CHAPTER 23 How do instruments make music?
Contents
Introduction
Longitudinal waves
Wave speed
The speed of sound
FS
The wave equation
Producing sound
O
Observing sound
PR
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The three-dimensional nature of sound
The power of sound
Intensity
Intensity and the inverse square law
G
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Sound intensity levels
Sound intensities and intensity levels
PA
The human ear’s response to sound
Loudness level curves
Reflection of waves
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The phon
Reflection of transverse waves in strings
Standing waves
EC
Reflection of sound waves
Transverse standing waves in strings or springs
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Sound and standing waves
R
Making music
O
Resonance
Stringed instruments
C
Wind and brass instruments
N
Timbre of instruments
U
The human voice — speaking and singing
Diffraction
Sound detection
Subjective sound
The structure of the human ear
Chapter review
Summary
Questions
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
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Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
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CHAPTER 23 How do instruments make music?
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This sound barrier on Melbourne’s CityLink reflects traffic noise away from nearby high-rise apartments. Sound is
reflected off the solid left wall of the barrier and travels up and to the right, away from residential areas.
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REMEMBER
■
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Before beginning this chapter, you should be able to:
describe waves as a method of the transfer of energy from one place to another without any net transfer of
matter
■
describe periodic waves in terms of their wavelength (λ), amplitude (A), period (T) and frequency (f)
■
distinguish between transverse waves and longitudinal waves
■
understand how waves can interfere constructively and destructively.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
KEY IDEAS
After completing this chapter, you should be able to:
describe sound as the transmission of energy via longitudinal pressure waves
■
analyse sound using wavelength, frequency and speed of propagation of sound waves: v = fλ
■
distinguish between sound intensity (W m–2) and sound intensity level (dB)
■
calculate sound intensity at different distances from a source using an inverse square law
■
analyse a standing wave as the superposition of a travelling wave and its reflection
■
explain resonance and identify it as related to the natural frequency of an object
■
■
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analyse, for strings and open and closed resonant tubes, the fundamental and subsequent harmonics, and
apply this analysis to selected musical instruments
analyse the unique sound of an instrument as a consequence of multiple resonances created by the
instrument and described as timbre
investigate how the amount of diffraction around an obstacle varies with the size of the obstacle and the
wavelength of the sound
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investigate and explain a variety of musical instruments with reference to the similarities and differences of
sound production between instrument types (brass, string, woodwind and percussion) and how they compare
to the human voice
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■
investigate and explain the human voice box as a resonance chamber with vibration provided by vocal cords
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■
investigate factors that influence natural frequency, including shape and material, and explain how these
factors relate to instruments
EC
■
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■
describe the structure of the human ear with reference to the transfer and amplification of vibrations
■
interpret the frequency response curve of the human ear
■
differentiate between pitch, timbre and loudness
■
identify the representation of timbre as a combination of specific frequencies
■
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describe how particular musical intervals can be represented as ratios of their frequencies and how
consonant frequencies tend to have simple ratios
■
investigate the phenomenon of beats
■
investigate an aspect of contemporary research in psychoacoustics.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Introduction
Sound is a form of energy that is caused by vibrating objects. The vibrating objects cause waves in the
surrounding medium. The medium you are most familiar with is air, but sound can travel through other gases,
liquids and solids.
Sound waves transfer energy from the vibrating object.
Sound waves travels at different speeds through different media. It travels faster through liquids than gases. It
travels faster through solids than liquids.
FS
Longitudinal waves
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In a longitudinal wave, as opposed to a transverse wave, the oscillations are parallel to the direction the wave
is moving. Longitudinal waves can be set up in a slinky, as shown in part (a) below. Sound waves in air are also
longitudinal waves, as shown in part (b) below. They are produced as a vibrating object, such as the arm of a
tuning fork, first squashes the air, then pulls back creating a partial vacuum into which the air spreads.
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Longitudinal waves in (a) a slinky (b) air
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Longitudinal waves cause the medium to bunch up in places and to spread out in others. Compressions are
regions in the medium where the particles are closer together. Referring to sound waves in air, compressions are
regions where the air has a slightly increased pressure, as a result of the particles being closer together.
Rarefactions are regions in the medium where the particles are spread out. This results in a slight decrease in air
pressure in the case of sound waves.
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The wavelength (λ) for longitudinal waves is the distance between the centres of adjacent compressions (or
rarefactions). The amplitude of a sound wave in air is the maximum variation of air pressure from normal air
pressure.
Wave speed
At the beach the approaching wave is described as the crest of the wave that is about to crash. In the case of
sound a compression is approaching the ear drum. These approaching crests and compressions are called
wavefronts. The direction that a wave is travelling (and hence the direction of its speed) is always at right angles
to the wavefront. The speed of a wave is the rate at which a wavefront moves through the medium, not the speed
of individual particles of the medium.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
The speed of sound
The speed of sound in various media is shown in the table below.
Table 23.1 The speed of sound in various media
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Speed of sound (m s–1)
331
343
260
1005
1440
1560
≈4500
≈5100
Material
Air (0 °C)
Air (20 °C)
Carbon dioxide
Helium
Water
Sea water
Glass
Iron and steel
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The speed of sound in air depends on the temperature. The higher the temperature, the greater the speed of
sound.
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AS A MATTER OF FACT
The speed with which a wave moves through a medium depends on:
•
for a gas, on the pressure as the molecules are pushed together, and, for a solid, on the elastic forces
between atoms
the mass of the particles or how easy it is to move them or, more correctly, the density of the medium.
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•
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The greater the interaction between the particles, the faster waves will travel through it. The less dense the
medium, the greater the speed of the wave.
EC
The speed of a wave depends only on the medium through which it is travelling. It does not depend on how much
energy is being transferred by the wave nor on the frequency of the wave.
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The wave equation
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The speed, frequency and wavelength of both transverse and longitudinal waves are related by the following
formula:
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v  f .
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This is known as the universal wave formula and is derived from the fact that a wave will travel a distance of one
wavelength (λ) in the time it takes to produce one complete wave — that is, the period (T).
Applying the formula:
speed 
distance travelled
d
, or v 
time taken
t
where
d =λ
t=T
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
we get the useful formula:
v

T
.
By substituting f 
1
into this formula, the universal wave formula is obtained.
T
SAMPLE PROBLEM 23.1
FS
What is the speed of a sound wave if it has a period of 2.0 ms and a wavelength of 68 cm?
Solution:
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Step 1: Note down the known variables in their appropriate units. Time must be expressed in seconds and length
in metres.
T  2.0 ms
 2.0  10 3 s
  68 cm
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 0.68 m
v

T
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Step 2: choose the appropriate formula.
.
Step 4: substitute values and solve.
0.68 m
2.0  10 3 s
 340 ms 1
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v 
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Step 3: Transpose the formula. (Not necessary in this case.)
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SAMPLE PROBLEM 23.2
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What is the wavelength of a sound of frequency 550 Hz if the speed of sound in air is 335 m s–1?
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Solution:
f  550H z , v  335 m s 1
v f
 
v
f
335 m s 1
550 H z
 0.609 m

Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
REVISION QUESTION 23.2
A siren produces a sound wave with a frequency of 587 Hz. Calculate the speed of sound if the
wavelength of the sound is 0.571 m.
Producing sound
FS
Sound is produced by vibrating objects. The figure below shows how the vibrating membrane of a drum causes
compressions and rarefactions to travel outwards through the air.
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The frequency of the sound produced will be the same as the frequency of the vibrating object.
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Production of a sound wave
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The figure below shows how air pressure varies with distance at a large distance from the source of sound. This
type of graph shows the pressure variations in the air at a particular instant. As time goes by, the positions of the
compressions and rarefactions move away from the source at the speed of sound.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
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Air pressure variations at a distance from the source
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A pressure-versus-distance graph enables the wavelength of the sound wave to be found. It is the distance
between the centres of two adjacent compressions, or the length of one complete cycle.
SAMPLE PROBLEM 23.3
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Determine (a) the wavelength and (b) the frequency of the sound depicted in the above figure when the speed of
sound in air is 340 m s–1.
Solution:
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v

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f 
R
b. λ = 0.40 m, v = 340 m s–1
EC
a. Each division on the distance axis is 0.10 m. The length of one complete cycle is four divisions; therefore,
the wavelength is 0.40 m.
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340 m s 1

0.40 m
 850 Hz
Observing sound
A cathode ray oscilloscope (CRO) is an electronic device that visually represents how voltage varies with time at
a point in a circuit. A typical CRO is shown in figure (a) above. A CRO produces a trace by sweeping a beam of
electrons (cathode rays) across the screen. A graphical depiction of the screen is shown in figure (b) above. It is
divided into 1 cm squares with axes marked. The horizontal axis is a time axis for the trace. The vertical axis is a
voltage axis. Adjusting the timebase and voltage controls on the CRO changes the scales for the trace produced
on the screen. How a CRO operates is discussed in greater detail in chapter 9.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
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(a) A typical CRO and (b) a graphical depiction of a CRO screen
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A microphone is a device that converts sound energy into a voltage signal. The amplitude of the voltage signal
represents the pressure variation amplitude of the sound incident at the microphone. When a CRO is connected
to a microphone, the trace on the screen shows how pressure varies with time at the point in space where the
microphone is placed.
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SAMPLE PROBLEM 23.4
a. What is the period of the sound?
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The figure below right shows the trace on a CRO screen produced by a microphone detecting a sound. The time
scale is as follows: 1 cm equals 2 ms.
Sketch the trace produced by a sound with the original frequency, but with twice the pressure variation.
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c.
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b. Sketch the trace produced by a sound of twice the frequency.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Solution:
a. One complete cycle is 4 cm on the screen; multiplying this by the time scale gives a period of 8 ms.
b. Doubling the frequency halves the period, so the trace shown in the following figure (a) is obtained.
Doubling the pressure variation will double the amplitude of the trace, so the trace shown in figure (b) is
obtained.
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c.
Digital doc:
Investigation 23.1:
Observing sound
In this investigation sound is
observed and studied using a
cathode ray oscilloscope (CRO).
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Sound can be represented by a graph of how pressure varies with time at
a point near the sound source. (See, for example, the graph below.) This is
basically what is shown on the screen of a CRO. The period of the sound
is the time for one complete cycle and this can be read directly from the
graph. This enables the frequency to be calculated.
A pressure-versus-time graph
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 23.5
What is the frequency of the sound depicted in the previous figure? From the graph, T = 8.0 ms or 8.0 × 10–3 s.
Solution:
f 
1
T
1
8.0  10 3 s
 125 H z
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The three-dimensional nature of sound
FS

