Phase difference and superposition
Question 10W: Warm-up Exercise
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This question helps you check your understanding of phase difference and gives you further practice
in superposing waves.
Graphs of waves
A graph of wave displacement against position shows a wave 'frozen' in space at an instant of time.
Really, the waves are travelling along. The graph shows 'snapshots' of two waves, A and B.
position
1. What is the phase difference between A and B? Give your answers in fractions of a wavelength
and degrees. There are at least two correct answers to this question!
2. Sketch the superposition pattern of A and B.
position
The next diagram shows two more waves, C and D.
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position
3. What is the phase difference between C and D?
4. Sketch the superposition pattern of C and D.
position
5. What phase angle corresponds to a phase difference of 1/3 of a wavelength?
6. Sketch a diagram showing two waves of equal amplitude with a phase difference equal to 1/3 of a
wavelength.
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position
Superposition of waves:
A drawing exercise
Question 20W: Warm-up Exercise
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Skills
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This is a question on superposition of waves. You will need two pieces of A4 graph paper.
Superposing waves
The diagram below shows a snapshot of two waves approaching each other.
2 m s–1
2 m s–1
P
1m
position / m
Take a sheet of A4 graph paper and copy the initial graph, taking up about one-quarter of the sheet.
1. Draw three new sets of axes, one below the other, and draw the waveform observed after one
second, two seconds and three seconds. Label each!
2. You have drawn wave displacement against position graphs. On a separate sheet of graph paper
draw a displacement against time graph for point P over the three second period.
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Practice with phasors
Question 30W: Warm-up Exercise
Quick Help
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Phasors are useful tools which help you to understand how waves superpose at a point. They
become essential in chapter 7 so it is best to get really confident with them at this stage!
Superposition of waves using phasors
In each of the cases below, add the two phasors and draw (accurately!) the resulting phasor. Draw
the larger circles with 1 cm radius (2 cm diameter). Draw the smaller circle with 0.5 cm radius (1 cm
diameter).
1.
2 cm
+
=
+
=
+
=
+
=
2.
3.
4.
1 cm
Linking phasors to waves
Look at the sketch graph below:
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A
B
C
0
D

For points A, B, C and D draw the corresponding phasors in the circles below:
0
5.
A
B
C
D
How big is the angle in each case?
0: Zero
A:
B:
C:
D:
The diagram below shows waves A, B and C:
wave A
wave B
0
wave C
From the diagram, give the phase differences between:
6.
Waves A and B
7. Waves A and C
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8. Waves B and C
If the amplitude of wave A is 1 cm, give the amplitudes of:
9. Wave B:
10. Wave C:
You have seen
1. That spinning arrows, called phasors, can help you to understand waves and their relative
phases.
'Lloyd's mirror' for microwaves
Question 30S: Short Answer
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What to do
Read the text below and answer the questions. Show working where applicable.
Dr Lloyd's mirror experiment
In 1834 Dr Lloyd developed a demonstration of the superposition of visible light, with light coming to
the eye both directly and just grazing the surface of a mirror. His idea is now often used with
microwaves in laboratory demonstrations. Microwaves are used because they are of a much greater
wavelength than visible light which makes the task of the experimenter much easier.
A student performs the investigation with the transmitter and receiver separated by 0.7 m. When the
mirror was 0.29 m from the midpoint P a minimum signal was detected at the receiver. The student
thought that it was possible to measure to the nearest 5 mm. These are the initial results:
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distance TR = 0.700 m
distance MP = 0.290 m.
transmitter T
M
receiver R
P
1. What was the path difference between microwaves travelling directly to the receiver and those
reflecting off the mirror?
2. The student now slowly moved the mirror back until another minimum signal was detected at the
receiver. At this point distance TR = 0.700 m (as before) and distance MP = 0.310 m.
What was the new path difference between the direct and reflected waves?
3. The difference between the two path differences is the extra distance the reflected wave has to
travel until it will once again superpose with the direct wave to give a minimum. What is this
distance?
4. What does this length represent?
5. What will happen to the path difference if the mirror is placed very close to point P?
6. At this point the student observed a minimum signal at the receiver, not a maximum. Why was
this?
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Superposition of sound waves
Question 40S: Short Answer
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What to do
This question is about superposition of sound waves from two loudspeakers. Answer the questions
below, showing working where applicable.
The experimental set-up
Two loudspeakers separated by 3 m are connected to a signal generator. They are set up in the open
air, well away from buildings. A student stands 10 m from the midpoint of the speakers, hearing a
maximum amplitude sound and walks in front of them in the direction shown by OX until the sound
drops to a minimum level. This occurs at point X as shown in the diagram.
A
X
1.5 m
3m
10 m
1.5 m
B
1. How much further does sound from speaker B have to travel to reach the student than that from
speaker A?
2. The position X was the point where the student heard the first minimum. How many
wavelengths was the path difference between speaker A to X and speaker B to X?
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3. What was the wavelength of the sound?
4. The speed of sound on that day was 340 m s–1. What was the frequency of the note?
5. Why should this experiment be performed outside?
Interference of sound waves
Question 50S: Short Answer
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Standing where the sound is loud
Two loudspeakers S1 and S2 were connected to signal generator set at 360 Hz. In the space in front
of the speakers an interference pattern is created. Where the path difference from a given point to the
two speakers is a whole number of wavelengths the waves arrive in phase and a louder sound is
heard at that point. Students were asked to stand in positions where the sound heard was a
maximum.
Open the JPEG file
Open the JPEG file
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1. Take measurements from the photographs to work out the wavelength of the sound. Use
measurements from the speakers S1 and S2 to students A, B, C and D. What assumptions about
the scale on the photograph have you made in your calculation?
2. Work out a value for the speed of sound.
3. Students standing further from the speakers found it more difficult to identify the loud regions.
Why is this?
4. The connections to one of the speakers are reversed. What difference do the students hear?
5. The experiment is repeated using sound of a higher frequency. What is the effect on the pattern?
Partial reflection of waves
Question 60S: Short Answer
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Instructions
A transmitter, T, emits radiation, some of which is reflected from a partially reflecting screen, S , and
1
some which carries on to be reflected by a second screen, S . The radiation reflected from the
2
screens is detected by a receiver, R, placed alongside T. At a certain separation of S and S the
1
2
receiver records a minimum zero signal. S is then moved away from S . As S is being moved, the
2
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1
2
detector records a second minimum and S is moved on until a third minimum is recorded. The total
2
movement of screen S was 120 mm.
2
partial reflector
total
reflector
S1
S2
T
R
1. What is the wavelength of the radiation?
2. At the original separation the signal detected was very nearly zero; but after S had moved 120
2
mm the minimum signal was quite perceptible. Why?
Superposition and speed measurement
Question 70S: Short Answer
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This is a question about using the principle of superposition in radar speed measurement.
Radar measuring speed
A transmitter emitting a radio waves of wavelength 30 mm is placed alongside a receiver as shown.
Some of the output of the transmitter is fed (by the wire shown) directly to the receiver and is then
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compared with the reflected signal.
M
transmitter
receiver
A sheet of metal, M, is held in front of the device as shown, so that the intensity detected by the
receiver is a maximum. When the metal sheet is moved 7.5 mm towards the device the intensity falls
to a minimum. Explain, in words, why:
1. The receiver signal decreases.
2. The minimum is not necessarily zero.
The radiation is now directed at an approaching motor-car which is moving directly towards the
device at steady speed, and the receiver detects a signal fluctuating in strength with a frequency of
2.0 kHz.
3. Calculate the speed of the car.
4. The driver objects that the calculated speed of the car is only about one ten-millionth of the speed
of the radar waves, and that comparing the two speeds to this precision is obviously impossible.
Reply to the objection.
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Stationary waves in a string
Question 90S: Short Answer
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This is a question about stationary wave patterns set up in a vibrating string.
A vibrating string
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Quick Help
2m
vibration
generator
bench
Frequency Adjust
5
3
7
2
8
1
10Hz
100Hz
1kHz
10kHz
9
100kHz
Frequency range
1000
100
10
1
1000
10
100
Frequency
Wave
55 Hz
Outputs
A
power
signal generator
0
0
15
30
45
60
75
Frequency of vibration / Hz
An elastic string is clamped at both ends. Near one end a vibration generator is loosely attached. The
vibrator oscillates with fixed amplitude but with variable frequency. A graph of maximum amplitude
along the string against frequency of vibration is shown.
