UNIT 4 - Download as DOC

Report as inappropriate Your report has been received

Shared by: j2mxu1K

Tags

Stats

views:: 4
posted:: 12/7/2011
language:: English
pages:: 30

Public Domain

Document Sample

Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 46

Recording and Reproducing Sound
This detailed study extends student understanding of waves together with aspects of
electromagnetism to the recording and reproducing of sound. Ideas of sound and electromagnetism
combine to aid the understanding of the operation and use of microphones and speakers, control of
sound level, need for hearing protection and acoustic properties of recording and performance spaces.

Nature of waves wave equation frequency response
1
dB → Wm-2
r2

Reflection Microphones
Diffraction Electret-condenser
Superposition Crystal
Dynamic
Resonance and standing Velocity
waves
Loudspeakers
Dynamic

Frequency response

Fidelity

 describe sound as the transmission of energy via longitudinal pressure waves;

Longitudinal waves
In longitudinal waves the vibration of the waves is in the same direction as the line of travel, the
particles do not move forward, they vibrate around an equilibrium position.



The movement of particles is in this direction.
compression rarefaction
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 47

max. positive pressure variation
(compression) greater than normal air pressure

Pressure variation

max. negative pressure variation
(rarefaction) less than normal air pressure

normal air pressure is midway between compressions and rarefactions

1998 Examples
A student produces two different types of travelling waves by shaking the end of a long
slinky spring. The figure below shows snapshots of parts of the waves at an instant of time.

Question 1
Estimate the wavelength of travelling wave A.

Question 2
Which of the waves (A or B) is a better model for the propagation of sound waves? Your
answer must mention the similarities between the model you choose and sound waves.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

1999 Examples
A teacher uses longitudinal waves on a very long spring to demonstrate travelling sound
waves. The first part of the spring is shown below

Question 1
Estimate the wavelength of this wave.

Question 2
What is the speed of this wave if its frequency is 4.0 Hz? Give your answer in cm s-1.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 48

2001 Examples
The cone of a loudspeaker is turned on at time t = 0 s, and is driven back and forth such that
its position as a function of time is as shown below.

Question 1
What is the frequency of oscillation of the speaker cone?

Question 2
What is the wavelength of the sound transmitted through the air by the loudspeaker?
(speed of sound in air = 340 m s-1)

Question 3
Which one of the diagrams shown below (A – D) best represents the pressure variation at the
microphone, as a function of time? The time scale for each starts at t = 0 (when the speaker
commenced to oscillate). You must justify your answer.

(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 49


 analyse sound using wavelength, frequency and speed of propagation of sound waves,
v = fλ;

Wave equation

v= f 

Frequency and period

1
f= .
T

Loudness
The louder a sound is the greater the magnitude of
the variation from normal pressure.

Reflection, absorption and transmission of sound
When sound strikes a surface, some is reflected, some is transmitted and some is absorbed. The law of
reflection states that the angle of incidence = angle of reflection. When sound is reflected there is no
change in speed, wavelength or frequency. Reflected sound is often referred to as an echo; multiple
reflections contribute to what is known as reverberation. When sound is absorbed, sound wave
energy is usually transformed into heat energy within the absorbing material.

 analyse the differences between sound intensity (W m-2) and sound intensity level (dB);

Intensity of a sound wave
The loudness of a sound depends on how much energy is carried by the wave. The intensity of a
sound wave is used to give a measurement of the rate at which the sound wave is carrying energy.

Sound-intensity Levels
The ear can detect intensities from 1 to 10-12 Wm-2. The doubling of the loudness of a quiet whisper
would be very noticeable, but adding the same sound to a louder source would not be noticed at all.
The sound loudness scale measures the ratio of the intensity of a sound wave against a standard
intensity, called the “threshold of hearing” and is 10-12 Wm-2, I0.
I
The sound intensity level,  (in dB) = 10 log
I0
A tenfold increase in intensity will correspond to a level
increase of 10dB, - heard as a doubling of the loudness. A one
hundred fold increase is a level increase of 20dB, - heard as an
increase of ×4 in the loudness.

