IB PHYSICS AHL - Sciencepress the book includes This book provides questions and answers for each dot point in the IB Physics AHL syllabus from the International Baccalaureate Diploma

IB PHYSICS AHL

• Atit Bhargava • Brian Shadwick •

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of Science Press. ABN 98 000 073 861

© Science Press 2015First published 2015

Science PressPrivate Bag 7023 Marrickville NSW 1475 AustraliaTel: +61 2 9516 1122 Fax: +61 2 9550 [email protected] www.sciencepress.com.au

Contents

Words to Watch iv

Introduction v

Dot Points

Wave Phenomena vi

Fields vi

Electromagnetic Induction vi

Quantum and Nuclear Physics vi

Questions

Wave Phenomena 1

Fields 27

Electromagnetic Induction 73

Quantum and Nuclear Physics 123

Answers

Wave Phenomena 161

Fields 170

Electromagnetic Induction 183

Quantum and Nuclear Physics 198

Appendices

Data Sheet 210

Equations 211

Periodic Table 212

Index 213

Dot Point IB Physics AHL iii Contents

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account, account for State reasons for, report on,

give an account of, narrate a series of events or

transactions.

analyse Interpret data to reach conclusions.

annotate Add brief notes to a diagram or graph.

apply Put to use in a particular situation.

assess Make a judgement about the value of

something.

calculate Find a numerical answer.

clarify Make clear or plain.

classify Arrange into classes, groups or categories.

comment Give a judgement based on a given

statement or result of a calculation.

compare Estimate, measure or note how things are

similar or different.

conctruct Represent or develop in graphical form.

contrast Show how things are different or opposite.

create Originate or bring into existence.

deduce Reach a conclusion from given information.

define Give the precise meaning of a word, phrase or

physical quantity.

demonstrate Show by example.

derive Manipulate a mathematical relationship(s) to

give a new equation or relationship.

describe Give a detailed account.

design Produce a plan, simulation or model.

determine Find the only possible answer.

discuss Talk or write about a topic, taking into account

different issues or ideas.

distinguish Give differences between two or more

different items.

draw Represent by means of pencil lines.

estimate Find an approximate value for an unknown

quantity.

evaluate Assess the implications and limitations.

examine Inquire into.

explain Make something clear or easy to understand.

extract Choose relevant and/or appropriate details.

extrapolate Infer from what is known.

hypothesise Suggest an explanation for a group of facts or phenomena.

identify Recognise and name.

interpret Draw meaning from.

investigate Plan, inquire into and draw conclusions about.

justify Support an argument or conclusion.

label Add labels to a diagram.

list Give a sequence of names or other brief answers.

measure Find a value for a quantity.

outline Give a brief account or summary.

plan Use strategies to develop a series of steps or processes.

predict Give an expected result.

propose Put forward a plan or suggestion for consideration or action.

recall Present remembered ideas, facts or experiences.

relate Tell or report about happenings, events or circumstances.

represent Use words, images or symbols to convey meaning.

select Choose in preference to another or others.

sequence Arrange in order.

show Give the steps in a calculation or derivation.

sketch Make a quick, rough drawing of something.

solve Work out the answer to a problem.

state Give a specific name, value or other brief answer.

suggest Put forward an idea for consideration.

summarise Give a brief statement of the main points.

synthesise Combine various elements to make a whole.

Words to Watch

Dot Point IB Physics AHLiv

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Words to Watch

What the book includes

This book provides questions and answers for each dot point in the IB Physics AHL syllabus from the International Baccalaureate Diploma Programme for Physics:

• Wave Phenomena

• Fields

• Electromagnetic Induction

• Quantum and Nuclear Physics

Format of the book

The book has been formatted in the following way:

1.1 Subtopic from syllabus.

1.1.1 Assessment statement from syllabus.

1.1.1.1 First question for this assessment statement.

1.1.1.2 Second question for this assessment statement.

The number of lines provided for each answer gives an indication of how many marks the question might be worth in an examination. As a rough rule, every two lines of answer might be worth 1 mark.

How to use the book

Completing all questions will provide you with a summary of all the work you need to know from the syllabus. You may have done work in addition to this with your teacher as extension work. Obviously this is not covered, but you may need to know this additional work for your school exams.

