Chapter 17 Sound and the Human Ear...

Physics Including Human Applications

353

Chapter 17 Sound and the Human Ear

GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms and use it in an operational definition:

sound wave infrasonic intensity of sound wave ultrasonic amplitude distortion frequency

standing wave wavelength beats acoustic impedance Doppler effect decibel

Sound Wave Form Graph the amplitude of a sound wave in pressure versus space or time coordinates. Sound Level Calculate the sound level in decibels (dB). Sound Intensity Solve sound intensity and sound amplitude problems. Sound Application Explain the use of sound in selected medical applications. Hearing System Describe the properties of the human sound detection system. Doppler Effect Solve Doppler- effect and standing-wave problems.

PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 5, Energy, Chapter 10, Temperature and Heat, Chapter 15, Simple Harmonic Motion, and Chapter 16, Traveling Waves, respectively.


354

Chapter 17 Sound and the Human Ear

17.1 Introduction How does sound originate? Do you know what gives rise to sounds? Hold a ruler tight against your desk and strike the end that extends in the air. What happens? What can you hear? How can you change what you hear? If you put your head behind a book and strike the ruler again, can you still hear something? What does that tell you about the ability of sound to travel around objects? At sometime or other you probably have heard the old philosophical question - if a tree falls in the forest and no one is there, does the tree crashing to the ground make a sound? If you know the physical principles of sound, you can answer this question logically. This chapter is intended to introduce you to the basic physical principles of sound and of your ear. Sound is becoming increasingly useful in medical applications. Your understanding of sound will not only enable you to appreciate the properties of your ear as a sound detector, but may enable you to understand the design of a sound system for making medical diagnoses.

17.2 Sound The term sound has two distinct uses. The psychologist and physiologist think of sound as a sensation due to the stimulation of the auditory nervous centers. To the physicist it is a form of vibrational energy that produces the audio sensation. Sound is modeled as longitudinal wave motion which is transmitted by an elastic medium. A sound system has three components. First, there is a sound source consisting of a vibrating object, for example, a speaker. Second, there is the medium through which the sound is transmitted, air in a room, for example. And finally, there is the detector, the human ear. Sound is an example of interaction-at-a-distance. The vibrating source interacts with a detector some distance away. For the familiar case of a vibrating object in air, we can think of the vibrating source as sweeping out a small volume of space and causing fluctuations in the density (or pressure) of the air in the vicinity of the source. These density or pressure fluctuations are then transmitted to the detector as traveling waves in the air. We can define sound as a variation in pressure set up by a vibrating source. Then sound is seen as waves propagated through matter and characterized by a periodic variation of pressure (or density) in a medium. These periodic pressure variations occur in the direction of the wave propagation. Thus, sound is defined as a longitudinal wave as discussed in Chapter 16. A sound wave diagram is shown in Figure 17.1. The mathematical formula for the pressure in this sound wave being propagated in the x-direction is given by Equation 17.1.


355

P = Po sin (2πft - 2πx/λ) (17.1) Po = pressure amplitude (N/m2) at source x = position (m) or distance from source f = frequency (Hz) = number of complete pressure cycles per second λ = wavelength (m) = distance between adjacent equal pressure amplitudes t = time in seconds This equation expresses the periodicity of the pressure in both space (x) and time (t). You can see this periodicity, if you examine the equation with time constant or with position constant. EXAMPLES 1. Consider the following as an equation for a typical sound wave: P = (2 x 10-2N/m2) sin (1000 πt - πx/0.34) Graph P against x for two different times, t1 = 0.8 x 10-3 sec and t2 = 10-3 sec. What is the wavelength of this wave? What is its speed? Which direction is it traveling? See Figure 17.2.


356


357

2. Draw a graph P versus t for two different values of x, x1 = 0, and x2 = 0.17 m as shown in Figure 17.3. For x1 = 0, P1 = 2 x 10-2 sin 1000πt For x2 = 0.17 m P2 = 2 x 10-2 sin (1000πt/2)

What is the period of oscillation for this wave? T = 2 x 10-3 sec. Can you compare it to the frequency of some known sound? f = 500 Hz. ( Hint: middle C on a piano has a frequency of 256 Hz.) Could this sound be heard? Recall that the magnitude of the velocity, the speed, of a wave is given by fλ = v (Equation 16.2). Find the speed of the sound for this example. In the first case we have a sinusoidal wave in space (x -direction), and in the second case for a given x, the pressure varies sinusoidally in time as the wave goes by a given position. Such a wave transports energy from one place to another in the direction of the wave propagation. In Chapter 16 we learned that the velocity of a compressional wave such as sound in a solid is given by

where Y is Young's modulus for the solid and ρ is the density of the solid.

for a fluid

where B is the bulk modulus and ρ is the density of the fluid. And

for a gas

where γ = ratio of cp/c v, Po = undisturbed gas pressure, and ρ = the density of the gas. The velocity and the wavelength change in such a way that v s = f λ is always satisfied. When a sound wave passes from one medium to another, for example, from air to water, the frequency of the wave remains constant. In other words, the frequency is a property of the source and not of the transmitting medium. On the other hand, the velocity is a property of the elastic medium in which the wave is traveling. The wavelength results from the combination of these two properties.


