Additional context

This document delves into the physics of sound, a phenomenon central to our perception and interaction with the environment. Sound waves, as longitudinal waves, are a fundamental area of study in acoustics, building upon the established principles of wave mechanics and thermodynamics. This work likely draws from the extensive research on the human auditory system, exploring how our ears detect and interpret pressure fluctuations. Furthermore, the discussion of the Doppler effect connects to broader applications in fields like medical imaging and radar technology, where frequency shifts are used to determine the velocity and position of objects. The study of resonance in musical instruments builds on centuries of work in musical acoustics, while the analysis of overlapping sound waves relates to concepts in signal processing and noise control, areas critical in modern engineering.

Image summary: The image is a photograph. It depicts a man standing in a snowy, mountainous landscape, playing a very long wooden horn. The background features more snow-covered mountains and a clear sky. It is likely that the man is playing traditional music in a mountainous region.

Sound and Hearing

of all the mechanical waves that occur in nature, the most important in our everyday lives are longitudinal waves in a medium—usually air—called sound waves. The reason is that the human ear is tremendously sensitive and can detect sound waves even of very low intensity. The ability to hear an unseen nocturnal predator was essential to the survival of our ancestors, so it is no exaggeration to say that we humans owe our existence to our highly evolved sense of hearing.

Definition

Pressure Fluctuation: The amount by which the pressure in a sound wave differs from normal atmospheric pressure at a given point and time; it can be positive or negative, corresponding to compressions and rarefactions.

In Chapter 15 we described mechanical waves primarily in terms of displacement; however, because the ear is primarily sensitive to changes in pressure, it's often more appropriate to describe sound waves in terms of pressure fluctuations. We'll study the relationships among displacement, pressure fluctuation, and intensity and the connections between these quantities and human sound perception.

When a source of sound or a listener moves through the air, the listener may hear a frequency different from the one emitted by the source. This is the Doppler effect, which has important applications in medicine and technology.

16.1 Sound Waves

The most general definition of sound is a longitudinal wave in a medium. Our main concern is with sound waves in air, but sound can travel through any gas, liquid, or solid. You may be all too familiar with the propagation of sound through a solid if your neighbor's stereo speakers are right next to your wall.

The simplest sound waves are sinusoidal waves, which have definite frequency, amplitude, and wavelength. The human ear is sensitive to waves in the frequency range from about 20 to 20,000 Hz, called the audible range, but we also use the term “sound” for similar waves with frequencies above (ultrasonic) and below (infrasonic) the range of human hearing.

Learning Outcomes

In this chapter, you'll learn...

16.1 How to describe a sound wave in terms of either particle displacements or pressure fluctuations.

16.2 How to calculate the speed of sound waves in different materials.

16.3 How to calculate the intensity of a sound wave.

16.4 What determines the particular frequencies of sound produced by an organ or a flute.

16.5 How resonance occurs in musical instruments.

16.6 What happens when sound waves from different sources overlap.

16.7 How to describe what happens when two sound waves of slightly different frequencies are combined.

16.8 Why the pitch of a siren changes as it moves past you.

16.9 Why an airplane flying faster than sound produces a shock wave.

You'll need to review...

6.4 Power.

8.1 The impulse-momentum theorem.

Definition

Bulk Modulus: A measure of a substance's resistance to uniform compression; the ratio of pressure change to fractional volume change.

11.4 Bulk modulus and Young's modulus.

12.2 Gauge pressure and absolute pressure.

14.8 Forced oscillations and resonance.

15.1 to 15.8 Mechanical waves.

Figure 16.1 A sinusoidal longitudinal wave traveling to the right in a fluid. (Compare to figure 15.7.)

Figure 16.1 summary: The figure illustrates a sound wave propagating, showing displacement over time. It depicts a plunger moving in simple harmonic motion, creating a wave. The wave's progression is shown at different time intervals. The figure highlights the concept of wavelength and amplitude in relation to the wave's motion. The displacement of two particles in the medium, separated by one wavelength, is traced over time, showing their relative positions.

Longitudinal waves are shown at intervals of 1 over 8 times T for one period T.

Sound waves usually travel outward in all directions from the source of sound, with an amplitude that depends on the direction and distance from the source. We'll return to this point in the next section. For now, we concentrate on the idealized case of a sound wave that propagates in the positive x-direction only. As we discussed in Section 15.3, for such a wave, the wave function y(x, t) gives the instantaneous displacement y of a particle in the medium at position x at time t. If the wave is sinusoidal, we can express it by using equation (15.7):

Math summary: This equation models a sound wave moving in one direction. It calculates the displacement of a particle at a specific location and time, based on the wave's amplitude, wave number, angular frequency, position, and time.

Definition

Displacement Amplitude: The maximum displacement of a particle in a medium from its equilibrium position as a sound wave passes.

In a longitudinal wave the displacements are parallel to the direction of travel of the wave, so distances x and y are measured parallel to each other, not perpendicular as in a transverse wave. The amplitude A is the maximum displacement of a particle in the medium from its equilibrium position (Fig. 16.1). Hence A is also called the displacement amplitude.

Sound Waves as Pressure Fluctuations

We can also describe sound waves in terms of variations of pressure at various points. In a sinusoidal sound wave in air, the pressure fluctuates sinusoidally above and below atmospheric pressure p sub a with the same frequency as the motions of the air particles. The human ear operates by sensing such pressure variations.

A sound wave entering the ear canal exerts a fluctuating pressure on one side of the eardrum; the air on the other side of the eardrum, vented to the outside by the Eustachian tube, is at atmospheric pressure. The pressure difference on the two sides of the eardrum sets it into motion. Microphones and similar devices also usually sense pressure differences, not displacements.

Let p of x comma t be the instantaneous pressure fluctuation in a sound wave at any point 10 at time t. That is, p of x comma t is the amount by which the pressure differs from normal atmospheric pressure p sub a. Think of p of x comma t as the gauge pressure defined in Section 12.2; it can be either positive or negative. The absolute pressure at a point is then p sub a plus p of x comma t.

To see the connection between the pressure fluctuation p of x comma t and the displacement y of x comma t in a sound wave propagating in the plus x direction, consider an imaginary cylinder of a wave medium (gas, liquid, or solid) with cross-sectional area S and its axis along the direction of propagation (Fig. 16.2). When no sound wave is present, the cylinder has length delta x and volume 5 equals S times delta x, as shown by the shaded volume in figure 16.2. When a wave is present, at time t the end of the cylinder that is initially at x is displaced by y sub 1 equals y of x comma t, and the end that is initially at x plus delta x is displaced by y sub 2 equals y of x plus delta x comma t; this is shown by the red lines. If y sub 2 is greater than y sub 1, as shown in figure 16.2, the cylinder's volume has increased, which causes a decrease in pressure. If y sub 2 is less than y sub 1, the cylinder's volume has decreased and the pressure has increased. If y sub 2 equals y sub 1, the cylinder is simply shifted to the left or right; there is no volume change and no pressure fluctuation. The pressure fluctuation depends on the difference between the displacements at neighboring points in the medium.

Figure 16.2 summary: The figure is an illustration of a sound wave propagating along the x-axis, depicting how the left and right ends of a fluid cylinder experience different displacements. The figure shows an undisturbed cylinder of fluid with a specific cross-sectional area, length, and volume. The sound wave causes varying displacements at different points along the x-axis, indicating that the displacement is dependent on both position and time.

Quantitatively, the change in volume delta V of the cylinder is

Math summary: This expression calculates the change in volume of a cylinder. It multiplies the cross sectional area by the difference between the cylinder's height at a slightly displaced position and its height at the original position.

In the limit as delta x approaches 0, the fractional change in volume d V divided by V (volume change divided by original volume) is

Math summary: This expression calculates the fractional change in volume as the limit of the change in x approaches zero. It computes the partial derivative of a function y with respect to x, representing the infinitesimal change in y as x changes.

The fractional volume change is related to the pressure fluctuation by the bulk modulus B, which by definition is B equals negative p of x, t divided by dV over V. Solving for p of x, t, we have

Math summary: This expression calculates pressure fluctuation as the product of negative bulk modulus and the spatial derivative of a function. It determines how pressure changes based on material properties and the rate of change of displacement with respect to position.

The negative sign arises because when the partial derivative of y of x and t with respect to x is positive, the displacement is greater at x plus delta x than at x, corresponding to an increase in volume, a decrease in pressure, and a negative pressure fluctuation.

When we evaluate the partial derivative of y of x and t with respect to x for the sinusoidal wave of Equation sixteen point one, we find

Math summary: This expression calculates the pressure fluctuation of a sinusoidal sound wave. It takes the product of the bulk modulus, wave number, and amplitude, and then multiplies it by the sine of the wave number times position minus the angular frequency times time.

Definition

Compressions: Regions in a longitudinal wave where the particles of the medium are closer together than normal, resulting in increased pressure and density.

Figure 16.3 shows y(x, t) and p(x, t) for a sinusoidal sound wave at t = 0. It also shows how individual particles of the wave are displaced at this time. While y(x, t) and p(x, t) describe the same wave, these two functions are one-quarter cycle out of phase: At any time, the displacement is greatest where the pressure fluctuation is zero, and vice versa. In particular, note that the compressions (points of greatest pressure and density) and rara-efactions (points of lowest pressure and density) are points of zero displacement.

Figure 16.3 summary: The figure includes two line graphs and a particle diagram to describe a sound wave. The first graph illustrates the displacement of particles relative to their position, while the second graph depicts pressure fluctuations relative to position. The particle diagram displays the displacement of individual particles in a fluid. Regions of positive displacement in the first graph correspond to particles being displaced to the right in the particle diagram, while regions of negative displacement correspond to particles being displaced to the left. In the particle diagram, rarefaction zones, where particles are pulled apart, correlate with negative pressure in the second graph, and compression zones, where particles pile up, correlate with positive pressure.

Definition

Pressure Amplitude: The maximum amount by which the pressure in a sound wave differs from normal atmospheric pressure.

Equation (16.4) shows that the quantity B times k times A represents the maximum pressure fluctuation. We call this the pressure amplitude, denoted by p sub max:

Waves of shorter wavelength lambda (larger wave number k equals 2 times pi divided by lambda) have greater pressure variations for a given displacement amplitude because the maxima and minima are squeezed closer together. A medium with a large value of bulk modulus B is less compressible and so requires a greater pressure amplitude for a given volume change (that is, a given displacement amplitude).

Caution Graphs of a sound wave The graphs in figure 16.3 show the wave at only one instant of time. Because the wave is propagating in the +x-direction, as time goes by the wave patterns described by the functions y(x, t) and p(x, t) move to the right at the wave speed v = omega/k . The particles, by contrast, simply oscillate back and forth in simple harmonic motion as shown in figure 16.1.

Example 16.1 Amplitude of a Sound Wave in Air

In a sinusoidal sound wave of moderate loudness, the maximum pressure variations are about 3.0 times 10 to the power of negative 2 Pascals above and below atmospheric pressure. Find the corresponding maximum displacement if the frequency is 1000 Hertz. In air at normal atmospheric pressure and density, the speed of sound is 344 meters per second and the bulk modulus is 1.42 times 10 to the power of 5 Pascals.

Identify and Set Up This problem involves the relationship between two ways of describing a sound wave: in terms of displacement and in terms of pressure. The target variable is the displacement amplitude A. We are given the pressure amplitude p subscript max, wave speed v, frequency f, and bulk modulus B. Equation (16.5) relates the target variable A to p subscript max. We use omega equals v times k to determine the wave number k from v and the angular frequency omega equals 2 times pi times f. Then from Equation (16.5), the maximum displacement is

Math summary: This expression calculates the displacement amplitude by dividing the maximum pressure by the product of the bulk modulus and the wave number. It substitutes given values for maximum pressure, bulk modulus, and wave number to compute the displacement amplitude.

Evaluate This displacement amplitude is only about 1 over 100 the size of a human cell. The ear actually senses pressure fluctuations; it detects these minuscule displacements only indirectly.

Execute From equation (15.6), Keyconcept In a sound wave, the pressure amplitude (maximum pressure fluctuation) and displacement amplitude (maximum displacement of a particle in the medium) are proportional to each other. The proportionality constant depends on the wavelength of the sound and the bulk modulus of the medium.

Math summary: This expression calculates the wave number. It divides the angular frequency, calculated as two pi times the frequency, by the velocity to get the wave number, which in this case equals 18.3 radians per meter.

Definition

Ossicles: The three tiny bones in the middle ear (malleus, incus, and stapes) that transmit and amplify vibrations from the eardrum to the inner ear.

A sound wave that enters the human ear sets the eardrum into oscillation, which in turn causes oscillation of the ossicles, a chain of three tiny bones in the middle ear (Fig. 16.4). The ossicles transmit this oscillation to the fluid (mostly water) in the inner ear; there the fluid motion disturbs hair cells that send nerve impulses to the brain with information about the sound. The area of the moving part of the ear-drum is about 43 millimeters², and that of the stapes (the smallest of the ossicles) where it connects to the inner ear is about 3.2 millimeters². For the sound in Example 16.1, determine (a) the pressure amplitude and (b) the displacement amplitude of the wave in the fluid of the inner ear, in which the speed of sound is 1500 meters per second.

Figure 16.4 summary: The figure is a diagram of the human ear. The diagram shows the auditory canal leading to the eardrum. The ossicles, which are the malleus, incus, and stapes, connect the eardrum to the cochlea. The ear canal is located more externally, while the cochlea is located more internally. The ossicles are located between the eardrum and the cochlea.

Identify and Set Up Although the sound wave here travels in liquid rather than air, the same principles and relationships among the properties of the wave apply. We can ignore the mass of the tiny ossicles (about 58 milligrams = 5.8 × 10⁻⁵ kg), so the force they exert on the inner-ear fluid is the same as that exerted on the eardrum and ossicles by the incident sound wave. (In Chapters 4 and 5 we used the same idea to say that the tension is the same at either end of a massless rope.) Hence the pressure amplitude in the inner ear, p sub max (inner ear), is greater than in the outside air, p sub max (air), because the same force is exerted on a smaller area (the area of the stapes versus the area of the eardrum). Given p sub max (inner ear), we find the displacement amplitude A sub inner from equation (16.5).

Execute (a) From the area of the eardrum and the pressure amplitude in air found in Example 16.1, the maximum force exerted by the sound wave in air on the eardrum is F sub max equals p sub max of air times S sub eardrum. Then

Math summary: This calculates the maximum pressure in the inner ear. It divides the maximum force by the stapes area, which is equivalent to multiplying the maximum air pressure by the ratio of the eardrum area to the stapes area, resulting in a maximum inner ear pressure of 0.40 Pascals.

Figure 16.4 The anatomy of the human ear. The middle ear is the size of a small marble; the ossicles (incus, malleus, and stapes) are the smallest bones in the human body.

(b) To find the maximum displacement A sub inner ear, we use A equals p sub max divided by B times k as in Example 16.1. The inner-ear fluid is mostly water, which has a much greater bulk modulus B than air. From Table the compressibility of water (unfortunately also called k) is 45.8 times 10 to the power of negative 11 Pascals to the power of negative 1, so B sub fluid equals 1 divided by (45.8 times 10 to the power of negative 11 Pascals to the power of negative 1) equals 2.18 times 10 to the power of 9 Pascals.

The wave in the inner ear has the same angular frequency omega as the wave in the air because the air, eardrum, ossicles, and inner-ear fluid all oscillate together (see Example 15.8 in Section 15.8). But because the wave speed v is greater in the inner ear than in the air (1500 meters per second versus 344 meters per second), the wave number k = omega / v is smaller. Using the value of omega from Example 16.1, we find

Math summary: This expression calculates the wave number in the inner ear by dividing the angular frequency by the wave speed in the inner ear, resulting in a wave number of 4.2 radians per meter.

Putting everything together, we have

Math summary: This expression calculates the displacement amplitude in the inner ear. It divides the maximum pressure in the inner ear by the product of the fluid's bulk modulus and a constant, resulting in the displacement amplitude.

