SoundLongitudinal Waves InterferencePressure Graphs Standing Waves in a String: Two fixed endsSpeed of Sound Standing Waves in a Tube:Wavefronts One open endFrequency & Pitch Two open ends(human range) Musical InstrumentsThe Human Ear (and other complex sounds)Sonar & Echolocation BeatsDoppler Effect Intensity(and sonic booms) Sound Level (decibels)
Longitudinal WavesAs you learned in the unit on waves, in a longitudinal wave theparticles in a medium travel back & forth parallel to the wave itself.Sound waves are longitudinal and they can travel through most anymedium, so molecules of air (or water, etc.) move back & forth in thedirection of the wave creating high pressure zones (compressions) andlow pressure zones (rarefactions). The molecules act just like theindividual coils in the spring. The faster the molecules move back &forth, the greater the frequency of the wave, and the greater distancethey move, the greater the wave’s amplitude. molecule wavelength, λ rarefaction compression Animation
Sound Waves: Molecular ViewWhen sound travels through a medium, there are alternating regions ofhigh and low pressure. Compressions are high pressure regions wherethe molecules are crowded together. Rarefactions are low pressureregions where the molecules are more spread out. An individualmolecule moves side to side with each compression. The speed atwhich a compression propagates through the medium is the wavespeed, but this is different than the speed of the molecules themselves. wavelength, λ
Pressure vs. PositionThe pressure at a given point in a medium fluctuates slightly as soundwaves pass by. The wavelength is determined by the distance betweenconsecutive compressions or consecutive rarefactions. At each com-pression the pressure is a tad bit higher than its normal pressure. Ateach rarefaction the pressure is a tad bit lower than normal. Let’s callthe equilibrium (normal) pressure P0 and the difference in pressure fromequilibrium ∆ P. ∆ P varies and is at a max at a compression orrarefaction. In a fluid like air or water, ∆ Pmax is typically very smallcompared to P0 but our ears are very sensitive to slight deviations inpressure. The bigger ∆ P is, the greater the amplitude of the soundwave, and the louder the sound. wavelength, λ
Pressure vs. A: ∆ P = 0; P = P0 Position Graph B: ∆ P > 0; P = Pmax animation∆P C: ∆ P < 0; P = Pmin B x A C λ
Pressure vs. TimeThe pressure at a given point does not stay constant. If we onlyobserved one position we would find the pressure there variessinusoidally with time, ranging from: P0 to P0 + ∆ Pmax back to P0 then to P0 - ∆ Pmax and back to P0The cycle can also be described as:equilibrium → compression → equilibrium → rarefaction → equilibriumThe time it takes to go through this cycle is the period of the wave.The number of times this cycle happens per second is the frequencyof the wave in Hertz.Therefore, the pressure in the medium is a function of both positionand time!
Pressure vs. Time Graph ∆P T tRather than looking at a region of space at an instant in time, here we’relooking at just one point in space over an interval of time. At time zero,when the pressure readings began, the molecules were at their normalpressure. The pressure at this point in space fluctuates sinusoidally asthe waves pass by: normal → high → normal → low → normal. Thetime needed for one cycle is the period. The higher the frequency, theshorter the period. The amplitude of the graph represents the maximumdeviation from normal pressure (as it did on the pressure vs. positiongraph), and this corresponds to loudness.
Comparison of Pressure GraphsPressure vs. Position: The graph is for a snapshot in time anddisplays pressure variation for over an interval of space. Thedistance between peaks on the graph is the wavelength of the wave.Pressure vs. Time: The graph displays pressure variation over aninterval of time for only one point in space. The distance betweenpeaks on the graph is the period of the wave. The reciprocal of theperiod is the frequency.Both Graphs: Sound waves are longitudinal even though these graphslook like transverse waves. Nothing in a sound wave is actuallywaving in the shape of these graphs! The amplitude of either graphcorresponds to the loudness of the sound. The absolute pressurematters not. For loudness, all that matters is how much the pressuredeviates from its norm, which doesn’t have to be much. In real lifethe amplitude would diminish as the sound waves spread out.
