Resolve both into components (iv) Add similar components (v) Calculate magnitude and direction if resultant

c) By Unit Vector Notation

If given a and b in vector notation:

a = (3i + 4j) km and b = (8i – 6j) km

a + b =

Note: (i) Make sure to add only those quantities that go in the same direction.

(ii) Much, much easier!!!

Subtracting Vectors We note that a – b = a + (-b). Multiplying Vectors We will meet three types of multiplication involving vectors. a) Vector by Scalar = Vector

a x (scalar) = r

Example 4 r = 3 x (3i + 4j) km =

Note: F = ma is an example of such a multiplication!

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b) Vector . Vector = Scalar also known as Scalar or Dot Product

• If given the vector in terms of magnitudes and angles i.e. as on page 2 then: a.b = ab cosθ

Example 5

For the 2 vectors a and b on page 2 find the dot product.

Note: (i) If θ = 90o i.e vectors are perpendicular, a.b =

Implication:

(ii) Dot product ensures the 2 things multiplied are parallel!!!

(iii) a.b = b.a (iv) Example of Use: W = F. s or P = F . v or KE = ½m v.v

• If the two vectors are given in vector notation, then do as in example 6 below:

Example 6

Let a = (3i + 4j) km and b = (8i – 6j) km

a.b =

a �

b �

θ �

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c) Vector x Vector = Vector also known as Vector or Cross Product

• If given the vector magnitudes and angles i.e. as on page 2 then: Magnitude of the new vector a x b = ab sinθ

To find the direction of the new vector we use the Right Hand Rule.

Right Hand Rule: Place vectors tail to tail. Place palm perpendicular to plane of vectors and in line with the first vector. Sweep palm from 1st to 2nd vector through smallest angle. Thumb gives the direction.

Example 7

For the 2 vectors a and b on page 2 find the cross product.

Note: (i) the resultant vector is perpendicular to plane (and each) of the original vectors.

(ii) Order of cross product is important. a x b ≠ b x a

(iii) If θ = 0o i.e vectors are parallel a x b =

Implication:

(iv) Cross product ensures the 2 things multiplied are perpendicular.

(iii) Example of Use: ז = r x .F

• If the two vectors are given in vector notation, then do as in example 8 below. However we must realise that:

i x j = k ; j x i = -k j x k = i ; k x j = -i k x i = j ; i x k = -j

Example 8

Let a = (3i + 4j) km and b = (8i – 6j) km

a x b =

a �

b �

θ �

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Class Question 1 a) Find the work done by a force F= (2i+2j-k) N if it causes a displacements of s = (6i-3j+2k) m.

Find also the angle between F and s.

b) Determine the value of a constant y such that the vectors a = 2i+yj+k and b = 4i-2j-2k are

perpendicular.

c) Find the torque if F = (6i - 2j) N and r = (8i + j) km.

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2-D MOTION Now that we have developed skill in manipulating vectors, we can review motion but in 2 dimensions i.e. using vector notation. Displacement, s By displacement we mean the change in position of a body.

Displacement, s = ∆ r

=

Example 1: In 1-D: At t1, x1 = 15 m and at t2, x2 = 30 m.

... displacement, s = ∆x

= =

In 2-D: Same principle. At t1, r1 = (3i-7j) m and at t2, r2 = (5i+5j) m

... displacement, s = ∆r

=

=

Can you plot r1, r2 and s?

Velocity, v

• Average Velocity – The ratio of the displacement to the time taken.

v = s = ∆r

∆t ∆t

Note: (i) The over bar on the v => average (ii) Units = ms

-1

(iii) Since velocity is a vector, direction = direction of displacement vector. (iv) In 2-D example above, if time for change was 2 s, then

Average velocity =

• Instantaneous Velocity – The velocity at an instant.

v = ∆r = dr = d(xi+yj) = vxi+ vyj ∆t dt dt t→0

Note: (i) Velocity at an instant can be divided into components i.e. v = vxi+ vyj (ii) v = dr/dt (where r is given as a function of t)

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(iii) The direction of the instantaneous velocity is always tangential to the path and perpendicular to the position vector that describes the body at the same instant. (Can you see a dot product question?)

(iv) Units = ms-1

Acceleration, a

• Average Acceleration – The ratio of the change in velocity to the time taken.

a = ∆v = ∆vx i + ∆vy j + ∆vz k ∆t ∆t ∆t ∆t

Note: (i) Due to change in velocity, whether magnitude or direction (ii) Units = ms

-2

(iii) Direction = direction of resultant velocity vector.

• Instantaneous Acceleration – The acceleration at an instant or given time.

a = ∆v = dv = d(vx i + vy j) = axi+ ayj ∆t dt dt t→0

Note: (i) Acceleration at an instant can be divided into components i.e. a = axi+ ayj (ii) a = dv/dt

(iv) Units = ms-2

Class Question 2

The position of a particle is given by r = i+4t2j+tk. What is its a) position b) velocity and c) acceleration

after t = 4 sec? Give your answers in vector notation.

Equations of Motion True only for constant acceleration!! In 1-D:

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In 2-D: x-direction y-direction Note: Don’t mix x and y directions – The beauty of the method!!!

Class Question 3 A particle is initially located at r = (2i+2j) m and has initial velocity, u = (-2i+4j) ms

-1 . It undergoes a

constant acceleration of a = (-2i+3j) ms-2

. What is the particle’s velocity after 2 seconds? (Give in vector

notation). Also where is it located?

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Projectile Motion

A good example of how to put 2-D motion into practice is in projectile motion i.e. a particle/body

fired or projected in the air at some angle.

Consider a body projected at an angle θ to the horizontal with an initial velocity U.

1. Neglect resistance due to motion in air for now. !!!!

2. U can be described by a vector i.e. U = Uxi + Uyj

where Ux = Uy =

3. Its path is described by a parabola. Therefore after time t, it will have been displaced in both the x

direction and in the y direction. (Same time t to travel both x and y!)

4. After the same time t, the velocity of the projectile also changes with time. (Can you see?)

If we represent the new velocity by V, it will also have two components i.e. V = Vxi+Vyj.

5. Only thing speeding up or slowing down the body is:

Note we have divided every variable into components in 2 dimensions (x and y). As with the 2-D motion we did before, we can write equations for each direction. The horizontal and vertical motion can be considered independent of each other i.e. neither motion affects the other. (This is the beauty of physics). Horizontal Motion (v = u + at)

---(1)

(s = ut+½ a t2) ---(2)

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Vertical Motion (v = u + at)

----(3) (s = ut+½ a t2) ----(4) (v2 = u2 + 2as) ----(5) These five equations completely describe projectile motion. They are all we need to answer any projectile question.

Some Additional Concepts 1. Maximum Height, H

We can write a relationship for the maximum height achieved.

At the maximum height:

2. Horizontal Range, R

The horizontal range is:

At the range:

Note: R is a maximum when:

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3. Equation for the Path

The path travel is parabolic, therefore it most be of the form y = ax2 + bx = c Class Question 4 A projectile is fired at 60

o and with an initial velocity of 42 ms

-1 (see

figure). After 5.5 seconds it reaches a point X in its trajectory. Determine

(a) the height of X above the horizontal (b) the velocity of the projectile at

X, and (c) H, the maximum height achieved by the projectile.

4. Other Scenarios

a. Particle initially fired horizontally: b. Particle fired off a cliff. c. Velocity just before hitting ground zero?

X

H

42 m/s

M-13

What if we didn’t neglect Friction Originally we said we neglected friction. If however we didn’t then we could assume friction has both and x and y component and could be represented by f = fx i + fyj .

Implications: ax, friction = ; ay, friction = How does this affect the equations we wrote previously?

In x-direction:

In y-direction: UP DOWN

Uniform Circular Motion Another example of 2-D motion is uniform circular motion. Only doing a quick review, to remind of

principles you already know.

Consider a body moving around a circle with constant speed (e.g. stone on the end of a string).

Note:

(i) Constant speed. But is it constant velocity?

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(ii) Because the velocity is changing the body has an acceleration. The acceleration is called the centripetal acceleration, and

a = (Proof given in text)

(iii) Centripetal acceleration acts towards the centre of the circle (Is tangential to velocity).

(iv) If there is an acceleration then there is a Force that keeps body moving in a circle. It is called the centripetal force.

FC =

(v) The centripetal force acts to the center of the circle.

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WORK AND ENERGY Because we need a force(s) to do work, we begin with a review of forces. 1. Forces Some points to remember about forces are:

(i) From Newton’s 2nd Law we know that:

Which terms are vectors and which are scalar?

The equation is an example of what type of vector multiplication?

• If several forces act on a body, then we rewrite as:

• If a = 0 what does it imply for the version of the equation you just wrote?

Does it mean that there are no forces acting on the body?

(ii) Units:

2. Common Examples Let us examine some common scenarios for which you must be able to identify and label the forces present. a) Block resting on a surface b) Block suspended by a taut string c) Force applied to a block on a d) Force applied to a block on a rough surface (body stationary) rough surface (body moving)

M-16

e) Block on an inclined plane f) Particle on string swung in a vertical circle

g) Particle on string swung in a horizontal circle.

3. Work Now let us examine the concept of work.

• Work done by a constant Force If the force is constant (i.e. is not dependent on for example x or y) then

W = F.s where W � F �

s �

Note: (a) Work results from a dot product. What kind of quantity is work?

(b) s is displacement !!

(c) Since W = F.s we could expand to write:

(d) Consider the scenario below where F is at an angle θ and causes the body to move through s. What is W?

The dot product ensures that the force and the displacement multiplied are in the same direction.

i.e. Work = Force x displacement in

direction of force OR

Work = displacement x component of the Force in the direction of the displacement

θ

F

s

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(e) Consider, what if (i) F and s are perpendicular, then W = (ii) F and s are in opposite directions, then

(f) Units: Class Question 5 A force F acting on a particle can be represented by F = (6i+4j) N. Find the work done when the particle

moves from r1=(2i+4j) m to r2= (6i+7j) m.

Class Question 6 A block (mass = 15 kg) is acted upon by a force F and moves

up a frictionless slope with constant speed (see figure).

Determine (a) F, and (b) the work done by F on the crate.

• Work done by a Varying Force

If the Force is not constant but constantly changing, then to find the work done by the

changing force we first plot how the force varies with displacement.

For example, the diagram shows how the force, F varies with the displacement x, for a body. The work done in moving the body from x1 to x2,

= area under the graph, between x1 and x2

We can find the area either by: (i) Dividing into regular shapes (not befitting

a P14A student!!) OR

(ii) If we realise that this is the same as integrating i.e.

W =

D = 5.7 m

H = 2.5 m

F

F

x x1 x2

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Generalising, we can write: W = OR for 2-D motion we can write: W =

Class Question 7

A force F acting on a particle can be represented by F = (6xi+4j) N. Find the work done when the particle

moves from r1=(2i+4j) m to r2= (6i+7j) m.

4. A Special Work Case – The Spring

Consider a mass attached to a spring.

• If we displace the spring by a distance x, the spring will apply a restoring force i.e. in the opposite direction.

• Hooke’s Law tells us that:

• If we plotted F vs. x :

• To find the work done by the restoring force, F

• k � the spring constant. It is a measure of the stiffness of the spring.

• Its units are:

• Is F constant?

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5. Conservative Vs Non-Conservative Forces One more concept, to understand when dealing with work and forces.

A force is conservative if the work it does on a particle that moves through a closed path is

zero, otherwise the force is non-conservative.

OR A force is conservative if the work it does on a particle that moves between two points is the same for all paths connecting these two points Otherwise the force is non-conservative. Example: From the 1st definition, the restoring force of a spring is conservative. When it returns to its point of reference it has gone through a closed loop. Since the displacement is zero, the work is also zero (recall W= -½kx2). The spring force therefore satisfies conditions to be conservative force. Example: From the 2nd definition, weight or gravitational force is conservative as irrespective of how you raise a body the W = mgh, where h is the vertical distance travelled. Non-conservative forces would be those that dissipate energy (i.e. energy can never be recovered). Examples would be friction or air resistance. (If going from A to B by different paths, the particles would experience differing amounts of resistance, therefore the work can’t be the same). 6. Work-Kinetic Energy Theorem First let’s refresh ourselves about what Kinetic Energy, K is.

• K is the

• K =

• Units = Joules

The Work-Kinetic Energy Theorem states that the net work done on a body equals its change in kinetic energy provided all the forces acting on the body are conservative. Note: (i) If we know the change in speed we can find the work done or vice-versa. (ii) If the change in velocity = 0, then NO work done!!

Proof of Work-Kinetic Energy Theorem in 1 Dimension.

• Is KE a vector or scalar?

Extremely useful and well used formula. Don’t forget!!!

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7. Power Defined as the rate at which work is done.

a) Average Power:

b) Instantaneous Power:

Note: (i) Units = Watt, W = J s

-1

(ii) Power is a scalar.

Class Question 8 A body of mass 2 kg leaves a point O with velocity =

2ms-1

and follows the path shown in figure 3. Find its

velocity v at the 4 m mark. What is its power at this

point?

Class Question 9 A conservative force acts on a body of mass 4 kg such that its displacement in the x direction can be

represented by x = t + 2.0 t3, where t represents time. Find the work done by the force acting on the body

during the first 2 seconds of its motion.

F (N)

6

2 4 x (m) O

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CONSERVATION OF ENERGY, MECHANICAL ENERGY & POTENTIAL ENERGY 1. Introduction

• Energy is a measure associated with the state of one or more bodies.

• There exist different types of energy. Three which we are interested in are:

o Kinetic Energy, K

o Thermal Energy, ETH

o Potential Energy, U

• At any time a body can possess one or more of these forms of energy.

• Energy can be converted from one form to another.

• Units:

2. Mechanical Energy

Recall that a body can have different types of energy at the same time. If it has both Kinetic Energy and Potential Energy, then we can define:

Mechanical Energy, E =

Given, above we can also talk about the principle of Conservation of Mechanical Energy . It states that if only conservative forces act on an isolated system then the mechanical energy of the system cannot change or is conserved.

Implication:

Note: (i) As K increases, U decreases and vice versa (ii) condition for conservation is that forces must be conservative.