Sketch the instantaneous appearance of the string at:
1. 15 Hz.
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2. 30 Hz.
3. 45 Hz.
4. What is the wavelength of the waves on the string at these frequencies?
5. What is the speed of travel of the waves along the string?
Standing waves in pipes
Question 100S: Short Answer
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This is a question about standing waves in organ pipes and flutes
Organ pipes
An organ pipe closed at one end can allow standing waves which have a node (zero displacement) at
that end and an antinode (maximum displacement) at the other (neglecting a small 'end correction').
This is called the 'fundamental'.
node
anti node
One such pipe has a fundamental note of 64 Hz.
1. Draw a diagram showing the next possible standing wave pattern that could be set up in this pipe.
What would its frequency be?
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2. Draw another diagram showing the third possible standing wave pattern and calculate its
frequency.
Flutes
Flutes are an example of a pipe open at both ends. This means the fundamental note has an
antinode at either end as shown.
anti node
node
anti node
3. A flute open at both ends and an organ pipe closed at one end have the same length. The organ
pipe has a fundamental note of frequency 130Hz. What is the frequency of the flute's fundamental
note? Explain your answer.
Effect of temperature
The speed of sound in air is proportional to the square root of the temperature in kelvin. A flute player
begins playing with the temperature of air in the tube at 288 K. The speed of sound at this
temperature is 340 m s–1. After a few minutes playing the air temperature in the bore rises to 298 K.
4. What is the speed of sound at this temperature?
5. When the musician started playing the flute a certain length of tube gave a note of frequency
438 Hz. What would be the frequency of this length of tube when the flute had 'warmed up' to
298 K? What assumptions are you making?
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Measuring the speed of light
Question 110S: Short Answer
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Skills
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This is a set of questions related to the Reading 30T 'Historical attempts to measure the speed of
light'. You will need to use the reading to answer the questions.
Galileo's method
1. The reading includes Galileo's description of the attempt to measure the speed of light. Describe
his method in your own words.
2. Why did the experiment prove inconclusive?
3. In an attempt to repeat Galileo's experiment you station an assistant on a hill 5 km distant. You
open your lamp and observe your assistant's lantern open 0.5 seconds later. Use the modern
value of the speed of light (3  108 m s–1) to estimate the possible reaction time of your assistant.
Assuming you used a digital stopclock reading to 0.001 s, would your answer for the reaction time
be measurably different if light did travel instantaneously from you to your assistant?
Bradley's method
4. A train window is 1 metre high and 10 metres long. How long will it take a raindrop to roll down
the window if its vertical speed is 5 m s–1? This time will be independent of the motion of the train.
5. If the train is moving at 30 m s–1 how far will it move in the time it takes the raindrop to roll down
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the window?
6. Draw a scale diagram of the path of the raindrop, as seen from the ground, showing the angle it
makes with the vertical.
7. The radius of the Earth's orbit is 1.5  1011 m. Using a value of 365.25 days for a year find the
speed of the Earth in orbit in m s–1.
8. Bradley calculated that light was aberrated at 20 seconds of an arc. This shows the sensitivity of
his measurements as there are 60 minutes in 1 degree and 60 seconds in a minute. Bradley was
measuring an angular change of 5.6  10–3 degrees. Draw a right-angled triangle showing the
situation in which the base represents the speed of the Earth in orbit and the vertical represents
the speed of light. Calculate the speed of light from these data.
Fizeau's method
9. Imagine that Fizeau's apparatus had 360 cogs. Use c = 3.00  108 m s–1 to calculate how many
revolutions per second the wheel would have to make for the reflected light to be visible 'one cog
along' after a round trip of 15.0 km.
Does this seem a reasonable answer?
Michelson's method
10. In a Michelson's mirror arrangement the rotating mirror had 16 sides and the fixed mirror was
33 km away. The observer saw the reflection from the fixed mirror when the rotating mirror was
set at 300 revolutions per second. Use this information to calculate a value for the speed of light.
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Questions on the two-slit experiment
Question 150S: Short Answer
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What to do
These questions will give you practice in handling data about the two-slit experiment. You will need a
pen, paper and a calculator.
Finding the wavelength of sodium light
In a two-slit apparatus the slits are 0.3 mm apart. Fringes in sodium are observed at a distance of
1.2 m from the slits. The separation of the fringes is 2.4 mm.
1. What is the wavelength of sodium light?
2. The same light gives a fringe separation of 3.6 mm with a different pair of slits. What is the slit
separation if the distance between the slits and the fringes is the same?
Red light of wavelength 7.0  10–7 m is shone at right angles through two slits of separation 0.3 mm.
Fringes are formed at a distance of 1.3 m from the slits.
3. What is the fringe spacing?
4. The same light gives a fringe spacing of 2 mm when passed through a different pair of slits. What
is the slit separation if the distance between the slits and the fringes is the same?
In a two-slit apparatus the slits are 0.3 mm apart. White light passes through the slits and fringes are
observed at a distance of 2 m from the slits. Red light has a wavelength of 700 nm and blue light has
a wavelength of 400 nm.
5. Calculate the fringe spacing for each colour.
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6. Use your answers to explain the coloured fringes seen on the screen.
Using two-slit interference to measure exhaust velocity
Question 160S: Short Answer
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This question is about using two-slit interference to measure the speed of gases from a jet engine.
Velocity of exhaust gas
One technique for measuring exhaust velocities from jet engines uses laser light which is focused to
produce an interference pattern in the exhaust gas.
exhaust gas flow
laser
light
628 nm
interference
pattern
spacing 0.1 mm
0.6 m
double slits
The spacing of the two slits is 0.10 mm and the distance from slits to pattern is 0.60 m. The
wavelength of the light is 628 nm.
1. Calculate the spacing of the bright lines (fringes) in the pattern.
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A dust particle moving in the stream of invisible gas is used to measure the speed of the exhaust. As
the dust passes through each bright fringe it reflects light into a detector and the output of this
detector is fed into an oscilloscope.
With the time base set to 10 microseconds per division the display shown is obtained.
10 s
2. Calculate the speed of the exhaust gas stream.
3. Suggest and explain one advantage or disadvantage this method may have as compared with
conventional methods which involve inserting a metal probe into the gas.
Explaining two-slit interference
Question 170S: Short Answer
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This question asks you to consider the geometry of Young's slits and the effect that various changes
will have on the pattern that appears on the screen.
Young's slits
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1. Sketch the experimental arrangement used to produce the interference pattern known as Young's
slits. Label on your diagram:
the slit separation d
the fringe spacing x
the slit–screen separation L.
2. What is the relationship between the ratios / d and x / L?
3. The distance L is doubled but s and  do not change. Describe the change to the fringe pattern
and use your ideas about phasors to explain this change.
4. The slit separation d is doubled. Everything else about the arrangement stays the same. Again,
describe and explain what happens to the fringe pattern.
5. A more difficult question: One of the slits is doubled in width. Nothing else changes (including
the second slit). What will happen to the interference pattern? Explain your answer.
Two-source interference:
Some calculations
Question 180S: Short Answer
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Teaching Notes | Key Terms | Hints | Answers
Quick Help
These questions pose simple exercises to give you some confidence in handling calculations about
two-slit interference.
Two slits
1. Light of wavelength 450 nm falls on to two slits separated by 1.0 mm. 10 m away is a screen.
What is the fringe spacing on the screen?
Two loudspeakers
2. Two loudspeakers are placed 4 m apart for an open-air concert. They are playing back a flute
sounding a note of 680 Hz. Members of the audience sit in a row 20 m from the loudspeakers,
parallel to the line between the loudspeakers. Take the speed of sound as 340 m s–1. Describe,
as precisely as possible, what different people in the row will hear.
Two slits
3. Light from a colour filter is used to produce Young's double-slit fringes. The slit spacing is
0.4 mm. The distance between the slits and the screen on which the fringes are formed is 1.4 m
and the distance between successive dark spaces (or bright fringes) is 1.7 mm. Find the average
wavelength of the light used. Why must the answer be only an average?
Superposition of radio waves
Question 190S: Short Answer
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Teaching Notes | Key Terms | Hints | Answers | Key
Skills
Quick Help
Look at the diagram carefully and then answer the questions.
