Chapter 16 -­‐ solutions Constructive and Destructive Interference Conceptual Question
Description: Conceptual question on whether constructive or destructive interference occurs at various
points between two wave sources.
Two sources of coherent radio waves broadcasting in phase are located as shown below.
Each grid square is 0.5
square, and the radio sources broadcast at
.
Part A
At Point A is the interference between the two sources constructive or destructive?
Hint
A.1
Path-length difference
Since the two sources emit radio waves in phase, the only possible phase difference between the
waves at various points is due to the different distances the waves have traveled to reach those
points. The difference in the distances traveled by the two waves from source to point of interest is
termed the path-length difference.
If the path-length difference is an integer multiple of the wavelength of the waves, one wave will
pass through an integer number of complete cycles more than the other wave, placing the two
waves back in perfect synchronization, resulting in constructive interference. If the path-length
difference is a half-integer multiple of the wavelength, one wave will be one-half of a cycle, or 180
, out of phase, resulting in destructive interference.
Hint
A.2
Find the path-length difference
What is the distance from the left source to Point A,
source to Point A,
?
Enter the distances in meters separated by a comma.
ANSWER:
? What is the distance from the right
What is the distance from the left source to Point A,
? What is the distance from the right
source to Point A,
?
Enter the distances in meters separated by a comma.
ANSWER:
,
=
,
The wave that leaves the source on the left and the wave that leave the source on the right travel
equal distances to Point A.
ANSWER:
constructive
destructive
Part B
At Point B is the interference between the two sources constructive or destructive?
Hint B.1
Find the path-length difference
What is the distance from the left source to Point B,
? What is the distance from the right
source to Point B,
?
Enter the distances in meters separated by a comma.
ANSWER:
,
=
The path-length difference between the two waves is 3
ANSWER:
,
.
constructive
destructive
Part C
At Point C is the interference between the two sources constructive or destructive?
ANSWER:
constructive
destructive
Part D
At Point D is the interference between the two sources constructive or destructive?
ANSWER:
constructive
destructive
ANSWER:
constructive
destructive
Open Organ Pipe Conceptual Question
Description: A conceptual question in which the fundamental frequency of an organ pipe of a certain
length is given. Questions deal with the new fundamental frequency after the organ pipe is cut in half,
closed off, filled with helium. Also asks to compare closed and open pipes.
An open organ pipe (i.e., a pipe open at both ends) of length
has a fundamental frequency
.
Part A
If the organ pipe is cut in half, what is the new fundamental frequency?
Hint
A.1
Fundamental wavelength in an open pipe
For a wave in an open pipe, the fundamental wavelength is the longest wave that "fits" in the pipe,
with an antinode (point of maximum wave amplitude) at both open ends.
Hint
A.2
Fundamental frequency
For speed of sound
, wavelength
, and frequency
,
,
so the frequency of a wave is given by
.
Thus, the fundamental frequency and the fundamental wavelength are inversely proportional.
ANSWER:
Part B
After being cut in half in Part A, the organ pipe is closed off at one end. What is the new fundamental
frequency?
Hint B.1
Fundamental wavelength in a closed pipe
For a wave in a closed pipe, the fundamental wavelength is the longest wave that "fits" in the pipe,
with an antinode (point of maximum wave amplitude) at the open end and a node (point of zero
wave amplitude) at the closed end.
ANSWER:
Hint B.1
Fundamental wavelength in a closed pipe
For a wave in a closed pipe, the fundamental wavelength is the longest wave that "fits" in the pipe,
with an antinode (point of maximum wave amplitude) at the open end and a node (point of zero
wave amplitude) at the closed end.
ANSWER:
The fundamental frequency of a half-length closed pipe is equal to that of a full-length open pipe.
Part C
The air from the pipe in Part B (i.e., the original pipe after being cut in half and closed off at one end)
is replaced with helium. (The speed of sound in helium is about three times faster than in air.). What
is the approximate new fundamental frequency?
Hint C.1
The role of the speed of sound
In an organ pipe, the wave speed is the speed of sound. For speed of sound
wavelength
, frequency
, and
,
.
Thus, the frequency of a wave is directly proportional to the wave speed.
