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characteristic The sound of any instrument is referred to as the quality of that sound. What is it about the sound from the tuba that allows us to distinguish between it and the sound from a ute?
14
OUTLINE
CHAPTER
14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10 14.11 14.12 14.13
Producing a Sound Wave Characteristics of Sound Waves The Speed of Sound Energy and Intensity of Sound Waves Spherical and Plane Waves The Doppler Effect Interference of Sound Waves Standing Waves Forced Vibrations and Resonance Standing Waves in Air Columns Beats Quality of Sound The Ear
Patrick Ward/CORBIS
Sound
Sound waves are the most important example of longitudinal waves. In this chapter we discuss the characteristics of sound waves: how they are produced, what they are, and how they travel through matter. We then investigate what happens when sound waves interfere with each other. The insights gained in this chapter will help you understand how we hear.
14.1 PRODUCING A SOUND WAVE
Whether it conveys the shrill whine of a jet engine or the soft melodies of a crooner, any sound wave has its source in a vibrating object. Musical instruments produce sounds in a variety of ways. The sound of a clarinet is produced by a vibrating reed, the sound of a drum by the vibration of the taut drumhead, the sound of a piano by vibrating strings, and the sound from a singer by vibrating vocal cords. Sound waves are longitudinal waves traveling through a medium, such as air. In order to investigate how sound waves are produced, we focus our attention on the tuning fork, a common device for producing pure musical notes. A tuning fork consists of two metal prongs, or tines, that vibrate when struck. Their vibration disturbs the air near them, as shown in Figure 14.1. (The amplitude of vibration of the tine shown in the gure has been greatly exaggerated for clarity.) When a tine swings to the right, as in Figure 14.1a, the molecules in an element of air in front of its movement are forced closer together than normal. Such a region of high molecular density and high air pressure is called a compression. This compression
458
14.2
Characteristics of Sound Waves
459
moves away from the fork like a ripple on a pond. When the tine swings to the left, as in Figure 14.1b, the molecules in an element of air to the right of the tine spread apart, and the density and air pressure in this region are then lower than normal. Such a region of reduced density is called a rarefaction (pronounced rare a fak shun). Molecules to the right of the rarefaction in the gure move to the left. The rarefaction itself therefore moves to the right, following the previously produced compression. As the tuning fork continues to vibrate, a succession of compressions and rarefactions forms and spreads out from it. The resultant pattern in the air is somewhat like that pictured in Figure 14.2a. We can use a sinusoidal curve to represent a sound wave, as in Figure 14.2b. Notice that there are crests in the sinusoidal wave at the points where the sound wave has compressions and troughs where the sound wave has rarefactions. The compressions and rarefactions of the sound waves are superposed on the random thermal motion of the atoms and molecules of the air (discussed in Chapter 10), so sound waves in gases travel at about the molecular rms speed.
High-density region
(a) Low-density region
14.2 CHARACTERISTICS OF SOUND WAVES
As already noted, the general motion of elements of air near a vibrating object is back and forth between regions of compression and rarefaction. This back-andforth motion of elements of the medium in the direction of the disturbance is characteristic of a longitudinal wave. The motion of the elements of the medium in a longitudinal sound wave is back and forth along the direction in which the wave travels. By contrast, in a transverse wave, the vibrations of the elements of the medium are at right angles to the direction of travel of the wave.
(b)
Royalty-Free/Corbis
(c)
Categories of Sound Waves
Sound waves fall into three categories covering different ranges of frequencies. Audible waves are longitudinal waves that lie within the range of sensitivity of the human ear, approximately 20 to 20 000 Hz. Infrasonic waves are longitudinal waves with frequencies below the audible range. Earthquake waves are an example. Ultrasonic waves are longitudinal waves with frequencies above the audible range for humans and are produced by certain types of whistles. Animals such as dogs can hear the waves emitted by these whistles.
Figure 14.1 A vibrating tuning fork. (a) As the right tine of the fork moves to the right, a high-density region (compression) of air is formed in front of its movement. (b) As the right tine moves to the left, a low-density region (rarefaction) of air is formed behind it. (c) A tuning fork.
Applications of Ultrasound
Ultrasonic waves are sound waves with frequencies greater than 20 kHz. Because of their high frequency and corresponding short wavelengths, ultrasonic waves can be used to produce images of small objects and are currently in wide use in medical applications, both as a diagnostic tool and in certain treatments. Internal organs can be examined via the images produced by the reection and absorption of ultrasonic waves. Although ultrasonic waves are far safer than x-rays, their images dont always have as much detail. Certain organs, however, such as the liver and the spleen, are invisible to x-rays but can be imaged with ultrasonic waves. A P P L I C AT I O N Medical Uses of Ultrasound
(a)
(b)
Figure 14.2 (a) As the tuning fork vibrates, a series of compressions and rarefactions moves outward, away from the fork. (b) The crests of the wave correspond to compressions, the troughs to rarefactions.
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Direction of vibration Electrical connections Crystal Figure 14.3 An alternating voltage applied to the faces of a piezoelectric crystal causes the crystal to vibrate.
Medical workers can measure the speed of the blood ow in the body with a device called an ultrasonic ow meter, which makes use of the Doppler effect (discussed in Section 14.6). The ow speed is found by comparing the frequency of the waves scattered by the owing blood with the incident frequency. Figure 14.3 illustrates the technique that produces ultrasonic waves for clinical use. Electrical contacts are made to the opposite faces of a crystal, such as quartz or strontium titanate. If an alternating voltage of high frequency is applied to these contacts, the crystal vibrates at the same frequency as the applied voltage, emitting a beam of ultrasonic waves. At one time, a technique like this was used to produce sound in nearly all headphones. This method of transforming electrical energy into mechanical energy, called the piezoelectric effect, is reversible: If some external source causes the crystal to vibrate, an alternating voltage is produced across it. A single crystal can therefore be used to both generate and receive ultrasonic waves. The primary physical principle that makes ultrasound imaging possible is the fact that a sound wave is partially reected whenever it is incident on a boundary between two materials having different densities. If a sound wave is traveling in a material of density i and strikes a material of density t , the percentage of the incident sound wave intensity reected, PR, is given by PR
i i t t 2
100
An ultrasound image of a human fetus in the womb.
A P P L I C AT I O N Ultrasonic Ranging Unit for Cameras
This equation assumes that the direction of the incident sound wave is perpendicular to the boundary and that the speed of sound is approximately the same in the two materials. The latter assumption holds very well for the human body because the speed of sound doesnt vary much in the organs of the body. Physicians commonly use ultrasonic waves to observe fetuses. This technique presents far less risk than do x-rays, which deposit more energy in cells and can produce birth defects. First the abdomen of the mother is coated with a liquid, such as mineral oil. If this were not done, most of the incident ultrasonic waves from the piezoelectric source would be reected at the boundary between the air and the mothers skin. Mineral oil has a density similar to that of skin, and a very small fraction of the incident ultrasonic wave is reected when i t . The ultrasound energy is emitted in pulses rather than as a continuous wave, so the same crystal can be used as a detector as well as a transmitter. An image of the fetus is obtained by using an array of transducers placed on the abdomen. The reected sound waves picked up by the transducers are converted to an electric signal, which is used to form an image on a uorescent screen. Difculties such as the likelihood of spontaneous abortion or of breech birth are easily detected with this technique. Fetal abnormalities such as spina bida and water on the brain are also readily observed. A relatively new medical application of ultrasonics is the cavitron ultrasonic surgical aspirator (CUSA). This device has made it possible to surgically remove brain tumors that were previously inoperable. The probe of the CUSA emits ultrasonic waves (at about 23 kHz) at its tip. When the tip touches a tumor, the part of the tumor near the probe is shattered and the residue can be sucked up (aspirated) through the hollow probe. Using this technique, neurosurgeons are able to remove brain tumors without causing serious damage to healthy surrounding tissue. Ultrasound is also used to break up kidney stones that are otherwise too large to pass. Previously, invasive surgery was more often required. Another interesting application of ultrasound is the ultrasonic ranging unit used in some cameras to provide an almost instantaneous measurement of the distance between the camera and the object to be photographed. The principal component of this device is a crystal that acts as both a loudspeaker and a microphone. A pulse of ultrasonic waves is transmitted from the transducer to the object, which then reects part of the signal, producing an echo that is detected by the device. The time interval between the outgoing pulse and the detected echo is electronically converted to a distance, because the speed of sound is a known quantity.
Bernard Benoit/Photo Researchers, Inc.
14.3
The Speed of Sound
461
14.3 THE SPEED OF SOUND
The speed of a sound wave in a uid depends on the uids compressibility and inertia. If the uid has a bulk modulus B and an equilibrium density , the speed of sound in it is v
B
[14.1]
Speed of sound in a uid
Equation 14.1 also holds true for a gas. Recall from Chapter 9 that the bulk modulus is dened as the ratio of the change in pressure, P, to the resulting fractional change in volume, V/V : B P V/V [14.2]
B is always positive because an increase in pressure (positive P) results in a decrease in volume. Hence, the ratio P/ V is always negative. Its interesting to compare Equation 14.1 with Equation 13.18 for the speed of transverse waves on a string, v F/ , discussed in Chapter 13. In both cases, the wave speed depends on an elastic property of the medium (B or F ) and on an inertial property of the medium ( or ). In fact, the speed of all mechanical waves follows an expression of the general form v
elastic property inertial property
Another example of this general form is the speed of a longitudinal wave in a solid rod, which is v
Y
[14.3]
where Y is the Youngs modulus of the solid (see Eqn. 9.3), and is its density. This expression is valid only for a thin, solid rod. Table 14.1 lists the speeds of sound in various media. The speed of sound is much higher in solids than in gases, because the molecules in a solid interact more strongly with each other than do molecules in a gas. Striking a long steel rail with a hammer, for example, produces two sound waves, one moving through the rail and a slower wave moving through the air. A student with an ear pressed against the rail rst hears the faster sound moving through the rail, then the sound moving through air. In general, sound travels faster through solids than liquids and faster through liquids than gases, although there are exceptions. The speed of sound also depends on the temperature of the medium. For sound traveling through air, the relationship between the speed of sound and temperature is v (331 m/s)
TABLE 14.1
Speeds of Sound in Various Media
Medium Gases Air (0 C) Air (100 C) Hydrogen (0 C) Oxygen (0 C) Helium (0 C) Liquids at 25 C Water Methyl alcohol Sea water Solids Aluminum Copper Iron Lead Vulcanized rubber v (m/s) 331 386 1 290 317 972 1 490 1 140 1 530 5 100 3 560 5 130 1 320 54
T 273 K
[14.4]
where 331 m/s is the speed of sound in air at 0 C and T is the absolute (Kelvin) temperature. Using this equation the speed of sound in air at 293 K (a typical room temperature) is approximately 343 m/s.
Quick Quiz 14.1
Which of the following actions will increase the speed of sound in air? (a) decreasing the air temperature (b) increasing the frequency of the sound (c) increasing the air temperature (d) increasing the amplitude of the sound wave (e) reducing the pressure of the air.
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Applying Physics 14.1 The Sounds Heard During a Storm
How does lightning produce thunder, and what causes the extended rumble? Explanation Assume that youre at ground level, and neglect ground reections. When lightning strikes, a channel of ionized air carries a large electric current from a cloud to the ground. This results in a rapid temperature increase of the air in the channel as the current moves through it, causing a similarly rapid expansion of the air. The expansion is so sudden and so intense that a tremendous disturbance is produced in the air thunder. The entire length of the channel produces the sound at essentially the same instant of time. Sound produced at the bottom of the channel reaches you rst, because thats the point closest to you. Sounds from progressively higher portions of the channel reach you at later times, resulting in an extended roar. If the lightning channel were a perfectly straight line, the roar might be steady, but the zigzag shape of the path results in the rumbling variation in loudness, with different quantities of sound energy from different segments arriving at any given instant.
INTERACTIVE EXAMPLE 14.1 Sound Waves in Various Media
Goal Calculate and compare the speeds of sound in different media. Problem (a) If a solid bar of aluminum 1.00 m long is struck at one end with a hammer, a longitudinal pulse propagates down the bar. Find the speed of sound in the bar, which has a Youngs modulus of 7.0 1010 Pa and a density of 2.7 103 kg/m3. (b) Calculate the speed of sound in ethyl alcohol, which has a density of 806 kg/m3 and bulk modulus of 1.0 109 Pa. (c) Compute the speed of sound in air at 35.0 C. Strategy Substitute the given values into the appropriate equations. Solution (a) Compute the speed of sound in an aluminum bar. Substitute values into Equation 14.3: vAl
Y
7.0 1010 Pa 2.7 103 kg/m3
5 100 m/s, or about 11 000 mi/h ! (b) Compute the speed of sound in ethyl alcohol. Substitute values into Equation 14.1: (c) Compute the speed of sound in air at 35.0 C. Substitute values into Equation 14.4: v (331 m/s) v
B
1.0 109 Pa 806 kg/m3
1.1
103 m/s
(273 K 35.0 K) 273 K
352 m/s
Remark The speed of sound in aluminum is dramatically higher than in either liquid alcohol or air. Exercise 14.1 Compute the speed of sound in the following substances at 273 K: (a) lead (Y (B 2.8 1010 Pa), and (c) air at 15.0 C. Answers (a) 1.2 103 m/s (b) 1.4 103 m/s (c) 322 m/s 1010 Pa), (b) mercury
1.6
You can compare the speeds of sound through various media by logging into PhysicsNow at www.cp7e.com and going to Interactive Example 14.1.
14.4
Energy and Intensity of Sound Waves
463
14.4 ENERGY AND INTENSITY OF SOUND WAVES
As the tines of a tuning fork move back and forth through the air, they exert a force on a layer of air and cause it to move. In other words, the tines do work on the layer of air. The fact that the fork pours sound energy into the air is one of the reasons the vibration of the fork slowly dies out. (Other factors, such as the energy lost to friction as the tines bend, are also responsible for the lessening of movement.) The average intensity I of a wave on a given surface is dened as the rate at which energy ows through the surface, E/ t, divided by the surface area A: I 1 A E t [14.5]
where the direction of energy ow is perpendicular to the surface at every point. SI unit: watt per meter squared (W/m2) A rate of energy transfer is power, so Equation 14.5 can be written in the alternate form I power area [14.6]
Intensity of a wave
A
where is the sound power passing through the surface, measured in watts, and the intensity again has units of watts per square meter. The faintest sounds the human ear can detect at a frequency of 1 000 Hz have an intensity of about 1 10 12 W/m2. This intensity is called the threshold of hearing. The loudest sounds the ear can tolerate have an intensity of about 1 W/m2 (the threshold of pain). At the threshold of hearing, the increase in pressure in the ear is approximately 3 10 5 Pa over normal atmospheric pressure. Because atmospheric pressure is about 1 105 Pa, this means the ear can detect pressure uctuations as small as about 3 parts in 1010! The maximum displacement of an air molecule at the threshold of hearing is about 1 10 11 m a remarkably small number! If we compare this displacement with the diameter of a molecule (about 10 10 m), we see that the ear is an extremely sensitive detector of sound waves. The loudest sounds the human ear can tolerate at 1 kHz correspond to a pressure variation of about 29 Pa away from normal atmospheric pressure, with a maximum displacement of air molecules of 1 10 5 m.
Intensity Level in Decibels
The loudest tolerable sounds have intensities about 1.0 1012 times greater than the faintest detectable sounds. The most intense sound, however, isnt perceived as being 1.0 1012 times louder than the faintest sound, because the sensation of loudness is approximately logarithmic in the human ear. (For a review of logarithms, see Section A.3, heading G, in Appendix A.) The relative intensity of a sound is called the intensity level or decibel level, dened by 10 log I I0 [14.7]
Intensity level
The constant I 0 1.0 10 12 W/m2 is the reference intensity, the sound intensity at the threshold of hearing I is the intensity, and is the corresponding intensity