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3. Potential Energy, U

• Recall it is the energy associated with the configuration of a body/system in which a conservative force acts.

• We can derive a general formula for Potential Energy: If the forces are conservative:

We can use this generic formula to derive expressions for the Gravitational Potential Energy and the Elastic Potential Energy (Spring). • Elastic Potential Energy (Spring)

This is the energy associated with the state of compression or extension of an elastic (spring like) object.

Consider a mass attached to a spring as shown. The mass is displaced so that the spring is compressed (or extended) by x.

We already know how to derive the work done by the spring force in 1-dimension.

W = Therefore, ∆U =

• Gravitational Potential Energy

This is the energy associated with the state or separation between bodies that attract each other via the gravitational force.

We have an idea already what it should be, but we must know how to derive it from the generic formula.

Consider a body moved vertically upwards from y1 to y2.

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4. Conservation of Energy Recall that energy cannot be created or destroyed i.e. Total Energy is conserved. Let us examine the implications of this for an isolated system i.e. when no external forces act. a) If the Forces are Conservative If only conservative forces act in the system, then conservation of energy reduces to the conservation of mechanical energy. That is: We can use this idea to learn much about some common systems (scenarios). Let’s examine three such systems which we meet often. System 1 - A Body released from rest and falling What is (are) the forces acting?

Is (are) they conservative?

What does the conservation of energy principle tell us?

System 2 - A Pendulum released from rest and swinging

What is (are) the forces acting?

Is (are) they conservative?

What does the conservation of energy principle tell us?

System 3 - A Block going down a frictionless inclined plane toward a spring.

What is (are) the forces acting?

Is (are) they conservative?

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What does the conservation of energy principle tell us?

What if we consider a point D between B and C?

b) If the Forces are NOT Conservative

• If the forces are not conservative then some dissipative forces are acting e.g.

• We can use ∆EINT to represent the dissipative force effect on energy.

• We also know that Mechanical Energy, E, is not conserved as some energy is dissipated

usually in the form of heat. It means we can’t write:

• BUT since total energy is still conserved we can write:

• If the dissipative force is friction then we can relate ∆EINT (the energy lost by friction) to the

work done by the frictional force. We can write:

System 4 - A block sliding on a rough table to a spring

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Class Question 10

A 2 kg package slides along a horizontal floor and hits a spring with speed 4 m s-1

. Determine the

compression of the spring if (i) the floor is frictionless (ii) the floor has a coefficient of friction of 0.77. The

spring constant is 10,000 N/m and the package remains in contact with the floor while compressing the

spring.

5. Potential Energy Curves

Quick Aside:

Before we explore, recall what it means for a body to be in equilibrium.

Recall also that there are three types of equilibrium:

We will need this information in our examination of PE Curves!!

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For one dimensional systems, one consequence of the formula: ∆U = -∫F dx is that:

Fnet = -dU/dx -----(1)

If we therefore know how U varies with x, we know about the force acting on the system.

Some implications of equation (1) are:

1. If U as a function of x is known, we can determine a formula for F.

Example: If U = 3x2 + 4 what is the equation for F?

2. When dU/dx = 0, what can we say about the body in the system?

3. If given a plot of how U varies with x we can (i) sketch how F varies with x (ii) locate points

when the system is in equilibrium (iii) label the equilibrium points.

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CENTER OF MASS

By center of mass we mean:

We need to locate the center of mass for any body or any system of particles.

1. If the body is symmetric and of uniform density

2. If the body is irregularly shaped but of uniform density

3. If we have a system of particles

Class Question 11

Consider 3 particles A (4 kg), B (4 kg) and C (8 kg) arranged as shown and initially at rest. Locate the

center of mass.

6N

12 N

14 N

A (-2 ,3)

B (1,-2)

C (4,2)

45o

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4. If body is irregular and of uniform density BUT can be divided into regular shapes.

5. If the body is 3-dimensional, rigid and of uniform density

Motion of the Center of Mass

All of the above assumes that the particles are stationary. What however if the particles were

moving. What do you think would happen to the center of mass?

We can derive relationships to describe the motion of the center of mass.

i. Velocity

Recall from above that rcm =

Expanding:

Differentiating w.r.t. time:

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Class Question 12

An old Chrysler with mass 2400 kg is moving along a stretch of road at 80 km/h. It is followed by a Ford

with mass 1600 kg moving at 60 km/h. How fast is the center of mass of the 2 cars moving?

ii. Acceleration and Force

Recall from above that vcm =

Expanding:

Differentiating w.r.t. time:

Some implications of Σ Fext = Macm

1. Remember which quantities are vectors. The equation can be split into components.

2. It is an expression of Newton’s 2nd Law. We can therefore use Newton’s equations of motion for

the motion of the center of mass provided:

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Class Question 13

Refer to the three particles in Question 11. If they are acted on by the forces shown in the diagram,

determine the acceleration of the center of mass. Where is the center of mass located after 3 seconds?

3. What however if the sum of the external forces is zero or no external forces act on the system,

i.e. Σ Fext = 0?

It implies that acm = 0 !! What does this further imply?

Interpreting:

(i) If the center of mass was originally stationary, then regardless of the motion of the

particles in the system, once Σ Fext = 0, then the center of mass will remain in its original

position.

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Class Question 14

Two balloons touch each other as shown. One is inflated and the

other not. A thin membrane separates them. They are at rest and

no external forces act. At some time the membrane separating

them breaks and some of the air of the first is transferred to the

second. After time t they are of equal mass. At time t, has the

first balloon shifted i.e. to position A, B, C, or D? Explain.

(ii) If the center of mass was originally moving with a velocity vcm then after time t it will still be

moving with the same velocity i.e. vcm is constant.

B

C

D

A

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COLLISIONS A collision is an isolated event in which a relatively strong force acts on each of two or more

colliding bodies for a relatively short period of time.

To study collisions we must first review linear momentum. 1. Linear Momentum

• Linear momentum, p =

What kind of vector multiplication is it?

What kind of quantity is it?

• Units of momentum are:

• Newton’s 2nd Law can be written in terms of momentum

• If no net external force acts on the system i.e. Σ Fext = 0, then linear momentum is

conserved! Note the condition for conservation of linear momentum!!!

Proof:

We can now examine collisions.

2. Impulse

During a collision between two bodies, equal but opposite forces act on each body. The forces

vary with time and change the momentum of the respective bodies.

Impulse is a measure of the strength and duration of the collision force.

We use the symbol J to represent impulse.

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There are three methods which can be used to measure impulse:

• Method 1 – A Force versus time graph

• Method 2 – Linear Momentum Theorem

• Method 3 – From the equation including average force

Class Question 15

A body of mass 0.14 kg and initially moving with speed 39

m/s collides with a wall and recoils with speed 45 m/s in the

direction shown. What is the average force exerted during the

collision if the duration of the collision is 1.2 m/s?

Note: (i) What kind of quantity is J? (ii) What are the units of J?

45 m/s

39 m/s

30o

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3. Elastic vs Inelastic Collisions

• Momentum is conserved for all collisions between 2 bodies provided Σ Fext = 0.

That is we can write:

However this is not what differentiates elastic and inelastic collisions!!!

• If Kinetic Energy is also conserved in the collision then the collision is said to be ELASTIC,

otherwise it is INELASTIC.

Let’s examine each type of collision.

a) Elastic collisions

If the collision is elastic we know 2 things are conserved:

We use this knowledge to solve questions.

Class Question 16

A billiard ball moving at 5 m/s strikes a stationary ball of the same mass. After collision the first ball

moves at 4.33 m/s at an angle of 30o with respect to the original line of motion. Assuming an elastic

collision and ignoring friction and rotational motion, find the struck ball’s velocity.

NB. It is always useful when doing momentum questions to sketch a before and after picture, as linear momentum is conserved just before and just after the collision.

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b) Inelastic Collisions

• If the collision is inelastic we know at least one thing is conserved. What?

• Kinetic Energy is however not conserved i.e. Kbefore Kafter

So for inelastic collisions we often talk about the fractional change in Kinetic Energy.

• Inelastic collisions often (but not always) involve bodies that stick together, coalesce or

become embedded as a result of the collisions. Can you see why KE is not

conserved?

• Though Kinetic Energy is not conserved, the inelastic collision scenario will often

involve conservation of Total or Mechanical Energy, very often after or just prior to the

collision. See the example below.

Class Question 17

A bullet of mass 5 g is fired into a suspended block of wood of mass 1 kg. The block with the bullet

embedded is observed to rise 5 cm. Determine (a) the speed of the bullet and block immediately after the

collision (b) the initial speed of the bullet (c) the loss in energy due to the collision.

NB. It is again useful to sketch a diagram illustrating what is happening. Remember linear momentum is only conserved for the collision.

M-36

4. Final Thoughts on Linear Momentum

Remember linear momentum is conserved for the collision, i.e.

• So for example: A bullet is fired vertically upwards with velocity u and it travels through

height h before striking a stationary target. It becomes embedded and the bullet and target

move off together with a velocity v. Is the following equation true?

mbullet u = (mbullet + mtarget) v

• Or, a bullet has velocity u when it strikes and becomes embedded in a block at the edge of

a table. The bullet-block system flies off the table with velocity v’ and lands a distance x

away form the foot of the table. It’s velocity just before striking the ground is v. Which of

the following two statements is true?

A: mbullet u = (mbullet + mtarget) v’

B: mbullet u = (mbullet + mtarget) v

M-37

ROTATIONAL MOTION

So far we have only considered LINEAR or TRANSLATIONAL motion. We will change focus and

now begin to consider ROTATIONAL motion.

• By ROTATIONAL motion we mean a rigid body rotating about a fixed axis. Usually the axis of

rotation is located on the rotating body. This is illustrated below.

A good example is the earth rotating about its axis!

• Contrast this with CIRCULAR MOTION – a body moving in a circle, where the center of the

circle is not located on the body. A good example is the earth going around the sun.

• What two things therefore stand out about ROTATIONAL versus CIRCULAR MOTION?

• The beauty of ROTATIONAL MOTION however is that it is analogous to TRANSLATIONAL

MOTION. Know one and you know the other. We will utilise this concept.

Displacement

In TRANSLATIONAL MOTION: 1-D displacement is given by:

In ROTATIONAL MOTION: Displacement given by:

Note: 1. ∆θ called the angular displacement.

2. θ increasing is considered positive =

θ decreasing is considered negative =

3. θ is measured in radians. 1 revolution =

4. For n revolutions, the angular displacement ∆θ =

M-38

Velocity

In TRANSLATIONAL MOTION: Linear velocity is given by:

In ROTATIONAL MOTION: Angular velocity is given by:

Note: 1. Units of ω are:

2. When a body rotates with ω, all the particles on the body rotate with ω !!!!

Acceleration

In TRANSLATIONAL MOTION: Linear acceleration is given by:

In ROTATIONAL MOTION: Angular acceleration is given by:

Note: 1. Units of α are:

2. As with linear motion ω must change for there to be an α !!!!

Equations of Motion

Provided the angular acceleration is constant we can write equations of motion for rotational

motion.

TRANSLATIONAL MOTION ROTATIONAL MOTION

v = u + at

v2 = u2 + 2as

s = ut + ½ at2

Three Additional Equations

We consider three additional equations which are important as they link TRANSLATIONAL and

ROTATIONAL motion.

(1) θ = s/r

(2) v = r ω

Note: 1. θ �

s � r �

2. s is linear displacement which is linked to θ which is angular displacement.

Note: 1. v �

r � ω �

2. A particle on rotating body has a linear velocity, v. How is it directed?

3. Do all particles on a rotating body have the same v?

M-39

(3) A particle on a rotating body has a linear acceleration a which we can resolve into two linear

accelerations, ac � the centripetal acceleration and at � the tangential acceleration. How

would you draw them on the diagram below?

We can write equations for both linear accelerations:

ac =

at =

Why have we never met at before e.g. when we did CIRCULAR

motion?

Class Question 18

A record on a turntable rotates at the rate of 33 rev/min and takes 20.0s to come to rest. (a) What is the

angular acceleration of the record? (b) How many rotations does it make? (c) If the radius is 14 cm what is

at and ar at t=0? (c) What is the linear speed of a point 8cm from the center of the record?

Kinetic Energy

In TRANSLATIONAL MOTION: Kinetic Energy is given by:

We are going to derive a formula for the kinetic energy of a rotating body.

Consider a body rotating with angular velocity ω. Also consider the body as consisting on n

particles.

• Does each particle move with angular velocity ω?

Each particle on the body will possess a linear velocity.

• Are the linear velocities of each particle the same?

ω

ω

M-40

The KE of the rotating body, K = The sum of the KE for each particle

=

=

BUT for each particle, vi =

Therefore, K =

=

Let I =

Therefore, K =

Note: I is called the rotational inertia or moment of inertia of the rotating body. Let us examine in more

detail.

Rotational Inertia

• Rotational inertia, I, plays the same role in ROTATIONAL MOTION that mass, m, plays in

TRANSLATIONAL MOTION!! But I ≠ m !!!!

• The units of I are:

• For a system of particles rotating about a fixed axis: I =

Quick Example: 4 particles, each of mass 2 kg, are located as shown. If the system rotates about a vertical axis passing through particle P and perpendicular to the plane of the page, what is the rotational inertia of the system?

• For a rotating body of continuously distributed mass: I =

This formula is useful as it enables us to derive formulae for I for different common bodies.

See for example the diagrams on the extra handout. We will soon illustrate the use of this formula

to get the equation for the rod rotating about an axis passing through its center of mass.

o Note however that that if we combine 2 or more bodies. Their moments of inertia add.

ω

3 m

4 m

P

R Q

S

Quick Example: What is I of a rod of length L and mass, 2m, rotating about an axis through its center of mass, and with a particle of mass m attached at one end?

M-41

Class Question 19a

What is the moment of inertia about an axis perpendicular to a thin uniform rigid rod of length L and

mass M and passing through its center of mass?

Using the formula for I of a body of continuously distributed mass.

From the diagrams depicting I for different bodies, can you tell if I depends on where the axis of

rotation is located? Consider the rod. Why?

Does this mean we have to derive from scratch a new I every time we shift the axis of rotation?

NO! This would be too tedious. Instead we can use the Parallel Axis Theorem.