When working through the questions, write the answers you have to look up in a different colour to those you know without having to research the work. This will provide you with a quick reference for work needing further revision.

Introduction

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Introduction

Verbs to Watch

Wave Phenomena9.1 Simple harmonic motion 2

9.1.1 The defining equation of SHM. 29.1.2 Energy changes. 4

9.2 Single slit diffraction 7

9.2.1 The nature of single slit diffraction. 7

9.3 Interference 12

9.3.1 Young’s double slit experiment. 129.3.2 Modulation of two slit interference 15

pattern by one slit diffraction effect.9.3.3 Multiple slit and diffraction grating 15

interference patterns.9.3.4 Thin film interference. 17

9.4 Resolution 19

9.4.1 The size of a diffracting aperture. 199.4.2 The resolution of simple 20

monochromatic two-source systems.

9.5 The Doppler effect 22

9.5.1 The Doppler effect for soundwaves 22 and light waves.

Answers to Wave Phenomena 161

Fields10.1 Describing fields 28

10.1.1 Gravitational fields. 2810.1.2 Electrostatic fields. 3310.1.3 Electric potential and gravitational 39

potential.10.1.4 Field lines. 4110.1.5 Equipotential surfaces. 46

10.2 Fields at work 48

10.2.1 Potential and potential energy. 4810.2.2 Potential gradient. 5510.2.3 Potential difference. 5710.2.4 Escape speed. 5910.2.5 Orbital motion, orbital speed and 61

orbital energy.10.2.6 Forces and inverse-square 70

law behaviour.

Answers to Fields 170

Electromagnetic Induction11.1 Electromagnetic induction 74

11.1.1 Electromagnetic force (emf). 7411.1.2 Magnetic flux and magnetic 77

flux linkage.11.1.3 Faraday’s law of induction. 8011.1.4 Lenz’s law. 80

11.2 Power generation and 87 transmission

11.2.1 Alternating current (AC) generators. 8711.2.2 Average power and root mean square 93

(rms) values of current and voltage.11.2.3 Transformers. 9611.2.4 Diode bridges. 10811.2.5 Half-wave and full-wave rectification. 111

11.3 Capacitance 113

11.3.1 Capacitance. 11311.3.2 Dielectric materials. 11611.3.3 Capacitors in series and parallel. 11711.3.4 Resistor-capacitor (RC) series circuits. 12011.3.5 Time constant. 120

Answers to Electromagnetic Induction 183

Quantum and Nuclear Physics12.1 The interaction of matter with 124

radiation12.1.1 Photons. 12412.1.2 The photoelectric effect. 12812.1.3 Matter waves. 14012.1.4 Pair production and pair annihilation. 14412.1.5 Quantisation of angular momentum 146

in the Bohr model for hydrogen.12.1.6 The wave function. 14612.1.7 The uncertainty principle for 149

energy and time and position and momentum.

12.1.8 Tunnelling, potential barrier and 151 factors affecting tunnelling probability.

12.2 Nuclear physics 15212.2.1 Rutherford scattering and nuclear 152

radius.12.2.2 Nuclear energy levels. 15412.2.3 The neutrino. 15612.2.4 The law of radioactive decay and 157

the decay constant.

Answers to Quantum and Nuclear Physics 198

Dot Points

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Dot Points

DOT POINT

Wave Phenomena

AHL 9

1 AHL 9 Wave PhenomenaDot Point IB Physics AHL

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9.1 Simple harmonic motion.

9.1.1 The defining equation of SHM.

9.1.1.1 Describe the features of simple harmonic motion, referring to a specific example.

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9.1.1.2 State and discuss the mathematical descriptions of the following, referring to appropriate signs in each case.

(a) SHM. ��

(b) The restoring force. ��

9.1.1.3 The time period of a simple pendulum is given by TLg

= 2π . A simple pendulum has a time period of 1.0 s.

(a) Make L the subject of the equation. ��

(b) What is its length, assuming acceleration due to Earth’s gravity is 9.8 m s–2? ��

9.1.1.4 What is the time period of a pendulum with length 1.0 m? ��

9.1.1.5 In order to determine the acceleration due to Earth’s gravity, a student measures the time period of a simple pendulum made with a string and a small mass. Her results are shown in the table.