The human ear can only detect changes as small as 3dB, this
is equal to the doubling or halving of the sound’s intensity.
This means that to make the least noticeable increase in the
volume of a stereo, we need to double the power
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 50

Sound intensity and level.
Humans respond best to frequencies around 3000 Hz. Higher or lower frequency sounds need to have
greater sound intensity levels to be perceived as being as loud. The graph shows the sound intensity
that is required for a person with normal hearing to just hear each frequency. The low point on the
graph corresponds to the frequency that the ear is most sensitive to, so this frequency, it will appear
louder. Loudness and intensity are not the same, though they are related. Intensity is measurable, but
loudness is subjective.

1997 Example
The figure below represents the sensitivity curve for hearing for an average person; that is,
the lowest intensity sound that can be heard at a given frequency.

To test the sensitivity of hearing at different frequencies, a student with normal hearing sits
near a small speaker which is connected to a frequency generator. The frequency generator is
tuned to 200 Hz. The intensity is increased slowly from zero until the student can just hear
the sound.
The generator is now tuned to a frequency of 600 Hz without changing its output level.

Question 8
By how many dB must the sound intensity level be decreased so that the student can just
hear the sound?
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

2000 Examples
Kylie is rehearsing for a concert. She sings a pure note of a single frequency. The time
between successive compressions of the sound wave arriving at her ears is 5 ms.

Question 4
What is the frequency of the sound?

The intensity at a microphone situated 0.10 m from her mouth is 1.0 x 10-6 Wm-2.

Her manager is sitting 2.0 m away from the mouth of the singer, in the first row of the
theatre. At this time the sound system is not turned on.

Question 5
What is the sound intensity at the manager's location? Show your working.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 51

The sound system is now turned on, and the sound intensity at the position of the manager
increases by a factor of 1000.

Question 6
By how many decibels does the sound level increase for the manager, when the sound system
is turned on?

In an experiment to determine hearing sensitivity, a student uses a set of earphones and a
signal generator. For a range of frequencies, he determines the sound intensity (measured at
the ear) at which the sound just becomes inaudible. The figure below shows the sensitivity of
hearing of the student.

Question 7
At what frequency is the student's hearing most sensitive?

Question 8
What range of frequencies can the student hear if the sound intensity at the student's ear is
1.0 x 10-11 Wm-2 ?
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

2001 Examples
Mel and Jill attended an orchestral concert in a large cathedral with stone walls and ceiling.
They noticed that after the orchestra stopped playing they could still hear sound for some
time. They also observed that the low-frequency and high-frequency sound persisted for a
shorter time than sound in the middle-frequency range.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 52

Question 4
Explain
• why the sound persisted after the orchestra stopped playing
• why the high-frequency and low-frequency sounds were heard for a shorter time.
You may need to refer to the figure above which shows the minimum detectable sound as a
function of frequency of an average human ear.

Question 5
What is the frequency of the sound that they could hear longest after the orchestra stopped?
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

• calculate sound intensity at different distances from a source using an inverse square law
(knowledge of acoustic power is not required);

1
The intensity varies I . This means that if you double the distance the intensity will decrease by
r2
a factor of 4.

Sound intensities and sound intensity levels of common sounds (not on the course, but useful
background)
Sound Intensity (W m-2) Intensity level (dB)
Threshold of hearing 10-12 0
Rustle of leaves 10 -11 10
Whisper 10-10 20
Normal conversation 10 -6 50 - 65
Loud radio 10-4 80
Jackhammer 10 -2 100
Rock concert 1 120
Threshold of pain 1 120
Jet aircraft 100 140
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 53

1998 Examples
A loudspeaker is emitting sound of a fixed intensity which travels equally in all directions.
The figure below shows the pressure variation plotted against distance from the speaker, at a
particular instant of time.

Question 3
Explain why the amplitude of the pressure variation decreases as the distance from the
speaker increases.

Question 4
Circle the letter (A–D) of the graph below which shows the pressure variation as a function
of distance from the speaker, at a time that is ¼ cycle later than shown in Figure 2.

A B

C D

(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

1999 Examples
The siren at the headquarters of the Country Fire Authority is mounted on a high stand. It
emits sound with a frequency of 4000 Hz. The figure below shows the siren and two
volunteers.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 54

On a day when there is no wind, the intensity of the sound heard by the volunteer at point P,
a distance of 20 m from the siren, is 0.0040 W m-2.