358

Consider the problem of comparing the velocity of sound in hydrogen and deuterium. What are reasonable assumptions to make for these two gases? Since they are isotopes, it should be a good approximation to assume the same elastic constant (bulk modulus) for each. The velocity of sound is then primarily determined by the density of the gases according to the previous equation. Since the mass of a deuterium molecule is twice that of a hydrogen molecule, we find the magnitudes of the velocity of sound in the two gases are related as vs(H2) =vs(D2) x SQR RT(2) This result agrees with the actual measured values to within 2 percent.

17.3 Sound Intensity The intensity of a traveling wave is defined as the energy per unit time (that is, power) transported perpendicularly across a surface of unit area Figure 17.4. The units of wave intensity in the SI are watts/meter2. You should verify these as the correct units from the definition of intensity. In Chapter 16 on traveling waves, you were given an equation for the intensity of a traveling wave (Equation 16.7). The equation for the intensity of a sound wave follows by analogy. In the medium of density ρ, we have the following expression for the sound intensity. I = (½) ( Aω)2ρvs (watt/m2) (17.2) where vs = magnitude of the velocity of sound (m/sec), A = amplitude of motion (m), and ω = 2πf (rad/sec). The intensity of a sound wave can also be expressed in terms of the pressure amplitude Po, of the wave as follows:

(17.3)

where ρvs is the acoustical impedance of the medium, Za.

EXAMPLE Normal conversation has an intensity of about 1.00 x 10-6 watts/m2. The velocity of sound at room temperature is 345 m/sec and density of air in the room is 1.26 kg/m3.


359

What is the pressure amplitude?

Po = SQR RT( 2Iρvs) = SQR RT(2 x 10-6 x 1.26 x 345) = 29.5 x 10-3 N/m2

Normal atmospheric pressure is about 105 N/m2. The transmission and reflection of sound energy at an interface between two media is determined by the acoustic impedances of the media involved. Let ZA be the acoustic impedance of first medium and ZB, the acoustic impedance of second medium. For sound traveling perpendicular to the interface from medium A to medium B, it can be shown that the ratio of reflected intensity Ir to incident intensity Io is given by: Ir / Io = (ZA - ZB)2 / (ZA + ZB)2 (17.4) and the ratio of transmitted intensity It to incident intensity is given by: It / Io = 4ZA ZB / (ZA + ZB)2 (17.5) Note if the impedances match (ZA = ZB), no energy is reflected at the interface and It = Io. This is another example of maximum energy transfer when inertial properties match at the interface of two systems. EXAMPLE An application of ultrasound is its use in brain surgery. Let us determine the energy transmitted from water to bone and compare it with energy transmitted from air to bone. Za(water) = 1.43 x 106 (kg/m2-sec); Za(bone) = 6 x 106 (kg/m2-sec); Za(air) = 430 kg/m2-sec. Thus, we have I(bone)/I(water) = 4 x 1.43 x 106 x 6 x 106

/(7.43 x 106)2 = 0.62, or 62 percent of the energy is transmitted from the water to the bone. For I(bone)/I(air) = 4 x 430 x 6 x 106/(6.00043 x 106)2 = 0.29 x10-3, or .03 percent of the energy is transmitted from air to the bone. This illustrates why water is used to match an ultrasonic transducer to bone or brain tissue.


360

17.4 Threshold of Hearing We can calculate the amplitude of motion for the minimum intensity detected by a normal human ear from Equation 17.2. The minimum threshold of hearing occurs at a frequency of approximately 3500 Hz and has an intensity value of 1.00 x 10-12 watt/m 2 for the typical human being. The density of air is 1.25 kg/m3, and vs = 345 m/sec at 300øK. Thus the amplitude of motion is given by: I = 1.00 x 10-12 watts/m2 = A2 / 2 (3500 x 2 π)2 x 1.25 x 345 A2 = 2.00 x 10-12 / (1.25 x 345. x 12.25 x 106 x 4 x π2) = 2.00 x 10-18 / (1.25 x 345. x 12.25 x 4 x π2 ) = 9.60 x 10-24

A = 3.10 x 10-12 m

An ultrasonic scan showing multiple gall stones. This is a longitudinal scan through the right lobe of the liver, the finely distributed liver structure. The lumen of the gall bladder shows as a black reflection. [free area, with 4 light reflections at the gall stones, dorsally the sound shadow zone.](Siemens Corporation.)

