#set document(title: "17.1 Sound Waves", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 17.1#h(0.6em)Sound Waves The physical phenomenon of #strong[sound] is a disturbance of matter that is transmitted from its source outward. #strong[Hearing] is the perception of sound, just as seeing is the perception of visible light. On the atomic scale, sound is a disturbance of atoms that is far more ordered than their thermal motions. In many instances, sound is a periodic wave, and the atoms undergo simple harmonic motion. Thus, sound waves can induce oscillations and resonance effects . #figure(figph[Picture shows a photograph of a wine glass shattering into many little pieces.], alt: "Picture shows a photograph of a wine glass shattering into many little pieces.", caption: [This glass has been shattered by a high-intensity sound wave of the same frequency as the resonant frequency of the glass.]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ This #link("https://openstax.org/l/21waveswineglas")[video] shows waves on the surface of a wine glass, being driven by sound waves from a speaker. As the frequency of the sound wave approaches the resonant frequency of the wine glass, the amplitude and frequency of the waves on the wine glass increase. When the resonant frequency is reached, the glass shatters. ] A speaker produces a sound wave by oscillating a cone, causing vibrations of air molecules. In , a speaker vibrates at a constant frequency and amplitude, producing vibrations in the surrounding air molecules. As the speaker oscillates back and forth, it transfers energy to the air, mostly as thermal energy. But a small part of the speaker’s energy goes into compressing and expanding the surrounding air, creating slightly higher and lower local pressures. These #strong[compressions] (high-pressure regions) and #strong[rarefactions] (low-pressure regions) move out as longitudinal pressure waves having the same frequency as the speaker—they are the disturbance that is a sound wave. (Sound waves in air and most fluids are longitudinal, because fluids have almost no shear strength. In solids, sound waves can be both transverse and longitudinal.) (a) shows the compressions and rarefactions, and also shows a graph of gauge pressure versus distance from a speaker. As the speaker moves in the positive #emph[x]-direction, it pushes air molecules, displacing them from their equilibrium positions. As the speaker moves in the negative #emph[x]-direction, the air molecules move back toward their equilibrium positions due to a restoring force. The air molecules oscillate in simple harmonic motion about their equilibrium positions, as shown in part (b). Note that sound waves in air are longitudinal, and in the figure, the wave propagates in the positive #emph[x]-direction and the molecules oscillate parallel to the direction in which the wave propagates. #figure(figph[Figure A is a graph that shows the gauge pressure of the air versus the distance from the speaker. The gauge pressure is modeled with a sine function, where the crests of the function line up with the compressions and the troughs line up with the rarefactions. Figure B is the displacement of the air molecules versus the position. The displacement is modeled with a cosine function, where the zeroes are for the molecules in their equilibrium position and are centered at the compressions and rarefactions.], alt: "Figure A is a graph that shows the gauge pressure of the air versus the distance from the speaker. The gauge pressure is modeled with a sine function, where the crests of the function line up with the compressions and the troughs line up with the rarefactions. Figure B is the displacement of the air molecules versus the position. The displacement is modeled with a cosine function, where the zeroes are for the molecules in their