Time for 10 oscillations (s) 10.9 12.6 14.0 15.4 16.6 17.8 18.9 19.9

Length of the string (m) 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

(a) Use a graphical method to find the magnitude of g.

(b) Give two reasons why the value of g is not 9.8 m s–2. ��

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9.1.1.6 Which one of the following graphs best describes the relationship between the time period and length of a simple pendulum?

Tim

e p

erio

d

Tim

e p

erio

d

Tim

e p

erio

d

Tim

e p

erio

d

Length LengthLength Length

(A) (B) (C) (D)

2AHL 9 Wave Phenomena

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Dot Point IB Physics AHL

9.1.1.7 A light spring of length 30.0 cm is attached to a horizontal beam. A mass of 100 g is gently attached to it and upon doing so the spring is set into periodic oscillations as shown in the diagram with amplitude of 5.0 cm. Each image is 0.1 s apart. Note that the time period of such an oscillating mass,

m is T = Tmk

= 2p ,, where k is the spring constant.

(a) Using the values shown in the diagram estimate the spring constant. ��

(b) What is the time period of the oscillation? ��

(c) Which one of the following graphs best describes the displacement of the mass if the downward motion from the rest position is negative?

M

L

O

N P

M P

N

LL

N

M P

L

M P

O

N

Time

Dis

pla

cem

ent

Time

Dis

pla

cem

ent

Time

Dis

pla

cem

ent

Time

Dis

pla

cem

ent

(A) (B) (C) (D)

(d) List three locations where the mass has the maximum net force. Explain your answer.

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(e) What would be the effect on the time period of the oscillation if the mass is doubled?

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9.1.1.8 The gradient (derivative) of y = A sin Bx is dydx

AB Bx = cos = AB cos Bx and the gradient of y = P cos Qx is

dydx

PQ Qx = sin – . = −PQ sin Qx. With this knowledge of calculus, determine (a) the velocity and (b) acceleration

when the displacement of a linear SHM oscillator is given as x = x0 cos ωt.

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9.1.1.9 The motion of an object undergoing SHM is graphically shown. Using this information, find the quantities (a) to (f) for the SHM of the object using appropriate units.

(a) Time period. ��

(b) Amplitude. ��

(c) Angular frequency. ��

(d) Maximum speed. ��

(e) Maximum acceleration. ��

(f) Speed at time t = 2.2 s. ��

QO PNML

Spring

Meanposition5 cm

Mass = 100 g

−6−5

−3−2

−4

−10

23

1

56

4

0 0.5 1 1.5 2 2.5 3

Dis

pla

cem

ent

(cm

)

Time (s)

3 AHL 9 Wave Phenomena

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9.1.1.10 The displacement of an object of mass m kg undergoing SHM is given as x = Y sin ωt, given ω is the angular frequency = 2πf = 2π/T, where f = frequency and T is the time period.

(a) Derive an expression for the (i) velocity, (ii) acceleration and (iii) restoring force on the object at time t.

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(b) Given that the angular frequency, ω = 0.2π and the maximum amplitude at time t = 0 s is 0.3 m, (i) write an expression for the displacement of the object at time t (ii) find the frequency of the SHM and (iii) deduce the acceleration of the object at 1.1 s.

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9.1.1.11 The displacement of a mass undergoing SHM is given by x = x0 cos ωt.

(a) Show that v = ±ω (x − x )02 2 . ��

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(b) Comment on the significance of this expression. ��

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9.1.1.12 An ideal (frictionless) linear oscillator undergoes SHM and it is noted that at time t = 0 s, it has a displacement of −3.0 cm, velocity of −0.25 m s–1 and an acceleration of 12.1 m s–2. Determine its angular frequency giving your answer in appropriate units.

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9.1.1.13 A mass of 1.5 kg undergoes SHM with a frequency of 2.5 Hz and an amplitude of 0.5 m.

(a) What is the maximum restoring force on the body? ��

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(b) What is the magnitude of the restoring force when the mass is 0.25 m from its original position?

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9.1.1.14 An object is moving with SHM and has a time period of 0.1π s and an amplitude of 0.06 m. What is its velocty at the instant when it is at a distance of 0.05 m from its equilibrium position?

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9.1.2 Energy changes.

9.1.2.1 The diagrams show a linear harmonic oscillator and a pendulum. Indicate (a) a position of maximum kinetic energy and (b) a position of maximum potential energy. The positions of maximum displacement from the mean are shown.