Note that this amplitude of motion for the threshold of hearing is of the order of magnitude of the diameter of an atom (10-10

m). By using proportional reasoning, calculate the amplitude of an intensity of 1 watt/m2 (this corresponds to the threshold of pain) at the same frequency. The range of intensities that the human ear can successfully detect varies from 10-12 watt/m2 to 1 watt/m2. Because of this wide range of intensities, a logarithmic scale is used for defining sound level. The sound level in decibels (dB) is defined as follows: sound level (dB) = 10 log10 (I/Io) (17.6) where I is the sound intensity and Io is the standard intensity of 10-12 watt/m2. Ordinary conversation is approximately 60 dB. This corresponds to log I/I0 = 60/10, or I = I0 x 106 = 10-6 watts/m2.


361

EXAMPLES 1. Table 17.1 lists some sound levels as measured at the ear. Complete the table using a frequency of 2000 Hz. 2. Suppose that a snowmobile has a sound level of 80 dB at 30 m from the machine. If the maximum noise limit is 77 dB, how much must the intensity be reduced to meet this requirement, and what is the intensity at this noise limit? Using Equation 17.6 we have 80 dB = 10 log I10-12 watt/m2

Thus, I = 10-12 x 108 = 10-4 watt/m2

for 77 dB, I = 107.7/108 x 10-4 watt/m2

Thus for 77 dB, I = 10-4.3 watt/m2

a reduction of 0.5 x 10-4 watt/m2. Notice that doubling the intensity increases the sound level by 3 dB.

17.5 Infrasonics and Ultrasonics Recall the equation for traveling waves, v = f λ, that relates the magnitude of the velocity of the wave (v) to the frequency (f) and wavelength (λ). In phenomena involving sound, the frequency range is usually divided into three different regions, frequencies below 20 Hz are referred to as infrasonic, between 20 Hz and 20,000 Hz is the audio frequency range for the human ear, and frequencies above 20,000 Hz are referred to as ultrasonic. The effect of infrasound on living systems is not well understood, and it is a current area of research. There are several applications of ultrasonics in medical science. One such application is found in neurosonic surgery. Ultrasonic waves are focused on a particular


362

part of the brain, destroying only the tissue at the focal point without interfering with the normal activity of other parts of the brain. This application has proved to be effective in treating Parkinson's disease, which seems to have well-localized centers in the brain. What are the particular properties of ultrasound that make it useful for this kind of application? Try to design a model of interaction for the tissue and the ultrasonic wave.

17.6 Superposition Principle and Sound Thus far we have considered pure sinusoidal sound waves. Such waves are sometimes referred to as pure tones. Using the mathematical formulation known as Fourier analysis (see Section 16.12), it is possible to synthesize any wave shape by adding up a series of sinusoidal waves of the proper amplitudes, frequencies, and phases. Most natural sounds are complex in shape and involve many of these Fourier components, or pure sound waves. If the amplitude of the wave changes rapidly, the number of different frequency components necessary to synthesize the sound is large. For example, a square wave must have more high-frequency components than a wave with slowly varying amplitude. The relative amplitudes of the harmonics making up a sawtooth wave and a square wave of the same frequency are illustrated in Figure 17.5a. Such a component analysis of a sound into its harmonics is called a sound spectrum. The sound spectrum for a clarinet and a french horn playing the same note are shown in Figure 17.5b. Your voice can be analyzed when you say certain words, but your voice will have considerable variability. By using computers to make the Fourier analysis of voice prints, it is possible to detect changes that are due to the emotional state of the speaker. Such voice print analysis systems may become the "lie detectors" of the future. It has already been demonstrated that each of us has a voice print as unique as our fingerprints, and thus such prints can be used for identification.

Figure 17.5

The relative ratios of the fundamental and the overtones at a given constant frequency are shown for a saw tooth wave, a square wave, a clarinet, and a French horn.


363

17.7 Distortion Not all sound systems satisfy the linear approximation necessary for the superposition principle to be valid. When this linear condition is not satisfied by any part of an acoustical system, distortion is introduced. Distortion may take the form of new components added to the sound wave. Distortion is frequently encountered when working with large amplitude sound waves. Large amplitude waves may force the source, medium, or detector to operate outside the region where the deformation of a system is proportional to the force producing the deformation, that is, outside the region of Hooke's law behavior for the system.