Direct and indirect transmission
Medium-wave radio signals from a transmitter can reach a receiver by two routes – the ground wave
travels direct from transmitter to receiver, whilst the sky wave travels up to the ionosphere, and is
reflected from it down to the receiver.
ionosphere
h
sky wave
transmitter
ground wave
receiver
surface of Earth
d
This means that superposition may occur. At night the sky wave is particularly strong; its amplitude is
comparable with that of the ground wave. The receiver may receive a strong signal, or almost none at
all, depending on the effective height, h, of the ionosphere at that moment. This height tends to vary
over a period of a few seconds and so the signal will rise and fall – an irritating phenomenon called
'fading'.
1. What can you say about the path difference between the two routes when the signal is a
maximum?
2. If the signal now changes to a minimum what must have happened to the path difference
between the two routes?
Suppose that a maximum signal is received when the wavelength  = 250 m, the distance from
transmitter to receiver d = 120 km, and the height of the ionosphere h is effectively 80 km.
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3. By what distance would h have to change in order for this signal to become a minimum?
Grating calculations
Question 200S: Short Answer
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These questions give you practice in using the grating formula n  = d sinn.
A grating with 500 lines per millimetre
A grating is labelled '500 lines per mm'.
1. Calculate the spacing of the slits in the grating.
2. Monochromatic light is aimed straight at the grating and is found to give a first-order maximum at
15º. Calculate the wavelength of the light source.
3. Calculate the position of the first-order maximum when red light of wavelength 730 nm is shone
directly at the grating.
4. The longest visible wavelength is that of red light with  = 750 nm. The shortest visible
wavelength is violet where  = 400 nm. Use this information to calculate the width of the angle
into which the first-order spectrum is spread out when white light is shone onto the grating.
A grating with light of wavelength 550 nm
A grating is illuminated with a parallel beam of light of wavelength 550 nm. The first-order maximum is
in a direction making an angle of 20º with the straight-through direction.
5. Calculate the spacing of the grating slits.
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6. What would be the angle of the first-order maximum if a grating of slit spacing of 2.5  10–6 m
were used with the same light source?
7. Calculate the wavelength of light which would give a second-order maximum at  = 32º with a
grating of slit spacing 2.5  10–6 m.
Using diffraction gratings
Question 210S: Short Answer
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Instructions
The photograph shows a two-colour light-emitting diode (LED) viewed through a diffraction grating.
The grating has 300 lines per millimetre and is placed over the camera lens which is 400 mm from the
LED. The scale shown is in centimetres.
Open the JPEG file
1. How can you tell that the LED is giving out two distinct colours?
2. The grating has 300 lines per millimetre. What is the spacing between the lines?
3. Look at the first-order green image, and work out the angle at which the green light is being
diffracted by the grating.
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4. Calculate the wavelength of the green light.
5. Repeat the calculations for the first-order red image of the LED.
6. Looking at the second-order images, make further measurements to confirm your wavelength
calculations.
7. How many further orders could be created using this grating?
Open the JPEG file
8. This photograph uses a different grating but the same LED as before. Calculate the spacing
between the lines in this grating and the number of lines per millimetre on the grating.
Open the JPEG file
Look carefully at the image of a street scene taken through a diffraction grating. There are three
major light sources A, B, C. Two of these are street lights, the third is a domestic security floodlight.
9. By looking at the spectra produced by the three sources write down the two sources that are
similar. What is the difference in the spectra produced by the two types of light?
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Open the JPEG file
This closer view of the street light shows the detail of the spectrum more clearly.
10. How many distinct colours are present in the spectrum from the street light?
11. Other street lights produce a more yellow light from 'sodium vapour lamps' which consists of only
one colour of light. How would such a street light appear if viewed through a diffraction grating?
Phasors to oscillations
Question 220S: Short Answer
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Quick Help
You will draw out the trace of a sinusoidal oscillation of given frequency and amplitude, starting from a
rotating phasor.
How to draw oscillations
You will need a sheet of graph paper, ruler, pair of compasses and protractor. As in the diagram,
draw a circle on the left of the paper and graph axes of displacement against time to the right of the
circle. Draw the time axis through the centre of the circle.
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displacement
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2
time / s
rate of rotation 300° / s
Divide the circle into twelve 30° sectors, as shown. Divide the time axis of the graph into at least
twelve divisions, each representing 0.1 s.
1. You are going to use a phasor rotating by 30° every 0.1 s. How many degrees does it turn per
second? How long will the phasor take to turn through one complete revolution? How many
complete revolutions does it make per second?
displacement
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2
time / s
rate of rotation 300° / s
2. Choose any marked radius of the circle, and start the phasor off there at time t = 0. Copy across
to the graph axes the displacement of the tip of the phasor above the axis, and mark this
displacement at t = 0. This step is shown in the diagram. You may start at a different initial angle
to that chosen in the diagram.
3. Go forward in time by 0.1 s. Rotate the phasor anticlockwise (angle to axis increasing) by 30°.
Again copy across to the graph axes the new displacement of the tip of the phasor above the
axis, and mark this displacement at t = 0.1 s. This step is also shown in the diagram.
4. Continue to rotate the phasor at 30°. every 0.1 s, and mark the vertical displacement of its tip on
the graph, at the appropriate moment in time.
5. Having completed rotating the phasor through a complete circle, join up the graph points
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smoothly. What kind of curve have you drawn?
6. What is the periodic time of the curve you have drawn? What is its frequency?
Wave to phasor
Question 230S: Short Answer
Quick Help
Teaching Notes | Key Terms | Hints | Answers
Here you practice identifying the phasor that describes different times as a wave passes.
Find the phasor
displacement
1
6
4
2
5
time / s
3
Draw on the circle a phasor arrow for each of the points 1 to 6 marked on the displacement–time
graph.
Linear and angular speed
Question 240S: Short Answer
Teaching Notes | Key Terms | Answers
Quick Help
This question introduces you to the relationship between the linear speed at the edge of a rotating
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object and the angular speed at which the object rotates.
Spinning wheels
As you have learnt, phasors rotate and this rotation can represent oscillation in a wave. Sometimes it
is of interest to know the linear speed at which the tip of the phasor is moving. This has many
practical applications. For example, an engineer may need to calculate the stress in a gear rotating at
high speed knowing only the rotation speed in degrees turned every second. But the answer may
depend on how fast the edge of the gear moves. Think of a bicycle wheel with radial spokes.
Open the JPEG file
Source
This will make a good model for a phasor of amplitude equal to the radius of the wheel from axle to
the edge of the tyre. The frequency of rotation is equivalent to the number of rotations of the wheel in
one second.
1. What is the circumference if the wheel has a radius of 0.4 m?
2. Suppose the wheel rotates twice every second. If the wheel does not slip at the ground, how long
is the track made by one second of travel?
3. The centre of the wheel is always vertically above the contact point between the tyre and the
ground. How far does the centre move forward in one second?
4. So, what is the linear speed of the bicycle?
Now repeat the above calculations using symbols rather than numbers.
5. If the radius of the wheel is r , what is the circumference of the wheel?
6. How far does the bicycle move for each turn of the wheel?
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7. If the number of rotations every second is f, how far does the bicycle move in one second?
8. Write down the equation for the linear speed of the bicycle in terms of its wheel radius and
frequency.
Specifications for a lens
Question 250S: Short Answer
Teaching Notes | Key Terms | Answers | Key Skills
Quick Help
Checking
This question looks at the trip times for light passing through a lens. If light on separate paths is in
phase on the way into the lens then it also needs to be in phase as it arrives at the focus. So all the
trip times are the same. Use this to calculate the refractive index for the lens.
Use the following data: c = 3.0  1010 cm s–1, f = 6.0  1014 Hz.
Designed for a material
5 cm
5 cm
50 cm
0.5 cm
This lens design might be useful if a material of an appropriate refractive index can be found. Answer
these questions to find the required refractive index.
1. How long does it take the light to cover the 50 cm on the way to the lens?
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As all the beams use the same path we can now ignore this distance.
2. Calculate the trip time for the top, red, path from the lens to the focus.
3. Write down the total trip time for the middle, blue, path from when it reaches the lens to the focus.
4. Calculate the partial trip time for the section in air for the middle, blue, path.
5. Calculate the time the light must spend in the glass.
6. How many wavelengths must there be in this 0.5 cm?
7. What is the wavelength in glass?
8. What is the required refractive index?
9. See if you can find a material of this refractive index from the File 10D 'Materials database'.
10. How would you have to reshape the lens so that it could be made of a material of lower refractive
index?
