Homework 5 (Chapter 16)

Homework 5 (Chapter 16) Exercise 16.1 For sound waves in air with frequency 1000, a displacement amplitude of 1.2 10 8 m produces a pressure amplitude of. Use = 344. 3.0 10 2 Pa vsound What is the wavelength of these waves? Express your answer using two significant figures. λ = 0.34 m For 1000 waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa? Express your answer using two significant figures. A = 1.2 10 5 m Part C For what wavelength will waves with a displacement amplitude of of 1.5 10 3 Pa? Express your answer using two significant figures. 1.2 10 8 m produce a pressure amplitude λ = 6.9 m Part D For what frequency will waves with a displacement amplitude of 1.2 10 8 m produce a pressure amplitude of 1.5 10 3 Pa? 1/23

Express your answer using two significant figures. Homework 5 (Chapter 16) f = 50 The Hearing of a Bat Bats like the one shown in are mainly active at night. They have several senses that they use to find their way about, locate prey, avoid obstacles, and "see" in the dark. Besides the usual sense of vision, bats are able to emit high frequency sound waves and hear the echo that bounces back when these sound waves hit an object. This sonar like system is called echolocation. Typical frequencies emitted by bats are between 20 and 200 k. Note that the human ear is sensitive only to frequencies as high as 20 k. cm m k A moth of length 1.0 is flying about 1.0 from a bat when the bat emits a sound wave at 80.0. The temperature of air is about 10.0 C. To sense the presence of the moth using echolocation, the bat must emit a sound with a wavelength equal to or less than the length of the insect. The speed of sound that propagates in an ideal gas is given by v = γ γ = 1.4 T where is the ratio of heat capacities ( for air), is the absolute temperature in kelvins (which is equal to the Celsius temperature plus 273.15 C), M is the molar mass of the gas (for air, the average molar mass is M = 28.8 10 3 kg/mol), and R is the universal gas constant ( R = 8.314 J mol 1 K 1 ). γrt M, k Find the wavelength λ of the 80.0 wave emitted by the bat. Express your answer in millimeters. Hint 1. Relating wavelength, frequency, and speed of a wave In periodic waves, the speed at which the wave pattern travels is given by v = λf, where λ is the wavelength and f is the frequency of the wave. Hint 2. Find the speed of sound in air 2/23

Find the speed of sound v in air at 10.0. Express your answer in meters per second. C Homework 5 (Chapter 16) v = 338 λ = 4.23 mm Will the bat be able to locate the moth despite the darkness of the night? yes no Part C How long after the bat emits the wave will it hear the echo from the moth? Express your answer in milliseconds to two significant figures. Hint 1. How to approach the problem After emitting the high frequency sound, the bat waits for any echoes coming back from possible obstacles. Therefore the time needed to locate an obstacle depends on the speed of sound and the distance of the obstacle from the bat. Hint 2. Find the time needed for the sound wave to reach the moth How long does it take the sound wave to reach the moth? Express your answer in milliseconds to three significant figures. 2.96 ms 3/23

5.9 ms Exercise 16.11 m A 55.0 long brass rod is struck at one end. A person at the other end hears two sounds as a result of two longitudinal waves, one traveling in the metal rod and the other traveling in air. What is the time interval between the two sounds? (The speed of sound in air is 344 12.1 in the textbook for relevant information about brass.) Express your answer to two significant figures and include the appropriate units. t = 0.14 s ; see Tables 11.1 and Can the Firecracker Be Heard? The decibel scale is logarithmic in intensity:. I0 In this formula, I0 is a reference intensity. To convert the intensity of sound waves from SI units to db, I0 is taken to be. 10 12 W/ m 2 I β = 10 log 10 Once you know the intensity of a source in db as measured at some reference distance, the intensity at a new distance can be found by subtracting the change in db appropriate to the ratio of the new distance to the reference distance. The change in db corresponding to the change in distance may be found by replacing I0 in the formula given here with the intensity at the reference distance and replacing I with the intensity at the new distance. We apply the db scale to a small firecracker that has an intensity of 140 db at a distance of 1 m. db A child sets off the firecracker at a distance of 100 m from the family house. What is the sound intensity at the house? Express the sound intensity in decibels. β100 Hint 1. How to approach the problem First, use the dependency of sound intensity on distance to determine the factor by which the intensity at the house is smaller than the intensity 1 m from the firecracker. Then determine what change in db this change in intensity corresponds to. (Recall that because db is a logarithmic scale, a multiplicative factor in the intensity corresponds to an additive number of decibels.) Finally, add the change in db 4/23