equilibrium position and are centered at the compressions and rarefactions.", caption: [(a) A vibrating cone of a speaker, moving in the positive #emph[x]-direction, compresses the air in front of it and expands the air behind it. As the speaker oscillates, it creates another compression and rarefaction as those on the right move away from the speaker. After many vibrations, a series of compressions and rarefactions moves out from the speaker as a sound wave. The red graph shows the gauge pressure of the air versus the distance from the speaker. Pressures vary only slightly from atmospheric pressure for ordinary sounds. Note that gauge pressure is modeled with a sine function, where the crests of the function line up with the compressions and the troughs line up with the rarefactions. (b) Sound waves can also be modeled using the displacement of the air molecules. The blue graph shows the displacement of the air molecules versus the position from the speaker and is modeled with a cosine function. Notice that the displacement is zero for the molecules in their equilibrium position and are centered at the compressions and rarefactions. Compressions are formed when molecules on either side of the equilibrium molecules are displaced toward the equilibrium position. Rarefactions are formed when the molecules are displaced away from the equilibrium position.]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Models Describing Sound] Sound can be modeled as a pressure wave by considering the change in pressure from average pressure, #math.equation(block: true, alt: "Δ P equals Δ P sub max sin open parenthesis k x ∓ ω t plus ϕ close parenthesis .")[$"Δ" P = "Δ" P_("max") "sin" ( k x ∓ ω t + ϕ ) .$] This equation is similar to the periodic wave equations seen in Waves, where #math.equation(block: false, alt: "Δ P")[$"Δ" P$] is the change in pressure, #math.equation(block: false, alt: "Δ P sub max")[$"Δ" P_("max")$] is the maximum change in pressure, #math.equation(block: false, alt: "k equals the fraction 2 π over λ")[$k = frac(2 π, λ)$] is the wave number, #math.equation(block: false, alt: "ω equals the fraction 2 π over T equals 2 π f")[$ω = frac(2 π, T) = 2 π f$] is the angular frequency, and #math.equation(block: false, alt: "ϕ")[$ϕ$] is the initial phase. The wave speed can be determined from #math.equation(block: false, alt: "v equals the fraction ω over k equals the fraction λ over T .")[$v = frac(ω, k) = frac(λ, T) .$] Sound waves can also be modeled in terms of the displacement of the air molecules. The displacement of the air molecules can be modeled using a cosine function: #math.equation(block: true, alt: "s open parenthesis x , t close parenthesis equals s sub max cos open parenthesis k x ∓ ω t plus ϕ close parenthesis .")[$s ( x , t ) = s_("max") "cos" ( k x ∓ ω t + ϕ ) .$] In this equation, #emph[s] is the displacement and #math.equation(block: false, alt: "s sub max")[$s_("max")$] is the maximum displacement. ] Not shown in the figure is the amplitude of a sound wave as it decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. The intensity decreases as it moves away from the speaker, as discussed in Waves. The energy is also absorbed by objects and converted into thermal energy by the viscosity of the air. In addition, during each compression, a little heat transfers to the air; during each rarefaction, even less heat transfers from the air, and these heat transfers reduce the organized disturbance into random thermal motions. Whether the heat transfer from compression to rarefaction is significant depends on how far apart they are—that is, it depends on wavelength. Wavelength, frequency, amplitude, and speed of propagation are important characteristics for sound, as they are for all waves. === Summary - Sound is a disturbance of matter (a pressure wave) that is transmitted from