Evaluate In part (a) we see that the ossicles increase the pressure amplitude by a factor of (43 millimeters to the 2) / (3.2 millimeters to the 2) = 13 . This amplification helps give the human ear its great sensitivity. The displacement amplitude in the inner ear is even smaller than in the air. But pressure variations within the inner-ear fluid are what set the hair cells into motion, so what matters is that the pressure amplitude is larger in the inner ear than in the air.

Keyconcept When a sound wave travels from one medium into a different medium, the wave frequency and angular frequency remain the same. The wave number and wavelength can change, however, as can the pressure amplitude and displacement amplitude.

Perception of Sound Waves

The physical characteristics of a sound wave are directly related to the perception of that sound by a listener. For a given frequency, the greater the pressure amplitude of a sinusoidal sound wave, the greater the perceived loudness. The relationship between pressure amplitude and loudness is not a simple one, and it varies from one person to another. One important factor is that the ear is not equally sensitive to all frequencies in the audible range. A sound at one frequency may seem louder than one of equal pressure amplitude at a different frequency.

At 1000 Hz the minimum pressure amplitude that can be perceived with normal hearing is about 3 times 10 to the power of negative 5 Pascals; to produce the same loudness at 200 Hz or 15,000 Hz requires about 3 times 10 to the power of negative 4 Pascals. Perceived loudness also depends on the health of the ear. Age usually brings a loss of sensitivity at high frequencies.

The frequency of a sound wave is the primary factor in determining the pitch of a sound, the quality that lets us classify the sound as “high” or “low.” The higher the frequency of a sound (within the audible range), the higher the pitch that a listener will perceive. Pressure amplitude also plays a role in determining pitch. When a listener compares two sinusoidal sound waves with the same frequency but different pressure amplitudes, the one with the greater pressure amplitude is usually perceived as louder but also as slightly lower in pitch.

3 Definitions

Definition 1: Harmonic Content: The specific combination of frequencies and amplitudes that constitute a complex tone. A graph of harmonic content shows amplitude plotted as a function of frequency.

Definition 2: Fourier Analysis: A mathematical process that decomposes a complex waveform into a sum of simpler sinusoidal waves, revealing its harmonic content.

Definition 3: Fundamental Frequency: The lowest natural frequency of a vibrating object or system.

Musical sounds have wave functions that are more complicated than a simple sine function. Figure 16.5a shows the pressure fluctuation in the sound wave produced by a clarinet. The pattern is so complex because the column of air in a wind instrument like a clarinet vibrates at a fundamental frequency and at many harmonics at the same time. (In Section 15.8, we described this same behavior for a string that has been plucked, bowed, or struck. We'll examine the physics of wind instruments in Section 16.4.) The sound wave produced in the surrounding air has a similar amount of each harmonic—that is, a similar harmonic content. Figure 16.5b shows the harmonic content of the sound of a clarinet. The mathematical process of translating a pressure-time graph like figure 16.5a into a graph of harmonic content like figure 16.5b is called Fourier analysis.

Definition

Timbre: The subjective quality of a sound that distinguishes it from others of the same pitch and loudness, largely determined by the harmonic content and the attack and decay of the sound.

Two tones produced by different instruments might have the same fundamental frequency (and thus the same pitch) but sound different because of different harmonic content. The difference in sound is called timbre and is often described in subjective terms such as reedy, mellow, and tinny. A tone that is rich in harmonics, like the clarinet tone in Figs. 16.5a and 16.5b, usually sounds thin and “reedy,” while a tone containing mostly a fundamental, like the alto recorder tone in Figs. 16.5c and 16.5d, is more mellow and flute-like. The same principle applies to the human voice, which is another wind instrument; the vowels “a” and “e” sound different because of differences in harmonic content.

2 Definitions

Definition 1: Attack: The way a sound starts, including the characteristics of the initial onset.

Definition 2: Decay: The way a sound fades away, including the characteristics of the ending.

Another factor in determining tone quality is the behavior at the beginning (attack) and end (decay) of a tone. A piano tone begins with a thump and then dies away gradually. A harpsichord tone, in addition to having different harmonic content, begins much more quickly with a click, and the higher harmonics begin before the lower ones. When the key is released, the sound also dies away much more rapidly with a harpsichord than with a piano. Similar effects are present in other musical instruments.

B.I.O Application Hearing Loss from Amplified Sound Due to exposure to highly amplified music, many young musicians have suffered permanent ear damage and have hearing typical of persons 65 years of age. Headphones for personal music players used at high volume pose similar threats to hearing. Be careful!

Image summary: This is a photograph of a musician playing an electric bass guitar on a stage. The musician is positioned in front of a stack of Marshall amplifiers and speakers. The musician is actively performing, as evidenced by the motion of their hair and the position of their hands on the instrument. The presence of the instrument and amplifiers suggests a musical performance is taking place. The musician is likely playing a song, and the amplifiers are being used to project the sound of the instrument to a larger audience.

Definition

White Noise: A sound containing equal amounts of all frequencies across the audible range.

Unlike the tones made by musical instruments, noise is a combination of all frequencies, not just frequencies that are integer multiples of a fundamental frequency. (An extreme case is “white noise,” which contains equal amounts of all frequencies across the audible range.) Examples include the sound of the wind and the hissing sound you make in saying the consonant “s.”

Test Your Understanding of Section 16.1 You use an electronic signal generator to produce a sinusoidal sound wave in air. You then increase the frequency of the wave from 100 Hz to 400 Hz while keeping the pressure amplitude constant. What effect does this have on the displacement amplitude of the sound wave? (1) It becomes four times greater; (2) it becomes twice as great; (3) it is unchanged; (4) it becomes 1 over 2 as great; (v) it becomes 1 over 4 as great. number k equals omega divided by v equals 2 times pi times f divided by v also increases by a factor of 4. Since A is inversely proportional to k, the displacement amplitude becomes 1 divided by 4 as great. In other words, at higher frequency a smaller maximum displacement is required to produce the same maximum pressure fluctuation.

(5) From Equation sixteen point five, the displacement amplitude is A equals p sub max divided by B times k. The pressure amplitude p sub max and bulk modulus B remain the same, but the frequency f increases by a factor of 4. Hence the wave

Figure 16.5 summary: The figure includes a combination of images and plots. The plots represent the sound produced by a clarinet and an alto recorder using two different representations: pressure fluctuation versus time and harmonic content. The clarinet's pressure fluctuation plot shows a more complex waveform compared to the alto recorder's, which appears more sinusoidal. The harmonic content plot for the clarinet shows a broader range of frequencies with significant amplitudes, while the alto recorder's plot has a more concentrated frequency distribution with rapidly decreasing amplitudes.

16.2 Speed of Sound Waves

We found in Section 15.4 that the speed v of a transverse wave on a string depends on the string tension F and the linear mass density mu: v equals square root of F divided by mu. What, we may ask, is the corresponding expression for the speed of sound waves in a gas or liquid? On what properties of the medium does the speed depend?

We can make an educated guess about these questions by remembering a claim that we made in Section 15.4: For mechanical waves in general, the expression for the wave speed is of the form

Math summary: This expression calculates the speed of a wave as the square root of a ratio. The ratio is the restoring force divided by the inertia resisting the return to equilibrium.

Definition

Rarefactions: Regions in a longitudinal wave where the particles of the medium are farther apart than normal, resulting in decreased pressure and density.

A sound wave in a bulk fluid causes compressions and rarefactions of the fluid, so the restoring-force term in the above expression must be related to how difficult it is to compress the fluid. This is precisely what the bulk modulus B of the medium tells us. According to Newton's second law, inertia is related to mass. The “massiveness” of a bulk fluid is described by its density, or mass per unit volume, rho. Hence we expect that the speed of sound waves should be of the form v = square root of B/rho.

To check our guess, we'll derive the speed of sound waves in a fluid in a pipe. This is a situation of some importance, since all musical wind instruments are pipes in which a longitudinal wave (sound) propagates in a fluid (air) (Fig. 16.6). Human speech works on the same principle; sound waves propagate in your vocal tract, which is an air-filled pipe connected to the lungs at one end (your larynx) and to the outside air at the other end (your mouth). The steps in our derivation are completely parallel to those we used in Section 15.4 to find the speed of transverse waves.

Figure 16.6 summary: The figure is an image of a French horn. The image shows the instrument's structure, including its pipes and large bell. The sound produced by the instrument depends on the properties of the sound waves propagating through it.

Speed of Sound in a Fluid

Figure 16.7 shows a fluid with density rho in a pipe with cross-sectional area A. In equilibrium (Fig. 16.7a, the fluid is at rest and under a uniform pressure p. We take the x-axis along the length of the pipe. This is also the direction in which we make a longitudinal wave propagate, so the displacement y is also measured along the pipe, just as in Section 16.1 (see figure 16.2).

Figure 16.7 summary: The figure consists of two diagrams that illustrate the movement of a movable piston within a fluid-filled space. The first diagram depicts the fluid initially in equilibrium with equal pressure applied on both sides of the piston. The second diagram shows the piston moving, compressing the fluid. The pressure on the piston increases, and the fluid in front of the piston begins to move, while the fluid behind remains at rest. The diagrams illustrate how a change in pressure initiates movement and compression within the fluid.

At time t = 0 we start the piston at the left end moving toward the right with constant speed v sub y . This initiates a wave motion that travels to the right along the length of the pipe, in which successive sections of fluid begin to move and become compressed at successively later times.

Figure 16.7b shows the fluid at time t. All portions of fluid to the left of point P are moving to the right with speed v sub y , and all portions to the right of P are still at rest. The boundary between the moving and stationary portions travels to the right with a speed equal to the speed of propagation or wave speed v. At time t the piston has moved a distance v sub y times t, and the boundary has advanced a distance v times t. As with a transverse disturbance in a string, we can compute the speed of propagation from the impulse–momentum theorem.

The quantity of fluid set in motion in time t originally occupied a section of the cylinder with length vt, cross-sectional area A, volume vtA, and mass rho vtA . Its longitudinal momentum (that is, momentum along the length of the pipe) is

Math summary: This calculates longitudinal momentum of a fluid. It multiplies the fluid's density, velocity, time, and cross-sectional area by the fluid's vertical velocity.

Next we compute the increase of pressure, delta p, in the moving fluid. The original volume of the moving fluid, A times v sub t, has decreased by an amount A times v sub y times t. From the definition of the bulk modulus B, Equation 11.13 in Section 11.5,

Math summary: This calculates the bulk modulus as the ratio of the negative pressure change to the fractional volume change. It also calculates the change in pressure as the product of the bulk modulus and the ratio of the fluid's velocity change to its original velocity.

The pressure in the moving fluid is p plus delta p, and the force exerted on it by the piston is open parenthesis p plus delta p close parenthesis times A. The net force on the moving fluid (see figure 16.7b is delta p times A, and the longitudinal impulse is

Math summary: This equation calculates the longitudinal impulse. It states that the longitudinal impulse equals the change in momentum multiplied by the area and time, which is also equal to a constant B multiplied by the vertical velocity divided by the velocity, and then multiplied by the area and time.

Because the fluid was at rest at time t = 0, the change in momentum up to time t is equal to the momentum at that time. Applying the impulse–momentum theorem (see Section 8.1), we find

Math summary: This equation balances the change in momentum of a fluid with an applied impulse. It equates the product of a constant, a velocity ratio, area, and time to the product of density, velocity, time, area, and vertical velocity.

When we solve this expression for v, we get

Math summary: This expression calculates the speed of a longitudinal wave in a fluid. It takes the square root of the quotient of a property of the fluid divided by the density of the fluid to produce the wave speed.

which agrees with our educated guess.

While we derived equation (16.7) for waves in a pipe, it also applies to longitudinal waves in a bulk fluid, including sound waves traveling in air or water.

Speed of Sound in a Solid

When a longitudinal wave propagates in a solid rod or bar, the situation is somewhat different. The rod expands sideways slightly when it is compressed longitudinally, while a fluid in a pipe with constant cross section cannot move sideways. Using the same kind of reasoning that led us to equation (16.7), we can show that the speed of a longitudinal pulse in the rod is given by

Math summary: This calculates the speed of a longitudinal wave in a solid rod. It takes Young's modulus of the rod material and divides it by the density, then calculates the square root of the result to produce the wave speed.

We defined Young's modulus in Section 11.4.

Caution Solid rods versus bulk solids Equation (16.8) applies to only rods whose sides are free to bulge and shrink a little as the wave travels. It does not apply to longitudinal waves in a bulk solid because sideways motion in any element of material is prevented by the surrounding material. The speed of longitudinal waves in a bulk solid depends on the density, the bulk modulus, and the shear modulus.

Note that Eqs. (16.7) and (16.8) are valid for sinusoidal and other periodic waves, not just for the special case discussed here.

Table 16.1 lists the speed of sound in several bulk materials. Sound waves travel more slowly in lead than in aluminum or steel because lead has a lower bulk modulus and shear modulus and a higher density. Figure 16.7 A sound wave propagating in a fluid confined to a tube. (a) Fluid in equilibrium. (b) A time t after the piston begins moving to the right at speed v sub y , the fluid between the piston and point P is in motion. The speed of sound waves is v.

Table 16.1 summary: The speed of sound varies across different materials. Gases generally have a lower speed of sound compared to liquids, while solids tend to have the highest speed of sound. Within each state of matter, the speed of sound also varies, such as water at different temperatures.

Example 16.3 Wavelength of Sonar Waves

A ship uses a sonar system (Fig. 16.8) to locate underwater objects. Find the speed of sound waves in water using equation (16.7), and find the wavelength of a 262 Hz wave.

Figure 16.8 summary: The figure is a diagram that depicts a sonar system in operation. The diagram illustrates how sound waves are emitted from a ship to detect a submerged object. The sound waves travel through the water, reflect off the submerged object, and return to the ship, allowing the system to locate the object. The figure also illustrates the relationship between the speed of sound waves, wavelength, and the properties of the fluid medium. Sound waves of a certain frequency exhibit a longer wavelength in a medium characterized by a faster sound speed.

Identify and Set Up Our target variables are the speed and wavelength of a sound wave in water. In equation (16.7), we use the density of water, rho equals 1.00 times 10 to the power of 3 kilograms per meter cubed, and the bulk modulus of water, which we find from the compressibility (see Table 11.2). Given the speed and the frequency f equals 262 Hertz, we find the wavelength from v equals f times lambda.

Execute In Example we used Table to find B equals 2.18 times 10 to the power of 9 Pascals. Then

Math summary: First, the square root of the ratio of a value B to a density is calculated, resulting in a value of 1476 meters per second. Then, this result is divided by a frequency of 262 inverse seconds, yielding a final value of 5.64 meters.

and Evaluate The calculated value of v agrees well with the value in Table 16.1. Water is denser than air (rho is larger) but is also much more incompressible (B is much larger), and so the speed v equals square root of B divided by rho is greater than the 344 meters per second speed of sound in air at ordinary temperatures. The relationship lambda equals v divided by f then says that a sound wave in water must have a longer wavelength than a wave of the same frequency in air. Indeed, we found in Example 15.1 (Section 15.2) that a 262 Hertz sound wave in air has a wavelength of only 1.31 m.

Dolphins emit high-frequency sound waves (typically 100,000 Hz) and use the echoes for guidance and for hunting. The corresponding wavelength in water is 1.48 centimeters. With this high-frequency “sonar” system they can sense objects that are roughly as small as the wavelength (but not much smaller). Ultrasonic imaging in medicine uses the same principle; sound waves of very high frequency and very short wavelength, called ultrasound, are scanned over the human body, and the “echoes” from interior organs are used to create an image. With ultrasound of frequency 5 M.H.z = 5 × 10⁶ Hz, the wavelength in water (the primary constituent of the body) is 0.3 millimeters, and features as small as this can be discerned in the image (Fig. 16.9). Ultrasound is more sensitive than x rays in distinguishing various kinds of tissues and does not have the radiation hazards associated with x rays.

Figure 16.9 summary: This figure shows a three-dimensional ultrasound image of a fetus inside the womb. The image was created by combining multiple two-dimensional ultrasound scans. Ultrasound imaging can be applied to study heart valve action and detect tumors.