Parallel Axis Theorem: The rotational inertia of a body about any axis is equal to the rotational

inertia it would have about that axis if all its mass were concentrated at its center of mass plus the

rotational inertia about a parallel axis through the center of mass.

I = Icm + Mh2

Let us illustrate the use of this Theorem.

Icm �

I �

M �

h �

M-42

Class Question 19b

What is I if instead the axis of rotation (still perpendicular to the rod) is at one end of the rod?

Let us resume our comparison of TRANSLATIONAL and ROTATIONAL MOTION.

Torque

In TRANSLATIONAL MOTION: bodies move (in straight lines) due to Forces.

In ROTATIONAL MOTION: bodies rotate because of Torques. A torque is the tendency of a force

to rotate an object about an axis.

Note: 1. A Torque plays the same role in rotational motion that a Force plays in translational

motion. BUT a Torque ≠ Force.

2. Equation:

3. A Torque is the same thing as a moment!!! Units of Torque =

Quick Example 1: Consider a rod of length L, pivoted about its center of mass. A force F acts on one end as shown. 1) Any other forces on the rod? 2) Γ due to F = 3) Γ due to mg = 4) In what direction will the rod rotate?

Important!! If a force passes through the axis of rotation of a body, it does not contribute to the rotation of the body as its torque is zero!!!

4. By convention if the Torque causes anticlockwise motion it is positive and vice versa.

5. In the same way that we can find the net Force on a body, we can find the net Torque on a body.

F

M-43

Quick Example 2: Consider a disc of radius R with an axis of rotation through its center of mass. Four forces act on it as shown. The net torque on the disc is given by:

Quick Example 3: Consider a rod of length L rotating about one end. (a) Draw in all the forces contributing to the rotating torque at t = 0 and t = t.

At t = 0, Γ =

At t = t, Γ =

Newton’s 2nd Law

In TRANSLATIONAL MOTION:

In ROTATIONAL MOTION:

Work

In TRANSLATIONAL MOTION:

In ROTATIONAL MOTION:

Instantaneous Power

In TRANSLATIONAL MOTION:

In ROTATIONAL MOTION:

Work Kinetic Energy Theorem

In TRANSLATIONAL MOTION:

In ROTATIONAL MOTION:

F2

F1

F4

F3

θ

t = 0

t = t

M-44

One Special (Common Case)

Consider a rod of length L rotating about one end as

shown. After time t it has fallen through angle θ, Find

expression for (a) its angular acceleration α and (b) its

angular velocity ω after time t.

Class Question 20

A uniform disc whose mass M is 2.5kg and radius R is 20cm is mounted

on a fixed horizontal axle as shown in the figure. A block whose mass m

is 1.2kg hangs from a massless cord wrapped around the rim of the disk.

Find (a) the acceleration of the falling block (b) α (c) the tension in the

cord (the cord does not slip) (d) the kinetic energy of the disc after 2.5

seconds.

t = 0

t = t

M

m

M-45

ANGULAR MOMENTUM

In TRANSLATIONAL MOTION we discussed Linear momentum. Formula:

In ROTATIONAL MOTION we talk about ANGULAR MOMENTUM. Symbol:

There are two scenarios which we consider:

A) A particle moving with velocity, v, relative to an origin.

Consider a particle moving, with some velocity, v, and located at position vector, r, relative to an

origin, O (see diagram). It has an angular momentum about O, which is defined as:

Angular Momentum, L =

=

Magnitude of L =

=

Units: =

Direction:

Some things to note:

1. Though we are discussing rotational motion, the body does not have to be rotating to have

angular momentum!!! It just needs to move with velocity v relative to an origin!!!

Quick Example 1: Consider a particle of mass 2 kg located at distance r = 3m from a point O. It has velocity 4 m/s which can be directed along A, B, C, D, E or F as shown. What is its angular momentum in each case? Θ = 30

o.

2. If we change where the origin is located, then L must also change for the particle moving

with velocity v. Why?

3. If it is a system of i particles then: LNET = ΣLi

V

Origin

r

A

O

r θ

θ

B

C

D E

F

M-46

B) A rigid rotating body.

If we have a rigid rotating body, then the angular momentum, L =

Where, I =

ω =

(Compare this formula to the linear momentum formula!!)

Quick Example 2: (a) Consider a thin rod of length L and mass M = 2m, rotating about its center of mass with angular velocity ω. Write an expression for its L. (b) What would its L be if there were two particles of mass m attached at each end of the rod?

Newton’s 2nd Law and Momentum

Recall from TRANSLATIONAL MOTION that Newton’s 2nd Law can be written in terms of linear

momentum. It is given by:

Similarly, for ROTATIONAL MOTION Newton’s 2nd Law can be written in terms of angular

momentum. It is given by:

Conservation of Angular Momentum

Recall from TRANSLATIONAL MOTION that Linear Momentum is conserved if:

Similarly, for ROTATIONAL MOTION, Angular Momentum is conserved if:

Proof

Implication:

M-47

3 Common Conservation of Angular Momentum Scenarios

A) The Figure Skater

Which is bigger, ωo or ω?

Why does her angular velocity change?

B) Silly boy running with velocity v and jumps on to rim of stationary merry-go-round.

(You can treat the Merry-go-round like a disk of radius r, and the boy like a particle)

C) Silly boy walks from end of rotating merry-go-round to a point r/2 from center.

ωo ω

Before After

ω

Before After

ω

Before After

ωo

M-48

P

Vcm

VTOP

ROLLING Consider a wheel rolling smoothly (without slipping) on a flat surface. There are two ways to

consider its motion:

1) Pure Rotation about the point of contact, P, with the surface.

• Axis of rotation is through P and perpendicular to the plane of the paper.

• Rotation equations apply.

• If radius of the wheel is R, and wheel rotates with angular velocity ω, then VP = Vcm = VTOP =

2) Combination of Pure Rotation (about C.O.M.) and Pure Translational Motion. Kinetic Energy of Rolling If can appreciate the above, then KErolling = KErotation + KEtranslation

= + Proof: Consider a wheel rolling smoothly. Also consider rolling as pure rotation about P.

(P defined as before). K = BUT, Parallel Axis Theorem relates IP and ICM. How? Therefore, K =

VP = VP = VP =

VCM = VCM = VCM

VTOP = V VTOP = VTOP =

+ =

Pure Rotation

about COM

Pure

Translational

Rolling + =

M-49

BUT, VCM = Therefore, K =

Role of Friction in Rolling

If a wheel rolls at constant speed, it has no tendency to slide at point of contact P. However if a

net force acts on the body, to cause the rolling wheel to speed up or slow down, then it ensures

that acm or at exists. Note it also implies that the rate of rotation will also increase. Why?

These accelerations tend to make the wheel slide at point P.

If body is to not slide i.e. is to roll smoothly, then a frictional force must act on wheel to oppose the

tendency to slide!!!

Analogy: Box on incline

• Put in forces.

• Role of mgsinθ?

• Role then of fs?

Now consider body rolling smoothly down incline:

• Put in forces.

• Role of mgsinθ?

• First role of fs?

• But is there a net torque on the body?

• 2nd role of fs?

M-50

NOTE!! If body rolls smoothly on a flat surface i.e. without slipping => fs !!! If body rolls down a

string (e.g. yo-yo) no fs!!!

Question? What direction is fs for body rolling smoothly

on a flat surface with increasing rotational speed?

Class Question 22 An example of how to analyse rolling motion.

A uniform sphere of mass 2 kg is released from rest and rolls down an incline of height 0.4m and angle 8.1o

without slipping. (a) What is the acceleration of the sphere down the incline (b) the magnitude of the static

frictional force (c) the speed at the bottom of the incline. Find again the speed at the bottom using the

principle of conservation of energy.

• Draw in forces

• Divide motion into purely rotational (about c. of m) and purely translational. That is, when considering each type of motion, ignore the other.

• Use Newton’s 2nd Law

M-51

SIMPLE HARMONIC MOTION

• Another type of motion.

• Any motion that repeats itself periodically (i.e. oscillates) is called HARMONIC MOTION. a) Displacement

• We call it SIMPLE HARMONIC MOTION if we can represent its displacement with time by: x = A Cos(ωt + φ)

x = A = ω = f = T = φ =

Plotting the displacement: If φ = 0, then at t = 0, x = ; If φ ≠ 0, then at t = 0, x = x1

b) Velocity

• Since v = dx/dt =

• Note then that vmax =

• Plotting (for φ = 0) :

• v and x are out of phase by:

� At maximum displacement, velocity equals . When passing through equlbm position, velocity equals .

x

t

x

t

x

t

t

v

M-52

b) Acceleration

• Since a = dv/dt =

BUT, x = Therefore, a = IMPT!!! Is the defining equation for SHM!!!!! A body undergoes SHM if a proportional to –x!!

• Note then that amax =

• Plotting (for φ = 0) :

• a and x are out of phase by:

� At maximum displacement, acceleration is When passing through equilibrium position, Acceleration is

Question? Write equivalent SHM equation for Rotational Motion?

Class Question 23

The displacement of a body undergoing SHM is given by x = 4cos(πt + π/4). Determine (a) the

amplitude, frequency and period of the oscillations (b) relationships for velocity and acceleration

(c) the maximum velocity and acceleration (d) the displacement of the body between t = 0 and

t=1 second.

x

t

t

a

M-53

4 SHM Systems 1. Mass attached to a spring

Consider a mass m attached to spring of spring constant k, which is displaced from its equilibrium position by x.

• Since SHM Equation is a = -ω2x and for Mass-Spring a = -(k/m)x

ω2 = for mass-spring system!!

BUT, ω = 2π/T; Therefore, T = for mass-spring system!!

• All other SHM equation derived previously for x, v and a also true for Mass-spring system.

SHM Energy Considerations Can use mass spring-system to deduce the behaviour of PE and KE for SHM. For PE of Spring, U = = ------(1) Also we know KE, K = = = ------(2) Adding (1) and (2) gives, E = U + K (what is E?) = = Note: 1) E is constant!!!!

2) Could write equation: ω2A

2 = ω

2x

2 + v

2 (Can you prove?)

3) Plotting E vs. x

Restoring force due to spring. F =

From N’s 2nd Law and Hooke’s Law = Therefore, a =

=> SHM!!!!!!!

k M

M-54

2. Simple Pendulum

Consider a mass m attached to string of length l which is displaced from equilibrium by x.

• Since SHM Equation is a = -ω2x and for simple pendulum a = -(g/l)x

ω2 = for simple pendulum!!

BUT, ω = 2π/T; Therefore, T = for simple pendulum!!

• All other SHM equation derived previously for x, v and a also true. 3. Compound Pendulum

Not all masses suspended by string. Consider a rigid body suspended about a point on the body, e.g. a metre rule, and displaced through angle θ.

Note: (i) Pivot not necessarily at end of body!

(ii) d is distance from pivot to COM.

• Since SHM Equation is = -ω2 and for compound pendulum =

ω2 = for compound pendulum!!

BUT, ω = 2π/T; Therefore, T = for compound pendulum!!

Restoring force due to: F =

From N’s 2nd Law = If displacement is small: Therefore, a = => SHM!!!!!!!

θ

x

l

Restoring torque due to: Ґ =

From N’s 2nd Law and if angle small. = Therefore, = => SHM!!!!!!!

θ

Pivot d

M-55

• T depends on the type of body.

• As d � 0, what happens to T? Implications?

• All other SHM equation derived previously for x, v and a also true.

4. Torsion Pendulum

Not all pendulums swing back and forth. Consider a torsion pendulum displaced through angle θ.

• Therefore, T = for torsion pendulum!!

Class Question 24

An oscillator consists of a block attached to a spring (k= 400N/m). At some time t, the position (measured

from the system’s equilibrium location), velocity, and acceleration of the block are x = 0.100m,

v = -13.6m/s, and a = -123m/s2. Calculate (a) the frequency of oscillation, (b) the mass of the block, and

(c) the amplitude of the motion.

Restoring torque Ґ = - κθ

(κ is torsion constant. Is determined by l, diameter, and

material of suspension wire).

From N’s 2nd Law (no small angle approximation needed!) = Therefore, = => SHM!!!!!!!

ω

M-I

Some Rotational Inertias

M-II

DEPARTMENT OF PHYSICS MONA

P14 MATHEMATICAL NOTES

(Thanks to Dr. Patrick Chin for preparing these notes)

In the courses Introductory Physics A and B there are instances where simple calculus is required

and “HOW” it is done is summarized here.

1. DIFFERENTIATION

This is a mathematical operation which determines how one quantity varies when another is

changed.

Example: The rate of change of velocity v with time t is acceleration a.

This is written as a = dv/dt

OR Electric current i is the rate of flow of charge q.

This is written as i = dq /dt

How differentiation is done

When differentiating a function y with respect to a variable x, multiply each term in x with the

exponent (power) of x and subtract one from the exponent.

Example: Given y = 3x2

Differentiating y with respect to x gives dy/dx = 2 x 3x(2 - 1)

= 6 x

Example: Given y = 4 + 2x + 3x2

Differentiating y with respect to x gives dy/dx =0 + 2 + 6x

Example: Given y = 8x½

Differentiating y with respect to x gives dy/dx = ½ x 8x ½ - 1

= 4 x- ½

Differentiating trigonometric functions

θθ

θcos

)(sin=

d

d θ

θ

θsin

)(cos−=

d

d θ

θ

θ 2sec)(tan

=d

d

Differentiating exponential functions of x

The exponential function is unchanged but it is multiplied by the differential of the exponent with

respect to x .

Example: Given y = e 3x

dy/dx = 3 e 3x

Example: Given y=e -ax

where a is constant, dy/dx = -a e – ax

2. SOME USEFUL TRIGONOMETRIC RELATIONSHIPS

cos θ = sin (π/2 - θ) sin2 θ + cos

2 θ = 1 sec

2 θ = 1 + tan

2 θ tan θ = sin θ/cos θ

2sin θ cos θ = sin 2θ

sin (A + B) = sin A cos B + cos A sin B sin (A - B) = sin A cos B - cos A sin B

−

+=+

2cos

2sin2sinsin

BABABA

+

−=−

2cos

2sin2sinsin

BABABA

e+jθ

= cosθ + j sin θ e - jθ = cosθ - j sin θ

M-III

3. INTEGRATION

This mathematical process is the inverse of differentiation and is tantamount to a summation of the

function over the values of the independent variable.

