CHAPTER 10 Sound is a longitudinal mechanical wave.sph3u-bosnic.weebly.com/uploads/1/3/5/6/13569308/chapter_10.pdf · between two pins. Being a natural scientist, you likely noticed

LearningExpectations

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A thunder clap rattles your home and you feel the low rumble shake your body. Thunder is one of nature’s most dramatic demonstrations that sound is a mechanical wave. Sound waves

come to us from all directions — the low hum of a dishwasher, the raucous song of a blue jay or the drone of the TV. It is hard to imagine a world without sound. Not only does sound provide us with vital information about our environment, but it also enriches our lives through music and is used in numerous technological applications from diagnostic imaging in the body to non-destructive testing of materials (Figure 10.1).

The study of sound, also called acoustics, is an important branch of physics and of mechanical engineering. You may be interested in pursuing a career in this fascinating field.

Sound is a longitudinal mechanical wave.

C H A P T E R

10

By the end of this chapter, you will:

Relating Science to Technology, Society, and the Environment

● analyze the negative impact that mechanical waves and/or sound can have on society and the environment, and assess the effectiveness of a technology intended to reduce this impact

Developing Skills of Investigation and Communication

● plan and conduct inquiries to determine the speed of waves in a medium, compare theoretical and empirical values, and account for discrepancies

● analyze the relationship between a moving source of sound and the change in frequency perceived by a stationary observer

● predict the conditions needed to produce resonance in vibrating objects or air columns, and test their predictions through inquiry

● analyze the conditions required to produce resonance in vibrating objects and/or in air columns, and explain how resonance is used in a variety of situations

Understanding Basic Concepts

● explain and graphically illustrate the principle of superposition with respect to beat frequencies

● explain the conditions required for standing waves to occur

● explain the relationship between the speed of sound in various media and the particle nature of the media

Figure 10.1 Sound enriches our lives through music.

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Chapter 10 Sound is a longitudinal mechanical wave. 321©P

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Characteristics of Sound Waves

You may have made a toy guitar by stretching an elastic band tightly between two pins. Being a natural scientist, you likely noticed that as you changed the length or tension in the elastic band, the pitch of the sound changed. But how did the sound arise in the first place? The story of how sound is produced involves three important factors: the air that acts as the medium to transport wave motion, a vibrating string to disturb the medium, and the energy that you supplied by plucking the string (Figure 10.2). The elastic band vibrates back and forth to create a series of compressions and rarefactions in the air that race outward in all directions from the quivering elastic band.

Pressure Changes in a Sound WaveA sound wave is a travelling disturbance of compressions — regions in which air pressure rises — followed by rarefactions — regions where air pressure drops compared to quiet or still air. Figure 10.3 is an air pressure graph in which the compressions in the wave are matched with high pressure and rarefactions are matched with low pressure. You can see from this graph that sound is an example of a longitudinal mechanical wave because the wave moves in a direction parallel to the direction in which the medium moves.

The wavelength of a sound wave is the distance between successive compressions or rarefactions. This means that we can consider that each new compression or rarefaction coincides with a wave front of the sound wave (Figure 10.3). The amplitude of the sound wave is measured by how much the medium moves from its equilibrium state.

Section Summary

● Sound is an example of a longitudinal mechanical wave.

● A sound wave is a travelling disturbance of compressions and rarefactions.

● The wavelength of a sound wave is the distance between successive compressions or rarefactions.

● The speed of sound depends on the spacing between particles in a medium and the stiffness of the medium.

● Sound intensity is the energy per unit area that passes a point each second.

● In Ontario, the maximum continuous sound intensity level that workers can be exposed to for 8 h is 85 dB.

10.1

compressions

rarefactions

Figure 10.2 A sound wave is a disturbance that passes through a medium as a series of compressions (high-pressure regions) followed by rarefactions (low-pressure regions).

Figure 10.3 Sound waves are travelling regions of high and low pressure.


322 Unit D Waves and Sound ©P

The Sound SpectrumSounds come to us in an entire spectrum of frequencies. Normal sounds that you hear in conversation with your friends are in the range of 80 Hz to about 300 Hz. A trained opera singer can reach frequencies as high as 1100 Hz for brief periods of time. Musical instruments extend this range from as low as 20 Hz for the rumble of a pipe organ to as high as 5000 Hz for the shrill top notes of a piccolo. Bird songs can reach higher than 15 000 Hz. In general, your ears are able to respond to sounds in the range of 20 Hz to 20 000 Hz, but they are most sensitive to sounds in the range of about 2000 Hz to 5000 Hz. The highest and lowest frequencies that you can hear are called your upper and lower audible limit. The range from 20 Hz to 20 000 Hz is the human audible range for sounds.

Other animals have very different audible ranges. For example, whales and elephants can hear sounds of much lower frequency than humans whereas dogs and bats can hear much higher frequencies. To help identify sounds, physicists have developed a “sound spectrum” (Figure 10.4). • Infrasonic sounds have frequencies of less than 20 Hz. Rather than

being able to hear sounds in this range, you may “feel” them as a rumble that passes through your body. Elephants are able to hear sounds as low as 12 Hz.

• Audible sounds are in the range of 20 Hz to 20 kHz and can be detected by humans.

• Ultrasonic sounds have frequencies greater than 20 kHz. Dogs are able to hear sounds as high as 50 kHz.

Non-destructive TestingUltrasound is commonly used in engineering for non-destructive testing (NDT) of materials. This technique is used to detect hidden flaws within an object without damaging the object. For example, in the aircraft industry, ultrasound is beamed through a wing panel. Any small holes or fractures within the wing material scatter the sound. The scattering is detected by a microphone positioned either beneath the wing or beside the sound source. When sound is scattered, there is a detectable loss in energy received by the sound detector (Figure 10.5). Ultrasound is used because the wavelengths are small, which enables engineers to detect defects that are about the same size.

Gravity Waves Infrasound

FarInfrasound

NearInfrasound

UltrasoundAudible Sounds

(Limits of human hearing)

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Figure 10.4 The sound spectrum

Concept Check

1. What are the wavelengths of sounds in the audible range for humans?

2. How does the size of the average human ear compare to the wavelengths of sounds that are audible to humans?

3. Would you expect there to be a relationship between the audible range for an organism and the organism’s size? Give reasons for your answer.

sound source scansacross surface

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Figure 10.5 Non-destructive testing is used to detect tiny defects in materials without damaging the materials.



The Speed of SoundYou may have seen old cowboy movies in which the “good guy” puts his ear to the ground to detect approaching horses and riders. The vibrations created by the pounding hooves travel through the ground at a much higher rate than the sound that they would create in air. Why does sound have different speeds in different media?

The speed of sound through a medium depends on the physical characteristics of the medium. If you compare sound waves in a solid medium (such as steel) with those in a gaseous medium (air), you will find that they have very different speeds. The speed of sound in steel is almost 5800 m/s, whereas in air it is only about 350 m/s. At the atomic level, steel consists of many more particles that are packed close together, so they can propagate a wave very quickly. In air, the average spacing between particles is much farther apart than in steel, so sound travels at a slower speed.

Another factor that affects the speed of a wave through a medium is the forces between the particles in the medium. These forces determine the stiffness of a medium. In general, the stiffer the medium, the faster a wave will propagate in it. Stiffness is a measure of the forces that resist compression of the medium away from equilibrium. You know from having played with a party balloon that air has “stiffness.” If you compress the balloon, it bounces back. Metals have much higher stiffness than air and correspondingly higher wave speeds. Table 10.1 shows the speed of sound through different materials. In this table, the speed of sound for different gases is measured at atmospheric pressure.

Experiments show that the speed of sound in air depends on temperature and can be given by the formula

vsound � 331.6 � 0.606T

where T is the air temperature measured in degrees Celsius and the speed of sound is in m/s.

Suggested Activity● D6 Inquiry Activity Overview on

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PHYSICS•SOURCE

Medium Speed (m/s)

Air (0°C) 332

Air (20°C) 344

Carbon dioxide (0°C) 259

Hydrogen (0°C) 1284

Helium (0°C) 972

Water (25°C) 1493

Sea Water (25°C) 1533

Blood (37°C) 1570

Steel 5790

Copper 3560

Aluminium 5100

Lead 1322

Table 10.1 Speed of Sound in Various Media



Example 10.1

Early digital cameras emitted high-frequency sound pulses. The time taken for a pulse to bounce from a subject back to the camera was used to determine the distance of the subject and to adjust the focus accordingly. If an object is 6.56 m from the camera and sound travels at 344 m/s, determine the length of time it takes the emitted sound pulse to return to the camera.

Given �d � 6.56 m vsound � 344 m/s

RequiredTime for the pulse to travel to the target and back (�t)

Analysis and SolutionThe total distance travelled by the sound pulse is 2 � 6.56 m � 13.1 m.

Use the equation from chapter 1, v � �d _ �t

, to determine the time

required for the sound pulse to return to the camera.

�t � �d _ v

� 13.1 m __ 344 m/s

� 0.0381 s

ParaphraseA focusing pulse from the camera takes 0.0381 s or 38.1 ms to return to the camera.

Practice Problems1. How much time is required for

a sound pulse to travel 2.56 m and back in air at 20°C?

2. A sound returns to an autofocus camera 0.0112 s after it was emitted. How far away is the subject? Assume that the speed of sound is 344 m/s.

3. What is the speed of sound if a sound pulse travels 12.8 m in 0.0382 s?

Answers1. 0.0149 s

2. 1.93 m

3. 335 m/s

Concept Check

1. (a) How would the wavelength of a 1000-Hz sound wave change as the wave passes from air into water?

(b) Why would you expect a difference?

2. Why do you think the speed of sound depends on temperature?

3. How would increasing the temperature of air affect the speed of sound in air? Why?

Example 10.2

What is the wavelength of a sound of frequency 225 Hz that is produced in air at a temperature of 20.0°C?