Speed of SoundAs with all waves, the speed of sound depends on the mediumthrough which it is traveling. In the wave unit we learned that thespeed of a wave traveling on a rope is given by: F F = tension in rope Rope: v = µ µ = mass per unit length of ropeIn a rope, waves travel faster when the rope is under more tension andslower if the rope is denser. The speed of a sound wave is given by: B B = bulk modulus of medium Sound: v = ρ ρ = mass per unit volume (density)The bulk modulus, B, of a medium basically tells you how hard it isto compress it, just as the tension in a rope tells you how hard it isstretch it or displace a piece of it. (continued)
Speed of Sound (cont.) F Notice that each equation is in the form Rope: v = µ elastic property B v =Sound: v = inertial property ρThe bulk modulus for air is tiny compared to that of water, since airis easily compressed and water nearly incompressible. So, eventhough water is much denser than air, water is so much harder tocompress that sound travels over 4 times faster in water.Steel is almost 8 times denser than water, but it’s over 70 timesharder to compress. Consequently, sound waves propagate throughsteel about 3 times faster than in water, since (70 / 8) 0.5 ≈ 3.
Mach NumbersDepending on temp, sound travels around 750 mph, which wouldbe Mach 1. Twice this speed would be Mach 2, which is about themax speed for the F-22 Raptor.Speed Racer drives a car called “The Mach 5,” which would implyit can go 5 times the speed of sound.
Temperature & the Speed of Sound B Because the speed of sound is inversely proportional v = to the medium’s density, the less dense the medium, ρ the faster sound travels. The hotter a substance is, the faster its molecules/atoms vibrate and the more room they take up. This lowers the substance’s density, which is significant in a gas. So, in the summer, sound travels slightly faster outside than it does in the winter. To visualize this keep in mind that molecules must bump into each other in order to transmit a longitudinal wave. When molecules move quickly, they need less time to bump into their neighbors. The speed of sound in dry air is given by: v ≈ 331.4 + 0.60 T, where T is air temp in°C. Here are speeds for sound:Air, 0 °C: 331 m/s Air, 20 °C: 343 m/s Water, 25 °C: 1493 m/sIron: 5130 m/s Glass (Pyrex): 5640 m/s Diamond: 12 000 m/s
Wavefronts crest troughSome waves are one dimensional, like vibrations in a guitar string orsound waves traveling along a metal rod. Some waves are twodimensional, such as surface water waves or seismic waves travelingalong the surface of the Earth. Some waves are 3-D, such as soundtraveling in all directions from a bell, or light doing the same from aflashlight. To visualize 2-D and 3-D waves, we often drawwavefronts. The red wavefronts below could represent the crest ofwater waves on a pond moving outward after a rock was dropped inthe middle. They could also be used to represent high pressurezonesin sound waves. The wavefronts for 3-D sound waves would bespherical, but concentric circles are often used to simplify the picture.If the wavefronts are evenly spaced, then λ is a constant. Animation
Frequency & PitchJust as the amplitude of a sound wave relates to its loudness, thefrequency of the wave relates to its pitch. The higher the pitch, thehigher the frequency. The frequency you hear is just the number ofwavefronts that hit your eardrums in a unit of time. Wavelengthdoesn’t necessarily correspond to pitch because, even if wavefrontsare very close together, if the wave is slow moving, not manywavefronts will hit you each second. Even in a fast moving wavewith a small wavelength, the receiver or source could be moving,which would change the frequency, hence the pitch. Frequency ↔ Pitch Amplitude ↔ Loudness Listen to a pure tone (up to 1000 Hz) Listen to 2 simultaneous tones (scroll down)
The Human EarThe exterior part of the ear (the auricle, or pinna) is made of cartilageand helps funnel sound waves into the auditory canal, which has waxfibers to protect the ear from dirt. At the end of the auditory canal liesthe eardrum (tympanic membrane), which vibrates with the incomingsound waves and transmits these vibrations along three tiny bones(ossicles) called the hammer, anvil, and stirrup (malleus, incus, andstapes). The little stapes bone is attached to the oval window, amembrane of the cochlea.The cochlea is a coil that converts the vibrations it receives intoelectrical impulses and sends them to the brain via the auditory nerve.Delicate hairs (stereocilia) in the cochlea are responsible for this signalconversion. These hairs are easily damaged by loud noises, a majorcause of hearing loss!The semicircular canals help maintain balance, but do not aid hearing. Animation Ear Anatomy