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TIP 14.1 Intensity Versus Intensity Level
Dont confuse intensity with intensity level. Intensity is a physical quantity with units of watts per meter squared; intensity level, or decibel level, is a convenient mathematical transformation of intensity to a logarithmic scale.
level measured in decibels (dB). (The word decibel, which is one-tenth of a bel, comes from the name of the inventor of the telephone, Alexander Graham Bell (1847 1922). To get a feel for various decibel levels, we can substitute a few representative numbers into Equation 14.7, starting with I 1.0 10 12 W/m2 : 10 log 1.0 1.0 10 10
12
W/m2 12 W/m2
10 log(1)
0 dB
From this result, we see that the lower threshold of human hearing has been chosen to be zero on the decibel scale. Progressing upward by powers of ten yields 10 log 10 log 1.0 1.0 1.0 1.0 10 10 10 10
11 12 10 12
TABLE 14.2
Intensity Levels in Decibels for Different Sources
Source of Sound Nearby jet airplane Jackhammer, machine gun Siren, rock concert Subway, power mower Busy trafc Vacuum cleaner Normal conversation Mosquito buzzing Whisper Rustling leaves Threshold of hearing (dB) 150 130 120 100 80 70 50 40 30 10 0
W/m2 W/m2
10 log(10) 10 log(100)
10 dB 20 dB
W/m2 W/m2
Notice the pattern: Multiplying a given intensity by ten adds 10 db to the intensity level. This pattern holds throughout the decibel scale. For example, a 50-dB sound is 10 times as intense as a 40-dB sound, while a 60-dB sound is 100 times as intense as a 40-dB sound. On this scale, the threshold of pain (I 1.0 W/m2) corresponds to an intensity level of 10 log(1/1 10 12) 10 log(1012) 120 dB. Nearby jet airplanes can create intensity levels of 150 dB, and subways and riveting machines have levels of 90 to 100 dB. The electronically amplied sound heard at rock concerts can attain levels of up to 120 dB, the threshold of pain. Exposure to such high intensity levels can seriously damage the ear. Earplugs are recommended whenever prolonged intensity levels exceed 90 dB. Recent evidence suggests that noise pollution, which is common in most large cities and in some industrial environments, may be a contributing factor to high blood pressure, anxiety, and nervousness. Table 14.2 gives the approximate intensity levels of various sounds.
EXAMPLE 14.2 A Noisy Grinding Machine
Goal Work with watts and decibels. Problem A noisy grinding machine in a factory produces a sound intensity of 1.00 10 5 W/m2. Calculate (a) the decibel level of this machine, and (b) the new intensity level when a second, identical machine is added to the factory. (c) A certain number of additional such machines are put into operation alongside these two. When all the machines are running at the same time the decibel level is 77.0 dB. Find the sound intensity. Strategy Parts (a) and (b) require substituting into the decibel formula, Equation 14.7, with the intensity in part (b) twice the intensity in part (a). In part (c), the intensity level in decibels is given, and its necessary to work backwards, using the inverse of the logarithm function, to get the intensity in watts per meter squared. Solution (a) Calculate the intensity level of the single grinder. Substitute the intensity into the decibel formula: 10 log 70.0 dB (b) Calculate the new intensity level when an additional machine is added. Substitute twice the intensity of part (a) into the decibel formula: 10 log 2.00 1.00 10 10
5 12
1.00 1.00
10 10
W/m2 12 W/m2
5
10 log(107)
W/m2 W/m2
73.0 dB
14.5
Spherical and Plane Waves
465
(c) Find the intensity corresponding to an intensity level of 77.0 dB. Substitute 77.0 dB into the decibel formula and divide both sides by 10: 7.70 Make each side the exponent of 10. On the right-hand side, 10log u u , by denition of base ten logarithms. 77.0 dB log 10 log
12
I I0
10
I W/m2 I 10
107.70 I
5.01 5.01
107 10
5
1.00 W/m2
12
W/m2
Remark The answer is ve times the intensity of the single grinder, so in part (c) there are ve such machines operating simultaneously. Because of the logarithmic denition of intensity level, large changes in intensity correspond to small changes in intensity level. Exercise 14.2 Suppose a manufacturing plant has an average sound intensity level of 97.0 dB created by 25 identical machines. (a) Find the total intensity created by all the machines. (b) Find the sound intensity created by one such machine. (c) Whats the sound intensity level if ve such machines are running? Answers (a) 5.01 10
3
W/m2 (b) 2.00
10
4
W/m2 (c) 90.0 dB
Federal regulations now demand that no ofce or factory worker be exposed to noise levels that average more than 90 dB over an 8-h day. From a management point of view, heres the good news: one machine in the factory may produce a noise level of 70 dB, but a second machine, while doubling the total intensity, increases the noise level by only 3 dB. Because of the logarithmic nature of intensity levels, doubling the intensity doesnt double the intensity level; in fact, it alters it by a surprisingly small amount. This means that equipment can be added to the factory without appreciably altering the intensity level of the environment. Now heres the bad news: as you remove noisy machinery, the intensity level isnt lowered appreciably. In Exercise 14.2, reducing the intensity level by 7 dB would require the removal of 20 of the 25 machines! To lower the level another 7 dB would require removing 80% of the remaining machines, in which case only one machine would remain.
A P P L I C AT I O N OSHA Noise-Level Regulations
14.5 SPHERICAL AND PLANE WAVES
If a small spherical object oscillates so that its radius changes periodically with time, a spherical sound wave is produced (Fig. 14.4, page 466). The wave moves outward from the source at a constant speed. Because all points on the vibrating sphere behave in the same way, we conclude that the energy in a spherical wave propagates equally in all directions. This means that no one direction is preferred over any other. If av is the average power emitted by the source, then at any distance r from the source, this power must be distributed over a spherical surface of area 4 r 2, assuming no absorption in the medium. (Recall that 4 r 2 is the surface area of a sphere.) Hence, the intensity of the sound at a distance r from the source is I average power area
av av
A
4 r2
[14.8]
This equation shows that the intensity of a wave decreases with increasing distance from its source, as you might expect. The fact that I varies as 1/r 2 is a result of the
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Spherical wave front
Wave front r1 r2 Source
l
Ray
Figure 14.4 A spherical wave propagating radially outward from an oscillating sphere. The intensity of the wave varies as 1/r 2.
Figure 14.5 Spherical waves emitted by a point source. The circular arcs represent the spherical wave fronts concentric with the source. The rays are radial lines pointing outward from the source, perpendicular to the wavefronts.
Rays
assumption that the small source (sometimes called a point source) emits a spherical wave. (In fact, light waves also obey this so-called inverse-square relationship.) Because the average power is the same through any spherical surface centered at the source, we see that the intensities at distances r 1 and r 2 (Fig. 14.4) from the center of the source are
Wave fronts Figure 14.6 Far away from a point source, the wave fronts are nearly parallel planes and the rays are nearly parallel lines perpendicular to the planes. Hence, a small segment of a spherical wavefront is approximately a plane wave.
av av
I1
4 r 12
I2
4 r 22
The ratio of the intensities at these two spherical surfaces is I1 I2 r 22 r 12
[14.9]
Plane wave front
y
x v z
Figure 14.7 A representation of a plane wave moving in the positive x-direction with a speed v. The wavefronts are planes parallel to the yz-plane.
Its useful to represent spherical waves graphically with a series of circular arcs (lines of maximum intensity) concentric with the source representing part of a spherical surface, as in Figure 14.5. We call such an arc a wave front. The distance between adjacent wave fronts equals the wavelength . The radial lines pointing outward from the source and perpendicular to the arcs are called rays. Now consider a small portion of a wave front that is at a great distance (relative to ) from the source, as in Figure 14.6. In this case, the rays are nearly parallel to each other and the wave fronts are very close to being planes. At distances from the source that are great relative to the wavelength, therefore, we can approximate the wave front with parallel planes, called plane waves. Any small portion of a spherical wave that is far from the source can be considered a plane wave. Figure 14.7 illustrates a plane wave propagating along the x-axis. If the positive x-direction is taken to be the direction of the wave motion (or ray) in this gure, then the wave fronts are parallel to the plane containing the y- and z-axes.
14.6
The Doppler Effect
467
EXAMPLE 14.3 Intensity Variations of a Point Source
Goal Relate sound intensities and their distances from a point source. Problem A small source emits sound waves with a power output of 80.0 W. (a) Find the intensity 3.00 m from the source. (b) At what distance would the intensity be one-fourth as much as it is at r 3.00 m? (c) Find the distance at which the sound level is 40.0 dB. Strategy The source is small, so the emitted waves are spherical and the intensity in part (a) can be found by substituting values into Equation 14.8. Part (b) involves solving for r in Equation 14.8 followed by substitution (though Equation 14.9 can be used instead). In part (c), convert from the sound intensity level to the intensity in W/m2, using Equation 14.7. Then substitute into Equation 14.9 (though 14.8 could be used, instead) and solve for r 2. Solution (a) Find the intensity 3.00 m from the source. Substitute 14.8:
av
80.0 W and r
3.00 m into Equation
I
av
4 r2
80.0 W 4 (3.00 m)2
0.707 W/m2
(b) At what distance would the intensity be one-fourth as much as it is at r 3.00 m? Take I 14.8: (0.707 W/m2)/4, and solve for r in Equation r
av 1/2
4I
80.0 W 4 (0.707 W/m2)/4.0
1/2
6.00 m
(c) Find the distance at which the sound level is 40.0 dB. Convert the intensity level of 40.0 dB to an intensity in W/m2 by solving Equation 14.7 for I: 40.0 104.00 Solve Equation 14.9 for r 22, substitute the intensity and the result of part (a), and take the square root: I1 I2 r 22 r2 10 log I I0 r 22 r 12 : I I0 I r 22 : 4.00 log 1.00 I I0 10
8
104.00I 0 r 12
W/m2
:
(3.00 m)2 2.52
I1 I2 0.707 W/m2 1.00 10 8 W/m2
104 m
Remarks Once the intensity is known at one position a certain distance away from the source, its easier to use Equation 14.9 rather than Equation 14.8 to nd the intensity at any other location. This is particularly true for part (b), where, using Equation 14.9, we can see right away that doubling the distance reduces the intensity to onequarter its previous value. Exercise 14.3 Suppose a certain jet plane creates an intensity level of 125 dB at a distance of 5.00 m. What intensity level does it create on the ground directly underneath it when ying at an altitude of 2.00 km? Answer 73.0 dB
14.6 THE DOPPLER EFFECT
If a car or truck is moving while its horn is blowing, the frequency of the sound you hear is higher as the vehicle approaches you and lower as it moves away from you. This is one example of the Doppler effect, named for the Austrian physicist Christian Doppler (1803 1853), who discovered it. The same effect is heard if youre on a motorcycle and the horn is stationary: the frequency is higher as you approach the source and lower as you move away.
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Chapter 14
vO
Sound
v Observer O
lS
Source S vS = 0
Although the Doppler effect is most often associated with sound, its common to all waves, including light. In deriving the Doppler effect, we assume that the air is stationary and that all speed measurements are made relative to this stationary medium. The speed vO is the speed of the observer, vS is the speed of the source, and v is the speed of sound.
ACTIVE FIGURE 14.8 An observer moving with a speed vO toward a stationary point source (S ) hears a frequency fO that is greater than the source frequency fS .
Case 1: The Observer Is Moving Relative to a Stationary Source
In Active Figure 14.8 an observer is moving with a speed of vO toward the source (considered a point source), which is at rest (vS 0). We take the frequency of the source to be fS , the wavelength of the source to be S , and the speed of sound in air to be v. If both observer and source are stationary, the observer detects fS wave fronts per second. (That is, when vO 0 and vS 0, the observed frequency fO equals the source frequency fS .) When moving toward the source, the observer moves a distance of vO t in t seconds. During this interval, the observer detects an additional number of wave fronts. The number of extra wave fronts is equal to the distance traveled, vO t, divided by the wavelength S : Additional wave fronts detected vO t
S
Log into PhysicsNow at www.cp7e.com and go to Active Figure 14.8 to adjust the speed of the observer.
Divide this equation by the time t to get the number of additional wave fronts detected per second, vO/ S . Hence, the frequency heard by the observer is increased to fO
vO
fS
vO
S
Substituting
S
v/fS into this expression for fO we obtain fO fS v v vO [14.10]
Source S vS = 0
v
lS
Observer O
Figure 14.9 An observer moving with a speed of vO away from a stationary source hears a frequency fO that is lower than the source frequency fS .
When the observer is moving away from a stationary source (Fig. 14.9), the observed frequency decreases. A derivation yields the same result as Equation 14.10, but with v vO in the numerator. Therefore, when the observer is moving away from the source, substitute vO for vO in Equation 14.10.
Case 2: The Source Is Moving Relative to a Stationary Observer
Now consider a source moving toward an observer at rest, as in Active Figure 14.10. Here, the wave fronts passing observer A are closer together because the source is moving in the direction of the outgoing wave. As a result, the wavelength O measured by observer A is shorter than the wavelength S of the source at rest. During each vibration, which lasts for an interval T (the period), the source moves a distance v ST vS /f S and the wavelength is shortened by that amount. The observed wavelength is therefore given by
O S
TIP 14.2 Doppler Effect Doesnt Depend on Distance
The sound from a source approaching at constant speed will increase in intensity, but the observed (elevated) frequency will remain unchanged. The Doppler effect doesnt depend on distance.
vS fS v vS fS v v fS vS fS
Because
S
v/fS , the frequency observed by A is fO v
O S
v
or fO fS v vS [14.11]
As expected, the observed frequency increases when the source is moving toward the observer. When the source is moving away from an observer at rest, the minus sign in the denominator must be replaced with a plus sign, so the factor becomes (v vS ).
14.6
The Doppler Effect
469
Observer B S
O
vS
Observer A
(a)
(b)
ACTIVE FIGURE 14.10 (a) A source S moving with speed vS toward stationary observer A and away from stationary observer B. Observer A hears an increased frequency, and observer B hears a decreased frequency. (b) The Doppler effect in water, observed in a ripple tank. The source producing the water waves is moving to the right.
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 14.10 to adjust the speed of the source.
General Case
When both the source and the observer are in motion relative to Earth, Equations 14.10 and 14.11 can be combined to give fO fS v v vO vS [14.12]
Doppler shift equation observer and source in motion
In this expression, the signs for the values substituted for vO and vS depend on the direction of the velocity. When the observer moves toward the source, a positive speed is substituted for vO ; when the observer moves away from the source, a negative speed is substituted for vO. Similarly, a positive speed is substituted for vS when the source moves toward the observer, a negative speed when the source moves away from the observer. Choosing incorrect signs is the most common mistake made in working a Doppler effect problem. The following rules may be helpful: The word toward is associated with an increase in the observed frequency; the words away from are associated with a decrease in the observed frequency. These two rules derive from the physical insight that when the observer is moving toward the source (or the source toward the observer), there is a smaller observed period between wave crests, hence a larger frequency, with the reverse holding a smaller observed frequency when the observer is moving away from the source (or the source away from the observer). Keep the physical insight in mind whenever youre in doubt about the signs in Equation 14.12: Adjust them as necessary to get the correct physical result. The second most common mistake made in applying Equation 14.12 is to accidentally reverse numerator and denominator. Some nd it helpful to remember the equation in the following form: fO v vO v fS vS
The advantage of this form is its symmetry: both sides are very nearly the same, with Os on the left and Ss on the right. Forgetting which side has the plus sign and which has the minus sign is not a serious problem, as long as physical insight is used to check the answer and make adjustments as necessary.
Courtesy of Educational Development Center, Newton, MA.
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Quick Quiz 14.2
Suppose youre on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzzer over the side while the balloon is rising at constant speed, what can you conclude about the sound you hear as the buzzer falls toward the ground? (a) the frequency and intensity increase, (b) the frequency decreases and the intensity increases, (c) the frequency decreases and the intensity decreases, or (d) the frequency remains the same, but the intensity decreases.
Applying Physics 14.2 Out-of-Tune Speakers
Suppose you place your stereo speakers far apart and run past them from right to left or left to right. If you run rapidly enough and have excellent pitch discrimination, you may notice that the music playing seems to be out of tune when youre between the speakers. Why? Explanation When you are between the speakers, you are running away from one of them and toward the other, so there is a Doppler shift downward for the sound from the speaker behind you and a Doppler shift upward for the sound from the speaker ahead of you. As a result, the sound from the two speakers will not be in tune. A calculation shows that a world-class sprinter could run fast enough to generate about a semitone difference in the sound from the two speakers.