Question 8
What is the sound intensity level (dB) at this point P? Reflections at the ground may be
ignored.

Question 9
What is the intensity (W m-2) of the sound heard by the volunteer at point Q, which is 80 m
from the siren? The figure below shows a graph of the sensitivity of the ear of an average
person as a function of frequency.

Sound intensity levels below the threshold of hearing cannot be heard, while those above can
be heard.

Question 11
Explain, making use of the graph, why the frequency of the siren was chosen to be 4000 Hz.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

2001 Examples

In a demonstration of the perception of loudness, a teacher sets up a loudspeaker (P) on a
stand at the centre of the school oval as shown above. The loudspeaker emits sound equally
in all directions with a wavelength of 1.0 m. Ignore reflections from the ground.

A student, Xena, stands at the point X, 10 m from the loudspeaker, and measures the
intensity of the sound to be 9.0 × 10-8 W m-2. Xena now moves to a place further away from
the loudspeaker, and measures the intensity of the sound to be 2.25 × 10-8 W m-2.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 55

Question 8
How far is Xena from the loudspeaker at this new position?
Question 9
By how many decibels has the intensity level of the sound changed between the two
readings?
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

2004 Pilot Examples
Peter is recording a band rehearsal in the school hall. During one of the loud songs played by
the band Peter goes to the centre of the hall to measure the sound intensity. He measures a
sound intensity level of 120 dB at this location.
Question 2
Which one of the following best describes the maximum sound intensity at this location?
A. 1 × 10-12 W m-2 B. 1 W m-2 C. 10 W m-2 D. 120 W m-2

Peter is concerned that 120 dB is too loud for the parents. The band agrees to try and play
with half the sound intensity.
Question 3
Which one of the following best describes the anticipated maximum sound intensity level
under these conditions?
A. 119.5 dB B. 117 dB C. 114 dB D. 60 dB
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

2004 Sample Examples
Jenny is conducting some experiments to investigate the loudness of a fire alarm. Her tests
are aimed at verifying that the alarm will be heard above background noise and that it will
not endanger anyone in the classroom, including the teacher. The alarm produces a high-
pitched sound which the manufacturers claim is at least 120 dB.

Here is a copy of the relevant page from Jenny’s logbook. Notice that two entries are missing.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 56

Question 9
Calculate the missing entry in the logbook for the 100 cm measurement.
Jenny has plotted the data using a spreadsheet program.

Question 10
Draw a ‘best fit’ to the data below and use the fit to estimate the distance at which the
87.5 dB measurement was made.

Question 11
Use the trend in Jenny’s logbook or the graph to estimate the distance at which the sound
intensity level is 120 dB.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 57

 Explain resonance in terms of the superposition of a travelling wave and its reflection;

Superposition
The displacement of two waves combining with each other is calculated by the vector addition of the
two components. The two pulses pass through each other without being altered.

Constructive interference is when the two pulses pass through each other and superimpose and
reinforce each other to give a maximum disturbance of the medium; this results in a louder sound.
Destructive interference is when the two pulses pass through each other and superimpose and cancel
each other out to give minimum or zero disturbance of the medium; this results in a quieter sound.

Resonance and standing waves
Resonance is the condition when a natural vibrating system responds to an external driving
frequency, it occurs when a forcing frequency, the same as the natural frequency, is applied. Each
object has its own natural or resonant frequency.

Standing waves
resultant
direction of wave

direction of wave

resultant

resultant
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 58

Envelope of vibration

N N N N N

A A A A

If we have two identical waves travelling in the opposite directions in one medium we get a standing
or stationary wave. The superposition principle is used to obtain the waveform.
 Certain points marked N = node = point of zero displacement, a node or nodal point.
 Loops or antinodes, marked A, are points of maximum displacement, midway between the nodes.
 The wave does not progress through the medium.
 Wavelength is the same as that of the components.
 Maximum amplitude of the resultant wave is twice that of the components.
λ
 The distance between adjacent nodes or antinodes is .
2
 Any particles between any two successive nodes are in phase. Their motions correspond at any
instant. They have zero displacement and maximum displacement at the same instant.
 They can only be produced by the superposition of two waves of equal amplitude and frequency
travelling in the opposite direction
 They are the result of resonance and occur only at the natural frequencies of the vibration.