Given f � 225 Hz T � 20.0°C

Requiredwavelength (λ)

Explore More

How does the speed of sound in ice differ from the speed of sound in water?

PHYSICS•SOURCE



IntensityOur ears are marvellous organs. They can respond to the faintest of whispers or the roar of a jet engine. The ability to distinguish variations in loudness is an important environmental cue that humans and other animals use to navigate in their surroundings.

Imagine a wave front created by a disturbance in air. The disturbance could be a pin dropped onto a floor. A tiny fraction of the pin’s gravitational potential energy is turned into sound energy. As you saw in the last chapter, the energy of the sound wave is spread over an expanding spherical surface. The sound wave also occurs over an interval of time. Physicists combine the ideas of energy, time, and area into the concept of intensity. The intensity of a sound is the energy per unit area that passes a point each second. It has units of (J/m2)/s or J/s·m2, which is the same as W/m2.

Our ears can respond to sounds as faint as one-trillionth of a watt per square metre (1 � 10�12 W/m2) or as loud as 100 W/m2. Table 10.2 on the next page shows you how some common sounds are related to their intensities.

Analysis and SolutionYou are asked to find wavelength. Wavelength and frequency are related by the universal wave equation, v � fλ. You are given the frequency, but not the speed of sound.

The speed of sound in air as a function of temperature (and 1.0 atm of pressure) is given by the equation

vsound � 331.6 � 0.606T

First calculate vsound and then use the universal wave equation to calculate λ.

At 20.0°C,

vsound � 331.6 � 0.606(20.0) m/s � 343.7 m/s

v � fλ

λ � v _ f

� 343.7 m/s __ 225 Hz

� 1.53 m

ParaphraseAt 20.0°C, the wavelength of a 225-Hz sound wave in air is 1.53 m.

Practice Problems1. What is the speed of sound at

�20.0°C?

2. What is the wavelength of a 225-Hz wave produced in air at �20.0°C?

3. A 1000-Hz sound wave is found to have a wavelength of 38 cm. What is the temperature of the air propagating this wave?

Answers1. 319 m/s

2. 1.42 m

3. 80°C

Concept Check

1. Suppose a 1000-Hz wave travelling in air at 340 m/s enters a cooler layer of air.(a) How will the wavelength change?(b) Will the frequency change?(c) What factor(s) determine(s) the frequency of a wave?



The Decibel Scale Knowing how to measure sound intensity is an important skill. It has many useful applications, from providing a safe working environment to designing concert halls in which performers can be heard. Table 10.2 shows that most audible sounds carry very little energy — often only a few billionths of a watt per square metre. Even though measuring sound intensity in terms of watts per square metre emphasizes the actual energy carried by waves, these numbers are small and also awkward to use in real life. A more common measurement is sound intensity level using the decibel scale. Decibels (dB) are named in honour of Alexander Graham Bell, the Canadian inventor of the telephone.

Concept Check

1. What happens to the area of a spherical wave front if you double its radius? (Hint: If necessary, look up the formula for the surface area of a sphere.)

2. If there is no absorption of sound energy by air, explain why the energy per unit area of the wave front must decrease.

3. Suppose that the intensity of a sound wave is 1 W/m2 when you are 5 m from the source. How will the intensity change when you move to a point 10 m from the source?


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PHYSICS•SOURCE

Source Intensity (W/m2) Intensity Level (dB)

Threshold of human hearing 10�12 0

Rustling leaves 10�11 10

Whisper 10�10 20

Empty theatre 10�9 30

Quiet residential area at night 10�8 40

Average home 10�7 50

Two-person conversation 10�6 60

Busy traffic — street level 10�5 70

Household vacuum cleaner 10�4 80

Loud stereo in average living room 10�3 90

Your MP3 player headphones at maximum volume

10�2 100

Front row of rock concert 10�1 110

Small, propeller aircraft taking off (30 m) 100 120

Pain threshold 101 130

Military jet taking off (50 m) 102 140

Table 10.2 Intensities and Intensity Levels of Common Sounds



The Greek letter beta, �, is commonly used to represent sound intensity levels (SIL) expressed in dB. The faintest sound that humans can hear represents the start of the decibel scale and is given a value of � � 0 dB. Table 10.2 shows how the idea of intensity (in W/m2) is related to the decibel scale. In this table, the intensity range is from 10�12 W/m2 to 102 W/m2 — a range of 100 trillion! This same range corresponds to a difference of 140 dB. Every 10-dB step in sound intensity level corresponds to a multiplication factor of 10 in actual sound intensity. For example, if two sounds differ by 20 dB, then they differ by two 10-dB steps or 10 � 10 � 100 times in sound intensity.

The decibel scale is a logarithmic scale that corresponds to how our ears perceive loudness. Figure 10.6 shows the sensitivity of the human ear to both sound intensity level and frequency. The thick wavy blue lines represent lines of equal perceived loudness. For example, sound at an intensity level of 80 dB and frequency of 50 Hz is perceived to have roughly the same loudness as a 1000-Hz sound at an intensity level of 60 dB. The shaded blue area represents the frequency and intensity range that corresponds to normal speech. This area is the range for which our ears are most sensitive.

Sound Intensity and Sound Intensity LevelFigure 10.7, on the next page, shows a graphical comparison to sound intensities measured in units of W/m2 and dB. In this graph you can see a very useful rule of thumb for relating sound intensity level in dB to the intensity measured in W/m2. A doubling in the intensity of sound results in an increase of 3 dB in SIL. If two sounds differ in intensity by 3 dB, then the louder sound is judged to be roughly twice as loud as the fainter sound.

In general, the smallest difference in loudness that can be detected by the human ear is 1 dB.

When using the decibel scale, every 3-dB increase in SIL is a doubling in intensity. A 10-dB increase increases the intensity by 10 times.

Figure 10.6 This figure shows the sensitivity of the human ear to both sound intensity level and frequency. The shaded blue area represents the frequency and intensity level range that corresponds to normal speech.

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Sound Intensity Level versus Frequency

PHYSICS INSIGHT

Sound intensity level (SIL) is measured in decibels (dB). Intensity is measured in W/m2.



1.00 � 10�11

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doublingintensity

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Sound Intensity Level versus Sound Intensity

Figure 10.7 The decibel scale versus sound intensity

Example 10.3

A firecracker explodes and makes a sound of intensity level � � 90 dB. How much more intense is this sound than that of a typical conversation? Assume that you are the same distance from the firecracker as you would be from a person in a normal conversation.Use Table 10.2 on page 326.

Given� � 90 dB

RequiredCompare the intensity of a 90-dB sound level with a normal conversation.

Analysis and SolutionTable 10.2 indicates that a normal conversation has an intensity level of 60 dB. The firecracker is 30 dB louder. For every 10-dB difference, the multiplication factor is 10. Therefore, the 90-dB-level sound is 10 � 10 � 10 � 1000 times more intense than the 60-dB-level sound.

ParaphraseThe sound wave created by a firecracker has about 1000 times the intensity of a normal conversation if each sound is heard at the same distance.

Practice Problems1. By what factor would the

sound intensity increase if the sound intensity level in an office increased from 68 dB to 77 dB?

2. A sound changes from an intensity of 5 � 10�6 W/m2 to 5 � 10�7 W/m2. Has the sound intensity level increased or decreased?

3. By how much has the sound intensity level changed in question 2? Express your answer in decibels.

Answers1. 8

2. decreased

3. 10 dB

The decibel scale is a very useful way to describe the intensity of sound, but it is not equivalent to intensity. You cannot add decibel measures together when trying to determine the combined intensity of sounds. Consider the following example.



The terms “loudness” and “intensity” do not have the same meaning. Loudness is a measure of the ear’s response to sound. Two sounds can have equal intensity, but you may hear one sound as louder than another because your ears can detect it better. The upper threshold of human hearing is 20 000 Hz, so humans cannot hear a sound of frequency 25 000 Hz. This sound may have an intensity level of 100 dB, but to us, its loudness is zero.

Noise, Music, and Hearing LossThe distinction between “noise” and “music” is a subjective one and one person’s music may be another person’s noise. Noise or music both represent energy carried by sound waves from a source to your ear. So, whether it’s noise or music, you are surrounded by sound. In some cases, the sound intensity level can be so great that it becomes a health concern.

The Ontario Occupational Health and Safety Act (1990) sets the limits on the exposure of workers to continuous sound levels at no more than 85 dB per daily eight-hour shift. All persons will experience a natural hearing loss over their lifetimes, called presbycusis. However, any exposure greater than 85 dB per eight-hour period each day can lead, over time, to permanent hearing loss, called noise-induced hearing loss.

Imagine that you are using a chain saw (Figure 10.8). A chain saw produces an average sound intensity level of 110 dB. So, without ear protection, it is harmful to your hearing to use this machine for more than a few minutes each day. Some doctors have described noise-induced hearing loss as a national epidemic.

Example 10.4

Your portable stereo is producing a sound intensity of 90 dB in the classroom. How would the sound intensity change in the classroom if your friend turned on an identical portable stereo?

Given�1 � 90 dB�2 � 90 dB

RequiredCombined sound intensity level

Analysis and SolutionCompare the intensity of the sound when one portable stereo is playing and when two portable stereos are playing.

If you combine two portable stereos, the total intensity will double. According to the decibel doubling rule, doubling the intensity of a sound increases its decibel reading by 3 dB. So, the final SIL is 90 � 3 � 93 dB.

ParaphraseTwo portable stereos, each capable of producing a 90-dB sound level, will produce a sound level of 93 dB when played together.

Practice Problems1. If one blue jay produces a

sound intensity level of 75 dB, what sound intensity level would four blue jays produce?

2. Suppose you had a choir of 10 blue jays, each with an SIL of 75 dB. What SIL would they all produce?

Answers1. 81 dB

2. 85 dB

Explore More

How is it possible for a very loud sound (140 dB) to shatter a window?