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So far we have considered only the motion of particles and the variations of air pressure for sound waves
travelling in one direction. In reality, sound travels in three dimensions; it spreads out in all directions from its
source. Figure (a) below shows a volume of air with no sound and figure (b) shows the three-dimensional nature
of sound waves. Note that the particles oscillate in the same direction as the wave direction at all points and that
this is at right angles to the wavefront.
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A point source of sound is an idealised way of looking at how sound is produced. It assumes that the vibrating
object is very small, spherical in shape and causes compressions and rarefactions in all directions at the same
time.
(a) Particle distribution in undisturbed air and (b) sound waves travelling through air
The power of sound
Sound waves transfer energy. Energy is produced by a sound source and transmitted by longitudinal waves
through the medium. This medium is usually air.
It is not meaningful to talk about the amount of energy produced by a sound source. This is because a small
loudspeaker or button earpiece from a cassette player may eventually produce as much total energy as a
thunderclap if it is in use for a long enough period of time. It is more meaningful to talk about the acoustical
power of the sound source, as this is the amount of sound energy produced by a source every second.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
REMEMBER THIS
Power (P) is the rate of doing work or of transferring energy. Power, measured in watts (W), equals the energy
transferred, measured in joules (J), divided by the time taken, measured in seconds (s).
Power 
energy
time
FS
AS A MATTER OF FACT
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Loudspeakers for entertainment systems are rated in watts. For example, a system might be fitted with 40 W
speakers. In this case, 40 W does not refer to the acoustical power produced by the speakers, but to the
maximum electrical power dissipated by the speakers. Under normal operating conditions, the acoustical power
produced will be much less than the stated power rating.
Intensity
I 
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As sound travels away from a source, it spreads out through the air. The acoustical power produced by the
source is spread out over a larger area the further it travels, and so the sound energy becomes less intense. The
intensity (I) of a sound at a point is a measure of the amount of power (P) passing through a unit of area (A) at
that point. The area here is taken at right angles to the direction of propagation of the sound. The relevant
equation is:
P
.
A
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Power is measured in watts (W) and area is measured in square metres (m2), so intensity is measured in watts
per square metre (W m–2).
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The area is at right angles to the direction of propagation.
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SAMPLE PROBLEM 23.6
What is the intensity of a sound if 6.0 × 10–3 W of acoustical power passes through an open window that has an
area of 0.30 m2?
Solution:
P
A
6.0  10 3 W

0.30 m 2
I 
 2.0  10 2 W m 2
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 23.7
How much energy is transferred to a human eardrum which has an area of 5.0 × 10–5 m2 by a sound of intensity 2.0
× 10–2 W m–2 in 20 s?
Solution:
First we have to find the power of the sound.
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P  IA
 2.0  10 2 W m 2  5.0  10 5 m 2
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 1.0  10 6 W
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Energy  Pt
 1.0  10 6 W  20 s
 2.0  10 5 J
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If a source produces an intensity I at some distance, then n identical sources operating simultaneously will
produce an intensity n times greater at that same distance.
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SAMPLE PROBLEM 23.8
If an alarm bell produces a sound intensity of 5.0 × 10–6 W m–2 at a distance of 150 m, what will be the sound
intensity produced at a distance of 150 m by five identical bells ringing at the same time in the same place?
EC
Solution:
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The total intensity will be five times the intensity of one bell. The final intensity will be 2.5 × 10–5 W m–2.
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REVISION QUESTION 23.8
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A window has an area of 0.50 m2. 4.5 × 10–4 J of energy passes through the window in 30 seconds.
Calculate (a) the acoustic power of the sound and (b) the sound intensity at the window.
AS A MATTER OF FACT
Loudness is a subjective quality of sound and is related to the physical quality of intensity. The greater the
intensity of the sound, the louder it will seem to be, for a constant frequency.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Intensity and the inverse square law
The inverse square law for sound applies to small (point) sources of sound that produce sound uniformly in all
directions. It assumes that the medium through which the sound travels is uniform and that the sound does not
reflect from, nor is absorbed by, any surfaces or the air.
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Under these conditions, sound will travel as a spherical wave. It will spread out evenly in all directions and its
wavefronts will form the surface of a sphere.
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A sound wave travelling away from a source has a spherical shape. Two compressions (wavefronts) are shown
here, with radii r1 and r2.
P
4 r 2
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I 
EC
The power of the source is therefore spread out over the surface of a sphere. When finding the intensity of a
sound at a distance of r m from the source, the sphere will have a surface area of 4πr2. Since the intensity is the
power per unit area, if you divide the power of the source by the area of the sphere, you will calculate the intensity
at a distance of r from the source.
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SAMPLE PROBLEM 23.9
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Karen measures the sound intensity at a distance of 5.0 m from a lawn-mower to be 3.0 × 10–2 W m–2. Assuming
that the lawn-mower acts as a point sound source and ignoring the effects of reflection and absorption, what is the
total acoustical power of the mower?
Solution:
P
 4 r 2 I
 4  (5.0 m )2  3.0 10 2 W m 2
 9.4 W
P
, it can be seen that, for a
4 r 2
particular sound source, the sound intensity it produces is inversely proportional to the square of the distance
from the source.
Referring back to the formula for the sound intensity produced by a source, I 
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
I
1
r2
This is the inverse square law which can be restated as: the intensity of sound is inversely proportional to the
square of the distance from the source.
When comparing the sound intensities at two distances r1 and r2 from a source, it should be remembered that the
power of the source is a constant. Therefore, P = 4πr12I1 = 4πr22I2. This relationship then gives the following useful
formula:
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I 2 r12

I 1 r22
PR
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SAMPLE PROBLEM 23.10
If the sound intensity 3.0 m from a sound source is 4.0 × 10–6 W m–2, what is the intensity at (a) 1.5 m and (b) 12 m
from the source?
3.0 m
4.0  10 6 W m 2
1.5 m
?
I2
I1

r12
r22
 I2

I 1r12
r22
PA




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r1
I1
r2
I2
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a.
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Solution:
4.0 10 6 W m 2  (3.0 m )2
(1.5 m )2
5
 1.6 10 W m 2
 3.0 m
 4.0 10 6 W m 2
r2
I2
 12 m
 ?
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r1
I1
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b.
R
R