Nodes are a result of destructive interference, there is little or no variation in air pressure, so it is
perceived by the listener as a region of soft sound. Antinodes are a result of constructive interference,
so there will be maximum variation of air pressure, which will be perceived as a region of loud sound.

Harmonics
The standing wave frequencies are referred to as harmonics. The simplest mode of vibration, which
has only one antinode, is called the fundamental. Higher level harmonics are referred to as overtones.
The first overtone is the second harmonic.

1998 Examples
Students use a narrow tube of length 0.432 m open at both ends to model a flute. By varying
the frequency of sound emitted from a small loudspeaker placed near one end, as shown
below they observe resonances at several frequencies.

The wavelengths of the sound at which the resonances with the three lowest frequencies
occur are 0.864 m, 0.432 m and 0.288 m. The speed of sound in air is 340 m s-1.

Question 10
What is the lowest frequency at which resonance is observed by the students? Show your
working.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 59

The students then fill the 0.432 m tube with helium. The speed of sound in helium is 1000
ms–1, compared to 340 ms–1 for air.

Question 11
The students find the longest wavelength of sound which produces resonance in the tube
filled with helium. Is this wavelength the same, longer or shorter than the longest
wavelength when the tube was filled with air?
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

 Analyse, for strings and open and closed resonant tubes, the fundamental as the first
harmonic, and subsequent harmonics;

Reflection in strings
When a wave reaches a free end, or yielding boundary, it will reflect with crests as crests and troughs
as troughs. Strings in musical instruments are always fixed at both ends.

2L nv
The wavelength of the standing waves corresponding to the natural harmonics is n = or f = .
n 2L
All harmonics (n = 1,2,3, …) may be present, the ratio of frequencies f1 : f2 : f3 = 1 : 2 : 3.
The fundamental frequency of a stretched wire depends on: length, tension and mass per unit length.
1 T
This gives f = where: f = fundamental frequency in Hertz (Hz), L = length in metre (m),
2L m
T = tension in Newton (N), and m = mass per unit length in kg/m.

Reflections in tubes
Wind instruments rely on reflection of sound waves. When a compression arrives at the end of the
tube this high pressure this will reflect back down the tube resulting in a compression being reflected
as a rarefaction. A rarefaction arriving at an open end creates an area of low pressure. This will draw
free air from beyond the end into the tube creating an area of higher pressure, ie a compression

In a pipe organ or a flute the wave travels down the pipe as a series of rarefactions and compressions
Pressure variation (Open at both ends)
First harmonic (fundamental frequency)
L
λ 1 = 2L
v
f1 = N A N
2L

Second harmonic (first overtone)
2L
2 = =L
2
2v
f1 = = 2f1 N A N A N
2L
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 60

Pressure variation (Closed at one end)
First harmonic (fundamental frequency)
L
 1 = 4L
v
f1 = N A
4L

third harmonic (first overtone, second resonant frequency)
4L
2 =
3
3v
f3 = = 3f1 N A N A
4L

Overtones
The term overtone is applied to harmonics other than the fundamental frequency.
The main features of standing waves in air columns can be summarised
 at open ends of the tube there is always an antinode
 at closed ends there is always a node
 the wavelength of the sound must fit the length of the pipe. The length of the air column that
vibrates is slightly longer than the length of the pipe, since antinodes form just outside the end of
the pipe
 the fundamental vibration in a closed pipe has a wavelength twice as long as the fundamental in
an open pipe of the same length. This makes the frequency of the sound produced by a closed pipe
half that of the same length of an open pipe.

1999 Examples
A parasaurolophus dinosaur made sounds from a hollow horn on its head.

This horn can be modelled as a 2.80 m pipe that is open at both ends.

The figure above shows a graph of pressure variation of the air inside this pipe when a
sound of fundamental frequency f0 is being emitted. The two lines represent the maximum
and minimum values of the pressure variation.

Question 4
What is the wavelength of the sound produced under these conditions?
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 61

Question 5
What is the frequency f0 of this sound? Take the speed of sound in air to be 340 m s-1.

The figure below shows the pressure variation as a function of time at the middle of the pipe
(point Q in the figure above).