9.1.2.2 For the motion of a simple pendulum, outline the changes to potential and kinetic energy during one oscillation making reference to maximum and minimum values.

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Linear harmonic oscillator Simple pendulum


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9.1.2.3 What is the total energy of a linear oscillator at (a) minimum displacement and (b) maximum displacement?

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9.1.2.4 The graph shows the variation of kinetic, potential and total energy of a mass attached to the end of a pendulum experiencing SHM and starting from a position of maximum displacement.

(a) Annotate on the graph the variation of potential, kinetic and total energy.

(b) Under what condition would you observe the graph of total energy to become non-linear?

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9.1.2.5 The restoring force on a linear harmonic oscillator when displaced from its equilibrium position is F = −kx = ma, where k is the spring constant.

(a) Derive an expression for time period, T of the oscillator in terms of mass, m and spring constant, k.

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(b) Express displacement, x at any time, t in terms of amplitude, x0, k and m.

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(c) By using the principle of conservation of energy show that maximum velocity, v0 = ωx0 where ω is the angular frequency.

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(d) Deduce an expression for kinetic energy Ek, at a displacement, x. ��

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9.1.2.6 A 10.0 g mass is suspended by a spring of spring constant = 108 N m–1. A student pulls the mass to exactly 10.0 cm and lets go so that the mass executes SHM. Evaluate the magnitudes of the quantities in the table giving appropriate units.

Physical quantity Magnitude, unit

(a) Total mechanical energy, ET

(b) Maximum kinetic energy of the mass, Ek,max

(c) Ek at a displacement of 1.0 cm from equilibrium

(d) Speed at a displacement of 1.0 cm from equilibrium

9.1.2.7 A block of mass 110 g is attached to a spring of k = 50.0 N m–1 and then pulled horizontally a distance of x = 5.00 cm setting the system in SHM on a frictionless surface.

(a) Determine the (i) angular frequency ω, (ii) frequency f, (iii) time period T. ��

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Maximumdisplacement

Displacement

Energy

+


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(b) Calculate the magnitude of the (i) maximum speed of the mass, v0 and (ii) maximum acceleration, amax.

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(c) What is the total mechanical energy of the system? ��

(d) Deduce the EP and EK at a displacement of x = 1.5 cm. ��

9.1.2.8 The table shows some data collected for a linear harmonic oscillator. Use the given information to evaluate the missing information and complete the table.

Quantity Data

Total mechanical energy ET 3.1 J

Amplitude x0 11.3 cm

Maximum speed v0 2.1 m s–1

Spring constant k

Mass m

Angular frequency ω

Frequency f

Time period T

Maximum potential energy EP,max

Maximum kinetic energy EK,max

Potential energy at displacement of 5.0 cm EP,5.0cm

Kinetic energy at displacement of 5.0 cm EK,5.0cm

9.1.2.9 In a class experiment the following data was collected for a mass of 2.00 kg undergoing SHM as a linear oscillator.

Velocity v (m s–1) 0 1.38 1.7 1.86 1.95 1.99 2 1.99 1.95 1.86 1.7 1.38

Displacement x (cm) –6 –5 –4 –3 –2 –1 0 1 2 3 4 5

Kinetic energy EK (J)

(a) Sketch a graph of EK versus x.

(b) Estimate k, the spring constant.

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(c) Determine the maximum elastic potential energy EP,max.

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(d) Evaluate EP at x = +4.0 cm.

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9.2 Single slit diffraction.

9.2.1 The nature of single slit diffraction.

9.2.1.1 An electromagnetic wave is incident on a single slit opening as shown in the figure.

The distance between wavefronts is the wavelength.

Incident electromagnetic wave

Screen

X

Illustrate on the diagram how the intensity of the pattern would change along the screen.

9.2.1.2 Explain why a diffraction pattern is seen with a single slit. ��

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9.2.1.3 The intensity of interference fringe as a function of angle θ is shown in the diagram. The angle θ is the angle between the centre of the slit to the location of the fringe.

–30 –20 –10 10 20 30

q (degrees)

Intensity

(a) Explain why the intensity of the peaks reduces on either side of the central maximum.