ANSWER:
Based on your answer to this part, can you explain why you have a high pitched voice after inhaling
the helium from a helium balloon?
± Harmonics of a Piano Wire
Description: ± Includes Math Remediation. Calculate the fundamental frequency of a piano wire under
tension, then find the highest number harmonic audible by a human.
A piano tuner stretches a steel piano wire with a tension of 765
0.700
and a mass of 5.25
. The steel wire has a length of
.
Part A
What is the frequency
Hint
A.1
of the string's fundamental mode of vibration?
How to approach the problem
Find the mass per unit length for the wire. Then apply the equation for the fundamental frequency
of the wire.
Hint
A.2
Find the mass per unit length
What is the mass per unit length for the wire?
Express your answer numerically in kilograms per meter using three significant figures.
ANSWER:
=
Hint
A.3
Equation for the fundamental frequency of a string under tension
The fundamental frequency of a string under tension is given by
,
where
length.
is the length of the string,
is the tension in the string, and
is its mass per unit
Express your answer numerically in hertz using three significant figures.
ANSWER:
=
Part B
What is the number
of the highest harmonic that could be heard by a person who is capable of
hearing frequencies up to
Hint
B.1
= 16 kHz?
Harmonics of a string
The harmonics of a string are given by
, where
is the
th harmonic of a string
with fundamental frequency
. Be careful if you get a noninteger answer for
can only be integer multiples of the fundamental frequency.
, as harmonics
Hint
B.1
Harmonics of a string
The harmonics of a string are given by
, where
is the
th harmonic of a string
with fundamental frequency
. Be careful if you get a noninteger answer for
can only be integer multiples of the fundamental frequency.
, as harmonics
Express your answer exactly.
ANSWER:
=
When solving this problem, you may have found a noninteger value for
be integer multiples of the fundamental frequency.
, but harmonics can only
What Is a Sound Wave?
Description: Conceptual questions about sound waves including the properties: frequency, wavelength,
loudness, pitch, and timbre.
Learning Goal: To understand the nature of a sound wave, including its properties: frequency,
wavelength, loudness, pitch, and timbre.
Sound is a phenomenon that we experience constantly in our everyday life. Therefore, it is important to
understand the physical nature of a sound wave and its properties to correct common misconceptions
about sound propagation.
Most generally, a sound wave is a longitudinal wave that propagates in a medium (i.e., air). The
particles in the medium oscillate back and forth along the direction of motion of the wave. This
displacement of the particles generates a sequence of compressions and rarefactions of the medium.
Thus, a sound wave can also be described in terms of pressure variations that travel through the
medium. The pressure fluctuates at the same frequency with which the particles' positions oscillate.
When the human ear perceives sound, it recognizes a series of pressure fluctuations rather than
displacements of individual air particles.
Part A
Based on the information presented in the introduction of this problem, what is a sound wave?
ANSWER:
Propagation of sound particles that are different from the particles that
comprise the medium
Propagation of energy that does not require a medium
Propagation of pressure fluctuations in a medium
Propagation of energy that passes through empty spaces between the
particles that comprise the medium
Part B
Having established that a sound wave corresponds to pressure fluctuations in the medium, what can
you conclude about the direction in which such pressure fluctuations travel?
ANSWER:
The direction of motion of pressure fluctuations is independent of the
direction of motion of the sound wave.
Pressure fluctuations travel perpendicularly to the direction of propagation
of the sound wave.
Pressure fluctuations travel along the direction of propagation of the sound
wave.
Part C
Does air play a role in the propagation of the human voice from one end of a lecture hall to the other?
Hint C.1
Sound propagates in a medium
Sound is a mechanical wave. As such, it needs a medium to travel. Air is the medium through
which the human voice usually propagates.
ANSWER:
yes
no
Part D
The graphs shown here
represent pressure
variation versus time recorded by a microphone. Which could correspond to a sound wave?
Hint
D.1
Sound as propagation of pressure fluctuations
Sound propegation is the displacement of particles along the direction of motion of the sound wave
which generates a sequence of compressions and rarefactions of a medium. Thus, a sound wave can
be described in terms of pressure variations or fluctuations. It follows that a steady increase or
variation versus time recorded by a microphone. Which could correspond to a sound wave?
