[{"data":1,"prerenderedAt":918},["ShallowReactive",2],{"layer:sound:extend":3},{"layer":4,"contentHash":897,"dependencyHashes":898,"approval":911,"releaseId":917},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":40,"sourceIds":892,"reviewStatus":893,"authoring":894},1,"sound","en","extend","Doppler shifts, digital recording and listening to the Earth","The physics of a passing siren, why your recorded voice sounds strange, and how earthquakes get located","Work out how much a siren's pitch shifts as it passes, find out why your recorded voice sounds strange (a real anatomical reason), and see why 44,100 Hz was not an arbitrary choice. Try two projects, solve combined puzzles, and use sound's own reasoning to locate an earthquake.",[13,14,15,16,17],"Calculate the Doppler-shifted frequency of an approaching or receding source, and explain why a source's own listener hears no shift.","Explain redshift and blueshift as light's Doppler effect, and name one everyday and one astronomical use.","Explain digital sampling and why 44,100 Hz was chosen, and trace the round trip from voice to file and back.","Explain bone conduction and why a recorded voice differs from how a speaker hears themselves.","Use P and S wave speeds to find the distance to an earthquake, and explain why three stations are needed to locate it.",45,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Extend",{"label":26,"value":27},"Reading time","≈ 45 minutes",{"label":29,"value":30},"Prior knowledge","All four earlier layers",{"label":32,"value":33},"Chapters","11",{"label":35,"value":36},"Labs","Sort game, match game, two projects, one animation",{"label":38,"value":39},"Units used","Hz, km\u002Fh, m\u002Fs, km, s",[41,45,51,57,60,65,75,88,123,128,146,162,196,201,206,249,254,257,261,267,280,292,297,302,305,309,321,325,348,359,398,403,406,409,414,419,423,428,453,457,462,465,489,494,499,510,521,533,546,551,555,558,562,573,577,582,614,617,621,626,631,634,639,681,867,880],{"id":42,"type":43,"markdown":44},"e-intro","prose","Four layers in, you can explain almost any everyday sound question with real numbers. Extend takes you past the school syllabus into the physics of a passing siren, the engineering choice behind every music file you own, a strange fact about your own voice, two real research careers, and a handful of puzzles designed to make you combine everything you know at once.",{"id":46,"type":47,"variant":48,"title":49,"markdown":50},"e-how-to-read","callout","observation","How to use this lesson","Some of this layer goes beyond what most schools teach at this age, on purpose. Take the projects seriously: recording a passing vehicle and measuring its pitch shift yourself is far more convincing than reading about someone else doing it.",{"id":52,"type":53,"title":54,"eyebrow":55,"navLabel":56},"e-ch01","chapter","The Doppler effect, precisely","Chapter 01","1 The Doppler effect",{"id":58,"type":43,"markdown":59},"e-ch1-p1","A motorbike approaches, honking steadily, and passes you. The note is not constant: it sounds **higher** as it approaches and drops to something noticeably **lower** the instant it passes, even though the rider's hand never touches the horn button differently. This is the **Doppler effect**, named after the Austrian physicist Christian Doppler, who proposed it in 1842.",{"id":61,"type":47,"variant":62,"title":63,"markdown":64},"e-def-doppler","definition","The Doppler effect","The **Doppler effect** is the change in the frequency (and so the pitch) an observer hears when there is relative motion between a sound source and the observer.