Speed of Sound in a Gas

Most of the sound waves that we encounter propagate in air. To use equation (16.7) to find the speed of sound waves in air, we note that the bulk modulus of a gas depends on pressure: The greater the pressure applied to compress a gas, the more it resists further compression and hence the greater the bulk modulus. (That's why specific values of the bulk modulus for gases are not given in Table 11.1.) The expression for the bulk modulus of a gas for use in equation (16.7) is

Math summary: This expression calculates the bulk modulus of a gas. It multiplies the ratio of heat capacities by the equilibrium pressure to produce the bulk modulus.

Definition

Ratio of Heat Capacities: A dimensionless number (γ) that characterizes the thermal properties of a gas, appearing in the formula for the speed of sound in a gas; it is the ratio of the heat capacity at constant pressure to the heat capacity at constant volume.

where p sub 0 is the equilibrium pressure of the gas. The quantity gamma (the Greek letter gamma) is called the ratio of heat capacities. It is a dimensionless number that characterizes the thermal properties of the gas. (We'll learn more about this quantity in Chapter 19.) As an example, the ratio of heat capacities for air is gamma equals 1.40. At normal atmospheric pressure p sub 0 equals 1.013 times 10 to the power of 5 Pascals, so B equals (1.40) times (1.013 times 10 to the power of 5 Pascals) equals 1.42 times 10 to the power of 5 Pascals. This value is minuscule compared to the bulk modulus of a typical solid (see Table 11.1), which is approximately 10 to the power of 10 to 10 to the power of 11 Pascals. This shouldn't be surprising: It's simply a statement that air is far easier to compress than steel.

The density rho of a gas also depends on the pressure, which in turn depends on the temperature. It turns out that the ratio B/rho for a given type of ideal gas does not depend on the pressure at all, only the temperature. From equation (16.7), this means that the speed of sound in a gas is fundamentally a function of temperature T :

Math summary: This expression calculates the speed of sound. It takes the ratio of heat capacities, the gas constant, and the absolute temperature, divides by the molar mass, and then takes the square root of the result to determine the speed of sound.

2 Definitions

Definition 1: Molar Mass: The mass of one mole of a substance, typically expressed in grams per mole (g/mol) or kilograms per mole (kg/mol).

Definition 2: Gas Constant: A physical constant (R) that relates the energy scale to temperature and quantity of substance; it has the same value for all ideal gases.

This expression incorporates several quantities that we'll study in Chapters 17, 18, and 19. The temperature T is the absolute temperature in kelvins (K), equal to the Celsius temperature plus 273.15; thus 20.00 degrees Celsius corresponds to T = 293.15 K. The quantity M is the molar mass, or mass per mole of the substance of which the gas is composed. The gas constant R has the same value for all gases. The current best numerical value of R is

Math summary: This expression defines the gas constant. It states that the gas constant equals approximately eight point three one four joules per mole kelvin.

which for practical calculations we can write as 8.314 joules per mole·K.

For any particular gas, gamma , R, and M are constants, and the wave speed is proportional to the square root of the absolute temperature. We'll see in Chapter 18 that equation (16.10) is almost identical to the expression for the average speed of molecules in an ideal gas. This shows that sound speeds and molecular speeds are closely related.

Example 16.4 Speed of Sound in Air

Find the speed of sound in air at T equals 20 degrees Celsius, and find the range of wavelengths in air to which the human ear (which can hear frequencies in the range of 20 to 20000 Hertz) is sensitive. The mean molar mass for air (a mixture of mostly nitrogen and oxygen) is M equals 28.8 times 10 to the power of negative 3 kilograms per mole and the ratio of heat capacities is gamma equals 1.40.

Identify and Set Up We use equation (16.10) to find the sound speed from gamma, T, and M, and we use v equals f times lambda to find the wavelengths corresponding to the frequency limits. Note that in equation (16.10) temperature T must be expressed in kelvins, not Celsius degrees. Using this value of v in lambda equals v divided by f, we find that at 20 degrees Celsius the frequency f equals 20 Hz corresponds to lambda equals 17 m and f equals 20,000 Hz to lambda equals 1.7 centimeters. Evaluate Our calculated value of v agrees with the measured sound speed at T equals 20 degrees Celsius.

Keyconcept The speed of sound in a gas is determined by the temperature of the gas, its molar mass, and its ratio of heat capacities.

Execute At T = 20 degrees Celsius = 293 K, we find

Math summary: This expression calculates the velocity by taking the square root of a ratio. The ratio consists of a numerator with three terms: a constant, the ideal gas constant, and the temperature, divided by a denominator representing the molar mass, resulting in the velocity.

A gas is actually composed of molecules in random motion, separated by distances that are large in comparison with their diameters. The vibrations that constitute a wave in a gas are superposed on the random thermal motion. At atmospheric pressure, a molecule travels an average distance of about 10 sup -7 m between collisions, while the displacement amplitude of a faint sound may be only 10 sup -9 m. We can think of a gas with a sound wave passing through as being comparable to a swarm of bees; the swarm as a whole oscillates slightly while individual insects move about within the swarm, apparently at random.

Test Your Understanding of Section 16.2 Mercury is 13.6 times denser than water. Based on Table 16.1, at 20 degrees Celsius which of these liquids has the greater bulk modulus? (1) Mercury; (2) water; (3) both are about the same; (4) not enough information is given to decide.

Answer

Math summary: This expression calculates a scaled value. It takes a scaling factor of thirteen point six and multiplies it by the square of the ratio of fourteen hundred fifty-one to fourteen hundred eighty-two, resulting in thirteen point zero.

(1451 meters per second versus 1482 meters per second), but the density of mercury is greater than that of water by a large factor (13.6). Hence the bulk modulus of mercury is greater than that of water by a factor of (i) From Equation sixteen point seven, the speed of longitudinal waves (sound) in a fluid is v equals square root of B divided by rho. We can rewrite this to give an expression for the bulk modulus B in terms of the fluid density rho and the sound speed v: B equals rho times v squared. At 20 degrees Celsius the speed of sound in mercury is slightly less than in water.

16.3 Sound Intensity

Traveling sound waves, like all other traveling waves, transfer energy from one region of space to another. In Section 15.5 we introduced the wave intensity I, equal to the time average rate at which wave energy is transported per unit area across a surface perpendicular to the direction of propagation. Let's see how to express the intensity of a sound wave in a fluid in terms of the displacement amplitude A or pressure amplitude p sub max .

Definition

Particle Velocity: The instantaneous velocity of a particle in a medium as it oscillates due to the passage of a wave; it is distinct from the wave velocity.

Let's consider a sound wave propagating in the +x-direction so that we can use our expressions from Section 16.1 for the displacement y (x, t) and pressure fluctuation p (10, t) . In Section we saw that power equals the product of force and velocity. So the power per unit area in this sound wave equals the product of p (10, t) (force per unit area) and the particle velocity v y (x, t) , which is the velocity at time t of that portion of the wave medium at coordinate x. Using Eqs. (16.1) and (16.4), we find

Math summary: This expression calculates the instantaneous power per unit area of a sound wave. It multiplies the pressure fluctuation and the particle velocity, simplifying the result to a constant multiplied by the square of a sine function involving position and time.

Caution Wave velocity versus particle velocity Remember that the velocity of the wave as a whole is not the same as the particle velocity. While the wave continues to move in the direction of propagation, individual particles in the wave medium merely slosh back and forth, as shown in figure 16.1. Furthermore, the maximum speed of a particle of the medium can be very different from the wave speed.

The intensity is the time average value of the power per unit area p of x,t times v sub y of x,t. For any value of x the average value of the function sine squared of k times x minus omega times t over one period T equals 2 pi divided by omega is 1/2, so

Math summary: This expression calculates the intensity. It multiplies one half by a coefficient, angular frequency, wave number, and the square of the amplitude to produce the intensity.

Using the relationships omega equals v times k and v equals square root of B divided by rho, we can rewrite equation (16.11) as

Math summary: This expression calculates the intensity of a sinusoidal sound wave in a fluid. It takes the density of the fluid, the bulk modulus of the fluid, the square of the angular frequency, and the square of the displacement amplitude as inputs and computes the sound wave intensity.

It is usually more useful to express I in terms of the pressure amplitude p sub max. Using Equations sixteen point five and sixteen point twelve and the relationship omega equals v times k, we find

Math summary: This equation calculates the intensity of a wave. It takes the angular frequency and maximum pressure squared, divides by two times the bulk modulus and wave number, and equates it to the wave speed times the maximum pressure squared divided by two times the bulk modulus.

By using the wave speed relationship v equals square root of B divided by rho, we can also write equation (16.13) in the alternative forms

Math summary: This equation calculates the intensity of a sinusoidal sound wave. It takes the square of the pressure amplitude and divides it by two times the product of the fluid density and the speed of sound, or equivalently, two times the square root of the product of the fluid density and the bulk modulus.

You should verify these expressions. Comparison of Eqs. (16.12) and (16.14) shows that sinusoidal sound waves of the same intensity but different frequency have different displacement amplitudes A but the same pressure amplitude p sub max . This is another reason it is usually more convenient to describe a sound wave in terms of pressure fluctuations, not displacement.

The total average power carried across a surface by a sound wave equals the product of the intensity at the surface and the surface area, if the intensity over the surface is uniform. The total average sound power emitted by a person speaking in an ordinary conversational tone is about 10 sup -5 W, while a loud shout corresponds to about 3 times 10 sup -2 W. If all the residents of New York City were to talk at the same time, the total sound power would be about 100 W, equivalent to the electric power requirement of a medium-sized light bulb. On the other hand, the power required to fill a large auditorium or stadium with loud sound is considerable (see Example 16.7).

If the sound source emits waves in all directions equally, the intensity decreases with increasing distance r from the source according to the inverse-square law: The intensity is proportional to 1/r sup 2 . The intensity can be increased by confining the sound waves to travel in the desired direction only (Fig. 16.10), although the 1/r sup 2 law still applies.

Figure 16.10 summary: The figure is a photograph. The photograph shows a person cupping their hands around their mouth as if shouting. Cupping your hands directs sound waves, allowing sound to travel further.

The inverse-square relationship also does not apply indoors because sound energy can reach a listener by reflection from the walls and ceiling. Indeed, part of the architect's job in designing an auditorium is to tailor these reflections so that the intensity is as nearly uniform as possible over the entire auditorium.

Problem-Solving Strategy 16.1 Sound Intensity

Identify the relevant concepts: The relationships between the intensity and amplitude of a sound wave are straightforward. Other quantities are involved in these relationships, however, so it's particularly important to decide which is your target variable.

Set Up the problem using the following steps:

1. Sort the physical quantities into categories. Wave properties include the displacement and pressure amplitudes A and p sub max . The frequency f can be determined from the angular frequency omega , the wave number k, or the wavelength lambda . These quantities are related through the wave speed v, which is determined by properties of the medium (B and rho for a liquid, and gamma , T, and M for a gas). 2. List the given quantities and identify the target variables. Find relationships that take you where you want to go.

Execute the solution: Use your selected equations to solve for the target variables. Express the temperature in kelvins (Celsius temperature plus 273.15) to calculate the speed of sound in a gas.

Evaluate your answer: If possible, use an alternative relationship to check your results.

Example 16.5 Intensity of a Sound Wave in Air

Find the intensity of the sound wave in Example 16.1, with p sub max equals 3.0 times 10 to the power of negative 2 Pascals. Assume the temperature is 20 degrees Celsius so that the density of air is rho equals 1.20 kilograms per meter cubed and the speed of sound is v equals 344 meters per second.

Identify and Set Up Our target variable is the intensity I of the sound wave. We are given the pressure amplitude p sub max of the wave as well as the density rho and wave speed v for the medium. We can determine I from p sub max, rho, and v from Equation 16.14.

With Variation Problems

Execute From equation (16.14),

Math summary: This expression calculates the intensity of a sound wave. It squares the maximum pressure, divides by two times the density of the medium times the speed of sound, and outputs the intensity in watts per square meter.

Evaluate This seems like a very low intensity, but it is well within the range of sound intensities encountered on a daily basis. A very loud sound wave at the threshold of pain has a pressure amplitude of about 30 Pa and an intensity of about 1 W/m². The pressure amplitude of the faintest sound wave that can be heard is about 3 times 10 sup -5 Pa, and the corresponding intensity is about 10 sup -12 W/m². (Try these values of p sub max in equation (16.14) to check that the corresponding intensities are as we have stated.)

Keyconcept The intensity (power per unit area) of a sound wave is proportional to the square of the pressure amplitude of the wave. The proportionality constant depends on the density of the medium and the speed of sound in the medium.

Example 16.6 Same Intensity, Different Frequencies

What are the pressure and displacement amplitudes of a 20 Hz sound wave with the same intensity as the 1000 Hz sound wave of Examples 16.1 and 16.5?

Identify and Set Up In Examples and we found that for a 1000 Hz sound wave with p sub text max equals 3.0 times 10 to the power of negative 2 Pascals, A equals 1.2 times 10 to the power of negative 8 meters and I equals 1.1 times 10 to the power of negative 6 Watts per meter squared. Our target variables are p sub text max and A for a 20 Hz sound wave of the same intensity I. We can find these using Eas. (16.14) and (16.12). respectively.

Execute We can rearrange Equations sixteen point fourteen and sixteen point twelve as p sub max squared equals two times I times square root of rho times B and omega squared times A squared equals two times I divided by square root of rho times B, respectively. These tell us that for a given sound intensity I in a given medium (constant rho and B), the quantities p sub max and omega times A (or, equivalently, f times A) are constants that don't depend on frequency. From the first result we immediately have p sub max equals three point zero times ten to the power of negative two Pascals for f equals twenty Hertz, the same as for f equals one thousand Hertz. If we write the second result as f sub twenty times A sub twenty equals f sub one thousand times A sub one thousand, we have Evaluate Our result reinforces the idea that pressure amplitude is a more convenient description of a sound wave and its intensity than displacement amplitude.

Math summary: This expression calculates the displacement amplitude at 20 hertz by scaling the displacement amplitude at 1000 hertz. It multiplies the amplitude at 1000 hertz by the ratio of the 1000 hertz frequency to the 20 hertz frequency, resulting in the displacement amplitude at 20 hertz.

Keyconcept If two sound waves in a given medium have the same intensity but different frequencies, the wave with the higher frequency has the greater displacement amplitude. The two waves have the same pressure amplitude, however.

Example 16.7 "Play it Loud!"

With Variation Problems

For an outdoor concert we want the sound intensity to be 1 W/m² at a distance of 20 m from the speaker array. If the sound intensity is uniform in all directions, what is the required average acoustic power output of the array?

Identify, Set Up, and Execute This example uses the definition of sound intensity as power per unit area. The total power is the target variable; the area in question is a hemisphere centered on the speaker array. We assume that the speakers are on the ground and that none of the acoustic power is directed into the ground, so the acoustic power is uniform over a hemisphere 20 m in radius. The surface area of this hemisphere is one half, times, 4 pi, times, 20 meters, squared, or about 2500 meters squared. The required power is the product of this area and the intensity: 1 Watt per meter squared, times, 2500 meters squared, equals 2500 Watts equals 2.5 kiloWatts.

Evaluate The electrical power input to the speaker would need to be considerably greater than 2.5 kilowatt, because speaker efficiency is not very high (typically a few percent for ordinary speakers, and up to 25% for horn-type speakers).

Keyconcept To find the acoustic power output of a source of sound, multiply the area over which the emitted sound wave is distributed by the average intensity of the sound over that area.

The Decibel Scale

Definition

Sound Intensity Level: A logarithmic measure of the intensity of a sound wave, expressed in decibels (dB), relative to a reference intensity (typically the threshold of human hearing).

Because the ear is sensitive over a broad range of intensities, a logarithmic measure of intensity called sound intensity level is often used:

Math summary: This expression calculates the sound intensity level in decibels. It takes the logarithm base ten of the ratio between the sound wave's intensity and a reference intensity, then multiplies the result by ten decibels.

The chosen reference intensity I 0 in equation (16.15) is approximately the threshold of human hearing at 1000 Hz. Sound intensity levels are expressed in decibels, abbreviated decibel. A decibel is 1 over 10 of a bel, a unit named for Alexander Graham Bell (the inventor of the telephone). The bel is inconveniently large for most purposes, and the decibel is the usual unit of sound intensity level.