Range of Human HearingThe maximum range of frequencies for most people is from about20 to 20 thousand hertz. This means if the number of high pressurefronts (wavefronts) hitting our eardrums each second is from 20 to20 000, then the sound may be detectable. If you listen to loudmusic often, you’ll probably find that your range (bandwidth) willbe diminished.Some animals, like dogs and some fish, can hear frequencies that arehigher than what humans can hear (ultrasound). Bats and dolphinsuse ultrasound to locate prey (echolocation). Doctors make use ofultrasound for imaging fetuses and breaking up kidney stones.Elephants and some whales can communicate over vast distanceswith sound waves too low in pitch for us to hear (infrasound). Hear the full range of audible frequencies (scroll down to speaker buttons)
Echoes & ReverberationAn echo is simply a reflected sound wave. Echoes are morenoticeable if you are out in the open except for a distant, largeobject. If went out to the dessert and yelled, you might hear adistant canyon yell back at you. The time between your yell andhearing your echo depends on the speed of sound and on thedistance to the to the canyon. In fact, if you know the speed ofsound, you can easily calculate the distance just by timing the delayof your echo.Reverberation is the repeated reflection of sound at close quarters.If you were to yell while inside a narrow tunnel, your reflectedsound waves would bounce back to your ears so quickly that yourbrain wouldn’t be able to distinguish between the original yell andits reflection. It would sound like a single yell of slightly longerduration. Animation
Sonar SOund NAvigation and RangingIn addition to locating prey, bats and dolphins use sound wavesfor navigational purposes. Submarines do this too. Theprinciple is to send out sound waves and listen for echoes. Thelonger it takes an echo to return, the farther away the object thatreflected those waves. Sonar is used in commercial fishing boatsto find schools of fish. Scientists use it to map the ocean floor.Special glasses that make use of sonar can help blind people byproducing sounds of different pitches depending on how close anobstacle is.If radio (low frequency light)waves are used instead of soundin an instrument, we call it radar(radio detection and ranging).
Doppler EffectA tone is not always heard at the same frequency at which it isemitted. When a train sounds its horn as it passes by, the pitch ofthe horn changes from high to low. Any time there is relativemotion between the source of a sound and the receiver of it, there isa difference between the actual frequency and the observedfrequency. This is called the Doppler effect. Click to hear effect:The Doppler effect applied to electomagnetic waves helpsmeteorologists to predict weather, allows astronomers to estimatedistances to remote galaxies, and aids police officers catch youspeeding.The Doppler effect applied to ultrasound is used by doctors tomeasure the speed of blood in blood vessels, just like a cop’s radargun. The faster the blood cell are moving toward the doc, the greaterthe reflected frequency. Animation (click on “The Doppler Effect”, then click on the button marked:
Sonic Booms When a source of sound is moving at the speed of sound, the wavefronts pile up on top of each other. This makes their combined amplitude very large, resulting in a shock wave and a sonic boom. At supersonic speeds a “Mach cone” is formed. The faster the source compared to sound, the smaller the shock wave angle will be.Wavefront Animations Another cool animationAnimation with sound (click on “The Doppler Effect”, then click onthe button marked:Movie: F-18 Hornet breaking the sound barrier (click on MPEGmovie)
Doppler f L = frequency as heard by a listener Equation f S = frequency produced by the source v = speed of sound in the medium v ± vL fL = fS ( v ± vS ) vL = speed of the listener v S = speed of the sourceThis equation takes into account the speed of the source of thesound, as well as the listener’s speed, relative to the air (orwhatever the medium happens to be). The only tricky part is thesigns. First decide whether the motion will make the observedfrequency higher or lower. (If the source is moving toward thelistener, this will increase f L, but if the listener is moving awayfrom the source, this will decrease f L.) Then choose the plus orminus as appropriate. A plus sign in the numerator will make f Lbigger, but a plus in the denominator will make f L smaller.Examples are on the next slide.
v ± vL Doppler Set-ups fL = fS ( v ± vS )The horn is producing a pure 1000 Hz tone. Let’s find the frequency asheard by the listener in various motion scenarios. The speed of soundin air at 20 °C is 343 m/s. 343 ( f L = 1000 343 - 10 ) still = 1030 Hz 10 m/s f L = 1000 ( 343 + 10 343 ) still = 1029 Hz 10 m/s Note that these situation are not exactly symmetric. Also, in real life a horn does not produce a single tone. More examples on the next slide.