11. Explain why this reshaping would be necessary.
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Making estimates about waves
Question 100E: Estimate
Teaching Notes | Key Terms | Answers
Quick Help
In these questions you are expected to make sensible estimates of quantities related to wave
behaviour, and then combine them to calculate the required answer.
Give your answer to a justifiable number of significant figures. Show the units of your estimated
quantities and of your final answer.
Write the answers in the spaces provided.
For advice on estimation, see Making rough estimates in Data and Measurement Skills.
Estimate the distance of a thunderstorm
1. Remember the last time you experienced a thunderstorm. Estimate how far away it was.
Informed guesses for relevant data:
Calculation:
Estimate Galileo’s reaction time
2. Galileo wanted to try to measure the speed of light by flashing a light on one hill, to have the flash
answered by a lamp on another hill. Estimate how quickly Galileo would have had to react, if any
delay were to be detected.
Informed guesses for relevant data:
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Calculation:
Estimate the wavelength of an FM radio station
3. Estimate the wavelength of an FM radio station.
Informed guesses for relevant data:
Calculation:
Estimate the spreading of a laser beam
4. Estimate how much the light from a laser beam spreads if you shine it through a pinhole onto a
wall.
Informed guesses for relevant data:
Calculation:
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Estimate the spacing of an ultrasound grating
5. Suppose you want to make a diffraction grating for ultrasound travelling in water. Estimate the
grating spacing required.
Informed guesses for relevant data:
Calculation:
Superposition outside the laboratory
Question 80X: Explanation–Exposition
Teaching Notes | Key Terms | Key Skills
Quick Help
Finding out about a phenomenon involving superposition
You have seen a number of examples of superposition in the laboratory. These will have introduced
the idea that when waves pass through one another their amplitudes add in such a manner that 'more
can be less and less can be more'.
This is an activity in which you find out and report on a phenomenon involving superposition. Your
report should not be more than two sides of A4 paper but must include a description of the effect you
have chosen and an explanation of the physics involved. Diagrams are very helpful.
Superposition is the basis of more complex phenomena such as interference, diffraction and standing
waves. These phenomena are observed in all forms of wave motion and can be used to explain a
very wide range of effects. Other phenomena involving superposition are the colours of soap bubbles
and oil films and the 'blooming of lenses'. Musical instruments such as guitars and flutes involve
waves superposing to make standing waves. But these are just examples, the tip of the iceberg.
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What to do
1. Think of (or find) an application of the principle of superposition that particularly interests you.
This application could be chosen from the list given below or could be your own idea.
2. Research your chosen application. You should use the usual sources of information (books,
journals, the internet and this CD-ROM.
3. Write a brief description of the application, including diagrams where appropriate.
4. Describe the physics of the situation, starting from the principle of superposition. Explain any
scientific terminology you use.
Possible areas of study
This list is not exhaustive, but gives you an idea of the range of phenomena you might consider:
irridescence in nature
colours on an oil film
blooming lenses
room acoustics
musical instruments
interference, radio reception and 'ionospheric fading'
standing waves in microwave ovens
'hundred year' waves
An example to help you
This short piece of writing and the comments that follow it will give you an idea how to go about this
activity. It was written by a student taking AS-level.
Lens blooming
When you look at the lens of a camera or telescope you see a blue / purple colour. This colour is due
to a thin coating on the surface of the lens called bloom. This is a layer of magnesium fluoride (MgFl2)
which is applied to the lens by vaporising the MgFl2 and allowing it to condense on the surface of the
lens. This layer is designed to reduce reflection of light thereby maximising the light intensity reaching
the film, but how does it do this?
The thickness of the bloom is a quarter of a wavelength () of the middle part of the visible band of
light.
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R
R1
air
bloom
lens
T
T1
As the incident ray reaches the surface of the bloom some is reflected and some is transmitted. The
transmitted ray travels / 4 to the surface of the glass. At the lens surface again some of the light is
reflected and some is transmitted (T). The reflected light travels back a further / 4 to the edge of the
bloom where again some light is transmitted and some reflected. At each reflection, phase changes
can occur. Ignoring these for the moment, the transmitted light (R1), due to having travelled a further 
/ 2 than the original reflected light is now in antiphase with it and so due to destructive interference of
waves they cancel each other out. This means that much less energy is reflected back from the lens.
Instead the reflected light is then transmitted out of the bloom into the lens (T1) after travelling a
further / 4, making it in phase again with the first transmitted light in to the lens (T). This results in
constructive interference of light which increases the intensity of the transmitted beam.
This is very clearly shown in the photograph taken by Fritz Goro. You can see just how effective lens
blooming can be. The phase changes on reflection complicate the story a bit, but these are the
essentials of how it works.
Open the JPEG file
Source
Open the JPEG file
Source
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This classic photograph taken in 1944 shows the multiple reflections from an uncoated lens. As with
all light entering a more dense medium the light is refracted when it enters the bloom and again when
it enters the lens. Due to the exaggerated angle of incidence on the diagram these can be seen
clearly. This means the wave is slowed and so the wavelength decreases. This means that the / 4
has to apply to the wavelength of the middle part of the visible spectrum when the light is travelling
through magnesium fluoride rather than in a vacuum.
The purple colour of the lens occurs because the bloom has a thickness of / 4 of the middle part of
the spectrum, i.e. green. This means that no green light is reflected but some blue and some red is,
so by addition of colours a purple colour results. Modern blooms often look red, indicating a shift in
the wavelength chosen for good transmission.
Our comment
This student has made a pretty good attempt at this piece on superposition. But the most useful and
interesting parts are the diagrams and photographs. Although he has included a lot of information it is
very densely written. More diagrams and fewer words would really help.
You should have
1. Found out about an example of superposition which interests you.
2. Practised presenting information clearly and effectively.
Checking out a mirror
Question 120X: Explanation–Exposition
Teaching Notes | Key Terms | Hints | Answers
Quick Help
A possible focus
This is a drawing exercise that relates the trip time for different paths that light might take to the
curvature and focus of the mirror.
A possible design
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Parabolic mirror
y = 0.05x2
0.75 m
5m
This mirror has been suggested as a design that will focus 0.75 m from the bottom of the mirror.
1. Check, using a scale drawing on graph paper, whether this design does show the focus correctly
placed.
2. Use your ideas about phasors and trip times to write an account of how your scale drawing
justifies your answer.
3. Suggest how the true focus can be found for any future design.
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Calculating wavelength in two-slit interference
Question 130X: Explanation–Exposition
Quick Help
Teaching Notes | Key Terms | Answers
This question takes you through the steps leading to the result / d = x / L which allows you to
calculate the wavelength  in a Young's two-slit experiment.
The experiment
The two-slit experiment looks like this:
Young’s two-slit interference experiment
two slits: 1 mm
spacing or less
narrow
source
several metres
screen
bright and
dark fringes
several metres
It ought to be drawn much wider, of course. The light from one slit is very nearly parallel to the light
from the other: an angle amounting to less than 1 mm in more than 1 m. Try drawing such an angle if
you aren't persuaded.
Making fringes
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Bright fringe at centre
to central bright
fringe on screen
d

waves in phase:
path difference = 0
1. Use this diagram to explain why there is a bright fringe at the centre of the pattern on the screen.
Light going to first dark fringe
to dark fringe on screen


d
d sin  = /2
waves in anti-phase:
path difference = /2
path difference = d sin 
2. Explain why the path difference is half a wavelength.
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3. Explain why the path difference is d sin
First bright fringe off centre
to bright
fringe on
screen

d

d sin  = 
waves in phase:
path difference = 
path difference = d sin 
4. Explain why the path difference is now equal to one wavelength.
5. Show that / d = sin 
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Geometry of two-slit experiment
fringe
spacing x
slit-screen distance L

light combines
at distant screen
from
source d

x/L = sin  approximately
6. Use the diagram to explain why, approximately, sin = x / L.
7. Show that d = x / L.
Another way of looking at the fact that the slit to screen distance is very nearly the same as the
sloping light path is that:
x
 sin 
lightpath
while
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x
 tan .
L
The angle  is very small, less than 1. For these small angles, sin  is almost the same as tan .
8. Use a calculator to complete the table. Use a level of precision that shows the difference in each
case – your answers will need to become steadily more precise.