\n\nA source **approaching** you compresses its own sound waves into a shorter distance ahead of it, so more wave crests reach you each second: a **higher** frequency. A source **receding** stretches its waves out behind it, so fewer crests reach you each second: a **lower** frequency. The source's own note, and what a rider next to the horn hears, never changes at all.",{"id":66,"type":67,"items":68},"e-formulas-doppler","formulas",[69,72],{"expression":70,"caption":71},"f' = f × v ÷ (v − vₛ)","Frequency heard as the source approaches. f is the source's true frequency, v is the speed of sound, vₛ is the source's speed.",{"expression":73,"caption":74},"f' = f × v ÷ (v + vₛ)","Frequency heard as the source recedes: the sign in the bracket flips.",{"id":76,"type":77,"title":78,"problem":79,"steps":80,"help":86},"e-we-doppler-scooter","worked_example","A scooter horn, approaching and receding","A scooter's horn sounds at 300 Hz. The scooter moves at 30 km\u002Fh. What frequency do you hear as it approaches, and as it drives away? (Speed of sound = 343 m\u002Fs.)",[81,82,83,84,85],"Convert the speed to metres per second: 30 km\u002Fh = 30,000 ÷ 3,600 = **8.33 m\u002Fs**.","Approaching: f' = 300 × 343 ÷ (343 − 8.33) = 300 × 343 ÷ 334.67 = **307.5 Hz**.","Receding: f' = 300 × 343 ÷ (343 + 8.33) = 300 × 343 ÷ 351.33 = **292.9 Hz**.","The shift both ways is small (about 2.5%) because 8.33 m\u002Fs is slow compared with 343 m\u002Fs.","Sense check: the approaching frequency is higher than 300 Hz and the receding one is lower, exactly matching what you hear as a vehicle passes.",{"simplerExplanation":87},"Convert km\u002Fh to m\u002Fs first. Approaching divides by a slightly smaller number (v minus the speed), which raises the frequency; receding divides by a slightly bigger number, which lowers it.",{"id":89,"type":90,"caption":91,"columns":92,"rows":98},"e-table-doppler-examples","table","How much the Doppler effect shifts a few real, moderate speeds (speed of sound = 343 m\u002Fs)",[93,94,95,96,97],"Source","True note","Speed","Heard approaching","Heard receding",[99,105,111,117],[100,101,102,103,104],"Scooter horn","300 Hz","30 km\u002Fh","307.5 Hz","292.9 Hz",[106,107,108,109,110],"Car horn","400 Hz","72 km\u002Fh","424.8 Hz","377.5 Hz",[112,113,114,115,116],"Train horn","350 Hz","80 km\u002Fh","374.2 Hz","328.7 Hz",[118,119,120,121,122],"Ambulance siren","700 Hz","120 km\u002Fh","775.3 Hz","638.0 Hz",{"id":124,"type":47,"variant":125,"title":126,"markdown":127},"e-nuance-bigger-shift","nuance","Why faster sources shift more","Look down the table: the ambulance, the fastest vehicle listed, has by far the biggest gap between its approaching and receding frequencies (775.3 − 638.0 = 137.3 Hz), while the scooter's gap is tiny (307.5 − 292.9 = 14.6 Hz).\n\nThe reason is in the formula: vₛ (the source's speed) sits right next to v (the speed of sound, 343 m\u002Fs) in the bottom of the fraction. A slow vₛ barely changes that bottom number, so f' barely moves. A vₛ that is a much bigger fraction of 343 m\u002Fs changes the bottom number, and so the frequency, far more. If a source could ever reach the speed of sound itself, the approaching formula would divide by zero: an extreme case (a sonic boom) far beyond anything in this table.",{"id":129,"type":130,"prompt":131,"options":132,"explanation":145},"e-predict-doppler-source","prediction","A rider sitting on the motorbike, right next to its own horn, listens as the bike accelerates from a stop up to 80 km\u002Fh. What frequency change does the rider hear from their own horn?",[133,136,139,142],{"id":134,"label":135},"a","The same rise and fall a bystander would hear",{"id":137,"label":138},"b","None: the horn's frequency relative to the rider never changes",{"id":140,"label":141},"c","Only a rise, never a fall",{"id":143,"label":144},"d","A much bigger shift than a bystander hears","**None.