(which is negative for an attenuation) to the db of the original intensity. Hint 2. Find the intensity decrease resulting from changing the distance by a factor If you increase the distance from the source to the point of measurement by a factor of 100, by what factor will the sound level change? f I Express the change as a decimal number, noting that a number less than unity corresponds to a decrease in the intensity. Hint 1. How to set up the problem You need to determine the value of in the equation. Hint 2. Variation in intensity with distance Recall that the intensity of a sound wave, as a function of distance from the source, is proportional to (i.e., ). r 2 I(r) 1/r 2 f I I(100 m) = I(1 m) f I f I = I(100 m) I(1 m) = 0.0001 Hint 3. Calculate the db change when intensity decreases by a factor What is the change in decibels Δβ that corresponds to a factor of 0.0001 change in intensity? Express the change in decibels, noting that less intensity corresponds to a negative change. Hint 1. What are I and I0? I0 I = 0.0001I0 Take as the reference intensity; then. Now use the formula from the problem introduction. Δβ = 40 db 100 db 5/23

The child's parents are inside the house talking. A typical house attenuates (reduces the intensity of) outside sounds by a factor of around 40 db more at high frequencies (when double pane windows are closed) and less at low frequencies. Also, a typical conversation has an average sound level of 65 db. Will the firecracker sound louder than the parents' conversation? yes no Exercise 16.16 m A fan at a rock concert is 50.0 from the stage, and at this point the sound intensity level is 115. Sound is detected when a sound wave causes the tympanic membrane (the eardrum) to vibrate. Typically, the diameter of this membrane is about 8.40 in humans mm db How much energy is transferred to her eardrums each second? E = 17.5 μj How fast would a 2.00 mg mosquito have to fly to have this much kinetic energy? v = 4.19 Part C How fast a typical 2.0 mosquito would have to fly (in ) to have an amount of energy delivered to the eardrum each second when someone whispers (20 ) a secret in your ear? mg db m 6/23

v = 0.0744 m Part D Compare the mosquito's speeds found in parts (B) and (C). Please enter your answer as a number (factor) without the units. v = 5.62 10 4 v Exercise 16.24 If two sounds differ by 9.00 db, find the ratio of the intensity of the louder sound to that of the softer one. 7.94 If one sound is 130 times as intense as another, by how much do they differ in sound intensity level (in decibels)? Δb = 21.1 db Part C If you increase the volume of your stereo so that the intensity doubles, by how much does the sound intensity level increase? 7/23

Δb = 3.01 db Introduction to Wind Instruments The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed. Throughout the problem, take the speed of sound in air to be 343. Consider a pipe of length 80.0 open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe? Express your answer in hertz. cm Hint 1. How to approach the problem The lowest frequency that can be produced is the fundamental frequency of the standing wave in the pipe. Hint 2. Frequencies of standing waves in an open open pipe The frequencies possible in an open open pipe of length L are given by the formula f m where v is the speed of sound in the air. v = m, m = 1,2,3,4, 2L, f = 214 If your pipe were a flute, this frequency would be the lowest note that can be produced on that flute. This frequency is also known as the fundamental frequency or first harmonic. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now Hint 1. How to approach the problem 8/23

Since the hole opens the pipe to the pressure of the surrounding air, the standing wave created in the pipe has an antinode near the hole. In other words, the presence of a hole in the pipe reduces the length of the column of air that can oscillate in the pipe. Consider the formula used in and use the fact that the length of the vibrating column of air is now shorter. the same as before. lower than before. higher than before. By opening successive holes closer and closer to the opening used to blow air into the pipe, the pipe can be made to produce sound at higher and higher frequencies. This is what flutists do when they open the tone holes on the flute. Part C If you take the original pipe in and drill a hole at a position half the length of the pipe, what is the fundamental frequency f of the sound that can be produced in the pipe? Express your answer in hertz. Hint 1. How to approach the problem Repeat the same calculation you did in, this time using half the length of the pipe. f = 429 Part D What frequencies, in terms of the fundamental frequency of the original pipe in, can you create when blowing air into the pipe that has a hole halfway down its length? Hint 1. How to approach the problem Recall from the discussion in that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present. 9/23