PHYSICS•SOURCE

Figure 10.8 Using a chain saw for longer than a few minutes without ear protection leads to hearing loss.



Example 10.5

You have been asked to work on a very noisy assembly line. Being concerned about your own health and the health of fellow workers, you bring a sound level meter to work and discover that the sound level is 97 dB. According to the Ontario Health and Safety Act, what is the maximum amount of time you can be exposed to sound of this intensity?

Given� � 97 dB

RequiredMaximum time of daily exposure (t)

Analysis and SolutionThis sound level is above the threshold of 85 dB, so apply the 3-dB exchange rate.

The sound level is 97 dB � 85 dB � 12 dB above the level for safe exposure in an eight-hour period. There are four steps of 3-dB increase (4 � 3 dB � 12 dB) in going from 85 dB to 97 dB. Exposure time should be cut in half for each 3-dB rise. The following table summarizes this relationship.

ParaphraseThis working environment is very unsafe. Without ear protection, you and your co-workers may, by law, only be subjected to sound of this level for a maximum of 30 minutes per day.

Practice Problems1. How long can you work

safely without ear protection in an environment producing a sound intensity level of 94 dB?

2. How long would you be able to work in a noisy shop producing a sound intensity level of 108 dB if you wear ear plugs capable of reducing the sound intensity level by 20 dB?

3. By what factor is the sound intensity reduced if ear protectors can reduce sound levels by 40 dB?

Answers1. 1 h

2. 4 h

3. 10 000

Table 10.3 Sound Intensity Level and Maximum Exposure Time

Sound Intensity Level (dB) Maximum Daily Exposure Time (h)

85 8

85 � 3 � 88 4

88 � 3 � 91 2

91 � 3 � 94 1

94 � 3 � 97 0.5

Take It Further

Did sound waves trigger the formation of the galaxies? The Wilkinson Microwave Anisotropy Probe (WMAP) has created an image of what the universe was like about 300 000 years after the Big Bang. This image shows that the universe had regions of different density. Astronomers think that these differences were caused by pressure waves in the early universe that are equivalent to sound waves. Find out more about the WMAP project.

PHYSICS•SOURCE

If workers are exposed to sound intensity levels greater than 85 dB, then the allowable time of exposure to that sound level drops. Most provinces in Canada adopt the following rule: For every 3-dB noise increase above 85 dB, workers must cut their exposure time in half. This rule is called the three-decibel exchange rate.

Besides hearing loss, excessive exposure to high levels of noise can produce other negative effects on health, such as sleep disturbance and high blood pressure.



QuestionHow fast does sound travel through air at room temperature and standard pressure?

Activity OverviewIn this activity, you will determine an average value for the speed of sound using echoes created with a microphone and a digital interface.

Prelab QuestionsConsider the questions below before beginning this activity.

1. What variables affect the speed of sound in a medium?

2. How do these variables affect the speed of sound?

Measuring the Speed of Sound by Echo Timing

Figure 10.9 Experimental setup

Inquiry Activity PHYSICS•SOURCE

REQUIRED SKILLS■ Analyzing patterns■ Designing an experimental procedure

D6

Is Your Music Too Loud?


REQUIRED SKILLS■ Analyzing patterns■ Drawing conclusions

D7

Figure 10.11 An MP3 player

QuestionAre the average sound levels produced by the portable MP3 players and ear buds used by students in your school safe?

Activity OverviewIn this activity, you will measure the sound levels produced by various students’ MP3 players by using a sound transducer and then plot a graph to determine the average sound level at which students listen to their MP3 players. You will then apply the three-decibel exchange rate to determine the maximum length of time you can safely listen to your MP3 player at this sound level. Is the sound level of your MP3 player safe?


1. What is the normal range of hearing for humans?

2. At what intensities are sounds considered loud?

Figure 10.10 A portable sound level meter


Check and Reflect


16. In an experiment to determine the speed of sound in water, the echo method was used. A loud sound pulse was created and the time for the return echo to be detected was measured. The sound pulse was created 25 m above a lake bed and the echo was detected 34.8 ms later. Determine the speed of sound in the water.

17. A bat is using echolocation to find insects at night. If the bat emits a high-frequency chirp and hears an echo 0.023 s later, how far away is the insect? The speed of sound is 344 m/s.

18. What is the maximum daily sound exposure (in hours) for a worker in a factory if the average sound intensity level is 88 dB?

19. T-waves are exceptionally pure low-frequency tones that travel through the ocean. These tones are produced by active volcanoes on the ocean floor and can produce bursts of sound that can be detected by underwater microphones. What is the wavelength of a 10-Hz T-wave if the speed of sound in water is 1530 m/s?

Reflection

20. Identify a concept that you studied in this section that you would like to learn more about.

Key Concept Review

1. Are sound waves able to travel through metals? Explain.

2. What is the normal audible frequency range for human beings?

3. In order to determine the frequency of a 2.0-m sound wave, what other piece of information would you require?

4. What is a compression? What is a rarefaction?

5. What is the sound intensity level (� ) in dB for a sound of intensity 10�8 W/m2?

6. A sound of intensity 2 � 10�8 W/m2 increases to 4 � 10�8 W/m2. How big of a change is this increase on the decibel scale?

7. At a distance of 125 m from a sound source, the intensity is 4.0 � 10�6 W/m2. What will the intensity (in W/m2) be at a distance of 375 m from the source?

8. What is noise-induced hearing loss?

9. In your own words, define sound intensity, sound intensity level, and loudness. Give the units of each term, if applicable.

Connect Your Understanding

10. What is the speed of sound in air at a temperature of 10°C?

11. What is the wavelength of a 512-Hz sound wave passing through air at room temperature (20°C)?

12. Medical ultrasounds routinely use sound waves with frequencies in the megahertz range. What is the wavelength of a 10-MHz sound wave passing through water at 25°C?

13. On a very cold day (�30°C), an echo returned in 4.00 s. How far away was the reflecting surface?

14. A cliff is located 200 m away from a climber. If the climber shouts, how long will it take the echo to return if air temperature is 20°C?

15. How long will it take sound to travel 150 m through air at 10°C?

10.1

For more questions, go to PHYSICS•SOURCE

25 m25255 m

Question 16



Resonance, Standing Waves, and Musical Sound

Section Summary

● Standing waves in a vibrating air column can produce musical notes.

● When resonance is produced in an air column closed at one end, there is an antinode at the open end.

● The fundamental frequency is the lowest frequency produced by a particular instrument.

● Overtones are frequencies above the fundamental frequency that may exist simultaneously with the fundamental frequency.

● Overtones are called harmonics when they are whole multiples of the fundamental frequency.

● Beating is the periodic pattern of constructive and destructive interference that produces a periodic change in the intensity of two combining waves.

10.2

If you sing in the shower, why does your voice sound so much richer while lathering up and belting out your favourite song? Even though you may have a lovely voice, there is something about the shower that makes your voice sound much better, much more resonant. The major reason is that sound, like all mechanical waves, can undergo resonance. In an enclosed space such as the bathroom shower, sound can produce standing waves.

This also explains why taller people usually have deeper, more resonant voices. The sound of the human voice depends, among other things, on the length of the air column or your wind-pipe. Longer wind-pipes produce lower-frequency resonances.

Standing Waves in a Vibrating Air ColumnAs a young child you likely discovered that you could make a musical note by blowing across the top of a soda-pop bottle (Figure 10.12). As you drank more of the pop and the level of pop in the bottle dropped, the note would also drop in frequency.


page 344

PHYSICS•SOURCE

Figure 10.12 The tone produced when you blow across the top of an open bottle depends on the length of the air column.



Nodes and Antinodes in Closed-Pipe Resonance In the air column, nodes are located every half-wavelength from the end at which the wave is reflected, just as they are in a standing wave in a spring (Figure 10.13(a) and (c)).

Antinodes in the air column are located one quarter-wavelength from the end of the pipe where reflection occurs, and then every half-wavelength from that point (Figure 10.13(a) and (c)). Thus, resonance isheard when the pipe is 1 _ 4 λ, 3 _ 4 λ, 5 _ 4 λ, … long. This information can be used

to measure the wavelength of sound in gases. If the frequency of the sound is known, then the wavelength can be used to calculate the speed of sound in the gas.

R

R resonance heardNR no resonance heard

NR

NR

R

Figure 10.13 A tuning fork sets up a standing wave in the air column. The volume of the sound one hears will vary depending on whether there is resonance (an antinode), (a) and (c), or no resonance, (b) and (d), in the pipe.

Concept Check

1. Look carefully at Figure 10.13 and especially at how sound waves are depicted in this diagram. What is really being shown in this figure? (Hint: Is sound a longitudinal or a transverse wave?)

2. If a sound wave is a disturbance in air, where is the compression of air maximum in this diagram? Does this area represent a node or an antinode? Explain.

3. Redraw Figure 10.13(c) in your notebook using longitudinal rather than transverse wave representations. Which representation do you prefer? Why?

Explore More

How does a pipe organ produce its sound?

PHYSICS•SOURCE When a wave source is held at the open end of a pipe, it sends down a wave that reflects from the closed end of the pipe. If the length of the pipe is just right, the reflected wave can combine with the incident wave to establish a standing wave pattern. The sound you hear depends on the length of the air column in the pipe relative to the length of the standing wave. If an antinode occurs at the open end of the pipe (Figure 10.13(a) and (c)), a point of resonance (resulting from constructive interference) occurs at the open end of the pipe and the sound appears to be amplified. This phenomenon is known as closed-pipe or closed-tube resonance. However, if the open end of the pipe does not coincide with the position of an antinode, then little sound can be heard because the incident wave (from the tuning fork) and reflected wave do not combine to produce resonance (Figure 10.13(b) and (d)).

(a) (b) (c) (d)



Example 10.6

A tuning fork with a frequency of 384 Hz is held above an air column. As the column is lengthened, a closed-pipe resonant point is found when the length of the air column is 67.5 cm. What are possible wavelengths for this data? If the speed of sound is known to be slightly greater than 300 m/s, what is the actual wavelength, and the actual speed of sound?