17.8 Standing Waves We will now consider other applications of the superposition principle. First, we will apply this principle to the addition of waves in a medium within a specific volume with well-defined boundaries. When a sound wave meets such a boundary, it undergoes reflection. The determining factor for the amplitude and phase shift of the reflected wave is the acoustical impedance of the boundary layer. The acoustic impedance is defined as the density of the medium times the velocity of sound in the medium, Za = ρvs (17.7) You will recall that this parameter appeared in the expression for the intensity of a sound wave. Under certain boundary conditions the addition of reflected and incident waves in a defined volume yields a standing wave pattern. The sound in this volume is not described as a traveling wave, but instead we find an equation such as the following: P = 2Po sin kx cos ωt (17.8) which results from the addition of P1 = P0 sin (ωt + kx) and P2 = P0 sin (ωt - kx). You will recall that standing waves are possible only for the wavelengths that satisfy the conditions of nodes at fixed ends and antinodes at free ends. For both ends fixed, L = n (λ/2) where n = 1, 2, .... For one end free, L = m(λ/4), where m = 1, 3, 5, ... (odd integers only). An example of such a standing wave is the motion of a string fixed at each end and plucked in its center. Such standing waves are the basis of sound production for many musical instruments. The lowest frequency standing wave of a system is called the fundamental or first harmonic of the system. Other possible frequencies of standing waves for a system involve integral multiples of the fundamental frequency. If all the boundaries of a system are the same, the enclosed system will support standing waves of all integral harmonics. If all boundaries are not the same, that is, with the pipe opened at one end and closed at the other, only odd harmonics will be supported by the system. Some possible representations for standing waves for open and closed pipes are shown in Figure 17.6. Please note that for an open pipe, L = n(λ/2) where n is an integer and for a closed pipe, L = n(λ/4) where n is an odd integer.


364

17.9 Beat Frequency Another result of the application of the superposition principle is the production of beats when two pure sound waves are generated together. Beats are the pulsation of the amplitude of the sum of two waves with slightly different frequencies. The beat frequency is equal to the difference between the frequencies of the two waves involved. An example of such a beat phenomenon is shown in Figure 17.7. Beats are heard as periodic variation in the loudness of the sound. The human ear can detect beats to about seven beats per second. Higher beat frequencies do not yield distinct beats, and the sound is characterized by dissonance. See the enrichment section of this chapter for a mathematical treatment of beats.

Figure 17.7

Two waves of different frequencies add to produce a beat frequency equal to the difference of the frequency of the two waves.

17.10 The Doppler Effect The Doppler effect involves the change in frequency of a wave due to the relative motion of the source and observer. First, consider the case where the sound source is approaching an observer at rest. Let v designate the velocity of the source toward the observer and vs the magnitude of the velocity of sound in air. The motion of the source toward the observer means that more waves will be crowded into each meter than would be present if both source and observer were at rest. This means that the apparent wavelength is decreased (see Figure 17.8).


365

Figure 17.8

The Doppler Effect

Since during one cycle the source will move v/f, this wavelength will be reduced by v/f, so the apparent wavelength is λ' = λ - v/f =vs/f - v/f = vs - v/f (17.9) and the frequency of the approaching source will appear to be f ' = velocity/wavelength = vs/(vs - v)/f or f ' = f vs /( vs –v) If the source moves away from the observer at the same speed, the same argument yields a frequency f ' =fvs /(vs + v). If the observer moves toward a source at rest with a speed of vo, the apparent velocity of the sound waves relative to the observer is vs + vo and, thus, the apparent frequency of the sound is given by the ratio of the velocity of the wave to its wavelength, f ' = vs + vo /λ = f (vs +vo) / vs (17.10) If the observer moves away from the source at the same speed, the result is a frequency of f ' = f (vs -vo) /vs We can combine these results into an equation for the relative motion of approach and an equation for the relative motion of separation. For relative approach: f ' = f (vs + vo) / (vs - v) (17.11) For relative separation: f ' = f (vs - vo) / (vs + v) where f = frequency with no relative motion, v s = speed of sound, vo = speed of observer, and v = speed of source.


366

The Doppler effect is used in medicine by measuring the frequency shift of an ultrasonic wave that is reflected off a moving organ inside the body. Such a technique can be used to study the heart beat of the fetus early in pregnancy. A small ultrasonic transducer is placed on the mother's abdomen. Some of the ultrasound waves are reflected back by the beating fetus' heart. The fetal heart serves as a moving reflector and provides a Doppler shift of the ultrasound when received. When the heart is moving away the frequency is lower and when the heart moves toward the receiver the frequency of the detected signal is increased. EXAMPLE An observer is standing near the edge of a state highway, watching an automobile traveling on the highway at 20 m/sec. The automobile approaches the observer, and then recedes from him. The natural frequency of the automobile's horn is 540 Hz. If the horn is sounded on the approach and on the separation from the observer, what frequency does the observer hear? On this particular day the velocity of sound in air is 340 m/sec. On the automobile's approach, the horn's sound seems to have a frequency greater than the natural frequency. f ' = f vs/(vs -v) where the velocity of sound in air = vs = 340 m/sec. And the velocity of the car (source) = v = 20 m/sec. f ' = 540 x 340/(340 - 20) = 540 x 340/320 = 574 Hz As the automobile moves away, the frequency of the horn's sound is heard at less than the natural frequency:


367

f ' = f vs/(vs +v) = 540 x 340/(340 + 20) = 510 Hz Now suppose the car is stationary and you are in a car moving along the highway under the conditions stated above. As you approach the horn, you hear a frequency greater than the natural frequency. f ' = f vs + v/vs = 540 x (340 + 20)/340 = 573 Hz Moving away from the horn you hear a frequency less than the natural frequency: f ' = f vs - v/vs = 540 (340 - 20 /340) = 508 Hz