Only the odd multiples of the fundamental frequency Only the even multiples of the fundamental frequency All integer multiples of the fundamental frequency Part E as the open What length of open closed pipe would you need to achieve the same fundamental frequency f open pipe discussed in? Hint 1. Frequencies on an open closed pipe The frequencies possible in an open closed pipe of length L are given by f m where v is the speed of sound in the air. v = m, m = 1,3,5, 4L, Half the length of the open open pipe Twice the length of the open open pipe One fourth the length of the open open pipe Four times the length of the open open pipe The same as the length of the open open pipe Part F What is the frequency f pipe described in Part E? of the first possible harmonic after the fundamental frequency in the open closed Express your answer in hertz. Hint 1. How to approach the problem Recall that possible frequencies of standing waves that can be generated in an open closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third harmonic. f = 643 10/23

Exercise 16.32 cm g N You have a stopped pipe of adjustable length close to a taut 62.0, 7.25 wire under a tension of 4910. You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude. How long should the pipe be? Express your answer with the appropriate units. L = 5.49 cm Exercise 16.37 Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 641. You are standing between the speakers, along the line connecting them and are at a point of constructive interference. How far must you walk toward speaker B to move to reach the first point of destructive interference? Take the speed of sound to be 344. x = 0.134 m Exercise 16.39 Two small stereo speakers are driven in step by the same variable frequency oscillator. Their sound is picked up by a microphone arranged as shown in the figure. Assume that the speed of sound is 344. 11/23

For what frequencies does their sound at the speakers produce constructive interference? Express your answer in terms of n (an integer factor). f cons = 344n 0.42 For what frequencies does their sound at the speakers produce destructive interference? Express your answer in terms of n (a positive integer factor). f dest = 344(n 0.5) 0.42 Two Traveling Waves Beating Together Learning Goal: To see how two traveling waves of nearly the same frequency can create beats and to interpret the superposition as a "walking" wave. Consider two similar traveling transverse waves, which might be traveling along a string for example: y 1 (x, t) = Asin( k1x ω1 t) and y 2 (x, t) = Asin( k2x ω2t). They are similar because we assume that k1 and k2 are nearly equal and also that ω1 and ω2 are nearly equal. Which one of the following statements about these waves is correct? Both waves are traveling in the Both waves are traveling in the +i^ i^ +i^ +i^ direction. direction. Only wave y 1 is traveling in the direction. Only wave y 2 is traveling in the direction. 12/23

The principle of superposition states that if two waves each separately satisfy the wave equation then the sum (or difference) also satisfies the wave equation. This follows from the fact that every term in the wave equation is linear in the amplitude of the wave. Consider the sum of the two waves given in the introduction, that is, (x, t) = (x, t) + (x, t) y sum y 1 y 2. These waves have been chosen so that their sum can be written as follows: (x, t) = C (x, t) (x, t), y sum y envelope y carrier where C is a constant, and the functions y envelope and y carrier are trigonometric functions of x and t. This form is especially significant because the first function, called the envelope, is a slowly varying function of both position ( x) and time ( t), whereas the second varies rapidly with both position ( x) and time ( t). Traditionally, the overall amplitude is represented by the constant C, while the functions y envelope and y carrier are trigonometric functions with unit amplitude. Find,, and y carrier. Express your answer in terms of A, k1, k2, x, t, ω1, and ω2. Separate the three parts of your response with commas. Recall that yenvelope (the second term) varies slowly whereas ycarrier (the third term) varies quickly. Both yenvelope and ycarrier should be trigonometric functions of unit amplitude. C y envelope (x, t) (x, t) Hint 1. A useful trigonometric identity Hint 2. Which is the envelope and which is the carrier wave? Recall that the carrier signal varies much more quickly over time and space than does the envelope signal. Given a choice between sin(ax) and sin(bx), where a > b, which wave oscillates faster in space? sin(ax) sin(bx) α+β α β sin(α)+sin(β) = 2sin( )cos( ) 2 2 13/23