** The Doppler effect is about **relative** motion between source and listener. The rider is not moving relative to the horn at all, whatever speed the bike reaches, so the rider always hears the horn's true, unshifted frequency. The shift only exists for someone the bike is moving *relative to*, such as a bystander on the pavement.",{"id":147,"type":148,"itemId":149,"prompt":150,"check":151,"hints":156,"feedback":159},"e-prac-doppler-train","practice","sound.extend-doppler-receding","A train horn sounds at 350 Hz. The train recedes from you at 80 km\u002Fh (about 22.22 m\u002Fs). Using f' = f × v ÷ (v + vₛ), what frequency do you hear, to one decimal place?",{"kind":152,"answer":153,"tolerance":154,"unit":155},"number",328.7,0.3,"Hz",[157,158],"Convert 80 km\u002Fh to m\u002Fs first: 80,000 ÷ 3,600.","Add this speed to 343 in the bottom of the fraction, since the train is receding.",{"correct":160,"incorrect":161},"Right: f' = 350 × 343 ÷ (343 + 22.22) = 350 × 343 ÷ 365.22 ≈ **328.7 Hz**.","Convert to m\u002Fs (about 22.22 m\u002Fs), add it to 343 for a receding source, then compute 350 × 343 ÷ 365.22 ≈ 328.7 Hz.",{"id":163,"type":164,"component":165,"componentVersion":5,"config":166,"objective":194,"textAlternative":195},"e-lab-match-doppler-terms","interactive","match-pairs",{"prompt":167,"mode":168,"pairs":169},"Match each term from this layer to its meaning.","connect",[170,173,176,179,182,185,188,191],{"a":171,"b":172},"Doppler effect","A frequency change caused by relative motion between source and listener",{"a":174,"b":175},"Redshift","Light shifted towards red because its source is receding",{"a":177,"b":178},"Sampling rate","How many times a second an audio signal's height is measured and stored",{"a":180,"b":181},"Nyquist-Shannon theorem","You must sample at more than twice the highest frequency you want to capture",{"a":183,"b":184},"Bone conduction","Sound reaching the inner ear through the skull rather than the air",{"a":186,"b":187},"P wave","The faster, longitudinal earthquake wave",{"a":189,"b":190},"S wave","The slower, transverse earthquake wave",{"a":192,"b":193},"Triangulation","Combining distance circles from several stations to find one exact location","Connect eight terms from this layer to their meanings.","A matching game with eight pairs covering this layer's core vocabulary: the Doppler effect, redshift, sampling rate, the Nyquist-Shannon theorem, bone conduction, P waves, S waves and triangulation, each matched to a plain-English meaning.",{"id":197,"type":47,"variant":198,"title":199,"markdown":200},"e-misconception-source-changes","misconception","\"The Doppler effect changes the source's own note\"","It is easy to picture an approaching ambulance's siren as somehow *actually* rising in pitch as it drives. It does not. The siren itself, and anyone riding inside the ambulance, hears exactly the same steady note the whole time.\n\nWhat changes is only what an **outside listener** receives, because the source and the listener are moving relative to each other. The Doppler effect is a property of the whole approaching-or-receding *situation*, never a property of the source by itself. This is exactly why a rider next to their own horn hears no shift at all, whatever speed the bike reaches.",{"id":202,"type":47,"variant":203,"title":204,"markdown":205},"e-example-sonic-boom","example","What happens if a source reaches the speed of sound itself","The approaching Doppler formula divides by (v − vₛ). As a source's speed vₛ gets closer and closer to the speed of sound v, that bottom number shrinks towards zero, and the formula predicts an ever more extreme, and finally undefined, frequency.\n\nReal supersonic aircraft do not defy this; they simply leave the ordinary wave picture behind. Instead of spreading outward smoothly ahead of the plane, the compressions pile up into a single, very strong pressure wave trailing behind it in a cone shape: the **sonic boom**, heard on the ground as one sharp crack rather than a smoothly rising note. It is a genuinely different regime of physics from anything else in this topic, mentioned here only so the formula's own breaking point makes sense.",{"id":207,"type":164,"component":208,"componentVersion":5,"config":209,"objective":247,"textAlternative":248},"e-lab-sort-doppler","sort-game",{"prompt":210,"bins":211,"items":221,"seconds":246},"Sort each situation by whether the heard pitch rises, falls, or stays the same.",[212,215,218],{"id":213,"label":214},"rises","Pitch rises",{"id":216,"label":217},"falls","Pitch falls",{"id":219,"label":220},"same","No