Givenf � 384 Hz L � 67.5 cm � 0.675 m

Requiredwavelength and speed of sound

Analysis and SolutionThe resonant point might represent 1 _

4 λ, 3 _

4 λ, 5 _

4 λ, …, etc., for this

tuning fork. Assume that 67.5 cm is the first resonant point; that

means 67.5 cm is 1 _ 4 λ. Calculate the wavelength and the speed

of sound from that data.

Assume that L � 1 _ 4 λ. Therefore,

λ � 4L v � fλ � 4(0.675 m) � (384 Hz)(2.70 m) � 2.70 m � 1037 m/s � 1.04 � 103 m/s

This value is larger than the speed of sound in air.

If the speed of sound is not of the proper order of magnitude, then assume that the resonant point is the second point of

resonance and that 67.5 cm is 3 _ 4

λ. Calculate the wavelength

and the speed of sound from that data.

Assume that L � 3 _ 4 λ. Therefore,

λ � 4L _ 3

� 4(0.675 m)

___ 3

� 0.900 m

v � fλ � (384 Hz)(0.900 m) � 345.6 m/s � 346 m/s

This is a reasonable speed for sound in air.

Complete the analysis by assuming that L � 5 _ 4

λ. Therefore,

λ � 4L _ 5

� 4(0.675 m)

___ 5

� 0.540 m

v � fλ � (384 Hz)(0.540 m) � 207.4 m/s � 207 m/s

This value is less than the speed of sound in air.

ParaphraseThe calculations for the speed of sound indicate that the data must have been for the second point of resonance. This assumption gives the speed for sound of 346 m/s. The assumption that the pipe length is for the first resonant point results in a speed about three times that of sound. The assumption that the pipe length is for the third resonant point produces a speed less than 300 m/s.

Practice Problems1. A tuning fork of frequency 512 Hz

is used to generate a standing wave pattern in a closed pipe, 0.850 m long. A strong resonant note is heard indicating that an antinode is located at the open end of the pipe. (a) What are the possible

wavelengths for this note and the corresponding speed of sound?

(b) Which wavelength will give the most reasonable value for the calculation of the speed of sound in air?

2. A tuning fork with a frequency of 256 Hz is held above a closed air column while the column is gradually increased in length. At what lengths for this air column would the first four resonant points be found, if the speed of sound is 330 m/s?

3. A standing wave is generated in a spring that is stretched to a length of 6.00 m. The standing wave pattern consists of three antinodes. If the frequency used to generate this wave is 2.50 Hz, what is the speed of the wave in the spring?

4. When a spring is stretched to a length of 8.00 m, the speed of waves in the spring is 5.00 m/s. The simplest standing wave pattern for this spring is that of a single antinode between two nodes at opposite ends of the spring. (a) What is the frequency that

produces this standing wave? (b) What is the next higher

frequency for which a standing wave exists in this spring?

Answers1. (a) 3.40 m @ 1.74 � 103 m/s; 1.13 m

@ 580 m/s; 0.680 m @ 348 m/s; 0.486 m @ 249 m/s

(b) 0.680 m

2. 0.322 m, 0.967 m, 1.61 m, 2.26 m

3. 10.0 m/s

4. 0.313 Hz, 0.625 Hz



fundamental frequencyequilibrium position

Music and Resonance in Stringed InstrumentsComplex modes of vibration give instruments their distinctive sounds and add depth to the musical tones they create. A string of a musical instrument is simply a tightly stretched spring for which the simplest standing wave possible is a single antinode with a node at either end. For this pattern, the length of the string equals one-half a wavelength and the frequency produced is called the fundamental frequency (Figure 10.14).

The fundamental frequency is the lowest frequency produced by a particular instrument. But other standing wave patterns can exist in the string at the same time as it oscillates at its fundamental frequency. By plucking or bowing a string nearer its end than its middle, the string is encouraged to vibrate with multiple frequencies. The frequencies above the fundamental frequency that may exist simultaneously with the fundamental frequency are called overtones. Figures 10.15 and 10.16 show the shape of a string vibrating in its first and second overtones, respectively. Figure 10.17 shows a violinist bowing and fingering the strings of her violin to produce notes.

Sometimes overtones are whole multiples of the fundamental note. In this case they are also called harmonics. A harmonic frequency is a resonant frequency that is a whole multiple of the fundamental resonant frequency. Vibrating strings and air columns usually produce harmonic overtones.

Figure 10.14 The fundamental frequency of a vibrating string oscillates as a standing wave with an antinode at the centre of the string.

fundamental frequency1st overtone with thefundamental frequencyequilibrium position

1st overtone without thefundamental frequencyequilibrium position

Figure 10.15 The first overtone has the form of a standing wave with two antinodes. A node exists at the midpoint of the string. The lower portion of the diagram shows a string vibrating with both the fundamental frequency and the first overtone simultaneously.



fundamental frequency2nd overtone with thefundamental frequencyequilibrium position

2nd overtone without thefundamental frequencyequilibrium position

Figure 10.16 The vibration that produces the second overtone has three antinodes. The lower portion of the diagram shows a string vibrating with both the fundamental frequency and second overtone.

Figure 10.17 The violinist’s fingering technique changes the length of the string and thus changes the fundamental frequency of vibration.

Example 10.7

A string of length 0.860 m is stretched tightly between two fixed ends. The string is plucked and a vibration is created that travels with a speed of 125 m/s along the string. (a) Sketch the fundamental wave and first two overtones produced in

the string. (b) Use your sketches to determine the fundamental frequency and

frequencies of the overtones. (c) Are the overtones harmonics of the fundamental frequency?

GivenL � 0.860 m v � 125 m/s

Required(a) sketch of the wave vibrating with the fundamental frequency and

waves vibrating at the first two overtones (b) fundamental frequency, overtone frequencies (c) Test if the overtones are harmonics.

Analysis and Solution(a) For the fundamental wave, the string is fixed at both ends, so you

know that nodes must be located at the ends. The fundamental wave has only one antinode (Figure 10.18). The next possible resonant wave pattern (first overtone) has two antinodes and one more node (Figure 10.19). The third resonant wave pattern has three antinodes and two nodes between the fixed ends (Figure 10.20).

�1

�2

0

1

2

0.6 0.80.2 0.4

A

x

L � λ12

length (m)

ampl

itude

Amplitude versus Length

Figure 10.18 The fundamental wave



(b) Apply the universal wave equation to determine frequencies.Fundamental wave: One-half of the fundamental wave fits in the length L (Figure 10.18). Therefore,

L � 1 _ 2

λ

λ � 2L � 2(0.860 m) � 1.72 m

v � fλ

f � vλ

� 125 m/s __ 1.72 m

� 72.7/s � 72.7 Hz

(c) To test if the overtones are harmonics, check if the they are whole multiples of the fundamental frequency.

The first overtone (second harmonic) is 2 � 72.7 Hz � 145 Hz. The second overtone (third harmonic) is 3 � 72.7 Hz � 218 Hz. The overtones are whole-number multiples of the fundamental frequency, so they are also harmonics.

ParaphraseThe string will vibrate with a fundamental frequency of 72.7 Hz, with the first two overtones vibrating at 145 Hz and 218 Hz.

Practice Problems1. What is the wavelength of

a wave vibrating with the fundamental frequency in a string that is stretched between two fixed points 2.40 m apart?

2. If the wave in question 1 travels with a speed of 25 m/s, determine the fundamental frequency for the wave.

3. What is the frequency of the third harmonic wave in question 1?

Answers1. 4.80 m

2. 5.2 Hz

3. 15.6 Hz

�1

�2

0

1

2

0.6 0.80.2 0.4

A

A

Nx

L � λ

length (m)

ampl

itude


Figure 10.19 The first overtone

�1

�2

0

1

2

0.6 0.80.2 0.4

A

A A

N Nx

L � λ32

length (m)

ampl

itude


Figure 10.20 The second overtone

First overtone: One full wave fits in the length L (Figure 10.19). Therefore,

λ � L � 0.860 m

f � vλ

� 125 m/s __ 0.860 m

� 145/s � 145 Hz

Second overtone: One and one-half waves fit in the length L (Figure 10.20). Therefore,

L � 3 _ 2

λ

λ � 2 _ 3

L

� 2 _ 3

(0.860 m)

� 0.573 m

f � vλ

� 125 m/s __ 0.573 m

� 218/s � 218 Hz

PHYSICS INSIGHT

Assume that the fundamental frequency is f. In physics and in music, the frequency 2f is called the first overtone; 3f is the second overtone, and so on. These frequencies are said to form a harmonic series. Thus, physicists may also refer to the fundamental frequency (f ) as the first harmonic, the frequency 2f as the second harmonic, the frequency 3f as the third harmonic, and so on.



The actual form of a vibrating string can be very complex as many overtones can exist simultaneously with the fundamental frequency. The actual wave form for a vibrating string is the result of the constructive and destructive interference of the fundamental wave with all the existing overtones that occur in the string. For example, Figure 10.21 shows the wave trace on an oscilloscope for the sound of a violin.

Wind InstrumentsWind instruments produce different musical notes by changing the lengths of air columns (Figure 10.22). For a pipe that is closed at one end and open at the other end, resonance will occur when a node is present at the closed end and an antinode is present at the open end. For a pipe that is closed at one end, the longest wavelength that can resonate is four times as long as the pipe (Figure 10.23).

If the pipe is open at both ends, then the wavelengths for which resonance occurs must have antinodes at both ends of the open pipe or open tube (Figure 10.24). The distance from one antinode to the next is one-half a wavelength. Thus, the longest wavelength that can resonate in an open pipe is twice as long as the pipe.

Figure 10.21 The interference of the fundamental frequency with the overtones produced by a bowed string creates the wave form that gives the violin its unique sound. The wavelength of the fundamental frequency is the distance between the tall sharp crests.