Hint
D.1
Sound as propagation of pressure fluctuations
Sound propegation is the displacement of particles along the direction of motion of the sound wave
which generates a sequence of compressions and rarefactions of a medium. Thus, a sound wave can
be described in terms of pressure variations or fluctuations. It follows that a steady increase or
decrease in pressure does not correspond to a sound wave.
Enter the letters of all the correct answers in alphabetical order. Do not use commas. For example, if
you think all three graphs could represent sound waves, enter ABC.
ANSWER:
BC
Part E
The next graph
shows a sound wave consisting of a sinusoidal displacement of air particles versus time, as recorded
at a fixed location. For sinusoidal waves, it is possible to identify a specific frequency (rate of
oscillation) and wavelength (distance in space corresponding to one complete cycle).
Taking the speed of sound in air to be 344
the sound wave shown in the graph?
Hint
E.1
, what are the frequency
and the wavelength
of
Definition of frequency
The frequency of a periodic wave is the number of complete oscillations per unit time.
Hint
E.2
Definition of wavelength
In a periodic wave pattern, the distance from one crest to the next (or alternatively from one trough
to the next one or from any point to the corresponding point on the next oscillation cycle) is called
wavelength and it is typically denoted by
propagates is given by
. In addition, the speed
,
at which the wave pattern
In a periodic wave pattern, the distance from one crest to the next (or alternatively from one trough
to the next one or from any point to the corresponding point on the next oscillation cycle) is called
wavelength and it is typically denoted by
propagates is given by
. In addition, the speed
at which the wave pattern
,
where
is the frequency of the wave.
Express your answers in, respectively, hertz and meters to three significant figures. Separate the two
answers with a comma.
ANSWER:
,
=
Part F
A certain sound is recorded by a microphone. The same microphone then detects a second sound,
which is identical to the first one except that the amplitude of the pressure fluctuations is larger. In
addition to the larger amplitude, what distinguishes the second sound from the first one?
Hint F.1
Amplitude of a sound wave
For a given frequency, the greater the amplitude of a sound wave, the louder it is perceived.
ANSWER:
It is perceived as higher in pitch.
It is perceived as louder.
It has a higher frequency.
It has a longer wavelength.
Part G
A certain sound is recorded by a microphone. The same microphone then detects a second sound,
which is identical to the first one except that it has twice the frequency. In addition to the higher
frequency, what distinguishes the second sound from the first one?
Hint G.1
Frequency of a sound wave
The frequency of a sound wave is what mainly determines the pitch of that sound. The higher the
frequency of a sound wave, the higher in pitch it is perceived.
ANSWER:
It is perceived as higher in pitch.
It is perceived as louder.
When we double the frequency of a sound wave, we produce a sound that is said to be an octave
above the first.
It has a higher amplitude.
It has a longer wavelength.
When we double the frequency of a sound wave, we produce a sound that is said to be an octave
above the first.
Part H
What varies between two tones that are different in timbre, that is, two tones that have the same
fundamental frequency but are produced, say, by different musical instruments?
Note that Figures (b) and (c) from Part D could represent tones with different timbre.
Hint H.1
Timbre
The quality of a sound is related to the harmonic content of that sound. Harmonics are higher
frequency components of a sound that are integer multiples of a sound's fundamental frequency. A
tone that is rich in harmonics has a different quality from one that contains mainly the fundamental
frequency, even if the fundamental frequencies of the two sounds are the same. This difference is
also called tone color.
ANSWER:
the pitch
the harmonic content
nothing
Tactics Box 16.1 Identifying Constructive and Destructive Interference
Description: Knight/Jones/Field Tactics Box 16.1 Identifying constructive and destructive interference
is illustrated.
Learning Goal: To practice Tactics Box 16.1 Identifying constructive and destructive interference.
For any two wave sources, this Tactics Box sums up how to determine whether the interference at a
point is constructive or destructive.
TACTICS BOX 16.1 Identifying constructive and destructive interference
1. Identify the path length from each source to the point of interest. Compute the path-length
difference.