If the intensity of a sound wave equals I sub 0 or 10 to the power of negative 12 Watts per meter squared, its sound intensity level is beta equals 0 decibel. An intensity of 1 Watt per meter squared corresponds to 120 decibel. Table 16.2 gives the sound intensity levels of some familiar sounds. You can use equation (16.15) to check the value of beta given for each intensity in the table.

Table 16.2 summary: The table presents sound intensity levels from various sources, ranging from the threshold of hearing to the sound of a military jet. The sound intensity and intensity levels vary significantly across different sources; the sound intensity is highest for a military jet and lowest at the threshold of hearing.

Definition

dBA Scale: A weighted scale for measuring sound levels that deemphasizes low and very high frequencies to better reflect the human ear's sensitivity.

Because the ear is not equally sensitive to all frequencies in the audible range, some sound-level meters weight the various frequencies unequally. One such scheme leads to the so-called dBA scale; this scale deemphasizes the low and very high frequencies, where the ear is less sensitive.

Example 16.8 Temporary—Or Permanent—Hearing Loss

A 10 min exposure to 120 decibel sound will temporarily shift your threshold of hearing at 1000 Hz from 0 decibel up to 28 decibel. Ten years of exposure to 92 decibel sound will cause a permanent shift to 28 decibel. What sound intensities correspond to 28 decibel and 92 decibel?

Identify and Set Up We are given two sound intensity levels beta ; our target variables are the corresponding intensities. We can solve equation (16.15) to find the intensity I that corresponds to each value of beta .

Execute We solve Equation sixteen point fifteen for I by dividing both sides by ten decibel and using the relationship ten to the power of log x equals x :

Math summary: This expression calculates a value I. It takes an initial value I subscript 0, multiplies it by 10 raised to the power of beta divided by 10 dB.

For beta equals 28 decibel and beta equals 92 decibel, the exponents are beta divided by the quantity 10 decibel equals 2.8 and 9.2, respectively, so that

Math summary: This expression calculates two intensity values. It multiplies a base value of ten to the power of two point eight and nine point two respectively, by a scaling factor of ten to the negative twelfth power to obtain the intensity.

Evaluate If your answers are a factor of 10 too large, you may have entered 10 times 10 to the power of negative 12 in your calculator instead of 1 times 10 to the power of negative 12. Be careful!

Keyconcept The sound intensity level (in decibels, or decibel) is a logarithmic measure of the intensity of a sound wave. Adding 10 decibel to the sound intensity level corresponds to multiplying the intensity by a factor of 10.

Example 16.9 A Bird Sings in a Meadow

Consider an idealized bird (treated as a point source) that emits constant sound power, with intensity obeying the inverse-square law (Fig. 16.11). If you move twice the distance from the bird, by how many decibels does the sound intensity level drop?

Figure 16.11 summary: The figure is an illustration depicting sound waves emanating from a point source, represented by a bird singing. The sound waves are shown as concentric circles spreading outwards. The figure shows two observers at different distances from the sound source. The observer farther away from the source experiences a lower sound intensity level compared to the closer observer.

Identify and Set Up The decibel scale is logarithmic, so the difference between two sound intensity levels (the target variable) corresponds to the ratio of the corresponding intensities, which is determined by the inverse-square law. We label the two points P 1 and P 2 (Fig. 16.11). We use equation (16.15), the definition of sound intensity level, at each point. We use equation (15.26), the inverse-square law, to relate the intensities at the two points.

Execute The difference beta sub 2 minus beta sub 1 between any two sound intensity levels is related to the corresponding intensities by

Math summary: This calculates the difference between two sound intensity levels. It involves taking the logarithm of the ratio of the intensities to a reference intensity, scaling the result, and then simplifying the expression.

With Variation Problems

For this inverse-square-law source, equation (15.26) yields I sub 2 divided by I sub 1 equals r sub 1 squared divided by r sub 2 squared equals 1 divided by 4, so

Math summary: This expression calculates the difference in sound intensity levels in decibels. It takes the base 10 logarithm of the ratio of two sound intensities, multiplies the result by ten decibels, and finds the difference to be negative six decibels.

Evaluate Our result is negative, which tells us (correctly) that the sound intensity level is less at P 2 than at P 1 . The 6 decibel difference doesn't depend on the sound intensity level at P 1 ; any doubling of the distance from an inverse-square-law source reduces the sound intensity level by 6 decibel.

Note that the perceived loudness of a sound is not directly proportional to its intensity. For example, most people interpret an increase of 8 decibel to 10 decibel in sound intensity level (corresponding to increasing intensity by a factor of 6 to 10) as a doubling of loudness.

Keyconcept The difference between the sound intensity levels of two sounds is proportional to the logarithm of the ratio of the intensities of those sounds.

Test Your Understanding of Section 16.3 You Double the Intensity of a Sound

16.3 You double the intensity of a sound wave in air while leaving the frequency unchanged. (The pressure, density, and temperature of the air remain unchanged as well.) What effect does this have on the displacement amplitude, pressure amplitude, bulk modulus, sound speed, and sound intensity level?

Equations (16.9) and (16.10) show that the bulk modulus B and sound speed v remain the same because the physical properties of the air are unchanged. From Eqs. (16.12) and (16.14), the intensity is proportional to the square of the displacement amplitude or the square of the pressure amplitude. Hence doubling d density means that A and p sub max both increase by a factor of square root of 2. Example 16.9 shows that d factor of 2 (I sub 2 divided by I sub 1 equals 2) corresponds to adding to the sound d A and p sub max increase by a factor of square root of 2, B and v are unchanged, beta increases by 3.0 decibel

16.4 Standing Sound Waves and Normal Modes

When longitudinal (sound) waves propagate in a fluid in a pipe, the waves are reflected from the ends in the same way that transverse waves on a string are reflected at its ends. The superposition of the waves traveling in opposite directions again forms a standing wave. Just as for transverse standing waves on a string (see Section 15.7), standing sound waves in a pipe can be used to create sound waves in the surrounding air. This is the principle of the human voice as well as many musical instruments, including woodwinds, brasses, and pipe organs.

2 Definitions

Definition 1: Displacement Node: A point in a standing wave where the displacement of the medium is always zero.

Definition 2: Displacement Antinode: A point in a standing wave where the displacement of the medium is maximum.

Transverse waves on a string, including standing waves, are usually described only in terms of the displacement of the string. But, as we have seen, sound waves in a fluid may be described either in terms of the displacement of the fluid or in terms of the pressure variation in the fluid. To avoid confusion, we'll use the terms displacement node and displacement antinode to refer to points where particles of the fluid have zero displacement and maximum displacement, respectively.

We can demonstrate standing sound waves in a column of gas using an apparatus called a Kundt's tube (Fig. 16.12). A horizontal glass tube a meter or so long is closed at one end and has a flexible diaphragm at the other end that can transmit vibrations. A nearby loudspeaker is driven by an audio oscillator and amplifier; this produces sound waves that force the diaphragm to vibrate sinusoidally with a frequency that we can vary. The sound waves within the tube are reflected at the other, closed end of the tube.

Figure 16.12 summary: The figure is a diagram of a Kundt's tube setup used to demonstrate standing sound waves. The Kundt's tube contains a gas and fine powder, with a speaker at one end to generate sound waves. The sound waves create areas of high and low displacement within the tube, known as antinodes and nodes, respectively. The powder within the tube collects at the nodes, where there is minimal displacement. This demonstrates that sound waves of a specific frequency produce standing waves, and the powder distribution shows the location of nodes and antinodes.

We spread a small amount of light powder uniformly along the bottom of the tube. As we vary the frequency of the sound, we pass through frequencies at which the amplitude of the standing waves becomes large enough for the powder to be swept along the tube at those points where the gas is in motion. The powder therefore collects at the displacement nodes (where the gas is not moving). Adjacent nodes are separated by a distance equal to lambda/2 .

Figure 16.13 shows the motions of nine different particles within a gas-filled tube in which there is a standing sound wave. A particle at a displacement node (N) does not move, while a particle at a displacement antinode (A) oscillates with maximum amplitude. Note that particles on opposite sides of a displacement node vibrate in opposite phase.

Figure 16.13 summary: The figure is a depiction of a standing wave at different time intervals. The standing wave is shown at different time intervals for a complete period. The figure illustrates how the wave oscillates between its maximum and minimum displacement over time. The nodes remain stationary, while the antinodes exhibit maximum displacement. Displacement nodes correspond to pressure antinodes, and displacement antinodes correspond to pressure nodes.

When these particles approach each other, the gas between them is compressed and the pressure rises; when they recede from each other, there is an expansion and the pressure drops. Hence at a displacement node the gas undergoes the maximum amount of compression and expansion, and the variations in pressure and density above and below the average have their maximum value. By contrast, particles on opposite sides of a displacement antinode vibrate in phase; the distance between the particles is nearly constant, and there is no variation in pressure or density at a displacement antinode.

2 Definitions

Definition 1: Pressure Node: A point in a standing sound wave at which the pressure and density do not vary.

Definition 2: Pressure Antinode: A point in a standing sound wave at which the variations in pressure and density are greatest.

We use the term pressure node to describe a point in a standing sound wave at which the pressure and density do not vary and the term pressure antinode to describe a point at which the variations in pressure and density are greatest. Using these terms, we can summarize our observations as follows:

A pressure node is always a displacement antinode, and a pressure antinode is always a displacement node.

Figure 16.12 depicts a standing sound wave at an instant at which the pressure variations are greatest; the blue shading shows that the density and pressure of the gas have their maximum and minimum values at the displacement nodes.

When reflection takes place at a closed end of a pipe (an end with a rigid barrier or plug), the displacement of the particles at this end must always be zero, analogous to a fixed end of a string. Thus a closed end of a pipe is a displacement node and a pressure antinode; the particles do not move, but the pressure variations are maximum. An open end of a pipe is a pressure node because it is open to the atmosphere, where the pressure is constant. Because of this, an open end is always a displacement antinode, in analogy to a free end of a string; the particles oscillate with maximum amplitude, but the pressure does not vary. (The pressure node actually occurs somewhat beyond an open end of a pipe. But if the diameter of the pipe is small in comparison to the wavelength, which is true for most musical instruments, this effect can safely be ignored.) Thus longitudinal sound waves are reflected at the closed and open ends of a pipe in the same way that transverse waves in a string are reflected at fixed and free ends, respectively.

Conceptual Example 16.10 The Sound of Silence

A directional loudspeaker directs a sound wave of wavelength lambda at a wall (Fig. 16.14). At what distances from the wall could you stand and hear no sound at all?

Figure 16.14 summary: The figure is a diagram illustrating a standing sound wave created by the interference of a sound wave and its reflection off a wall. The diagram depicts the spatial distribution of nodes and antinodes in the standing wave. The distance between adjacent nodes and antinodes varies. The distance from the speaker to the first antinode is shorter than the distance to the second antinode.

Solution Your ear detects pressure variations in the air; you'll therefore hear no sound if your ear is at a pressure node, which is a displacement antinode. The wall is at a displacement node; the distance from any node to an adjacent antinode is lambda divided by 4, and the distance from one antinode to the next is lambda divided by 2 (Fig. 16.14). Hence the displacement antinodes (pressure nodes), at which no sound will be heard, are at distances d equals lambda divided by 4, d equals lambda divided by 4 plus lambda divided by 2 equals 3 times lambda divided by 4, d equals 3 times lambda divided by 4 plus lambda divided by 2 equals 5 times lambda divided by 4, and so on from the wall. If the loudspeaker is not highly directional, this effect is hard to notice because of reflections of sound waves from the floor, ceiling, and other walls.

Keyconcept In a standing sound wave, a pressure node is a displacement antinode, and vice versa. The sound is loudest at a pressure antinode; there is no sound at a pressure node.

Organ Pipes and Wind Instruments

Definition

Normal Modes: The characteristic vibration patterns of a system, each with a specific frequency, in which all parts of the system oscillate in phase.

The most important application of standing sound waves is the production of musical tones. Organ pipes are one of the simplest examples (Fig. 16.15). Air is supplied by a blower to the bottom end of the pipe (Fig. 16.16). A stream of air emerges from the narrow opening at the edge of the horizontal surface and is directed against the top edge of the opening, which is called the mouth of the pipe. The column of air in the pipe is set into vibration, and there is a series of possible normal modes, just as with the stretched string.

Figure 16.15 summary: The figure shows a photograph of organ pipes. The organ pipes have varying sizes. The size of the organ pipes has an impact on the frequency of the tones produced; larger pipes produce lower frequency tones, while smaller pipes produce higher frequency tones.

Figure 16.16 summary: The figure shows cross sections of an organ pipe at two different times. The figure depicts the displacement nodes and antinodes within the pipe, indicating areas of maximum and zero pressure variation, respectively. Airflow from a blower enters the pipe, creating pressure variations that produce sound. The position of nodes and antinodes changes over time, demonstrating the fluctuation of pressure within the pipe.

The mouth acts as an open end; it is a pressure node and a displacement antinode. The other end of the pipe (at the top in figure 16.16) may be either open or closed.

Definition

Open Pipe: A pipe or tube that is open at both ends.

In figure 16.17, both ends of the pipe are open, so both ends are pressure nodes and displacement antinodes. An organ pipe that is open at both ends is called an open pipe. The fundamental frequency f 1 corresponds to a standing-wave pattern with a displacement antinode at each end and a displacement node in the middle (Fig. 16.17a The distance between adjacent antinodes is always equal to one half-wavelength, and in this case that is equal to the length L of the pipe; lambda/2 = L . The corresponding frequency, obtained from the relationship f = v/lambda , is

Math summary: This equation calculates the fundamental frequency of an open pipe. It divides the speed of sound by twice the length of the pipe.

Figure 16.17 summary: The figure illustrates a cross section of an open pipe, depicting the first three normal modes of sound waves within the pipe. The figure shows the pressure variations and displacement along the pipe axis. The figure also indicates displacement nodes and antinodes. The relationship between the length of the pipe and the wavelength of the sound waves is shown, demonstrating how different modes correspond to different fractions of a wavelength fitting within the pipe's length. The frequency of the second harmonic is twice the frequency of the first harmonic.

Figures 16.17b and 16.17c show the second and third harmonics; their vibration patterns have two and three displacement nodes, respectively. For these, a half-wavelength is equal to L divided by 2 and L divided by 3, respectively, and the frequencies are twice and three times the fundamental, respectively: f sub 2 equals 2 times f sub 1 and f sub 3 equals 3 times f sub 1. For every normal mode of an open pipe the length L must be an integer number of half-wavelengths, and the possible wavelengths lambda sub n are given by

Math summary: This equation relates the length of an open pipe to the wavelength of the sound produced, where the length equals an integer multiple of half-wavelengths. Alternatively, the wavelength is calculated as twice the length of the pipe divided by an integer, where the integer represents the mode number.

The corresponding frequencies f sub n are given by f sub n equals v divided by lambda sub n, so all the normal-mode frequencies for a pipe that is open at both ends are given by Standing waves, open pipe:

Math summary: This equation calculates the frequency of a standing wave in an open pipe. It multiplies the harmonic number by the speed of sound, then divides by twice the length of the pipe.

Definition

Overtone: A resonant frequency above the fundamental frequency.

The value n = 1 corresponds to the fundamental frequency, n = 2 to the second harmonic (or first overtone), and so on. Alternatively, we can say

Math summary: This expression calculates the nth frequency as a multiple of the fundamental frequency. It multiplies the fundamental frequency by an integer n, where n represents the harmonic number.

with f sub 1 given by Equation sixteen point sixteen.

Definition

Stopped Pipe: A pipe or tube that is open at one end and closed at the other.

Figure 16.18 shows a stopped pipe: It is open at the left end but closed at the right end. The left (open) end is a displacement antinode (pressure node), but the right (closed) end is a displacement node (pressure antinode). Figure 16.18a shows the lowest-frequency f sub 1 equals v divided by 2 times L Figure 16.18 A cross section of a stopped pipe showing the first three normal modes as well as the displacement nodes and antinodes. Only odd harmonics are possible.