Angle,  /
degrees
sin
tan
sin/ tan
10
5
2
1
Colours in thin films
Question 140X: Explanation–Exposition
Teaching Notes | Key Terms | Hints | Answers
Quick Help
What to do
Look at the following images of colours produced by thin films, and answer the questions about what
you can see.
Coloured bands in soap films
Open the JPEG file
Source
Open the JPEG file
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Source
Open the JPEG file
Source
Look at the above sequence of images showing a soap film stretched over a wire loop. The coloured
bands are formed as light reflecting from the front and rear surfaces interferes. For example, where
blue bands are formed the phase difference between these two sets of waves is such that blue light
emerges in phase and so combines constructively. A little further down the film the phase difference
is constructive for red light. In moving down the soap film from one band of a certain colour to the next
band of the same colour, the path difference increases by one wavelength of light. The path
difference is twice the thickness of the film, since light reflected from the back surface goes twice
through the film.The bands are rather like contour lines on a map but here they show regions of
uniform thickness rather than height.
Open the JPEG file
Source
Open the JPEG file
Source
Open the JPEG file
Source
These images have been modified to show the intensities of red, green and blue light in different
regions of the soap film. The different positions for the dark bands can clearly be seen.
1. Why are the blue bands closer together than the red bands?
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Open the JPEG file
Source
Open the JPEG file
Source
Open the JPEG file
Source
Open the JPEG file
Source
This sequence of images is taken in light from a sodium lamp. This is a monochromatic source with a
wavelength of 589 nm. We can use the pattern of dark bands to give us information on how the
thickness of the film is changing.
Assume that the thickness of the film is almost zero for the dark band at the top of the film. Also
remember that the refractive index of water will reduce the wavelength of the light in the film by a
factor of about 1.33, in this case to 589 / 1.3 nm, i.e. 443 nm.
2. Using the first image, plot a graph of position up the soap film (y-axis) against thickness of film
(x-axis). Remember that each band represents increments of (in this case) half of 443 nm.
3. Repeat for the next two images. How is the profile of the soap film changing over time?
Look at the fourth image. The soap film is still stretched completely over the wire loop but the upper
portion is completely black. For this extremely thin section the path difference is virtually zero and yet
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you do not see a bright band. The phase difference must be half a cycle to produce the destructive
interference you see. This half-cycle phase difference happens as the light undergoes reflection from
the rear surface inside the film.
Reflection on lenses
Open the JPEG file
Many optical instruments have 'antireflection coatings' designed to maximise the light transmitted
through the optical element by minimising the reflections from its surfaces. The wavelength chosen
for zero reflection in these examples is in the middle of the visible spectrum (green).
4. If the path difference is one whole wavelength of green light what can you say about the path
difference for red or blue light? How does this account for the characteristic purple colour of the
coating? Recently, coated lenses have become red in colour. What does this say about the colour
of light they transmit best?
The Nautilus loudspeaker
Reading 10T: Text to Read
Teaching Notes | Key Terms
Quick Help
The Nautilus speaker helps reproduce high-fidelity sound and looks remarkable. But its looks are not
only to create an impression – they are based on clear physical principles that you will be able to
understand. The short passage below is written by Laurence Dickie, one of the designers of the
loudspeaker.
See Display Material 10P 'Nautilus loudspeaker' for images of this speaker.
The Nautilus
The principle of operation of the Nautilus loudspeaker is rather simple. A conventional loudspeaker
cone radiates low frequencies very badly, because the waves from the front and back of the cone are
in antiphase. That is, the rearward acoustic wave from a vibrating diaphragm is out of phase with that
from the front. If both front and back are allowed to radiate freely, there is no radiation sideways from
the diaphragm, in its own plane, because the two waves cancel. Along the axis of the cones, the
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power radiated is proportional to frequency, so that low-frequency efficiency is poor.
The most common solution to this problem is to enclose the rear of the speaker in a box and prevent
the rear wave escaping. Unfortunately the box has a major shortcoming. The enclosed volume of air
resonates at frequencies where the internal dimensions of the box are equal to integer multiples of
half wavelengths. These resonances are radiated back through the loudspeaker cone which is not a
perfect acoustic barrier, resulting in a frequency response with many peaks and troughs to which the
human ear is very sensitive.
An alternative solution is to mount the oscillating cone – the driver – on the end of an infinite tube.
The rear wave can then propagate down the tube without ever returning to do mischief. The problem
is to make an infinite tube! The novel step introduced in the Nautilus was to exploit the fact that an
exponentially tapered waveguide behaves identically to a constant area tube above a certain 'cut-off'
frequency. However, the volume of an infinitely long exponential horn is finite, so resonance-free
performance can be obtained in a practical enclosure which is a finite size.
The area A of the cross section of such a tube is given by
A  A0 e  mx
where
m
4f
c
in which A0 is the initial area of the tube, x is the distance along the tube, f is the cut-off frequency and
c is the speed of sound in air. The volume of such a tube comes out to be A0 / m.
A further imperfection commonly encountered in traditional cabinets is the sharp edge around the
front panel on which the drivers are mounted. The sound wave coming from the driver is reflected by
this edge and interferes with the direct signal causing further irregularities in the response. In Nautilus
the cabinet edge is as rounded as possible to minimise this effect.
Lastly, the flat wooden panels themselves are prone to resonances which are usually in the frequency
range of the loudspeaker. The round fibre glass tubes and surface of Nautilus are intrinsically stiff
shapes and have resonant frequencies well above the pass band of each driver.
Acoustics of rooms
Reading 20T: Text to Read
Teaching Notes | Key Terms
Quick Help
The sound you hear is not the same as the sound that was made. Reflections from walls, floor and
ceiling make a difference to what you hear.
Sound intensities
Any vibrating object which creates sound waves radiates power. The greater the energy radiated, the
louder you hear the sound. However, the loudness of a sound decreases the further away it is. The
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amount of power arriving at the point where it is perceived is the intensity of the sound. It is expressed
in watts per square metre – that is to say, the number of watts of power passing through an imaginary
window of one square metre at a given distance from the source. The relationship between the
acoustic power P of the source, and the intensity I of the resultant sound, at a distance r metres from
the source is given by
I 
P
4 r 2 .
Consider an example of a sound source which is radiating 0.4 W of acoustic power equally in all
directions.
loudspeaker
A
3m
6m
B
The intensity at point A is then
I 
0 .4 W
4 3
2
 0.003 54 W m  2 .
A doubling of the distance to 6 metres results in an intensity at point B of
I 
0 .4 W
4 6 2
 0.000 88 W m  2 .
This is a reduction in intensity by a factor of 4. In general, and in what is known as a free-field
environment, with no boundaries or obstacles to impede the sound wave, the decrease in intensity
is inversely proportional to the square of the distance. This is known as the inverse square law.
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Much of the time, however, we hear sounds in enclosed environments – ones with walls and ceilings.
The shape and size of the room, together with the materials which make it up, affect the sound wave
in a number of ways, and consequently have an impact on the perceived quality of the sound. The
sound of a voice or a musical instrument heard within a cathedral has a very different quality from the
same sound heard within a normal living room, or outside in the street.
Sound reflection off surfaces
Iincident
Wall
Ireflected
When a sound wave, with intensity Iincident in watts per square metre, encounters a surface, such as a
wall, it will be reflected, as shown. However, some of its energy will be lost, absorbed within the fabric
of the wall; the intensity of the reflected wave, Ireflected, will therefore be less than that of the incident
wave.
The amount of energy absorbed, Iincident, is the difference between Iincident and Ireflected, and varies with
the material used. Concrete and brick, for example, absorb relatively little of the sound energy,
whereas plasterboard and wood panelling absorb rather more.
We can define a value for a given material, called its absorption coefficient, for which the symbol 
is commonly used. This can be calculated as:

I absorbed
.
I incident
The absorption coefficient will be somewhere between 0 (no absorption at all) and 1 (complete
absorption). It will also vary with frequency. Most materials absorb high-frequency sound more readily
than low-frequency sound, and this needs to be taken into account when considering the fabric of a
room in which sound quality is of importance. Below is a table of typical building materials, together
with their absorption coefficients at different frequencies (in Hz).