2. Find the wavelength, if it is not specified.
3.
4.
If the path-length difference is a whole number of wavelengths
, crests are
aligned with crests and there is constructive interference.
If the path-length difference is a whole number of wavelengths plus a half wavelength
, crests are aligned with troughs and there is destructive
interference.
Follow these steps to solve this problem: Two identical loudspeakers, speaker 1 and speaker 2, are 2.0
apart and are emitting 1700-
sound waves into a room where the speed of sound is 340
.
Is the point 4.0
in front of speaker 1, perpendicular to the plane of the speakers, a point of maximum
constructive interference, a point of perfect destructive interference, or something in between?
Part A
Compute the path-length difference
Hint
A.1
.
Find the path length from speaker 1 to the point of interest
What is the path length
from speaker 1 to the point of interest?
Express your answer in meters to three significant figures.
ANSWER:
=
Hint
A.2
Find the path length from speaker 2 to the point of interest
What is the path length
from speaker 2 to the point of interest?
Express your answer in meters to three significant figures.
ANSWER:
=
Express your answer in meters to two significant figures.
ANSWER:
=
Part B
What is the wavelength
Hint
B.1
of the sound waves emitted by the speakers?
An equation for wavelength
The wavelength
of a wave that travels with speed
is given by
,
where
is the frequency of the wave.
Express your answer in meters to two significant figures.
ANSWER:
=
Now, determine whether the path-length difference is a whole number of wavelengths or a whole
number of wavelengths plus a half wavelength, or something in between.
Part C
Hint C.1
Find the ratio of the path-length difference to the wavelength
Compute the ratio of the path-length difference to the wavelength,
Express your answer to three significant figures.
.
ANSWER:
=
ANSWER:
maximum constructive interference.
The point of interest is
a point of
perfect destructive interference.
neither maximum constructive interference nor
perfect destructive interference.
Two Loudspeakers in an Open Field
Description: Determine if constructive interference occurs at a certain point and then find the shortest
distance you need to walk in order to experience destructive interference.
Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier
so that they emit sound waves in phase at 688
. Take the speed of sound in air to be 344
.
Part A
If you are 3.00
from speaker A directly to your right and 3.50
from speaker B directly to your
left, will the sound that you hear be louder than the sound you would hear if only one speaker were in
use?
Hint
A.1
How to approach the problem
The perceived loudness depends on the amplitude of the sound wave detected by your ear. When
two sound waves arrive at the same region of space they overlap, and interference occurs. The
resulting wave has an amplitude that can vary depending on how the two waves interfere. If
destructive interference occurs, the total wave amplitude is zero and no sound is perceived; if
constructive interference occurs, the total wave amplitude is twice the amplitude of a single wave,
and sound is perceived as louder than what it would be if only one wave reached your ear.
Hint
A.2
Constructive and destructive interference
Constructive interference occurs when the distances traveled by two sound waves differ by a
integer number of wavelengths. If the difference in paths is equal to any half-integer number of
wavelengths, destructive interference occurs.
Hint A.3
Find the wavelength of the sound
What is the wavelength
Hint
A.3.1
of the sound emitted by the loudspeakers?
Relationship between wavelength and frequency
In a periodic wave, the product of the wavelength
and the frequency
is the speed at which
What is the wavelength
Hint
A.3.1
of the sound emitted by the loudspeakers?
Relationship between wavelength and frequency
In a periodic wave, the product of the wavelength
the wave pattern travels; that is,
and the frequency
is the speed at which
.
Express your answer in meters.
ANSWER:
=
The sound's wavelength is equal to the difference between the distances traveled by the sound
waves from the two speakers.
ANSWER:
yes
no
Because the path difference is equal to the wavelength of the sound, the sound originating at the two
speakers will interfere constructively at your location and you will perceive a louder sound.
Part B
What is the shortest distance
speakers?
Hint
B.1
you need to walk forward to be at a point where you cannot hear the
How to approach the problem
You will not be able to hear the speakers if you are at a point of destructive interference. At a point
of destructive interference, the lengths of the paths traveled by the sound waves differ by a halfinteger number of wavelengths. Therefore, you can find the shortest distance you need to walk in
the forward direction by determining the difference in distance from the two speakers that
corresponds to the smallest possible half-integer multiple of the wavelength. Then, figure out how
far forward you need to walk to obtain this path-length difference.