Question 6
Circle the letter (A–D) of the graph that best shows the pressure variation at point P as a
function of time.
A B

C D

Sound of only a few frequencies could be generated by this dinosaur. The pressure variation
for the lowest possible frequency f0 is shown above

Question 7
Circle the letter (A–D) of the graph that shows the maximum and minimum values of the
pressure variation along the tube for the next higher frequency f1 above f0.

A B

C D

(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 62

2000 Examples
The vocal tract can be modelled by a tube of length L that is open at one end and closed at
the other, as shown below. The fundamental frequency of this tube is 500 Hz.

Vocal cords (V)

Question 1
Calculate the length L of the tube, given that the speed of sound in air is 340 m s-1.

Question 2
With which one or more of the following frequencies (A-E) could the tube also resonate?
Write your response in the box provided.
A. 250 Hz B. 1500Hz C. 1000 Hz
D. 2000 Hz E. 2500 Hz

Question 3
A singer emits a pure sound of frequency 500 Hz. Which one of the diagrams (A-E) best
shows the maximum pressure variation above and below normal atmospheric pressure (P0),
along the tube which models the vocal tract? The letters M and V indicate the location of the
mouth and vocal cords as shown above.

(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 63

2001 Examples
Nigel is studying standing waves that are set up in a narrow glass tube. He has an audio
signal generator and a small speaker that is near one end of the tube, and adjusts the
frequency to set up the resonances. The tube is filled with fine dust so that when a resonance
is formed the dust indicates the positions of the pressure nodes and antinodes. Although he
can see the entire tube, shields prevent him from seeing whether an end is open or closed.

At a particular frequency of 680 Hz, he observes that there are 5 nodes and 5 antinodes.

Question 6
How many open ends does the tube have? Include a diagram to justify your answer.

Question 7
What is the length of the tube? Take the speed of sound in air to be 340 m s-1.

 describe in terms of electrical and electromagnetic effects, the operation of
- microphones, including electret-condenser, crystal, dynamic and velocity microphones
- dynamic loudspeakers;

Microphones
Microphones are transducers that transform sound energy into electrical energy that can be magnified
and then changed back to sound. The electrical signal has the same frequency as the sound. The
amplitude of the electrical signal should ideally be proportional to the amplitude of the variations in
air pressure of the sound

Electret condenser microphone
This type of microphone is made with the diaphragm as one plate of a parallel plate capacitor and can
also be referred to as an electrostatic microphone.

An electret is a piece of dielectric material that is permanently
polarised. One side of an electret is permanently positive, the
other side is permanently negative. The front plate is very thin
and usually covered in a very fine layer of gold.
The sound pressure causes the front plate to vibrate, this changes
the spacing between the diaphragm and the stationary back plate.
This causes a change in capacitance. A voltage is supplied to the
plates and thus the amount of charge on the plates varies,
producing a current.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 64

The diaphragm of a condenser microphone can be very light compared with the dynamic microphone
and, consequently, can respond
more quickly and at higher
frequencies. Condenser
microphones usually have a greater
frequency range and a better
linearity than the dynamic
microphones.

Crystal microphone
The diaphragm is attached to a thin piece of piezoelectric crystal such
as quartz. Piezoelectric crystals produce a current when subjected to
changes in pressure. Sound causes the diaphragm to vibrate. This
produces a changing pressure on the crystal. This produces an electric
signal current whose size is proportional to the air pressure variations
of the sound.

Dynamic microphone
Dynamic microphones have a
coil attached to a diaphragm
suspended in the magnetic field of a magnet. Pressure changes
(sound) causes the diaphragm to vibrate and the coil to move
backwards and forwards over the pole of the magnet. This, in
turn, causes the magnetic flux in the coil to change, inducing a
current in the
coil. The size,
frequency and
waveform of
the induced
current are proportional to the pressure variations of
the sound.

The dynamic microphone has good frequency
linearity and is relatively strong. Good quality
dynamic microphones are used for recording
purposes.

Dynamic microphones have peaks designed to gain clarity with stage vocals

Velocity microphone/Ribbon microphone
The ribbon microphone does not have a diaphragm, it has a corrugated
ribbon of aluminium alloy suspended in a strong magnetic field. Sound
makes the aluminium ribbon vibrate. The ribbon is corrugated to make it less
stiff (so that it responds better to low frequencies), and it is thin so that it is
easily moved by air particles.