v ± vL Doppler Set-ups (cont.) fL = fS ( v ± vS )The horn is still producing a pure 1000 Hz tone. This time both thesource and the listener are moving with respect to the air. (343 - 3 f L = 1000 343 - 10 ) = 1021 Hz 10 m/s 3 m/s f L = 1000 ( 343 + 3 343 - 10 ) = 1039 Hz 10 m/s 3 m/s Note the when they’re moving toward each other, the highest frequency possible for the given speeds is heard. Continued . . .
v ± vL Doppler Set-ups (cont.) fL = fS ( v ± vS )The horn is still producing a pure 1000 Hz tone. Here are the finaltwo motion scenarios. f L = 1000 ( 343 - 3 343 + 10 ) = 963 Hz 10 m/s 3 m/s f L = 1000 ( 343 + 3 343 + 10 ) = 980 Hz 10 m/s 3 m/s Note the when they’re moving toward each other, the highest frequency possible for the given speeds is heard. Continued . . .
Doppler ProblemMr. Magoo & Betty Boop are heading toward each other. Mr. Magoodrives at 21 m/s and toots his horn (just for fun; he doesn’t actually seeher). His horn sounds at 650 Hz. How fast should Betty drive so thatshe hears the horn at 750 Hz? Assume the speed o’ sound is 343 m/s. v ± vL fL = fS ( v ± vS ) 750 = 650 ( 343 + v L 343 - 21 ) vL = 28.5 m/s 21 m/s vL
InterferenceAs we saw in the wave presentation, waves can passes through eachother and combine via superposition. Sound is no exception. Thepic shows two sets of wavefronts, each from a point source of sound.(The frequencies are the same here, but this is not required forinterference.) Wherever constructive interference happens, a listenerwill here a louder sound. Loudness is diminished where destructiveinterference occurs. AA: 2 crests meet; B constructive interferenceB: 2 troughs meet; constructive interference CC: Crest meets trough; destructive interference
Interference: Distance in WavelengthsWe’ve got two point sources emitting the same wavelength. If thedifference in distances from the listener to the point sources is amultiple of the wavelength, constructive interference will occur.Examples: Point A is 3 λ from the red center and 4 λ from the greencenter, a difference of 1 λ. For B, the difference is zero. Since 1 and0 are whole numbers, constructive interference happens at thesepoints. If the difference in distance is an odd multiple of half the wavelength, destructive A interference occurs. Example: Point C is 3.5 λ from the green B center and 2 λ from the red center. The difference is 1.5 λ , so destructive interference occurs there. C Animation
Interference: Sound DemoUsing the link below you can play the same tone from each ofyour two computer speakers. If they were visible, thewavefronts would look just as it did on the last slide, exceptthey would be spheres instead of circles. You can experiencethe interference by leaning side to side from various places inthe room. If you do this, you should hear the loudnessfluctuate. This is because your head is moving through pointsof constructive interference (loud spots) and destructiveinterference (quiet regions, or “dead spots”). Turning onespeaker off will eliminate this effect, since there will be nointerference. Listen to a pure tone (up to 1000 Hz)
Interference: Noise ReductionThe concept of interference is used to reduce noise. Forexample, some pilots where special headphones that analyzeengine noise and produce the inverse of those sounds. Thiswaves produced by the headphones interfere destructivelywith the sound waves coming from the engine. As a result,the noise is reduced, but other sounds can still be heard, sincethe engine noise has a distinctive wave pattern, and onlythose waves are being cancelled out.Noise reduction graphic (Scroll down to “NoiseCancellation” under the “Applications of Sound” heading.)
Acoustics Acoustics sometimes refers to the science of sound. It can also refer to how well sounds traveling in enclosed spaces can be heard. The Great Hall in the Krannert Center is an example of excellent acoustics. Chicago Symphony Orchestra has even recorded there. Note how the walls and ceiling are beveled to get sound waves reflect in different directions. This minimizes the odds of there being a “dead spot” somewhere in the audience.scroll down to zoom in on the Great Hall pic.