Hint
B.2
If
Find the path-length difference at a point of destructive interference
is the distance between you and speaker A and
B, by how much does
differ from
destructive interference?
Hint
is the distance between you and speaker
if you are now at the closest possible point of
Condition for destructive interference
Destructive interference occurs if the difference in paths traveled by sound waves is equal to any
half-integer number of wavelengths. Therefore, the closest possible point of destructive
interference corresponds to a path-length difference of half a wavelength.
Express your answer in meters.
ANSWER:
B.2.1
Destructive interference occurs if the difference in paths traveled by sound waves is equal to any
half-integer number of wavelengths. Therefore, the closest possible point of destructive
interference corresponds to a path-length difference of half a wavelength.
Express your answer in meters.
ANSWER:
=
Now combine this result with the Pythagorean Theorem and solve for
.
Find your distance from speaker A
Hint
B.3
If initially you were 3.00
from speaker A and then you walked forward the shortest possible
distance needed to experience destructive interference, what is your new distance
same speaker?
Hint
B.3.1
from that
Geometrical considerations
Geometrically, your initial distance from one speaker and the distance you walked north
represent the legs of a right triangle, whose hypothenuse is your new distance from that same
speaker. If you apply the Pythagorean Theorem twice, you can write an expression that links
to
. This equation, combined with the relation previously found by imposing the
condition of destructive interference, allows you to find
Express your answer in meters to four significant figures.
ANSWER:
=
Express your answers in meters to three significant figures.
ANSWER:
=
.
Q16.21. Reason: The earliest (and only) time that will equal 2 mm at the point
is when the top of
the triangular peak on the left moves right to the point
That wave pulse is moving at
take 0.5 s to move 3 cm.
The correct choice is A.
Assess: Because the pulse moving from right to left is a negative pulse, it will make
so it will
at
but
not a positive
Q16.22. Reason: Sketch (a) below shows each pulse at t = 0 s and again at t = 1.5 s. Sketch (b) shows the
resultant interference as the two pulses pass through each other. Note that at x = 10 cm the resultant amplitude is
–1.0 mm.
The correct choice is B.
Assess: The ability to inspect a figure and determine what will occur is a skill that physics will help you develop.
Q16.23. Reason: The maximum displacement will be the sum of the contributions from the two traveling
waves at each point and at each time; however, you are not guaranteed to be watching a point where a crest will
meet a crest. It is true that you might be watching an antinode where the maximum displacement would be
but it is also possible that you might be watching a node that doesn’t move at all—or any place in between.
The correct choice is D.
Assess: If you watch the whole string, you will see points whose maximum displacement is
and other points
whose maximum displacement is 0 and other points in between.
Problems
P16.13. Prepare: A string fixed at both ends forms standing waves. Three antinodes means the string are
vibrating as the m = 3 standing wave. The wavelengths of standing wave modes of a string of length L are given
by Equation 16.1.
Solve: (a) The frequency is f3 = 3f1, so the fundamental frequency is f1
(420 Hz) = 140 Hz. The fifth harmonic
will have the frequency f5 = 5f1 = 700 Hz.
(b) The wavelength of the fundamental mode is λ1 = 2L = 1.20 m. The wave speed on the string is v = λ1 f1 =
(1.20 m)(140 Hz) = 168 m/s. Alternatively, the wavelength of the n = 3 mode is λ3 = (2L) = 0.40 m, from
which v = λ 3 f3 = (0.40 m)(420 Hz) = 168 m/s. The wave speed on the string, given by Equation 15.2, is
Assess: You must remember to use the linear density in SI units of kg/m. Also, the speed is the same for all
modes, but you must use a matching λ and f to calculate the speed.
P16.14. Prepare: A string fixed at both ends forms standing waves.