change",[222,226,230,234,238,242],{"id":223,"label":224,"bin":213,"why":225},"ambulance-approach","An ambulance siren approaching you","An approaching source compresses its waves ahead of it: a higher heard frequency.",{"id":227,"label":228,"bin":216,"why":229},"ambulance-recede","The same ambulance driving away after passing","A receding source stretches its waves out behind it: a lower heard frequency.",{"id":231,"label":232,"bin":219,"why":233},"rider-own-horn","A rider listening to their own horn while riding","No relative motion between the rider and the horn, so no Doppler shift for the rider.",{"id":235,"label":236,"bin":216,"why":237},"train-passing","A train horn, the instant it passes you","The moment of passing marks the switch from approaching (higher) to receding (lower).",{"id":239,"label":240,"bin":219,"why":241},"parked-car-horn","A parked car's horn, heard by someone walking past","The source is not moving at all, so there is no Doppler shift, whoever is moving.",{"id":243,"label":244,"bin":213,"why":245},"cricket-ball","A cricket ball's sound as it is thrown towards a wicketkeeper","A moving sound source approaching a listener always shifts higher, whatever the source.",0,"Sort six situations into pitch rises, pitch falls, or no change, using the Doppler effect.","A sorting game with six cards and three bins: **pitch rises**, **pitch falls** and **no change**.\n\nRises: an ambulance approaching, and a thrown cricket ball approaching a wicketkeeper. Falls: the same ambulance after it passes, and a train horn the instant it passes you. No change: a rider listening to their own horn (no relative motion), and a parked car's horn heard by a passing walker (the source itself is not moving).\n\nThe test behind every card: is the **source** moving relative to the **listener**, and if so, in which direction?",{"id":250,"type":53,"title":251,"eyebrow":252,"navLabel":253},"e-ch02","Doppler beyond sound","Chapter 02","2 Doppler beyond sound",{"id":255,"type":43,"markdown":256},"e-ch2-p1","Christian Doppler's original 1842 proposal was actually about **light and colour**, not sound; sound was confirmed to show the same effect within a few years, and it is sound where most people first meet the idea. But the underlying reasoning, waves bunching up ahead of an approaching source and stretching out behind a receding one, applies to any wave at all, including light.",{"id":258,"type":47,"variant":203,"title":259,"markdown":260},"e-example-redshift","Redshift: the whole universe is (mostly) receding","Light from almost every distant galaxy reaches Earth slightly shifted towards the red end of the spectrum, a **redshift**, the light equivalent of a receding source's lower pitch. Astronomers use exactly this to work out that most distant galaxies are moving away from us, and that the further away a galaxy is, the faster it appears to recede: strong evidence that the universe is expanding.\n\nCloser to home, the same idea inside a **radar speed gun** bounces radio waves (also a form of light) off a moving car and measures the tiny shift in the returned frequency to calculate its speed. A weather radar uses the same trick on raindrops to see how fast, and in which direction, a storm is moving.",{"id":262,"type":263,"conceptId":264,"relation":265,"explanation":266},"e-conn-light","connection","light","related_to","Redshift and blueshift are light's own Doppler effect: the same reasoning about waves bunching up or stretching out, applied to light instead of sound.",{"id":268,"type":77,"title":269,"problem":270,"steps":271,"help":278},"e-we-radar","How a police radar gun measures speed from a Doppler shift","An X-band radar gun sends out a 10.525 GHz radio wave. It reflects off an approaching car and returns with its frequency shifted by 2,106.5 Hz. Given that the shift is Δf = 2 × f₀ × v ÷ c (the factor of 2 because the wave makes a there-and-back trip), find the car's speed.",[272,273,274,275,276,277],"Rearrange for v: v = Δf × c ÷ (2 × f₀).","v = 2,106.5 × 299,792,458 ÷ (2 × 