Standing Waves: 2 Fixed EndsWhen a guitar string of length L is plucked, only certain frequenciescan be produced, because only certain wavelengths can sustainthemselves. Only standing waves persist. Many harmonics canexist at the same time, but the fundamental (n = 1) usually dominates. As we saw in the wave presentation, a standing wave occurs when awave reflects off a boundary and interferes with itself in such a wayas to produce nodes and antinodes. Destructive interference alwaysoccurs at a node. Both types occur at an antinode; they alternate.n = 1 (fundamental) n=2 Node Antinode Animation: Harmonics 1, 2, & 3
Wavelength λ=2L Formula: n=1 2 Fixed Ends(string of length L) λ=LNotice the pattern is of n=2the form: 2L λ = n λ=2L 3where n = 1, 2, 3, …. n=3 Thus, only certain wave- lengths can exists. To obtain tones corresponding 1 λ= L to other wavelengths, one 2 must press on the string n=4 to change its length.
Vibrating String ExampleSchmedrick decides to build his own ukulele. One of the fourstrings has a mass of 20 g and a length of 38 cm. By turning thelittle knobby, Schmed cranks up the tension in this string to 300 N.What frequencies will this string produced when plucked? Hints:1. Calculate the string’s mass per unit length, µ: 0.0526 kg / m2. How the speed of a wave traveling on this string using the formula v = F / µ from last chapter: 75.498 m / s3. Calculate several wavelengths of standing waves on this string: 0.76 m, 0.38 m, 0.2533 m4. Calculate the corresponding frequencies: 99 Hz, 199 Hz, 298 Hz Hear what a ukulele sounds like. (Scroll down.)
Standings Waves in a Tube: 2 Open EndsLike waves traveling on a string, sound waves traveling in a tubereflect back when they reach the end of the tube. (Much of the soundenergy will exit the open tube, but some will reflect back.) If thewavelength is right, the reflected waves will combine with the originalto create a standing wave. For a tube with two open ends, there will bean antinode at each end, rather than a node. (A closed end wouldcorrespond to a node, since it blocks the air from moving.) The picshows the fundamental. Note: the air does not move like a guitar stringmoves; the curve represents the amount of vibration. Maximumvibration occurs at the antinodes. In the middle is a node where the airmolecules don’t vibrate at all.Harmonics animation1st, 2nd, and 3rd Harmonics n = 1 (fundamental)
Wavelength λ=2L Formula: n=1 2 Open Ends(tube of length L) λ=LAs with the string, thepattern is: n=2 2L λ = 2 n λ= L 3where n = 1, 2, 3, …. n=3Thus, only certain wave-lengths will reinforce eachother (resonate). To obtain λ=1Ltones corresponding to other 2wavelengths, one must n=4change the tube’s length.
Standings Waves in a Tube: 1 Open EndIf a tube has one open and one closed end, the open end is a regionof maximum vibration of air molecules—an antinode. The closedend is where no vibration occurs—a node. At the closed end, onlya small amount of the sound energy will be transmitted; most willbe reflected. At the open end, of course, much more sound energyis transmitted, but a little is reflected. Only certain wavelengths ofsound will resonate in this tube, which depends on it length.Harmonics animation1st, 3rd, and 5thHarmonics animations (scroll down) n = 1 (fundamental)
Wavelength L= 1 λ 4 Formula: n=11 Open End(tube of length L) L= 3 λThis time the pattern is 4different: n=3 nλ L = 4 5or, L= λ 4 4L λ = n=5 nwhere n = 1, 3, 5, 7, ….Note: only odd harmonics L= 7 λ 4exist when only one end isopen. n=7
Tuning Forks & ResonanceTuning forks produce sound when struck because, as the tines vibrateback and forth, they bump into neighboring air molecules. (A speakerworks in the same way.) AnimationTouch a vibrating tuning fork to the surface of some water, and you’llsee the splashing. The more frequently the tines vibrate, the higher thefrequency of the sound. The harmonics pics would look just like thosefor a tube with one open end. Smaller tuning forks make a high pitchsound, since a shorter length means a shorter wavelength.If a vibrating fork (A) is brought near one that is not vibrating (B), Awill cause B to vibrate only if they made to produce the samefrequency. This is an example of resonance. Ifthe driving force (A) matches the natural fre-quency of B, then A can cause the amplitude of A BB to increase. (If you want to push someone ona swing higher and higher, you must push at thenatural frequency of the swing.)