its source outward. Hearing is the perception of sound. - Sound can be modeled in terms of pressure or in terms of displacement of molecules. - The human ear is sensitive to frequencies between 20 Hz and 20 kHz. === Conceptual Questions What is the difference between sound and hearing? #solutionbox[ Sound is a disturbance of matter (a pressure wave) that is transmitted from its source outward. Hearing is the human perception of sound. ] You will learn that light is an electromagnetic wave that can travel through a vacuum. Can sound waves travel through a vacuum? Sound waves can be modeled as a change in pressure. Why is the change in pressure used and not the actual pressure? #solutionbox[ Consider a sound wave moving through air. The pressure of the air is the equilibrium condition, it is the change in pressure that produces the sound wave. ] === Problems Consider a sound wave modeled with the equation#math.equation(block: false, alt: "s open parenthesis x , t close parenthesis equals 4.00 nm")[$s ( x , t ) = 4.00 #h(0.2em) "nm"$] #linebreak() #math.equation(block: false, alt: "cos open parenthesis 3.66 m to the power −1 x minus 1256 s to the power −1 t close parenthesis .")[$#h(0.2em) "cos" ( 3.66 #h(0.2em) "m"^(−1) #h(0.2em) x − 1256 #h(0.2em) "s"^(−1) t ) .$] What is the maximum displacement, the wavelength, the frequency, and the speed of the sound wave? #solutionbox[ #math.equation(block: true, alt: "s sub max equals 4.00 nm , λ equals 1.72 m , f equals 200 Hz , v equals 343.17 m/s")[$s_("max") = 4.00 #h(0.2em) "nm" , #h(0.5em) λ = 1.72 #h(0.2em) "m" , #h(0.5em) f = 200 #h(0.2em) "Hz" , #h(0.5em) v = 343.17 #h(0.2em) "m/s"$] ] Consider a sound wave moving through the air modeled with the equation#math.equation(block: false, alt: "s open parenthesis x , t close parenthesis equals 6.00 nm")[$s ( x , t ) = 6.00 #h(0.2em) "nm"$] #linebreak() #math.equation(block: false, alt: "cos open parenthesis 54.93 m to the power −1 x minus 18.84 times 10 cubed s to the power −1 t close parenthesis .")[$#h(0.2em) "cos" ( 54.93 #h(0.2em) "m"^(−1) x − 18.84 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "s"^(−1) t ) .$] What is the shortest time required for an air molecule to move between 3.00 nm and –3.00 nm? Consider a diagnostic ultrasound of frequency 5.00 MHz that is used to examine an irregularity in soft tissue. (a) What is the wavelength in air of such a sound wave if the speed of sound is 343 m/s? (b) If the speed of sound in tissue is 1800 m/s, what is the wavelength of this wave in tissue? #solutionbox[ a. #math.equation(block: false, alt: "λ equals 68.60 μ m;")[$λ = 68.60 #h(0.2em) μ "m;"$] b. #math.equation(block: false, alt: "λ equals 360.00 μ m")[$λ = 360.00 #h(0.2em) μ "m"$] ] A sound wave is modeled as #math.equation(block: false, alt: "Δ P equals 1.80 Pa sin open parenthesis 55.41 m to the power −1 x minus 18,840 s to the power −1 t close parenthesis .")[$"Δ" P = 1.80 #h(0.2em) "Pa" #h(0.2em) "sin" ( 55.41 #h(0.2em) "m"^(−1) #h(0.2em) x − 18,840 #h(0.2em) "s"^(−1) t ) .$] What is the maximum change in pressure, the wavelength, the frequency, and the speed of the sound wave? A sound wave is modeled with the wave function #math.equation(block: false, alt: "Δ P equals 1.20 Pa sin open parenthesis k x minus 6.28 times 10 to the power 4 s to the power −1 t close parenthesis")[$"Δ" P = 1.20 #h(0.2em) "Pa" #h(0.2em) "sin" ( k x − 6.28 #h(0.2em) × #h(0.2em) 10^(4) "s"^(−1) t )$] and the sound wave travels in air at a speed of #math.equation(block: false, alt: "v equals 343.00 m/s .")[$v = 343.00 #h(0.2em) "m/s" "."