Explore More

When musicians tune stringed instruments, such as guitars, what are they doing? How does increasing the tension in a string affect frequency and wavelength?

PHYSICS•SOURCE

Figure 10.22 The trumpeter produces different notes by opening valves to change the instrument’s overall pipe length.

antinode

node

L � λ14 Figure 10.23 In a closed

pipe, the longest possible resonant wavelength is four times the length of the pipe.

Figure 10.24 In an open pipe, the longest possible resonant wavelength is twice the length of the pipe.

antinode

antinode

L � λ12



Wind instruments are generally open pipes. The length of the pipe determines the wavelength of the resonant frequency (Figure 10.25). In a clarinet or oboe, for example, the effective length of the pipe is changed by covering or uncovering holes at various lengths down the side of the pipe. The strongest or most resonant frequency will be the wave whose length is twice the distance from the mouthpiece to the first open hole. Overtones are also generated but the note you hear is that with the longest wavelength. As with stringed instruments, the overtones contribute to the wind instrument’s characteristic sound.

If the speed of sound in air never varied, then a given wavelength would always be associated with the same frequency. But the speed of sound changes slightly with air temperature and pressure. Thus, in the case of resonance in a pipe, the length of the pipe must be increased or decreased as the speed of sound increases or decreases to ensure that the frequency is that of the desired note.

Example 10.8

(a) Sketch the fundamental wave and the first two overtones for a vibrating closed-end air column produced in a tube that is 0.860 m long.

(b) Find the wavelengths and frequencies for these resonances. (c) Are the overtones harmonics? Use 344 m/s as the speed of sound in this case.

GivenL � 0.860 mv � 344 m/s

Required(a) Sketch the fundamental wave and the first two overtones.(b) frequencies of the fundamental wave and the first two overtones (f)(c) Test if the overtones are harmonics.

Analysis and Solution(a) Since the air column is closed at one end, you know that a node

must be located at the end. The fundamental wave has only one antinode (Figure 10.26).

�1

�2

0

1

2

0.6 0.80.2 0.4

A

N x

L � λ14

length (m)

ampl

itude


Figure 10.26 The fundamental wave

Figure 10.25 A variety of wind instruments: (a) saxophone, (b) clarinet, (c) flute

The next possible resonant wave pattern (first overtone) has two antinodes and one more node (Figure 10.27). The third resonant wave pattern (second overtone) has three antinodes and two more nodes (Figure 10.28).

(a) (b) (c)



(b) Determine the wavelength of the fundamental wave and first two overtones. Then apply the universal wave equation to determine frequencies.Fundamental wave: One-quarter of the wave fits in the length L (Figure 10.26). Therefore,

L � 1 _ 4 λ

λ � 4L � 4(0.860 m) � 3.44 m

v � fλ

f � vλ

� 344 m/s __ 3.44 m

� 100/s � 100 Hz

(c) Check if the overtones are whole multiples of the fundamental frequency. The first overtone is 300 Hz, which is 3 � 100 Hz � 300 Hz, and the second overtone is 5 � 100 Hz � 500 Hz. Therefore, the overtones are also harmonics.

ParaphraseThe air column will resonate with a fundamental frequency of 100 Hz and the first two overtones will resonate at 300 Hz and 500 Hz.

�1

�2

0

1

2

0.6 0.80.2 0.4

A

A

N Nx

L � λ34

length (m)

ampl

itude


Figure 10.27 The first overtone

�1

�2

0

1

2

0.6 0.80.2 0.4

AA

A

NN N

x

L � λ54

length (m)

ampl

itude


Figure 10.28 The second overtone

First overtone: Three-quarters of the wave fits in the length L (Figure 10.27). Therefore,

L � 3 _ 4 λ

λ � 4 _ 3 L

� 4 _ 3 (0.860 m)

� 1.15 m

v � fλ

f � vλ

� 344 m/s __ 1.15 m

� 300/s � 300 Hz

Second overtone: Five- quarters of the wave fits in the length L (Figure 10.28). Therefore,

L � 5 _ 4

λ

λ � 4 _ 5

L

� 4 _ 5

(0.860 m)

� 0.688 m

v � fλ

f � vλ

� 344 m/s __ 0.688 m

� 500/s � 500 Hz

Practice Problems1. What is the fundamental

frequency produced by a vibrating air column 2.50 m long and open at one end? The speed of sound is 344 m/s.

2. What is the frequency of the first overtone that will be produced by the air column in question 1?

3. What harmonic is the overtone in question 2?

Answers1. 34.4 Hz

2. 103 Hz

3. third



Beats and Beat FrequencyYou may have watched a piano tuner make delicate adjustments to the strings by first striking a tuning fork and then listening carefully to the same note being played on the piano. By listening and adjusting the tension in the strings, the piano tuner is bringing the piano back into peak performance condition. But what exactly is the piano tuner doing?

The tuner is exploiting one of the most basic phenomena that arises when waves of slightly different frequencies combine. This is the phenomenon of beats.

Imagine two sound waves: tone A with a frequency of 100 Hz and tone B with a frequency of 101 Hz, arriving at the same location — your ear. Further, imagine that at a specific instant of time (t � 0 s), the wave crests (compressions) of both waves match up to produce constructive interference. You will hear a sound of maximum loudness (Figure 10.30(a)).

What happens a half-second later? By this time, 1 _ 2 (100) � 50 complete waves from tone A have passed by and again a wave crest or compression from tone A is at your ear. For tone B, 1 _ 2 (101) � 50.5 waves have passed by and now a rarefaction is at your ear (Figure 10.30(b)). Now the waves are exactly a half-wavelength out of phase with each other and destructive interference occurs. You now hear a minimum or little sound. In another half-second, the two waves will again be in phase, undergo constructive interference, and produce a maximum combined sound intensity.

This pattern of minimum followed by maximum loudness is called a beat. In this case, the beat repeats once per second. Beating is the periodic pattern of constructive and destructive interference that produces a periodic change in the intensity of two combining waves.

Concept Check

1. Figure 10.29 represents a resonant wave in a closed-end air column. (a) Copy the diagram into your notebook. Indicate the places where the air is

able to move the maximum amount around equilibrium. Are these places nodes or antinodes?

(b) On the same sketch, show where the air pressure is greatest. (c) Physicists distinguish between displacement antinodes and pressure antinodes.

Based on your diagram, how do you think these two ideas are related?

�1

�2

0

1

2

6 8 102 4

length (m)

ampl

itude


Figure 10.29



t � 0.0 s

50.0 waves

50.5 waves

t � 0.5 s

Figure 10.31 This graph shows the addition of two waves and the resulting beat pattern.

Take It Further

Our Sun vibrates in millions of subtle overtones. Astrophysicists use these vibrations as a sensitive probe to understand what is happening throughout the entire Sun, from surface to centre. This image shows just one of these modes. The red areas are regions of the Sun that are sinking inward and the blue areas are moving outward. Find out more about overtones in our Sun and what they tell us.

PHYSICS•SOURCE

(a) (b)

Figure 10.30 (a) Waves from two sources arrive first in phase to produce a sound of maximum amplitude and then, (b) a half-second later, arrive completely out of phase to produce a sound of minimum amplitude.

You can summarize the phenomenon of beats in the following way:

If two waves of equal amplitude and slightly different frequency, f1 and f2, combine, they will produce a beat that has a frequency given by the expression

fbeat � |f1 � f2|

Figure 10.31 illustrates the formation of beats.



Example 10.9

Two flutists play what should be the same note — an A at 440 Hz. Unfortunately, one of the flutes is slightly out of tune and produces a tone at 443 Hz. Describe the combined sound produced by the two flutes.

Givenf1 � 440 Hzf2 � 443 Hz

Requiredcombined sound produced by the two flutes (fbeat)

Analysis and SolutionThe two notes will combine to produce a beat frequency given by the formula fbeat � |f1 � f2|.

The beat frequency will be fbeat � |440 Hz � 443 Hz|

� 3 Hz

ParaphraseWhen sounded together, the two flutes will produce a beat of 3 Hz. The sound will rise and fall in amplitude three times per second.

Practice Problems1. Two sounds of very nearly the

same frequency are sounded together. The combined sound rises and falls in intensity every 5 s. What beat frequency does this imply?

2. Two singers sing slightly off key. One sings a note of 260.0 Hz while the other sings 262.1 Hz. What is the beat frequency of their song?

3. A sound of frequency 550 Hzis combined with another sound to produce a beat of 5 Hz. What are two possible frequencies for the second sound?

Answers1. 0.2 Hz

2. 2.1 Hz

3. 555 Hz, 545 Hz

QuestionPart 1What is the speed of sound in air?

Part 2Do interference patterns exist for two in-phase sound sources?

Activity OverviewIn Part 1 of this activity, you will use resonance from a tuning fork to calculate the speed of sound and then compare your experimentally obtained value to a theoretical value. In Part 2 of this activity, you will design a lab to measure the wavelengths of sounds of known frequencies using interference patterns. You will then compare which method is better for measuring wavelengths of sound waves and determining the speed of sound.


1. What is resonance?

2. How can you calculate the speed of sound from the length of an air column?

Measuring the Speed of Sound Using Closed-pipe Resonance and Interference

tuningfork

open-ended pipetall cylinder

water

metre-stick

Figure 10.32 Activity setup for Part 1


REQUIRED SKILLS■ Analyzing patterns■ Designing an experimental procedure

D8

DI Key Activity


Chapter 10 Sound is a longitudinal mechanical wave. 345

Check and Reflect

©P

Key Concept Review

1. In terms of the length of an air column, what is the longest standing wavelength that can exist in an air column that is

(a) closed at one end? (b) open at both ends?

2. An air column is said to be closed if it is closed at one end. Consider a pipe of length L. For a standing wave in this pipe, what are the three longest wavelengths for which an antinode exists at the open end of the pipe?