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(b) Discuss how the central maximum is different from other fringes. ��

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9.2.1.4 An electromagnetic wave is incident on a single slit opening as shown in the diagram.

Incident electromagnetic

wave

Screen

X

The distance between wavefronts is the wavelength.

(a) Explain why there is a finite intensity of the electromagnetic wave recorded at point X even though it is not directly in front of the opening.

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(b) Explain the effect on the image seen on the screen as the slit opening is made wider.

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9.2.1.5 Which of the following statements is true about diffraction of water waves through an opening? One or more answers may be true.

(A) Diffraction is more when the wavelength is greater than the size of the opening.

(B) Diffraction is more when the wavelength is comparable with the size of the opening.

(C) Diffraction is the least when the size of the opening is larger than the wavelength.

(D) If the opening is much larger than the wavelength, the wave will pass through it without significant diffraction.

9.2.1.6 What conditions would cause the most diffraction of waves through an opening?

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9.2.1.7 What conditions would cause the most diffraction of waves around an obstacle?

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9.2.1.8 Upon diffraction through an opening, which of the following does not change: speed, wavelength, frequency? Explain your answer.

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9.2.1.9 What is the effect on the diffraction of a wave through an opening when (a) the size of the opening is reduced or (b) the wavelength of the wave is increased or (c) the frequency of the wave is increased? Explain your answer.

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9.2.1.10 Estimate a size of slit which would give significant diffraction for red light.

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9.2.1.11 As red light is replaced with blue light, explain what would be the effect on diffraction through a narrow slit.

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9.2.1.12 Which electromagnetic radiation, X-rays or ultraviolet, is more likely to show diffraction in atomic crystals such as sodium chloride? Explain your answer.

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9.2.1.13 What is radar? Outline how radar can provide an image of a terrain. ��

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9.2.1.14 Which of the following statements about diffraction through a narrow gap are correct?

(A) As the gap narrows further, the extent of diffraction increases, provided w » l.

(B) For an existing diffraction pattern, as the frequency of an incident wave increases further, the extent of diffraction also increases.

(C) For an existing diffraction pattern, as the frequency of an incident wave increases further, the extent of diffraction decreases.

(D) For an existing diffraction pattern, replacing a blue light with red will increase the extent of diffraction.

(E) The extent of diffraction through a narrow gap is independent of the wavelength of the incident wave.

(F) Diffraction through a narrow gap occurs because light is a wave.

(G) The speed of a wave slows down after diffracting through a narrow gap.

(H) The width of the central maximum is about twice the size of the width of secondary maxima.

(I) All bands are of equal thickness.

(J) If a double slit is replaced with a single slit, then the entire diffraction fringe pattern of the double slit is less than the width of the central maximum of the single slit.

9.2.1.15 A monochromatic red light source is incident on a single narrow (w » l) slit and a screen is placed quite far from the slit.

(a) Which of the patterns, A, B or C is closest to your expectation? Explain the reasons for your selection.

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A B C

Intensity

Intensity Intensity


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(b) Various changes are now made to the experimental set-up. Determine what the effects will be on the diffraction pattern as compared to that obtained from a monochromatic light source, and enter your answers in the table.

Changes Effect on the diffraction pattern

(i)Red light source is replaced with a blue monochromatic source.

(ii) Red light source is replaced with a source of white light.

(iii) Width of the slit is doubled.

(iv) The screen is moved further away from the slit.

9.2.1.16 The diagram shows the location of the 1st order dark fringe (first minimum).

(a) What is the relationship between path difference and wavelength for the first minimum?

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(b) With reference to the figure, derive a relationship between θ, l and b for the first minimum.

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(c) Write the condition for the nth order dark fringe. ��

(d) Write the condition for the nth order bright fringe, not including the central maximum.

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9.2.1.17 Consider a single slit pattern with a monochromatic light source as shown in the diagram.

(a) For L to be much larger than b and θ to be a small angle, show that for the nth order

dark fringe xL

nb

=λ, where

b and λ are of comparable magnitude and x is half the width of the central maximum.

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(b) Show that for the nth order bright fringe ( )x

L

n

b=

+ 12

λ. Note that the relationship is opposite to that for

double slit interference.