17.11 Sound Detectors We will now consider some common sources and detectors of sound. Two important characteristics of a sound source or detector are efficiency and fidelity. The efficiency of a sound source is a measure of its effectiveness in converting the power supplied to the source into power of the generated sound wave. The efficiency is dependent upon the coupling of the source to the medium of propagation as well as the transduction efficiency of the source. The fidelity of a sound source is a measure of how faithfully the source reproduces the frequencies supplied to it. The speakers of electronic sound reproduction systems have efficiencies and fidelities that are dependent partly on their size. A single speaker is not equally efficient at both low and high frequencies. A good high- fidelity sound system requires different size speakers for different frequency ranges. (Can you figure out the relation between speaker size and frequency for best sound reproduction?) When considering a sound detector we are interested in its sensitivity to different frequencies, which is sometimes referred to as its frequency response. There are several different kinds of microphones available. The common ones are the condenser, the crystal, and the electromagnetic, or dynamic, microphones. Microphones have varying sensitivities and display frequency responses that vary over the audible frequency range (20 Hz to 20,000 Hz).

17.12 The Human Ear The human ear is a superb sound transducer coupled to a computer center, the human brain.Figure 17.9 shows a block diagram of the human sound detection system.


368

The ear has the following typical specifications: frequency response from 20 Hz to 20,000 Hz, with nonuniform sensitivity in this range; sound level response from 0 dB to 120 dB see Figure 17.10 ; and threshold amplitude response for molecular vibrations with amplitudes less than the size of an atom. The ear is a detector that transduces the incoming pressure wave information into an electrical signal that is transmitted to the brain via nerve impulses. A cross section of the ear is illustrated in Figure 17.11.


369

For our analysis of the ear we consider three main parts, the external canal, the middle ear, and the inner ear. The external canal can be thought of as a closed-end organ pipe. The closed end is formed by the tympanic membrane, or ear drum, that separates the external canal from the middle ear. Do you know that your minimum threshold of hearing occurs for sounds of frequencies near 3.5 kHz? Compare this frequency with the fundamental frequency of a 2.5-cm outer ear canal. The middle ear cavity contains three small bones forming a lever system attached to the ear drum. The purpose of the middle ear is to transmit the vibrations in the air due to sound stimulus to the fluid inside of the inner ear. The middle ear mechanism is a nonlinear coupling system. At high sound levels the middle ear produces harmonics of the frequencies exciting it. One of the main functions of the middle ear is that of impedance matching between the sound input and the inner ear. The middle ear has been studied in detail, and it has been shown that the middle ear provides a pressure amplification of 17 at low sound levels. At high sound levels the middle ear action resembles that of an automatic volume control on a radio. The middle ear reduces the energy fed into the inner ear through a feedback control system that attempts to maintain a constant input to the inner ear. The inner ear consists primarily of the cochlea, a spiral canal inside the bone of the skull. The cochlea is divided into three channels by fibrous membranes. These channels are filled with fluid. The fluid transmits the sound energy from the middle ear. The basilar membrane contains the endings of the auditory nerve. These nerve cells have hairlike projections that are anchored in the tectorial membrane. The basilar membrane has a varying thickness as it proceeds toward the apex of the spiral. Relative motion between the tectorial and basilar membranes produces an electrical potential in the nerve cell. The varying thickness of the basilar membrane causes different frequencies to stimulate maximum signals at different locations along the canal. The sound information is coded into an electrical signal that is transmitted to the brain. The signal has all of the frequency and amplitude information that is available from the input


370

sound stimulus. The brain, through a process as yet not well understood, interprets this message and supplies the necessary feedback controls and output action orders. There is still much to learn about the entire hearing process. Some of the questions yet to be resolved are: How is the phase information transduced and analyzed? What determines the limit of the ears' ability to hear beat frequencies? How is it possible to ignore loud noises and detect much lower sound levels? If you are intrigued by such questions, you may wish to pursue studies in this area of biophysics and psychoacoustics.

17.13 Medical Applications of Acoustic Holography A hologram is a photographic record of an interference pattern generated by a set of waves and a set of reference waves. One can make holograms using sound waves provided that the interference pattern can be recorded, Figure 17.12 . Many parts of the human body react to high-frequency sound waves in a much different manner than they do to other forms of waves such as x rays. The nonbony parts of the human body are essentially transparent to x rays. However, these parts of the body absorb and reflect the sound waves in different ways. Hence the application of holographic techniques are being developed. This form of sound imaging provides a differentiation between the various types of tissues. It is being predicted that the detection of tumors in the human body will be one of the greatest contributions of acoustical holography to medicine.

Figure 17.12

Schematic sketch of method to produce an ultrasonic hologram. The interference pattern produced by the reference beam and the object beam is produced on a liquid surface. An optical system is needed to photograph this pattern. If one wishes to produce a visual image, the camera is replaced by a TV camera and a TV monitor.