EXAMPLE 14.4 Listen, but Dont Stand on the Track
Goal Solve a Doppler shift problem when only the source is moving. Problem A train moving at a speed of 40.0 m/s sounds its whistle, which has a frequency of 5.00 102 Hz. Determine the frequency heard by a stationary observer as the train approaches the observer. The ambient temperature is 24.0 C. Strategy Use Equation 14.4 to get the speed of sound at the ambient temperature, then substitute values into Equation 14.12 for the Doppler shift. Because the train approaches the observer, the observed frequency will be larger. Choose the sign of vS to reect this fact. Solution Use Equation 14.4 to calculate the speed of sound in air at T 24.0 C: v (331 m/s) (331 m/s) The observer is stationary, so vO 0. The train is moving toward the observer, so vS 40.0 m/s (positive). Substitute these values and the speed of sound into the Doppler shift equation: v v
T 273 K (273 24.0)K 273 K 345 m/s
fO
fS
vO vS 102 Hz) 345 m/s 345 m/s 40.0 m/s
(5.00 566 Hz
Remark If the train were going away from the observer, vS
40.0 m/s would have been chosen instead.
Exercise 14.4 Determine the frequency heard by the stationary observer as the train recedes from the observer. Answer 448 Hz
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The Doppler Effect
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INTERACTIVE EXAMPLE 14.5 The Noisy Siren
Goal Solve a Doppler shift problem when both the source and observer are moving.
Problem An ambulance travels down a highway at a speed of 75.0 mi/h, its siren emitting sound at a frequency of 4.00 102 Hz. What frequency is heard by a passenger in a car traveling at 55.0 mi/h in the opposite direction as the car and ambulance (a) approach each other and (b) pass and move away from each other? Take the speed of sound in air to be v 345 m/s. Strategy Aside from converting from mi/h to m/s, this problem only requires substitution into the Doppler formula, but two signs must be chosen correctly in each part. In part (a), the observer moves toward the source and the source moves toward the observer, so both vO and vS should be chosen to be positive. Switch signs after they pass each other. Solution Convert the speeds from mi/h to m/s: 0.447 m/s 1.00 mi/h 0.447 m/s 1.00 mi/h
vS vO
(75.0 mi/h) (55.0 mi/h)
33.5 m/s 24.6 m/s
(a) Compute the observed frequency as the ambulance and car approach each other. Each vehicle goes toward the other, so substitute vO 24.6 m/s and vS 33.5 m/s into the Doppler shift formula: fO fS v v vO vS 102 Hz) 345 m/s 345 m/s 24.6 m/s 33.5 m/s 475 Hz
(4.00 (b) Compute the observed frequency as the ambulance and car recede from each other. Each vehicle goes away from the other, so substitute vO 24.6 m/s and vS 33.5 m/s into the Doppler shift formula: fO fS v v
vO vS 102 Hz) 345 m/s 345 m/s ( 24.6 m/s) ( 33.5 m/s)
(4.00 339 Hz
Remarks Notice how the signs were handled: In part (b), the negative signs were required on the speeds because both observer and source were moving away from each other. Sometimes, of course, one of the speeds is negative and the other is positive.
Exercise 14.5 Repeat this problem, but assume the ambulance and car are going the same direction, with the ambulance initially behind the car. The speeds and the frequency of the siren are the same as in the example. Find the frequency heard by the observer in the car (a) before and (b) after the ambulance passes the car. [Note: The highway patrol subsequently gives the driver of the car a ticket for not pulling over for an emergency vehicle!] Answers (a) 411 Hz (b) 391 Hz
You can alter the relative speeds of two submarines and observe the Doppler-shifted frequency by logging into PhysicsNow at www.cp7e.com and going to Interactive Example 14.5.
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Conical wave front 0 1 vt
vS
1973 Kim Vandiver & Harold E. Edgerton/Courtesy of Palm Press, Inc.
2 Sn S1 S2
S0
vS t
(a)
(b)
Figure 14.11 (a) A representation of a shock wave, produced when a source moves from S0 to Sn with a speed vs that is greater than the wave speed v in that medium. The envelope of the wave fronts forms a cone with half-angle of sin v/vs . (b) A stroboscopic photograph of a bullet moving at supersonic speed through the hot air above a candle.
Shock Waves
What happens when the source speed vS exceeds the wave velocity v? Figure 14.11a describes this situation graphically. The circles represent spherical wave fronts emitted by the source at various times during its motion. At t 0, the source is at point S 0 , and at some later time t, the source is at point Sn . In the interval t, the wave front centered at S0 reaches a radius of vt. In this same interval, the source travels to Sn , a distance of vs t. At the instant the source is at Sn , the waves just beginning to be generated at this point have wave fronts of zero radius. The line drawn from Sn to the wave front centered on S 0 is tangent to all other wave fronts generated at intermediate times. All such tangent lines lie on the surface of a cone. The angle between one of these tangent lines and the direction of travel is given by v vs
sin
Figure 14.12 The V-shaped bow wave of a boat is formed because the boat travels at a speed greater than the speed of the water waves. A bow wave is analogous to a shock wave formed by an airplane traveling faster than sound.
The ratio vs /v is called the Mach number. The conical wave front produced when vs v (supersonic speeds) is known as a shock wave. Figure 14.11b is a photograph of a bullet traveling at supersonic speed through the hot air rising above a candle. Notice the shock waves in the vicinity of the bullet. Another interesting example of a shock wave is the V-shaped wave front produced by a boat (the bow wave) when the boats speed exceeds the speed of the water waves (Fig. 14.12). Jet aircraft and space shuttles traveling at supersonic speeds produce shock waves that are responsible for the loud explosion, or sonic boom, heard on the ground. A shock wave carries a great deal of energy concentrated on the surface of the cone, with correspondingly great pressure variations. Shock waves are unpleasant to hear and can damage buildings when aircraft y supersonically at low altitudes. In fact, an airplane ying at supersonic speeds produces a double boom, because two shock waves are formed one from the nose of the plane and one from the tail (Fig. 14.13).
1994, Comstock
14.7
Interference of Sound Waves
473
Pressure
Keith Lawson/Bettmann/Corbis
Atmospheric pressure (a) (b) Figure 14.13 (a) The two shock waves produced by the nose and tail of a jet airplane traveling at supersonic speed. (b) A shock wave due to a jet traveling at the speed of sound is made visible as a fog of water vapor. The large pressure variation in the shock wave causes the water in the air to condense into water droplets.
Quick Quiz 14.3
As an airplane ying with constant velocity moves from a cold air mass into a warm air mass, does the Mach number (a) increase, (b) decrease, or (c) remain the same?
14.7 INTERFERENCE OF SOUND WAVES
Sound waves can be made to interfere with each other, a phenomenon that can be demonstrated with the device shown in Figure 14.14. Sound from a loudspeaker at S is sent into a tube at P, where there is a T-shaped junction. The sound splits and follows two separate pathways, indicated by the red arrows. Half of the sound travels upward, half downward. Finally, the two sounds merge at an opening where a listener places her ear. If the two paths r 1 and r 2 have the same length, waves that enter the junction will separate into two halves, travel the two paths, and then combine again at the ear. This reuniting of the two waves produces constructive interference, so the listener hears a loud sound. If the upper path is adjusted to be one full wavelength longer than the lower path, constructive interference of the two waves occurs again, and a loud sound is detected at the receiver. We have the following result: If the path difference r 2 r 1 is zero or some integer multiple of wavelengths, then constructive interference occurs and r2 r1 n (n 0, 1, 2, . . .) [14.13] Suppose, however, that one of the path lengths, r 2 , is adjusted so that the upper path is half a wavelength longer than the lower path r1. In this case, an entering sound wave splits and travels the two paths as before, but now the wave along the upper path must travel a distance equivalent to half a wavelength farther than the wave traveling along the lower path. As a result, the crest of one wave meets the trough of the other when they merge at the receiver, causing the two waves to
r2 S Figure 14.14 An acoustical system for demonstrating interference of sound waves. Sound from the speaker enters the tube and splits into two parts at P. The two waves combine at the opposite side and are detected at R. The upper path length is varied by the sliding section.
Condition for constructive interference
P r1 Speaker
R Receiver
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Condition for destructive interference
cancel each other. This phenomenon is called totally destructive interference, and no sound is detected at the receiver. In general, if the path difference r 2 r1 is 11 1 2 , 12 , 2 2 . . . wavelengths, destructive interference occurs and r2 r1 (n
1 2)
(n
0, 1, 2, . . .)
[14.14]
A P P L I C AT I O N Connecting Your Stereo Speakers
TIP 14.3 Do Waves Really Interfere?
In popular usage, to interfere means to come into conict with or to intervene to affect an outcome. This differs from its use in physics, where waves pass through each other and interfere, but dont affect each other in any way.
Nature provides many other examples of interference phenomena, most notably in connection with light waves, described in Chapter 24. In connecting the wires between your stereo system and loudspeakers, you may notice that the wires are usually color coded and that the speakers have positive and negative signs on the connections. The reason for this is that the speakers need to be connected with the same polarity. If they arent, then the same electrical signal fed to both speakers will result in one speaker cone moving outward at the same time that the other speaker cone is moving inward. In this case, the sound leaving the two speakers will be 180 out of phase with each other. If you are sitting midway between the speakers, the sounds from both speakers travel the same distance and preserve the phase difference they had when they left. In an ideal situation, for a 180 phase difference, you would get complete destructive interference and no sound! In reality, the cancellation is not complete and is much more signicant for bass notes (which have a long wavelength) than for the shorter wavelength treble notes. Nevertheless, to avoid a signicant reduction in the intensity of bass notes, the color-coded wires and the signs on the speaker connections should be carefully noted.
EXAMPLE 14.6 Two Speakers Driven by the Same Source
Goal Use the concept of interference to compute a frequency. Problem Two speakers placed 3.00 m apart are driven by the same oscillator (Fig. 14.15). A listener is originally at point O, which is located 8.00 m from the center of the line connecting the two speakers. The listener then walks to point P, which is a perpendicular distance 0.350 m from O, before reaching the rst minimum in sound intensity. What is the frequency of the oscillator? Take the speed of r1 sound in air to be vs 343 m/s. 1.15 m 0.350 m
P
Strategy The position of the rst minimum in sound intensity is given, which is a point of destructive interference. We can nd the path lengths r 1 and r 2 with the Pythagorean theorem and then use Equation 14.14 for destructive interference to nd the wavelength . Using v f then yields the frequency. Solution Use the Pythagorean theorem to nd the path lengths r 1 and r 2: Substitute these values and n solving for the wavelength: 0 into Equation 14.14,
3.00 m r2 8.00 m
O
1.85 m
Figure 14.15 (Example 14.6) Two loudspeakers driven by the same source can produce interference.
r1 r2
(8.00 m)2 (8.00 m)2
r2 r1 8.08 m 343 m/s 0.26 m
(1.15 m)2 (1.85 m)2 (n
1 2)
8.08 m 8.21 m
8.21 m f v
0.13 m 1.3 kHz
/2
:
0.26 m
f for the frequency f and substitute the Solve v speed of sound and the wavelength:
Remark For problems involving constructive interference, the only difference is that Equation 14.13, r 2 would be used instead of Equation 14.14.
r1
n,
Exercise 14.6 If the oscillator frequency is adjusted so that the location of the rst minimum is at a distance of 0.750 m from O, what is the new frequency? Answer 0.642 kHz
14.8
Standing Waves
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14.8
STANDING WAVES
Vibrating blade Figure 14.16 Standing waves can be set up in a stretched string by connecting one end of the string to a vibrating blade. When the blade vibrates at one of the natural frequencies of the string, large-amplitude standing waves are created.
Standing waves can be set up in a stretched string by connecting one end of the string to a stationary clamp and connecting the other end to a vibrating object, such as the end of a tuning fork, or by shaking the hand holding the string up and down at a steady rate (Fig. 14.16). Traveling waves then reect from the ends and move in both directions on the string. The incident and reected waves combine according to the superposition principle. (See Section 13.10.) If the string vibrates at exactly the right frequency, the wave appears to stand still hence its name, standing wave. A node occurs where the two traveling waves always have the same magnitude of displacement but the opposite sign, so the net displacement is zero at that point. There is no motion in the string at the nodes, but midway between two adjacent nodes, at an antinode, the string vibrates with the largest amplitude. Figure 14.17 shows snapshots of the oscillation of a standing wave during half of a cycle. The pink arrows indicate the direction of motion of different parts of the string. Notice that all points on the string oscillate together vertically with the same frequency, but different points have different amplitudes of motion. The points of attachment to the wall and all other stationary points on the string are called nodes, labeled N in Figure 14.17a. From the gure, observe that the distance between adjacent nodes is one-half the wavelength of the wave: d NN
1 2
N (a) t=0
N
N
(b)
t = T/ 8
Consider a string of length L that is xed at both ends, as in Active Figure 14.18. For a string, we can set up standing-wave patterns at many frequencies the more loops, the higher the frequency. Three such patterns are shown in Active Figures 14.18b, 14.18c, and 14.18d. Each has a characteristic frequency, which we will now calculate. First, the ends of the string must be nodes, because these points are xed. If the string is displaced at its midpoint and released, the vibration shown in Active Figure 14.18b can be produced, in which case the center of the string is an antinode, labeled A. Note that from end to end, the pattern is N A N. The distance from a node to its adjacent antinode, N A, is always equal to a quarter wavelength, 1/4. There are two such segments, N A and A N, so L 2( 1/4) 2L. The frequency of this vibration is therefore 1/2, and 1 f1 v
1
(c)
t = T/4
(d)
t = 3T/ 8
v 2L
(e)
[14.15]
t = T/ 2
Recall that the speed of a wave on a string is v F/ , where F is the tension in the string and is its mass per unit length (Chapter 13). Substituting into Equation 14.15, we obtain f1 1 2L
Figure 14.17 A standing-wave pattern in a stretched string, shown by snapshots of the string during one-half of a cycle. In part (a), N denotes a node.
F
[14.16]
N L f2 n=2 A
A
N
A
N ACTIVE FIGURE 14.18 (a) Standing waves in a stretched string of length L xed at both ends. The characteristic frequencies of vibration form a harmonic series: (b) the fundamental frequency, or rst harmonic; (c) the second harmonic; and (d) the third harmonic. Note that N denotes a node, A an antinode.
(a) A N N
(c) N A N
L = 2 A
N
N
f1
f3 Log into PhysicsNow at www.cp7e.com and go to Active Figure 14.18 to choose the mode number and see the corresponding standing wave.
n=1 (b)
1 L = 1 2
n=3 (d)
L = 3 3 2
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Richard Megna, Fundamental Photographs
(a)
(b)
(c)
Multiash photographs of standing-wave patterns in a cord driven by a vibrator at the left end. The single-loop pattern in (a) represents the fundamental frequency (n 1), the two-loop pattern in (b) the second harmonic (n 2), and the three-loop pattern in (c) the third harmonic (n 3).
This lowest frequency of vibration is called the fundamental frequency of the vibrating string, or the rst harmonic. The rst harmonic has nodes only at the ends the points of attachment, with node antinode pattern of N A N. The next harmonic, called the second harmonic (also called the rst overtone) can be constructed by inserting an additional node antinode segment between the endpoints. This makes the pattern N A N A N, as in Active Figure 14.18c. We count the node antinode pairs: N A, A N, N A, and A N, four segments in all, each representing a quarter wavelength. We then have L 4( 2/4) 2, and the second harmonic (rst overtone) is f2 v
2
v L
2
v 2L
2f 1
This frequency is equal to twice the fundamental frequency. The third harmonic (second overtone) is constructed similarly. Inserting one more N A segment, we obtain Active Figure 14.18c, the pattern of nodes reading N A N A N A N. There are six node antinode segments, so L 6( 3/4) 3( 3/2), which means that 3 2L/3, giving f3 v
3
3v 2L
3f1
All the higher harmonics, it turns out, are positive integer multiples of the fundamental: fn nf1 n 2L
F
n
1, 2, 3 . . .
[14.17]
The frequencies f1, 2f 1, 3f 1, and so on form a harmonic series.
Quick Quiz 14.4
Which of the following frequencies are higher harmonics of a string with fundamental frequency of 150 Hz? (a) 200 Hz (b) 300 Hz (c) 400 Hz (d) 500 Hz (e) 600 Hz. When a stretched string is distorted to a shape that corresponds to any one of its harmonics, after being released it vibrates only at the frequency of that harmonic. If the string is struck or bowed, however, the resulting vibration includes different amounts of various harmonics, including the fundamental frequency. Waves not in the harmonic series are quickly damped out on a string xed at both ends. In ef-