The aluminium ribbon responds to the air velocity of the sound waves and
not to any pressure variations. An EMF is induced across the ends of the ribbon and this creates an
electric current signal that is proportional to variations in air pressure caused by the sound.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 65

This is a very sensitive microphone, but it cannot be used in places where it would be subjected to
mechanical shocks. These microphones will pick up sounds coming from both sides of the
microphone.

2004 Sample Examples

Choose one of the microphones labelled M1 or M2

Using the microphone of your choice, answer Questions 1 and 2.
Question 1
Identify the type of microphone which you have chosen.
A. electret condensor B. crystal C. velocity D. dynamic

Question 2
A sound pressure wave is incident on the microphone. Describe how the microphone of your
choice detects the wave and produces the signal output.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

Dynamic loudspeakers
The alternating current makes the coil around the fixed magnet move backwards and forwards,
creating compressions and rarefactions in the air in front of the cone. The larger the current that’s
moving through the wire, the larger the induced magnetic field will be, and hence a greater force of
magnetic attraction or repulsion will be exerted by the permanent magnet. The speaker cone will
move in and out through a greater amplitude, transferring more energy to the surrounding air
molecules and thus creating a louder sound.

Loudspeakers come in a wide variety of sizes, roughly matching the range of frequencies they have
been designed to best produce: ‘woofers’ for frequencies from 30–500 Hz, mid-range loudspeakers for
500–4000 Hz, and the aptly named small ‘tweeters’ for the high frequencies from 4 kHz – 20 kHz.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 66

1997 Examples
A loudspeaker consists of a cardboard cone that acts as a diaphragm which is moved
backwards and forwards to produce a travelling sound wave in air. Such a speaker is
mounted on a wall as shown below. The figure also shows the horizontal displacement of the
speaker cone as a function of time.

Question 1
What is the frequency of the sound?

In the air are fine dust particles which are floating at rest. After the loudspeaker is turned on,
the particles will be forced to move by the pressure variations associated with the sound
wave.

Question 2
For the dust particle at point P, directly in front of the loudspeaker, which of the statements
(A–D) below best describes its motion?
A. It vibrates vertically up and down at the frequency of the sound wave.
B. It vibrates horizontally backwards and forwards at the frequency of the sound wave.
C. It moves horizontally forwards, travelling with the sound wave.
D. It remains at rest.

Question 3
Explain the reason for your choice in Question 2.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

2004 Sample Examples
In the air are fine dust particles, which are floating at rest. After
the loudspeaker is turned on, the particles will be forced to
move by the pressure variations associated with the sound
wave.

Question 5
For the dust particle at point P, directly in front of the loudspeaker, which of the following
statements (A–D) best describes its motion?
A. lt vibrates vertically up and down at the frequency of the sound wave.
B. It vibrates horizontally back and forwards at the frequency of the sound wave.
C. It moves horizontally forwards, travelling with the sound wave.
D. It remains at rest.

Question 6
Explain the reason for your answer to Question 5.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 67

 describe the effects of baffles and enclosures for loudspeakers in terms of the interference
of sound waves due to phase difference,

Enclosures
When a speaker cone moves forward, the front surface sends out a compression wave. But at the same
time the rear of the cone is creating a rarefaction. At low frequencies (less than 200 Hz), diffraction
effects cause the sound waves from the back of the loudspeaker to bend around the outer rim of the
speaker and cancel out the sound waves from the front surface.

Usually the various speakers making up the left channel of a stereo output are mounted together in
one enclosure, and those making up the right in another, although the higher frequencies don’t suffer
as much from diffraction effects. Some modern systems are now keeping speakers separate to allow
more ‘tuning’ of the listening environment. Either way, the supplied frequencies are filtered and each
range is directed to the appropriate speaker: the high-frequency signals are sent to the tweeters and
the low frequencies to the woofer or sub-woofer.

Baffles and ports
Designers aim to stop the unwanted sound from the back of the speaker superimposing with that
from the front. Placing the loudspeaker in a large baffle will always improve the production of low
frequencies, because of the increased distance from the back to the front of the speaker. A large, or
even infinite, distance—termed an infinite baffle—is desirable but hardly practical.