I2
I1

r12
r22
 I2

I 1r12
r22

4.0  10 6 W m 2  (3.0 m )2
(12 m )2
 2.5  10 7 W m 2
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
The examples on the previous page show the following general rules of thumb: if you halve the distance, the
intensity is multiplied by 4; if you double the distance, the intensity is divided by 4.
Sound intensity levels
The human ear is a very sensitive organ. It can detect a range of sound intensities starting at less than 1.0 × 10–12
W m–2 at certain frequencies and having an upper limit of around 1.0 W m–2.
The threshold of hearing — which is defined as being a sound intensity of 1.0 × 10–12 W m–2 when the frequency
of the sound is 1000 Hz — is the minimum intensity that an average young person can detect.
FS
When the sound intensity is greater than 1.0 W m–2, it is painful. Consequently, this intensity is sometimes called
the threshold of pain.
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The sound intensity at the threshold of pain is about 1012 times greater than the sound intensity at the threshold of
hearing. Using sound intensities to describe sound is too awkward for everyday use, so a logarithmic scale is
used instead. This scale uses the logarithm to the base 10 of the ratio of the intensity of the sound to the intensity
at the threshold of hearing. This value is given the unit of the bel (B), named after Alexander Graham Bell who
was once credited with inventing the telephone.
The bel as a unit is too big for convenient use, so the decibel (dB) unit is more often used. One decibel is onetenth of a bel.
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1 dB  10 1 B
The sound intensity level (L) of any sound is measured in decibels and is defined in terms of its intensity.
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I0
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L (in dB)  10 log10
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where
I0 is a reference sound intensity, usually taken to be the threshold of hearing.
EC
Therefore I0 = 1.0 × 10–12 W m–2 at 1000 Hz.
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The sound intensity level is measured with a sound level meter. The one shown in the figure at right has two
ranges, 40–80 dB and 80–120 dB.
A sound level meter
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© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 23.11
What is the sound intensity level of a sound of intensity 2.6 × 10–7 W m–2?
L
 10 log10
I
I0
 10 log10
2.6 10 7 W m 2
1.0 10 12 W m 2
FS
Solution:
PR
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 10 log10 2.6 105
 54 dB
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REVISION QUESTION 23.11
The sound intensity level at a point is 64 dB. Calculate the sound intensity at that point.
AS A MATTER OF FACT
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D
PA
When finding the sound intensity that corresponds to a sound intensity level, the formula I = 10(L/10 – 12) can be
applied.
EC
The average human being can distinguish a difference in sound intensity levels of only about 1 dB.
I2
I
 10 log10 1
I0
I0
C
 10 log10
R
 L 2  L1
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L
R
The formula for calculating sound intensity levels can be adapted to give the change in intensity level between
two sounds of different intensities: the intensity of a sound to the reference intensity of the threshold of hearing.
The change in intensity level, (∆L), between two sound intensities, I2 and I1, can be found as follows:
I 2 I1
 )
I0 I0
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 10 log10 (
 10 log10 (
 10 log10
I2 I0
 )
I 0 I1
I2
.
I1
Therefore, L  10 log10
I2
I1
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 23.12
What is the change in intensity level when a sound intensity is doubled? In this case, I2 = 2I1.
Solution:
L
 10 log10
I2
I1
 10 log10
2I 1
I1
FS
 10 log10 2
O
 3.01dB
PR
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Sample problem 23.12 gives a useful rule of thumb: if the sound intensity doubles, the sound intensity level
increases by 3 dB; if the sound intensity halves, the sound intensity level decreases by 3 dB. In fact, each 3 dB
increase in the sound intensity level requires a factor of 2 increase in the sound intensity.
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SAMPLE PROBLEM 23.13
PA
What is the change in intensity level when a sound intensity is multiplied by 10? In this case, I2 = 10I1.
 10 log10
I2
I1
 10 log10
10 I 1
I1
EC
L
TE
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Solution:
 10 log10 10
R
 10 dB
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Sample problem 23.13 gives another rule of thumb: if the sound intensity is multiplied by 10, the sound intensity
level increases by 10 dB; if the sound intensity is divided by 10, the sound intensity level decreases by 10 dB. In
fact, for each addition of 10 dB to the sound intensity level, the sound intensity is increased by a factor of 10.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Sound intensities and intensity levels
Table 23.2 Sound intensities and intensity levels of some everyday sounds
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Beats
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Intensit
(W m–2)
1 × 10–12
1 × 10–11
1 × 10–10
1 × 10–9
1 × 10–8
3.2 × 10–6
1 × 10–5
1 × 10–3
1 × 10–2
1 × 10–1
1
10
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Threshold of hearing
Rustle of leaves
Purring cat, at a distance of 1 m
Library reading room
Inside a car with the engine on
Conversation, at a distance of 1 m
Busy street traffic
Jackhammer, at a distance of 10 m
Siren, at a distance of 30 m
Loud thunder
Threshold of pain
Jet engine, at a distance of 50 m
Intensity level
(dB)
0
10
20
30
40
65
70
90
100
110
120
130
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Source of the sound
Digital doc:
Investigation 23.2: Sound
intensities and intensity levels
In this investigation a sound level
meter is used to construct a table
of sound intensities and produce
a graph of sound intensity versus
distance from the source.
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When two notes are played together, the sound waves superimpose or
add together. If the two notes are close together in frequency, a
throbbing sound is produced: the combined sound gets louder and
softer. When a maximum of one sound wave coincides with a maximum
from the other sound wave, the resultant amplitude is bigger and the
sound is much louder. When a maximum of one sound wave coincides
with a minimum from the other sound wave, the resultant amplitude is
smaller and the sound is much softer.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
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(a) and (b) are two notes that are close in frequency. When the two notes are played together, they form a
composite sound wave. (c) The resulting sound wave shows how beats occur. Just after the 1 second point, the
maximum of one sound wave occurs at the same time as the other, resulting in a large amplitude and a louder
sound. At the 1.5 second point, the maximum of wave (a) coincides with the minimum of wave (b), producing a
softer sound.
Two musical instruments are in tune if they produce exactly the same frequency when they play the same note.
There will be no beats.
This effect can be used to tune a stringed instrument such as a guitar. If no beats are heard when two strings are
struck, then both strings are producing the same frequency. If beats are heard, they are producing different
frequencies.
The beat frequency (the number of loud beats produced per second) is obtained by subtracting the two
frequencies:
f beat  f 2  f1
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 23.22
A string on a guitar produces a note with a frequency of 256 Hz. Another string is plucked at the same time and a
beat frequency of 4 Hz is heard. What are the possible frequencies of the second string?
Solution:
4  f  256
f  250 H z
Or
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4  256  f
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f  252
Consonance
Sometimes, if two musical notes are played together, they can produce an unpleasant effect. If this happens, the
notes are said to be dissonant. If the effect is pleasing, the notes are said to be consonant.
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Frequency ratio
1:2
2:3
3:4
4:5
3:5
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Interval
Octave
Fifth
Fourth
Major third
Major sixth
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Pythagoras (c. 570–495 BC) observed that two notes were consonant if the ratio of their frequencies could be
expressed as two small whole numbers.
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If the beat frequency when two notes are played together is less than 6, the sound is generally consonant or
pleasing.
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Research investigation into an aspect of psychoacoustics
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Psychoacoustics is the study of the perception of sound. Why is this important? Many universities throughout the
world have psychoacoustic laboratories. Find out the type of research that the laboratories carry out and
summarise the methods and findings.
the Shepard illusion
phantom words
the McGurk effect.
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Research auditory or audio or sound illusions. Find three different illusions and explain how they work. Examples
include:
The human ear’s response to sound
The response of the human ear to sound depends on the frequency and intensity, and therefore on the intensity
level, of the sound involved.
The figure below shows the average range of sound intensity levels for human hearing. Note how the threshold of
hearing depends on the frequency of the sound. Low-frequency sounds, for example, must have a greater sound
intensity level to be detected.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
The figure below also shows the approximate range of frequencies and intensity levels associated with music and
speech. The average human can hear sounds from about 25 Hz to 20 000 Hz. This range is affected by factors
such as age and ear defects.
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It should be noted that the frequency axis for this figure is logarithmic in nature.
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The sensitivity of the average human ear to different frequencies of sound
SAMPLE PROBLEM 23.21
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a. What is the range of frequencies that the average human ear can detect at a sound intensity level of 20
dB?
b. What is the lowest sound intensity level at which a sound of frequency of 10 kHz can be heard?
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Solution:
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a. This problem is solved by finding the lowest and highest frequencies where the threshold of hearing line
cuts the 20 dB line. This gives an approximate range of 200 Hz to 12 000 Hz.
b. From the graph, the threshold of hearing is 10 dB.
REVISION QUESTION 23.21
Refer to the figure above. What is the sound intensity level of the threshold of hearing at 100 Hz?
Loudness level curves
Loudness is the subjective quality of sound related to sound intensity. Remember that the loudness of a sound
depends both on the amplitude of the wave and on the response of the human ear to the relevant frequency of
the sound. The figure above indicates that loudness depends on the frequency of the sound. A 30 dB sound at
1000 Hz is easily heard, but a 30 dB sound at 100 Hz is below the threshold of hearing.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
For a person to perceive the same loudness at different frequencies of sound means that the frequencies must
have different sound intensity levels. The diagram shown in the figure below shows a series of lines joining
frequencies and intensity levels that have the same loudness. (These are called equal loudness curves.) This
diagram was developed by testing a large number of people.
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The equal loudness curves show that a sound of 1000 Hz at 20 dB is as loud as a 50 Hz sound at about 55 dB.
Equal loudness curves for pure tones
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The phon
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The phon is the unit of equivalent loudness of a sound. It is a measure of how loud a sound is relative to a
reference sound. The reference sound is usually the threshold of hearing for a sound with a frequency of 1000
Hz. All points on the individual curves in the figure above have the same value in phons. The value of equivalent
loudness along a line is given by the sound intensity level of the line at 1000 Hz. For example, the line that has an
intensity of 70 dB at 1000 Hz is the 70 phon line.
Characteristics of sound
Loudness is a subjective quality of sound. A subjective quality depends on the interpretation of the observer. The
loudness of a sound is related to the amount of energy transferred by the wave and this, in turn, depends on the
amplitude of the wave. The greater the amplitude, the louder the sound.
The loudness of a sound depends on its frequency. The human ear is more responsive to some frequencies than
to others.
Pitch is another subjective quality of sound. Pitch is related to the frequency of a sound wave. High-pitched
sounds, like those from a piccolo, have a high frequency; low-pitched sounds, like those from a bass guitar, have
a low frequency.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
AS A MATTER OF FACT
In musical scales, doubling the frequency of a note produces a sound one octave higher than the original sound.
Halving the frequency produces a sound one octave lower. When musical notes or tones are one octave apart,
they can be perceived to be the same note, even though they have a different pitch. Try playing two notes an
octave apart on a keyboard or some other instrument and comment on how they sound.
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Timbre (pronounced ‘tamber’) is a subjective quality that distinguishes musical notes of the same frequency
played on different instruments. Timbre depends on the ‘shape’ of the wave (known as the waveform) as can be
observed on a cathode ray oscilloscope. Different instruments produce differently shaped waves, as illustrated in
the figure below.
The waveforms produced by (a) a pipe organ, (b) a piano and (c) a clarinet
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Reflection of waves
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When waves arrive at a barrier, reflection occurs. Reflection is the returning of the wave into the medium in which
it was originally travelling. When a wave strikes a barrier, or comes to the end of the medium in which it is