Image summary: The image is an illustration depicting the second harmonic in a pipe. The illustration shows a pipe with an air stream entering from the left, creating a standing wave inside. The standing wave exhibits three antinodes and three nodes. The length of the pipe is equal to three halves of the wavelength.

Figure 16.18 summary: The figure is a schematic diagram. It illustrates the third harmonic in a pipe closed at one end. The diagram depicts a pipe with one end closed and the other end connected to a mouthpiece. The standing wave pattern inside the pipe shows three-quarters of a wavelength, with an antinode at the open end and a node at the closed end. The length of the pipe is related to the wavelength, specifically, the length is three-quarters of the wavelength. The third harmonic frequency is three times the fundamental frequency.

(a) Fundamental: f sub 1 equals v divided by 4 times L mode; the length of the pipe is the distance between a node and the adjacent antinode, or a quarter-wavelength (L equals lambda sub 1 divided by 4). The fundamental frequency is f sub 1 equals v divided by lambda sub 1, or

Math summary: This equation calculates the fundamental frequency of a stopped pipe. It divides the speed of the wave by four times the length of the pipe to determine the fundamental frequency.

This is one-half the fundamental frequency for an open pipe of the same length. In musical language, the pitch of a closed pipe is one octave lower (a factor of 2 in frequency) than that of an open pipe of the same length. Figure 16.18b shows the next mode, for which the length of the pipe is three-quarters of a wavelength, corresponding to a frequency 3f 1 . For figure 16.18c, L = 5lambda/4 and the frequency is 5f 1 . The possible wavelengths are given by

Math summary: This equation calculates the possible wavelengths in a stopped pipe. It determines wavelength based on the length of the pipe and a mode number which can only be an odd integer.

The normal-mode frequencies are given by f sub n equals v divided by lambda sub n, or Standing waves, stopped pipe:

Math summary: This calculates the frequency of a harmonic in a stopped pipe. It takes the harmonic number and the speed of sound in the pipe as inputs, multiplies them, and then divides by four times the length of the pipe to produce the harmonic frequency.

or

Math summary: This formula calculates the frequencies produced by a stopped pipe. It multiplies the fundamental frequency by odd integer values to find the allowed higher frequency harmonics.

with f sub 1 given by Equation (16.20). We see that the second, fourth, and all even harmonics are missing. In a stopped pipe, the fundamental frequency is f sub 1 equals v divided by 4 times L, and only the odd harmonics in the series (3 times f sub 1, 5 times f sub 1, and so on) are possible.

A final possibility is a pipe that is closed at both ends, with displacement nodes and pressure antinodes at both ends. This wouldn't be of much use as a musical instrument because the vibrations couldn't get out of the pipe.

Example 16.11 A Tale of Two Pipes

On a day when the speed of sound is 344 meters per second, the fundamental frequency of a particular stopped organ pipe is 220 Hz. (a) How long is this pipe? (b) The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe. How long is the open pipe?

With Variation Problems

Identify and Set Up This problem uses the relationship between the length and normal-mode frequencies of open pipes (Fig. 16.17) and stopped pipes (Fig. 16.18). In part (a), we determine the length of the stopped pipe from equation (16.22). In part (b), we must determine the length of an open pipe, for which equation (16.18) gives the frequencies.

Execute (a) For a stopped pipe f sub 1 equals v divided by 4 times L, so

Math summary: This expression calculates the length of a stopped pipe. It divides the speed of sound by four times the fundamental frequency, resulting in the length of the stopped pipe.

(b) The frequency of the second overtone of a stopped pipe (the third possible frequency) is f sub 5 equals 5 times f sub 1 equals 5 times (220 hertz) equals 1100 hertz. If the wavelengths for the two pipes are the same, the frequencies are also the same. Hence the frequency of the third harmonic of the open pipe, which is at 3 times f sub 1 equals 3 times (v divided by 2L, equals 1100 Hertz. Then

Math summary: The equation calculates the length of an open pipe by setting 1100 hertz equal to three times the quantity of 344 meters per second divided by twice the length of the open pipe, and the length of the open pipe is 0.469 meters.

Evaluate The 0.391 m stopped pipe has a fundamental frequency of 220 Hz; the longer (0.469 m) open pipe has a higher fundamental frequency, (1100 H.z)/3 = 367 Hz. This is not a contradiction, as you can see if you compare Figs. 16.17a and 16.18a

Keyconcept For a pipe open at both ends (an “open pipe”), the normal-mode frequencies of a standing sound wave include both even and odd multiples of the pipe's fundamental frequency. For a pipe open at one end and closed at the other (a “stopped pipe”), the only normal-mode frequencies are the odd multiples of the pipe's fundamental frequency. The fundamental frequency of a stopped pipe is half that of an open pipe of the same length.

2 Definitions

Definition 1: Resonance Curve: A graph showing the amplitude of oscillation of a system as a function of the driving frequency.

Definition 2: Driving Frequency: The frequency of an external force or source that causes a system to oscillate.

Figure 16.19 (a) The air in an open pipe is forced to oscillate at the same frequency as the sinusoidal sound waves coming from the loudspeaker. (b) The resonance curve of the open pipe graphs the amplitude of the standing sound wave in the pipe as a function of the driving frequency. In an organ pipe in actual use, several modes are always present at once; the motion of the air is a superposition of these modes. This situation is analogous to a string that is struck or plucked, as in figure 15.28. Just as for a vibrating string, a complex standing wave in the pipe produces a traveling sound wave in the surrounding air with a harmonic content similar to that of the standing wave.

Figure 16.19 summary: The figure contains an illustration and a line graph. The illustration depicts a speaker emitting a sound into an open organ pipe, with the air inside the pipe oscillating at the speaker's frequency. The wave amplitude is dependent on the frequency. The line graph plots wave amplitude against frequency. The graph shows peaks at specific frequencies, which are multiples of a fundamental frequency. The wave amplitude is higher at lower integer multiples of the fundamental frequency and lower at higher integer multiples of the fundamental frequency.

A very narrow pipe produces a sound wave rich in higher harmonics; a fatter pipe produces mostly the fundamental mode, heard as a softer, more flutelike tone. The harmonic content also depends on the shape of the pipe's mouth.

(b) Resonance curve: graph of amplitude A versus driving frequency f. Peaks occur at normal-mode frequencies of the pipe: We have talked about organ pipes, but this discussion is also applicable to other wind instruments. The flute and the recorder are directly analogous. The most significant difference is that those instruments have holes along the pipe.

Opening and closing the holes with the fingers changes the effective length L of the air column and thus changes the pitch. Any individual organ pipe, by comparison, can play only a single note. The flute and recorder behave as open pipes, while the clarinet acts as a stopped pipe (closed at the reed end, open at the bell).

Equations (16.18) and (16.22) show that the frequencies of any wind instrument are proportional to the speed of sound v in the air column inside the instrument. As equation (16.10) shows, v depends on temperature; it increases when temperature increases. Thus the pitch of all wind instruments rises with increasing temperature. An organ that has some of its pipes at one temperature and others at a different temperature is bound to sound out of tune.

Test Your Understanding of Section 16.4 If you connect a hose to one end of a metal pipe and blow compressed air into it, the pipe produces a musical tone. If instead you blow compressed helium into the pipe at the same pressure and temperature, will the pipe produce (i) the same tone, (2) a higher-pitch tone, or (3) a lower-pitch tone?

(2) Helium is less dense and has a lower molar mass than air, so sound travels faster in helium than in air. The normal-mode frequencies for a pipe are proportional to the sound speed v, so the frequency and hence the pitch increase when the air in the pipe is replaced with helium. Answer

16.5 Resonance and Sound

Many mechanical systems have normal modes of oscillation. As we have seen, these include columns of air (as in an organ pipe) and stretched strings (as in a guitar; see Section 15.8). In each mode, every particle of the system oscillates with simple harmonic motion at the same frequency as the mode. Air columns and stretched strings have an infinite series of normal modes, but the basic concept is closely related to the simple harmonic oscillator, discussed in Chapter 14, which has only a single normal mode (that is, only one frequency at which it oscillates after being disturbed).

Suppose we apply a periodically varying force to a system that can oscillate. The system is then forced to oscillate with a frequency equal to the frequency of the applied force (called the driving frequency). This motion is called a forced oscillation.

We talked about forced oscillations of the harmonic oscillator in Section 14.8, including the phenomenon of mechanical resonance. A simple example of resonance is pushing Cousin Throckmorton on a swing. The swing is a pendulum; it has only a single normal mode, with a frequency determined by its length. If we push the swing periodically with this frequency, we can build up the amplitude of the motion. But if we push with a very different frequency, the swing hardly moves at all.

Resonance also occurs when a periodically varying force is applied to a system with many normal modes. In figure 16.19a an open organ pipe is placed next to a loudspeaker that emits pure sinusoidal sound waves of frequency f, which can be varied by adjusting the amplifier. The air in the pipe is forced to vibrate with the same frequency f as the driving force provided by the loudspeaker.

In general the amplitude of this motion is relatively small, and the air inside the pipe will not move in any of the normal-mode patterns shown in figure 16.17. But if the frequency f of the force is close to one of the normal-mode frequencies, the air in the pipe moves in the normal-mode pattern for that frequency, and the amplitude can become quite large. Figure 16.19b shows the amplitude of oscillation of the air in the pipe as a function of the driving frequency f. This resonance curve of the pipe has peaks where f equals the normal-mode frequencies of the pipe. The detailed shape of the resonance curve depends on the geometry of the pipe.

If the frequency of the force is precisely equal to a normal-mode frequency, the system is in resonance, and the amplitude of the forced oscillation is maximum. If there were no friction or other energy-dissipating mechanism, a driving force at a normal-mode frequency would continue to add energy to the system, the amplitude would increase indefinitely, and the peaks in the resonance curve of figure 16.19b would be infinitely high. But in any real system there is always some dissipation of energy, or damping, as we discussed in Section 14.8; the amplitude of oscillation in resonance may be large, but it cannot be infinite.

The “sound of the ocean” you hear when you put your ear next to a large seashell is due to resonance. The noise of the outside air moving past the seashell is a mixture of sound waves of almost all audible frequencies, which forces the air inside the seashell to oscillate. The seashell behaves like an organ pipe, with a set of normal-mode frequencies; hence the inside air oscillates most strongly at those frequencies, producing the seashell's characteristic sound. To hear a similar phenomenon, uncap a full bottle of your favorite beverage and blow across the open top. The noise is provided by your breath blowing across the top, and the “organ pipe” is the column of air inside the bottle above the surface of the liquid. If you take a drink and repeat the experiment, you'll hear a lower tone because the “pipe” is longer and the normal-mode frequencies are lower.

Resonance also occurs when a stretched string is forced to oscillate (see Section 15.8). Suppose that one end of a stretched string is held fixed while the other is given a transverse sinusoidal motion with small amplitude, setting up standing waves. If the frequency of the driving mechanism is not equal to one of the normal-mode frequencies of the string, the amplitude at the antinodes is fairly small. However, if the frequency is equal to any one of the normal-mode frequencies, the string is in resonance, and the amplitude at the antinodes is very much larger than that at the driven end. The driven end is not precisely a node, but it lies much closer to a node than to an antinode when the string is in resonance. The photographs of standing waves in figure 15.23 were made this way, with the left end of the string fixed and the right end oscillating vertically with small amplitude.

It is easy to demonstrate resonance with a piano. Push down the damper pedal (the right-hand pedal) so that the dampers are lifted and the strings are free to vibrate, and then sing a steady tone into the piano. When you stop singing, the piano seems to continue to sing the same note. The sound waves from your voice excite vibrations in the strings that have natural frequencies close to the frequencies (fundamental and harmonics) present in the note you sang.

A more spectacular example is a singer breaking a wine glass with her amplified voice. A good-quality wine glass has normal-mode frequencies that you can hear by tapping it. If the singer emits a loud note with a frequency corresponding exactly to one of these normal-mode frequencies, large-amplitude oscillations can build up and break the glass (Fig. 16.20). B.I.O Application Resonance and the Sensitivity of the Ear The auditory canal of the human ear (see figure 16.4) is an air-filled pipe open at one end and closed at the other (eardrum) end. The canal is about 2.5 centimeters = 0.025 m long, so it has a resonance at its fundamental frequency f sub 1 = v/4L = (344 meters per second) / [4(0.025 m)] = 3440 Hz . The resonance means that a sound at this frequency produces a strong oscillation of the eardrum. That's why your ear is most sensitive to sounds near 3440 Hz.

Figure 16.20 summary: The figure is a photograph. It depicts a trumpet playing a note that matches the resonant frequency of a glass goblet. The goblet is shown shattering due to the amplitude of the vibrations. The inference is that when the frequency of a sound matches the natural frequency of an object, resonance can occur, leading to vibrations of sufficient amplitude to cause the object to break.

Example 16.12 An Organ-Guitar Duet

With Variation Problems

A stopped organ pipe is sounded near a guitar, causing one of the strings to vibrate with large amplitude. We vary the string tension until we find the maximum amplitude. The string is 80% as long as the pipe. If both pipe and string vibrate at their fundamental frequency, calculate the ratio of the wave speed on the string to the speed of sound in air.

Identify and Set Up The large response of the string is an example of resonance. It occurs because the organ pipe and the guitar string have the same fundamental frequency. If we let the subscripts a and s stand for the air in the pipe and the string, respectively, the condition for resonance is f sub 1a = f sub 1s . Equation (16.20) gives the fundamental frequency for a stopped pipe, and equation (15.32) gives the fundamental frequency for a guitar string held at both ends. These expressions involve the wave speed in air ( v a ) and on the string ( v s ) and the lengths of the pipe and string. We are given that L s = 0.80L a ; our target variable is the ratio v s/v a .

Continued Execute From Eqs. (16.20) and (15.32), f sub 1a equals v sub a divided by 4 times L sub a and f sub 1s equals v sub s divided by 2 times L sub s. These frequencies are equal, so

Math summary: This equation equates two frequency expressions. It states that the ratio of a first velocity to four times a first length is equal to the ratio of a second velocity to two times a second length.

Substituting L sub s equals 0.80 times L sub a and rearranging, we get v sub s divided by v sub a equals 0.40.

Evaluate As an example, if the speed of sound in air is 344 meters per second, the wave speed on the string is 0.40 times 344 meters per second equals 138 meters per second. Note that while the standing waves in the pipe and on the string have the same frequency, they have different wavelengths lambda equals v divided by f because the two media have different wave speeds v. Which standing wave has the greater wavelength?

Keyconcept If you force or drive a mechanical system (such as a guitar string or the air in a pipe) to vibrate at a frequency f, the system will oscillate with maximum amplitude (or resonate) if f equals one of the normal-mode frequencies of the system.

Test Your Understanding of Section 16.5 A stopped organ pipe of length L has a fundamental frequency of 220 Hz. For which of the following organ pipes will there be a resonance if a tuning fork of frequency 660 Hz is sounded next to the pipe? (There may be more than one correct answer.) (i) A stopped organ pipe of length L; (2) a stopped organ pipe of length 2L; (3) an open organ pipe of length L; (4) an open organ pipe of length 2L.

Answer

(1) and (4) There will be a resonance if 660 Hz is one of the pipe's normal-mode frequencies. A stopped organ pipe has normal-mode frequencies that are odd multiples of its fundamental frequency [see equation (16.22) and figure 16.18]. Hence pipe (i), which has fundamental frequency 220 Hz, also has a normal-mode frequency of 3 (220 Hz) = 660 Hz. Pipe (2) has twice the length of pipe (i); from equation (16.20), the fundamental frequency of a stopped pipe is inversely proportional to the length, so pipe (2) has a fundamental frequency of (1 over 2) (220 Hz) = 110 Hz . Its other normal-mode frequencies are 330 Hz, 550 Hz, 770 Hz,... so a 660 Hz tuning fork will not cause resonance. Pipe (3) is an open pipe of the same length as pipe (i), so its fundamental frequency is twice as great as for pipe (i) [compare Eqs. (16.16) and (16.20)], or 2 (220 Hz) = 440 Hz. Its other normal-mode frequencies are integer multiples of the fundamental frequency [see equation (16.19)], or 880 Hz, 1320 Hz,..., none of which match the 660 Hz frequency of the tuning fork. Pipe (4) is also an open pipe but with twice the length of pipe (3) [see equation (16.18)], so its normal-mode frequencies are one-half those of pipe (3): 220 Hz, 440 Hz, 660 Hz,..., so the third harmonic will resonate with the tuning fork.