Solve: A simple string sounds the fundamental frequency f1 = v/(2L). Initially, when the string is of length
LA = 32.8 cm, the note has the frequency f1A = v/(2L)A . For a different length, f1C = v/(2L)C . Taking the ratio
of each side of these two equations gives
We know that the second frequency is desired to be f1B = 523 Hz. The string length must be
The question is not how long the string must be, but where must the violinist place his finger. The full string is
32.8 cm long, so the violinist must place his finger 5.2 cm from the end.
Assess: A fingering distance of 5.2 cm from the end is reasonable.
P16.22. Prepare: For an open-closed organ pipe, the length of the pipe is an integral number of quarter
wavelengths
where m = 1, 3, 5, etc. The expression
and frequency.
Solve: A general expression for the allowed frequiences is
connects speed with the wavelength
where m = 1, 3, 5, etc
For the fundamental (m=1) we have
Since all other frequencies are multiples of this frequency we have
and
Assess: These are reasonable frequencies and they are multiples of the fundamental frequency.
P16.25. Prepare: Follow Example 16.6 very closely. Assume the ear canal is an open-closed tube, for which
we need Equation 16.7
where we are given that
Take the audible range as
Solve: Plug in various values of
We assume that
in the warm ear canal.
and obtain the corresponding frequencies.
Higher frequencies are odd multiples of this fundamental.
Already
is out of the audible range. So the only one in the audible range is
P16.27. Prepare: First let’s agree that the temperature of the air in the vocal tract is a little over
C and
hence the speed of sound is 350 m/s. The relationship between speed, wavelength, and frequency for a
traveling wave disturbance in any medium is
The frequency of vibration in air is caused by and is the
same as the frequency of vibration of the vocal cords. The length of the vocal tract is an integral number of
half-wavelengths
The length of the vocal tract and hence the wavelengths that cause standing
wave resonance do not change as the diver descends. However, since the speed of the sound waves changes,
the frequency will also change.
Solve: When a sound of frequency 270 Hz is coming out of the vocal tract, the wavelength of standing waves
established in the vocal tract associated with this frequency is
As the diver descends, the vocal tract does not change length and hence this wavelength for standing wave
resonance will not change. However since the sound is now travelling through a helium-oxygen mixture with a
speed of 750 m/s, the frequency of the sound will change to
Going through the same procedure for sound at a frequence of 2300 Hz we get the frequency in the heliumoxygen mixture to be
Assess: We are aware that the sound should be at a higher frequency and the frequencies obtained have higher
values.
P16.33. Prepare: Knowing the following relationships for a vibrating string,
and
we can establish what happens to the frequency as the tension is increased.
Knowing the relationship
, we can determine the frequency of the string with the increased
tension. Solve: Combining the first three expressions above, obtain
which allows us to determine that the
frequency will increase as the tension increases. Using the expression for the beat frequency, we know that the
difference frequency between the two frequencies is 3 Hz. Combining this with our knowledge that the frequency
increases, we obtain
Assess: f1 is larger than f2 because the increased tension increases the wave speed and hence the frequency.
P16.41. Prepare: The wave on a stretched string with both ends fixed is a standing wave. We must distinguish
between the sound wave in the air and the wave on the string. The violin string oscillates at the same frequency,
because each oscillation of the string causes one oscillation of the air. But the wavelength of the standing wave
on the string is very different because the wave speed on the string is not the same as the wave speed in air.
Solve: The listener hears a sound wave of wavelength λsound = 40 cm = 0.40 m. Thus, the frequency is
Bowing a string produces sound at the string’s fundamental frequency, so the wavelength of the string is
The tension in the string is found using Equation 15.2 as follows:
Assess:
This is a relatively large tension, but still not unusually large.
P16.48. Prepare: According to Figure P16.48, a half wavelength is 400 km. From the study of standing wave
resonance we know that a standing wave goes from maximum displacement to normal water level in one fourth
of a period, which in this case is 3 hours. Finally we know that the velocity, wavelength, and frequency of a
traveling wave disturbance are related by
Solve: According to Figure P16.48, a half wavelength is 400 km, so the wavelength is =800 km. Knowing that
a standing wave goes from maximum displacement to normal water level in one fourth of a period, which in this
case is 3 hours, we get the period to be
or a frequency of
Using the relationship
we obtain a wave speed of
Assess:
m/s.