travelling, at least a part of the wave is reflected.
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A wave’s speed depends only on the medium, so the speed of the reflected wave is the same as for the original
(incident) wave. The wavelength and frequency of the reflected wave will also be
the same as for the incident wave.
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Reflection of transverse waves in strings
When a string has one end fixed so that it is unable to move (for example, when
it is tied to a wall or is held tightly to the ‘nut’ at the end of a stringed instrument),
the reflected wave will be inverted. This is called a change of phase. If the end is
free to move, the wave is reflected upright and unchanged, so there is no
change of phase. These situations are illustrated in the figure below.
Chapter 23 How do instruments make music?
Digital doc:
Investigation 23.3:
Reflection of pulses in
springs
This investigation explores
the transmission and
reflection of pulses in a
spring.
© John Wiley & Sons Australia, Ltd
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Reflection of sound waves
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Reflection of a transverse pulse on a string when (a) and (b) the end of the string is fixed (as in a guitar), and
when (c) and (d) the end of the string is free to move (as with a loop supported by a retort stand)
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Sound waves can similarly reflect off a fixed (e.g. a solid wall) end and also a free end (e.g. a gap in a barrier).
However, sound is a longitudinal pressure wave, so its behaviour is different. For a sound wave, reflection at a
fixed end occurs when sound echoes from a wall. When a compression hits an immovable object such as a wall it
is reflected as a compression. There is no change of phase. This is the opposite of what happens with a
transverse wave at a fixed end. Of course a rarefaction also does not change phase and is reflected from a wall
as a rarefaction.
A sound wave reflects (a) from a solid wall and (b) from the open end of a trumpet.
A free end for a sound wave is the opening at the end of a trumpet that projects the sound into the room. Some
sound reflects off this free or open end and travels back up the instrument.
A compression travelling down the trumpet spreads out at the opening and transfers energy to the outside air.
This creates a rarefaction behind it, which then moves back up the trumpet. Sound waves at an open end
experience a change of phase. When this rarefaction hits the mouthpiece, it is reflected as it is hitting a fixed end
and is reflected unchanged.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Standing waves
Standing waves are an example of what happens when two waves pass through the same point in space. They
can either interfere constructively or destructively. Interference is explained in chapter 13. Standing waves are an
example of interference is a confined space. The restriction may be a guitar string tied down at both ends, or a
trumpet closed at the mouthpiece and open at the other end, or even a drum skin stretched tightly and secured at
its circumference.
The questions are, How and where do the nodes and antinodes form? and What does this imply about what we
hear?
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Transverse standing waves in strings or springs
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When two symmetrical periodic waves of equal amplitude and frequency (and therefore wavelength), but
travelling in opposite directions, are sent through an elastic one-dimensional medium like a string, spring or a
rope, constructive interference and destructive interference occur. In fact, destructive interference occurs at
evenly spaced points along the medium and it happens all the time at these points. The medium at these points
never moves. Such points in a medium where waves cancel each other at all times are called nodes. In between
the nodes are points where the waves reinforce each other to give a maximum amplitude of the resultant
waveform. This is caused by constructive interference. Such points are called antinodes.
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When this effect occurs the individual waves are undetectable. All that is observed are points where the medium
is stationary and others where the medium oscillates between two extreme positions. There seems to be a wave,
but it has no direction of motion. When this occurs, it is said to be a stationary or standing wave.
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The figure at left shows how standing waves are formed in a string by two continuous periodic waves travelling in
opposite directions. It is important to note that the wavelength of the waves involved in the standing wave is twice
the distance between adjacent nodes (or adjacent antinodes).
The formation of a standing wave
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
The figure at left shows a photograph of a standing wave in a spring taken with an extended exposure time. This
enables the extreme positions of the spring to be shown. The positions of the nodes and antinodes are easily
observed. This standing wave was created by having a periodic wave reflect from the fixed end of the spring. The
reflected wave then interfered with the incoming wave and a standing wave was created.
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The figure below shows the motion of a spring as it carries a standing wave. It shows the shape of the spring as it
T
3T
and at t 
the medium
completes one cycle. The time taken to do this is one period (T). Note that (i) at t 
4
4
is momentarily undisturbed at all points and (ii) that adjacent antinodes are opposite in phase — when one
antinode is a crest, those next to it are troughs.
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A standing wave in a spring
A standing wave over one cycle
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 23.14
Use the preceding photo to answer the following questions.
How many nodes are there?
How many antinodes are there?
If the students are 8.0 m apart what is the wavelength of the wave?
If the student is moving her hand at a frequency of 4.0 Hz, what is the speed of the wave?
At what frequency would the student need to move her hand to have only one antinode?
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a.
b.
c.
d.
e.
Solution:
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a. There are four nodes, three in the picture and one at the elbow.
b. There are three antinodes.
c. The distance between nodes is given by 803 . The wavelength is twice this distance and equal to:
2  8.0
 5.3 m.
3
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d. Using v = f λ, speed = 4.0 × 5.3 = 21.3 m s–1 = 21 m s–1.
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REVISION QUESTION 23.14
213
 1.3 H z .
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e. The speed is unchanged at 21 m s–1 and the wavelength is now 16 m, so the frequency 
The spring is now tightened so that the speed is 30 m s–1.
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a. What frequency will be needed to reproduce the pattern in the picture?
b. A pattern is produced that has a wavelength of 8.0 m. Describe the pattern of nodes and
antinodes. What is the new frequency?
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Sound and standing waves
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Sound can also produce standing waves. In an air column such as an organ pipe, a whistle or a trumpet, we have
seen that sound reflects at each end. In the stretched string at the fixed end there is a node because cancellation
was caused by a change of phase, a trough is reflected as a crest. So when a longitudinal sound wave reflects
with a change of phase at an open end, there will also be a node, a pressure node. Whereas at a closed end in
an air column, there is no change of phase, so at this end there must be a pressure antinode.
At the nodes there will be little or no variation in air pressure. The antinodes will be distinct points where the
sound is the loudest and the variation of air pressure is the greatest. The figure below shows the related
variations in air pressure for a region of space where a standing wave is occurring over one cycle.
The wavelength of the sound is twice the distance between adjacent nodes or antinodes.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
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Particle distribution and pressure variations over one cycle for a standing sound wave
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Nodes and antinodes can be detected by ear, but a more reliable method is to
use a microphone connected to a cathode ray oscilloscope (CRO). At the
antinodes the trace has a maximum amplitude. At the nodes the trace has a
minimum amplitude. The figure below shows two loudspeakers facing each
other and connected to an audio amplifier and signal generator. The
microphone and CRO are used to detect the nodes and antinodes. As there is
an antinode just in front of each speaker, the sound will be loud and the
microphone will pick up a strong signal there.
Chapter 23 How do instruments make music?
Digital doc:
Investigation 23.4: Finding
the speed of sound in air
In this investigation the
speed of sound is calculated
by observing antinodes and
measuring the distance
between them.
© John Wiley & Sons Australia, Ltd
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Investigating a standing sound wave with a CRO
SAMPLE PROBLEM 23.15
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Two loudspeakers emitting sound with a frequency of 1000 Hz are set up 10 m apart and facing each other. A
microphone and CRO are used to detect points of loudest sound along a line between the loudspeakers. These
points occur at distances of 5.00 m, 5.17 m and 5.34 m from one of the loudspeakers.
PA
a. What is the wavelength of the sound waves?
b. Calculate the speed of sound in the region between the loudspeakers.
Solution:
  2  0.17 m
 0.34 m
v 
f
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 340 m s 1
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 1000 H z  0.34 m
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b. f = 1000 Hz, λ = 0.34 m, v = ?
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a. The wavelength is twice the distance between adjacent antinodes.
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Making music
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Noise is generally considered to be an undesired sound. In contrast, a musical note or tone is sound with a
definite frequency or pitch that is produced by a musical instrument or voice.
Musical instruments can be classified as either wind, string or percussion. Examples of each type are named in
table 23.3.
Table 23.3 Examples of musical instruments
String
instruments
Guitar
Violin
Piano
Harp
Wind
instruments
Pipe organ
Clarinet
Flute
Saxophone
Chapter 23 How do instruments make music?
Percussion
instruments
Drum
Xylophone
Triangle
Bell
© John Wiley & Sons Australia, Ltd
All musical instruments consist of three parts:
1.
2.
3.
the principal vibrator
the exciter that starts and, if necessary, maintains the vibrations of the principal vibrator
resonators that reinforce and sometimes modify the vibrations of the principal vibrator.
Resonance
Resonance is the condition where a medium responds to a periodic external force by vibrating with the same
frequency as the force.
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Every object has one or more natural frequencies of vibration. For example, when a crystal wine glass is struck
with a spoon, a distinct pitch of sound is heard. If the resonant frequency is produced by a sound source near the
glass, the glass will begin to vibrate. In this case, the alternating driving force is provided by the variations in air
pressure at the surface of the glass due to the sound produced by the sound source.
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If the intensity of the external sound is increased, it is possible to increase the amplitude of the vibrations in the
crystal wine glass until the crystal lattice falls apart and the glass shatters. Note, however, that resonance does
not necessarily mean that something will break!
In a musical instrument, the resonator vibrates at the same frequency as the principal vibrator. This has the effect
of making the sound of the instrument much louder.
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Stringed instruments
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For stringed instruments, the principal vibrator is a string. The string is excited by plucking (guitar), bowing (violin)
or striking (piano). The resonator can be a sounding board (piano), the ‘body’ of the instrument or the air inside
the body (guitar or violin). Vibrations are transferred from the string to the sounding board or body by the bridge,
which also raises the strings above the instrument.
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When a string is caused to vibrate, the transverse waves travel along the string and they are reflected from the
fixed ends of the string. The reflected waves interfere with the incoming waves to set up standing waves in the
string.
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For any particular length of string, there are many possible standing waves that can exist. All the standing waves
have nodes at the fixed ends. The figure below shows the first five standing waves that are possible in a string of
length L.
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This figure also shows the extreme positions that the string can have for the first five possible standing waves.
The solid line shows the position of the string at a time of 0, T, 2T, 3T, and so on, where T is the period of the
T 3T 5T
,
and so on. In this way, the
standing wave. The dotted line shows the position of the wave at a time of ,
2 2 2
location of nodes and antinodes can easily be seen for each standing wave.
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The actual position of the string throughout one period is illustrated in the figure overleaf. Note that this figure
corresponds to the situation in figure below.
In practice, when a string of a musical instrument is excited, more than just one standing wave is created. The
frequencies of these standing waves and their relative amplitudes contribute to the characteristic sound of the
instrument — its timbre.
The speed of a transverse wave in a string depends only on the tension of the string, its thickness and the type of
material it is made from. The speed of the wave is independent of the wavelength and frequency of the wave.
The wavelength for a standing wave is twice the distance between adjacent nodes. Remember that the frequency
v
of a wave is given by the expression f  .

Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
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Some possible standing waves in a string
v 

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Figure (a) on the previous page shows the lowest frequency of a standing wave possible in a string fixed at both
ends. The wavelength of this standing wave is 2L. As the frequency of this wave is the lowest possible, it is called
the fundamental frequency and is given the symbol f0. The speed of this wave (and all other waves in this string)
is given by:
f
f 0  2L
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 2 f0L .
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Overtones are other frequencies above the fundamental frequency that are produced by a musical instrument.
The position of a string throughout one period of a standing wave
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
AS A MATTER OF FACT
The frequencies produced by a musical instrument are known as partials. Overtones are more commonly referred
to as upper partials — they have frequencies above the fundamental frequency. A tone is a note containing no
upper partials.
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If the frequency of the overtone is a whole number multiple of the fundamental frequency, it is known as a
harmonic. Harmonics form an ordered sequence of frequencies and hence they are given numbers. If the
harmonic is n times the fundamental, it is known as the nth harmonic. The fundamental frequency is therefore
called the first harmonic, as it is one times the fundamental frequency. Note that all harmonics are overtones, but
not all overtones are harmonics.
v
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f 
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The second possible standing wave for a string fixed at both ends is shown in figure (b) on page 445. The
frequency associated with this standing wave is the next highest frequency produced by the string and it is given
L
the symbol f1. The wavelength of the standing wave is L as the distance between adjacent nodes is . Since the
2
speed of the wave in the string is 2f0L, it is possible to derive the value of f1 in terms of f0.

2 f0L
, since v  2 f 0 L for this string
L
 f1  2 f 0
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 f1 
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Therefore, this standing wave has twice the fundamental frequency and
produces the second harmonic.
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Similar analysis shows that the standing wave shown in figure (c) has a
frequency three times the fundamental and therefore produces the third
harmonic,
f2  3 f0
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f3  4 f 0
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and so on. For strings, fn is the frequency of the (n + 1)th harmonic.
Digital doc:
Investigation 23.5:
Standing waves in springs
In this investigation standing
waves in springs are
created and measured.
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In general, for strings fixed at both ends, all harmonics are present.
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SAMPLE PROBLEM 23.16
What is the frequency of the third harmonic of a string if the fundamental frequency is 250 Hz?
Solution:
The third harmonic, by definition, has a frequency three times the fundamental frequency. Therefore, the answer
is 750 Hz.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
REVISION QUESTION 23.16
A string produces a sound that has a second harmonic of 700 Hz. Calculate:
a. the fundamental frequency
b. the frequency of the fourth harmonic of this string.
Wind and brass instruments
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For wind instruments the exciter can be a vibrating reed (clarinet, saxophone, bagpipes), a pair of reeds (oboe,
bassoon), eddies of air on either side of an edge of the instrument’s body (organ pipes, recorder, flute), or
vibrating lips (trumpet, tuba, didgeridoo). These methods are illustrated in the figure below.
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Exciting vibrations in wind and brass instruments: (a) clarinet, (b) oboe, (c) pipe organ and (d) trumpet
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Standing waves and the position of nodes and antinodes in wind and brass instruments can be formed in two
different ways. Every instrument has an opening to allow the sound to pass into the air outside the instrument.
EC
Particles of a gas have maximum freedom of movement at the open end of a pipe. They move into and out of the
open end with maximum amplitude, creating a displacement antinode. As the particles move in and out of the
open end, they maintain the same separation when in the pipe as when outside it. There is little variation in the
pressure at an open end; therefore there is always a pressure node at the open end of a pipe.
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Particles of a gas have virtually no freedom of movement at the closed end of a pipe. There is always a
displacement node at this end. As a longitudinal wave is reflected from the closed end, compressions are turned
back on themselves, as are rarefactions. The different pressure regions constructively interfere with themselves,
creating pressure antinodes. Therefore, there is always a pressure antinode at the closed end of a pipe. It can
generally be stated that pressure nodes occur at the same points in a pipe as displacement antinodes, and that
pressure antinodes occur at the same points as displacement nodes. Standing waves in pipes can be
represented as either variations in the displacement of particles along the pipe, or variations in air pressure along
the pipe. In this book they will be represented as pressure variations.
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There is an opening at one end to transfer sound energy to the outside air. So every instrument has a pressure
node at that end.
However the other end has two possibilities. In the case of instruments such as the trumpet and the clarinet
where the mouth closes off the end, there is a fixed or closed end and so, there is a pressure antinode located
there. This means the standing wave patterns inside these instruments are not symmetrical as in a stringed
instrument.
The other possibility, strange as it may seem, is that the other end is also open. Instruments of this nature are
tubular bells and wind chimes, which produce sound by being hit on the outside. Surprisingly, a flute behaves as if
both ends are open.
In this case the flute player does not cover the opening with their mouth so it is open to the air. For these
instruments there is a pressure node at each end and so the standing wave patterns are symmetrical and similar
to those of stretched strings.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Standing waves in pipes open at both ends
There are many possible standing waves that can exist in any pipe. The standing wave with the lowest frequency
is called the fundamental or the first harmonic. The fundamental standing wave in a pipe open at both ends has a
pressure node at each end and an antinode in the middle. This situation is shown in the figure at left.
FS
The fundamental standing wave for a pipe open at both ends
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O
The figure at left depicts the variations of air pressure in the pipe as it supports a standing wave. Imagine a
distance axis down the centre of the pipe, with a variation in pressure axis at right angles to this. The diagram
shows the maximum variations in pressure along the pipe. At the antinode the air pressure fluctuates periodically
between its maximum and minimum values.
U
N
C
O
R
R
EC
TE
D
PA
G
E
The speed of the wave is the same as the speed of sound in air because the medium is air. The wavelength is
twice the distance between successive antinodes. The figure at left shows a pipe of length L. The wavelength of
this standing wave is therefore 2L. The fundamental frequency is given the symbol f0, so the speed of the wave is
2Lf0. The figure below shows the first five possible standing waves for a pipe open at both ends.
Standing waves in a pipe open at both ends
The frequency associated with the standing wave in figure (b) is the first resonant frequency for the pipe above
the fundamental. It is given the symbol f1. The wavelength is L, and the speed is 2Lf0.
f1 
v

2L f 0
L
 2 f0

Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
The first resonant frequency, f1, above the fundamental for this pipe is the second harmonic since it has twice the
frequency of the fundamental. A similar method can be used to show that the second resonant frequency, f2,
above the fundamental is the third harmonic and so on.
Pipes open at both ends can sustain all the harmonics.
If fn is the frequency of the nth resonant frequency above the fundamental for a pipe open at both ends, then:
f n  ( n  1) f 0 .
The ratio f 0 : f1 : f 2 : f 3 :  equals 1 : 2 : 3 : 4 
FS
SAMPLE PROBLEM 23.17
a. the fundamental frequency
b. the third resonant frequency above the fundamental for this pipe.
L  40 cm  0.40 m
f0
 2L
 0.80 m
v

PA


TE
D
a.
G
E
Solution:
PR
O
O
A pipe open at both ends has a length of 40 cm. The speed of sound in air is 340 m s–1. Determine:
EC
340 m s 1

0.80 m
 425H z
R
 4 f0
 4  425H z
 1700 H z
U
N
C
O
f3
R
b. For a pipe open at both ends, the third resonant frequency above the fundamental is the fourth harmonic.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 23.18
PR
O
O
FS
The figure below shows the pressure variation in a pipe open at both ends. At the instant shown in the figure, the
pressure is at its maximum variation from normal pressure. The speed of sound in air is 340 m s–1. The pipe has a
length of 0.80 m.
G
E
a. Mark the position of any pressure nodes and antinodes in the pipe.
i. at the instant shown in the diagram
ii. one-quarter of a period later
c.
TE
D
iii. one-half period later.
PA
b. Sketch a graph showing the variation of air pressure as a function of distance along the pipe:
What is the wavelength of this standing wave?
EC
d. What are the frequency and the period of this standing wave?
What is the fundamental frequency of this pipe?
R
f.
R
e. What harmonic is this standing wave?
O
Solution:
U
N
C
a. Nodes occur at points where the air pressure is normal, antinodes occur where the air pressure is a
maximum or a minimum. These points are shown below.
b. i.
The pressure variations are at a maximum. This situation is shown in the following figure (a).
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
ii. A quarter of a period later, the air pressure will be normal all along the pipe, as shown in figure (b).
PA
G
E
PR
O
O
FS
iii. One half a period after the first instant, the pressure variations will again be at a maximum, but they
will be the opposite of their original values, as shown in figure (c).
The wavelength is twice the distance between adjacent nodes, or the distance between alternate nodes.
So λ = 0.80 m.
v
d. f  
340 m s 1
 0.80 m
 425H z
1
T  f
1
 425H z
3
 2.35 10 s
C
O
R
R
EC
TE
D
c.
f.
U
N
e. This is the first standing wave above the fundamental, hence it is the second harmonic. (Refer back to the
figure at the bottom of page 448.)
The second harmonic is twice the fundamental frequency.
f0
425H z
2
 212.5H z

Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
REVISION QUESTION 23.18
A pipe that is open at both ends has a length of 60 cm. It produces a sound that has a second harmonic of
550 Hz. Calculate:
a. the fundamental frequency
b. the speed of sound in air
c. the frequency of the third overtone of this pipe.
FS
Other instruments that behave like pipes open at both ends include the piccolo, open organ pipe (the top of the
pipe is open, as illustrated in the figure on page XXX) and the recorder.
O
Standing waves in pipes closed at one end
TE
D
PA
G
E
PR
O
For instruments such as the trumpet and clarinet, the standing wave will have a pressure antinode at the mouth
end and a pressure node at the open end. The longest standing wave possible in a pipe of length L that is closed
at one end is shown in the figure at left. The frequency associated with this standing wave is the lowest possible,
therefore it is called the fundamental and is given the symbol f0. The wavelength for this standing wave is 4L and
the speed is 4Lf0. The first five possible standing waves for a pipe closed at one end are shown in the figure
below.
U
N
C
O
R
R
EC
The longest wavelength standing wave for a pipe closed at one end
Standing waves in pipe closed at one end
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
The second longest wavelength standing wave has a wavelength of
4L
. This is the first resonant frequency
3
above the fundamental for the pipe and is assigned a frequency f1. Since v is constant at 4Lf0, f1 has a value of
v