ENRICHMENT 17.14 Velocity of Sound in an Ideal Gas The equation for the velocity of sound in a gas as a function of the elastic modulus and


371

the density of the gas can be derived as follows: The elastic modulus is the bulk modulus of the gas, B, which is defined as

(17.12)

where ΔP is the pressure change producing volume change ΔV to an initial volume V. In calculus notation this becomes B = -V dP/dV (17.13) It can be shown that the sound wave passes through a gas in an essentially adiabatic process. How would you show that this is so? The gas law for adiabatic processes is P = (constant)V-γ where γ = cp/cv (Chapter 12). Thus dP/dV = (constant)(-γV-(γ+1)) Then B = -V(constant)(-γ V-(γ+1)) = γ(constant)V-γ Therefore, B = γP. Thus, the expression for the speed of sound in an ideal gas becomes vs = SQR RT(γP/ρ) = SQR RT(γRT/m) where R is ideal gas constant and m is the molecular mass of the gas. While this derivation is based on the ideal gas law, it gives good results for many gases at room temperature.

17.15 Derivation of Beat Frequencies The production of beats can be analyzed mathematically as follows: Consider two waves given by y1 A sin (2πx/λ1 -2πf1t) and y2 A sin (2πx/λ2 - 2 πf2t) The superposition principles give y = y1 + y2 = 2A sin [(2π/λ1 + 2π/λ2)x -(2πf1 + 2πf2/2 )t] x cos [(2π/λ1 -2π/λ2)x -(2πf1 -2πf2/2)t] since sin θ1 + sin θ2 = 2 sin (½)(θ1 + θ2) cos (½)(θ1 - θ2). This result can be written as follows noting that v = f1λ 1 = f2λ2. y = 2A sin [2π(f1 + f2)(x/v - t) / 2] cos [2π(f1 - f2)(x/v - t) / 2] This is the equation of a product of two waves, one of frequency fc = (f1 + f2)/2 and the other fm = (f1 - f2)/2. The first is the carrier frequency of the waveform and the latter is the modulation frequency. A beat occurs each time the amplitude as governed by the modulation frequency is a maximum -that is whenever cos [2π(f1 - f2)(x/v - t) / 2] = +1 or -1 Since this happens two times each cycle, the beat frequency = f1 - f2.


372

SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of the summary with the number of the section where you can find related content material. Definitions 1. The amplitude of a sound wave is measured in what SI units? 2. The pulsations in wave amplitude are called ______ and arise from ______. 3. The sound level of a sound is 40 dB; this means the intensity of the sound is ______

watts/m2. 4. Infrasound has frequency ______ than audio range. 5. Ultrasonic waves have frequencies ______ than audio range. 6. Standing waves are set up in organ pipes whenever the driving frequency produces

______. 7. Matching acoustic impedance at boundaries assures ______. 8. The Doppler effect is characterized by a change in ______ due to relative ______. If the

observer and source approach each other, the frequency is ______; if the source and observer retreat from each other, the frequency ______.

Waveforms 9. Graph the amplitude of the sound wave P = 30 sin 2π (440t - x/.77) N/m2.

a. as a function of time for x = 0 b. as function of x for t = 1/440 sec

Sound Level and Intensity 10. What is the intensity of a sound wave whose pressure amplitude is 2 N/m2? What is

the intensity level in dB? See Example on page 376. 11. What is the difference between intensity levels in dB if the intensity of one wave is

twice that of another? Applications 12. Name possible applications of sound in dental or medical professions. Hearing System 13. What are the essential parts of the human sound detecting system? Doppler Effect


373

14. A train is moving toward a listener with a speed of 20 m/sec. a. If the whistle of the train has a frequency of 330 Hz, what frequency does the

listener hear? b. After the train passes the listener, what frequency does he hear? Standing Waves 15. An open pipe resonates at 400 Hz (fundamental). At what frequency will it resonate

when one end is plugged? Answers

1. N/m2 (Section 17.2) 2. beats, superposition of two waves of nearly the same frequency (Section 17.9) 3. 10-8 watts/m2 (Section 17.4) 4. lower (Section 17.5) 5. greater (Section 17.5) 6. wavelengths satisfying boundary conditions (Section 17.8) 7. maximum power transfer (Section 17.3) 8. frequency, motion of observer and source, increased, decreased (Section 17.10) 9. a. sine curve with amplitude of 30 N/m2 and period of 1/440 sec, i.e., y = 30 sin 2π

(440t) N/m2

b. sine curve of amplitude of 30 N/m2 and wavelength of 0.77 m. P = 30 sin 2π [1 - x/.77)] (Section 17.2)

10. 0.01 watts/m2, 100 dB (Section 17.4) 11. 3 dB (Section 17.4) 12. ultrasonic cleaning for cleaning instruments, ultrasonic drilling or cleaning of

teeth, Doppler shift, heart beat embryo, acoustical hologram (Sections 17.5, 17.6, 17.13)

13. See Figure 17.11 (Section 17.12) 14. a. 351 Hz b. 312 Hz (Section 17.10) 15. 800 Hz (Section 17.8)

ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve single-concept problems.