3. Determine the beat frequency that you hear when a 256-Hz sound wave combines with a 260-Hz sound wave.

4. Provide the wavelengths for the first three resonant frequencies produced by a vibrating string that is 0.85 cm long.


5. A standing wave is generated in an air column closed at one end by a source that has a frequency of 768 Hz. The speed of sound in air is 325 m/s. What is the shortest column for which resonance will occur?

6. (a) A flute is essentially a tube that is open at both ends. Sketch the first and fourth standing wave patterns that could form in a flute of length 67.5 cm.

(b) What frequencies would these waves have if the flute was played at room temperature (20°C)?

7. Two tubes, each closed at one end, are sounded at the same time and each produces its fundamental note. The first tube is 1.20 m long whereas the second tube is 1.22 m long. Assume that the speed of sound is 343 m/s. Describe the sound that you will hear. (Hint: You will hear two very different sounds.)

8. Metal rods and tubes can also vibrate as standing waves. The metal tube at right is vibrating in a standing wave pattern. Although this standing wave is more complex than a standing wave in a string or air column, there are some key similarities.

(a) Redraw the figure in your notebook and locate the nodes and antinodes on this vibrating tube.

(b) Where should you hold the vibrating tube so that you can hear the sound it produces as it rings?

9. You pick up two metal rods that look the same and appear to have the same mass. However, when you strike the first rod, it produces a sound that has a much higher frequency than the second rod. What does this result tell you about the speed of sound in the two rods?

10. A source vibrating with a frequency of 150 Hz creates a standing wave in a string of length 73 cm. There are nine nodes in the string, including one at each end.

(a) What is the wavelength of the waves in the string?

(b) What is the speed of the waves in the string?

11. A trained opera singer can, by singing the right note, shatter a crystal wine glass. Explain the physics behind this effect and why it is important that the singer have a well-trained voice.

12. (a) You are holding a vibrating 1024-Hz tuning fork at the mouth of a cardboard tube that has a plunger and piston inserted into it. As you draw the plunger into the tube, the sound intensity suddenly increases and then drops again as you move past this point. Why does this happen? Assume that you are doing this experiment in a room at 20°C.

(b) Determine the position at which the sound will reach its first point of maximum loudness. Where will the next point of maximum loudness be?

Reflection

13. Which concept in this section required the greatest change in your thinking? Explain how your thinking changed.

10.2


Question 8



Motion Affects Waves

Section Summary

● The Doppler effect describes the increase in pitch and frequency of sounds moving towards you and the decrease in pitch and frequency of sounds moving away from you.

● A shock wave is formed when an object travels through a medium at a speed equal to or greater than the speed of sound in the medium.

● A sonic boom is the sound heard when a shock wave is produced.

● Aircraft that travel at supersonic speeds create shock waves.

10.3

Walking to school one morning you are deep in thought. Absentmindedly, you begin to cross the street without looking. A car horn from a passing vehicle warns you back to safety. If you listen carefully to the sound of the horn as the car approaches, you will detect a very interesting phenomenon. As the car passes you, you will hear the pitch of the horn drop rapidly. This phenomenon was explained by the Austrian physicist Christian Doppler (1803–1853) and is called the Doppler effect. The Doppler effect is the effect of motion on the frequency of sound waves from a source that is moving relative to us.

Wavelength and Frequency of a Source at RestAssume that the frequency of a source is 100 Hz and the speed of sound is 350 m/s (Figure 10.33). According to the universal wave equation, if this source is at rest, the wavelength of the sound is 3.50 m.

v � fλ

λ � v _ f

� 350 m/s __ 100 Hz

� 3.50 m


page 353

PHYSICS•SOURCE

Explore More

How does ultrasound medical imaging apply the Doppler effect?

PHYSICS•SOURCE

λ � 3.50 m

vw � 350 m/s

Figure 10.33 When a wavelength of 3.50 m travels toward you at a speed of 350 m/s, you hear sound that has a frequency of 100 Hz (diagram not to scale).



You hear the sound at a frequency of 100 Hz because at a speed of 350 m/s, the time lapse between crests that are 3.50 m apart is 1 _ 100 s.

If, however, the wavelengths that travel toward you were 7.0 m long, the time lapse between successive crests would be 1 _ 50 s — a frequency equal to 50 Hz.

v � fλ

f � vλ

� 350 m/s __ 7.0 m

� 50 Hz

Wavelength and Frequency of a Moving Source Imagine that the source generating the 100-Hz sound is moving toward you at a speed of 70 m/s. Assume that the source is at point A when it generates a crest (Figure 10.34). While the first crest moves a distance of 3.5 m toward you, the source also moves toward you. The distance the source moves while it generates one wavelength is the distance the source travels in 1 _ 100 s at 70 m/s, or 0.7 m. Because of the motion of the source, the next crest is generated (at point B) only 3.5 m � 0.7 m � 2.8 m behind the first crest. As long as the source continues at the speed of 70 m/s toward you, the crests travelling in your direction will be only 2.8 m apart. Hence, for a car moving toward you, the sound waves emitted by the car will be “squashed together” and thus reach you more frequently than if the car were stationary.

If waves that are 2.8 m long travel toward you at a speed of 350 m/s, then the frequency of the sound arriving at your ear will be 125 Hz. The pitch of the sound that you hear will be increased because the source is moving toward you.

f � vλ

� 350 m/s __ 2.80 m

� 125 Hz

directionof motion

2.8 m

0.7 m

A B C D

4.2 m

wave front generatedby source at position Aby source at position Bby source at position Cby source at position D

Figure 10.34 When a sound source moves toward you, the wavelengths in the direction of motion are decreased.



At the same time, along a line in the direction opposite to the motion of the source, the wavelengths are increased by the same amount that the waves in front of the source are shortened. For the 100-Hz sound source moving at 70 m/s, the waves behind the source are increased by 0.7 m, to a length of 3.5 m � 0.7 m � 4.2 m. The time lapse between these crests, which are 4.2 m apart and travelling at 350 m/s, is 0.012 s. Therefore, the perceived frequency in the direction opposite to the motion of the source is about 83 Hz. The pitch of the sound has been lowered.

Analysis of the Doppler EffectIf the velocity of the sound waves in air is vw, then the wavelength (λs) that a stationary source (s) with a frequency of fs generates is given by

λs � vw _ f s

The key to Doppler’s analysis is to calculate the distance the source moves in the time required to generate one wavelength — the period, Ts, of the source. If the source is moving at speed vs, then in the period Ts, the source moves a distance �ds that is given by

�ds � vsTs

By definition, Ts � 1 _ fs

, so

�ds � vs _ fs

Sources Moving Toward YouFor sources that are moving toward you, �ds is the distance by which the wavelengths are shortened. Subtracting �ds from λs gives the lengths of the waves (λd) that reach the listener. Therefore,

λd � λs � �ds

Replacing λs and �ds by their equivalent forms gives

λd � vw _ fs

� vs _ fs

λd � vw � vs __

fs

This wavelength is the apparent wavelength (Doppler wavelength) of the sound generated by a source that is moving toward you at a speed vs. Dividing the speed of the waves (vw) by the Doppler wavelength (λd) produces the Doppler frequency (fd) of the sound that you hear as the source approaches you. Therefore,

fd � vw

λd

� vw __

(vw � vs) __

fs

� vw ( fs __ vw � vs )

� ( vw __ vw � vs ) fs

is the Doppler frequency when the source is approaching the listener.



Concept Check

Figures 10.35(a) and (b) show data collected by a student performing an experimental test of the Doppler effect. In the experiment, the student attached a very loud buzzer producing a 3410-Hz tone to a long piece of string and then swung the buzzer rapidly overhead in a horizontal circular path. A microphone attached to a computer interface was used to record the sound of the buzzer. The student heard a rapid variation in the pitch of the buzzer. The graphs shown below are plots of the frequency or frequencies produced by the buzzer. The x-axis shows frequency and the y-axis shows the relative strength of the frequency or frequencies present in the sound. Figure 10.35(a) shows the frequency recorded when the buzzer was at rest and Figure 10.35(b) shows the frequencies when the buzzer was moving.

1. Figure 10.35(a) shows just one very distinct frequency at 3410 Hz whereas Figure 10.35(b) shows frequencies in a range from 3280 Hz to 3450 Hz. (a) Why are the frequencies spread out in Figure 10.35(b) but not in Figure 10.35(a)?(b) What parts of Figure 10.35(b) correspond to times when the buzzer was

approaching the microphone?(c) If the student were to swing the buzzer even faster, how would the

appearance of Figure 10.35(b) change?

3200 3300 3400Frequency (Hz)

Amplitude versus Frequency

Ampl

itude

3500 3600 3200 3300 3400Frequency (Hz)

Amplitude versus Frequency

Ampl

itude

3500 3600

Figure 10.35(a) Figure 10.35(b)

(a) (b)

Sources Moving Away from YouIf the source is moving away from the listener, the value of �ds is added to the value of λs, giving

λd � λs � �ds

If you replace λs and �ds by their equivalent forms and substitute into

the equation fd � vw

λd to find fd, the Doppler frequency for a sound where

the source moves away from the listener is given by

fd � ( vw __ vw � vs ) fs

The equations for the Doppler effect are usually written as a single equation of the form

fd � ( vw __ vw � vs ) fs



Example 10.10

A train is travelling at a speed of 30.0 m/s. Its whistle generates a sound wave with a frequency of 224 Hz. You are standing beside the tracks as the train passes you with its whistle blowing. What change in frequency do you detect for the pitch of the whistle as the train passes, if the speed of sound in air is 330 m/s?