$] (a) What is the wave number of the sound wave? (b) What is the value for #math.equation(block: false, alt: "Δ P open parenthesis 3.00 m , 20.00 s close parenthesis")[$"Δ" P ( 3.00 #h(0.2em) "m" , 20.00 #h(0.2em) "s" )$]? #solutionbox[ a. #math.equation(block: false, alt: "k equals 183.09 m to the power −1 ;")[$k = 183.09 "m"^(−1) ;$] #linebreak() b. #math.equation(block: false, alt: "Δ P equals −1.11 Pa")[$"Δ" P = −1.11 #h(0.2em) "Pa"$] ] The displacement of the air molecules in sound wave is modeled with the wave function #math.equation(block: false, alt: "s open parenthesis x , t close parenthesis equals 5.00 nm")[$s ( x , t ) = 5.00 #h(0.2em) "nm" #h(0.2em)$] #linebreak() #math.equation(block: false, alt: "cos open parenthesis 91.54 m to the power −1 x minus 3.14 times 10 to the power 4 s to the power −1 t close parenthesis")[$"cos" ( 91.54 #h(0.2em) "m"^(−1) x − 3.14 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "s"^(−1) t )$]. (a) What is the wave speed of the sound wave? (b) What is the maximum speed of the air molecules as they oscillate in simple harmonic motion? (c) What is the magnitude of the maximum acceleration of the air molecules as they oscillate in simple harmonic motion? A speaker is placed at the opening of a long horizontal tube. The speaker oscillates at a frequency #emph[f], creating a sound wave that moves down the tube. The wave moves through the tube at a speed of #math.equation(block: false, alt: "v equals 340.00 m/s .")[$v = 340.00 #h(0.2em) "m/s" "."$] The sound wave is modeled with the wave function #math.equation(block: false, alt: "s open parenthesis x , t close parenthesis equals s sub max cos open parenthesis k x minus ω t plus ϕ close parenthesis .")[$s ( x , t ) = s_("max") "cos" ( k x − ω t + ϕ ) .$] At time #math.equation(block: false, alt: "t equals 0.00 s")[$t = 0.00 #h(0.2em) "s"$], an air molecule at #math.equation(block: false, alt: "x equals 3.5 m")[$x = 3.5 #h(0.2em) "m"$] is at the maximum displacement of 7.00 nm. At the same time, another molecule at #math.equation(block: false, alt: "x equals 3.7 m")[$x = 3.7 #h(0.2em) "m"$] has a displacement of 3.00 nm. What is the frequency at which the speaker is oscillating? #solutionbox[ #math.equation(block: true, alt: "s sub 1 equals 7.00 nm, s sub 2 equals 3.00 nm, k x sub 1 plus ϕ equals 0 rad; k x sub 2 plus ϕ equals 1.128 rad; k open parenthesis x sub 2 minus x sub 1 close parenthesis equals 1.128 rad, k equals 5.64 m to the power −1; λ equals 1.11 m, f equals 306.31 Hz")[$s_(1) = 7.00 #h(0.2em) "nm," #h(0.5em) s_(2) = 3.00 #h(0.2em) "nm," #h(0.5em) k x_(1) + ϕ = 0 #h(0.2em) "rad" \ k x_(2) + ϕ = 1.128 #h(0.2em) "rad" \ k ( x_(2) − x_(1) ) = 1.128 #h(0.2em) "rad," #h(0.5em) k = 5.64 #h(0.2em) "m"^(−1) \ λ = 1.11 #h(0.2em) "m," #h(0.5em) f = 306.31 #h(0.2em) "Hz"$] ] A 250-Hz tuning fork is struck and begins to vibrate. A sound-level meter is located 34.00 m away. It takes the sound #math.equation(block: false, alt: "Δ t equals 0.10 s")[$"Δ" t = 0.10 #h(0.2em) "s"$] to reach the meter. The maximum displacement of the tuning fork is 1.00 mm. Write a wave function for the sound. A sound wave produced by an ultrasonic transducer, moving in air, is modeled with the wave equation #math.equation(block: false, alt: "s open parenthesis x , t close parenthesis equals 4.50 nm cos open parenthesis 9.15 times 10 to the power 4 m to the power −1 x minus")[$s \( x , t \) = 4.50 #h(0.2em) "nm" #h(0.2em) "cos" \( 9.15 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "m"^(−1) x −$] #linebreak() #math.equation(block: false, alt: "2 π open parenthesis 5.00 MHz close parenthesis t close parenthesis .")[$2 π \( 5.00 #h(0.2em) "MHz" \) t \) .$] The transducer is to be used in nondestructive testing to test for fractures in steel beams. The speed of sound in the steel beam is #math.equation(block: false, alt: "v equals 5950 m/s .")[$v = 5950 #h(0.2em) "m/s" "."