10,525,000,000).","v = 2,106.5 × 299,792,458 ÷ 21,050,000,000.","v ≈ **30 m\u002Fs**.","Convert to more familiar units: 30 m\u002Fs × 3.6 = **108 km\u002Fh**.","A K-band gun at 24.15 GHz would measure a bigger shift, about 4,833.3 Hz, for the exact same car and speed, simply because its own frequency is higher; both guns still report the same 108 km\u002Fh once the arithmetic is done.",{"simplerExplanation":279},"The radar reflects off the moving car and comes back Doppler-shifted, twice (out and back). Rearranging the shift formula turns the measured frequency shift back into a speed.",{"id":281,"type":77,"title":282,"problem":283,"steps":284,"help":290},"e-we-redshift-speed","Estimating a galaxy's recession speed from redshift","A particular spectral line normally at a wavelength of 500 nanometres is measured, from a distant galaxy's light, at 500.5 nanometres: a shift of 0.5 nm towards red. Using the slow-speed approximation v ≈ c × Δλ ÷ λ, estimate the galaxy's recession speed.",[285,286,287,288,289],"v ≈ c × Δλ ÷ λ, with c = 299,792.458 km\u002Fs (using km\u002Fs makes the answer easier to read).","v ≈ 299,792.458 × 0.5 ÷ 500.","v ≈ 299,792.458 × 0.001.","v ≈ **299.8 km\u002Fs**.","This approximation is only valid because 299.8 km\u002Fs is a tiny fraction of light's own speed; for galaxies receding at a meaningful fraction of light speed, the full relativistic formula is needed instead, exactly the model limit noted earlier in this chapter.",{"simplerExplanation":291},"Work out what fraction the shift is of the original wavelength (0.5 out of 500, or 0.001), then multiply the speed of light by that same fraction.",{"id":293,"type":47,"variant":294,"title":295,"markdown":296},"e-model-limit-light-doppler","model_limit","Light's Doppler shift is not identical maths to sound's","It is tempting to swap v (343 m\u002Fs) for the speed of light in the sound formula and call it done. Real Doppler shift for light needs Einstein's special relativity once speeds become a meaningful fraction of light's own enormous speed, and the everyday sound formula in this layer is not built for that.\n\nWhat carries over cleanly, and is genuinely the same idea, is the **direction** of the effect: approaching sources or observers shift towards higher frequency (blueshift for light, higher pitch for sound); receding ones shift towards lower frequency (redshift for light, lower pitch for sound). The exact arithmetic differs; the underlying reasoning about waves bunching up or stretching out does not.",{"id":298,"type":53,"title":299,"eyebrow":300,"navLabel":301},"e-ch03","How sound becomes a file","Chapter 03","3 Sound becomes a file",{"id":303,"type":43,"markdown":304},"e-ch3-p1","A microphone, from the Investigate layer, turns sound into a smoothly, continuously changing electric current. A digital recording cannot store something smooth and continuous; a computer only stores numbers. So somewhere between the microphone and the file on your phone, that smooth signal has to be chopped into a very long list of numbers, a process called **sampling**.",{"id":306,"type":47,"variant":62,"title":307,"markdown":308},"e-def-sampling","Sampling and sampling rate","**Sampling** measures the exact height of an audio signal at a huge number of evenly spaced instants and stores each measurement as a number.\n\nThe **sampling rate** is how many of these measurements are taken every second, in hertz. Standard compact-disc audio samples at **44,100 times a second** (44.1 kHz); many modern recordings use 48,000 or even 96,000 times a second.",{"id":310,"type":77,"title":311,"problem":312,"steps":313,"help":319},"e-we-nyquist","Why 44,100, and not some smaller, tidier number?","Human hearing reaches about 20,000 Hz. The Nyquist-Shannon sampling theorem says you must sample at more than twice the highest frequency you want to capture. What is the minimum sampling rate this theorem allows for full-range human hearing, and how does the CD standard of 44,100 Hz compare?",[314,315,316,317,318],"Twice the highest