• In mechanics, the work done by a force causing motion in the x direction is given by the

summation W = ∫ F. dx .

• In heat, the work done by a gas in expanding against pressure forces is given by the summation

W = ∫ p dV

How integration is done

When integrating a function y with terms containing powers of the independent variable x, add one

to the exponent and divide the new function by the new exponent.

Example: ∫ 6x dx = ½ x 6x 1+1

= 3x2

Example: ∫ (2 + 6x) dx = 2x + 3x2

Integrating the reciprocal of linear functions of x

A linear function of x contains terms that are either constant or include x 1

.

The integral will be a natural logarithm of the linear function.

Example: ∫ = xdxx

ln1

Example: ∫ +=+

)23ln(2

1

23

1xdx

x

Integrating trigonometric functions

∫ cos θ = sin θ ∫ sin θ = - cos θ ∫ sec2 θ = tan θ

Integrating simple exponential functions of x

The exponential function is unchanged but it is divided by the differential of the exponent (power)

with respect to x.

Example: ∫ e3x dx = (1/3) e

3x

Example: ∫ e-ax dx = (-1/a) e

-ax where a is constant.

When the limits of the variable are specified

Sometimes the range of values of the variable x is specified. The upper limit is placed at the top of

the integral sign and the lower limit at the bottom.

Do the integration first as normal. After the integration is done, substitute the upper limit of x into

the new function and subtract from it the value when the lower limit of x is substituted into the new

function.

Example: 45)1(3)4(3]3[6 22241

4

1=−==∫ xxdx

WAVES

&

OPTICS

W-1

WAVES – PART I

What is a wave? - A disturbance which enables information and energy to move from one point to

another without the need for a material object to travel the distance.

How is this different from a particle?

CLASSIFYING WAVES

There are different ways to classify waves. Let consider 3 ways below:

By Type

1) A Mechanical Wave - Waves that exist in a material medium and require the medium for

propagation.

� Examples:

� Ones most familiar with. Will use these to learn wave properties.

� Can apply Newton’s laws.

2) Electromagnetic Waves – Require no material medium to exist or to propagate. Can travel

through a vacuum.

� Examples:

� All have the same speed in a vacuum, c =

� Will look later on some of these.

3) Matter Waves – Waves associated with atomic particles e.g. electrons, protons and other

fundamental particles.

� Think of particles as matters, hence the name.

� We will not examine.

We will concentrate on mechanical waves, but the principles dealt with will be generally applicable to all types.

By the Motion of the Particles

Consider a mechanical wave e.g. on a string or on a spring. When the wave (disturbance) is created

and sent through the medium, the displaced particles of the medium will oscillate about their

immediate positions as the wave travels through the medium, but won’t propagate the length of the

medium. How did we know this again?

They can oscillate in two ways with respect to the motion of the disturbance.

1) Transverse Wave – The displacement of the elements/particles of the medium is

to the direction of travel of the wave.

•••• Examples:

•••• Remember particles

displaced up and down.

W-2

2) Longitudinal Wave - The displacement of the elements/particles of the medium is

to the direction of travel of the wave.

•••• Examples:

•••• Remember particle going side to side.

SOUND

SPRING

Note:

•••• In both cases, the particles are moving, and the wave is moving. (Remember this. It will

become important soon!) The particles therefore have a speed and the wave also has a

speed. Do you think the two speeds are the same?

•••• We will concentrate on transverse waves.

By the Motion of the Wave Form

1) Traveling Waves – If the wave form moves from one point to the

other it is called a traveling wave. We can further distinguish

between a pulse and a periodic wave.

a. Pulse – a (single) distortion/disturbance which moves

through the medium.

b. Periodic – Continuous. A wave consisting of cycles or

patterns that are repeated.

2) Standing Waves – If the wave form appears to be stationary, i.e.

you do not see a propagating wave. It is called a standing wave.

(We will return to this later!).

W-3

Now that we know about different types of waves, let’s apply some physics. Can we write an equation to characterize or describe a wave?

MATHEMATICAL DESCRIPTION OF A WAVE

We are going to make three provisions:

a) We are only considering transverse waves along a taut string, purely for ease of use. (Easy

to visualize). But the arguments are the same for longitudinal waves and for waves in other

media as well.

b) We are only considering propagation along an ideal string. So frictional effects that would

attenuate the wave as it travels are absent in the string. What would happen if we did not

ignore these effects?

c) We are assuming the string is long enough to ignore the rebounding (reflecting) of the wave

at the far end.

• The Equation:

Think of a wave on string. We want one equation that describes the wave. It therefore has to

describe the displacement (y position) of all the string particles (located in the x direction) and

how it changes with time (t) to allow the wave to propagate.

If you agree, the equation therefore has to be of the form:

where, y �

h �

If you were to choose a mathematical function that describes the shape of the wave you would

likely choose:

Therefore we write as the equation representing a traveling wave:

----- (1)

where, y(x,t) �

ym �

We’ll soon explain the other terms but let’s see why this is a really good equation for describing the wave.

Note:

• The units of y are:

• Sin (kx ± ωt) varies between and ; so y can vary between and .

• If the wave is moving to the right we use ; if it is moving to the left we use .

W-4

• The Beauty of this Equation:

If we fixed t, then => A SNAPSOT of the wave at that time.

What would this plot look like?

If we fixed x, then we would be looking at how the displacement (y) of a single wave particle

(located at x) varies with time.

What would this plot look like?

What about the other terms in the equation? Let’s examine.

• Wavelength and Angular Wave Number

Let us consider a wave moving to the right, at time t = 0, (note we have fixed t). A plot of this

wave would look like below. Do you agree?

1) We can define the wavelength, as the distance, λ, parallel to the direction of the wave’s

travel, between successive repetitions of the wave pattern. Units:

2) In diagram above, A and B are a wavelength apart, which also implies: yA yB

3) From equation (1) for these conditions (i.e. t = 0), we can write:

----(2)

4) We call k the angular wave number. Units:

A B

W-5

• Period and Angular Frequency

Let us consider a single particle (located at x = 0) on a wave moving to the right. (Note we have

now fixed x). A plot of this wave would look like below. Do you agree?

1) We can define the period, T, as:

Units:

2) An aside: We can also talk about the frequency, f, of a wave:

Units:

3) In diagram above, what can you say about the displacement of the particle at t1 and t2 ?

4) From equation (1) for these conditions (i.e. x = 0), we can write:

----(3)

5) We call ω the angular wave frequency. Units:

• Final Note on Wave Equation

The equation we have chosen to use i.e. equation (1) is a simple function that can be used to

represent a traveling wave. It is not the only possible representation however. The general wave

function is given by: y = f(x – vt).

t1 t2

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SAMPLE QUESTION 1

The wave function for a harmonic wave on a string is y(x,t) = (0.03m)sin(2.2x – 3.5t).

(a) In what direction is the wave traveling? (b) What is the amplitude, wavelength, frequency, and period of the wave? (c) Determine two locations on the string which have maximum displacement at t = 2

seconds.

SUMMARY

General wave equation: y(x,t) = ymsin(kx ± ωt)

Where, y(x, t) �

ym �

k �

ω �

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WAVES PART 2

Let’s now consider speeds associated with the propagating wave.

1. PHASE SPEED (in terms of other wave equation variables)

Consider (2 snapshots) used to illustrate a wave propagating to the right as shown.

It will have a speed, v,

called the phase speed of

the wave.

(Note: This is different

from the speed of the

particles).

Can we find an equation for v?

Note two things from the diagram above:

1) Every point on the wave form will be displaced ∆x in time ∆t.

. . . The average speed of the wave v =

. . . Instantaneous speed of the wave, (i.e. t � 0), v =

(Hmmm…. if we had a relationship telling us how x of the wave varies with time t we could find v).

2) The wave travels without a shape change. . . . Any given point on the wave retains its y

displacement. Example: Point A at t = 0 and at t = ∆t have the same y displacement.

(Note: Points on the string however do not retain their displacement).

From our wave equation:

For the displacements to be the same:

Oh, so now we have relationship for x as a function of time.

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So, from point 1):

---(1)

Note: a. The units are dimensionally consistent. True?

b. The fact that (kx-ωt) = constant implies that as t increases, x has to increase.

Can you see? The choice of –ωt for motion to the right is justified!!

[Prove for yourself that if we started with a wave going to the left we would

have had (kx+ ωt) = constant. Therefore as t increases, x would have to

decrease i.e. the wave would be moving to the left.]

c. We also know that: ω = and k =

Therefore, v = ----(2)

2. PHASE SPEED (in terms of the properties of the medium)

It shouldn’t be hard to see that the speed of a propagating wave should also be related to the

properties of the medium it is traveling in. After all:

o As the wave passes through the medium it must cause the particles in the medium to

oscillate by imparting energy (kinetic and potential energy).

o KE and PE associated with properties of the medium e.g. mass and elasticity.

o . . . Wave speed in one medium may not be the same as that in another medium having

different physical properties.

• For a wave on string: µτ /=v ----(3)

τ �

µ �

• For other medium we get similar equations.

For speed of sound: ρ/Bv = ----(4)

B �

ρ �

So speed depends on string properties i.e. how taut and how dense the string is!!

vsound in air (20oC) = 343 m/s

vsound in water (20oC)

= 1482 m/s

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Proof of Equation for Taut String

• Consider a pulse (or a segment of a wave) traveling from left to right with speed v.

• View the pulse and the string from a reference frame in which the pulse is at rest. In this frame,

the pulse is at rest and the string appears to move from right to left with speed v.

• Also now consider:

a. A very small length of the string, ∆l, of mass ∆m, at the top of the pulse.

b. Since the string is curved, the small element, ∆l, can be considered as moving through a

part of an arc of a larger circle of radius R.

c. But anything moving in a circle is subject to a centripetal force, such that;

Fcentripetal =

Therefore for string element, Fcentripetal = -----(a)

If we can find Fcentripetal then we have an equation for v.

• Fcentripetal is the net force acting to the center of the circle.

Force acting on the element is: It acts on both ends.

On each side it can be resolved into horizontal component =

And a vertical component toward the centre O =

What happens to the two horizontal components?

. . . FNET to center =

. . . Fcentripetal =

Assuming small angles: Fcentripetal =

From our knowledge of radians: 2θ =

. . . Fcentripetal = ----(b)

• Equating (a) and (b):

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2. SPEED OF THE OSCILLATING PARTICLES/SEGMENTS OF THE MEDIUM

TRANSPORTING THE WAVE (STRING)

We know that string elements are displaced in the y-direction according to the wave equation:

We also know that instantaneous speed, v =

-----(5)

Is v for the particle constant?

What is the maximum value of v?

3. ENERGY AND POWER OF A WAVE

Consider a Snapshot of a traveling wave

• When a wave is set up on a stretched

string, energy is provided for the

oscillatory motion of the string

elements.

• As the wave moves away, it transports

that energy as both kinetic, K, and

elastic potential energy, U.

• At b, K and U are at maximum. Why?

• At a, K and U are at minimum. Why?

• Change in Kinetic Energy, dK associated with a string element of mass dm is given by;

dK =

• Rate of Change of the Kinetic Energy:

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• Average Rate at which Kinetic Energy is transported:

Can you see that this also an expression for average power due to KE?

• Average Power transmitted by the wave

Remember that energy transported by both U and K

But it turns out that average KE = average PE in an ideal oscillating system.

. . . Paverage = (dK/dt)avg + (dU/dt)avg

= 2(dK/dt)avg

= ----(6)

Note: Energy and power are proportional to Amplitude2

CLASS QUESTION 2

Consider the wave function given before for a wave on a string i.e. y(x,t) = (0.03m)sin(2.2x – 3.5t).

(a) What is the phase speed of the wave?

(b) Derive the equation for the speed of a very short string segment.

(c) What is the maximum speed of a short string segment?

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Illustrating

WAVES PART 3

Interference and Superposition

Suppose two or more waves pass simultaneously through the same region. Example: Two

waves on a string or sound waves from two instruments. What will be expected to happen?

The phenomenon of combining waves is referred to as the interference.

Is there any way that we can find the resultant wave?

Yes (because physics simplifies things). But before we do so let us note some things!

• Phase Constant The resultant wave will depend on the extent of shift of one wave with respect to the other

i.e. on the extent to which the waves are in phase or out of phase with respect to each other.

Consider the simplest case of two identical waves, located in the same space at a given time.

We can describe them by: y1 = ym sin(kx – ωt) and y2 = ym sin(kx – ωt + φ)

How did we know the waves were identical?

In the 2nd wave equation, φ, is called the phase constant. Since it is the only difference

between the 2 waves it tells how much y2 is out of phase with y1 i.e. it is the phase

difference between the 2 waves. φ is an angle usually given in radians.

Scenario One: φ = 0 rad (or 0o)

Physical meaning:

• 2 waves are exactly in

phase

• Known as fully

constructive

interference

Implication for the

resultant amplitude:

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Illustrating

Illustrating

Scenario Two: φ = π rad (or 180o)

Scenario Three: φ = 2π/3 rad (or 120o)

Back to our original question: Can we write an equation for the resultant wave? Of course we can!

Physical meaning:

• 2 Waves are exactly out

of phase

• Known as fully

destructive

interference

Implication for the

resultant amplitude:

Physical meaning:

• Neither crests and

crests or crests and

troughs exactly overlap.

• Known as intermediate

interference

Implication for the

resultant amplitude:

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CASE ONE:

If we have two identical waves (same amplitude, frequency and wavelength) traveling the

same space in the same direction, with a phase difference of φ between them.

We can represent our two waves by:

We can find an equation for the resultant wave using the Principle of Superposition.

Principle of Superposition: When several effects occur simultaneously, the resultant effect

is the sum of the individual effects.

So for 2 waves in the same space: yR = y1 + y2 Principle of Superposition

. . . yR = y1 + y2

=

=

=

=

Note that {2 ym cos½ φ} is a constant. Why?