16.6 Interference of Waves

Wave phenomena that occur when two or more waves overlap in the same region of space are grouped under the heading interference. As we have seen, standing waves are a simple example of an interference effect: Two waves traveling in opposite directions in a medium can combine to produce a standing-wave pattern with nodes and antinodes that do not move.

Figure 16.21 shows an example of another type of interference that involves waves that spread out in space. Two speakers, driven in phase by the same amplifier, emit identical sinusoidal sound waves with the same constant frequency. We place a microphone at point P in the figure, equidistant from the speakers. Wave crests emitted from the two speakers at the same time travel equal distances and arrive at point P at the same time; hence the waves arrive in phase, and there is constructive interference. The total wave amplitude that we measure at P is twice the amplitude from each individual wave.

Figure 16.21 summary: The figure is a diagram illustrating wave interference from two speakers connected to the same amplifier. The diagram depicts two points, P and Q, where sound waves from the speakers interfere. At point P, the distances from the speakers are equal, resulting in constructive interference. Conversely, at point Q, the distances differ by half a wavelength, leading to destructive interference. The diagram demonstrates that the path length difference significantly impacts the type of interference observed.

Now let's move the microphone to point Q, where the distances from the two speakers to the microphone differ by a half-wavelength. Then the two waves arrive a half-cycle out of step, or out of phase; a positive crest from one speaker arrives at the same time as a negative crest from the other. Destructive interference takes place, and the amplitude measured by the microphone is much smaller than when only one speaker is present. If the amplitudes from the two speakers are equal, the two waves cancel each other out completely at point Q, and the total amplitude there is zero.

Caution Interference and traveling waves The total wave in figure 16.21 is a traveling wave, not a standing wave. In a standing wave there is no net flow of energy in any direction; by contrast, in figure 16.21 there is an overall flow of energy from the speakers into the surrounding air, characteristic of a traveling wave. The interference between the waves from the two speakers simply causes the energy flow to be channeled into certain directions (for example, toward P) and away from other directions (for example, away from Q). You can see another difference between figure 16.21 and a standing wave by considering a point, such as Q, where destructive interference occurs.

Such a point is both a displacement node and a pressure node because there is no wave at all at this point. In a standing wave, a pressure node is a displacement antinode, and vice versa.

Constructive interference occurs wherever the distances traveled by the two waves differ by a whole number of wavelengths, 0, lambda, 2 times lambda, 3 times lambda, and so on; then the waves arrive at the microphone in phase (Fig. 16.22a If the distances from the two speakers to the microphone differ by any half-integer number of wavelengths, lambda divided by 2, 3 times lambda divided by 2, 5 times lambda divided by 2, and so on, the waves arrive at the microphone out of phase and there will be destructive interference (Fig. 16.22b In this case, little or no sound energy flows toward the microphone. The energy instead flows in other directions, to where constructive interference occurs.

Figure 16.22 summary: The figure is a combination of two illustrations. Both illustrations depict a scenario with two speakers connected to an amplifier and a microphone. The illustrations describe the concept of wave interference. In the first illustration, the path length difference between the speakers and the microphone is equal to a wavelength, resulting in constructive interference and a louder sound detected by the microphone. Conversely, in the second illustration, the path length difference is half a wavelength, leading to destructive interference and a much quieter sound or no sound at all detected by the microphone. In essence, the figure demonstrates how the path length difference affects the type of interference and the sound intensity perceived by the microphone.

Example 16.13 Loudspeaker Interference

Two small loudspeakers, A and B (Fig. 16.23), are driven by the same amplifier and emit pure sinusoidal waves in phase. (a) For what frequencies does constructive interference occur at point P? (b) For what frequencies does destructive interference occur? The speed of sound is 350 meters per second.

Figure 16.23 summary: The figure is a schematic diagram. It depicts two speakers, labeled A and B, positioned at different distances from a central point, with a receiver labeled P located at a distance from speaker B. The distance between speaker A and the central point is shorter than the distance between speaker B and the central point. The distance between speaker B and the receiver P is greater than the distance between speaker A and the central point. The receiver P is positioned further away from speaker B than speaker A is from the central point.

Identify and Set Up The nature of the interference at P depends on the difference d in path lengths from point A to P and from point B to P. We calculate the path lengths using the Pythagorean theorem. Constructive interference occurs when d equals a whole number of wavelengths, while destructive interference occurs when d is a half-integer number of wavelengths. To find the corresponding frequencies, we use v = flambda .

Execute The A-to-P distance is, open square bracket, open parenthesis 2.00 meters, close parenthesis, squared, plus, open parenthesis 4.00 meters, close parenthesis, squared, close square bracket, to the power of one half, equals 4.47 meters, and the B-to-P distance is, open square bracket, open parenthesis 1.00 meters, close parenthesis, squared, plus, open parenthesis 4.00 meters, close parenthesis, squared, close square bracket, to the power of one half, equals 4.12 meters. The path difference is d equals 4.47 meters minus 4.12 meters equals 0.35 meters.

(a) Constructive interference occurs when d equals 0, lambda, 2 times lambda, and so on or d equals 0, v over f, 2 times v over f, and so on, equals n times v over f. So the possible frequencies are

Math summary: This calculates possible frequencies based on the ratio of an integer multiple of velocity to distance. It computes a series of frequencies, specifically 1000 hertz, 2000 hertz, 3000 hertz, and so on.

(b) Destructive interference occurs when d equals lambda divided by 2, 3 times lambda divided by 2, 5 times lambda divided by 2, and so on, or d equals v divided by 2 times f, 3 times v divided by 2 times f, 5 times v divided by 2 times f, and so on. The possible frequencies are

Math summary: This calculates possible frequencies based on the mode number, sound velocity, and distance. It multiplies the mode number by the ratio of the sound velocity to twice the distance, resulting in a series of frequencies.

Evaluate As we increase the frequency, the sound at point P alternates between large and small (near zero) amplitudes, with maxima and minima at the frequencies given above. This effect may not be strong in an ordinary room because of reflections from the walls, floor, and ceiling.

Keyconcept Two sound waves of the same frequency interfere constructively at a certain point if the waves arrive there in phase. If the waves arrive at that point out of phase, they interfere destructively.

Figure 16.24 This aviation headset uses destructive interference to minimize the amount of noise from wind and propellers that reaches the wearer's ears.

Figure 16.24 summary: This is a photograph of a woman inside the cockpit of an aircraft. The woman is wearing a headset with a microphone and sunglasses. Based on the equipment, it is likely the woman is either a pilot or a passenger communicating during flight.

Interference is the principle behind active noise-reduction headsets, which are used in loud environments such as airplane cockpits (Fig. 16.24). A microphone on the headset detects outside noise, and the headset circuitry replays the noise inside the headset shifted in phase by one half-cycle. This phase-shifted sound interferes destructively with the sounds that enter the headset from outside, so the headset wearer experiences very little unwelcome noise.

Test Your Understanding of Section 16.6 Suppose that speaker A in figure 16.23 emits a sinusoidal sound wave of frequency 500 Hz and speaker B emits a sinusoidal sound wave of frequency 1000 Hz. What sort of interference will there be between these two waves? (1) Constructive interference at various points, including point P, and destructive interference at various other points; (2) destructive interference at various points, including point P, and constructive interference at various points; (3) neither (1) nor (2).

Answer

(destructive interference).

(3) Constructive and destructive interference between two waves can occur only if the two waves have the same frequency. In this case the frequencies are different, so there are no points where the two waves always reinforce each other (constructive interference) or always cancel each other.

16.7 Beats

In Section 16.6 we talked about interference effects that occur when two different waves with the same frequency overlap in the same region of space. Now let's look at what happens when we have two waves with equal amplitude but slightly different frequencies. This occurs, for example, when two tuning forks with slightly different frequencies are sounded together, or when two organ pipes that are supposed to have exactly the same frequency are slightly "out of tune."

Consider a particular point in space where the two waves overlap. In figure 16.25a we plot the displacements of the individual waves at this point as functions of time. The total length of the time axis represents 1 second, and the frequencies are 16 Hz (blue graph) and 18 Hz (red graph). Applying the principle of superposition, we add the two displacement functions to find the total displacement function.

The result is the graph of figure 16.25b At certain times the two waves are in phase; their maxima coincide and their amplitudes add. But at certain times (like t = 0.50 s in figure 16.25) the two waves are exactly out of phase. The two waves then cancel each other, and the total amplitude is zero.

Figure 16.25 summary: The figure includes two line graphs illustrating sound waves and their superposition. The upper graph displays two individual sound waves with slightly different frequencies. The lower graph shows the resultant wave formed by the superposition of the two individual waves. The superposition of two sound waves with slightly different frequencies produces fluctuations in amplitude, known as beats. The amplitude of the resultant wave varies depending on whether the individual waves are in phase or out of phase. When the waves are in phase, they interfere constructively, resulting in a higher amplitude. Conversely, when the waves are out of phase, they interfere destructively, resulting in a lower amplitude.

2 Definitions

Definition 1: Beats: Periodic variations in amplitude (loudness) that occur when two sound waves of slightly different frequencies are superimposed.

Definition 2: Beat Frequency: The number of amplitude maxima (or minima) per unit time that equals the difference in frequency between two interfering waves.

The resultant wave in figure 16.25b looks like a single sinusoidal wave with an amplitude that varies from a maximum to zero and back. In this example the amplitude goes through two maxima and two minima in 1 second, so the frequency of this amplitude variation is 2 Hz. The amplitude variation causes variations of loudness called beats, and the frequency with which the loudness varies is called the beat frequency. In this example the beat frequency is the difference of the two frequencies. If the beat frequency is a few hertz, we hear it as a waver or pulsation in the tone.

We can prove that the beat frequency is always the difference of the two frequencies f sub a and f sub b. Suppose f sub a is larger than f sub b; the corresponding periods are T sub a and T sub b, with T sub a less than T sub b. If the two waves start out in phase at time t equals 0, they are again in phase when the first wave has gone through exactly one more cycle than the second. This happens at a value of t equals T sub beat, the period of the beat. Let n be the number of cycles of the first wave in time T sub beat; then the number of cycles of the second wave in the same time is (n minus 1), and we have the relationships

Math summary: These equations calculate the beat period. The beat period equals the number of cycles, n, times the period of the first wave, and it also equals n minus one, times the period of the second wave.

Eliminating n between these two equations, we find

Math summary: This expression calculates the beat period. It divides the product of two input periods by the difference between those same two input periods.

The reciprocal of the beat period is the beat frequency, f sub beat equals 1 divided by T sub beat, so

Math summary: This calculates the beat frequency as the difference between the inverse of two periods. Specifically, it subtracts the inverse of the second period from the inverse of the first period to determine the beat frequency.

and finally

Math summary: This calculates the beat frequency between two waves. It subtracts the frequency of wave b from the frequency of wave a, with the result being the beat frequency.

As claimed, the beat frequency is the difference of the two frequencies.

An alternative way to derive Equation sixteen point twenty four is to write functions to describe the curves in Figure sixteen point twenty five a and then add them. Suppose that at a certain position the two waves are given by y sub a of t equals A times sine of two times pi times f sub a times t and y sub b of t equals negative A times sine of two times pi times f sub b times t. We use the trigonometric identity

Math summary: This expression calculates the difference between the sine of a first value and the sine of a second value. It outputs this difference by computing two intermediate values based on the sum and difference of the inputs, then scaling the product of the sine and cosine of those intermediate values by two.

We can then express the total wave y of t equals y sub a of t plus y sub b of t as

Math summary: This expression sums two wave functions to produce a combined wave. It calculates the combined wave as the product of a sine function, which modulates the amplitude, and a cosine function, which represents the average frequency of the original waves.

The amplitude factor (the quantity in brackets) varies slowly with frequency one half times the quantity f sub a minus f sub b. The cosine factor varies with a frequency equal to the average frequency one half times the quantity f sub a minus f sub b. The square of the amplitude factor, which is proportional to the intensity that the ear hears, goes through two maxima and two minima per cycle. So the beat frequency f sub beat that is heard is twice the quantity one half times the quantity f sub a minus f sub b, or just f sub a minus f sub b, in agreement with Equation sixteen point twenty four.

Beats between two tones can be heard up to a beat frequency of about 6 or 7 Hz. Two piano strings or two organ pipes differing in frequency by 2 or 3 Hz sound wavery and “out of tune,” although some organ stops contain two sets of pipes deliberately tuned to beat frequencies of about 1 to 2 Hz for a gently undulating effect. Listening for beats is an important technique in tuning all musical instruments. Avoiding beats is part of the task of flying a multiengine propeller airplane (Fig. 16.26).

Figure 16.26 summary: This figure is a photograph. The photograph depicts an airplane flying in the sky. The propellers of the airplane must be precisely synchronized to avoid loud and annoying sounds. The propellers can be synced either electronically or by ear.

At frequency differences greater than about 6 or 7 Hz, we no longer hear individual beats, and the sensation merges into one of consonance or dissonance, depending on the frequency ratio of the two tones. In some cases the ear perceives a tone called a difference tone, with a pitch equal to the beat frequency of the two tones. For example, if you listen to a whistle that produces sounds at 1800 Hz and 1900 Hz when blown, you'll hear not only these tones but also a much lower 100 Hz tone. Caution Beat frequency tells you only the difference in frequency between two sound waves. The beat frequency for two sound waves always equals the higher frequency minus the lower frequency, so its value alone doesn't tell you which wave frequency is higher. For example, if you hear a beat frequency of 1 Hz and you know one of the sounds has frequency 256 Hz, the frequency of the other sound could be either 257 Hz or 255 Hz.

Test Your Understanding of Section 16.7 One Tuning Fork Vibrates at 440 Hz.

while a second tuning fork vibrates at an unknown frequency. When both tuning forks are sounded simultaneously, you hear a tone that rises and falls in intensity three times per second. What is the frequency of the second tuning fork? (1) 434 Hz; (2) 437 Hz; (3) 443 Hz; (4) 446 Hz; (v) either 434 Hz or 446 Hz; (6) either 437 Hz or 443 Hz. distinguish between the two possibilities by comparing the pitches of the two tuning forks sounded one at a time: The frequency is 437 Hz if the second tuning fork has a lower pitch and 443 Hz if it (6) The beat frequency is 3 Hz, so the difference between the two tuning fork frequencies is also 3 Hz. Hence the second tuning fork vibrates at a frequency of either 443 Hz or 437 Hz. You can

16.8 The Doppler Effect

When a car approaches you with its horn sounding, the pitch seems to drop as the car passes. This phenomenon, first described by the 19th-century Austrian scientist Christian Doppler, is called the Doppler effect. When a source of sound and a listener are in motion relative to each other, the frequency of the sound heard by the listener is not the same as the source frequency. A similar effect occurs for light and radio waves; we'll return to this later in this section.

To analyze the Doppler effect for sound, we'll work out a relationship between the frequency shift and the velocities of source and listener relative to the medium (usually air) through which the sound waves propagate. To keep things simple, we consider only the special case in which the velocities of both source and listener lie along the line joining them. Let v S and v 50 be the velocity components along this line for the source and the listener, respectively, relative to the medium. We choose the positive direction for both v S and v 50 to be the direction from the listener L to the source S. The speed of sound relative to the medium, v, is always considered positive.

Moving Listener and Stationary Source

Let's think first about a listener L moving with velocity v sub L toward a stationary source S (Fig. 16.27). The source emits a sound wave with frequency f sub S and wavelength lambda equals v divided by f sub S. The figure shows four wave crests, separated by equal distances lambda. The wave crests approaching the moving listener have a speed of propagation relative to the listener of, open parenthesis, v plus v sub L, close parenthesis. So the frequency f sub L with which the crests arrive at the listener's position (that is, the frequency the listener hears) is

Math summary: This equation calculates the frequency detected by a listener. It divides the sum of the wave speed and listener velocity by the wavelength, which is equivalent to dividing the sum of the wave speed and listener velocity by the wave speed divided by the source frequency.

or

Math summary: This expression calculates the frequency perceived by a moving listener when the sound source is stationary. It multiplies the original frequency by a scaling factor, which is one plus the ratio of the listener's velocity to the speed of sound.