Material
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250 Hz
500 Hz
1 kHz
2 kHz
4 kHz
Material
125 Hz
250 Hz
500 Hz
1 kHz
2 kHz
4 kHz
Concrete floor
0.01
0.01
0.015
0.02
0.02
0.02
Wooden floor
0.15
0.11
0.10
0.07
0.06
0.07
Heavy carpet
0.08
0.27
0.39
0.34
0.48
0.63
Window glass
0.35
0.25
0.18
0.12
0.07
0.04
Plaster on brick
0.013
0.015
0.02
0.03
0.04
0.05
Source
The absorption coefficient tells us the proportion of the incident energy absorbed by a material.
Clearly, however, ten square metres of, say, plasterboard will absorb more sound than five square
metres of the same material. We need to know exactly how much sound energy will be absorbed by a
surface, based on its size and absorption coefficient. For this, the standard unit is the metric Sabin
named after the American pioneer Wallace Sabine.
One metric Sabin is defined as the amount of sound absorption provided by one square metre of a
material whose absorption coefficient is 1.0.
Thus the absorption of a material whose area is S square metres and whose absorption coefficient is
 is S metric Sabin.
The table above shows that, for a sound wave of 500 Hz, ten square metres of heavy carpet will
absorb 10 m2  0.39 = 3.9 metric Sabin.
The total absorption A of the surfaces of a room will be the sum of the absorption of all the materials
used on those surfaces. A room with S1 square metres of material with coefficient 1, S2 square
metres of material with coefficient 2, and so on, is:
A  S1 1  S 2  2  ...etc.
The amount of sound absorption provided by a room is one of the most important factors to be
considered when designing the room for sound.
Reverberation
Reverberation is the audible persistence of sound caused by multiple reflections off the walls, floor
and ceiling of an enclosed space. The length of time that the sound remains audible depends on the
size of the room, and also on the absorbing qualities of the surfaces of the room.
Let us take as an example a room in which a short, sharp sound, such as a handclap is made. The
diagram below shows five of the theoretically infinite number of paths that the sound wave can take.
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3
4
1
2
5
10 m
The sound radiates away from the source, encountering and reflecting off all the surfaces in the room.
The listener will initially receive the sound wave whose path is direct (labelled 1 in the diagram). If we
assume that this path is 6.7 metres, and that the wave travels at 344 metres per second, the normal
speed of sound in air, the time taken for it to arrive at the listener will be:
6 .7 m
344 m s 1
 19 .48 ms.
The intensity (shown by the vertical line marked 1) of the sound along the direct path, arriving at time
19.48 ms will be determined by the inverse square law.
1
2
3
reverberant sound
4
5
0
10
20
30
40
50
60
time / ms
A very short time after that, the sound wave represented by path 2 arrives at the listener. It has had
further to travel (7.34 metres), and therefore takes 21.34 milliseconds to arrive, a delay of 1.86
milliseconds. Its intensity is also less, for two reasons. Firstly, it has travelled further. Secondly, some
of its energy has been absorbed by the floor off which it reflected.
In the same way, the sound waves represented by wave paths 3, 4 and 5 arrive at staggered intervals
over about 10 milliseconds. This phase then gives way to one in which the listener is very quickly
receiving multiple reflections, and reflections of reflections, resulting in a 'smearing' of the sound
known as reverberant sound.
What happens if the absorbency of all the surfaces in the room is increased?
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1
2
reverberant sound
3
4
5
0
10
20
30
40
50
60
time / ms
The amount of energy absorbed, and therefore removed from each sound wave path on each
reflection is much greater. The intensity of the reverberant sound is considerably less. A useful
measure for estimating reverberation is reverberation time T. This is defined as the time in seconds
required for the intensity of the reverberant sound to decay to one-millionth of its original intensity.
Expressed in decibels, this is equivalent to a decrease in intensity of 60 dB. For this reason acoustic
engineers give the name RT60 to the reverberation time.
An equation for calculating reverberation time, first put forward by Wallace Sabine, shows the
relationship between the volume of a room, and its overall absorption properties:
T 
0.16V
A
where V is the volume of the room in cubic metres, and A is the total absorption, in metric Sabins, of
all the surfaces in the room.
Take an example: a room whose length, width and height are 20 metres, 15 metres and 10 metres
respectively, has a ceiling made of acoustic tiles, whose absorption coefficient at 500 Hz is 0.66. The
floor is covered in thick carpet (absorption coefficient 0.39 at 500 Hz) and the walls are of plaster on
brick (absorption coefficient 0.02 at 500 Hz).
Area of surface
Absorption coefficient
Absorption in met
Area of floor
20 m  15 m = 300 m2
0.39
300 m2  0.39 = 117
Area of ceiling
20 m  15 m = 300 m2
0.66
300 m2  0.66 = 198
Area of two end walls
15 m  10 m 2 = 300 m2
0.02
300 m2  0.02 = 6
Area of two side walls
20 m  10 m  2 = 400 m2
0.02
400 m2  0.02 = 7
Total absorption of ro
The volume of the room is 20 m  10 m  15 m = 3000 m3.
The reverberation time T for this room is therefore predicted to be:
T  0.161  3000 m 3 / 329 metric Sabin  1.47 s.
Note that this is only at 500 Hz – the different values of absorption coefficient at other frequencies for
a given material suggest that reverberation time must vary with frequency, and this is in fact the case.
Furthermore, for high frequencies, the absorption properties of air need to be taken into account; at
54 Advancing Physics
frequencies in the region of 5000 Hz, the absorption of sound by air may be comparable to that of the
surfaces of the room.
Clearly, the reverberation time for a room intended primarily for speech needs to be quite short, of the
order of 0.4 to 0.8 seconds or so, if the speaker is to be intelligible. Concert halls, however, need a
rather longer reverberation time, dependent on the type of music. Symphonic music of the nineteenth
century requires a rather longer reverberation time than does baroque music of the seventeenth and
eighteenth centuries.
Artificial reverberation
There is a range of audio equipment available to the sound engineer which uses digital techniques
designed to simulate reverberation. The reverberation appropriate to various room sizes – room, hall,
etc – can be re-created. In essence, the technique involves splitting the signal in such a way that part
of it is held back for a predefined period of time, and then mixed in with the direct signal to form the
output, as shown.
gain control
delay
input
+
output
adder
artificial reverberation
Sound localisation
The human hearing apparatus provides us with a very sophisticated ability to localise sound – to
determine the direction from which a sound is coming. Our binaural hearing is one of the most
important mechanisms we have for this – people with hearing in only one ear have severe difficulties
in localising sound.
There seem to be two distinctly but complementary processes at work. At high frequencies (from
about 1500 Hz upwards) the head itself casts a shadow, resulting in a higher intensity on the side of
the head from which the sound is coming. This provides us with an important perceptual cue in
localising that sound.
However, this mechanism operates less effectively below 1500 Hz. Low-frequency sound, with long
wavelengths, bends, or diffracts, around an obstacle to a greater extent than high-frequency sound,
and therefore the head casts less of a sound 'shadow'. In this frequency band, it is the very slight
difference in the arrival time of a sound wave at each ear which is crucial.
Given that a listener in a room with reflective walls is receiving not only the direct sound, but also a
number of reflections coming from other directions, how does the hearing mechanism integrate and
fuse all this information into one perceptual signal?
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Imagine that we place two loudspeakers so that they are equidistant from a listener sitting in front of
them. If the sounds from the speakers are identical and equal in intensity, the listener will perceive the
sound as coming from a point directly between them. If the intensity is gradually increased in the
left-hand speaker, the sound will appear to move towards that speaker, until eventually all the sound
appears to come from it. This illusion of sound localisation is the way sound engineers generate a
stereo image from a multitrack recording.
Now suppose that the intensity of sound coming from the speakers is equal, but we introduce a time
delay of about 20 to 30 milliseconds into the signal from one of them. The listener will perceive the
sound as coming from the speaker from which the sound arrived first. This is called the precedence
or Haas effect. It explains why it is the direct sound, rather than the first or any subsequent
reflections, which enables the localisation of a sound source.
If the delay is increased to more than about 35 milliseconds, however, the auditory mechanism is
unable to fuse the two signals into one perceptual whole. The reflection will then be heard as a
distinctly separate sound, or echo – something which is undesirable in the concert hall.