Resonance: Shattering a GlassCan sound waves really shatter a wine glass? Yes, if the frequency ofthe sound matches the natural frequency of the glass, and if theamplitude is sufficient. The glass’s natural frequency can be determined by flicking the glass with your finger and listening to the tone it makes. If the glass is being bombarded by sound waves of this freq- uency, the amplitude of the vibrating glass with grow and grow until the glass shatters.
Standing Waves: Musical InstrumentsAs we saw with Schmedrick’s ukulele, string instruments make useof vibrations on strings where each end is a vibrational node. Thestrings themselves don’t move much air. So, either an electricalpickup and amplifier are needed, or the strings must transmitvibrations to the body of the instrument in which sound waves canresonate.Other instruments make use of standing waves in tubes. A flute forexample can be approximated as cylindrical tubes with two openends. A clarinet has just one open end. (The musician’s mouthblocks air in a clarinet, forming a closed end, but a flutist blows airover a hole without blocking the movement of air in and out.)Other instruments, like drums, produce sounds via standing waves ona surface, or membrane.Hear and See a Transverse Flute Standing Waves on aDHear a Clarinet, etc. (scroll down)
Complex SoundsReal sounds are rarely as simple as the individual standing wavepatterns we’ve seen on a string or in a tube. Why is it that twodifferent instruments can play the exact same note at the samevolume, yet still sound so different? This is because many differentharmonics can exist at the same time in an instrument, and the wavepatterns can be very complex. If only fundamental frequencies couldbe heard, instruments would sound more alike. The relative strengthsof different harmonics is known as timbre (tam’-ber). In other words,most sounds, including voices, are complexmixtures of frequencies. The sound made by aflute is predominately due to the first & second fluteharmonics, so it’s waveform is fairly simple.The sounds of other instruments are more pianocomplicated due to the presence of additionalharmonics. Combine Harmonics Create a Complex Sound violin
Octaves & RatiosSome mixtures of frequencies are pleasing to the ear;others are not. Typically, a harmonious combo of soundsis one in which the frequencies are in some simple ratio.If a fundamental frequency is combined with the 2ndharmonic, the ratio will be 1 : 2. (Each is the samemusical note, but the 2nd harmonic is one octave higher.In other words, going up an octave means doubling thefrequency.)Another simple (and thereforeharmonious) ratio is 2 : 3. This canbe produced by playing a C note(262 Hz) with a G note (392 Hz).
BeatsWe’ve seen how many frequencies can combine to produce acomplicated waveform. If two frequencies that are nearly the samecombine, a phenomenon called beats occurs. The resultingwaveform increases and decreases in amplitude in a periodic way,i.e., the sound gets louder and softer in a regular pattern. Hear BeatsWhen two waves differ slightly in frequency, they are alternately inphase and out of phase. Suppose the two original waves havefrequencies f1 and f2. Then their superposition (below) will havetheir average frequency and will get louder and softer with afrequency of f| f- f f2| |. f beat = | 1 1 - 2 Beats Animation (click on “start simulation”) f combo = ( f1 + f2 ) / 2 soft loud
Beats ExampleMickey Mouse and Goofy are playing an E note. Mickey’sguitar is right on at 330 Hz, but Goofy is slightly out of tuneat 332 Hz.1. What frequency will the audience hear? 331 Hz, the average of the frequencies of the two guitars.2. How often will the audience hear the sound getting louder and softer? They will hear it go from loud to soft twice each second. (The beat frequency is 2 Hz, since the two guitars differ in frequency by that amount.)