frequency of interest: 2 × 20,000 = **40,000 Hz** is the bare minimum the theorem allows.","44,100 Hz is indeed greater than 40,000 Hz, satisfying the theorem with some room to spare.","That extra room, called a transition band, gives engineers a practical margin, since no real filter can cut off frequencies with mathematical perfection exactly at 20,000 Hz.","Half of 44,100 is 22,050 Hz: the highest frequency this sampling rate can faithfully represent, comfortably above the top of human hearing.","So 44,100 Hz was not an arbitrary round number; it was chosen to sit safely above the Nyquist minimum for the full range of human hearing, with early technical constraints of 1970s-80s recording equipment shaping the exact figure.",{"simplerExplanation":320},"You need to sample faster than twice the highest pitch you want to capture. Twice 20,000 Hz is 40,000 Hz, and 44,100 clears that with a small safety margin.",{"id":322,"type":47,"variant":125,"title":323,"markdown":324},"e-nuance-compression","Sampling is not the whole story: compression too","A CD-quality recording, sampled 44,100 times a second with each sample stored precisely, takes up a lot of storage space: roughly 10 megabytes per minute for stereo sound. Most music and voice files you actually use are also **compressed**, using clever mathematics to store an almost-identical-sounding file in a fraction of the space, often by discarding parts of the sound that human hearing is worst at noticing, such as very quiet sounds masked by louder ones nearby in pitch or time.\n\nThis is a separate step from sampling itself, and it is why two files sampled at exactly the same rate can still sound different, or take up very different amounts of storage, depending on how aggressively they were compressed afterwards.",{"id":326,"type":90,"caption":327,"columns":328,"rows":331},"e-table-sampling-rates","Sampling rates in everyday and specialist use",[329,177,330],"Use","Highest frequency captured",[332,336,340,344],[333,334,335],"Old analogue telephone calls","8,000 Hz","4,000 Hz: enough for speech to be understood, not for music",[337,338,339],"CD-quality audio","44,100 Hz","22,050 Hz: comfortably above the top of human hearing",[341,342,343],"Professional studio recording","96,000 or 192,000 Hz","48,000 or 96,000 Hz: far more headroom than hearing needs",[345,346,347],"Recording a bat's ultrasonic calls","250,000 Hz or higher","125,000 Hz or higher: needed to capture calls up to about 120,000 Hz",{"id":349,"type":77,"title":350,"problem":351,"steps":352,"help":357},"e-we-bat-recording","Why bat-recording equipment needs such a high sampling rate","A bat researcher wants to record calls up to 120,000 Hz without losing information. Using the Nyquist-Shannon rule (sample at more than twice the highest frequency), what minimum sampling rate is needed, and why would an ordinary 44,100 Hz voice recorder be useless for this?",[353,354,355,356],"Minimum sampling rate = 2 × 120,000 = **240,000 Hz** (240 kHz), and in practice a bit higher for a safety margin.","An ordinary voice recorder sampling at 44,100 Hz can only faithfully capture frequencies up to half of that: 22,050 Hz.","120,000 Hz is more than five times higher than what a 44,100 Hz recorder can capture at all, so it would not just sound poor; it would miss the bat's call almost entirely.","This is exactly why specialist bioacoustic recorders, built for bats, dolphins and other ultrasonic callers, use sampling rates of 250,000 Hz or more, far beyond anything needed for human speech or music.",{"simplerExplanation":358},"Double the highest frequency you want to capture. A bat's 120,000 Hz call needs at least 240,000 samples a second, over five times what a phone voice recorder manages.",{"id":360,"type":361,"component":362,"componentVersion":5,"config":363,"textAlternative":397},"e-anim-recording-pipeline","animation","process-steps",{"title":364,"diagram":365,"steps":366},"From your voice to a saved file, and back