Bathrooms and concert halls
Just as waves in an organ pipe can form standing waves, so can waves in an enclosed space. This
can result in an undesirably uneven distribution of sound in a room – patches of loudness and relative
quiet. The Royal Albert Hall in London suffered very badly from this until acoustic engineers hung
sound absorbers from the ceiling. Secondly, frequencies at which standing waves can be established
will be noticeably louder than other frequencies, and take longer to decay, resulting in a characteristic
'boom' or ringing. This is why singing in a bathroom which has tiled walls can be such a rewarding
experience. The response of the room to your voice gives the overall sound greater body and
fullness. For a recording studio, or small concert hall, however, this can cause unwanted 'colouring' of
the sound, and it is for this reason that sound absorbers of various kinds will be strategically placed in
order to damp out unwanted standing waves.
Summary
1. The intensity of sound in a free-field environment is determined by the inverse square law, in
which the decrease in intensity is inversely proportional to the square of the distance.
2. In a closed environment, a sound wave will reflect off surfaces – walls, ceilings, etc – that it
encounters, having lost a proportion of its energy on the way – a proportion which is determined
by the absorption coefficients of the surfaces.
3. Reverberation is caused by multiple reflections of the sound from the boundaries of the
enclosure.
4. The time taken for the reverberating sound to reduce in intensity by 60 dB is called the
reverberation time, and is determined by the volume of the room, and the absorption properties of
the surfaces, objects and air within the room.
5. Our ability to localise the whereabouts of a source of sound arises from the perception of intensity
and phase differences in the sound coming from it.
6. The auditory mechanism is able to fuse direct sound and early reflections into one perceptual
whole, thus enabling the localisation of sound even in a reverberant environment.
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Historical attempts to measure the speed of light
Reading 30T: Text to Read
Teaching Notes | Key Terms
Quick Help
This short reading will introduce you to some of the problems investigators found in attempting to
measure the speed of light and the imaginative solutions developed.
Measurements of the velocity of light from Galileo to Michelson
In everyday life we assume that light travels instantaneously from one place to the next. Shine a torch
into a darkened room and the far wall is 'instantly' illuminated. Common sense tells us that light either
instantly crosses distances or travels so fast that it cannot be measured.
The philosopher Descartes (1596–1650) supported the idea of instantaneous travel in which light
takes no time at all to cover vast distances. The Italian astronomer and mathematician, Galileo
(1564–1632) held the contrary view and claimed that light travelled at a finite though very high speed.
Galileo attempted to support his belief through experiment and so made the first serious recorded
attempt to measure the speed of light.
Galileo flashes a lamp (around 1600)
Galileo publicised many of his scientific ideas through fictional 'dialogues' in which the participants
discuss the new ideas. As the fictional characters discuss the theories and gradually develop their
understanding the reader also follows the argument and reaches a similar level of knowledge.
At the beginning of Galileo's masterpiece 'Dialogues Concerning Two New Sciences' (1638) three
people, Sagredo, Salviatti and Simplicio, discuss the problem of the speed of light. It is worthwhile to
eavesdrop on their conversation.
Sagredo: But of what kind and how great must we consider this speed of light to be? Is it
instantaneous or momentary or does it like other motions require time? Can we not decide this by
experiment?
Simplicio: Everyday experience shows that the propagation of light is instantaneous; for when we see
a piece of artillery fired at a great distance, the flash reaches our eyes without lapse of time; but the
sound reaches our ears only after a noticeable interval.
How would you reply to Simplicio's point? What does the observation really show us? Perhaps you'll
agree with Sagredo's response.
Sagredo: Well, Simplicio, the only thing I am able to infer from this familiar bit of experience is that
sound in reaching our ears, travels more slowly than light; it does not inform me whether the coming
of light is instantaneous or whether, although extremely rapid, it still occupies time.
Salviatti now describes what is, in fact, Galileo's attempt to measure light's velocity.
Salviatti : Let each of two persons take a light contained in a lantern, or other receptacle, such that by
the interposition of the hand, the one can shut off or admit light to the vision of the other. Let them
stand opposite one another and practice until they gain such skill in uncovering and covering their
lights that the instant one sees the light of his companion he will uncover his own. Let the two
experimenters take up positions separated by a distance of two or three miles and perform the same
experiment at night, noting carefully whether the exposures and occultations [where the light is
covered] occur in the same manner as at short distance; if they do we may safely conclude that the
propagation of light is instantaneous; but if time is required at a distance of three miles which,
considering the going of one light and the coming of the other really amounts to six, then the delay
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ought to be easily observable….
Sagredo: This experiment strikes me as a clever and reliable invention. But tell us what you conclude
from the results.
Salviatti : In fact I have tried the experiment only at a short distance, less than a mile, from which I
have not been able to ascertain with certainty whether the appearance of the light was instantaneous
or not; but if not instantaneous it is extraordinarily rapid.
(From: Dialogues Concerning Two New Sciences translated by Crew and de Salvio (Dover Books
1954).)
So, the first attempted measurement of light's speed proved inconclusive. It was too fast to measure
on the ground because the distances were too small. Three-quarters of a century later Römer looked
to the movement of Jupiter's four brightest moons to show the finite speed of light. Huygens used his
results to deduce a value of 2  108 m s–1. The next attempt was made by the English astronomer
James Bradley in the eighteenth century.
Bradley watches light fall like raindrops (1727)
You have probably experienced the dilemma of how fast to run in a rain storm. When rain is coming
down vertically you only get soaked on your head and shoulders if you walk slowly. Start running and
the rain slaps you in the face. You may also have watched raindrops running down a train window. If
the train is stationary during the storm you will see water drops running vertically down the window,
but as soon as the train begins to pick up speed the drops run at an angle. The faster the train moves
the greater the angle from vertical.
If you knew the speed of the train you could take a few simple measurements to find the velocity of
the rain.
Now Bradley, of course, would not have had the pleasure of sitting in a train on a rainy day, but he
employed a similar technique to measure the speed of light. He calculated the speed of the Earth in
its orbit and then observed the precise vertical angle of a star near the zenith (in other words, nearly
directly upwards). He observed the same star six months later and found that the apparent position
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had changed. He called this effect 'aberration'. Just as the raindrops take a different path down the
windows of the train when it reverses, so the beams of light fell down the telescope at a different
angle when the Earth reversed direction. The light was moving vertically but the movement of the
Earth made it appear to move at a slight angle to the vertical.
to star
light appears
to come at
an angle.
light now appears
to come at this
angle
six months
later.....
telescope moving
telescope moving
in opposite
direction
Earth’s speed in orbit
1
speed of light
10 000
1
Angle ~
radian
10 000
–3
~ 5.7×10
~
Bradley found that the star's position moved 20 seconds away from the vertical on each occasion.
This took careful measurement as there are 60 minutes in one degree and 60 seconds in a minute.
Bradley was measuring an angular change of 5.6  10–3 degrees. From this he calculated a value of
3.04  108 m s–1 for the speed of light.
Terrestrial methods:
Fizeau spins a cog-wheel (1849)
As methods of timing became more precise experimenters looked at ways to measure the speed of
light without looking up to the skies. These are the so-called 'terrestrial methods'.
In 1849 Hippolyte Louis Fizeau (1819–96), a French physicist, measured the speed of light using a
toothed wheel.
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at Senlis, near Paris
mirror
M2
in Paris
bright source
toothed wheel
telescope
angled mirror,
partly reflecting
M1
Light from the source S reflects off the half-silvered mirror M1. The light then travelled to a very distant
mirror M2 and back. Some of the reflected light will be transmitted through the half-silvered mirror and
seen by the observer. Fizeau had his source of light in Paris and the distant mirror in Senlis (a small
town to the north of Paris, where tolls on the autoroute start nowadays).
The toothed wheel acts to chop up the light beam, letting pulses through as it rotates. The experiment
is simple; the observer looks at the reflection and revolves the wheel faster and faster until the
reflection disappears. What is happening? A gap in the toothed wheel lets light through, but by the
time the light has travelled from Paris to Senlis and back, the gap has moved on and a tooth has
taken its place – no light gets through to the observer.
Fizeau used a distance of 8630 metres between the mirrors and a wheel with 720 cogs. He reached a
value for the speed of light of 3.13 m  108 m s–1.
Michelson spins a mirror (first attempt 1880)
The American physicist Albert Michelson (1852–1931) used an adapted version of Fizeau's
experiment with a rotating mirror substituting for the toothed wheel. This greatly improved the
precision of the experiment and Michelson took many measurements using it over a period of 50
years.