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L

x

bq

L

L

b

xL

q

Path difference

P, nth order dark fringe

Central maximum


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AHL 9 Wave Phenomena

9.1.1.1 SHM is a periodic motion where the amplitude does not reduce with time; the acceleration is proportional to displacement and the force on the object is a restoring force directed towards the equilibrium position. Such a motion can be achieved in a simple pendulum oscillating without friction and a spring oscillating under the action of a mass without friction. In both these examples, the conditions described here are met.

9.1.1.2 (a) Since acceleration, a is proportional to displacement, x and is directed towards the equilibrium position, the mathematical equation is a = −ω2x, where ω is the angular frequency.

(b) Since the restoring force, F is proportional to displacement and directed towards the equilibrium position, F = −kx, where k is the spring constant. In each case the negative sign indicates acceleration or force directed towards the equilibrium position.

9.1.1.3 (a) L = gT 2

24� (b) 0.248 m

9.1.1.4 T = 2.01 s

9.1.1.5 Gradient = 1.99 = 2π/ g . Therefore g = 9.97 m s–2.

Time for 10 oscillations (s) 10.9 12.6 14.0 15.4 16.6 17.8 18.9 19.9

Time period (s) 1.09 1.26 1.4 1.54 1.66 1.78 1.89 1.99

Length of the string (m) 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

L 0.548 0.632 0.707 0.775 0.837 0.894 0.949 1

0 0.2 0.4 0.6 0.8 1 1.2

0

0.5

1

1.5

2

2.5

y = 1.992x – 0.003 T (s)

ÖL (m )1_2

(b) Variation on Earth’s surface due to the Earth not being a perfect sphere (refer to F = GMm/R2); dampening effects due to air friction; and the formula only works for small angles.

9.1.1.6 C Refer to T = 2�Lg

9.1.1.7 (a) F = −kx. Therefore k = F/x = 0.1 × 9.8/0.05 = 19.6 N m–1

(b) T = 0.2 s

(c) B

(d) M, N and P. Maximum force is where there is maximum direction, the two being in the opposite direction.

(e) T = 2� mk

. If the mass is doubled, the time period will increase by a factor of 1.414.

9.1.1.8 (a) v = DD

xt = −xoω sin ωt

(b) a = DD

vt = −xoω

2 cos ωt = −ω2x

9.1.1.9 (a) T = 1 s

(b) Amplitude = 4 cm

(c) Angular frequency = 2πf = 2π rad/s

(d) Maximum speed = ωxo = 2π × 4 = 25.1 cm s–1

(e) Maximum acceleration = ω2xo = (2π)2 × 4 = 16π2 = 157.9 cm s–2 = 1.58 m s–2

(f) Since v = vo cos ωt and x2.2 s = 2 cm (from graph), v2.2 s = 25.1 cos (2π × 2.2) = 7.76 cm s–1 or use v x x= ( )02 2ω −


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9.1.1.10 (a) (i) v = ωY cos ωt

(ii) a = −Yω2 cos ωt

(iii) F = −mYω2 cos ωt

(b) (i) x = 0.3 sin (0.2πt)

(ii) f = = .ω2

0 1π

Hz

(iii) a = −0.3(0.2π)2 sin (0.2πt) = −0.012π2 sin (0.2πt). At t = 1.1 s, a = 0.0755 m s–2.

9.1.1.11 (a) Since x = x0 cos ωt, v = DD

xt

= −x0ω sin ωt.

Since sin2 θ + cos2 θ = 1, v = x0ω 1 120

002 2– – –cos = =ω ω ωt x

xx

x x±

(b) This expression shows that velocity is minimum (zero) when displacement is maximum, which is a condition for SHM.

9.1.1.12 Since x = x0 cos ωt and a = −x0 ω2 cos ωt, the two equations can be solved to give ω = =

..

= .ax

12 10 03

20 1 rad/s

9.1.1.13 (a) amax = ω2x0 = (2π × 2.5)2 × 0.5 = 123.37 m s–1. Therefore Fmax = 123.37 × 1.5 = 185 N

(b) First estimate the time at 0.25 m using x = x0 sin ωt; t = 0.033 s. Now use a = −ω2x0 sin ωt = 61.7 m s–2. Therefore F0.25m = 92.5 N.

9.1.1.14 First find time at 0.05 m displacement using x = x0 sin ωt; t = 2.82 s. Now use v = ωx0 cos ωt = 20 × 0.06 cos (20 × 2.82) = 0.66 m s–1.