IntesityAll waves carry energy. In a typical sound wave the pressure doesn’tvary much from the normal pressure of the medium. Consequently,sound waves don’t transmit a whole lot of energy. The more energy asound wave transmits through a given area in a givenamount of time, the more intensity it has, and thelouder it will sound. That is, intensity is powerper unit area: P I = A 1 m2Suppose that in one second the greenwavefronts carry one joule of soundenergy through the one square meteropening. Then the intensity at the redrectangle is1 W / m 2. (1 Watt = 1 J / s.) wavefronts
Intensity ExampleIf you place your alarm clock 3 times closer to your bed, howmany times greater will the intensity be the next morning? answer:Since the wavefronts are approximatelyspherical, and the area of a sphere isproportional to the square of its radius(A = 4 π r 2), the intensity is inverselypropotional to the square of thedistance (since I = P / A). So, cuttingthe distance by a factor of 3 will makethe intensity of its ring about ninetimes greater. However, our ears donot work on a linear scale. The clockwill sound less than twice as loud.
Threshold IntensityThe more intense a sound is, the louder it will be. Normal sounds carrysmall amounts of energy, but our ears are very sensitive. In fact, we canhear sounds with intensities as low as 10-12 W / m 2 ! This is called thethreshold intensity, I 0. I 0 = 10 -12 W / m 2This means that if we had enormous ears like Dumbo’s, say a fullsquare meter in area, we could hear a sound delivering to this area an energy of only one trillionth of a jouleeach second! Since our earsare thousands of times smaller, the energy our ears receive in a second is thousands of times less.
Sound Level in Decibels The greater the intensity of a sound at a certain place, the louder it will sound. But doubling the intensity will not make it seem twice as loud. Experiments show that the intensity must increase by about a factor of 10 before the sound will seem twice as loud to us. A sound with a 100 times greater intensity will sound about 4 times louder. Therefore, we measure sound level (loudness) based on a logarithmic scale. The sound level in decibels (dB) is given by: β = 10 log I (in decibels) I0Ex: At a certain distance from a siren, the intensity of the soundwaves might be 10 –5 W / m 2 . The sound level at this location wouldbe: 10 log (10 –5 / 10 –12) = 10 log (10 7 ) = 70 dBNote: According to this definition, a sound at the intensity levelregisters zero decibels: 10 log (10 –12 / 10 –12) = 10 log (1 ) = 0 dB
The Decibel ScaleThe chart below lists the approximate sound levels of various sounds. The loudness of a given sound depends, of course, on the power ofthe source of the sound as well as the distance from the source. Note: Listening to loud music will gradually damage your hearing! Source DecibelsAnything on the verge 0 of being audible Whisper 30 ConstantNormal Conversation 60 exposure Busy Traffic 70 leads to } Niagara Falls 90 permanent Train 100 Construction Noise 110 hearing loss. Rock Concert 120 Pain Machine Gun 130 Damage Jet Takeoff 150 Rocket Takeoff 180
β = 10 log IIntensity & Sound Level I0Every time the intensity of a sound is increased by a factor of 10,the sound level goes up by 10 dB (and the sound seems to us to beabout twice as loud). Let’s compare a 90 dB shout to a 30 dBwhisper. The shout is 60 dB louder, which means its intensity is10 to the 6th power (a million) times greater. Proof:60 = β1 - β2 = 10 log (I 1 / I 0 ) - 10 log(I 2 / I 0 ) = 10 log I 1 / I 0 I2 / I0 60 = 10 log (I 1 / I 2 ) 6 = log (I 1 / I 2 ) 10 6 = I 1 / I 2 answers:Compare intensities: 80 dB vs. 60 dB factor of 100Compare intensities: 100 dB vs. 75 dB factor of 316 (10 2.5 = 316)Compare sound levels: differ by 30 dB ( I’s 4.2 · 10 –4 W / m 2 vs. 4.2 · 10 –7 W / m 2 differ by 3 powers of 10 )
Decibel ExampleSuppose a 75 g egg is dropped from 50 m up onto the sidewalk. Thesplat takes 0.05 s. Nearly all of the gravitational potential energy theegg had originally is converted into thermal energy, but a very smallfraction goes into sound energy. Let’s say this fraction is only6.7582 · 10 –11. How loud is the splat heard from the point at whichthe egg was dropped? Hints: Answers:1. How much energy does the egg originally have? 36.75 J2. How much of that energy goes into sound? 2.4836 · 10 –9 J3. Calculate sound power output of the egg. 4.9673 · 10 –8 W4. Figure intensity at 50 m up. 3.1623 · 10 –12 W / m 2 (Assume the hemispherical wavefronts.)5. Compute sound level in decibels. 5 dB, very faint