C yenvelope (x, t) (x, t),, = Homework 5 (Chapter 16) ( ycarrier 2A,cos( k1 k2 )x ( ω1 ω2 ),sin( ) )t ( k1 + k2 )x ( ω1 + ω2 )t 2 2 Part C Which of the following statements about ycarrier(x, t) is correct? It is a rapidly oscillating wave traveling in the It is a rapidly oscillating wave traveling in the It is slowly oscillating in time but is standing still. It is traveling rapidly but oscillating slowly. +i^ i^ direction. direction. Part D Which of the following statements about is correct if you assume that δ k k1 k2 and δ ω = ω1 ω2 are both positive? It is a slowly oscillating wave traveling in the It is a slowly oscillating wave traveling in the It is slowly oscillating in time but is standing still. It is traveling rapidly but oscillating slowly. yenvelope (x, t) = +i^ i^ direction. direction. Part E The envelope function can be written simply in terms of and δ ω ω1 ω2. If you do so, what is vgroup, the velocity of propagation of the envelope? δ ω Express your result in terms of δk and. δ k = k1 k2 = Hint 1. Substituting δk and Substitute and into the expression for yenvelope. Express y envelope in terms of δk and δ ω. δ ω δ k = k1 k2 δ ω = ω1 ω2 (x, t) (x, t) 14/23

δ x t yenvelope(x, t) = cos( k δ ω ) 2 Homework 5 (Chapter 16) Hint 2. Traveling wave velocity What is the velocity v of a traveling wave? Express your answer in terms of k and ω. y(x, t) = sin(kx ωt) v = ω k vgroup = δω δ k vgroup is called the group velocity because this ratio is the velocity at which the group of waves under a maximum of the envelope function appears to travel. It is technically defined as a derivative (although you found it for a finite difference): dω vgroup. dk This contrasts with the velocity of propagation of the waves themselves: v phase ω. k Information (e.g., applied as amplitude modulation of the wave, that is, as a variation in the envelope) travels at the group velocity. No information can be sent at the phase velocity of a wave, which therefore can exceed the speed of light. (Einstein's Special Theory of Relativity implies that neither a physical body nor information can travel faster than light.) = = Exercise 16.43 Two organ pipes, open at one end but closed at the other, are each 1.06 long. One is now lengthened by 2.70. m cm Find the frequency of the beat they produce when playing together in their fundamental. 15/23

fbeat = 2.02 Doppler Shift Learning Goal: To understand the terms in the Doppler shift formula. The Doppler shift formula gives the frequency f L at which a listener L hears the sound emitted by a source S at frequency f S : v+vl f L = f S, v+vs where v is the speed of sound in the medium, vl is the velocity of the listener, and vs is the velocity of source. The velocity of the source is positive if the source is. Note that this equation may not use the sign convention you are accustomed to. Think about the physical situation before answering. Hint 1. Relating the frequency and the source velocity The figure below shows that wave crests emitted by a moving source are crowded together in front of the source and stretched out behind it. This means that the frequency heard by a listener in front of the source will be greater than the frequency of the source, and similarly, the frequency heard by a listener behind the source will be less than the frequency of the source. Use this information to determine the sign convention for the source velocity that, along with the formula given in the introduction, produces the correct frequency behavior. traveling in the +x direction traveling toward the listener traveling away from the listener 16/23

The velocity of the source is measured with respect to the. medium (such as air or water) listener Part C The velocity of the listener is positive if the listener is. Hint 1. Relating the frequency and the listener's velocity A listener moving toward the source will run into more wave crests in a given time period than were emitted by the source (in the same period). So a listener moving toward the source will hear a higher frequency than the emitted frequency. Similarly, a listener moving away from the source will hear a lower frequency than the emitted frequency. Use this information to find the correct sign convention for the listener's velocity that goes with the supplied formula. traveling in the +x direction traveling toward the source traveling away from the source Part D The velocity of the listener is measured with respect to the. source medium Here are two rules to remember when using the Doppler shift formula: 1. Velocity is measured with respect to the medium. 2. The velocities are positive if they are in the direction from the listener to the source. 17/23