A rotating mirror was positioned 35 km from a fixed mirror. When the mirror is stationary you can see
that light will simply travel from the source to the mirror and back to the observer. Source and mirror
were placed on two adjacent mountaintops.
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fixed
mirror
fast spinning
octagonal
mirror
bright source
this face ready to
reflect light on the
way back.
this face
reflects
light out
If the mirror is set rotating, when the light comes back the mirror will have turned and will not send the
light back in its original direction. But when the mirror goes fast enough, the next face can have got
into just the place of the one in front which reflected the light out. So the light coming back is reflected
just as if the mirror were not turning at all.
When this is the case the time taken for light to travel the 70 km to the distant mirror and back must
be one-eighth of the time it takes an eight-sided mirror to rotate. Using this method and mirrors of 8,
12 and 16 sides Michelson determined the speed of light as 2.997 96  108 m s–1 with an uncertainty
of 0.000 04  108 m s–1.
The story continues
This is not the end of the story. Measurements of the velocity of light are made to increasing degrees
of precision to test theories. After centuries of attempts, the speed of light is now known to an
astonishing level of precision.
In 1973 a laser was first used to determine a value 2.997 924 574  108 m s–1 with an uncertainty of
only 1 m s–1.
Finally, in recent years the speed of light became a defined constant. From then on, a 'measurement
of the speed of light' became a measurement of the distance the light travelled.
Beats:
Continually varying phase difference
Reading 40T: Text to Read
Teaching Notes | Key Terms
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Quick Help
Here you can read about the superposition of two different frequencies.
Combining two notes
Two notes of almost the same frequency produce an interesting new effect, which piano tuners or
orchestral players can use to get the two notes exactly in tune. When the frequencies are a little
different, they superpose to make a sound which slowly pulses in intensity. The effect is called 'beats'.
One way to think about this is to plot out the two oscillations, and see how they go regularly in and out
of step.
Beats between different frequencies:
oscillation picture
frequency f1
frequency f2
in phase
in antiphase
in phase
in antiphase
in phase
beat frequency
f1 - f2
Another way to think about it is to go back to the phasor description. If one oscillation goes a little
faster than the other its phasor arrow spins a little faster. Like the speedier of two long-distance track
runners or of two Grand Prix drivers, the faster one is bound to lap the slower one from time to time.
Every time one arrow laps the other, they are pointing in the same direction, and the combined
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amplitude is large. Half way between, the two phasor arrows point in opposite directions, and give
only a small resultant amplitude.
Suppose the faster one is in phase with the slower at one moment, and comes round again to be in
phase with the slower one once again after a time T. The phasor arrow for the faster one must have
gone round just once more than the slower one, in the same time T. If the two frequencies are f2 and
f1, then:
turns (cycles) made by slower phasor  f1 T
turns (cycles) made by faster phasor  f2 T  f1 T  1.
From this the frequency 1/T of the beats is easily shown to be just the difference between the two
frequencies:
f2 T  f1 T  1
1 / T  f2  f1.
The rotating arrow or phasor picture just helps to make this result look simple and obvious. But any
way of looking at things which makes part of a problem look simple is welcome; it frees your mind to
think about other parts of the problem. Never mind that there is no real spinning arrow. The idea
helps.
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Beats between different frequencies:
phasor picture
phasor arrows start together
slow red arrow
fast blue arrow
slow component
fast component
resultant
in phase large resultant
resultant getting smaller
in antiphase small resultant
resultant getting larger
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Radio waves:
Fading and interference
Reading 50T: Text to Read
Teaching Notes | Key Terms
Quick Help
Radio waves reaching your receiver by two different paths can interfere with one another. This
reading illustrates how wave interference can lead to problems with radio reception.
Signals from abroad
Radio and television bands are limited and space on them is at a premium. There is no difficulty in
assigning two stations to the same frequency providing that the stations are well apart and an aerial
intended for one of the stations can never receive the signal from the other. Unfortunately, the best
laid plans can go astray as those living near the south coast of England will know. When weather
conditions are right, often at the end of a spell of stable, high-pressure conditions over Britain,
television signals from France can be received in England. The problem is that some of the French
stations have the same frequency as some of the British ones. So the waves combine (superpose)
and the picture on the screen can become distorted with wavy, fringe-like markings. When the effect
is particularly bad, ghost images can appear and the picture becomes hard to follow.
Fading is another effect that can make viewing and listening difficult. It is often experienced by those
who live near to an airport or those who travel along some motorways (the M4 is notorious for this
problem). The effect shows up as a slow fading and swelling in the volume of a radio station over tens
of seconds, or as a rapid flutter two or three times every second. The diagram shows what can
happen near an airport.
waves reflected by aircraft
waves direct to car
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The diagram shows how driving a car with an aeroplane going overhead can cause the rapid flutter of
intensity. The metal body of the aircraft reflects the signal to the receiving aerial of the car, and two
signals are picked up: one direct, one by a longer route. These two signals can arrive in phase, in
antiphase, or anywhere in between. The aircraft is moving, usually climbing or descending as well as
going forwards, and so the phase relationship between the direct signal and the reflected one is
constantly changing. This results in the strength of the received signal varying rapidly at the receiver.
signal from transmitter to NE
signal from transmitter to south
The slower variation – the M4 problem – is caused by the car radio picking up two signals from
different transmitters both working on the same frequency. The diagram shows that as the car moves,
the phase relationship again varies and the two signals can interfere constructively or destructively as
the car moves along. This leads to the annoying effect where the volume in the car is constantly
changing from very loud (in phase) to very quiet (in antiphase).
Revision Checklist
I can show my understanding of effects, ideas and relationships by
describing and explaining:
how standing waves are formed by sets of wave travelling in opposite directions
e.g. by drawing diagrams to show what happens
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A–Z References: amplitude, frequency, wavelength and wave speed, travelling and standing
waves
Summary Diagrams: Standing waves, Standing waves on a guitar, Standing waves in pipes
how waves passing through two slits combine and interfere (superpose) to produce a wave /
no-wave pattern
A–Z References: interference, path difference, double-slit interference, superposition
Summary Diagrams: Two-slit interference
what happens when waves pass through a single narrow gap (diffraction)
A–Z References: diffraction, phase and phasors
Summary Diagrams: Diffraction
how a diffraction grating works in producing a spectrum
A–Z References: gratings and spectra, phase and phasors
Summary Diagrams: A transmission grating
I can use the following words and phrases accurately when
describing effects and observations:
wave, standing wave, frequency, wavelength, amplitude, phase, phasor
A–Z References: amplitude, frequency, wavelength and wave speed, travelling and standing
waves, phase and phasors
Summary Diagrams: Phase and angle, Oscillations in phase, Oscillations in antiphase
path difference, interference, diffraction, superposition, coherence
A–Z References: interference, path difference, double-slit interference, diffraction, superposition,
coherence
Summary Diagrams: Two-slit interference, Coherence
I can sketch and interpret diagrams:
illustrating the propagation of waves
A–Z References: amplitude, frequency, wavelength and wave speed, travelling and standing
waves
showing how waves propagate in two dimensions using Huygens' wavelets
A–Z References: Huygen's wavelets
showing how waves that have travelled to a point by different paths combine to produce the
wave amplitude at that point
i.e. by adding together the different phases of the waves, using phasors
A–Z References: interference, double-slit interference, diffraction, gratings and spectra
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Summary Diagrams: Two-slit interference, A transmission grating
I can calculate:
wavelengths, wave speeds and frequencies by using (and remembering) the formula v = f 
A–Z References: amplitude, frequency, wavelength and wave speed
calculating the wavelengths of standing waves
e.g. in a string or a column of air
A–Z References: travelling and standing waves
Summary Diagrams: Standing waves, Standing waves on a guitar, Standing waves in pipes
path differences for waves passing through double slits and diffraction gratings
A–Z References: interference, double-slit interference, diffraction, gratings and spectra
Summary Diagrams: Two-slit interference, A transmission grating
the unknown quantity when given other relevant data in using the formula n  = d sin 
A–Z References: gratings and spectra
Summary Diagrams: A transmission grating
I can show my ability to make better measurements by:
measuring the wavelength of light
A–Z References: accuracy, systematic error, uncertainty
I can show an appreciation of the growth and use of scientific
knowledge by:
commenting on how and why ideas about the nature of light have changed
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