In a ported enclosure, the closed box is modified by the inclusion of a carefully designed opening in
the front. Through the ‘vented’ or ‘ported’ enclosure (also referred to as a bass-reflex monitor), sound
from the back of the speaker can be added to that from the front without cancelling it. If the port is
carefully designed, it acts like a second diaphragm driven by the backside of the speaker. It can add
an octave or more to the system’s low end frequency response.

One key to the successful design of a port is to make sure the enclosure resonance matches that of the
speaker itself. The process reverses the phase of the backwave, resulting in radiated sound that is in
phase with the sound from the front of the speaker.

 Interpret frequency response curves of microphones, speakers, simple sound systems and
hearing, including loudness (phon);

Sound Systems
Human ears do not respond equally to all frequencies. Generally they respond best to frequencies
around 3000 Hz. The graph shows the sound intensity that is required for a person with normal
hearing to just hear each frequency. This means that the low point on the graph corresponds to the
frequency that the ear is most sensitive to. Because the ear is sensitive to this frequency, it will appear
louder.
The curved line is a line representing all the
frequencies that sound as loud as each other. So on this
graph, a 100 Hz sound will seem equally as loud as a
1000 Hz sound. This indicates that our ears are less
sensitive to 100 Hz sounds, because we need more
energy/m2 to hear the 100 Hz sound.

You need to have a very good understanding of this
graph
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 68

Phon
The phon is the unit of equivalent loudness. It
measures how loud a sound is perceived to be
compared to a reference sound – normally at
1kHz (1000Hz). Graphs can be drawn for an
average human ear/brain combination. The
phon level is read from the graph below. Hence
60 phons means “as loud as a 60dB, 1000Hz
tone”.

To find what intensity level is required for a
sound of frequency 5kHz to have a loudness of
80phons, read along the 80-phon curve until it
intersects with 5kHz. 80dB

Audibility range for the human ear

Frequency response curves

Microphone frequency response curves (or characteristics) show how well microphones respond to
sounds of the same sound intensity level at different frequencies. They usually have a vertical axis in
decibels (dB) and a horizontal axis in hertz (Hz). The graph enables you to read the variations in the
power gain or loss of the microphone with the frequency of the sound.

Frequency response curve for (a) a woofer and (b) a tweeter
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 69

 Evaluate the fidelity of microphones and loudspeakers in terms of
- the intended purpose of the device
- the frequency response of the system
- physical construction (qualitative)

Fidelity of microphones and loudspeakers
Fidelity is the degree to which a sound reproduction system accurately reproduces the original
recorded sound. A microphone is high fidelity if it responds equally well to most frequencies in the
human range (approximately 20 Hz to 20 kHz). The electrical signal produced by the microphone
should be accurately proportional to the original sound.

High fidelity Low fidelity

2004 Sample Examples
One of the world’s most popular microphones is the Shure SM58. The microphone is shown
together with a graph detailing its frequency response.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 70
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 71

Question 3
Which one of the following (A–D) best describes the range of frequencies that can be reliably
detected by this microphone?
A. frequencies below 200 Hz
B. frequencies above 10 000 Hz
C. frequencies below 200 Hz and above 10 000 Hz
D. frequencies between 200 Hz and 10 000 Hz

A synthesiser produces two pure sine tones, one at a frequency of about 1000 Hz, and
another at a higher frequency of 4000 Hz. The SM58 microphone is used to capture these
sounds. The synthesiser has been adjusted to ensure both tones have the same amplitude at
the microphone. An oscilloscope is used to observe the signal output of the microphone.

Question 4
Which one of the following diagrams (A–D) best represents the waveform observed?

(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

 describe diffraction as the directional spread of various frequencies in terms of different
gap width or obstacle size, including the significance of the magnitude of the /w ratio
 identify and apply safe and responsible practices when working with sound sources and
sound equipment

Diffraction
Sound waves can travel around corners, they spread out when they come through an open door.
Diffraction is the bending (or spreading) of waves around obstacles in the path of the waves, or as
waves pass through narrow openings. When sound is diffracted there is no change in wave speed,
wavelength or frequency.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 72

Diffraction around obstacles.

long wavelengths, low frequencies short wavelengths, high frequencies

When the obstacle is small compared with the wavelength of the sound, there is very little
disturbance. Larger sound 'shadows' occur when the obstacle is much larger than the wavelength of
the incident wave.