Let yRm = {2 ym cos½ φ}

. . . yR = -----(1)

Note the following:

1. We could only use this method (i.e. the Trig Identity) because we could factor out ym.

2. We get back a wave! Equation for the resultant looks like our previous wave equations.

3. The resultant for this case has the same wavelength and frequency as the two interfering

waves, and travels in the same direction. How did we know this?

4. In comparison to the first wave the resultant is phase shifted by

5. The resultant has a new amplitude given by: yRm =

i. Suppose φ = 0 rad i.e.

yR =

True for Φ = n2π, n = 0, 1, 2 etc., an integer

ii. Suppose Φ = π rad i.e.

yR =

True for Φ = nπ, n = 1, 3, 5 etc., odd integer

Two Trig Identities

2cos

2sin2sinsin

BABABA

−+=+

cos(-A) = cos(A)

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CLASS QUESTION 3

A wave given by y1 = 0.2 sin(4πx - 60 πt) interferes with a wave given by y2 = 0.2 sin(4πx - 60 πt + �/2). What is the equation of the resultant wave?

CASE TWO:

If we have two waves (same frequency and wavelength) traveling the same space in the

same direction, with a phase difference of φ but with different amplitudes!!!

We can represent our two waves by:

Can’t use previous method i.e. the Trig Identity! Why?

We use the Phasor Method.

• Says that all waves can be represented by a phasor.

• A phasor is a vector that has a magnitude equal to the amplitude of the wave.

• A phasor also has an angle above the x-axis which equals to the argument of the sin

function in the wave equation.

• A phasor rotates about the x and y axes with an angular speed equal to the angular

frequency of the wave.

We can therefore represent y1 and y2 as vectors drawn below.

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We can now easily get the resultant wave yR by adding the two vectors.

CLASS QUESTION 4

A wave given by y1 = 3 sin(4�x - 60 �t) interferes with a wave given by y2 = 4 sin(4�x - 60 �t + �/2). What is the equation of the resultant wave?

CASE THREE (Standing Waves):

If we have two identical waves (same amplitude, frequency and wavelength) traveling the

same space but in opposite directions. (For simplicity there is no phase difference).

We can represent our two waves by:

We know the resultant will be a

wave and therefore will have an

equation given by:

yR = yRm sin (kx –ωt + β).

Label on the diagram:

• yRm

• β

The resultant wave still has the

same k and ω of the original

waves!

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We can again find an equation for the resultant wave using the Principle of Superposition.

Why can we go back to this method i.e. using the Trig Identities?

. . . yR = y1 + y2

-----(2)

This is an interesting equation! Compare to original wave equation.

• There is no ± ωt in the sin term =>

• Called a Standing Wave. Looks like below.

• y” is the amplitude of oscillation of a particle located at x. Do all particles of a

standing wave have the same amplitude?

• Consider what happens when kx = nπ, where n = 0, 1, 2, 3, etc. At these points

(irrespective of the value of t) :

y” =

So points that satisfy kx = nπ are always at rest. These points are called nodes.

And nodes occur when x = nπ/k

Now, k = 2π/λ,

. . . nodes occur when x = -----(3)

Nodes occur every half of a wavelength.

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• Consider also points where kx = (n + ½ )π, where n = 0, 1, 2, 3, etc.

y” =

These points that can achieve amplitude 2ym are called antinodes. (Note this can

also be stated as: The amplitude of the antinodes = 2ym).

So antinodes satisfy kx = (n + ½ )π

Now, k = 2π/λ

. . . antinodes occur when x =

-----(4)

Antinodes also occur every half of a wavelength.

• Because the wave is not propagating, the positions of nodes and antinodes never

change!

• A standing wave on a stretched string (string with clamped ends) can be set up by

letting a traveling wave be reflected from the far end of the string. In this situation

the reflected wave travels back with the same amplitude, wavelength and combines

with the incident wave to produce a standing wave pattern.

Resonance

Before we leave the principle of interference, let us consider the case of a string stretched

between two clamps (so there is tension in the string), and a continuous sinusoidal wave of

a certain frequency is sent along the string. So for example as below:

The wave travels from one end to the other,

gets reflected and returns to the first end. It then

gets reflected from the first end and then

reflected again at the second end, and the

process continues. At the same time, the source

is also generating more sinusoidal waves. In a

short time there will be many waves of the

same amplitude going back and forth. The waves interfere and (as expected) produce a

standing wave pattern, which is also called an oscillation mode.

If we vary the frequencies of the waves (e.g. by varying the frequency of the source) we

will notice that at certain frequencies, the interference produces a standing wave pattern

with clear (very distinct) nodes and antinodes, with the antinodes also having very large

amplitudes. When this occurs we say that the string is in resonance with the applied

frequencies. So at resonance we get a distinct standing wave pattern with large amplitudes

for the antinodes.

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Question: What happens at other non resonant frequencies? Think of the lab!

There are many resonant frequencies. Three are shown below:

Note the following:

1. In (a), the first oscillation mode for resonance is shown. (Remember the two ends

are fixed and therefore are constrained to be nodes). The first resonance mode has

only one antinode at the centre.

. . . The wavelength has to satisfy: L = ½λ . Can you see why?

2. In (b), the second oscillation mode is shown. There are two antinodes.

. . . The wavelength has to satisfy: L = 2 (½λ) = λ

3. In (c), the third oscillation mode is shown. There are three antinodes.

. . . The wavelength has to satisfy: L = 3 (½λ) = (3/2)λ

4. Generalising, for the nth oscillation mode, there are n antinodes.

. . . The wavelength has to satisfy:

. . . The wavelength of the standing wave for the nth oscillation mode is given by:

λ =

Now, f =

. . . Resonant frequencies are given by:

-----(5)

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5. For a given stretched string, you can change the frequency at which a resonance

mode will occur by altering the tension in the string. Can you see why?

6. For a given stretched string, the lowest resonant frequency is achieved when n =1. It

is f = v/2L. Other resonant frequencies are integer multiples of this lowest

frequency.

7. The oscillation mode at this lowest frequency is termed the fundamental mode, or

the first harmonic (associated with musical instruments).

8. Therefore n=2 corresponds to the second harmonic; n=3 is the third harmonic, etc.

CLASS QUESTION 5

An oscillator of frequency 300 Hz sets up standing waves on a taut string which is clamped at both ends. The wave speed for the string is 200 m/s. The standing wave pattern was observed to have 5 loops and the amplitude was measured as 4.0 mm. (i) Determine the wavelength of the wave and the length of the string? (ii) Also obtain an equation for the displacement of the string as a function of position and time.

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Wavefronts, Rays and the Doppler Effect

Wavefronts and Rays

So far we have used a wave on a string to illustrate most of the wave principles. We will

now look at representing a wave via wavefronts and rays. This will be useful when we

examine sound waves and optic phenomena.

Wavefronts: Surfaces over which the oscillations of the particles of the medium

transporting the wave (e.g. air) have the same value.

Rays: Directed lines perpendicular to the wavefronts that indicate the direction of

travel of the wavefronts.

a) Spherical Wavefronts: Consider a point source S. It emits (for e.g. sound) in all

directions. We represent the direction of travel and the spread of the waves by the

diagram shown below.

o The spherical lines are the wavefronts.

o Wavefronts are like contours of a

specified value of displacement of the

particle e.g. the amplitude.

o Therefore distance between wavefronts

=

o Because of the 3 dimensional nature, the

wavefronts are spherical.

o The farther from the source the curvature gets less.

o Rays are perpendicular to the wavefronts.

o If the wave is a sound wave, the particles of air oscillate parallel to the direction

of the ray.

b) Planar Wavefronts: Consider 5th

point above. Sufficiently far away the source, the

curvature becomes so gradual that the wavefront is like a straight line. Wavefronts

represented by straight lines are called planar wavefronts.

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Doppler Effect

• Consider a police car with a blaring siren. You are standing at your gate. What do you

hear:

(i) When the car is far away?

(ii) As the car approaches?

(iii) After the car passes you and is driving away.

• The Doppler Effect is the change/modification in the observed frequency of a wave that

occurs when the source (e.g. the car) or the detector, or both move relative to the

transmitting medium (air). (However: If the transmitting medium is stationary then it

can be seen as due to relative motion between the detector and source).

The difference between the actual frequency and the observed frequency is called the

Doppler Shift.

The Doppler Effect holds for sound waves, electromagnetic waves including

microwaves, radio waves, and even visible light. Practical uses: Astronomy, and more

importantly in speed traps.

• We are only going to consider sound waves.

• Can we write an equation for the frequency we hear relative to the actual frequency?

In writing this equation we have assumed that the transmitting medium is air and air is

at rest with respect to ground.

How do we use the equation?

a) Note first that the detector controls the numerator. The source controls the

denominator.

b) Now consider the motion of the detector and source independently (i.e. irrespective

of what the other is doing) and its effect on f to determine the correct sign.

c) So (for example) consider first the detector:

• If the detector is moving toward the source, we expect the frequency of the

sound heard to be getting higher/lower i.e. f ‘ is increasing/decreasing.

Detector can only achieve this by sign in numerator. Why?

• If the detector is moving away from the source, we expect the frequency of the

sound heard to be higher/lower i.e. f ‘ is increasing/decreasing.

Detector can only achieve this by sign in numerator. Why?

f’ � f �

vD � vS �

v �

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d) Consider now the source:

• If the source is moving toward the detector, we expect the frequency of the

sound heard to be higher/lower i.e. f ‘ is increasing/decreasing.

Source can only achieve this by sign in denominator. Why?

• If the source is moving away from the detector, we expect the frequency of the

sound heard to be higher/lower i.e. f ‘ is increasing/decreasing

Source can only achieve this by sign in denominator. Why?

Note then that toward => Doppler shift up to higher frequencies. Away from => Doppler

shift down to lower frequencies.

CLASS QUESTION 6

Write the equation to describe the scenarios below. (Remember consider what detector is doing irrespective of what the source is doing and its effect on f and vice versa).

Source Detector Equation Source Detector Equation

(a)

0 speed

(d)

(b)

0 speed

(e)

(c)

(f)

Using Wavefront analysis to prove the Doppler Equation

(a) Detector and Source Stationary

• No Doppler Effect. Why?

• The detector is represented by an ear.

• Wavefronts are drawn one wavelength apart.

• Waves move with speed v.

• vt/λ is the no. of wavefronts moving to right over time t

• In time, t, the frequency is the rate at which

wavefronts are intercepted by the detector i.e. f = no of wavefronts/ t

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=

=

Did you expect this solution?

(b) Detector Moving to Stationary Source

• In time t, the detector intercepts waves moving to

right plus an additional amount of wavelengths due

to its motion to the left.

• Additional amount = vD t/λ.

• Therefore, frequency is now the new rate of

interception of wavelengths:

i.e f’ =

Could do the same if detector had been moving away from the source i.e. would

intercept less wavefronts so: f’ =

Generalizing: v

vvff D±

=' -----(1)

(c) Source Moving to Stationary Detector

• Source moving from position S1 to S7

• Detector stationary, source moving with vS

• W1 emitted when at S1

• Consider one period, T , later.

• By then source moves to S2 and emits W2.

• One more period later the source moves to S3

by which time it emits W3.

• What do you notice about distance between

wavefronts on right side (i.e. near to

detector)?

• But distance between wavefronts is the

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• The motion of S is changing the wavelength of the sound waves heard by the detector

and therefore the frequency heard by the detector.

• The new wavelength λ’ = distance between wavefronts

= vT - vST

Hence new frequency f’ = v /λ’

Could do the same if source had been moving away from the detector (put detector on left

side of diagram). Therefore motion of S causes wavelengths to increase:

f’ = fv/(v+vS)

Generalizing: Svv

vff

±='

-----(2)

Combining two scenarios we get the general equation for the Doppler Effect which we

noted before.

S

D

vv

vvff

±

±='

Reading Assignment: Read on “beats” from your text or from any other text.

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OPTICS 1

We turn our attention to optics and will spend the rest of our time examining light. We will

look at a couple of phenomenon which show that light is a wave. We start with Refraction.

Refraction

• What is it? When light travels through an interface/surface that separates two media

(e.g. air and water) it is refracted. Unless the incident ray was perpendicular to the

surface, refraction is evident by the bending of the light ray. The bending occurs at the

surface and the light travels in a straight line after being bent.

• Refractive Index: We assign different media a number which we will refer to as its

index of refraction, n. It is the factor by which the phase velocity is slowed down in the

material.

n = c/v c �

v �

Some common values for n

Vacuum 1 Air 1.00029 ≈ 1

Water 1.33 Sodium chloride 1.54

Crown glass 1.52 Diamond 2.42

• Illustrating Refraction: Consider the three scenarios below. Note for each we can

identify an incident ray, a refracted ray, and the normal. What is the normal?

(a) If n1 = n2

(b) If n1 < n2

(c) If n1 > n2

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• Law of Refraction: -----(1)

Can you see that it satisfies the three scenarios illustrated above?

• Effect of Refraction on Frequency and Wavelength

o Note that refraction does not change the frequency of the light wave. So the

frequency of light is the same in both media!

o HOWEVER, n1/n2 =

o Therefore, not only is the speed less when you go into a more optically dense

medium (i.e. n2 > n1) but the wavelength is also !!!

o If medium 1 is air (often the case):

Okay, so we know about Refraction. We need to be able to prove the Law of Refraction. To do so however he need to know Huygens’ Principle.

Huygens’ Principle

• Dutch physicist, Christian Huygens (1678) suggested a principle which allows us to

account for refraction in terms of wave motion i.e. cements that light is also wavelike.

• Principle: All points on a wavefront serve as point sources of spherical secondary

wavelets. After a time, t, the new position of the wavefront will be that of a surface

tangent to the secondary wavelets.

1

2

3

4

• Line ab is a wavefront

• Points on line ab maybe considered

as secondary sources (labeled

1,2,3,4…).

• After time ∆t, the wave from

secondary source 1 will spread out

(spherical wavefront called a

wavelet).

• Same for secondary sources 2, 3,

4… (after same time ∆t)

• After time ∆t we can draw a line

tangential to each wavelet (line de)

• That line is the new position of the

wavefront after time ∆t.

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Using Huygen’s Principle to Prove the Law of Refraction

Note: See diagram.