14.8
Standing Waves
477
fect, when disturbed, the string selects the standing-wave frequencies. As well see later, the presence of several harmonics on a string gives stringed instruments their characteristic sound, which enables us to distinguish one from another even when they are producing identical fundamental frequencies. The frequency of a string on a musical instrument can be changed by varying either the tension or the length. The tension in guitar and violin strings is varied by turning pegs on the neck of the instrument. As the tension is increased, the frequency of the harmonic series increases according to Equation 14.17. Once the instrument is tuned, the musician varies the frequency by pressing the strings against the neck at a variety of positions, thereby changing the effective lengths of the vibrating portions of the strings. As the length is reduced, the frequency again increases, as follows from Equation 14.17. Finally, Equation 14.17 shows that a string of xed length can be made to vibrate at a lower fundamental frequency by increasing its mass per unit length. This is achieved in the bass strings of guitars and pianos by wrapping them with metal windings.
A P P L I C AT I O N Tuning a Musical Instrument
INTERACTIVE EXAMPLE 14.7 Guitar Fundamentals
Goal Apply standing-wave concepts to a stringed instrument. Problem The high E string on a certain guitar measures 64.0 cm in length and has a fundamental frequency of 329 Hz. When a guitarist presses down so that the string is in contact with the rst fret (Fig. 14.19a), the string is shortened so that it plays an F note that has a frequency of 349 Hz. (a) How far is the fret from the nut? (b) Overtones can be produced on a guitar string by gently placing the index nger in the location of a node of a higher harmonic. The string should be touched, but not depressed against a fret. (Given the width of a nger, pressing too hard will damp out higher harmonics as well.) The fundamental frequency is thereby suppressed, making it possible to hear overtones. Where on the guitar string relative to the nut should the nger be lightly placed so as to hear the second harmonic? The fourth harmonic? (This is equivalent to nding the location of the nodes in each case.) Strategy For part (a) use Equation 14.15, corresponding to the fundamental frequency, to nd the speed of waves on the string. Shortening the string by playing a higher note doesnt affect the wave speed, which depends nut only on the tension and linear density of the 1st fret string (which are unchanged). Solve Equation bridge 14.15 for the new length L, using the new fun2nd fret damental frequency, and subtract this length (a) (b) from the original length to nd the distance Figure 14.19 (Example 14.7) (a) Playing an F note on a guitar. (b) Some parts of from the nut to the rst fret. In part (b), re- a guitar. member that the distance from node to node is half a wavelength. Calculate the wavelength, divide it in two, and locate the nodes, which are integral numbers of half-wavelengths from the nut. [Note: The nut is a small piece of wood or ebony at the top of the fret board. The distance from the nut to the bridge (below the sound hole) is the length of the string. (See Fig. 14.19b.)] Solution (a) Find the distance from the nut to the rst fret. Substitute L 0 0.640 m and f1 329 Hz into Equation 14.15, nding the wave speed on the string: f1 v Solve Equation 14.15 for the length L, and substitute the wave speed and the frequency of F. Subtract this length from the original length L 0 to nd the distance from the nut to the rst fret: L v 2L 0 2L 0 f1 v 2f x L0 2(0.640 m)(329 Hz) 421 m/s 2(349 Hz) L 64.0 cm 0.603 m 421 m/s 60.3 cm
Charles D. Winters
60.3 cm
3.7 cm
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(b) Find the locations of nodes for the second and fourth harmonics. The second harmonic has a wavelength L 0 64.0 cm. The distance from nut to node 2 corresponds to half a wavelength. The fourth harmonic, of wavelength 4 1 L 0 2 32.0 cm, has three nodes between the endpoints: x
1 2 2 1 2L0
32.0 cm
x x
1 2
4
16.0 cm , x 48.0 cm
2( 4/2)
32.0 cm ,
3( 4/2)
Remarks Placing a nger at the position x 32.0 cm damps out the fundamental and odd harmonics, but not all the higher even harmonics. The second harmonic dominates, however, because the rest of the string is free to vibrate. Placing the nger at x 16.0 cm or 48.0 cm damps out the rst through third harmonics, allowing the fourth harmonic to be heard. Exercise 14.7 Pressing the E-string down on the fret board just above the second fret pinches the string rmly against the fret, giving an F sharp, which has frequency 3.70 102 Hz. (a) Where should the second fret be located? (b) Find two locations where you could touch the open E-string and hear the third harmonic.
7.1 cm from the nut and 3.4 cm from the rst fret. Note that the distance from the rst to the second fret isnt the same as from the nut to the rst fret. (b) 21.3 cm and 42.7 cm from the nut.
Explore this situation by logging into PhysicsNow at www.cp7e.com and going to Interactive Example 14.7.
EXAMPLE 14.8 Harmonics of a Stretched Wire
Goal Calculate string harmonics, relate them to sound and combine them with tensile stress. Problem (a) Find the frequencies of the fundamental, second, and third harmonics of a steel wire 1.00 m long with a mass per unit length of 2.00 10 3 kg/m and under a tension of 80.0 N. (b) Find the wavelengths of the sound waves created by the vibrating wire for all three modes. Assume the speed of sound in air is 345 m/s. (c) Suppose the wire is carbon steel with a density of 7.80 103 kg/m3, a cross-sectional area A 2.56 10 7 m2, and an elastic limit of 2.80 108 Pa. Find the fundamental frequency if the wire is tightened to the elastic limit. Neglect any stretching of the wire (which would slightly reduce the mass per unit length). Strategy (a) Its easiest to nd the speed of waves on the wire then substitute into Equation 14.15 to nd the rst harmonic. The next two are multiples of the rst, given by Equation 14.17. (b) The frequencies of the sound waves are the same as the frequencies of the vibrating wire, but the wavelengths are different. Use vs f , where vs is the speed of sound in air, to nd the wavelengths in air. (c) Find the force corresponding to the elastic limit, and substitute it into Equation 14.16. Solution (a) Find the rst three harmonics at the given tension. Use Equation 13.17 to calculate the speed of the wave on the wire: Find the wires fundamental frequency from Equation 14.15: Find the next two harmonics by multiplication: v
F v 2L 2f1 2.00
2.00
80.0 N 10 3 kg/m
2.00
102 m/s
f1 f2
2.00 102 m/s 2(1.00 m) 102 Hz , f3
1.00 3f1
102 Hz 3.00 102 Hz
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Forced Vibrations and Resonance
479
(b) Find the wavelength of the sound waves produced. Solve vs f for the wavelength and substitute the frequencies.
1 2 3
vs /f1 vs /f2 vs /f3
(345 m/s)/(1.00 (345 m/s)/(2.00 (345 m/s)/(3.00
102 Hz) 102 Hz) 102 Hz)
3.45 m 1.73 m 1.15 m
(c) Find the fundamental frequency corresponding to the elastic limit. Calculate the tension in the wire from the elastic limit: F A F Substitute the values of F, , and L into Equation 14.16: f1 f1 elastic limit (2.80 1 2L : F (elastic limit)A 10
7
108 Pa)(2.56
m2)
71.7 N
F
1 2(1.00 m)
2.00
71.7 N 10 3 kg/m
94.7 Hz
Remarks From the answer to part (c), it appears we need to choose a thicker wire or use a better grade of steel with a higher elastic limit. The frequency corresponding to the elastic limit is smaller than the fundamental! Exercise 14.8 (a) Find the fundamental frequency and second harmonic if the tension in the wire is increased to 115 N. (Assume the wire doesnt stretch or break.) (b) Using a sound speed of 345 m/s, nd the wavelengths of the sound waves produced. 1.20 102 Hz, 2.40 102 Hz (b) 2.88 m, 1.44 m
14.9 FORCED VIBRATIONS AND RESONANCE
In Chapter 13 we learned that the energy of a damped oscillator decreases over time because of friction. Its possible to compensate for this energy loss by applying an external force that does positive work on the system. For example, suppose an object spring system having some natural frequency of vibration f0 is pushed back and forth by a periodic force with frequency f. The system vibrates at the frequency f of the driving force. This type of motion is referred to as a forced vibration. Its amplitude reaches a maximum when the frequency of the driving force equals the natural frequency of the system f0, called the resonant frequency of the system. Under this condition, the system is said to be in resonance. In Section 14.8 we learned that a stretched string can vibrate in one or more of its natural modes. Here again, if a periodic force is applied to the string, the amplitude of vibration increases as the frequency of the applied force approaches one of the strings natural frequencies of vibration. Resonance vibrations occur in a wide variety of circumstances. Figure 14.20 illustrates one experiment that demonstrates a resonance condition. Several pendulums of different lengths are suspended from a exible beam. If one of them, such as A, is set in motion, the others begin to oscillate because of vibrations in the exible beam. Pendulum C, the same length as A, oscillates with the greatest amplitude because its natural frequency matches that of pendulum A (the driving force). Another simple example of resonance is a child being pushed on a swing, which is essentially a pendulum with a natural frequency that depends on its length. The
A
D B C
Figure 14.20 Resonance. If pendulum A is set in oscillation, only pendulum C, with a length matching of that A, will eventually oscillate with a large amplitude, or resonate. The arrows indicate motion perpendicular to the page.
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A P P L I C AT I O N Shattering Goblets with the Voice
Ben Rose/The IMAGE Bank/Getty Images
swing is kept in motion by a series of appropriately timed pushes. For its amplitude to increase, the swing must be pushed each time it returns to the persons hands. This corresponds to a frequency equal to the natural frequency of the swing. If the energy put into the system per cycle of motion equals the energy lost due to friction, the amplitude remains constant. Opera singers have been known to set crystal goblets in audible vibration with their powerful voices, as shown in Figure 14.21. This is yet another example of resonance: The sound waves emitted by the singer can set up large-amplitude vibrations in the glass. If a highly amplied sound wave has the right frequency, the amplitude of forced vibrations in the glass increases to the point where the glass becomes heavily strained and shatters. The classic example of structural resonance occurred in 1940, when the Tacoma Narrows bridge in the state of Washington was set in oscillation by the wind (Fig. 14.22). The amplitude of the oscillations increased rapidly and reached a high value until the bridge ultimately collapsed (probably because of metal fatigue). In recent years, however, a number of researchers have called this explanation into question. Gusts of wind, in general, dont provide the periodic force necessary for a sustained resonance condition, and the bridge exhibited large twisting oscillations, rather than the simple up-and-down oscillations expected of resonance. A more recent example of destruction by structural resonance occurred during the Loma Prieta earthquake near Oakland, California, in 1989. In a mile-long section of the double-decker Nimitz Freeway, the upper deck collapsed onto the lower deck, killing several people. The collapse occurred because that particular section was built on mud ll while other parts were built on bedrock. As seismic waves pass through mud ll or other loose soil, their speed decreases and their amplitude increases. The section of the freeway that collapsed oscillated at the same frequency as other sections, but at a much larger amplitude.
Courtesy of Professor Thomas D. Rossing/Getty Images Northern Illinois University
14.10 STANDING WAVES IN AIR COLUMNS
Standing longitudinal waves can be set up in a tube of air, such as an organ pipe, as the result of interference between sound waves traveling in opposite directions. The relationship between the incident wave and the reected wave depends on whether the reecting end of the tube is open or closed. A portion of the sound wave is reected back into the tube even at an open end. If one end is closed, a node must exist at that end because the movement of air is restricted. If the end is open, the elements of air have complete freedom of motion, and an antinode exists. Figure 14.23a shows the rst three modes of vibration of a pipe open at both ends. When air is directed against an edge at the left, longitudinal standing waves are formed and the pipe vibrates at its natural frequencies. Note that, from end to end, the pattern is A N A, the same pattern as in the vibrating string, except node and antinode have exchanged positions. As before, an antinode and its adjacent node, A N, represent a quarter-wavelength, and there are two, A N and N A, so L 2( 1/4) 2L. The fundamental frequency of the 1/2 and 1 pipe open at both ends is then f1 v/ 1 v/2L. The next harmonic has an addi-
Figure 14.21 (Top) Standing-wave pattern in a vibrating wineglass. The glass will shatter if the amplitude of vibration becomes too large. (Bottom) A wineglass shattered by the amplied sound of a human voice.
United Press International Photo
Figure 14.22 The collapse of the Tacoma Narrows suspension bridge in 1940 has been cited as a demonstration of mechanical resonance. High winds set up standing waves in the bridge, causing it to oscillate at one of its natural frequencies. Once established, the resonance may have led to the bridges collapse (although this interpretation is currently being challenged by mathematicians and physical scientists).
14.10
Standing Waves in Air Columns
481
L A N A
l1 = 2L v v f1 = = l1 2L l2 = L v f2 = = 2f1 L
2 l3 = L 3 f3 = 3v = 3f1 2L (a) Open at both ends
First harmonic
ANA
NA
Second harmonic
Figure 14.23 (a) Standing longitudinal waves in an organ pipe open at both ends. The natural frequencies f1, 2f 1, 3f 1 . . . form a harmonic series. (b) Standing longitudinal waves in an organ pipe closed at one end. Only odd harmonics are present, and the natural frequencies are f1, 3f 1, 5f 1, and so on.
A N A N A NA
Third harmonic
A
N
l1 = 4L v v f1 = = l1 4L
4 l3 = L 3 f3 = 3v = 3f1 4L 4 l5 = L 5 f5 = 5v = 5f1 4L (b) Closed at one end, open at the other
TIP 14.4 Sound Waves Are Not Transverse
First harmonic The standing longitudinal waves in Figure 14.23 are drawn as transverse waves only because its difcult to draw longitudinal displacements theyre in the same direction as the wave propagation. In the gure, the vertical axis represents either pressure or horizontal displacement of the elements of the medium.
A
N
A
N
Third harmonic
ANA NA N
Fifth harmonic
tional node and antinode between the ends, creating the pattern A N A N A. We count the pairs: A N, N A, A N, and N A, making four segments, each with length 2/4. We have L 4( 2/4) 2, and the second harmonic (rst overtone) is f2 v/ 2 v/L 2(v/2L) 2f1. All higher harmonics, it turns out, are positive integer multiples of the fundamental: fn n v 2L nf1 n 1, 2, 3, . . . [14.18]
Pipe open at both ends; all harmonics are present
where v is the speed of sound in air. Notice the similarity to Equation 14.17, which also involves multiples of the fundamental. If a pipe is open at one end and closed at the other, the open end is an antinode while the closed end is a node (Fig. 14.23b). In such a pipe, the fundamental frequency consists of a single antinode node pair, A N, so L 1/4 and 1 4L. The fundamental harmonic for a pipe closed at one end is then f1 v/ 1 v/4L. The rst overtone has another node and antinode between the open end and closed end, making the pattern A N A N. There are three antinode node segments in this pattern (A N, N A, and A N), so L 3( 3/4) and 3 4L/3. The rst overtone, therefore, has frequency f3 v/ 3 3v/4L 3f1. Similarly, f5 5f1. In contrast to the pipe open at both ends, there are no even multiples of the fundamental harmonic. The odd harmonics for a pipe open at one end only are given by fn n v 4L nf1 n 1, 3, 5, . . . [14.19]
Pipe closed at one end; only odd harmonics are present
Quick Quiz 14.5
A pipe open at both ends resonates at a fundamental frequency fopen. When one end is covered and the pipe is again made to resonate, the fundamental frequency is fclosed. Which of the following expressions describes how these two resonant frequencies compare? (a) fclosed fopen, (b) fclosed 3 f open, (c) fclosed 2 fopen, 2 (d) fclosed 1 fopen, (e) none of these. 2
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Quick Quiz 14.6
Balboa Park in San Diego has an outdoor organ. When the air temperature increases, the fundamental frequency of one of the organ pipes (a) increases (b) decreases (c) stays the same (d) impossible to determine. (The thermal expansion of the pipe is negligible.)
Applying Physics 14.3 Oscillations in a Harbor
Why do passing ocean waves sometimes cause the water in a harbor to undergo very large oscillations, called a seiche (pronounced saysh)? Explanation Water in a harbor is enclosed and possesses a natural frequency based on the size of the harbor. This is similar to the natural frequency of the enclosed air in a bottle, which can be excited by blowing across the edge of the opening. Ocean waves pass by the opening of the harbor at a certain frequency. If this frequency matches that of the enclosed harbor, then a large standing wave can be set up in the water by resonance. This situation can be simulated by carrying a sh tank lled with water. If your walking frequency matches the natural frequency of the water as it sloshes back and forth, a large standing wave develops in the sh tank.
Applying Physics 14.4 Why Are Instruments Warmed Up?
Why do the strings go at and the wind instruments go sharp during a performance if an orchestra doesnt warm up beforehand? Explanation Without warming up, all the instruments will be at room temperature at the beginning of the concert. As the wind instruments are played, they ll with warm air from the players exhalation. The increase in temperature of the air in the instruments causes an increase in the speed of sound, which raises the resonance frequencies of the air columns. As a result, the instruments go sharp. The strings on the stringed instruments also increase in temperature due to the friction of rubbing with the bow. This results in thermal expansion, which causes a decrease in tension in the strings. With the decrease in tension, the wave speed on the strings drops, and the fundamental frequencies decrease, so the stringed instruments go at.