Diffraction through gaps.
When sound travels through a narrow opening, such as a door, the waves bend around both sides of
the opening and are diffracted into the region beyond the barriers
on both sides of the doorway. So a narrow gap acts just like to
obstacles. The amount of diffraction is given by the value of the

ratio λ where w is either the width of the object or the opening.
w

As the value of λ increases, so does the amount of diffraction
w

(bending). If λ << 1, very little diffraction occurs, if the ratio λ ≥ 1, then it is complete diffraction,
w w
i.e. bending through 1800.

Locating sound sources
We use this concept of diffraction to locate sound sources. For high frequencies (short wavelengths)
the head acts as an obstacle. One ear will hear a greater intensity than the other, from this we are able
to locate the direction of the sound. Low frequencies are much more difficult to locate, because the
long wavelengths will diffract around the head and both ears will hear approximately the same sound
intensity. To locate low frequency sounds the brain also needs to pick the time delay between the
signals to identify the direction.

1997 Examples
Mary and John attend an orchestral concert. Unfortunately they have poor seats, and are
sitting in the balcony behind a column, such that John cannot see the double bass section of
the orchestra and Mary cannot see the flutes. The situation is illustrated below.

At the end of the concert Mary comments that she could not hear the flutes well. John,
however, says that to him the orchestra sounded balanced, and he could hear every
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 73

instrument including the double basses. Assume that the frequency played by the flutes was
about 2000 Hz, and that of the double basses was about 200 Hz.

Question 11
If the speed of sound in air is 340 m s-1, what are the wavelengths of the waves at the
frequencies of the flutes and the double basses?

Question 12
Explain why John could hear the double basses even though he could not see them, and why
Mary could not clearly hear the flutes which she could not see.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

1999 Examples
Alexandra and Gary arrived at a pop concert that had already started. While waiting at point
X in a queue, Alexandra commented that the sound quality was poor. Although they could
hear sound of low frequency, high frequency sound was relatively much weaker. They were
pleased to find that the sound quality improved when they reached point Y in front of the
hall entrance.

Question 13
Circle the letter (A–D) of the wave phenomenon that best explains their observations.

A. reflection B. refraction C. resonance D. diffraction

Question 14
Explain why the sound quality was poorer at X, but improved at Y.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 74

2000 Examples
Max and Michelle are buying loudspeakers for their hi-fi. They are choosing between two
models, a 35-cm diameter speaker (P), and a 5-cm one (Q). Both speakers operate equally well
over the complete audible frequency range. Max is standing in front of the speakers, and
Michelle is the same distance away, but to one side, as shown below.

As a test of the speakers they play sounds of 10 000 Hz and 200 Hz, and compare the
intensity that they each hear.

Question 11
The wavelength of sound of frequency 200 Hz is 1.65 m. What is the wavelength of sound of
frequency 10 000 Hz?
Max comments that the intensity of the sound of both frequencies seems the same from either
speaker. However Michelle says that for the larger speaker, the 10 000-Hz sound is
significantly softer than the low frequency sound.

Question 12
Explain these observations. Include relevant calculations and/or diagrams to support your
reasoning.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

2001 Examples
John and Maria are discussing how we determine the direction from which a sound comes.
They are considering the figure below.

John says that we can tell the direction of the sound from the speaker because the brain can
detect the time difference between the arrival time of the peak of a wave at each ear.
Colin Hopkins: colinhop@bigpond.net.au Physics Unit 4 2009 75

Maria says this cannot be so: we tell the direction of sound because diffraction causes the
intensity at the more distant ear to be lower, and the brain detects this.

Question 11
Discuss to what extent, and under what conditions, their explanations are correct. You may
use diagrams to explain your answer.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)

2004 Sample Examples
Louise and Leo like to listen to music while working. They have a new enclosure for their
loudspeaker. Louise has chosen to orient the enclosure as shown below, pointing mainly
towards her desk as she is further away from the speakers than Leo.

Louise comments that the music sounds great through the new speaker, but Leo complains
that the higher frequency sounds seem very soft.

Question 7
Which of the wave phenomena (A–D) best explains their differing observations?
A. reflection B. refraction C. resonance D. diffraction

Question 8
Explain why Louise and Leo hear the sounds differently.
(Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.)