Given fs � 224 Hz vw � 330 m/s vs � 30.0 m/s

Required(a) Doppler frequency for the whistle as the train approaches (fd)(b) Doppler frequency for the whistle as the train moves away (fd)(c) change in frequency (�f)

Analysis and SolutionUse the equations for Doppler shifts to find the Doppler frequencies of the whistle.(a) For the approaching whistle, (b) For the receding whistle,

fd � ( vw __ vw � vs ) fs fd � ( vw __ vw � vs

) fs

fd � ( 330 m _ s ___

330 m _ s � 30.0 m _ s ) 224 Hz fd � ( 330 m _ s

___ 330 m _ s � 30.0 m _ s

) 224 Hz

� ( 330 m _ s __

300 m _ s ) 224 Hz � ( 330 m _ s

__ 360 m _ s

) 224 Hz

� 246.4 Hz � 205.3 Hz � 246 Hz � 205 Hz

(c) The change in pitch is the difference in the two frequencies. Therefore, the pitch change is

�f � 246.4 Hz � 205.3 Hz � 41.1 Hz

ParaphraseAs the train passes, the pitch of its whistle is lowered by a frequency of 41.1 Hz.

The Moving Listener What happens if the sound source remains fixed and you move? Imagine that you are sitting in a high-speed train that is moving at 70 m/s away from a 100-Hz sound source. The wave crests from the source have to catch up to you. Assume that the speed of sound in still air is 350 m/s. As the wave crests reach you, they will still be 3.5 m apart. However, their pitch will be lower because the effective speed of the wave crests is not 350 m/s but 350 m/s � 70 m/s � 280 m/s.

Practice Problems1. You are crossing in a

crosswalk when an approaching driver blows his horn. If the true frequency of the horn is 264 Hz and the car is approaching you at a speed of 60.0 km/h, what is the apparent (or Doppler) frequency of the horn? Assume that the speed of sound in air is 340 m/s.

2. An airplane is approaching at a speed of 360 km/h. If you measure the pitch of its approaching engines to be 512 Hz, what must be the actual frequency of the sound of the engines? The speed of sound in air is 345 m/s.

3. An automobile is travelling toward you at a speed of 25.0 m/s. When you measure the frequency of its horn, you obtain a value of 260 Hz. If the actual frequency of the horn is known to be 240 Hz, calculate vw, the speed of sound in air.

4. As a train moves away from you, the frequency of its whistle is determined to be 475 Hz. If the actual frequency of the whistle is 500 Hz and the speed of sound in air is 350 m/s, what is the train’s speed?

Answers1. 278 Hz

2. 364 Hz

3. 325 m/s

4. 18.4 m/s



If you apply the universal wave equation,

v � fλ

f � v _ λ

� 280 m/s __ 3.5 m

� 80 Hz

If you compare this frequency with the moving source example discussed earlier, you can see that the Doppler shift (83 Hz) is almost the same.

The Sound Barrier and Sonic BoomsTo help make doing the supper dishes a bit more interesting, you may occasionally snap your tea towel with a rapid back-and-forth flick. The distinct snap you hear happens because the tip of the tea towel is moving faster than the speed of sound. Similarly, in the case of the much louder crack of a whip, high-speed photography confirms that the tip of the whip is travelling at over 400 m/s. Both the tea towel and the whip are producing shock waves. A shock wave is formed whenever an object travels through a medium at a speed equal to or greater than the speed of sound in the medium. The sound that you hear when a shock wave is produced is also called a sonic boom.

Why Shock Waves OccurAir is an elastic medium and any disturbance will quickly ripple away as a sound wave. In the previous section on the Doppler effect, you saw that if a source is moving, the wave crests begin to “bunch up” in the direction of motion. Figure 10.36 shows what happens if the source is now moving at the same speed as sound or faster than the speed of sound.

When a wave source moves at a speed equal to or greater than the speed of sound, the wave crests begin to pile up. Constructive interference occurs. The energy from the wave crests is added together to form a wave front that contains much more energy than each of the many wave fronts from which it forms. The resulting wave is a shock wave. When a source moves much faster than the speed of sound, the shock wave spreads out in a cone that travels along with the source (Figure 10.36(c)).

Slower than speed of sound:Pressure waves move out around plane.

At speed of sound:Pressure waves at noseform a shock wave.

At supersonic speed:Shock waves form a cone,resulting in a sonic boom.

(a) (b) (c)

Figure 10.36 Wave fronts from a moving source: (a) The source is moving at less than the speed of sound. (b) The source is moving at the speed of sound. (c) The source is moving faster than the speed of sound.

Explore More

How do some animals use sound to hunt, and as a weapon?

PHYSICS•SOURCE

Suggested Activity● D10 Decision-Making Analysis

Overview on page 353

PHYSICS•SOURCE

PHYSICS INSIGHT

Mach number refers to speed that is a multiple of the speed of sound. An object moving at Mach 1 is moving at the speed of sound, an object moving at Mach 2 is moving at twice the speed of sound, etc.



Some jet aircraft are capable of travelling at supersonic speeds. Supersonic speeds are speeds greater than the speed of sound. As the jet approaches the speed of sound, the pressure in front of the jet increases as the shock wave begins to form (Figure 10.37). In the 1940s, when jet aircraft capable of travelling at the speed of sound were being developed, some people speculated that this increased pressure would destroy the aircraft. The term “sound barrier” was coined. Chuck Yeager, flying the Bell X-1, became the first human to break the sound barrier (Figure 10.38). Although the pressure does increase as the shock wave forms, it has a relatively minor effect on the aircraft. What is startling is the loud sonic boom that occurs when the aircraft reaches the sound barrier.

Sonic booms are more than just loud noises. If a sonic boom from a jet aircraft were to occur at low altitude over a populated area, the shock wave produced would be capable of shattering windows and causing structural damage to buildings. For this reason, Transport Canada prohibits aircraft from flying at supersonic speeds in Canadian airspace.

Figure 10.37 When conditions are right, the change in pressure produced by the airplane’s wings can cause sufficient cooling of the atmosphere so that a cloud forms. The extreme conditions present when a jet is travelling near the speed of sound often result in the type of cloud seen in this photo.

Take It Further

Angiodynography is a sophisticated diagnostic test used by doctors to measure the rate of blood flow in arteries. Ultrasonic sound waves are directed into arteries and the reflected sound waves are captured by a sensitive microphone. Find out more about this technique and how it uses the physics of the Doppler effect.

PHYSICS•SOURCE

Anatomy of a Sonic BoomSonic booms are created by the piling up of sound waves. Sound waves consist of wave crests, or regions of high pressure, followed by rarefactions, or regions of low pressure. A sonic boom consists of a series of one or more pairs of high-pressure wave crests (compressions) followed by low-pressure rarefactions. The air pressure suddenly rises by 100 N/m2 or 100 Pa above normal air pressure (1.0 � 105 Pa) as the compressions form and very rapidly drops to 100 N/m2 below normal air pressure as the rarefactions pass by. If the sonic boom occurs near a window, the window will experience a sudden inward pressure of 100 N/m2 followed by a sudden outward pressure of the same magnitude. Such sudden changes in pressure can easily shatter the window. The pressure difference that can shatter a window is only 1/1000 of normal air pressure.

Figure 10.38 In 1947, flying the experimental Bell X-1 aircraft (nick-named Glamourous Glennis in honour of his wife), Chuck Yaeger became the first person to break the sound barrier.



QuestionHow would the Doppler effect change the frequency of the sound of a buzzer that is moving in a circular path relative to a listener?

Activity OverviewYou will demonstrate the Doppler effect by spinning an electric buzzer and recording the sounds using a computer interface.


1. Where relative to the microphone would you expect the frequency of the spinning buzzer to be the greatest? the least?

2. How can you calculate the speed at which the buzzer is spinning?

Demonstrating the Doppler Effect


REQUIRED SKILLS■ Analyzing patterns■ Drawing conclusions

D9

IssueExposure to loud sounds for long periods of time is a serious health concern. Workplaces as well as schools must conform to strict guidelines when it comes to sound levels. How safe is your school? Are there some areas of your school, such as the cafeteria, industrial arts room or music room, in which students or staff could be exposed to unsafe noise levels? These are some of the questions that you should consider as you decide whether or not your school has provided you with a safe acoustic environment.

Activity OverviewIn this activity, you will carry out sound level measurements in areas of your school where you think sound levels may be too high. You will then decide whether the problem can be corrected by • changing the location of equipment or people,• installing sound absorbing materials, or• using protective devices such as sound cancellation headphones

or ear plugs.

You will then report your findings to your school’s administration.


1. Which areas of your school do you suspect may be acoustically unsafe?

2. How could these areas be made acoustically safer?

Is Your School an Acoustically Safe Environment?

Decision-Making Analysis PHYSICS•SOURCE

REQUIRED SKILLS■ Gathering information■ Stating a conclusion

D10

Figure 10.40 How acoustically safe is your school?

Figure 10.39 Student spinning a buzzer overhead in a horizontal circular path. The microphone positioned at the left records the sound of the buzzer.


Check and Reflect


9. In an ultrasound test, a doctor directs a 2.0-MHz sound wave at the heart of a fetus. The tiny heart is beating with a maximum speed of 0.25 m/s. Explain how the Doppler effect could be used to measure the speed with which the heart beats.

10. Astronomers have discovered that the light from some stars has been “stretched” in wavelength, or shifted toward the red end of the light spectrum. This phenomenon is called red shift and is usually explained using the Doppler effect.

(a) If the light from a star is red shifted, what can you conclude about the motion of the star?

(b) Sometimes the light from stars is shifted toward the blue end of the spectrum (blue shifted). What can you conclude about the motion of stars that are blue shifted?

11. Are beats the result of the Doppler effect or does the formation of beats have more to do with interference? Give reasons for your answer.

12. As you go upward in the troposphere, the temperature drops. At an altitude of 10 km(typical for commercial jets), the air temperature is around �50°C. At what speed would a plane break the sound barrier if it flew at this altitude?

13. The speed of sound through water is around 1500 m/s, yet waves called bow waves form in front of motor boats travelling only a few metres per second. Why does this suggest that there are two different kinds of waves involved?