So:

v

 v
1

 4L f 0 
3
4L
FS
f1
G
E
PR
O
This means that the first resonant frequency above the fundamental for a
pipe closed at one end is the third harmonic. Similar calculations will show
that only the odd numbered harmonics are possible for a pipe closed at
one end. The second resonant frequency above the fundamental is the fifth
harmonic, the third resonant frequency above the fundamental is the
seventh harmonic and so on. Pipes closed at one end can sustain only the
odd numbered harmonics.
O
 3 f0 .
f n  (2n  1) f 0 .
Investigation 24.6: Resonant
frequencies in a tube
In this investigation the
resonant frequencies, and
their harmonics, are observed
and analysed.
TE
D
The ratio f0 : f1 : f2 : f3 … equals 1 : 3 : 5 : 7 …
PA
If fn is the frequency of the nth resonant frequency above the fundamental,
then:
Digital doc:
SAMPLE PROBLEM 23.19
EC
a. What is the fundamental frequency for a pipe closed at one end if it is 0.80 m long and the speed of
sound in air is 340 m s–1?
R
R
b. What is the frequency of the third resonant frequency above the fundamental for this pipe?
O
Solution:
N
 4  0.80 m
 3.2 m
v

U

C
a. For the fundamental frequency, use λ = 4L.
f0

340 m s 1
3.2 m
 106.25H z

Therefore, the fundamental frequency is 1.1 × 102 Hz.
b. The third resonant frequency above the fundamental is the seventh harmonic.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
f3
 7 f0
 7  106.25H z
 743.75H z
Therefore, the third resonant frequency above the fundamental has a frequency of 7.4 × 102 Hz.
REVISION QUESTION 23.19
FS
A pipe that is closed at one end has a length of 40 cm. It produces a sound that has a first overtone of 600
Hz. Calculate:
PR
O
O
a. the fundamental frequency
b. the speed of sound in air
c. the frequency of the third overtone of this pipe.
Other instruments that behave like pipes closed at one end include the oboe, bagpipes, the didgeridoo and a
stopped or closed organ pipe (where the top of the pipe is closed off).
PA
G
E
The length of resonating pipe for many instruments is controlled by covering holes along the body of the
instrument. The first open hole down the body sets the length of the pipe. Holes can be covered either with the
fingers (recorder) or by pads controlled by keys attached to the body (saxophone, clarinet).
Timbre of instruments
TE
D
When a musical instrument is sounded, it simultaneously produces a range of harmonics at different intensities.
The sound waves associated with each harmonic superimpose to produce a composite waveform which gives the
instrument its characteristic sound or timbre. The period of the composite waveform is the same as the period for
the fundamental frequency of the instrument.
R
EC
A sound spectrum is a line graph that shows the frequencies produced by an instrument and their relative
intensities or amplitudes. Figure below shows the waveforms and sound spectra for a piano and a clarinet.
R
AS A MATTER OF FACT
U
N
C
O
Some harmonics sound horrible when sounded together. Stringed instruments can suppress unwanted harmonics
by exciting (hitting, bowing or plucking) the string at a point where the unwanted harmonic has a node. For the
violin and other stringed instruments, the undesired harmonic is the seventh.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
TE
D
EC
R
R
O
C
N
U
The waveforms and sound spectra for (a) a clarinet and (b) a piano
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
The human voice — speaking and singing
The human voice tract can be modelled as a tube closed at one end. This is similar to a trumpet.
When you make a low frequency sound, the voice box (the larynx) lowers in the throat. This makes the tube
longer, so its resonant frequency is smaller.
There are three main parts involved in speaking and singing. These are:
the lungs as a source of air
2.
the vocal cords, which act as the exciters and can be modelled as a vibrating string
3.
the principal vibrator. This consists of the air in the lower throat or pharynx, the mouth and the nasal
cavity.
FS
1.
O
These parts are shown in figure below.
O
R
R
EC
TE
D
PA
G
E
PR
O
The human vocal tract can be modelled as a pipe closed at the end where the vocal cords are located. The length
of the pipe is controlled by raising and lowering the larynx in the throat.
N
C
The human vocal tract
U
Locate your larynx with a finger (on the outside of your throat!). Sing a low
and a high note, noticing the movement of the larynx up and down your
throat as you do so.
Digital doc
Investigation 23.1 Producing
high and low pitch sounds
doc-16172
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
AS A MATTER OF FACT
Infants are unable to produce articulate sounds until they are about one year old, when the larynx descends into
the throat.
As with musical instruments, the vocal tract produces and resonates at more harmonics than just the fundamental
frequency. The harmonics and their relative intensities contribute to the distinctive voice of each individual.
FS
AS A MATTER OF FACT
Why does the pitch of your voice become higher when you breathe in helium gas?
The speed of sound in helium is greater than the speed of sound in air:
•
•
The speed of sound in air at 20 °C is 343 m s–1.
The speed of sound in helium is 1005 m s–1.
G
E
v f
PR
O
O
The human voice tract can be modelled as a tube closed at one end. The wavelength of the fundamental
frequency is four times the length of the tube.
PA
This means that if the vocal tract is full of helium, the fundamental wavelength stays the same. As the speed of
sound increases, so does the fundamental frequency.
TE
D
Your voice is not pure. You produce many frequencies at the same time, giving your voice its own timbre. When
your vocal tract is full of Helium, it is the higher frequencies that resonate and dominate the pitch of your voice.
EC
The ear canal can be modelled as a pipe that is closed at the ear drum. The ear drum resonates like a drum top.
Both of these factors determine the range of frequencies that the ear best responds to.
R
SAMPLE PROBLEM 23.20
R
In an adult, the ear canal is about 2.7 cm long. What is the lowest frequency at which the ear canal will resonate?
Assume that the speed of sound in air is 340 m s–1.
C
O
Solution:
f0
 2.7  10 2 m, so   1.08  10 1 m
v

U
L
N
The lowest resonant frequency is the fundamental frequency of the ear canal. For the fundamental frequency of a
pipe closed at one end, λ = 4L.

340 m s 1
1.08  10 1 m
 3148H z

So the lowest resonant frequency is 3.1 × 103 Hz.
Note that this frequency corresponds to the frequency at which the human ear is most sensitive.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Diffraction
You can hear around corners
Diffraction is the bending of waves as they pass around an object, through a gap in a barrier or around the edge
of a barrier. Diffraction occurs because waves spread out as they travel. It is best seen with water waves in a
ripple tank. Figure below shows the way water waves diffract in various situations. The diagrams apply equally
well to the diffraction of sound waves.