Applying Physics 14.5 How Do Bugles Work?
A bugle has no valves, keys, slides, or nger holes. How can it be used to play a song? Explanation Songs for the bugle are limited to harmonics of the fundamental frequency, because there is no control over frequencies without valves, keys, slides, or nger holes. The player obtains different notes by changing the tension in the lips as the bugle is played, in order to excite different harmonics. The normal playing range of a bugle is among the third, fourth, fth, and sixth harmonics of the fundamental. Reveille, for example, is played with just the three notes G, C, and F. And Taps is played with these three notes and the G one octave above the lower G.
EXAMPLE 14.9 Harmonics of a Pipe
Goal Find frequencies of open and closed pipes. Problem A pipe is 2.46 m long. (a) Determine the frequencies of the rst three harmonics if the pipe is open at both ends. Take 343 m/s as the speed of sound in air. (b) How many harmonic frequencies of this pipe lie in the audible range, from 20 Hz to 20 000 Hz? (c) What are the three lowest possible frequencies if the pipe is closed at one end and open at the other? Strategy Substitute into Equation 14.18 for part (a) and Equation 14.19 for part (c). All harmonics, n 1, 2, 3 . . . are available for the pipe open at both ends, but only the harmonics with n 1, 3, 5, . . . for the pipe closed at one end. For part (b), set the frequency in Equation 14.18 equal to 2.00 104 Hz.
14.10
Standing Waves in Air Columns
483
Solution (a) Find the frequencies if the pipe is open at both ends. Substitute into Equation 14.18, with n 1: f1 f2 v 2L 2f 1 343 m/s 2(2.46 m) 139 Hz f3 69.7 Hz 3f 1 209 Hz
Multiply to nd the second and third harmonics: (b) How many harmonics lie between 20 Hz and 20 000 Hz for this pipe? Set the frequency in Equation 14.18 equal to 2.00 and solve for n: 104
fn n
n
v 2L
n
343 m/s 2 2.46 m
2.00
104 Hz
This works out to n 286.88, which must be truncated down (n 287 gives a frequency over 2.00 104 Hz). (c) Find the frequencies for the pipe closed at one end. Apply Equation 14.19 with n 1:
286
f1 f3
v 4L 3f1
343 m/s 4(2.46 m) 105 Hz f5
34.9 Hz 5f1 175 Hz
The next two harmonics are odd multiples of the rst:
Exercise 14.9 (a) What length pipe open at both ends has a fundamental frequency of 3.70 102 Hz? Find the rst overtone. (b) If the one end of this pipe is now closed, what is the new fundamental frequency? Find the rst overtone. (c) If the pipe is open at one end only, how many harmonics are possible in the normal hearing range from 20 to 20 000 Hz? 0.464 m, 7.40 102 Hz (b) 185 Hz, 555 Hz (c) 54
EXAMPLE 14.10 Resonance in a Tube of Variable Length
Goal Understand resonance in tubes and perform elementary calculations.
Problem Figure 14.24a shows a simple apparatus for demonstrating resonance in a tube. A long tube open at both ends is partially submerged in a beaker of water, and a vibrating tuning fork of unknown frequency is placed near the top of the tube. The length of the air column, L, is adjusted by moving the tube vertically. The sound waves generated by the fork are reinforced when the length of the air column corresponds to one of the resonant frequencies of the tube. Suppose the smallest value of L for which a peak occurs in the sound intensity is 9.00 cm. (a) With this measurement, determine the frequency of the tuning fork. (b) Find the wavelength and the next two air-column lengths giving resonance. Take the speed of sound to be 345 m/s. Strategy Once the tube is in the water, the setup is the same as a pipe closed at one values for v and L into Equation 14.19 with n 1 and nd the fref=? /4 3 /4 quency of the tuning fork. (b) The next resonance maximum occurs 5 /4 L when the water level is low enough First resonance to allow a second node, which is another half-wavelength in distance. Second The third resonance occurs when resonance (third Water the third node is reached, requirharmonic) Third ing yet another half-wavelength of (a) resonance distance. The frequency in each (fifth harmonic) case is the same, because its gener(b) ated by the tuning fork. end. For part (a), substitute
Figure 14.24 (Example 14.10) (a) Apparatus for demonstrating the resonance of sound waves in a tube closed at one end. The length L of the air column is varied by moving the tube vertically while it is partially submerged in water. (b) The rst three resonances of the system.
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Solution (a) Find the frequency of the tuning fork. Substitute n 1, v 345 m/s, and L1 into Equation 14.19: 9.00 10
2
m
f1
v 4L1
345 m/s 4(9.00 10 2 m)
958 Hz
(b) Find the wavelength and the next two water levels giving resonance. Calculate the wavelength, using the fact that, for a tube open at one end, 4L for the fundamental. Add a half-wavelength of distance to L1 to get the next resonance position: Add another half-wavelength to L 2 to obtain the third resonance position: L2 L3 4L1 L1 L2 4(9.00 10
2
m)
0.360 m
/2 /2
0.0900 m 0.270 m
0.180 m 0.180 m
0.270 m 0.450 m
Remark This experimental arrangement is often used to measure the speed of sound, in which case the frequency of the tuning fork must be known in advance. Exercise 14.10 An unknown gas is introduced into the aforementioned apparatus using the same tuning fork, and the rst resonance occurs when the air column is 5.84 cm long. Find the speed of sound in the gas. Answer 224 m/s
14.11 BEATS
The interference phenomena we have been discussing so far have involved the superposition of two or more waves with the same frequency, traveling in opposite directions. Another type of interference effect results from the superposition of two waves with slightly different frequencies. In such a situation, the waves at some xed point are periodically in and out of phase, corresponding to an alternation in time between constructive and destructive interference. In order to understand this phenomenon, consider Active Figure 14.25. The two waves shown in Active Figure 14.25a were emitted by two tuning forks having slightly different frequencies; Active Figure 14.25b shows the superposition of these waves. At some time ta the waves are in phase and constructive interference occurs, as demonstrated by the resultant curve in Active Figure 14.25b. At some later time, however, the vibrations of the two forks move out of step with each other. At time tb , one fork emits a compression while the other emits a rarefaction, and destructive interference occurs, as demonstrated by the curve shown. As time passes, the vibrations of the two forks move out of phase, then into phase again, and so on. As a consequence, a listener at some xed point hears an alternation in loudness, known as beats. The number of beats per second, or the beat frequency, equals the difference in frequency between the two sources: fb f2 f1 [14.20]
A P P L I C AT I O N Using Beats to Tune a Musical Instrument
where fb is the beat frequency and f1 and f2 are the two frequencies. The absolute value is used because the beat frequency is a positive quantity and will occur regardless of the order of subtraction. A stringed instrument such as a piano can be tuned by beating a note on the instrument against a note of known frequency. The string can then be tuned to the desired frequency by adjusting the tension until no beats are heard.
14.11
Beats
485
y
ta
tb t
(a)
y
ACTIVE FIGURE 14.25 Beats are formed by the combination of two waves of slightly different frequencies traveling in the same direction. (a) The individual waves heard by an observer at a xed point in space. (b) The combined wave has an amplitude (dashed line) that oscillates in time.
(b)
t
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 14.25 to choose two frequencies and see the corresponding beats.
Quick Quiz 14.7
You are tuning a guitar by comparing the sound of the string with that of a standard tuning fork. You notice a beat frequency of 5 Hz when both sounds are present. As you tighten the guitar string, the beat frequency rises steadily to 8 Hz. In order to tune the string exactly to the tuning fork, you should (a) continue to tighten the string (b) loosen the string (c) impossible to determine from the given information.
EXAMPLE 14.11 Sour Notes
Goal Apply the beat frequency concept. Problem A certain piano string is supposed to vibrate at a frequency of 4.40 102 Hz. In order to check its frequency, a tuning fork known to vibrate at a frequency of 4.40 102 Hz is sounded at the same time the piano key is struck, and a beat frequency of 4 beats per second is heard. (a) Find the two possible frequencies at which the string could be vibrating. (b) Suppose the piano tuner runs toward the piano, holding the vibrating tuning fork while his assistant plays the note, which is at 436 Hz. At his maximum speed, the piano tuner notices the beat frequency drops from 4 Hz to 2 Hz (without going through a beat frequency of zero). How fast is he moving? Use a sound speed of 343 m/s. (c) While the piano tuner is running, what beat frequency is observed by the assistant? [Note: Assume all numbers are accurate to two decimal places, necessary for this last calculation.] Strategy (a) The beat frequency is equal to the absolute value of the difference in frequency between the two sources of sound and occurs if the piano string is tuned either too high or too low. Solve Equation 14.20 for these two possible frequencies. (b) Moving toward the piano raises the observed piano string frequency. Solve the Doppler shift formula, Equation 14.12, for the speed of the observer. (c) The assistant observes a Doppler shift for the tuning fork. Apply Equation 14.12. Solution (a) Find the two possible frequencies. Case 1: f2 f1 is already positive, so just drop the absolute-value signs: Case 2: f2 f 1 is negative, so drop the absolute-value signs, but apply an overall negative sign: (b) Find the speed of the observer if running toward the piano results in a beat frequency of 2 Hz. Apply the Doppler shift to the case where frequency of the piano string heard by the running observer is fO 438 Hz: fO 438 Hz vO fS v v vO vS 343 m/s vO 343 m/s (343 m/s) 1.57 m/s fb f2 fb f2 f2 f1 : 4 Hz f2 4.40 102 Hz
444 Hz ( f2 436 Hz f1) : 4 Hz ( f2 4.40 102 Hz)
(436 Hz)
438 Hz 436 Hz 436 Hz
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(c) What beat frequency does the assistant observe? Apply Equation 14.12. Now the source is the tuning fork, so fS 4.40 102 Hz. fO fS v v vO vS 102 Hz) 442 Hz 343 m/s 343 m/s 1.57 m/s 436 Hz 6 Hz 442 Hz
(4.40 Compute the beat frequency. fb f2 f1
Remarks The assistant on the piano bench and the tuner running with the fork observe different beat frequencies. Many physical observations depend on the state of motion of the observer, a subject discussed more fully in Chapter 26, on relativity. Exercise 14.11 The assistant adjusts the tension in the same piano string, and a beat frequency of 2.00 Hz is heard when the note and the tuning fork are struck at the same time. (a) Find the two possible frequencies of the string. (b) Assume the actual string frequency is the higher frequency. If the piano tuner runs away from the piano at 4.00 m/s while holding the vibrating tuning fork, what beat frequency does he hear? (c) What beat frequency does the assistant on the bench hear? Use 343 m/s for the speed of sound. Answers (a) 438 Hz, 442 Hz (b) 3 Hz (c) 7 Hz
Tuning fork
(b) Flute
(c) Clarinet
Figure 14.26 Waveforms produced by (a) a tuning fork, (b) a ute, and (c) a clarinet, all at approximately the same frequency. Pressure is plotted vertically, time horizontally.
Adapted from C. A. Culver, Musical Acoustics, 4th ed., New York, McGraw-Hill, 1956
(a)
14.12 QUALITY OF SOUND
The sound-wave patterns produced by most musical instruments are complex. Figure 14.26 shows characteristic waveforms (pressure is plotted on the vertical axis, time on the horizontal axis) produced by a tuning fork, a ute, and a clarinet, each playing the same steady note. Although each instrument has its own characteristic pattern, the gure reveals that each of the waveforms is periodic. Note that the tuning fork produces only one harmonic (the fundamental frequency), but the two instruments emit mixtures of harmonics. Figure 14.27 graphs the harmonics of the waveforms of Figure 14.26. When the note is played on the ute (Fig. 14.26b), part of the sound consists of a vibration at the fundamental frequency, an even higher intensity is contributed by the second harmonic, the fourth harmonic produces about the same intensity as the fundamental, and so on. These sounds add together according to the principle of superposition to give the complex waveform shown. The clarinet emits a certain intensity at a frequency of the rst harmonic, about half as much intensity at the frequency of the second harmonic, and so forth. The resultant superposition of these frequencies produces the pattern shown in Figure 14.26c. The tuning fork (Figs. 14.26a and 14.27a) emits sound only at the frequency of the rst harmonic.
Relative intensity
Relative intensity
Relative intensity
Tuning fork
Clarinet Flute
Adapted from C. A. Culver, Musical Acoustics, 4th ed., New York, McGraw-Hill, 1956
123456 Harmonics (a) Figure 14.27
1234567 Harmonics (b)
123456789 Harmonics (c)
Harmonics of the waveforms in Figure 14.26. Note their variation in intensity.
14.13
The Ear
487
In music, the characteristic sound of any instrument is referred to as the quality, or timbre, of the sound. The quality depends on the mixture of harmonics in the sound. We say that the note C on a ute differs in quality from the same C on a clarinet. Instruments such as the bugle, trumpet, violin, and tuba are rich in harmonics. A musician playing a wind instrument can emphasize one or another of these harmonics by changing the conguration of the lips, thereby playing different musical notes with the same valve openings.
TIP 14.5 Pitch is Not The Same as Frequency
Although pitch is related mostly (but not completely) to frequency, they are not the same. A phrase such as the pitch of the sound is incorrect, because pitch is not a physical property of the sound. Frequency is the physical measurement of the number of oscillations per second of the sound. Pitch is a psychological reaction to sound that enables a human being to place the sound on a scale from high to low or from treble to bass. Frequency is the stimulus and pitch is the response.
a c, Royalty-Free/Corbis
Each musical instrument has its own characteristic sound and mixture of harmonics. (See Figures 14.26 and 14.27.) Instruments shown are (a) the tuning fork, (b) the ute, and (c) the clarinet.
(a)
(b)
(c)
Applying Physics 14.6 Why Does the Professor Sound Like Donald Duck?
A professor performs a demonstration in which he breathes helium and then speaks with a comical voice. One student explains, The velocity of sound in helium is higher than in air, so the fundamental frequency of the standing waves in the mouth is increased. Another student says, No, the fundamental frequency is determined by the vocal folds and cannot be changed. Only the quality of the voice has changed. Which student is correct? Explanation The second student is correct. The fundamental frequency of the complex tone from the voice is determined by the vibration of the vocal folds and is not changed by substituting a different gas in the mouth. The introduction of the helium into the mouth results in harmonics of higher frequencies being excited more than in the normal voice, but the fundamental frequency of the voice is the same only the quality has changed. The unusual inclusion of the higher frequency harmonics results in a common description of this effect as a high-pitched voice, but that description is incorrect. (It is really a quacky timbre.)
14.13 THE EAR
The human ear is divided into three regions: the outer ear, the middle ear, and the inner ear (Fig. 14.28, page 488). The outer ear consists of the ear canal (which is open to the atmosphere), terminating at the eardrum (tympanum). Sound waves travel down the ear canal to the eardrum, which vibrates in and out in phase with the pushes and pulls caused by the alternating high and low pressures of the waves. Behind the eardrum are three small bones of the middle ear, called the hammer, the anvil, and the stirrup because of their shapes. These bones transmit the vibration to the inner ear, which contains the cochlea, a snail-shaped tube about 2 cm long. The cochlea makes contact with the stirrup at the oval window and is divided along its length by the basilar membrane, which consists of small hairs (cilia) and nerve bers. This membrane varies in mass per unit length and in tension along its length, and different portions of it resonate at different frequencies. (Recall that the natural frequency of a string depends on its mass per unit length
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Figure 14.28 The structure of the human ear. The three tiny bones (ossicles) that connect the eardrum to the window of the cochlea act as a double-lever system to decrease the amplitude of vibration and hence increase the pressure on the uid in the cochlea.
Hammer Anvil Stirrup
Semicircular canals (for balance) Oval window
Vestibular nerve Cochlear nerve Cochlea
Eardrum (tympanum) Eustachian tube
Ear canal
and on the tension in it.) Along the basilar membrane are numerous nerve endings, which sense the vibration of the membrane and in turn transmit impulses to the brain. The brain interprets the impulses as sounds of varying frequency, depending on the locations along the basilar membrane of the impulse-transmitting nerves and on the rates at which the impulses are transmitted. Figure 14.29 shows the frequency response curves of an average human ear for sounds of equal loudness, ranging from 0 to 120 dB. To interpret this series of graphs, take the bottom curve as the threshold of hearing. Compare the intensity level on the vertical axis for the two frequencies 100 Hz and 1 000 Hz. The vertical axis shows that the 100-Hz sound must be about 38 dB greater than the 1 000-Hz sound to be at the threshold of hearing, which means that the threshold of hearing is very strongly dependent on frequency. The easiest frequencies to hear are around 3 300 Hz; those above 12 000 Hz or below about 50 Hz must be relatively intense to be heard. Now consider the curve labeled 80. This curve uses a 1 000-Hz tone at an intensity level of 80 dB as its reference. The curve shows that a tone of frequency 100 Hz would have to be about 4 dB louder than the 80-dB, 1 000-Hz tone in order to sound as loud. Notice that the curves atten out as the intensities levels of the sounds increase, so when sounds are loud, all frequencies can be heard equally well.