Figure 16.27 summary: This is an illustration depicting the Doppler effect with a listener moving towards a stationary sound source. The figure illustrates the wavefronts emitted by the source and their interaction with a moving listener. The listener's velocity is directed towards the source, while the source remains at rest. The wavelength of the sound waves is also shown. The listener perceives a higher frequency than the source emits because the relative speed between the listener and the wave is greater than the wave speed. This is a result of the listener intercepting more wavefronts per unit of time compared to when both the listener and source are stationary.

• Speed of sound wave = v

So a listener moving toward a source, open parenthesis v sub L greater than 0 close parenthesis, as in figure 16.27, hears a higher frequency (higher pitch) than does a stationary listener. A listener moving away from the source, open parenthesis v sub L less than 0 close parenthesis hears a lower frequency (lower pitch).

Moving Source and Moving Listener

Now suppose the source is also moving, with velocity v S (Fig. 16.28). The wave speed relative to the wave medium (air) is still v t ; it is determined by the properties of the medium and is not changed by the motion of the source. But the wavelength is no longer equal to v/f S . Here's why. The time for emission of one cycle of the wave is the period T = 1/f S . During this time, the wave travels a distance vT = v/f S and the source moves a distance v St = v S/f S . The wavelength is the distance between successive wave crests, and this is determined by the relative displacement of source and wave. As figure 16.28 shows, this is different in front of and behind the source. In the region to the right of the source in figure 16.28 (that is, in front of the source), the wavelength is

Math summary: This expression calculates the wavelength in front of a moving source. It subtracts the source's velocity divided by the source's frequency from the wave's velocity divided by the source's frequency, resulting in the difference between the wave's and source's velocities divided by the source's frequency.

Figure 16.28 summary: This is a wave diagram. It depicts a sound source moving from one location to another. The diagram illustrates the effect of the source's movement on the emitted sound waves. The sound waves in front of the source appear more compressed compared to those behind the source. The listener perceives a higher frequency as the source approaches and a lower frequency as it moves away.

In the region to the left of the source (that is, behind the source), it is

Math summary: This equation calculates the wavelength behind a moving source. It divides the sum of the wave velocity and the source velocity by the source frequency to determine the resulting wavelength.

The waves in front of and behind the source are compressed and stretched out, respectively, by the motion of the source.

To find the frequency heard by the listener behind the source, we substitute equation (16.28) into the first form of equation (16.25):

Math summary: This equation calculates the frequency perceived by a moving listener due to the Doppler effect, considering both the listener's and the source's velocities. It divides the sum of the wave's velocity and the listener's velocity by the ratio of the wave's velocity plus the source's velocity to the original frequency emitted by the source.

Figure 16.29 The Doppler effect explains why the siren on a fire engine or ambulance has a high pitch ( f L greater than f S ) when it is approaching you ( v S less than 0 ) and a low pitch ( f L less than f S ) when it is moving away ( v S greater than 0 ). Although we derived it for the particular situation shown in figure 16.28, equation (16.29) includes all possibilities for motion of source and listener (relative to the medium) along the line joining them. If the listener happens to be at rest in the medium, v 50 is zero. When both source and listener are at rest or have the same velocity relative to the medium, v L = v S and f L = f S . Whenever the direction of the source or listener velocity is opposite to the direction from the listener toward the source (which we have defined as positive), the corresponding velocity to be used in equation (16.29) is negative.

Figure 16.29 summary: This is a photograph of an ambulance. The ambulance is traveling on a road at night. The ambulance has its lights on. The ambulance is likely responding to an emergency.

As an example, the frequency heard by a listener at rest v sub L equals 0 is f sub L equals open bracket v divided by open parenthesis v plus v sub S close parenthesis close bracket times f sub S. If the source is moving toward the listener (in the negative direction), then v sub S is less than 0, f sub L is greater than f sub S, and the listener hears a higher frequency than that emitted by the source. If instead the source is moving away from the listener (in the positive direction), then v sub S is greater than 0, f sub L is less than f sub S, and the listener hears a lower frequency. This explains the change in pitch that you hear from the siren of an ambulance as it passes you (Fig. 16.29).

Problem-Solving Strategy 16.2 Doppler Effect

Identify the relevant concepts: The Doppler effect occurs whenever the source of waves, the wave detector (listener), or both are in motion.

Set Up the problem using the following steps:

1. Establish a coordinate system, with the positive direction from the listener toward the source. Carefully determine the signs of all relevant velocities. A velocity in the direction from the listener toward the source is positive; a velocity in the opposite direction is negative. All velocities must be measured relative to the air in which the sound travels.

2. Use consistent subscripts to identify the various quantities: S for source and L for listener.

3. Identify which unknown quantities are the target variables.

Execute the solution as follows:

1. Use equation (16.29) to relate the frequencies at the source and the listener, the sound speed, and the velocities of the source and

the listener according to the sign convention of step 1. If the source is moving, you can find the wavelength measured by the listener using equation (16.27) or (16.28).

2. When a wave is reflected from a stationary or moving surface, solve the problem in two steps. In the first, the surface is the “listener”; the frequency with which the wave crests arrive at the surface is f sub L . In the second, the surface is the “source,” emitting waves with this same frequency f sub L . Finally, determine the frequency heard by a listener detecting this new wave.

Evaluate your answer: Is the direction of the frequency shift reasonable? If the source and the listener are moving toward each other, f L greater than f S ; if they are moving apart, f L less than f S . If the source and the listener have no relative motion, f L = f S .

Example 16.14 Doppler Effect I: Wavelengths

A police car's siren emits a sinusoidal wave with frequency f S = 300 Hz. The speed of sound is 340 meters per second and the air is still. (a) Find the wavelength of the waves if the siren is at rest. (b) Find the wavelengths of the waves in front of and behind the siren if it is moving at 30 meters per second.

Identify and Set Up In part (a) there is no Doppler effect because neither source nor listener is moving with respect to the air; v = lambda f gives the wavelength. Figure 16.30 shows the situation in part (b): The source is in motion, so we find the wavelengths using Eqs. (16.27) and (16.28) for the Doppler effect.

Figure 16.30 summary: This is a diagram illustrating the Doppler effect with a moving source. The diagram shows a police car moving at a specified velocity. The diagram depicts the wavelength of the waves emitted by the police car both in front of and behind the car. The wavelength in front of the car is shorter, while the wavelength behind the car is longer, demonstrating the Doppler effect.

With Variation Problems

Execute (a) When the source is at rest,

Math summary: This calculates wavelength by dividing the speed of sound by the source frequency. The result is the wavelength, which in this case equals 1.13 meters.

(b) From equation (16.27), in front of the siren

Math summary: This expression calculates the wavelength in front of a siren. It divides the difference between the speed of sound and the speed of the source by the frequency of the source, resulting in a wavelength of 1.03 meters.

Figure 16.30 Our sketch for this problem.

From equation (16.28), behind the siren

Math summary: This expression calculates the wavelength behind a siren. It divides the sum of the speed of sound and the speed of the source by the frequency of the source, resulting in the wavelength.

Evaluate The wavelength is shorter in front of the siren and longer behind it, as we expect.

Keyconcept If a source of sound is moving through still air, a listener behind the source hears a sound of increased wavelength. A listener in front of the source hears a sound of decreased wavelength.

Example 16.15 Doppler Effect I.I: Frequencies

If a listener L is at rest and the siren in Example 16.14 is moving away from L at 30 meters per second, what frequency does the listener hear?

Identify and Set Up Our target variable is the frequency f sub L heard by a listener behind the moving source. Figure 16.31 shows the situation. We have v sub L = 0 and v sub S = +30 meters per second (positive, since the velocity of the source is in the direction from listener to source).

Figure 16.31 summary: This is a diagram that illustrates a scenario involving a stationary listener and a moving police car. The diagram depicts the listener at rest and the police car moving at a certain velocity towards the listener. The diagram is designed to help determine the frequency perceived by the listener. The listener will perceive a different frequency than the actual frequency emitted by the police car due to the Doppler effect, as the police car is moving towards the listener.

Execute From equation (16.29),

Math summary: This expression calculates the observed frequency. It divides the speed of sound by the sum of the speed of sound and the speed of the source, and then multiplies the result by the source frequency to obtain the observed frequency.

Evaluate The source and listener are moving apart, so f L less than f S . Here's a check on our numerical result. From Example 16.14, the wavelength behind the source (where the listener in figure 16.31 is located) is 1.23 m. The wave speed relative to the stationary listener is v = 340 meters per second even though the source is moving, so

Math summary: This calculation determines the frequency observed by a listener. It divides the speed of sound, three hundred forty meters per second, by the wavelength, one point two three meters, resulting in a frequency of two hundred seventy six hertz.

Keyconcept If a source of sound is moving through still air, a listener behind the source hears a sound of decreased frequency. A listener in front of the source hears a sound of increased frequency.

Example 16.16 Doppler Effect I.I.I: A Moving Listener

If the siren is at rest and the listener is moving away from it at 30 meters per second, what frequency does the listener hear?

Identify and Set Up Again our target variable is f L , but now L is in motion and S is at rest. Figure 16.32 shows the situation. The velocity of the listener is v L = -30 meters per second (negative, since the motion is in the direction from source to listener).

Figure 16.32 summary: The figure is a sketch illustrating a scenario involving the Doppler effect. It depicts a listener in a car moving towards a stationary police car. The listener's velocity is given, and the task is to determine the frequency perceived by the listener. The police car is at rest, indicated by a velocity of zero. The listener is approaching the source, so the perceived frequency should be higher than the emitted frequency.

Execute From equation (16.29),

Math summary: This equation calculates the observed frequency when a listener is moving relative to a sound source. It takes the speed of sound, adds the speed of the listener, divides by the speed of sound, and multiplies by the source frequency to find the shifted frequency perceived by the listener.

With Variation Problems

Evaluate Again the source and listener are moving apart, so f sub L is less than f sub S. Note that the relative velocity of source and listener is the same as in Example 16.15, but the Doppler shift is different because v sub S and v sub L are different.

Keyconcept If a listener is moving away from a source of sound, the listener hears a sound of decreased frequency. If the listener is moving toward the source, the listener hears a sound of increased frequency.

Example 16.17 Doppler Effect I.V: Moving Source, Moving Listener

The siren is moving away from the listener with a speed of 45 meters per second relative to the air, and the listener is moving toward the siren with a speed of 15 meters per second relative to the air. What frequency does the listener hear?

With Variation Problems

Identify and Set Up Now both L and S are in motion (Fig. 16.33). Again our target variable is f L . Both the source velocity v S = +45 meters per second and the listener's velocity v L = +15 meters per second are positive because both velocities are in the direction from listener to source.

Figure 16.33 summary: The figure is a diagram that illustrates a scenario involving the Doppler effect. The diagram depicts a listener in a car moving at a certain velocity and a police car, which is the source, moving at a higher velocity in the same direction. The diagram indicates that the listener is approaching the source. The objective is to determine the frequency observed by the listener, considering the relative motion between the listener and the source.

Execute From equation (16.29),

Math summary: This expression calculates the observed frequency. It divides the sum of the speed of sound and the speed of the listener by the sum of the speed of sound and the speed of the source, and then multiplies the result by the source frequency.

Evaluate As in Examples 16.15 and 16.16, the source and listener again move away from each other at 30 meters per second, so again f sub L is less than f sub S. But f sub L is different in all three cases because the Doppler effect for sound depend on how the source and listener are moving relative to the air, not simply on how they move relative to each other.

Keyconcept When a listener and source of sound are both moving relative to the air, the listener hears a sound of decreased frequency if the listener and source are moving apart. The listener hears a sound of increased frequency if the listener and source are moving closer together.

Example 16.18 Doppler Effect V: A Double Doppler Shift

The police car is moving toward a warehouse at 30 meters per second. What frequency does the driver hear reflected from the warehouse?

Identify This situation has two Doppler shifts (Fig. 16.34). In the first shift, the warehouse is the stationary "listener." The frequency of sound reaching the warehouse, which we call f sub w , is greater than 300 Hz because the source is approaching. In the second shift, the warehouse acts as a source of sound with frequency f sub w , and the listener is the driver of the police car; she hears a frequency greater than f sub w because she is approaching the source.

Figure 16.34 summary: The figure illustrates a scenario involving the Doppler effect with a moving source and a stationary listener, and the reverse. The diagram depicts a car moving towards a warehouse, emitting a sound that is received by a stationary listener inside the warehouse. The other diagram depicts the reverse; the car is stationary and the sound source is at the warehouse. The reflected sound heard by the driver has a higher frequency than the sound heard by a stationary listener in the warehouse.

Set Up To determine f sub W, we use Equation sixteen point twenty nine with f sub L replaced by f sub W. For this part of the problem, v sub L equals v sub W equals zero (the warehouse is at rest) and v sub S equals negative thirty meters per second (the siren is moving in the negative direction from source to listener).

To determine the frequency heard by the driver (our target variable), we again use equation (16.29) but now with f S replaced by f W . For this second part of the problem, v S = 0 because the stationary warehouse is the source and the velocity of the listener (the driver) is v L = +30 meters per second. (The listener's velocity is positive because it is in the direction from listener to source.)

Execute The frequency reaching the warehouse is

Math summary: This expression calculates the frequency reaching the warehouse. It divides the speed of sound by the sum of the speed of sound and the speed of the source, then multiplies the result by the original frequency to output the frequency at the warehouse.

Then the frequency heard by the driver is

Math summary: This calculates the frequency observed by a listener. It takes the speed of sound plus the speed of the listener, divides by the speed of sound, and multiplies the result by the original frequency, yielding the observed frequency.

Figure 16.34 Two stages of the sound wave's motion from the police car to the warehouse and back to the police car.

(a) Sound travels from police car's siren (source S) to warehouse ("listener" 50).

(b) Reflected sound travels from warehouse (source S) to police car (listener 50).

Keyconcept If a source of sound moves relative to a wall (or other reflecting surface), there are two shifts in the frequency of the sound: The frequency received by and reflected from the wall is shifted compared to the sound emitted by the source, and the frequency received back at the source is shifted compared to the sound reflected from the wall. Both shifts increase the frequency if the source is approaching the wall, and both shifts decrease the frequency if the source is moving away from the wall.

Doppler Effect for Electromagnetic Waves

In the Doppler effect for sound, the velocities v sub L and v sub S are always measured relative to the air or whatever medium we are considering. There is also a Doppler effect for electromagnetic waves in empty space, such as light waves or radio waves. In this case there is no medium that we can use as a reference to measure velocities, and all that matters is the relative velocity of source and receiver. (By contrast, the Doppler effect for sound does not depend simply on this relative velocity, as discussed in Example 16.17.)

To derive the expression for the Doppler frequency shift for light, we have to use the special theory of relativity. We'll discuss this in Chapter 37, but for now we quote the result without derivation. The wave speed is the speed of light, usually denoted by c, and it is the same for both source and receiver. In the frame of reference in which the receiver is at rest, the source is moving away from the receiver with velocity v. (If the source is approaching the receiver, v is negative.) The source frequency is again f S . The frequency f R measured by the receiver R (the frequency of arrival of the waves at the receiver) is then

Math summary: This equation calculates the observed frequency of light due to the Doppler effect. It scales the source frequency by a factor that depends on the speed of light and the relative velocity between the source and the receiver.

When v is positive, the source is moving directly away from the receiver and f sub R is always less than f sub S; when v is negative, the source is moving directly toward the receiver and f sub R is greater than f sub S. The qualitative effect is the same as for sound, but the quantitative relationship is different.

A familiar application of the Doppler effect for radio waves is the radar device mounted on the side window of a police car to check other cars' speeds. The electromagnetic wave emitted by the device is reflected from a moving car, which acts as a moving source, and the wave reflected back to the device is Doppler-shifted in frequency. The transmitted and reflected signals are combined to produce beats, and the speed can be computed from the frequency of the beats. Similar techniques ("Doppler radar") are used to measure wind velocities in the atmosphere.