1. Let n2 > n1.

2. AB is a wavefront. From Huygen’s

Principle, A and B can be considered

secondary sources.

3. In time t, the secondary wavelet from

B will travel to D (a distance v1t).

4. In the same time t, the secondary

wavelet from A will travel to C (a

distance v2t).

5. v2t < v1t. Why?

6. But Huygen’s Principle says we can find the new position of the wavefront at a later

time by drawing the tangent to the wavelets. The new wavefront after time t is given by

CD. Can you see that the wavefront has to be bent?

7. Since rays are perpendicular to wavefronts, the refracted ray has also bent.

8. Proving Equation (1)

a. Consider triangle ABD:

Angle BAD = θ1 Go and Prove!

Therefore sinθ1 = ----(i)

b. Consider triangle ACD: Go and Prove!

Angle ADC = θ2

Therefore sinθ2 = ----(ii)

c. Eliminating AD from (i) and (ii)

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Revisiting Phase Difference

Knowledge of Refraction allows us to revisit phase difference. When did we meet before?

Consider two rays of monochromatic light entering two media as illustrated below. The

refractive indices are not the same, but the length L is the same for both media.

When the rays emerge from the media, the

two waves may not be in phase. Why?

Let’s assume wavelength in air is λ and also

that n2 > n1 .

To find the phase difference between the two waves:

Determine the no. of wavelengths in medium 1, N1 =

Determine the no. of wavelengths in medium 2, N2 =

Subtract the smaller from the larger:

Phase difference =

Note:

a) In this case the phase difference will come out to a decimal e.g. 43.6. The decimal

represents the number of wavelengths one wave is ahead of the other. So phase

difference can be expressed in terms of wavelength.

b) We often speak of the effective phase difference (BUT only when phase difference

is being expressed in terms of wavelength). For the above example in a) the

effective phase difference is 0.6. Why?

c) We can convert a phase difference written in terms of wavelength to radians to

degrees. How?

n1

n2

L

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CLASS QUESTION 7

In the same figure above, let the light rays have wavelength 550.0 nm before entering the media. If medium 1 is now air and medium 2 is a transparent plastic layer of index of refraction 1.600 and thickness 2.600 µm, determine (a) the wavelength of the wave in the plastic (b) the phase difference of the emerging waves in wavelengths and in radians. Also suggest what type of interference the waves would produce at a point on the screen if they were slightly angled so that they met.

Diffraction

If a wave encounters a barrier that has an opening of dimensions similar to the wavelength,

the part of the wave that passes through the opening will flare (spread) out into regions

beyond the barrier. The flaring out of the wave is referred to as diffraction.

(Note then that we cannot form a ray of light by passing it through a very small hole, as the

smaller the hole the greater the diffraction). We will meet again later. But we need to

appreciate for principle we are examining next.

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YOUNG’S DOUBLE SLIT EXPERIMENT

In 1801, Tomas Young experimentally proved that light is a wave. He did this by proving

that light waves can interfere. We will examine his experiment and note 7 important things

about the experiment. Let’s go!

1. Important Thing One – The Experiment

Below is an experimental set up (similar to but not exactly like Young’s experiment) to

study interference. Let us identify the parts numbered 1 through 5.

1 ���� S is a monochromatic source of light i.e.

2 ���� Planar wavefronts.

3 ���� Two narrow slits (S1 and S2).

What do you know about the light waves reaching S1 and S2?

What do you know about the light waves leaving S1 and S2?

4 ���� Two light rays, one from each slit, heading to a distant point P.

Do the two light rays above travel the same distance?

S1

S2

P P

1

2

3

4

5

L

Screen

W-32

5 ���� Screen located a distance L away from the slits.

The pattern seen on the screen is shown in the diagram to the right. Describe it:

2. Important Thing Two – Explaining what we see on the Screen

The bands are referred to as fringes. The pattern of alternating bright and dark fringes on the

screen is called the interference pattern. How do we get these fringes?

a) Consider first the central bright fringe.

b) Now consider another bright fringe i.e. not located at the center of the screen.

c) Now consider a dark fringe on the screen.

• The waves from the two slits interfere at the centre of the

screen.

• Were they initially in phase?

• Do they travel the same distance?

• Are they in phase at the center of the screen?

• Hence what do you expect to see?

L

DL

• The waves from the two slits interfere on the screen.

• Were they initially in phase?

• Do they travel the same distance?

• Are they in phase at the point of their interference on the

screen?

• What can you say about the extra distance (or path

difference) traveled by the wave from S2?

L

Dark (destructive interference)

(c)

• The waves from the two slits interfere on the screen.

• Were they initially in phase?

• Do they travel the same distance?

• Are they in phase at the point of their interference on the

screen?

• What can you say about the extra distance (or path

difference) traveled by the wave from S2?

W-33

3. Important Thing Three – An Equation (Physics Making Life Easy!)

Consider again the experimental set up shown below.

• For constructive interference (i.e. bright fringes): b =

But from triangle S1XS2 : path difference, b =

Therefore for constructive interference:

• For destructive interference (i.e. dark fringes): b =

But from triangle S1XS2 : path difference, b =

Therefore for destructive interference:

A Note on the Order of the Fringes: The index m above is known as the order of the

fringes. It describes the order/sequence of fringes on the screen. The central fringe or central

maximum (where the central axis meets the screen) corresponds to m= 0 (zero order). The

next bright fringe on either side of the central one corresponds to m =1. They are also

known as the first order bright fringes/maxima. The next pair (m = 2) are the second order

fringes and so on. In a similar manner we can identify 1st order minima, 2

nd order etc.

θ

d

S1

S2

P

L

b = path difference

1

2

O

Central axis

X

Note:

• d is the distance

between the slits.

• b is the path

difference.

• L is the slit to

screed distance.

• L >> d (Under this

condition, ray 1,

ray 2, and the line

OP are taken as

parallel).

• Line S1X is

perpendicular to

ray 2.

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A Note on Phase Difference: Remember we said we could define phase difference in terms of wavelengths or radians (or

degrees). In the above diagram, for rays 1 and 2:

Phase difference in wavelengths =

Therefore, phase difference in radians, φ =

For maxima: φ = =

For minima: φ = =

4. Important Thing Four – Another Equation but for the Fringe Separation

The distance between adjacent bright (or dark) fringes is constant. Can we come up with an

equation for the distance between the fringes? Of course we can! Look again on the

diagram.

For the mth bright fringe:

From triangle POQ: tanθ =

Recall we also knew from Important Thing Three that:

d sinθ =

θ

d

S1

S2

P

L

b = path difference

1

2

O

Central axis

X

Q

Ym

Note also:

• Let P be the mth

bright fringe.

• Let Ym be the

distance from the

central axis to P.

• From geometry,

angle POQ is also

equal to θ

W-35

If we assume that the angle θ is small (reasonable if L >> d), then

tanθ ≈ sinθ

-----(1)

• By the same logic, we can write for the m+1 bright fringe:

• Therefore the separation of the fringes, ∆Y, is given by:

-----(2)

Note:

1. We would have gotten the same answer by considering the mth and (m+1)th dark

fringes.

2. Equation 3 is an important equation. It tells us the position of the mth bright fringe

on the screen. However we can only use it if the small angle approximation is true

i.e. L >> d!!

If the question does not suggest this or explicitly say so (i.e. you can’t assume the

small angle approximation), could you still find Ym?

3. The equivalent equation to (1) for the mth dark fringe is:

4. Note the difference between Ym in equation (1) and ∆Y in equation (2).

CLASS QUESTION 8

A screen containing two slits 0.100 mm apart is 1.20 m from the viewing screen. Light of wavelength 500 nm falls on the slits from a distance source. Approximately how far apart will the bright interference be on the screen? Use the small angle approximation.

W-36

CLASS QUESTION 9

Red light of wavelength 664 nm (in vacuum/air) is used in Young’s double slit experiment. The slits separation is 0.120 mm. The screen is located at a distance of 2.75 m from the slits. Find the distance Y on the screen between the central maxima and the 3rd order bright fringe.

5. Important Thing Five – This thing called Coherence

Suppose the phase difference of the 2 light waves reaching the screen at a point P was

constantly changing with time (i.e. from in phase to out of phase back to in phase, etc.).

What would you expect to see on the screen?

Therefore, a condition required to obtain fringes (interference pattern) is the existence of a

phase difference that does not vary with time. When the phase difference does not vary with

time, the light from the slits S1 and S2 are said to be completely coherent.

Note then that we ensured coherency by making the light from S1 and S2 come from the

same wavefront. Light from an incandescent light bulb is not coherent. So we couldn’t

achieve fringes by replacing the slits with two light bulbs.

6. Important Thing Six – White Light as the Source

Suppose S is a source of white light. Interference is still possible because white light from a

suitable source satisfies the requirement of coherence. However white light contains all the

various colours of the spectrum (ROYGBIV), and each colour corresponds to a different

wavelength.

What does each colour produce individually on the screen?

Is the fringe separation the same for each colour (λred ≈ 660 nm, λviolet ≈ 410 nm)?

W-37

What would you expect to see at the center of the screen?

What would you expect to see on the screen?

On either side of the central maximum, there is one

group of coloured fringes for m =1, and another for m

= 2 so on. Inside any group of coloured fringes, red is

farthest from the central maximum, and violet is closest

to the central maximum. Why?

7. Important Thing Seven –Intensity

Though we say alternating bright and dark fringes, it is really bright fading into dark, back

to bright, etc. How bright will the central bright be in comparison to the light from one

source? Can we write an equation for intensity? Of course we can!!

First however we must digress and examine briefly light as an electromagnetic (EM) wave.

Light waves are electromagnetic waves (EM waves).

• They do not require a medium to propagate. They can travel in vacuum.

• All EM waves have the speed c = 299 792 458 m/s (≈ 3.0 x 108 m/s) in vacuum.

• EM waves consist of an electric component (E field) and a magnetic component (B

field). The two components are mutually perpendicular. They co-exist.

• E field is in the y

direction. The B

field is in the z

direction. The

wave propagates

in the x direction.

• E = Em sin(kx –

wt)

• B = Bm sin(kx –

wt)

• Em and Bm are

the amplitudes of

the fields.

• From EM theory. Intensity is the measure of energy per unit time per unit area. For an

EM wave it is proportional to the E field’s amplitude squared.

y

x

z

W-38

Back to Intensity in double slit interference.

We assume that the slits are very narrow in comparison to the wavelength. When this is the

case the intensity of light from a single source is essentially uniform over the region of the

screen in which we wish to examine the pattern. That is, especially, around the central

region.

• Consider for point P on the screen where the two light rays meet.

Let the E-field for light ray 1 be given by: E1 = Eo sin (kx - wt)

Let the E-field for light ray 2 be given by: E2 = Eo sin (kx - wt + φ)

Why the same amplitude Eo?

Why φ for light ray 2?

Then at P, the resultant E wave is given by:

Suppose the E waves didn’t have the same amplitude, how would you determine ER?

Hence, the amplitude of the E wave at P is given by:

-----(1)

• Now we know that Intensity is proportional to the E-wave amplitude squared.

So: Intensity due to one source, Io ∝ Eo2

And: Intensity due to the resultant, I ∝ ERm2

∴ I / Io = ERm2/ Eo

2

-----(2)

Note: This is an idealized result and true really for the interference of any 2 light rays of

same amplitude i.e. irrespective of how they meet.

W-39

For Maxima: φ = m2π

φ /2 =

∴ I = Imax =

• Note: We could therefore have written our intensity equation in terms of Imax, i.e.

For Minima: φ = (m+½)2π

φ /2 =

∴ I = Imin =

• Also don’t forget that for Young’s Double Slit Experiment φ = 2πd(sinθ) / λ

φ /2 =

• General intensity pattern:

W-40

CLASS QUESTION 10

In a Young’s double slit experiment the slits-to-screen distance is 2.2 m and the slit separation is 0.3 mm. The bright lines on the screen are found to be 4.4 mm apart. (a) Determine the colour of the lines (b) What is the relative intensity (in terms of Imax) on the screen at a point 1.1 mm from the central

maximum?

W-41

Thin-Film Interference

In optics so far, we have been examining interference. As we have seen, interference results

from two light waves meeting that have a path (or phase) difference between them.

So far we have seen that we can get a path difference between two light rays by:

1) Having one ray travel through a different medium. The wavelength of the ray

changes due to the medium. When the two rays meet they have traveled through a

different no. of wavelengths.

2) Having one travel a greater distance than the other in the same medium (e.g.

Young’s Double Slit).

It turns out that we can also cause a phase change by reflecting light (but some conditions

apply). We need to know this to examine thin film interference. So let’s examine first!!!!

The Effect of Reflection on Light Rays

• Light is reflected at the interface between two media.

• If the light is moving from n1 to n2 and n1 < n2 then the

reflected light ray undergoes a phase shift of π radians

or half a wavelength.

o If it were a wave on string meeting a hard

surface (e.g. wall), the reflected wave would

be inverted at the point of reflection.

• If the light is moving from n1 to n2 and n1 > n2 then the

reflected light ray does not undergo a phase shift.

Quick Question

Phase Shift Case

Ray1 Ray 2

n1<n2 & n2 < n3

n1<n2 & n2 > n3

n1>n2 & n2 > n3

1

2

n1

n2

n3

W-42

Thin Films

Now we are ready to examine thin film interference.

• Referring to a thin transparent film of uniform thickness, L, and refractive index n2.

• Film usually exists in between two other media n1 and n3 e.g. oil film on water (air is the

third medium).

• Film is illuminated by bright light of wavelength λ from a distant point source.

A) Case One: n1 < n2 & n2 > n3

• In the diagram which is the thin film?

• Trace the light rays through the top layer and

the thin film only!!

• Rays 1 and 2 on trace will interfere.

• Assume that the angle of incidence is very

small ≈ 0. Therefore, Path difference between

rays 1 and 2 due to travel in thin film (i.e. not

considering reflection) = 2L

• Will observer see constructive (bright) or

destructive (dark) when looking down into the

thin film (i.e. between rays 1 and 2)?

Answer depends on the path difference between rays 1 and 2 + the phase shift in

light rays on reflection!!!