$] Find the wave function for the sound wave in the steel beam. #solutionbox[ #math.equation(block: true, alt: "k equals 5.28 times 10 cubed m; s open parenthesis x , t close parenthesis equals 4.50 nm cos open parenthesis 5.28 times 10 cubed m to the power −1 x minus 2 π open parenthesis 5.00 MHz close parenthesis t close parenthesis")[$k = 5.28 #h(0.2em) × #h(0.2em) 10^(3) "m" \ s ( x , t ) = 4.50 #h(0.2em) "nm" #h(0.2em) "cos" ( 5.28 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "m"^(−1) x − 2 π ( 5.00 #h(0.2em) "MHz" ) t )$] ] Porpoises emit sound waves that they use for navigation. If the wavelength of the sound wave emitted is 4.5 cm, and the speed of sound in the water is #math.equation(block: false, alt: "v equals 1530 m/s,")[$v = 1530 #h(0.2em) "m/s,"$] what is the period of the sound? Bats use sound waves to catch insects. Bats can detect frequencies up to 100 kHz. If the sound waves travel through air at a speed of #math.equation(block: false, alt: "v equals 343 m/s,")[$v = 343 #h(0.2em) "m/s,"$] what is the wavelength of the sound waves? #solutionbox[ #math.equation(block: true, alt: "λ equals 3.43 mm")[$λ = 3.43 #h(0.2em) "mm"$] ] A bat sends of a sound wave 100 kHz and the sound waves travel through air at a speed of #math.equation(block: false, alt: "v equals 343 m/s .")[$v = 343 #h(0.2em) "m/s" "."$] (a) If the maximum pressure difference is 1.30 Pa, what is a wave function that would model the sound wave, assuming the wave is sinusoidal? (Assume the phase shift is zero.) (b) What are the period and wavelength of the sound wave? Consider the graph shown below of a compression wave. Shown are snapshots of the wave function for #math.equation(block: false, alt: "t equals 0.000 s")[$t = 0.000 #h(0.2em) "s"$] (blue) and #math.equation(block: false, alt: "t equals 0.005 s")[$t = 0.005 #h(0.2em) "s"$] (orange), one half of an oscillation later. What are the wavelength, maximum displacement, velocity, and period of the compression wave? #figure(figph[Figure is a graph that shows a compression wave. The wave consists of two sinusoidal functions. The function shown with a blue color has maxima at 5, 11 and minima in 2, 8, 14. The function shown with red color has maxima at 2, 8, 14 and minima in 5, 11.], alt: "Figure is a graph that shows a compression wave. The wave consists of two sinusoidal functions. The function shown with a blue color has maxima at 5, 11 and minima in 2, 8, 14. The function shown with red color has maxima at 2, 8, 14 and minima in 5, 11.", caption: none) #solutionbox[ #math.equation(block: true, alt: "λ equals 6.00 m; s sub max equals 2.00 mm; v equals 600 m/s; T equals 0.01 s")[$λ = 6.00 #h(0.2em) "m" \ \ s_("max") = 2.00 #h(0.2em) "mm" \ \ v = 600 #h(0.2em) "m/s" \ \ T = 0.01 #h(0.2em) "s"$] ] Consider the graph in the preceding problem of a compression wave. Shown are snapshots of the wave function for #math.equation(block: false, alt: "t equals 0.000 s")[$t = 0.000 #h(0.2em) "s"$] (blue) and #math.equation(block: false, alt: "t equals 0.005 s")[$t = 0.005 #h(0.2em) "s"$] (orange). Given that the displacement of the molecule at time #math.equation(block: false, alt: "t equals 0.00 s")[$t = 0.00 #h(0.2em) "s"$] and position #math.equation(block: false, alt: "x equals 0.00 m")[$x = 0.00 #h(0.2em) "m"$] is #math.equation(block: false, alt: "s open parenthesis 0.00 m , 0.00 s close parenthesis equals 1.08 mm,")[$s ( 0.00 #h(0.2em) "m" , 0.00 #h(0.2em) "s" ) = 1.08 #h(0.2em) "mm,"$] derive a wave function to model the compression wave. A guitar string oscillates at a frequency of 100 Hz and produces a sound wave. (a) What do you think the frequency of the sound wave is that the vibrating string produces? (b) If the speed of the sound wave is #linebreak() #math.equation(block: false, alt: "v equals 343 m/s,")[$v = 343 #h(0.2em) "m/s,"$] what is the wavelength of the sound wave? #solutionbox[ (a) #math.equation(block: false, alt: "f equals 100 Hz,")[$f = 100 #h(0.2em) "Hz," #h(0.5em)$] (b) #math.equation(block: false, alt: "λ equals 3.43 m")[$λ = 3.43 #h(0.2em) "m"$] ]