The wavelength is large and the frequency is small, however, they result in a reasonable wave velocity.
P16.50. Prepare: The standing wave is a combination of two traveling waves, which are reflecting from the
ends of the tub.
A preliminary calculation shows that
Solve:
(a) The description leads us to believe that this is the fundamental mode,
with only one node. The
wavelength will be twice the length of the tub:
(b) Use the fundamental relationship for periodic waves:
Assess: These numbers all seem to be realistic. The frequency of the earthquake waves could be 0.5 Hz, the tub
could easily be 1.4 m, and 1.4 m/s is a reasonable speed.
P16.65. Prepare: The superposition of two slightly different frequencies gives rise to beats.
Solve:
(a) The third harmonic of note A and the second harmonic of note E are
The beat frequency between the first harmonics is
The beat frequency between the second harmonics is
The beat frequency between
frequency of 2 Hz.
It therefore emerges that the tuner looks for a beat
(b) If the beat frequency is 4 Hz, then the second harmonic frequency of the E string is
Note that the second harmonic frequency of the E string could also be
fIE
This higher frequency can be ruled out because the tuner started with low tension in the E string and we know that
Assess: It would be impossible to tune a piano without a good understanding of beat frequency and harmonics.
P16.71. Prepare: We will need concepts from Chapter 15 to solve this problem. The speed of ultrasound
waves in human tissue is given in Table 15.1 as
The frequency of the reflected wave must be
that is,
and
There are a couple of mathematical paths we could take, but it is probably easiest to carefully review
Example 15.13 and use the mathematical result there, as it is very similar to our problem and gives an
expression for the speed of the source
Solve:
Assess: We may not have a good intuition about how fast heart muscles move, but 0.40 m/s seems neither too
fast nor too slow for a maximum speed.
P16.72. Prepare: Knowing that the frequency of G is 392 Hz, we can determine the frequency of its second
harmonic. Knowing that the frequency of C is 262 Hz, we can determine the frequency of its third harmonic.
Knowing these two frequencies, we can determine the beat frequency.
Solve: The frequency of the second harmonic of G is
The frequency of the third harmonic of C is
The beat frequency is
The correct choice is B.
Assess: If you have access to a set of tuning forks or a piano you can experimentally verify this result.
P16.73. Prepare: The harmonics of the G-flat will be the integer multiples of 370 Hz: 370 Hz, 740 Hz,
1110 Hz, 1480 Hz, 1850 Hz, etc.
Solve: Comparing these harmonics with those given for C in the figure shows that some of them have
differences in the range that produces dissonance. For example, 740 Hz (the second harmonic of G-flat) is 46 Hz
away from 786 Hz (the third harmonic of C), and 1110 Hz is 62 Hz (the third harmonic of G-flat) away from 1048 Hz
(the fourth harmonic of C). Those differences may not be quite maximally dissonant, but most people agree that
that musical interval is a dissonant one.
The correct answer is B.
Assess: It should be obvious that G-flat and G sounded together would also be dissonant since each
corresponding harmonic is just different enough from the corresponding harmonic for the other note to produce
dissonance.
Acoustics in general and music in particular are fascinating areas of physics. It should in no way reduce our
enjoyment of the aesthetic pleasures of music to understand the physics of it all; much rather the understanding
should increase our appreciation for the beauty of the music.
P16.74. Prepare: Knowing the fundamental frequency of G (392 Hz) and the speed of sound, we can
determine the wavelength. Knowing that when the air in the pipe is resonating at the fundamental frequency and
that the wavelength is twice the length of the pipe, we can determine the length of pipe.
Solve: The wavelength is
The length of the pipe is
Assess: This is a reasonable length of pipe. The correct choice is A.
P16.75. Prepare: A pipe that is open on one end and closed on the other only produces the odd harmonics.
See Equation 16.7 and the preceding discussion.
Solve: The odd harmonics are 262 Hz, 786 Hz, and 1310 Hz, corresponding to
and
The correct answer is B.
Assess: Because only the odd harmonics are present in an open-closed pipe, it sounds different from an openopen pipe sounded at the same fundamental frequency.