• If due to reflection alone, rays 1 and 2 would be out of phase by 0.5 wavelength.

Agree?

Therefore, for Constructive interference:

Path difference due to travel in thin film =

OR =

OR =

And for Destructive interference:

Path difference due to travel in thin film =

OR =

OR =

n1

n2

n3

L

W-43

• These are not what we are used to seeing for constructive and destructive interference.

• These are not universal equation to be applied to all thin film questions!!!

B) Case Two: n1 < n2 & n2 < n3

• Diagram would be identical above.

• Will observer see constructive (bright) or destructive (dark) when looking down into the

thin film (i.e. between rays 1 and 2)?

Again the answer depends on the path difference between rays 1 and 2 + the

phase shift in light rays on reflection!!!

• If due to reflection: rays 1 and 2 will be in phase.

Agree?

Therefore, for Constructive interference:

Path difference due to travel in thin film =

OR =

OR =

Therefore, for Destructive interference:

Path difference due to travel in thin film =

OR =

OR =

• With thin films, you must first examine what reflection does, then formulate the

appropriate equations for constructive and destructive interference!!!

Interference from a thin wedge of air (wedge film):

We can use the concepts of thin films to explain the interference pattern seen when you

have a thin wedge of air. (You should remember this from the lab).

Setup:

• There are two glass plates touching at one end and separated by a sheet of paper (or a

wire) at the other end.

• The air between the glass plates is the thin film. It is between two glass layers.

• Same arguments apply as above e.g. incidence again is nearly perpendicular.

W-44

What do we expect to see?

Explaining the pattern

• If we trace the rays they look like below. This is very similar to Case A.

• So if we examine rays 1 and 2

due to reflection alone, do we

expect them to be in or put of

phase?

Therefore, for constructive interference:

And for destructive interference:

• BUT the thickness, t, of the film changes as we move from the left edge where the two

glass plates meet. Therefore as we move from left edge out, the condition for destructive

and constructive interference will be alternately satisfied, hence the pattern seen.

• Note that the mth order bright fringe is actually the (m+1)th bright fringe. Why?

• Note we could write the above equations in terms of l and θ, where l is the distance from

the point of meeting of the two glass slides to the bright or dark fringe, and θ is the

angle made by the glass slide with the horizontal.

Equations:

t

W-45

CLASS QUESTION 11 A broad beam of light of wavelength 683 nm is sent directly downward through the top plate of a pair of glass plates as shown. The refractive index of glass = 1.5. The plates are 120 mm long, touch at the left end, and are separated by a wire of diameter 0.048 mm at the right end. The air between the plates acts as a thin film. (i) Calculate the maximum order possible for the

bright fringes. (ii) How many bright fringes occur between the wire and the place where the two plates touch? (iii) What is the nature of the fringe (bright or dark) at the place of contact of the plates? Give reasons.

Newton’s Rings

This is another example of thin film interference (you also have met this in the lab).

Setup:

• As shown in the diagram. Curved glass surface (lens) on a glass block.

• This is very similar to the wedge of air example just done. Again air acts like a thin

film, between two glass surfaces.

• R is the radius of curvature of the lens.

• The thickness d of the air film changes as you move away from center.

• We assume d <<< R

• Expect to see interference pattern, but as concentric circles (due to curvature).

1 2

W-46

Explanation

• Due to reflection alone, do we expect rays 1 and 2 to be in or out of phase?

• Therefore, conditions for interference are as follows:

Constructive interference:

Destructive interference: m = 0, 1, 2, 3……….

• We can determine the radius for the mth bright fringe (ring).

Let its radius be rm

From the geometry: (R-d)2 + rm

2 = R

2

∴ For Bright Rings: rm = [(m + ½)λR]½

• For Dark rings: we would have used 2d = mλ

Therefore: rm = [m λR]½

• Separation of the fringes:

Bright rings: ∆r = rm+1 - rm

=

Dark rings: ∆r = rm+1 - rm

=

For large m i.e. m >> 1

∆r = ½{λR/m}½

W-47

CLASS QUESTION 12

For a Newton’s rings setup, R = 10 m and the diameter of the lens is 4 cm.

(a) How many bright fringes would you see if the apparatus was illuminated by yellow sodium light (λ = 590 nm)?

(b) What would be the diameter of the 6th bright fringe? (c) If glass has an index of refraction, n = 1.5 and water (n = 1.33) is placed between the two

pieces of glass, what change will take place in the bright fringe pattern?

W-48

DIFFRACTION

• We met diffraction before! We said it was the spreading, flaring or bending of light waves when they meet a narrow aperture. What condition?

• It turns out that the light that is getting diffracted can undergo interference and produce a pattern called the diffraction patterndiffraction patterndiffraction patterndiffraction pattern on a screen suitably placed at some distance from the obstacle.

• The diffraction pattern is not unlike the interference pattern (see below). Consists of bright and dark fringes.

• Phenomenon can be explained using Huygens’ Principle. • Light can also be diffracted by edges (i.e. does not have to go through a slit).

Single SSingle SSingle SSingle Slit diffraction:lit diffraction:lit diffraction:lit diffraction:

• Consider monochromatic light

passing through a narrow slit of

width a. (Just like example c

above)

• The viewing screen is far away

from the slit; L >> a

(a) (b) (c)

Diffraction patterns due to a (a) solid disc (b) razor blade (c) narrow slit.

a

L

θ

P

W-49

• Diffraction pattern is shown in the previous figure (i.e. figure ‘c’). How would you

describe it?

Pattern consists of a fairly broad maximum, and after that alternate dark, and bright

fringes of intensity less than the central maximum. The maxima of low intensity are

known as secondary/side maxima. Secondary maxima are located approximately

half way between the minima.

How does this differ from the Young’s Double Slit pattern?

• The typical intensity variation (as described above) is given by the figure below:

We can write an equation to first describe

the phenomena. Can you see that if we

write an equation for the location of the

minima we have essentially described the

phenomena?

It turns out (and we will prove below) that

for minima:

a sinθ = mλ ; m = 1, 2, 3…

----(1)

a �

θ �

If we assume the small angle approximation, then: a sinθ ≈ tanθ

----(2)

• We can also write an equation (will not prove) for the intensity of the light at any point

on the screen:

2

sin

=

α

αmII ; where

λ

θπφα

sin

2

a==

L

y

Im �

W-50

Let us try to explain how the single slit diffraction pattern arises.

The figure below illustrates the different possible scenarios that we can hypothesize.

(a) � • Represents the scenario corresponding to the central broad maximum.

• All waves/rays passing are in phase. Rays are nearly parallel. Thus they

interfere constructively and produce a bright fringe at the center.

• However not all the waves are exactly in phase, in fact some have path

difference of 0.000λ as opposed to exactly zero (near the edges). In this case,

interference is very nearly constructive but not full, but we can’t tell the

difference). These tend to produce less bright fringes around the center of the

screen and cause the decrease in the intensity as one move away from the

center.

• Thus a somewhat broad central maximum is seen.

(b) � • Represents the scenario corresponding to the first minimum.

• The deviation angle (deviation of rays from the dotted central line) θ is such

that the path difference for the ray from the top and the bottom is exactly one

wavelength (λ) when they reach the screen. So how dark?

• Mentally divide the slit into two. The path difference between the ray passing

through the center of the slit and the one at the bottom is ½λ . They interfere

destructively.

• Similarly, the next ray above the bottom one will interfere destructively with

the next ray above the central ray. This pairing continues and the result is a

dark fringe on the screen for this scenario.

• Similar arguments hold for the first dark fringe on the right hand side.

• The geometry suggests: a sinθ = λ

(c) � • Represents the scenario corresponding to the first secondary maximum on

the left hand side.

• Mentally divide the slit into 3-half wavelength zones. So, the angle θ is such

that the top ray travels (3/2)λ farther than the bottom ray.

• Consider the middle third and the bottom third. Their waves are out of phase

by half a wavelength. They interfere destructively for each pair.

a a

(a) Central broad maximum, θ = 0

(b) 1st Dark sinθ = λ/a

(c) 1st bright sinθ = 3λ/2a

(d) 2nd Dark sinθ = 2λ/a

W-51

• However, the rays from the top third (top zone) will not find partners to

interfere destructively. They reach the screen interfere to some extent and

produce a fringe of much less intensity than the central maximum, which is

the first secondary maximum

• The geometry suggests: a sinθ = 3λ/2

(d) � • Represents the scenario corresponding to the 2nd

dark fringe.

• The angle θ is such that the top ray travels 2λ farther than the bottom ray.

• Mentally divide the slit into 4 zones.

• Careful examination will show that the rays in one zone is exactly out of

phase (path difference = ½λ) with rays in the neighbouring zone. So

destructive interference occurs again.

• The geometry suggests: a sinθ = 2λ

DFFRACTION GRATING:

• A grating consists of a large number of equally spaced parallel slits. A grating of 10,000

lines per cm are common.

• The pattern consists of bright and dark fringes and

the analysis is similar to the double slit analysis.

There is however an important difference between

the patterns. The bright maxima of the grating are

sharper (b) and narrower than those of the double slit

pattern (a). They in essence look like a line (relate to

the lab).

• This property of the grating has made gratings very

useful in precise wavelength measurements, as

maxima can be referenced more precisely.

• Principal maxima are given by:

d sinθ = mλ; m = 0, 1, 2, 3,…

d � slit separation.

• Diffraction patterns can be obtained using more than one wavelength, and also white

light.

W-52

Class Question

In a single slit diffraction experiment a laser beam of wavelength 700 nm passes through a vertical slit 0.2 mm wide and hits a screen 6 m away. Find the width of the central maximum (a) in cm (b) in radians.

Class Question

Sodium light is incident on a diffraction grating with 12,000 lines per cm. At what angles in degrees will the 2 yellow lines (called sodium D lines) of wavelength 589.00 nm and 589.59 nm be seen in the first order.

HEAT

&

THERMODYNAMICS

H-1

Heat, Thermodynamics and (a little) Kinetic Theory

Some preliminaries:

Temperature

It goes without saying that the topic is very related to temperature. We will spend some time first

examining temperature.

• For thermodynamics we use the scale. It is the way we define

thermodynamic temperature.

• The scale: To set up any temperature scale we need a standard fixed point which we assign a

value.

E.g. for the Celsius scale our fixed point(s) are:

For the Kelvin scale we use:

• For the Kelvin scale, the standard thermometer against which all other thermometers are

calibrated is a constant volume thermometer.

gas

gas cv

T

Temperature to be measured

h

Gas filled bulb

Mercury manometer

Mercury reservoir

Gas filled bulb –

Mercury manometer –

Reservoir -

H-2

Using the above, the equation for temperature T is given by:

Note that m � 0. Why?

If we change the gas in the bulb, the temperature recorded is

slightly different. However it turns out that as we use smaller

and smaller amounts of gas (i.e. m � 0) the readings converge

to a single T, irrespective of the gas used. M � 0 therefore

implies a mass of negligible mass and density otherwise called an:

** Importantly, what we are doing is ensuring that the temperature scale is independent of the

property of the thermometer or the substance in the thermometer!!!

• We relate the Kelvin and Celsius scale by the formula:

Ideal Gases

As suggested above, an ideal gas is one of negligible mass and density i.e. The molecules are so

far apart they exert no forces on each other.

Experimentation shows that the following relationship of variables is true for an ideal gas:

IDEAL GAS LAW:

P = V =

R = T =

n =

T =

P =

P3 =

Note: (i) No such thing as an ideal gas, but all

real gases approach ideal gases at low enough

densities. The above equation however gives us

insights into the limiting behaviour of real gases.

(ii) If P and/or V change, then T must change ,

and vice versa. Also if given P. V, and n, then we

really know T.

Question: Determine the volume of 1.00 mol of

any gas at STP (i.e. T = 273 K and P = 1.0 atm =

1.013 x 105 N/m

2).

H-3

Heat, Q

A form of energy which can be transferred to or from a body.

The energy is called:

Relating Heat and Temperature

• Heat Capacity, C: For any substance (i.e. solid liquid or gas) which changes its temperature by

the addition or removal of heat, we can write the general equation:

C =

Units of C =

In differential form:

• Specific Heat Capacity, c: If we know the mass of the substance we can rewrite the formula as:

m =

c =

Definition of c:

Units of c:

• Molar Heat Capacity: We will deal with mostly gases. Very often we don’t know the mass of

the gas, but we know how many moles of the gas there are. In that case we rewrite our Heat

formula in terms of the molar heat capacity. Define molar heat capacity:

There are 2 forms of the heat formula with molar heat capacities:

For heat transfer at constant pressure:

For heat transfer at constant volume:

Therefore, Cv =

And, Cp =

Units of Cv and Cp are:

Heat and Change of State

It is important to realize that a body’s temperature does not have to change when Heat is added to it.

How so?

In that case the formula for heat becomes: m =

L =

Units of L =

There are 3 types of L: Lv =

Lf =

Ls =

Won’t use these much, but we must remember that addition or

removal of Heat does not have to change the temperature of a body!!!

H-4

THERMODYNAMICS

Thermodynamics is the study of interactions between Heat and other forms of energy. It is

applicable to any object or collection of objects (called a system) whether solid, liquid or gas.

We will concentrate on gaseous systems. Why? It turns out that that with a gas we can use Heat to

do work. How?

We will examine the relationship between, heat, work, energy, and the various variables found in the

ideal gas equation, using the laws of thermodynamics!!

First law of Thermodynamics

Stated by the formula:

Note that the law is really just an expression of the conservation of Energy!!!

In differential form:

Let us examine each term.

Heat, Q

• We already defined. Since we are looking at gaseous systems we will mostly use the equations:

• By convention: Q is positive when:

Q is negative when:

Work, W

• A gas can do work by

OR when work is done on a gas it causes the gas to

Therefore, work is related to change in

H-5

• We write a general formula for work: W =

V =

• If the gas changes volume while the pressure is constant:

• The above simple formula helps us see the sign conventions easily.

W is positive => =>

W is negative => =>

• What if the pressure not constant but the temperature is constant:

Internal energy, ∆U

The internal energy has to do with the kinetic and potential energies associated with the random

motion of the atoms and molecules of the gas. As we will later see, the internal energy is dependent

only on the state of the gas (i.e. its T, P and V) and is independent of the path a system takes.