The Doppler effect is also used to track satellites and other space vehicles. In figure 16.35 a satellite emits a radio signal with constant frequency f S . As the satellite orbits past, it first approaches and then moves away from the receiver; the frequency f R of the signal received on earth changes from a value greater than f S to a value less than f S as the satellite passes overhead.

Figure 16.35 summary: This is a diagram that shows a satellite passing a tracking station. The diagram illustrates the change in the velocity component along the line of sight of the satellite. As the satellite approaches the tracking station, the received frequency is higher. Conversely, as the satellite moves away, the received frequency becomes lower.

Test Your Understanding of Section 16.8 You are at an outdoor concert with a wind blowing at 10 meters per second from the performers toward you. Is the sound you hear Doppler-shifted? If so, is it shifted to lower or higher frequencies? is no Doppler shift.

Answer source. So both velocities are positive and v sub S equals v sub L equals plus 10 meters per second. The equality of these two velocities means that the numerator and the denominator in Equation 16.29 are the same, so f sub L equals f sub S and there The medium for sound waves) is moving from the source toward the listener. Hence, relative to the air, both the source and the listener are moving in the direction from listener to

16.9 Shock Waves

You may have experienced “sonic booms” caused by an airplane flying overhead faster than the speed of sound. We can see qualitatively why this happens from figure 16.36. Let v S denote the speed of the airplane relative to the air, so that it is always positive. The motion of the airplane through the air produces sound; if v S is less than the speed of sound v , the waves in front of the airplane are crowded together with a wavelength given by equation (16.27):

Figure 16.36 summary: The figure is an illustration of a sound source moving at a speed faster than the speed of sound. The illustration depicts the wave crests piling up in front of the source, creating a shock wave. The sound source overtakes the sound waves it emits, leading to the formation of a cone-shaped shock wave. As the speed of the sound source increases beyond the speed of sound, the wave crests compress, resulting in a more pronounced shock wave.

Math summary: This expression calculates the wavelength of sound waves in front of a moving object. It divides the difference between the speed of sound and the speed of the source by the frequency of the source.

As the speed v sub S of the airplane approaches the speed of sound v , the wavelength approaches zero and the wave crests pile up on each other (Fig. 16.36a The airplane must exert a large force to compress the air in front of it; by Newton's third law, the air exerts an equally large force back on the airplane. Hence there is a large increase in aerodynamic drag (air resistance) as the airplane approaches the speed of sound, a phenomenon known as the "sound barrier."

(a) Sound source S (airplane) moving at nearly the speed of sound (c) Shock waves around a supersonic airplane When v S is greater in magnitude than v , the source of sound is supersonic, and Eqs. (16.27) and (16.29) for the Doppler effect no longer describe the sound wave in front of the source. Figure 16.36b shows a cross section of what happens. As the airplane moves, it displaces the surrounding air and produces sound.

A series of wave crests is emitted from the nose of the airplane; each spreads out in a circle centered at the position of the airplane when it emitted the crest. After a time t the crest emitted from point S 1 has spread to a circle with radius v t , and the airplane has moved a greater distance v S t to position S 2 . You can see that the circular crests interfere constructively at points along the blue line that makes an angle alpha with the direction of the airplane velocity, leading to a very-large-amplitude wave crest along this line. This large-amplitude crest is called a shock wave (Fig. 16.36c

From the right triangle in figure 16.36b we can see that sine of alpha equals v times t divided by v sub St, or

Math summary: This expression calculates the sine of the shock wave angle produced by a sound source moving faster than sound. It divides the speed of sound by the speed of the source to determine the sine of the shock wave angle.

Definition

Mach Number: The ratio of the speed of an object to the speed of sound in the surrounding medium.

The ratio v S/v is called the Mach number. It is greater than unity for all supersonic speeds, and sin alpha in equation (16.31) is the reciprocal of the Mach number. The first person to break the sound barrier was Captain Chuck Yeager of the U.S. Air Force, flying the Bell X-1 at Mach 1.06 on October 14, 1947 (Fig. 16.37).

Figure 16.37 summary: The image is a photograph. It depicts the Bell X-1, which was the first supersonic airplane. The Bell X-1 has a shape similar to a bullet. The design of the Bell X-1 was influenced by the shape of a bullet, which was already known to surpass the speed of sound.

Shock waves are actually three-dimensional; a shock wave forms a cone around the direction of motion of the source. If the source (possibly a supersonic jet airplane or a rifle bullet) moves with constant velocity, the angle alpha is constant, and the shock-wave cone moves along with the source. It's the arrival of this shock wave that causes the sonic boom you hear after a supersonic airplane has passed by. In front of the shock-wave cone, there is no sound. Inside the cone a stationary listener hears the Doppler-shifted sound of the airplane moving away.

Caution Shock waves A shock wave is produced continuously by any object that moves through the air at supersonic speed, not only at the instant that it "breaks the sound barrier." The sound waves that combine to form the shock wave, as in figure 16.36b, are created by the motion of the object itself, not by any sound source that the object may carry. The cracking noises of a bullet and of the tip of a circus whip are due to their supersonic motion.

A supersonic jet airplane may have very loud engines, but these do not cause the shock wave. If the pilot were to shut the engines off, the airplane would continue to produce a shock wave as long as its speed remained supersonic.

Shock waves have applications outside of aviation. They are used to break up kidney stones and gallstones without invasive surgery, using a technique with the impressive name extracorporeal shock-wave lithotripsy. A shock wave produced outside the body is focused by a reflector or acoustic lens so that as much of it as possible converges on the stone. When the resulting stresses in the stone exceed its tensile strength, it breaks into small pieces and can be eliminated. This technique requires accurate determination of the location of the stone, which may be done using ultrasonic imaging techniques (see figure 16.9).

Example 16.19 Sonic Boom From a Supersonic Airplane

An airplane is flying at Mach 1.75 at an altitude of 8000 m, where the speed of sound is 320 meters per second. How long after the plane passes directly overhead will you hear the sonic boom?

Identify and Set Up The shock wave forms a cone trailing backward from the airplane, so the problem is really asking for how much time elapses from when the airplane flies overhead to when the shock wave reaches you at point 50 (Fig. 16.38). During the time t (our target variable) since the airplane traveling at speed v s passed overhead, it has traveled a distance v s t . Equation (16.31) gives the shock cone angle alpha ; we use trigonometry to solve for t. Execute From equation (16.31) the angle alpha of the shock cone is

Math summary: This expression calculates the angle alpha by taking the arcsine of one divided by one point seventy five. The result of this calculation is thirty four point eight degrees.

Figure 16.38 summary: The figure is an illustration depicting the formation of a sonic boom by an aircraft traveling at supersonic speed. The diagram shows the aircraft moving at Mach 1.75, generating a cone-shaped shock wave. The figure illustrates the position of a listener relative to the shock wave. The listener at location L is experiencing the sonic boom as the shock wave reaches them, while another listener to the right has not yet experienced it, and one to the left already has.

The speed of the plane is the speed of sound multiplied by the Mach number:

Math summary: This expression calculates the speed of an object. It multiplies a scaling factor of one point seven five by an input speed of three hundred twenty meters per second, resulting in an output speed of five hundred sixty meters per second.

From figure 16.38 we have

Math summary: First, the tangent of an angle is calculated by dividing eight thousand meters by the product of the plane's speed and time. Then, time is calculated by dividing eight thousand meters by the product of five hundred sixty meters per second and the tangent of thirty four point eight degrees, resulting in twenty point five seconds.

Evaluate You hear the boom 20.5 s after the airplane passes overhead, at which time it has traveled (560 meters per second) (20.5 s) = 11.5 kilometers since it passed overhead. We have assumed that the speed of sound is the same at all altitudes, so that alpha = arcsin v/v S is constant and the shock wave forms a perfect cone. In fact, the speed of sound decreases with increasing altitude. How would this affect the value of t ?

Keyconcept An object moving through the air faster than the speed of sound continuously produces a cone-shaped shock wave. The angle of the cone depends on the object's Mach number (the ratio of its speed to the speed of sound).

Test Your Understanding of Section 16.9 What would you hear if you were directly behind (to the left of) the supersonic airplane in figure 16.38? (1) A sonic boom; (2) the sound of the airplane, Doppler-shifted to higher frequencies; (3) the sound of the airplane, Doppler-shifted to lower frequencies; (4) nothing.

(3) Figure 16.38 shows that there are sound waves inside the cone of the shock wave. Behind the airplane the wave crests are spread apart, just as they are behind the moving source in figure 16.28. Hence the waves that reach you have an increased wavelength and a lower frequency. Answer

Chapter 16 Summary

Sound waves: Sound consists of longitudinal waves in a medium. A sinusoidal sound wave is characterized by its frequency f and wavelength lambda (or angular frequency omega and wave number k) and by its displacement amplitude A. The pressure amplitude p sub max is directly proportional to the displacement amplitude, the wave number, and the bulk modulus B of the wave medium. (See Examples 16.1 and 16.2.)

The speed of a sound wave in a fluid depends on the bulk modulus B and density rho . If the fluid is an ideal gas, the speed can be expressed in terms of the temperature T, molar mass M, and ratio of heat capacities gamma of the gas. The speed of longitudinal waves in a solid rod depends on the density and Young's modulus Y. (See Examples 16.3 and 16.4.)

Math summary: This expression calculates the maximum pressure of a sound wave. It multiplies the bulk modulus of the medium, the wave number, and the amplitude of the wave to determine the maximum pressure.

(sinusoidal sound wave)

Math summary: This expression calculates the speed of a sound wave in a fluid. It takes the square root of the bulk modulus divided by the density to produce the wave speed.

(longitudinal wave in a fluid)

Math summary: This expression calculates the speed of sound. It takes the square root of a value derived from the ratio of specific heats, the ideal gas constant, and the absolute temperature, all divided by the molar mass.

(sound wave in an ideal gas)

Math summary: This expression calculates the speed of a wave. It takes the square root of the ratio of a material property, such as Young's modulus, to its density to produce the wave speed.

(longitudinal wave in a solid rod)

Image summary: The figure includes two wave plots and a particle diagram. The upper wave plot shows displacement as a function of position, while the lower wave plot shows pressure variation as a function of position. The particle diagram illustrates regions of compression and rarefaction. The pressure wave is related to the displacement wave. Regions of compression in the particle diagram correspond to peaks in the pressure wave, while regions of rarefaction correspond to valleys.

Intensity and sound intensity level: The intensity I of a sound wave is the time average rate at which energy is transported by the wave, per unit area. For a sinusoidal wave, the intensity can be expressed in terms of the displacement amplitude A or the pressure amplitude p sub max . (See Examples 16.5 to 16.7.)

The sound intensity level beta of a sound wave is a logarithmic measure of its intensity. It is measured relative to I sub 0, an arbitrary intensity defined to be 10 to the power of negative 12 Watts per meter squared. Sound intensity levels are expressed in decibels (decibel). (See Examples 16.8 and 16.9.)

Math summary: This expression calculates sound intensity. It shows that sound intensity is proportional to the square of the maximum pressure divided by two times the square root of the product of the density and bulk modulus.

(intensity of a sinusoidal sound wave in a fluid)

Image summary: The image is an illustration. It depicts a bird singing, which is labeled as a point source of sound. The sound waves propagate outwards as concentric circles. Two people are standing at different distances from the bird. The person closer to the bird is labeled as P1 and the person farther away is labeled as P2. The image illustrates that the intensity of sound decreases with distance from the source. Therefore, the person closer to the bird will hear a louder sound than the person farther away.

Math summary: This expression calculates the sound intensity level. It takes the logarithm of the ratio between the sound intensity and a reference intensity, then scales the result by a factor of ten decibels to produce the sound intensity level.

(definition of sound intensity level)

Standing sound waves: Standing sound waves can be set up in a pipe or tube. A closed end is a displacement node and a pressure antinode; an open end is a displacement antinode and a pressure node. For a pipe of length L open at both ends, the normal-mode frequencies are integer multiples of the sound speed divided by 2L. For a stopped pipe (one that is open at only one end), the normal-mode frequencies are the odd multiples of the sound speed divided by 4L. (See Examples 16.10 and 16.11.)

Math summary: This equation calculates the normal mode frequencies of a pipe open at both ends. It takes the mode number, the sound speed, and the length of the pipe as inputs, and outputs the frequency, which is an integer multiple of the sound speed divided by twice the length.

A pipe or other system with normal-mode frequencies can be driven to oscillate at any frequency. A maximum response, or resonance, occurs if the driving frequency is close to one of the normal-mode frequencies of the system. (See Example 16.12.)

Math summary: This expression calculates the frequencies at which a pipe resonates. It takes the mode number, which must be an odd integer, multiplies it by the speed of sound, and then divides by four times the length of the pipe to determine the resonant frequency.

(stopped pipe)

Image summary: The image shows a diagram of an open pipe demonstrating different modes of resonance. The diagram illustrates the fundamental frequency and the first overtone in an open pipe, showing the positions of nodes and antinodes. The fundamental frequency has a longer wavelength and lower frequency, while the first overtone has a shorter wavelength and higher frequency, specifically twice the fundamental frequency.

Image summary: The figure displays diagrams of standing sound waves in a stopped pipe. The diagrams show the first and third harmonic frequencies. The third harmonic frequency is three times greater than the first harmonic frequency.

Interference: When two or more waves overlap in the same region of space, the resulting effects are called interference. The resulting amplitude can be either larger or smaller than the amplitude of each individual wave, depending on whether the waves are in phase (constructive interference) or out of phase (destructive interference). (See Example 16.13.)

Image summary: The image is a diagram that illustrates wave interference from two sources. The diagram depicts two speakers emitting waves, with points labeled P and Q representing locations where the waves are observed. At point P, the waves arrive in phase. However, at point Q, the waves arrive half a cycle out of phase. The path length from one speaker to point Q is greater than the path length from the other speaker to point Q by half of the wavelength. This path length difference leads to destructive interference at point Q, whereas the equal path lengths or path lengths differing by an integer multiple of the wavelength lead to constructive interference at point P.

Beats: Beats are heard when two tones with slightly different frequencies f sub a and f sub b are sounded together. The beat frequency f sub beat is the difference between f sub a and f sub b.

Math summary: This expression calculates the beat frequency. It subtracts the second frequency from the first frequency to determine the beat frequency.

(beat frequency)

Image summary: The figure is a line graph. It shows the displacement as a function of time for two individual waves and their superposition. The superposition of the two waves produces a beat pattern, where the amplitude of the combined wave varies periodically. The frequency of the individual waves appears higher than the frequency of the beat pattern.

Doppler effect: The Doppler effect for sound is the frequency shift that occurs when there is motion of a source of sound, a listener, or both, relative to the medium. The source and listener frequencies f S and f L are related by the source and listener velocities v S and v 50 relative to the medium and to the speed of sound v. (See Examples 16.14 to 16.18.)

Math summary: This equation calculates the listener frequency based on the Doppler effect. It scales the source frequency by a ratio involving the speed of sound, the listener's velocity, and the source's velocity.

(Doppler effect, moving source and moving listener)

Image summary: The image is an illustration depicting a physical phenomenon. It portrays a scenario involving a moving source emitting waves, and a stationary observer. The illustration shows how the wavelength of the waves appears to change due to the relative motion between the source and the observer. The waves appear compressed in the direction of the source's motion and stretched in the opposite direction. The observer perceives a different frequency than the source emits due to this change in wavelength. The wavelength is shorter in front of the source and longer behind the source.

Shock waves: A sound source moving with a speed v sub S greater than the speed of sound v creates a shock wave. The wave front is a cone with angle alpha. (See Example 16.19.) sine of alpha equals v divided by v sub S (shock wave)

Image summary: The figure is an illustration. It depicts an object moving at a speed faster than the wave speed it generates, resulting in a shock wave pattern. The illustration shows the wavefronts bunching up to form a cone-shaped shock wave. The object's velocity is higher than the wave velocity. The shock wave is formed due to the object's supersonic motion.

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