d
is important. As the value of this ratio
O
When waves diffract through a gap of width d in a barrier, the ratio
FS
The region where no waves travel is called a shadow. The amount of diffraction that occurs depends on the
wavelength of the waves. The longer the wavelength, the more diffraction occurs. As a general rule, if the
wavelength is less than the size of the object, there will be a significant shadow region.
C
O
R
R
EC
TE
D
PA
G
E
PR
O
increases, so, too, does the amount of diffraction that occurs.
U
N
Diffraction of water waves: (a) short wavelength around an object (b) long wavelength around the same object (c)
short wavelength through the same gap (e) short wavelength around the edge of a barrier and (f) long wavelength
around the edge of the same barrier
AS A MATTER OF FACT
Walls and fences built next to freeways are effective in protecting nearby residents from high-frequency sounds
as these have a short wavelength and undergo little diffraction. The low-frequency sounds from motors and tyres,
however, diffract around the barriers because of their longer wavelengths.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
FS
O
G
E
Directional spread of different frequencies
PR
O
The diffraction of low and high frequencies around a freeway barrier
PA
The opening at the end of a wind instrument such as a trumpet, the size of someone’s mouth and the size of a
loudspeaker opening all affect the amount of diffraction that occurs in the sound produced. High-frequency
sounds can best be heard directly in front of these devices.
U
N
C
O
R
R
EC
TE
D
When a trumpet plays a note, it produces more than one frequency. This gives the characteristic waveform of the
trumpet and is responsible for its timbre. Trumpets sound ‘bright’ because they produce many higher frequencies.
(The higher frequencies produced are called ‘overtones’ or ‘harmonics’ and will be dealt with in chapter 3.) Highfrequency sounds undergo less diffraction from the mouth of the trumpet than low-frequency sounds. If the
trumpet is heard at a large angle from a line directly in front of it, it will seem to lose its brightness due to
diffraction. The mouth of a trumpet is usually 0.15 m wide.
The diffraction of high and low frequencies from the mouth of a trumpet
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
This investigation should be carried out in an outdoors setting where there is
little interference from reflection.
FS
Investigation 23.2 Diffraction
and sound intensity levels for
sounds of different
frequencies
doc-0000
PR
O
Place the loudspeaker in the box facing the opening. Start the investigation
by using a gap of 0.20 m. Adjust the signal generator to a frequency of 500
Hz and adjust the amplifier to give a sound intensity level of 80 dB at a
distance of 5.0 m directly in front of the opening in the box. Maintaining a
distance of 5.0 m from the opening, measure the sound intensity at an angle
of 20, 40, 60 and 80 degrees on either side of the first reading.
Digital doc
O
You will need the following equipment: a sine wave signal generator, an
audio amplifier, a loudspeaker, a sound level meter and a soundproof box
(cardboard, lined with pillows) with an adjustable gap for sound to pass
through.
G
E
Adjust the signal generator to a frequency of 5000 Hz and adjust the amplifier to give a sound intensity level of 80
dB at a distance of 5.0 m directly in front of the opening in the box. Maintaining a distance of 5.0 m from the
opening, measure the sound intensity at angles of 20, 40, 60 and 80 degrees on either side of the first reading.
PA
1. Make a table of your results and draw graphs showing how the sound intensity level and the sound intensity
varied with the angle for each frequency.
2. Adapt this procedure to investigate how gap size affects the amount of diffraction.
TE
D
Set up a ripple tank with a plane wave generator. Introduce a barrier into the
ripple tank which allows waves to pass by its edge and investigate the effects
that wavelength has on the amount of diffraction that occurs. Sketch your
findings, clearly showing the shadow region.
Digital doc
Investigation 23.3
Diffraction of waves in a
ripple tank
R
doc-0000
O
Summarise your results.
R
EC
Now set up a barrier with a gap in its centre at right angles to the direction of
wave propagation. Study the effect that varying the gap size for a constant
wavelength has on the amount of diffraction. Then study the effect that
varying the wavelength (while keeping the gap size constant) has on the
amount of diffraction that occurs.
C
Where did that sound come from?
U
N
Binaural hearing involves the use of two ears to locate the source of sounds. This is possible because there are
differences in the sound reaching each ear.
The first difference is in the intensity of the sound reaching each ear. This is caused partly by the decrease in
intensity as distance from the source increases (the inverse square law). It is also caused by the diffraction effects
of sound passing around an obstacle. The diameter of the head is about 0.20 m. Sound with a wavelength greater
than 0.20 m will diffract significantly around the head. The drop in intensity due to diffraction is therefore more
significant for high frequencies.
Because the sound has travelled further to get to one ear compared with the other, it will be of a different phase
(that is to say, a different part of the compression-rarefaction cycle) in each ear.
There will also be a time delay between the sound reaching one ear and the other. This is useful for determining
the direction of short, sharp sounds and low frequencies.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
G
E
PR
O
O
FS
The brain uses the difference in phase, the time delay and loudness of the sound for each ear to determine where
the sound came from.
Locating a sound source using two ears
PA
Sound detection
TE
D
Subjective sound
EC
The sensation of sound is perceived by living organisms. Compression waves travel through the air from the
source (a vibrating object such as a loudspeaker). The waves enter the ear. The ear converts the sound waves to
electrical nerve impulses, which are interpreted by the brain as sound.
O
R
noise or musical (a tone)
high or low in pitch
loud or soft
having a timbre
pleasant or unpleasant.
C
•
•
•
•
•
R
Psychoacoustics is the study of how people perceive sounds. This involves both physical and psychological
factors. People can perceive a sound as being:
U
N
In this section you will learn about how the ear responds to sound and the physical characteristics of sound. You
will see how the physical characteristics of sound waves such as frequency and amplitude affect the perceived
characteristics such as pitch and loudness. You will then carryout your own research into an aspect of
psychoacoustics.
The structure of the human ear
The ear responds to sound using three processes. The first process is the collection of the sound energy from the
disturbances in the air. This happens in the outer ear. The outer ear is comprised of the pinna, the ear canal and
the eardrum. The second process is the mechanical amplification of the sound. This is carried out in the middle
ear. The middle ear is comprised of three hinged bones called the ossicles, which are the three smallest bones in
the body.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
The simplified structure of the human ear
EC
AS A MATTER OF FACT
TE
D
The ossicles conduct sound from the eardrum to the oval window, which is the opening to the inner ear. They help
to amplify the sound, although most of the amplification is caused by the relative sizes of the eardrum and the
oval window. When the sound is too loud, the ossicles have the ability to dampen (decrease) the vibrations to limit
any damage to the inner ear. The middle ear is maintained at normal air pressure by the eustachian tube which is
connected to the back of the throat and therefore to the outside air.
C
O
R
R
Having a head cold or sinus problems sometimes interferes with the body’s ability to balance the air pressure on
either side of the eardrum. This often results in a condition commonly called ‘blocked ears’. In scuba diving, where
there are considerably larger changes in pressure, it is essential that the diver has the ability to equalise the
pressure on either side of the eardrum, or the eardrum will burst. This equalising of pressure is achieved by
holding the nose and blowing gently, which opens the eustachian tube and allows air into the middle ear.
U
N
The third process of hearing is the conversion of the sound vibrations into nerve impulses that are sent to the
brain for interpretation. This conversion occurs in the inner ear. The inner ear is made up of the semicircular
canals and the cochlea. The semicircular canals can be thought of as three spirit levels arranged at 90 degrees to
each other so that the body can maintain its balance.
The cochlea is a liquid-filled coiled tube that has an arrangement of hairs called cilia on the inside. The vibrations
transmitted through the middle ear pass through the oval window and cause the liquid inside the tube to vibrate.
This excites specific hairs that correspond to the frequency of the vibrations. Once excited, the hairs generate and
transmit electrical impulses along the auditory nerve to the brain, where they are interpreted.
The eardrum is the first movable part in the body’s system for detecting sound. It can be thought of as a small
piece of thin plastic (a bit like a piece of Glad® Wrap), which has air on both sides. On the inside of the eardrum,
the air is at the same pressure as the normal undisturbed air pressure outside. When the air pressure on either
side of the eardrum is the same, then the shape of the eardrum remains flat (as in figure(a) below), as the same
force is applied to each side.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
FS
When the compression (increased pressure) part of the longitudinal wave reaches the eardrum, the pressure it
exerts is greater than that of the air inside the ear on the other side of the eardrum. As a result the eardrum
curves inwards, as in figure below(b).
O
How variations in pressure affect the shape of the eardrum
PR
O
The next part of the wave is at the same pressure as the surrounding wave, and as a result the eardrum returns
to its normal shape. This is then followed by a rarefaction (lower pressure). As a result, the air pressure on the
inside of the ear is greater compared with that on the outside, and the eardrum bulges out, as in figure above(c).
G
E
The next part of the wave is at the same pressure as the surrounding air and as a result the eardrum returns to its
normal shape, as in figure above(a). This completes the cycle.
PA
The range of frequencies that a young healthy ear can detect lies between about 25 Hz and 20 000 Hz. This
range decreases both with age and with exposure to loud sounds. Often the damage to the ear caused by loud
sounds does not become apparent until later in life.
U
N
C
O
R
R
EC
TE
D
When hearing is tested, the response to a range of frequencies is tested, producing an output called an
‘audiogram’. This is a chart that shows, among other things, the sound intensity at the threshold of hearing for a
range of frequencies. Figure below shows how the sound intensity levels at the threshold vary for different age
groups.
A chart showing how frequency response declines with age
The ear canal can be modelled as a pipe that is closed at the ear drum. The ear drum resonates like a drum top.
Both of these factors determine the range of frequencies that the ear best responds to.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Chapter review
Summary
■
■
Sound waves are longitudinal.
As longitudinal waves move through a medium, the particles of the medium vibrate parallel to the direction of
propagation.
The intensity (I) of a sound is the power per unit area at a point and it is measured in W m–2.
■
The intensity of a sound is inversely proportional to the distance from the source.
■
The intensity level (L) of a sound is measured in decibels (dB).
■
L  10 log10
O
PR
O
■
Resonance is the condition where a medium responds to a periodic external force by vibrating with the same
frequency as the force.
Standing waves are caused by the superposition of two wave trains of the same frequency travelling in
opposite directions.
G
E
■
I
I0
FS
■
The fundamental frequency of a string or pipe is the lowest frequency at which a standing wave occurs.
■
Harmonics are whole number multiples of the fundamental frequency.
Questions
Wave speed or velocity
TE
D
PA
■
EC
1. What is the speed of sound in air if it travels a distance of 996 m in 3.0 s?
2. How far does a wave travel in one period?
R
3. Do loud sounds travel faster than soft sounds? Justify your answer.
O
R
4. A marching band on the other side of a sports oval appears to be ‘out of step’ with the music. Explain why this
might happen.
C
5. You arrive late to an outdoor concert and have to sit 500 m from the stage. Will you hear high-frequency
sounds at the same time as low-frequency sounds if they are played simultaneously? Explain your answer.
U
N
6. A loudspeaker is producing a note of 256 Hz. How long does it take for 200 wavelengths to interact with your
ear?
7. During an electrical storm the thunder and lightning occur at the same time and place. Unless the centre of
the storm is directly above, you see the lightning flash before you hear the thunder. How far away is lightning
if it takes 5.0 s for the sound of thunder to reach you after the flash is seen? Assume the speed of sound in air
is 335 m s–1.
The wave equation
8. What is the wavelength of a sound that has a speed of 340 m s–1 and a period of 3.0 ms?
9. What is the speed of a sound if the wavelength is 1.32 m and the period is 4.0 × 10–3 s?