14. How would astronomers use the Doppler effect to detect the rotation of stars?

Reflection

15. Describe to a classmate one misconception you had about a concept in this section before you read this section. Explain what you know about this concept now.

Key Concept Review

1. A car horn emits a 400-Hz tone when it is at rest. How would the frequency of the car horn change from your perspective if the car was approaching you?

2. What causes the Doppler effect?

3. Two sound sources have the same frequency when at rest. If they are both moving away from you, how could you tell if one was travelling faster than the other?

4. Explain the cause of a sonic boom.


5. The siren of a police car has a frequency of 660 Hz. If the car is travelling toward you at 40.0 m/s, what do you perceive to be the frequency of the siren? The speed of sound in air is 340 m/s.

6. A police car siren has a frequency of 850 Hz. If you hear this siren to have a frequency that is 40.0 Hz greater than its true frequency, what was the speed of the car? The speed of sound is 350 m/s.

7. A jet, travelling at the speed of sound (Mach 1), emits a sound wave with a frequency of 1000 Hz. Use the Doppler effect equations to calculate the frequency of this sound as the jet first approaches you, then moves away from you. Explain what these answers mean in terms of what you would hear as the jet moved toward, then past, you.

8. Two trains are in a train yard. One train is stationary (as are you) while the other train is moving away from you with a speed of 30 km/h. Each train has a 400-Hz whistle. The trains blow their whistles at the same time.

(a) What frequency of sound will you hear from each train?

(b) Will there be a beat frequency? If so, what is it? (Assume that the speed of sound is 340 m/s in this case.)

10.3



Great CANADIANS in Physics Alexander Graham Bell

Physics CAREERS Medical Sonographer

Alexander Graham Bell (1847–1922) is one of Canada’s greatest scientists. Born in Scotland, he moved to Canada at the age of 22 to a homestead near what is now Brantford, Ontario. It was here that Bell set up his first laboratory, where he pursued a lifelong passion of understanding electricity, sound, and the human voice. During this time, he taught himself to speak the Mohawk language and helped devise written symbols for it. Bell also worked with the hearing impaired and helped develop a form of visible speech. He also worked with his father on devising educational programs for persons with hearing impairment. One of his most famous students was Helen Keller.

Although Bell is best known for his 1876 invention of the telephone, he also did pioneering work in energy generation (including the development of the forerunner to solar cells), the invention of the metal detector, hydrofoils, and in aeronautics, the development of the famed Silver Dart — the first Canadian aircraft. Bell held 18 patents for his inventions. He spent his later years living at Baddeck, Nova Scotia.

Figure 10.41 Alexander Graham Bell

To find out more, visit PHYSICS•SOURCE

Diagnostic medical imaging is a huge field and offers many different career paths. Ultrasound is the creation of diagnostic medical images through the use of sound waves. Knowledge of waves and sound is an important part of the training of a medical sonographer.

During an ultrasound scan, high-frequency sound waves are passed into the patient’s body. The reflected waves are collected by equipment that the sonographer operates. The images can be recorded as still images or as a video for later review by the treating physician. During the ultrasound, sonographers look for problem areas that they record for the physician. They also evaluate the quality of the images and edit them so the physician is sent the most relevant images.

Sonographers can specialize in a variety of different areas. Ophthalmologic sonographers take images of eyes. Abdominal sonographers take images of the pancreas, spleen, gallbladder, liver, and kidneys. Neurosonographers scan the brain. Obstetric and gynecologic sonographers scan the female reproductive system. Echocardiographers scan the heart and blood vessels.

To become a sonographer usually takes 2 –4 years of study at a college or university. Sonographers are employed at hospitals, private offices, and private imaging centres.

Figure 10.42 A medical sonographer at work



356 Unit D Waves and Sound

CHAPTER REVIEWC H A P T E R

©P

Key Concept Review

1. Why can a standing wave be generated only by what is defined as resonant frequency? k

2. How does the speed of sound through rock compare to the speed of sound through air? Why are the speeds different? k

3. Explain how two high-frequency sounds can be combined to produce a third and much lower frequency. k

4. How are the waves in the direction of a source’s motion affected as the speed of the source increases? k

5. Why is it incorrect to add SIL values directly? For example, why is 50 dB � 50 dB not equal to 100 dB? k

6. What is the intensity of a sound wave that delivers 1.50 � 10�7 W over an area of 15 m2? k

7. Which of the following sounds fall outside the range for human hearing? k

(a) whale “moan”, 12 Hz (b) an ocean T-wave, 5 Hz (c) sound of a fracture in rock, 5400 Hz (d) bat chirp, 17 000 Hz (e) acoustic distance finder, 44.1 kHz


8. Does the Doppler effect apply only to sound or can it apply to any form of wave motion? Explain. t

9. How much more intense is a 90-dB sound than a 70-dB sound? t

10. As measured from centre ice, a single wildly cheering hockey fan can produce a sound intensity level of 60 dB. What sound intensity level would be produced by 20 000 wildly cheering fans? t

11. A powerful explosion occurred at a petrochemical refinery. Home owners living several kilometres away reported feeling the tremor several seconds before hearing the explosion. Why? t

12. If sound waves travel through the ground with an average speed of 6150 m/s and a powerful explosion occurs 5.00 km away, how much time will elapse between when you feel the vibration from the explosion and hear the explosion? Use v � 344 m/s for the speed of sound in air. a

13. Why is the intensity of an echo always less than the sound that produces it? t

14. Ground crew at a large international airport may be subjected to sound intensity levels of 120 dB for as long as 2 h per day and must, therefore, wear ear protectors. What is the minimum acceptable sound absorption (in dB) that the ear protectors must provide for these workers? a

15. The term ultrasound means that the frequency is higher than the highest frequencies our ears can detect (about 20 kHz). Animals can often hear sounds that, to our ears, are ultrasound. For example, a dog whistle has a frequency of 22 kHz. If the speed of sound in air is 350 m/s, what is the wavelength of the sound generated by this whistle? a

16. A violin string is 33.0 cm long. The thinnest string on the violin is tuned to vibrate at a frequency of 659 Hz.

(a) What is the wave velocity in the string? a (b) If you place your finger on the string so

that its length is shortened to 28.0 cm, what is the frequency of the note that the string produces? a

17. (a) What is the shortest closed pipe for which resonance is heard when a tuning fork with a frequency of 426 Hz is held at the open end of the pipe? The speed of sound in air is 335 m/s. a

(b) What is the length of the next longest pipe that produces resonance? a

18. The horn on a car has a frequency of 290 Hz. If the speed of sound in air is 340 m/s and the car is moving toward you at a speed of 72.0 km/h, what is the apparent frequency of the sound? a

19. How fast is a sound source moving toward you if you hear the frequency to be 580 Hz when the true frequency is 540 Hz? The speed of sound in air is 350 m/s. Express your answer in km/h. a

10


Chapter 10 Review 357©P

20. (a) If the speed of sound in air is 350 m/s, how fast would a sound source need to travel away from you if the frequency that you hear is to be one-half the true frequency? a

(b) What would you hear if this sound source were moving toward you? a

21. When a police car is at rest, the wavelength of the sound from its siren is 0.550 m. If the car is moving toward you at a speed of 120 km/h, what is the frequency at which you hear the siren? Assume that the speed of sound is 345 m/s. a

22. (a) If the speed of sound in air is 350 m/s, how fast must a sound source move toward you if the frequency that you hear is twice the true frequency of the sound? a

(b) What frequency would you hear if this sound source had been moving away from you? a

23. For sound to reflect effectively from an object, the wavelength of the sound should be smaller than the size of the object. Bats are able to locate tiny insects by using echolocation. Estimate the frequency that a bat might use to locate insects. Would you expect this frequency to be in the audible range for humans? You will need to estimate the size of an insect to answer this question. t

24. What are the first three resonant frequencies that a tube with a length of 45 cm that is closed at both ends will produce in air at a temperature of 20°C? k

25. Suppose that you used the same tube in the previous question but used an unknown gas as the medium rather than air. Now the resonant frequencies become 333 Hz, 666 Hz, and 1000 Hz. What is the speed of sound in the mystery gas? t

26. You ordered large bottles of nitrogen, carbon dioxide, and helium for your lab. Unfortunately, when the bottles arrived, the labels had come off. How could you set up a simple experiment using the principle of resonance to determine which gas is which? Describe the test you could perform to distinguish between the gases. t

27. An acoustic range finder is a tool used in the lab or in industry to measure distances. The frequency typically used is 44.1 kHz.

(a) Explain why this frequency limits the precision of distance measurements to about 7 mm. t

(b) What effect would temperature have on the accuracy of the measurements? t

28. An explosion produces a sound pulse with an acoustic energy of 126 J. If this energy travelled outward as a spherical wave pulse, determine the following:

(a) the energy per square metre that the pulse would deliver at a distance of 100 m from the explosion (Recall that the formula for the surface area of a sphere is 4 r2, where r is the radius.) t

(b) If the explosion and resulting sound pulse happened within 0.01 s, what is the intensity of the sound at a distance of 100 m, in units of W/m2? t

(c) What is the SIL for the explosion at this distance? t

29. When radar (radio) waves were reflected from a swarm of insects, the frequency of the reflected waves was higher than the frequency of the original waves. What does this information tell you about the motion of the insects? a

Reflection

30. Which concept in this chapter did you find the most challenging? What would help you to understand this concept better? c

In the Unit Task, you will be measuring and comparing the

range over which you hear sound. What is the frequency

range of the sounds that most people will hear? What

is the relationship between the frequency and intensity

of a sound? What range of frequencies do people with

presbycusis typically have difficulty hearing?

Unit Task Link


ACHIEVEMENT CHART CATEGORIESk Knowledge and understanding t Thinking and investigation

c Communication a Application


Documents

CHAPTER 10 Sound is a longitudinal mechanical wave.sph3u-bosnic.weebly.com/uploads/1/3/5/6/13569308/chapter_10.pdf · between two pins. Being a natural scientist, you likely noticed