374

Equations P = Po sin (2πft - 2πx /λ) (17.1)

I = 1/2 ( Aω)2ρvs (watt/m2)= (17.2, 17.3)

sound level (dB) = 10 log10 I / Io (17.6) f ' = f (vs + vo) / (vs - v) (17.11) for closed pipe: L = n(λ/4) where n is an odd integer for open pipe: L = n(λ/2 where n is an integer Problems 1. Fill in the blanks in Table 17.1 in Example 1 of Section 17.4. 2. An automobile has a horn with a frequency of 510 Hz, and the velocity of sound in air

is 340 m/sec. What frequency does the observer hear if a. the car is approaching the observer with velocity of 20 m/sec when the horn is

sounded b. the car is receding from the observer with a velocity 20 m/sec at the time the horn

is sounded c. the observer is approaching the parked car with a velocity of 20 m/sec when the

horn is blown d. the observer is leaving the parked car with a velocity of 20 m/sec when the horn

is blown


375

3. A closed organ pipe has a length of 44 cm. The velocity of sound in air is 342 m/sec at

the temperature of the pipe. a. What is wavelength of the fundamental note? b. What is the frequency of the fundamental note? c. What is the wavelength of the first overtone? d. What is the frequency of the second overtone?

4. An open organ pipe has a length of 1 m. The velocity of sound in air at the temperature of the pipe is 340 m/sec.

a. What is the wavelength of the fundamental note? b. What is the frequency of the fundamental note? c. What is the wavelength of the second overtone? d. What is the frequency of the first overtone?

Answers

2. a. 542 HZ b. 482 HZ c. 540 Hz d. 480 Hz

3. a. 176 cm b. 194 Hz c. 59 cm d. 970 Hz

4. a. 2m b. 170 Hz c. 2/3 m d. 340 Hz

EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate the numerical answer is given in brackets at the end of the exercise. Section 17.2 1. Using appropriate values of B and ρ for air, calculate the velocity of sound in air.

Assume γ = 1.4 for air. [330 m/sec] 2. Using the value of vs from problem 1, determine the pressure amplitude associated

with a 1000 Hz sound wave of 10-12 watt/m2 intensity. (This corresponds to a barely audible sound.) Also determine the pressure amplitude for an intensity of 1 watt/m2. [2.9 x 10-5

5 N/m2, 2.9 x 10 N/m2] Section 17.3 3. Compare the acoustic impedance of water and air. If you are able to drive a sound

source with the same amplitude in air as in water, what is the ratio of the intensity of the sound generated in air and water by such a source? The velocity of sound in water is 1400 m/sec. [Iw/Ia = 1/3.20 x 103]


376

4. Compute the intensity of a compressional wave at 0°C and 760 mm Hg, if its frequency is 512 Hz and its amplitude is 2.0 x 10-3 cm. The speed of the wave in air is 331 m/sec. [8.6 x 10-1 watts/m2]

Section 17.4 5. If a 60-dB sound level is reduced to 1/10 its intensity the new sound level will be

what in decibels? [50 dB] 6. If one trumpet will produce a sound of 60 dB at the back row of the auditorium, a. How many dB will two trumpets produce? b. How many dB will four trumpets produce? c. How many dB will ten trumpets produce? d. How many dB will 100 trumpets produce? e. How many trumpets will be required to produce the discomfort threshold of 120

dB at the back row of the auditorium? [a. 63 dB; b. 66 dB; c. 70 dB; d. 80 dB; e. 106] 7. One person talking to you from a distance of one meter produces a sound of 40 dB.

How many people will it take simultaneously talking to you from 1 m to produce a sound of 80 dB? (This is sometimes called a stampede.) [104]

Section 17.8 8. Find the fundamental frequency and the first three overtones of a 0.5-m closed-end

organ pipe. Assume the velocity of sound is 340 m/sec. How many overtones could the average student hear? [170 Hz, 510 Hz, 850 Hz, 1190 Hz, about 59]

Section 17.10 9. At the Exploratorium Science Exhibit in San Francisco there is a whistle mounted on a

movable belt. By turning a crank you can cause the whistle (assume a natural frequency of 540 Hz) to move first away from you and then toward you. If you can make the whistle move at a speed of 34 m/sec, what frequencies will you hear? [receding 491 Hz, approaching 600 Hz]

PROBLEMS Each problem may involve more than one physical concept. The numerical answer to the problem is given in brackets at the end of the problem. A problem that requires material from the enrichment section is marked by a dagger. 10. What is the amplitude of vibration of air molecules for a 40- dB sound with a

frequency of 2000 Hz? The air has an acoustic impedance of about 440 kg/m2 sec. [5.3 x 10-10 m]