• We can write a formula for internal energy:

Note: (i) True even if volume is not constant!!!!!

(ii) If the temperature of a boy changes then ∆U must change and vice versa.

• Sign convention: ∆U is positive when

∆U is negative when

H-6

Summary

Q ∆U W

Formula

Sign Convention

Question:

(a) An amount of heat equal to 2500 J is added to a system, and 1800 J of work is done on the

system. What is the change in internal energy of the system? (b) What would be the internal energy

change if 2500 J of heat is added to the system and 1800 J of work is done by the system?

Indicator Diagrams

An indicator diagram is a P versus V diagram. It is extremely useful for illustrating thermodynamic

processes.

• E.g. Allows us to trace what happens to a gas if heat added to a system i.e. If have a gas with

initial conditions, Pi, Vi and heat added such that it now expands and changes pressure to Pf, Vf,

then we can plot this on a P-V diagram.

* Note the arrow indicates the

thermodynamic process that changed the

gases state.

* Could you find Ti and Tf?

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• Area under a P-V graph equals the Work done by that process!!!

Some Common Thermodynamic Processes

1. Isochoric =>

P-V Diagram Implication for Ideal gas Law

Work done 1st Law of Thermodynamics

2. Isobaric =>

P-V Diagram Implication for Ideal gas Law

Work done 1st Law of Thermodynamics

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3. Isothermal =>

P-V Diagram Implication for Ideal gas Law

Work done 1st Law of Thermodynamics

4. Adiabatic =>

P-V Diagram

Ideal Gas Law 1st Law of Thermodynamics

5. Cyclical Process =>

Examples of P-V Diagrams:

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Now, though we can apply the first law of thermodynamics to each stage of the cyclical process.

We can also apply it to the process as a whole.

* Implications of 1st Law for the process as a whole:

* Remember also that the work done for the cycle =

6. Free Expansion =>

Relating, Cp, Cv and R

We can deduce a relationship between Cp, Cv and R, by considering a constant pressure process.

1st Law of Thermodynamics:

H-10

Question:

Three moles of an ideal gas , initially at P = 5 x 104 N/m

2 and V = 0.2 m

3 is expanded isothermally

to twice its initial volume. The gas is then compressed at constant pressure and, after that heated at

constant volume so as to restore it to its initial state.

a. Sketch the above three processes on a P-V diagram.

b. Calculate the heat input and the work done by the gas during each process, if Cp for the

gas is 29.1 J K-1

mol-1

c. What is the net heat input, the net work, and the net change in internal for the cycle?

H-11

KINETIC THEORY

Kinetic theory is based on the concept that matter is made up of atoms which are in continual

random motion. Therefore the atoms of a gas have kinetic energy. The kinetic energy can be

translational kinetic energy or rotational kinetic energy or vibrational kinetic energy.

Translational Kinetic Energy

Can we derive an equation for translational KE?

Root Mean Square Speed

For an ideal gas, there exists a large no. of molecules, N, each of mass, m, moving in random

directions with a variety of speeds.

1. We define the root mean square speed of the molecules:

2. Note: that it is the square root of the average of the speeds squared!!

3. We can also relate vrms to the state of the gas by:

Kinetic Energy

With knowledge of vrms, we can define the average translational KE per molecule of a gas.

i.e. average K of 1 molecule of a gas:

Therefore, average K for the gas:

Note: This is the average KE of translational motion, where translational motion of

a body is such that all parts move in parallel directions through equal distances

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Monatomic Gases

If the gas is monatomic (What does that mean? ) then the gas has

only translational kinetic energy i.e. its internal energyat any instant, U, is entirely due to

translational KE.

Therefore for a monatomoic gas, U =

***Recall what we said about the relationship between U and temperature!!!

This also implies that the change in internal energy of the gas, ∆U =

***True only for monatomic gas!!!!

Relating Cv and R for a Monatomic Gas

Recall that we had a general formula for ∆U, i.e. ∆U =

From above we just learnt that; ∆U =

Equating the two equations we get for a monatomic gas: =

Relating Cp and R for a Monatomic Gas

Recall that we had a relationship between Cp , Cv and R:

Use above relationship:

Polyatomic Gases

What if the gas is not monatomic? Diatomic and polyatomic gases have translational KE as well as

rotational and vibrational KE. This means that the internal energy of a polyatomic gas, ∆U, is equal

to the translational KE + rotational KE + Vibrational KE. Can use to find the KE of rotation and

vibration. How?

Finding a Relationship for Cv for a Polyatomic Gas

This is important as if we know Cv we know ∆U.

We use the idea that molecules store energy in independent ways, and the no. of independent ways

that the energy can be stored is called the no. of degrees of freedom.

The Energy of each degree of freedom for one molecule = ½kT

If the molecule has f degrees of freedom, then energy of the molecule =

Therefore the internal energy, U, for the gas =

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As before, we know the general formula for ∆U:

Therefore for a polyatomic gas:

Note:

1. For a monatomic gas molecule there are 3 translational degrees of freedom:

2. For a diatomic gas molecule, there are 3 translational and 2 rotational degrees of freedom:

Question

Suppose 4.00 mol of an ideal diatomic gas, with molecular rotation but not oscillation, experiences a

temperature increase of 60.0 K under constant pressure conditions. (a) How much heat is added to

the gas? (b) By how much did the internal energy of the gas increase? (c) How much work was done

by the gas? (d) By how much did the internal translational kinetic energy of the gas increase? (e) By

how much did the rotational internal energy of the gas increase? (Use k = 1.38 x 10-23

J/K)

H-14

ADIABATIC EXPANSION OF AN IDEAL GAS

A) Recall for an adiabatic expansion, Q =

Therefore from 1st Law of Thermodynamics (in differential form):

If we assume the expansion is so rapid that pressure is constant, then W =

Hence from above, dU =

=

=

= -----(1)

B) Now for an ideal gas: =

=

=

=

= ----(2)

Equating (1) and (2)

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HEAT ENGINES and EFFICIENCIES

Heat Engine: A device which changes heat into work while operating in a cycle. E.g. petrol engine,

nuclear plant, etc. We know that for a cycle ∆Ucycle = 0

Therefore, Qcycle =

QNET =

How does a Heat Engine do Work?

Schematically

Representing a Typical Heat Engine On a P-V Diagram

Diagram

1. Gas is in contact with a hot surface. TH is the heat

source or Heat reservoir at temperature TH.

2. Gas extracts heat, QH from reservoir.

3. Gas expands and does work.

4. Gas delivers remaining heat QC to cold reservoir and

returns to original state via compression (work done on

gas).

5. TC is a heat sink or heat reservoir at temperature TC.

6. The following relationship is true:

Note: we are neglecting friction and other heat losses!!!

1. The engine is a cycle of processes.

Implication?

2. Usually 4 arms/processes with arrows indicating cycle.

3. In one arm, QH given to system.

4. In one arm, QC taken from system.

5. Work done is area under the graph.

6. We can write:

H-16

2nd

Law of Thermodynamics (First form)

It is not possible to change heat completely to work with no other change taking place.

Implications:

Efficiency

We define thermal efficiency as:

Note:

1. Theoretical range for efficiency, e is from:

2. For e = 1, |W| = |QH| Is this possible?

Implication?

Maximum Efficiency of an Engine

We compare the efficiency of all engines to the efficiency of a Carnot Engine. A Carnot Engine

undergoes four specific processes (which we will describe). We use it as the reference due to its

historical significance (it was the first engine used to analyze efficiency). Properties of the Carnot

Engine include:

1. It is an ideal engine

2. No unwanted heat losses e.g. by frictional parts, or loss due to conduction from sides of

cylinders.

3. The working substance is an ideal gas (e.g. not a mixture of air and gas vapor).

4. Processes are carried out in infinitesimal steps to maintain equilibrium.

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5. Each process/arm is reversible (can be retraced exactly).

Cycles P-V Diagram

Efficiency Equation for a Carnot Engine

Recall that efficiency is given by:

For a Carnot Engine we can write:

Note:

1. Temperatures are in Kelvin

2. Only need to know hottest (TH) and coldest (TC) temperatures to deduce efficiency.

3. Given TH and TC for any engine (not necessarily Carnot), the formula above gives the

maximum possible value for its efficiency. In reality the efficiency will be les than that

calculated using formula above.

4. There can never be a perfect engine, as to get e = 1:

Proof of Efficiency Equation for a Carnot Engine

A

C

B

D

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H-19

REFRIGERATOR

Reverse of an Engine. This means that we use work to transfer energy from a low temperature

reservoir (TC) to a high temperature reservoir (TH) in a cycle. E.g. the compressor of the refrigerator

extracts heat from the inside of the fridge and releases it outside.

Schematic First Law Implications

Coefficient of Performance

We can define a coefficient of performance:

Since refrigerator is like a Carnot engine in reverse:

2nd

Law of Thermodynamics (Second form)

The Coefficient of performance is limited by the 2nd

Law of Thermodynamics, which states: It si not

possible for heat to be transferred from one body to another that is at a higher temperature without

any other changes taking place.

Implication: No perfect refrigerators!!!!

compressor

Freezing Compartment

H-20

Question

The boiler of a steam engine has a temperature of 520 oC while its condenser has a temperature of

100 oC. What is the maximum possible efficiency of the engine?

Question

An ideal refrigerator of coefficient of performance K = 4.7 extracts heat from the cold chamber at

the rate of 250 J/cycle. (a) How much work per cycle is required to operate the refrigerator? (b) How

much heat is discharged to the room?

H-21

ENTROPY

Irreversible Processes

� One way processes that cannot be reversed by small changes in their environment

are called irreversible.

� Irreversible processes are so common that if they were to occur spontaneously (on

their own) we would be surprized.

� For example if you placed your hands around a cup of tea, you would be surprized if

your hands got cooler and the cup got warmer.

� That would clearly be the wrong way for the energy transfer.

� However the total energy of the closed system (hands + cup of tea) would be the

same as the total energy if the processes had run the right way.

� Thus, changes in energy within closed system do not set the direction of an

irreversible process. This is defined by what we will call the “change in entropy”

(∆S) of the system.

� If an irreversible process occurs in a closed system, the entropy of the system

always increases; it never decreases.

� Entropy differs from energy in that entropy does not obey

a conservation law.

� The energy in a closed system is conserved; it always remains

the same.

� For irreversible processes, the entropy of a closed

system always increases.

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Figure 21.1

Figure 21.2

Figure 21.3 Figure 21.4

� For an isothermal expansion:

∫=−=∆f

iif dQ

TSSS

1

Change in Entropy

� Consider the free expansion shown in Figure 21.1.

The initial and final states (Pi,Vi) and (Pf,Vf) are

shown in the PV diagram

� During the free expansion the temperature does not

change.

� We define the change in entropy Sf - Si of a

system during a process that takes the system from

an initial state i to a final state f as

∫=−=∆f

ifi

T

dQSSS

� S.I. Units for S: J/K

Like P, V and T, S is a state property – i.e. a property

that depends only on the state of the gas and not on how

it reached that state.

Remember: We cannot plot a free expansion!

� In the free expansion in the example Ti =Tf.

� We can therefore replace the free expansion

with an isothermal expansion that connect

states (Pi,Vi) and (Pf,Vf).

� See Figures 21.3 and 21.4

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� Because QdQ =∫ , where Q is the total energy transferred as heat during the process we

have:

T

QSSS if =−=∆ (change in entropy, isothermal process)

Entropy as a State Function

� We have assumed that entropy, like pressure, energy and temperature, is a property of the

state of the system and is independent of how that state is reached.

� However we can prove it is a state function for the case in which an ideal gas is taken

through a reversible process.

� To make the process reversible it is done in very small steps.

� Therefore from the 1st Law of Thermodynamics we have:

PdVdTnCdWdUdQ V +=+=

Substituting for P using the ideal gas law we have

dTnCdVV

nRTdQ V+=

T

dTnC

V

dVnR

T

dQV+=

Integrating each term between some initial state i and final state f:

∫∫∫ +=f

iV

f

i

f

i T

dTnC

V

dVnR

T

dQ

i

f

V

i

f

T

TnC

V

VnRS lnln +=∆

� Therefore the change in entropy depends only on the properties of the initial state (Vi and

Ti) and final states (Vf and Tf)

� It does not depend on how the system changes between the two states.

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The Second Law of Thermodynamics

� Consider now the reverse of the isothermal process shown on the previous page.

� In this case the gas will give back heat Q to the reservoir at temperature T.

� The second law of thermodynamics can be written as:

0≥∆S

Entropy in the Real World: Engines

Carnot Engine

i

f

The change in entropy of

the gas:

gasT

QS −=∆

The change in entropy of

the reservoir:

T

QS rev =∆

The net change in entropy

of the gas-reservoir system:

0=∆+∆=∆ revgas SSS

In a process that occurs in a

closed system, the entropy

of the system increases for

irreversible processes and

remains constant for

reversible processes. It

never decreases.

A → B: Isothermal expansion Gas in contact with reservoir at temperature, TH

QH given to gas to keep at constant temperature

B → C: Adiabatic expansion to temperature, TC

No heat lost in this step.

Gas cools i.e. temperature falls

C → D: Isothermal compression

Gas in contact with reservoir at temperature, TC

QC removed from gas to keep at constant temperature

D → A: Adiabatic compression to temperature, TH No heat lost again

Gas warms i.e. temperature rises

H-25

Question A Carnot engine operates between temperature TH = 850 K and TL = 300K. Then engine performs

1200 J of work each cycle, which takes 0.25s.

(a) What is the efficiency of this engine?

(b) What is the average power of this engine?

(c) How much energy is extracted as heat from the high temperature reservoir every cycle?

(d) How much energy is delivered as heat to the low temperature reservoir every cycle?

(e) By how much does the entropy of the working substance change as a result of the energy

transferred to it from the high temperature reservoir? From it to the low temperature reservoir?

� In the T-S diagram, the temperature is plotted as a

function of the entropy S during one cycle of the

Carnot engine.

� The change in entropy

0=−=∆L

L

H

H

T

Q

T

QS

L

L

H

H

T

Q

T

Q=⇒ or

H

L

H

L

T

T

Q

Q=