120 120 100 Intensity (dB) 80 60 40 20 20 0 20 50 100 Threshold of hearing 0 100 80 60 40
Threshold of pain
Figure 14.29 Curves of intensity level versus frequency for sounds that are perceived to be of equal loudness. Note that the ear is most sensitive at a frequency of about 3 300 Hz. The lowest curve corresponds to the threshold of hearing for only about 1% of the population.
500 1 000 Frequency(Hz)
5 000 10 000
Summary
489
The small bones in the middle ear represent an intricate lever system that increases the force on the oval window. The pressure is greatly magnied because the surface area of the eardrum is about 20 times that of the oval window (in analogy with a hydraulic press). The middle ear, together with the eardrum and oval window, in effect acts as a matching network between the air in the outer ear and the liquid in the inner ear. The overall energy transfer between the outer ear and the inner ear is highly efcient, with pressure amplication factors of several thousand. In other words, pressure variations in the inner ear are much greater than those in the outer ear. The ear has its own built-in protection against loud sounds. The muscles connecting the three middle-ear bones to the walls control the volume of the sound by changing the tension on the bones as sound builds up, thus hindering their ability to transmit vibrations. In addition, the eardrum becomes stiffer as the sound intensity increases. These two events make the ear less sensitive to loud incoming sounds. There is a time delay between the onset of a loud sound and the ears protective reaction, however, so a very sudden loud sound can still damage the ear. The complex structure of the human ear is believed to be related to the fact that mammals evolved from seagoing creatures. In comparison, insect ears are considerably simpler in design, because insects have always been land residents. A typical insect ear consists of an eardrum exposed directly to the air on one side and to an air-lled cavity on the other side. Nerve cells communicate directly with the cavity and the brain, without the need for the complex intermediary of an inner and middle ear. This simple design allows the ear to be placed virtually anywhere on the body. For example, a grasshopper has its ears on its legs. One advantage of the simple insect ear is that the distance and orientation of the ears can be varied so that it is easier to locate sources of sound, such as other insects. One of the most amazing medical advances in recent decades is the cochlear implant, allowing the deaf to hear. Deafness can occur when the hairlike sensors (cilia) in the cochlea break off over a lifetime or sometimes because of prolonged exposure to loud sounds. Because the cilia dont grow back, the ear loses sensitivity to certain frequencies of sound. The cochlear implant stimulates the nerves in the ear electronically to restore hearing loss that is due to damaged or absent cilia.
SUMMARY
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14.4 Energy and Intensity of Sound Waves
The average intensity of sound incident on a surface is dened by I power area A [14.6]
14.2
Characteristics of Sound Waves
Sound waves are longitudinal waves. Audible waves are sound waves with frequencies between 20 and 20 000 Hz. Infrasonic waves have frequencies below the audible range, and ultrasonic waves have frequencies above the audible range.
14.3
The Speed of Sound
where the power is the energy per unit time owing through the surface, which has area A. The intensity level of a sound wave is given by 10 log I I0 [14.7]
The speed of sound in a medium of bulk modulus B and density is v
B
[14.1]
The constant I 0 1.0 10 12 W/m2 is a reference intensity, usually taken to be at the threshold of hearing, and I is the intensity at level , measured in decibels (dB).
The speed of sound also depends on the temperature of the medium. The relationship between temperature and the speed of sound in air is v (331 m/s)
14.5 Spherical and Plane Waves
The intensity of a spherical wave produced by a point source is proportional to the average power emitted and inversely proportional to the square of the distance from the source: I
av
T 273 K
[14.4]
where T is the absolute (Kelvin) temperature and 331 m/s is the speed of sound in air at 0 C.
4 r2
[14.8]
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14.6
The Doppler Effect
The change in frequency heard by an observer whenever there is relative motion between a source of sound and the observer is called the Doppler effect. If the observer is moving with speed vO and the source is moving with speed vS , the observed frequency is fO fS v v vO vS [14.12]
where F is the tension in the string and unit length.
is its mass per
14.9 Forced Vibrations and Resonance
A system capable of oscillating is said to be in resonance with some driving force whenever the frequency of the driving force matches one of the natural frequencies of the system. When the system is resonating, it oscillates with maximum amplitude.
where v is the speed of sound. A positive speed is substituted for vO when the observer moves toward the source, a negative speed when the observer moves away from the source. Similarly, a positive speed is substituted for vS when the sources moves toward the observer, a negative speed when the source moves away.
14.10 Standing Waves in Air Columns
Standing waves can be produced in a tube of air. If the reecting end of the tube is open, all harmonics are present and the natural frequencies of vibration are fn n v 2L nf 1 n 1, 2, 3, . . . [14.18]
14.7
Interference of Sound Waves
When waves interfere, the resultant wave is found by adding the individual waves together point by point. When crest meets crest and trough meets trough, the waves undergo constructive interference, with path length difference r2 r1 n (n 0, 1, 2, . . .) [14.13]
If the tube is closed at the reecting end, only the odd harmonics are present and the natural frequencies of vibration are fn n v 4L nf 1 n 1, 3, 5, . . . [14.19]
When crest meets trough, destructive interference occurs, with path length difference r2 r1 (n
1 2)
14.11 Beats
The phenomenon of beats is an interference effect that occurs when two waves with slightly different frequencies combine at a xed point in space. For sound waves, the intensity of the resultant sound changes periodically with time. The beat frequency is fb f2 f1 [14.20]
(n
0, 1, 2, . . .)
[14.14]
14.8
Standing Waves
Standing waves are formed when two waves having the same frequency, amplitude, and wavelength travel in opposite directions through a medium. The natural frequencies of vibration of a stretched string of length L, xed at both ends, are fn nf 1 n 2L
where f 2 and f 1 are the two source frequencies.
F
n
1, 2, 3, . . .
[14.17]
CONCEPTUAL QUESTIONS
1. (a) You are driving down the highway in your car when a police car sounding its siren overtakes you and passes you. If its frequency at rest is f 0, is the frequency you hear while the car is catching up to you higher or lower than f0? (b) What about the frequency you hear after the car has passed you? 2. A crude model of the human throat is that of a pipe open at both ends with a vibrating source to introduce the sound into the pipe at one end. Assuming the vibrating source produces a range of frequencies, discuss the effect of changing the pipes length. 3. An autofocus camera sends out a pulse of sound and measures the time taken for the pulse to reach an object, reect off of it, and return to be detected. Can the temperature affect the cameras focus? 4. To keep animals away from their cars, some people mount short, thin pipes on the fenders. The pipes give out a high-pitched wail when the cars are moving. How do they create the sound? 5. Secret agents in the movies always want to get to a secure phone with a voice scrambler. How do these devices work? 6. When a bell is rung, standing waves are set up around its circumference. What boundary conditions must be satised by the resonant wavelengths? How does a crack in the bell, such as in the Liberty Bell, affect the satisfying of the boundary conditions and the sound emanating from the bell? 7. How does air temperature affect the tuning of a wind instrument? 8. Explain how the distance to a lightning bolt can be determined by counting the seconds between the ash and the sound of thunder. 9. You are driving toward a cliff and you honk your horn. Is there a Doppler shift of the sound when you hear the echo? Is it like a moving source or moving observer? What if the reection occurs not from a cliff, but from the forward edge of a huge alien spacecraft moving toward you as you drive?
Problems
491
10. Of the following sounds, state which is most likely to have an intensity level of 60 dB: a rock concert, the turning of a page in this text, a normal conversation, a cheering crowd at a football game, and background noise at a church? 11. Guitarists sometimes play a harmonic by lightly touching a string at its exact center and plucking the string. The result is a clear note one octave higher than the fundamental frequency of the string, even though the string is not pressed to the ngerboard. Why does this happen? 12. Will two separate 50-dB sounds together constitute a 100-dB sound? Explain. 13. An archer shoots an arrow from a bow. Does the string of the bow exhibit standing waves after the arrow leaves? If so, and if the bow is perfectly symmetric so that the arrow leaves from the center of the string, what harmonics are excited? 14. The radar systems used by police to detect speeders are sensitive to the Doppler shift of a pulse of radio waves. Discuss how this sensitivity can be used to measure the speed of a car. 15. As oppositely moving pulses of the same shape (one upward, one downward) on a string pass through each
other, there is one instant at which the string shows no displacement from the equilibrium position at any point. Has the energy carried by the pulses disappeared at this instant of time? If not, where is it? 16. A soft drink bottle resonates as air is blown across its top. What happens to the resonant frequency as the level of uid in the bottle decreases? 17. A blowing whistle is attached to the roof of a car that moves around a circular race track. Assuming youre standing near the outside of the track, explain the nature of the sound you hear as the whistle comes by each time. 18. Despite a reasonably steady hand, a person often spills his coffee when carrying it to his seat. Discuss resonance as a possible cause of this difculty, and devise a means for solving the problem. 19. An airplane mechanic notices that the sound from a twinengine aircraft varies rapidly in loudness when both engines are running. What could be causing this variation from loud to soft? 20. Why does a vibrating guitar string sound louder when placed on the instrument than it would if allowed to vibrate in the air while off the instrument?
PROBLEMS
1, 2, 3 = straightforward, intermediate, challenging = full solution available in Student Solutions Manual/Study Guide = coached problem with hints available at www.cp7e.com = biomedical application Section 14.2 Characteristics of Sound Waves Section 14.3 The Speed of Sound Unless otherwise stated, use 345 m/s as the speed of sound in air. 1. Suppose that you hear a clap of thunder 16.2 s after seeing the associated lightning stroke. The speed of sound waves in air is 343 m/s and the speed of light in air is 3.00 108 m/s. How far are you from the lightning stroke? 2. A dolphin located in sea water at a temperature of 25 C emits a sound directed toward the bottom of the ocean 150 m below. How much time passes before it hears an echo? 3. A sound wave has a frequency of 700 Hz in air and a wavelength of 0.50 m. What is the temperature of the air? 4. The range of human hearing extends from approximately 20 Hz to 20 000 Hz. Find the wavelengths of these extremes at a temperature of 27 C. 5. A group of hikers hears an echo 3.00 s after shouting. If the temperature is 22.0 C, how far away is the mountain that reected the sound wave? 6. A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.00 s later. Find the depth of the well if the air temperature is 10.0 C. 7. You are watching a pier being constructed on the far shore of a saltwater inlet when some blasting occurs. You hear the sound in the water 4.50 s before it reaches you through the air. How wide is the inlet? [Hint: See Table 14.1. Assume the air temperature is 20 C.] 8. The speed of sound in a column of air is measured to be 356 m/s. What is the temperature of the air? Section 14.4 Energy and Intensity of Sound Waves 9. The toadsh makes use of resonance in a closed tube to produce very loud sounds. The tube is its swim bladder, used as an amplier. The sound level of this creature has been measured as high as 100 dB. (a) Calculate the intensity of the sound wave emitted. (b) What is the intensity level if three of these sh try to imitate three frogs by saying Budweiser at the same time. 10. The area of a typical eardrum is about 5.0 10 5 m2. Calculate the sound power (the energy per second) incident on an eardrum at (a) the threshold of hearing and (b) the threshold of pain. 11. There is evidence that elephants communicate via infrasound, generating rumbling vocalizations as low as 14 hz that can travel up to 10 km. The intensity level of these sounds can reach 103 dB, measured a distance of 5.0 m from the source. Determine the intensity level of the infrasound 10 km from the source, assuming the sound energy radiates uniformly in all directions. 12. Two sounds have measured intensities of I1 100 W/m2 and I2 200 W/m2. By how many decibels is the level of sound 1 lower than that of sound 2? 13. A noisy machine in a factory produces sound with a level of 80 dB. How many identical machines could you add to the factory without exceeding the 90-dB limit? 14. A family ice show is held at an enclosed arena. The skaters perform to music playing at a level of 80.0 dB. This intensity level is too loud for your baby, who yells at 75.0 dB. (a) What total sound intensity engulfs you? (b) What is the combined sound level?
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15. Calculate the sound level in decibels of a sound wave that has an intensity of 4.00 W/m2. Section 14.5 Spherical and Plane Waves 16. An outside loudspeaker (considered a small source) emits sound waves with a power output of 100 W. (a) Find the intensity 10.0 m from the source. (b) Find the intensity level in decibels at that distance. (c) At what distance would you experience the sound at the threshold of pain, 120 dB? 17. A train sounds its horn as it approaches an intersection. The horn can just be heard at a level of 50 dB by an observer 10 km away. (a) What is the average power generated by the horn? (b) What intensity level of the horns sound is observed by someone waiting at an intersection 50 m from the train? Treat the horn as a point source and neglect any absorption of sound by the air.
23. Two trains on separate tracks move towards one another. Train 1 has a speed of 130 km/h, train 2 a speed of 90.0 km/h. Train 2 blows its horn, emitting a frequency of 500 Hz. What is the frequency heard by the engineer on train 1? 24. A bat ying at 5.0 m/s emits a chirp at 40 kHz. If this sound pulse is reected by a wall, what is the frequency of the echo received by the bat? 25. An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 442 Hz when the train is approaching him and 441 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he nd? 26. Expectant parents are thrilled to hear their unborn babys heartbeat, revealed by an ultrasonic motion detector. Suppose the fetuss ventricular wall moves in simple harmonic motion with amplitude 1.80 mm and frequency 115 per minute. (a) Find the maximum linear speed of the heart wall. Suppose the motion detector in contact with the mothers abdomen produces sound at precisely 2 MHz, which travels through tissue at 1.50 km/s. (b) Find the maximum frequency at which sound arrives at the wall of the babys heart. (c) Find the maximum frequency at which reected sound is received by the motion detector. (By electronically listening for echoes at a frequency different from the broadcast frequency, the motion detector can produce beeps of audible sound in synchrony with the fetal heartbeat.) 27. A tuning fork vibrating at 512 Hz falls from rest and accelerates at 9.80 m/s2. How far below the point of release is the tuning fork when waves of frequency 485 Hz reach the release point? Take the speed of sound in air to be 340 m/s. 28. A supersonic jet traveling at Mach 3 at an altitude of 20 000 m is directly overhead at time t 0, as in Figure P14.28. (a) How long will it be before the ground observer encounters the shock wave? (b) Where will the plane be when it is nally heard? (Assume an average value of 330 m/s for the speed of sound in air.)
18. A skyrocket explodes 100 m above the ground (Fig. P14.18). Three observers are spaced 100 m apart, with the rst (A) directly under the explosion. (a) What is the ratio of the sound intensity heard by observer A to that heard by observer B? (b) What is the ratio of the intensity heard by observer A to that heard by observer C?
P
100 m
A
100 m
B
100 m
C
Figure P14.18
19. Show that the difference in decibel levels 1 and 2 of a sound source is related to the ratio of its distances r 1 and r2 from the receivers by the formula
2 1
20 log
r1 r2
x
Section 14.6 The Doppler Effect 20. An airplane traveling at half the speed of sound (v 172 m/s) emits a sound of frequency 5.00 kHz. At what frequency does a stationary listener hear the sound (a) as the plane approaches? (b) after it passes? 21. A commuter train passes a passenger platform at a constant speed of 40.0 m/s. The train horn is sounded at its characteristic frequency of 320 Hz. (a) What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding? (b) What wavelength is detected by a person on the platform as the train approaches? 22. At rest, a cars horn sounds the note A (440 Hz). The horn is sounded while the car is moving down the street. A bicyclist moving in the same direction with one-third the cars speed hears a frequency of 415 Hz. What is the speed of the car? Is the cyclist ahead of or behind the car?
t=0 h h
Observer (a) Figure P14.28