377

11. The Super Sound Rock Group of Rolla, Missouri, uses two 150-watt, 15-inch (19-cm) radius speakers to put out sweet sounds to their admiring audiences. Assume that 75 percent of the power transmitted to the speakers is actually converted into sound energy.

a. What is the intensity of sound at the opening of one speaker (due to that speaker alone) when it is being operated at full power of 150 watts? How many dB is that?

b. Assuming that the intensity is inversely proportional to the square of the distance you are from the speakers, if the speakers are located together and are being operated at full power, how far away from them will you have to stand to have the sound be less than the pain of threshold of 120 dB? [a. 1000 watts/m2, 150 dB; b. 45 m]

12. If the sound level increases from 50 dB to 70 dB, find the ratio of the amplitudes of the second sound level to the first. [10]

13. A 2000-Hz sound can just be distinguished from another of the same frequency if it is about 1.00 dB louder than the other. (This is known as one JND = just noticeable difference.) Find:

a. the ratio of the JND sound intensities b. the pressure amplitude ratio associated with this JND pair [a. 1.26; b. 1.12] 14. It is well known that the high noise background of an industrial society causes your

threshold of hearing to increase. Suppose that your hearing threshold is being increased by 0.33 dB per year, and that your present threshold is 10 dB.

a. What will your hearing threshold be in 60 years? b. How much will the intensity of a barely audible sound have to be increased for

you to be able to hear it in 60 years? Assume that you can presently hear a sound intensity of 10-11 watts/m2.

c. How much will the pressure amplitude of a barely audible sound have to be increased for you to hear it in 60 years? [a. 30 dB; b. 100 times, 10-9 watts/m; c. 10 times]

15. As you are hitchhiking along the highway a passing motorist honks his horn (frequency Ä0) at you in a 60 km/hr speed zone. You note that his horn changes its apparent pitch by (1/10)f0 as he goes past (vs = 350 m/sec).

a. Does the pitch appear to increase or decrease? b. How fast is the motorist going? Should he be given a speeding ticket? [a. Δφ = f0

/10, decrease in pitch; b. v = 17.5 m/sec or 63 km/hr; give him a ticket] 16. Estimating the size of your external ear canal, and using the previously developed

ideas for a closed-end pipe, calculate the resonant frequency for your ear in air. The velocity of sound in water is about 5 times that in air. How does the resonant frequency of your ear canal change when your ears are full of water? What are you able to hear when you are swimming under water? [~3400 Hz, 5]

17. The fundamental frequency of the A string of a violin is 440 Hz. The vibrating portion of the string is 37.0 cm long and has a mass of 1.28 g. With what tension must it be stretched to be at the correct pitch? [367 N]


378

18. If the ear is assumed to be resonating as a closed pipe in problem 16, find the length

of the ear canal if 2000 Hz is its fundamental frequency. [4.3 cm] 19. What gas (in the ideal gas approximation) would give the greatest ratio of the

velocity of sound between any two of its isotopes? [hydrogen] 20. A recent report indicates that ultrasonic Doppler shift equipment is being used by

police to determine the speed of cars on freeways. Suppose that a 1.00 MHz sound was used, and a Doppler shift results in a detected signal of 0.88 MHz signal. What was the speed of the car as it receded from the policeman? (Be careful here. Remember that the source at 1.0 MHz is with the policeman at rest, and the speeding car reflects this sound as it moves away from the observer.) [22 m/sec, or 78 km/hr, or 49 mph]

21. A beat frequency of 6 beats/second is heard between two sources, one of which is at rest and the other moving toward the observer. If the frequency of the source at rest is 1000 Hz, find the speed of the moving source. [2m/sec]

22. An obstetrician uses a 22,000.00 Hz ultrasonic sound wave to detect the fetal heart motion. If she obtains Doppler frequencies of 22,002.84 and 21,997.16 on her detector, what is the velocity of the motion of the fetus' heart? The speed of sound in the body tissue is 1550 m/sec. See suggestion in problem 20. [0.1 m/sec]

23. Two whistles, Y and Z, each have a frequency of 500 Hz. Y is stationary, and Z is moving to the left, away from Y, with a velocity of 50 m/sec. An observer is between the two whistles and moving to the right at a velocity of 30 m/sec. What is the frequency of each whistle as heard by the observer? How many beats would he hear? Assume the velocity of sound in air to be 340 m/sec. [544 Hz, 398 Hz, 146 beats/sec which cannot be heard as beats]

24. Show that for speeds in which v << µ where v = velocity of source and µ = velocity of wave, the equation for Doppler shift may be expressed in an approximate form as Δλ/λ = v/µ where Δλ is the shift in wavelength.

25. Derive an equation that could be used to determine the bulk modulus of an unknown gas by comparing compressional wave intensity with that in a known gas using the same sound source. [Bx/Bs = SQR RT(Ix/Is)]

Chapter 17 Sound and the Human Ear...

Documents

Transcript of Chapter 17 Sound and the Human Ear...