9.1.2.1

9.1.2.2 When at maximum displacement from the centre, the mass has maximum potential energy and minimum kinetic energy, while at the centre position, it has maximum kinetic energy and minimum potential energy. If there is no dampening, mechanical energy is conserved.

9.1.2.3 The total energy is conserved and equals 12 kx2 at both (a) minimum displacement and (b) maximum displacement. It transforms

into all kinetic energy at the minimum potential energy point, which is the centre position and into all gravitational energy at the maximum displacement point. Total energy is conserved if there is no dampening.

9.1.2.4 (a)

(b) When there is loss (e.g. due to friction) or gain (e.g. due to reinforcing force) of energy.

9.1.2.5 (a) akm

x xkm

Tkm

= = . = =– –ω ω2 2 Therefore and �

(b) x xkm

t= sin ( )0 ×

(c) Maximum elastic potential energy Ep,max = 12

kx02 and gets converted to Ek,max =

12

mv02.

Using energy balance 12 kx0

2 = 12

mv02, v x

km

x0 0 0= = .ω

(d) Ek at x = ET – Ep at x, where ET = total mechanical energy.

Thus Ek at x = 12

mω2x0 – 12

mω2x = 12

mω2(x0 – x)

9.1.2.6 (a) ET = 12

kx02 = 0.540 J

(b) Ek,max = 0.54 J

(c) EP at x = 1.0 cm = 12

kx2 = 0.0054 J. Therefore

EK at x = 1.0 cm = 0.540 − 0.0054 = 0.535 J

(d) Using Ek,max = 12

mv02, vat x = 1.0 cm = 10.34 m s–1.

Linear harmonic oscillator Simple pendulum

Ek, max UG, maxUG, max

Ek, max

UG, max

MaximumDisplacement

Displacement

Energy

+

Total energy

Potential

Kinetic


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9.1.2.7 (a) (i)

ω = =.

= .km

500 11

21 32 rad s–1.

(ii) f = ω/(2π) = 3.39 Hz

(iii) T = 0.29 s.

(b) (i) v0 = ωx0 = 21.32 × 0.05 = 1.07 m s–1.

(ii) amax = ω2x0 = 21.232 × 0.05 = 22.7 m s–2.

(c) ET = 12

kx02 = 0.0625 J

(d) EP = 12

kx2 = 0.006 J, EK = ET − EP = 0.0563 J

9.1.2.8 k = 485.55 N m–1, m = 1.406 kg, ω = 18.584 rad s–1, f = 2.96 Hz, T = 0.338 s, EP,max = 3.1 J, EK,max = 3.1 J, EP at 0.05 m = 0.607 J, EK at 0.05 m = 2.49 J.

9.1.2.9 (a)

(b) kE

xp= =

.= .,2 2 4

0 062222 22

02 2

1 N mmax × –

(c) Ep,max = 4.0 J

(d) Ep at 4.0 cm = 12

kx2 = 1.78 J. Alternatively read EK from graph and subtract from ET .

9.2.1.1

9.2.1.2 Each point in the slit acts as a source of secondary wavelets (Huygens’ principle) which interfere to form regions of constructive and destructive interference, forming a diffraction pattern.

9.2.1.3 (a) Wavelets arrive at points from all positions along the slit. However, wavelets from one side of the slit will be out of phase to varying degrees except at the centre. The further one is from one side the more waves that are out of phase and the less the intensity of the maximum.

(b) The central maximum is wider (×2) and more intense than the other fringes as wavelets from all parts of the slit arrive in phase.

9.2.1.4 (a) Waves diffract around the opening to reach X. The opening acts as a source of secondary wavelets and spread out the wave as shown. Energy spreads out in accordance with the inverse square law.

(b) As the opening is made wider, the extent of diffraction becomes less and the pattern will spread less. Energy is less spread out and hence the brightness in the central region directly in front of the opening is more.

9.2.1.5 A, B and D.

–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7

0.5

1

0

1.5

2

2.5

3

3.5

Energy (J)

Displacement (cm)

4

Intensity

Incident electromagnetic wave

Screen

X


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IB PHYSICS AHL - Sciencepress the book includes This book provides questions and answers for each dot point in the IB Physics AHL syllabus from the International Baccalaureate Diploma