Observer hears the boom (b)
29. The now-discontinued Concorde ew at Mach 1.5, which meant the speed of the plane was 1.5 times the speed of sound in air. What was the angle between the direction of propagation of the shock wave and the direction of the planes velocity?
Problems
493
Section 14.7 Interference of Sound Waves 30. The acoustical system shown in Figure 14.14 is driven by a speaker emitting a 400-Hz note. If destructive interference occurs at a particular instant, how much must the path length in the U-shaped tube be increased in order to hear (a) constructive interference and (b) destructive interference once again? 31. The ship in Figure P14.31 travels along a straight line parallel to the shore and 600 m from it. The ships radio receives simultaneous signals of the same frequency from antennas A and B. The signals interfere constructively at point C, which is equidistant from A and B. The signal goes through the rst minimum at point D. Determine the wavelength of the radio waves.
A 800 m
B
antinodes; no other nodes or antinodes are present. What is the frequency of the resonance if the rod is 1.00 m long? Two speakers are driven by a common 37. oscillator at 800 Hz and face each other at a distance of 1.25 m. Locate the points along a line joining the speakers where relative minima of the amplitude of the pressure would be expected. (Use v 343 m/s.) 38. Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each xed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 60 Hz, what is the frequency of the second harmonic in the longer wire? 39. A 12-kg object hangs in equilibrium from a string of total length L 5.0 m and linear mass density 0.001 0 kg/m. The string is wrapped around two light, frictionless pulleys that are separated by the distance d 2.0 m (Fig. P14.39a). (a) Determine the tension in the string. (b) At what frequency must the string between the pulleys vibrate in order to form the standing-wave pattern shown in Figure P14.39b?
600 m
d
C D Figure P14.31
d
32. Two loudspeakers are placed above and below one another, as in Figure 14.15, and are driven by the same source at a frequency of 500 Hz. (a) What minimum distance should the top speaker be moved back in order to create destructive interference between the speakers? (b) If the top speaker is moved back twice the distance calculated in part (a), will there be constructive or destructive interference? 33. A pair of speakers separated by 0.700 m are driven by the same oscillator at a frequency of 690 Hz. An observer originally positioned at one of the speakers begins to walk along a line perpendicular to the line joining the speakers. (a) How far must the observer walk before reaching a relative maximum in intensity? (b) How far will the observer be from the speaker when the rst relative minimum is detected in the intensity? Section 14.8 Standing Waves 34. A steel wire in a piano has a length of 0.700 0 m and a mass of 4.300 10 3 kg. To what tension must this wire be stretched in order that the fundamental vibration correspond to middle C ( fC 261.6 Hz on the chromatic musical scale)? 35. A stretched string xed at each end has a mass of 40.0 g and a length of 8.00 m. The tension in the string is 49.0 N. (a) Determine the positions of the nodes and antinodes for the third harmonic. (b) What is the vibration frequency for this harmonic? 36. Resonance of sound waves can be produced within an aluminum rod by holding the rod at its midpoint and stroking it with an alcohol-saturated paper towel. In this resonance mode, the middle of the rod is a node while the ends are
g
m (a)
Figure P14.39
m (b)
40. In the arrangement shown in Figure P14.40, an object of mass m 5.0 kg hangs from a cord around a light pulley. The length of the cord between point P and the pulley is L 2.0 m. (a) When the vibrator is set to a frequency of 150 Hz, a standing wave with six loops is formed. What must be the linear mass density of the cord? (b) How many loops (if any) will result if m is changed to 45 kg? (c) How many loops (if any) will result if m is changed to 10 kg?
Vibrator P
L Pulley m
m Figure P14.40
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41. A 60.00-cm guitar string under a tension of 50.000 N has a mass per unit length of 0.100 00 g/cm. What is the highest resonant frequency that can be heard by a person capable of hearing frequencies up to 20 000 Hz? Section 14.9 Forced Vibrations and Resonance 42. Standing-wave vibrations are set up in a crystal goblet with four nodes and four antinodes equally spaced around the 20.0-cm circumference of its rim. If transverse waves move around the glass at 900 m/s, an opera singer would have to produce a high harmonic with what frequency in order to shatter the glass with a resonant vibration? Section 14.10 Standing Waves in Air Columns 43. The windpipe of a typical whooping crane is about 5.0 feet long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? Assume a temperature of 37 C. 44. The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find the frequency of the lowest note a piccolo can play, assuming the speed of sound in air is 340 m/s. (b) Opening holes in the side effectively shortens the length of the resonant column. If the highest note a piccolo can sound is 4 000 Hz, nd the distance between adjacent antinodes for this mode of vibration. 45. The human ear canal is about 2.8 cm long. If it is regarded as a tube that is open at one end and closed at the eardrum, what is the fundamental frequency around which we would expect hearing to be most sensitive? Take the speed of sound to be 340 m/s. 46. A shower stall measures 86.0 cm 86.0 cm 210 cm. When you sing in the shower, which frequencies will sound the richest (because of resonance)? Assume the stall acts as a pipe closed at both ends, with nodes at opposite sides. Assume also that the voices of various singers range from 130 Hz to 2 000 Hz. Let the speed of sound in the hot shower stall be 355 m/s. A pipe open at both ends has a funda47. mental frequency of 300 Hz when the temperature is 0 C. (a) What is the length of the pipe? (b) What is the fundamental frequency at a temperature of 30 C? 48. A 2.00-m-long air column is open at both ends. The frequency of a certain harmonic is 410 Hz, and the frequency of the next-higher harmonic is 492 Hz. Determine the speed of sound in the air column. Section 14.11 Beats 49. Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 523 Hz. The peg of one string slips slightly, and the tension in it drops to 196 N. How many beats per second are heard? 50. The G string on a violin has a fundamental frequency of 196 Hz. It is 30.0 cm long and has a mass of 0.500 g. While this string is sounding, a nearby violinist effectively shortens the G string on her identical violin (by sliding her nger down the string) until a beat frequency of 2.00 Hz is heard between the two strings. When this occurs, what is the effective length of her string? 51. Two train whistles have identical frequencies of 180 Hz. When one train is at rest in the station, sounding its whistle, a beat frequency of 2 Hz is heard from a moving train.
What two possible speeds and directions can the moving train have? 52. Two pipes of equal length are each open at one end. Each has a fundamental frequency of 480 Hz at 300 K . In one pipe, the air temperature is increased to 305 K . If the two pipes are sounded together, what beat frequency results? 53. A student holds a tuning fork oscillating at 256 Hz. He walks toward a wall at a constant speed of 1.33 m/s. (a) What beat frequency does he observe between the tuning fork and its echo? (b) How fast must he walk away from the wall to observe a beat frequency of 5.00 Hz? Section 14.13 The Ear 54. If a human ear canal can be thought of as resembling an organ pipe, closed at one end, that resonates at a fundamental frequency of 3 000 Hz, what is the length of the canal? Use a normal body temperature of 37 C for your determination of the speed of sound in the canal. 55. Some studies suggest that the upper frequency limit of hearing is determined by the diameter of the eardrum. The wavelength of the sound wave and the diameter of the eardrum are approximately equal at this upper limit. If the relationship holds exactly, what is the diameter of the eardrum of a person capable of hearing 20 000 Hz? (Assume a body temperature of 37 C.) ADDITIONAL PROBLEMS 56. A commuter train blows its horn as it passes a passenger platform at a constant speed of 40.0 m/s. The horn sounds at a frequency of 320 Hz when the train is at rest. What is the frequency observed by a person on the platform (a) as the train approaches and (b) as the train recedes from him? (c) What wavelength does the observer nd in each case? 57. A quartz watch contains a crystal oscillator in the form of a block of quartz that vibrates by contracting and expanding. Two opposite faces of the block, 7.05 mm apart, are antinodes, moving alternately towards and away from each other. The plane halfway between these two faces is a node of the vibration. The speed of sound in quartz is 3.70 km/s. Find the frequency of the vibration. An oscillating electric voltage accompanies the mechanical oscillation, so the quartz is described as piezoelectric. An electric circuit feeds in energy to maintain the oscillation and also counts the voltage pulses to keep time. 58. A owerpot is knocked off a balcony 20.0 m above the sidewalk and falls toward an unsuspecting 1.75-m-tall man who is standing below. How close to the sidewalk can the owerpot fall before it is too late for a warning shouted from the balcony to reach the man in time? Assume that the man below requires 0.300 s to respond to the warning. On a workday, the average decibel level 59. of a busy street is 70 dB, with 100 cars passing a given point every minute. If the number of cars is reduced to 25 every minute on a weekend, what is the decibel level of the street? 60. A variable-length air column is placed just below a vibrating wire that is xed at both ends. The length of the column, open at one end, is gradually increased from zero until the rst position of resonance is observed at
Problems
495
L 34.0 cm. The wire is 120 cm long and is vibrating in its third harmonic. If the speed of sound in air is 340 m/s, what is the speed of transverse waves in the wire? 61. A block with a speaker bolted to it is connected to a spring having spring constant k 20.0 N/m, as shown in Figure P14.61. The total mass of the block and speaker is 5.00 kg, and the amplitude of the units motion is 0.500 m. If the speaker emits sound waves of frequency 440 Hz, determine the lowest and highest frequencies heard by the person to the right of the speaker.
y
(x,y)
A x 9.00 m 10.0 m Figure P14.66
k
m
x Figure P14.61
62. A ute is designed so that it plays a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20.0 C. (a) Consider the ute to be a pipe open at both ends, and nd its length, assuming that the middle-C frequency is the fundamental frequency. (b) A second player, nearby in a colder room, also attempts to play middle C on an identical ute. A beat frequency of 3.00 beats/s is heard. What is the temperature of the room? 63. When at rest, two trains have sirens that emit a frequency of 300 Hz. The trains travel toward one another and toward an observer stationed between them. One of the trains moves at 30.0 m/s, and the observer hears a beat frequency of 3.0 beats per second. What is the speed of the second train, which travels faster than 30.0 m/s? 64. Many artists sing very high notes in ad lib ornaments and cadenzas. The highest note written for a singer in a published score was F-sharp above high C, 1.480 kHz, sung by Zerbinetta in the original version of Richard Strausss opera Ariadne auf Naxos. (a) Find the wavelength of this sound in air. (b) In response to complaints, Strauss later transposed the note down to F above high C, 1.397 kHz. By what increment did the wavelength change? 65. A speaker at the front of a room and an identical speaker at the rear of the room are being driven at 456 Hz by the same sound source. A student walks at a uniform rate of 1.50 m/s away from one speaker and towards the other. How many beats does the student hear per second? 66. Two identical speakers separated by 10.0 m are driven by the same oscillator with a frequency of f 21.5 Hz (Fig. P14.66). Explain why a receiver at A records a minimum in sound intensity from the two speakers. (b) If the receiver is moved in the plane of the speakers, what path should it take so that the intensity remains at a minimum? That is, determine the relationship between x and y (the coordinates of the receiver) such that the receiver will record a minimum in sound intensity. Take the speed of sound to be 344 m/s.
67. By proper excitation, it is possible to produce both longitudinal and transverse waves in a long metal rod. In a particular case, the rod is 150 cm long and 0.200 cm in radius and has a mass of 50.9 g. Youngs modulus for the material is 6.80 1010 Pa. Determine the required tension in the rod so that the ratio of the speed of longitudinal waves to the speed of transverse waves is 8. 68. A student stands several meters in front of a smooth reecting wall, holding a board on which a wire is xed at each end. The wire, vibrating in its third harmonic, is 75.0 cm long, has a mass of 2.25 g, and is under a tension of 400 N. A second student, moving towards the wall, hears 8.30 beats per second. What is the speed of the student approaching the wall? Use 340 m/s as the speed of sound in air. 69. Two ships are moving along a line due east. The trailing vessel has a speed of 64.0 km/h relative to a land-based observation point, and the leading ship has a speed of 45.0 km/h relative to the same station. The trailing ship transmits a sonar signal at a frequency of 1 200 Hz. What frequency is monitored by the leading ship? (Use 1 520 m/s as the speed of sound in ocean water.) 70. The Doppler equation presented in the text is valid when the motion between the observer and the source occurs on a straight line, so that the source and observer are moving either directly toward or directly away from each other. If this restriction is relaxed, one must use the more general Doppler equation fO v v vO cos( O) vS cos( S) fS
where O and S are dened in Figure P14.70a. (a) If both observer and source are moving away from each other along a straight line, show that the preceding equation yields the same result as Equation 14.12 in the text.
25.0 m/s vS Source uS
uO vO
Observer
(a) Figure P14.70
(b)
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Sound
(b) Use the preceding equation to solve the following problem: A train moves at a constant speed of 25.0 m/s toward the intersection shown in Figure P14.70b. A car is stopped near the intersection, 30.0 m from the tracks. If the trains horn emits a frequency of 500 Hz, what is the frequency heard by the passengers in the car when the train is 40.0 m to the left of the intersection? Take the speed of sound to be 343 m/s. 71. A rescue plane ies horizontally at a constant speed, searching for a disabled boat. When the plane is directly above the boat, the boats crew blows a loud horn. By the time the planes sound detector perceives the horns sound, the plane has traveled a distance equal to one-half its altitude above the ocean. If the sound takes 2.00 s to reach the plane, determine (a) the planes altitude and (b) its speed. 72. In order to determine her speed, a skydiver carries a tone generator. A friend on the ground at the landing site has equipment for receiving and analyzing sound waves. While the skydiver is falling at terminal speed, her tone generator emits a steady tone of 1.80 kHz. (Assume that the air is calm, that the speed of sound is 343 m/s, independent of altitude.) (a) If her friend on the ground (directly beneath the skydiver) receives waves of frequency 2.15 kHz, what is the skydivers speed of descent? (b) If the skydiver were also carrying sound-receiving equipment sensitive enough to detect waves reected from the ground, what frequency of waves would she receive? ACTIVITIES A.1. Use an empty 1-liter soft-drink container, blow over the open end, and listen to the sound that is produced. Add some water to the container to change the height of the air column, and repeat the procedure. How does the frequency that you hear change with the height of the air column? If you want to investigate this phenomenon in more detail, construct a musical instrument made up of several soft-drink bottles with different amounts of water in each. You can play your instrument as a wind instrument by blowing over the mouths of the bottles. A.2. Beats can easily be heard on a guitar. When a nger is placed at the fth fret of the second string, the note produced when the string is plucked should be identical to the note from the rst string when it is played without ngering. With your nger in position on the second string, pluck the two strings simultaneously. If one of the strings is slightly out of tune, a very pronounced beat frequency will be heard. What happens to the beat frequency as the string tension is changed in small increments from too low for the intended tuning to too high?
A.3. Attach a rope to a door and shake the other end to see how many of the standing-wave patterns in Figure 14.18 you can produce. When a pattern is formed, note that the amplitude of the ropes vibration is much larger than the movement of your hand. A.4. Snip off the corners of one end of a paper straw so that the end tapers to a point, as shown in Figure A14.4. Chew on this end to atten it, and you have created a doublereed instrument. Put your lips around the tapered end of the straw, press them together slightly, and blow through the straw. When you hear a steady tone, slowly snip off a piece of the straw at the other end. Be careful to keep about the same amount of pressure with your lips. How does the frequency of the sound change as the straw becomes shorter? Why does this change occur? You may be able to produce more than one tone for any given length of the straw. Why?
Figure A14.4
A.5. Inate a balloon just enough to form a small sphere. Measure its diameter. Use a marker to color in a 1-cm square on its surface. Now continue inating the balloon until it reaches twice the original diameter. Measure the size of the square now. Note how the color of the marked area has changed. Use the information in Section 14.5 to explain these results.
I love hearing that lonesome wail of the train whistle as the magnitude of the frequency of the wave changes due to the Doppler effect.
Sidney Harris