10. The speed of sound in air is 340 m s–1 and a note is produced that has a frequency of 256 Hz.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
a. What is its wavelength?
b. This same note is now produced in water where the speed of sound is 1.50 × 103 m s–1. What is the new
wavelength of the note?
11. Copy and complete table 23.4 by applying the universal wave formula.
Table 23.4
FS
λ (m)
0.67
25
0.30
2.5
1000
440
O
1500
60
340
260
f (Hz)
500
12
Producing sound
PR
O
v (m s–1)
12. Sketch a graph showing the variation of air pressure as a function of distance for a sound wave of wavelength
1.0 m.
G
E
13. Sketch a pressure–time graph to illustrate the following situations:
a. a 100 Hz sound of a certain loudness
b. a sound with the same frequency, but louder
PA
c. a sound with the same loudness as in (a) but twice the frequency
d. a sound that is an octave lower than in (a) and quieter.
U
N
C
O
R
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14. Figure (a) at the top of the facing page shows the pressure variation as a function of distance from a sound
source at an instant in time. The graph shown in figure (b) shows the pressure variation as a function of
distance from the sound source. The speed of sound in air is 340 m s–1.
a. What is the wavelength of this sound?
b. What is the period of this sound?
c. Using the same scale as shown in figure (b), sketch the distribution of particles one-sketch the pressure
variation one-quarter of a period later.
d. Using the same axes as shown in figure (b), sketch a graph of pressure variation versus distance one
quarter of a period later than shown in the graph.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
15. The figure below shows the variation of air pressure as a function of time near a sound source.
a. What is the period of this sound?
FS
b. What is the frequency of this sound?
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c. What is the wavelength of this sound if the speed of sound is 330 m s–1?
PR
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Observing sound
16. The figure below shows the trace on a CRO screen for a sound detected by a microphone. The time scale is 2
ms cm–1.
a. What is the period of this sound?
O
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PA
G
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b. What is the frequency of this sound?
C
c. If the wavelength is 5.28m, what is the speed of sound in this case?
N
d. Using the same scale shown in the figure, sketch the trace you would get under the following
circumstances:
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i. if the frequency is doubled, but the loudness stays the same
ii. if the frequency stays the same, but the loudness is increased.
The power of sound
17. What is the acoustical power of a source if it produces 2.0 J in 100 s?
18. What is the sound intensity if 4.0 × 10–8 W pass through an area of 0.080 m2?
19. Calculate the power passing through an area of 2.0 m2 if the sound intensity is 4.5 × 10–5 W m–2?
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
20. One siren produces a sound intensity of 3.0 × 10–3 W m–2 at a point which is 10 m away. What would be the
sound intensity produced at that point if five identical sirens sounded simultaneously at the same place as the
original?
21. If the sound intensity 4.0 m from a point sound source is 1.0 × 10–6 W m–2, what will be the sound intensity at
each of the following distances from the source?
a. 1.0 m
b. 2.0 m
c. 8.0 m
d. 40 m
FS
22. What are the sound intensity levels associated with the following sound intensities?
a. 5.0 × 10–10 W m–2
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b. 3.2 × 10–7 W m–2
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c. 4.9 × 10–3 W m–2
d. 1.8 × 10–9 W m–2
23. Calculate the sound intensities associated with the following sound intensity levels:
a. 7.0 dB
G
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b. 25 dB
c. 54 dB
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d. 115 dB.
24. Show that the rule of thumb stated in the text —halving the sound intensity reduces the intensity level by 3 dB
— is true.
TE
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25. The area of a human eardrum is approximately 5.0 × 10–5 m2. Sabiha is listening to music through a button
earphone. How much energy arrives at her eardrum during a song that lasts for 3 min if the average sound
intensity produced at the eardrum by the button earphone is 1.0 × 10–2 W m–2?
EC
The human ear’s response to sound
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26. What is meant by the expression ‘threshold of hearing’?
R
27. Refer to the figure on page XXX.
i. 100 Hz
C
ii. 500 Hz
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a. What is the sound intensity level at the threshold of hearing at the following frequencies?
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iii. 2000 Hz
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iv. 10 000 Hz
b. At what frequency is the average human ear the most sensitive?
c. What is the sound intensity level at the threshold of hearing at that frequency?
28. Refer to the the figure on page XXX.
a. Estimate the range of frequencies that an average human being can hear for sounds with a sound intensity
level of 10 dB.
b. Estimate the highest frequency used in the speech region.
c. What is the approximate range of frequencies that can be above 100 dB in performing music?
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Transverse standing waves in strings or springs
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FS
29. The figure below shows the positions of three sets of two pulses as they pass through each other. Copy the
diagram and sketch the shape of the resultant disturbances.
PA
30. What is the wavelength of a standing wave if the nodes are separated by a distance of 0.75 m?
TE
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31. The figure below shows a standing wave in a string. At that instant (t = 0) all points of the string are at their
maximum displacement from their rest positions.
EC
If the period of the standing wave is 0.40 s, sketch diagrams to show the shape of the string at the following
times:
a. t = 0.05 s
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b. t = 0.1 s
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c. t = 0.2 s
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d. t = 0.4 s.
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C
Sound and standing waves
32. Kim and Jasmine set up two loudspeakers in accordance with the following arrangements:
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• They faced each other.
• They were 10 m apart.
• The speakers are in phase and produce a sound of 330 Hz.
Jasmine uses a microphone connected to a CRO and detects a series of points between the speakers where
the sound intensity is a maximum. These points are at distances of 3.5 m, 4.0 m and 4.5 m from one of the
speakers.
a. What causes the maximum sound intensities at these points?
b. What is the wavelength of the sound being used?
c. What is the speed of sound on this occasion?
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
Stringed instruments
33. A standing wave is set up by sending continuous waves from opposite ends of a string. The frequency of the
waves is 4.0 Hz, the wavelength is 1.2 m and the amplitude is 10 cm.
a. What is the speed of the waves in the string?
b. What is the distance between the nodes of the standing wave?
c. What is the maximum displacement of the string from its rest position?
d. What is the wavelength of the standing wave?
e. How many times per second is the string straight?
FS
34. The speed of waves in a string is 250 m s–1. It has a length of 1.0 m.
b. What is the fundamental frequency for this string?
PR
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c. What is the frequency of the first resonant frequency above the fundamental?
O
a. What is the wavelength of the longest standing wave that can be produced in this string?
d. What harmonic corresponds to the second resonant frequency above the fundamental and what is its
frequency?
a. the first resonant frequency above the fundamental
b. the third resonant frequency above the fundamental
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c. the third harmonic
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35. If the fundamental frequency of a string is 240 Hz, find the frequencies of the following quantities:
d. the 22nd harmonic.
String
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Table 23.5
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C
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36. The following figure represents four standing waves for a string of length L that is fixed at both ends. Use this
figure to complete table 23.5.
Nodes
2
Antinodes
λ
Tone
f0
Harmonic
L
2
2
A
37. a. The first resonant frequency above the fundamental of a string is 500 Hz. What is the fundamental
frequency of this string?
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
b. The second harmonic of a string is 516 Hz. What is the fundamental frequency of the string?
38. The third harmonic of a string is 810 Hz. What is the fundamental frequency?
39. The fourth resonant frequency above the fundamental of a string is 1400 Hz. Find the following frequencies:
a. the fundamental frequency
b. the second harmonic
c. the second resonant frequency above the fundamental.
Wind and brass instruments
FS
40. If the speed of sound in air is 340 m s–1, find (i) the longest wavelength tone and (ii) the fundamental
frequencies of the following pipes:
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a. open at both ends, length 40 cm
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b. open at both ends, length 60 cm
c. open at both ends, length 1.21 m
d. open at both ends, length 1.00 m
e. closed at one end, length 0.50 m
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f. closed at one end, length 0.25 m
g. closed at one end, length 12.5 cm
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h. closed at one end, length 17 cm.
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41. The figure above right represents four standing waves in a pipe of length L. The pipe is open at both ends.
Use this figure to complete table 23.6.
Table 23.6
Pipe
B
Nodes
Antinodes
λ
Resonant Frequency
Harmonic
First
2L
3
5
42. The figure below represents four standing waves in a pipe of length L. The pipe is closed at one end. Use this
figure to complete table 23.7.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
O
FS
Pipe
Nodes
Antinodes
λ
Resonant Frequency
Harmonic
4L
5
G
E
C
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Table 23.7
2
Seventh
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PA
42. A graph of how air pressure varies with distance along a pipe at a particular instant is shown below. A
standing wave has been set up in the pipe and the graph represents the maximum variation from normal air
pressure. Assume that the speed of sound in air is 340 m s–1.
C
a. Is the pipe open at both ends or closed at one end? Explain.
N
b. What is the length of the pipe?
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c. What is the wavelength of the sound being produced?
d. What is the frequency of the sound being produced?
e. Which harmonic and resonant frequency above the fundamental is being produced?
f. What is the fundamental frequency for this pipe?
g. Sketch a graph showing the variation of air pressure from normal along the pipe at a time half a period
later than the instant shown in the diagram.
44. The figure below shows pressure variation in and around a pipe open at both ends as it is resonating at one
of its harmonics. Assume that the speed of sound in air is 340 m s–1.
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd
b. If the pipe is 0.85 m, what is the wavelength of the tone that the pipe is producing?
c. What is the frequency of the tone being produced?
FS
a. What harmonic is represented in the diagram?
PR
O
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d. Make a sketch to show the pressure variation in and around the pipe half a period later than the instant
shown.
e. Sketch the pressure variation in and around the pipe one quarter of a period later than the instant shown.
f. What is the period of the sound being produced by the pipe?
g. What is the frequency of the second resonant frequency above the fundamental for this pipe?
EC
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G
E
45. The following figure shows the pressure variation in and around a pipe closed at one end as it is resonating at
one of its harmonics. Assume that the speed of sound in air is 340 m s–1.
R
a. What harmonic is represented in the diagram?
R
b. If the pipe has a length of 50 cm, what is the wavelength of the tone that the pipe is producing?
O
c. What is the frequency of the tone being produced?
d. What is the fundamental frequency for this pipe?
N
C
e. What is the frequency of the third resonant frequency above the fundamental of this pipe? What harmonic
does this frequency correspond to?
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f. Make a sketch to show the pressure variation in and around the pipe half a period later than the instant
shown.
46. A student is tuning a guitar using a tuning fork with a frequency of 440 Hz. When the A string is struck at the
same time as the tuning fork, the string sounds higher in pitch and a beat frequency of 2 Hz is heard. What is
the frequency of the string?
47. The tone A has a frequency of 440 Hz. What is the frequency of the tone:
a. one octave above A
b. one octave below A?
Chapter 23 How do instruments make music?
© John Wiley & Sons Australia, Ltd