Part E Imagine that the source is to the right of the listener, so that the positive reference direction (from the listener to the source) is in the +x direction. If the listener is stationary, what value does fl approach as the source's speed approaches the speed of sound moving to the right? 0 1 2 f S 2f S It approaches infinity. Part F Now, imagine that the source is to the left of the listener, so that the positive reference direction is in the x direction. If the source is stationary, what value does fl approach as the listener's speed (moving in the +x direction) approaches the speed of sound? 0 1 2 f S 2f S It approaches infinity. Basically in this case the listener doesn't hear anything since the sound waves cannot catch up with him or her. Part G In this last case, imagine that the listener is stationary and the source is moving toward the listener at the speed of sound. (Note that it is irrelevant whether the source is moving to the right or to the left.) What is f L when the sound waves reach the listener? 0 1 2 f S 2f S It approaches infinity. 18/23

This case involves what is called a sonic boom. The listener will hear no sound ( fl ) until the sonic boom reaches him or her (just as the source passes by). At that instant, the frequency will be infinite. There is no time between the passing waves they are literally right on top of each other. That's a lot of energy to pass by the listener at once, which explains why a sonic boom is so loud. = 0 Exercise 16.45 On the planet Arrakis a male ornithoid is flying toward his mate at a speed of 20.0 of 1220. while singing at a frequency If the stationary female hears the sound at a frequency of 1260 atmosphere of Arrakis?, what is the speed of sound in the v = 630 Exercise 16.46 A railroad train is traveling at a speed of 26.0 whistle is 450. in still air. The frequency of the note emitted by the locomotive What is the wavelength of the sound waves in front of the locomotive? Use 344 for the speed of sound in air. λ = 0.707 m What is the wavelength of the sound waves behind the locomotive? Use 344 for the speed of sound in air. λ = 0.822 m 19/23

Part C What is the frequency of the sound heard by a stationary listener in front of the locomotive? Use 344 for the speed of sound in air. f = 487 Part D What is the frequency of the sound heard by a stationary listener behind the locomotive? Use 344 for the speed of sound in air. f = 418 Exercise 16.54 The siren of a fire engine that is driving northward at 36.0 emits a sound of frequency 2200. A truck in front of this fire engine is moving northward at 17.0. What is the frequency of the siren's sound that the fire engine's driver hears reflected from the back of the truck? f = 2460 What wavelength would this driver measure for these reflected sound waves? 20/23

λ = 0.155 m Exercise 16.58 The shock wave cone created by the space shuttle at one instant during its reentry into the atmosphere makes an angle of 59.0 with its direction of motion. The speed of sound at this altitude is 331. What is the Mach number of the shuttle at this instant? v Sv = 1.17 How fast (in ) is it traveling relative to the atmosphere? v S = 386 Part C How fast (in mi/h ) is it traveling relative to the atmosphere? v S = 864 mi/h Part D What would be its Mach number if it flew at the same speed but at low altitude where the speed of sound is 343? 21/23

v Sv = 1.13 Part E What would be the angle of its shock wave cone if it flew at the same speed but at low altitude where the speed of sound is 343? α = 62.7 Problem 16.59 A # A # kg/m 3 Pa A soprano and a bass are singing a duet. While the soprano sings an at 932, the bass sings an but three octaves lower. In this concert hall, the density of air is 1.12 and its bulk modulus is 1.34 10 5. In order for their notes to have the same sound intensity level, what must be the ratio of the pressure amplitude of the bass to that of the soprano. p B ps = 1 In order for their notes to have the same sound intensity level, what must be the ratio of the displacement amplitude of the bass to that of the soprano? A B A S = 8 Part C What displacement amplitude (in m) does the soprano produce to sing her at 74.0? 22/23 A # db

Homework 5 (Chapter 16) A = 6.15 10 8 m Part D What displacement amplitude (in ) does the soprano produce to sing her at 74.0? nm A # db A = 61.5 nm Problem 16.71 m rpm cm A turntable 2.00 in diameter rotates at 79.0. Two speakers, each giving off sound of wavelength 30.7, are attached to the rim of the table at opposite ends of a diameter. A listener stands in front of the turntable. What is the greatest beat frequency the listener will receive from this system? f = 53.9 Will the listener be able to distinguish individual beats? Yes, the listener will be able to distinguish individual beats. No, the listener will not be able to distinguish individual beats. 23/23