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University of Texas - PHY - 303K

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University of Texas - AST - 309

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University of Texas - AST - 309

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University of Texas - AST - 309

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University of Texas - AST - 309

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University of Texas - AST - 309

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University of Texas - PGE - 203

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Homework for MondayRecord the data in today's experiment on porosity. And calculate the porosity from the equation:porosity = pore volume/bulk volumeWrite a brief description of the work and calculation in such as way that you can understand it if you

University of Texas - PGE - 203

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University of Texas - GEO - 401

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University of Texas - GEO - 401

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University of Texas - GEO - 401

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University of Texas - GEO - 401

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University of Texas - GEO - 401

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University of Texas - GEO - 401

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At its maximum about 20 thousand years ago, an ice sheet covered most of Canada and part of the northernmost U. S. Today in the Northern Hemisphere an extensive ice sheet covers only Greenland.A few thousand years ago, rock debris embedded in an ice shee

University of Texas - GEO - 401

University of Texas - GEO - 401

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University of Texas - GEO - 401

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University of Texas - GEO - 401

GEO 303, Quiz 1 W Name (printed) _ October 3, 2006 100 points Page 1 of 5INSTRUCTION: (i) Put a "W" (for white exam) in the upper-right corner of your answer sheet. (ii) Answer the questions by blackening the "bubbles" on the provided special answer shee

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University of Texas - GEO - 401

University of Texas - GEO - 401

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University of Texas - GEO - 401

University of Texas - GEO - 401

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April 7, 2004 100 points Page 1 of 4WName (printed)_GEO 303, Quiz 2 INSTRUCTIONS: (i) Print and bubble in your name and student identification (not your social security number) on the answer sheet. (ii) Write "W" (for white exam) in the upper-right cor

University of Texas - GEO - 401

Answers to GEO 303 sample Quiz 2 1. A 2. A 3. D 4. C 5. D 6. C 7. B 8. A 9. B 10. C 11. B 12. B 13. D 14. C 15. A 16. C 17. D 18. D 19. A 20. A 21. C 22. D 23. C 24. B 25. A 26. A 27. B 28. C 29. D 30. C 31. B 32. A 33. B 34. D 35. D 36. B 37. C 38. D 39.

University of Texas - GEO - 401

Previously accumulated daughter argon is driven out of very hot basaltic lava during eruption.Basalt lava quickly cools and crystallizes, after which radiogenic Ar begins to accumulate once again.Magma comprising a batholith cools very slowly.Geologic

University of Texas - GEO - 401

Chapter 16: Seismicity Next lecture: Chapter 16 Seismic wavesmaking seismic waves: elastic rebound wave fronts and ray paths body (P and S) waves surface (L) waves: Love and RayleighEarthquake propagation exampleSan Francisco, 1906Seismometers and sei

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Today: Stratigraphy (Chapter 13) Next Time: Vertebrates Part I (Chapter 15)Chapter 13: Reading the strata Next lecture: Chapter 15 - VertebratesPrinciples of StratigraphyLaw of superposition Principle of original horizontality Principle of lateral cont

University of Texas - GEO - 401

Chapter 15: Vertebrates 1 Next lecture: More Chapter 15 Cladistics Classification and Characteristics Jawless Fish Jawed FishPlacoderms Cartilaginous fish Bony fish Ray-finned fish Lobe-finned fishLife on LandCladisticscladistics: classify living thin

University of Texas - GEO - 401

Today: Vertebrates, Part II (Chapter 15) Next Time: History of Geologic Thought (Review Ch. 11-15)Chapter 15: Vertebrates 2 Next lecture: Chapters 11-15 TetrapodsAmphibians Amniotes Reptiles (Diapsids) Dinosaurs Swimming reptiles Flying reptiles Birds S