again","none",[367,372,377,382,387,392],{"id":368,"label":369,"description":370,"highlight":371},"voice","You speak","Your vocal folds vibrate, and the resulting sound wave, compressions and rarefactions in air, reaches a microphone.",[],{"id":373,"label":374,"description":375,"highlight":376},"mic-signal","Microphone makes a current","The microphone's diaphragm and coil turn the arriving sound into a smoothly, continuously changing electric current, exactly matching the sound wave's shape.",[],{"id":378,"label":379,"description":380,"highlight":381},"sample","Sampling measures it 44,100 times a second","An analogue-to-digital converter measures the current's exact height 44,100 times every second (for CD-quality audio) and stores each measurement as a number.",[],{"id":383,"label":384,"description":385,"highlight":386},"store","The numbers are saved","Millions of these numbers, one after another, are saved to a file, sometimes compressed to take up less storage space.",[],{"id":388,"label":389,"description":390,"highlight":391},"playback","Playback rebuilds the current","A digital-to-analogue converter reads the saved numbers back out, 44,100 a second, and recreates a smoothly changing electric current from them.",[],{"id":393,"label":394,"description":395,"highlight":396},"speaker-out","A speaker turns it back into sound","The rebuilt current drives a loudspeaker's coil and cone, recreating compressions and rarefactions in the air: your recorded voice, played back.",[],"A six-step round trip. Your voice makes a sound wave, a microphone turns it into a matching electric current, an analogue-to-digital converter samples that current 44,100 times a second and stores the results as numbers, the numbers are saved to a file, a digital-to-analogue converter rebuilds a current from them on playback, and a loudspeaker turns that current back into sound. Every digital recording you have ever heard, on a phone, in a film or on a streaming service, follows this same six-step round trip.",{"id":399,"type":53,"title":400,"eyebrow":401,"navLabel":402},"e-ch04","Why your recorded voice sounds strange","Chapter 04","4 Your recorded voice",{"id":404,"type":43,"markdown":405},"e-ch4-p1","Almost everyone's first reaction to hearing their own recorded voice is the same: \"that doesn't sound like me.\" It is a completely genuine difference, not just unfamiliarity, and the reason is a second path sound takes that has nothing to do with air at all.",{"id":407,"type":47,"variant":62,"title":183,"markdown":408},"e-def-bone-conduction","**Bone conduction** is sound reaching your inner ear by travelling through the bones of your own skull, rather than through the air and your ear canal.\n\nWhen you speak, your vocal folds and throat vibrate the bones of your skull directly, and that vibration reaches your cochlea by this solid path, in addition to the ordinary air path everyone else hears you through.",{"id":410,"type":47,"variant":411,"title":412,"markdown":413},"e-aha-two-paths","aha","You hear two versions of your own voice; nobody else does","When you talk, sound reaches your own ears by **two** routes at once: through the air, exactly as it reaches anyone listening to you, and through the bones of your skull, bypassing the outer ear entirely.\n\nBone conducts low frequencies particularly well, so the bone-conduction path adds extra bass, richness and depth to the voice you hear inside your own head. Everyone else only ever gets the air-conducted version.\n\nA microphone, like everyone else's ears, only picks up the air-conducted sound. So a recording plays back **exactly what other people already hear you as**, missing only the private, bone-added richness you are used to hearing from the inside. The recording is not wrong; your own everyday impression of your voice was always the unusual one.",{"id":415,"type":47,"variant":416,"title":417,"markdown":418},"e-tryit-record-yourself","try_it","Prove it to your own ears","You need a phone or any device that can record and play back sound.\n\n1. Record yourself reading a sentence aloud, in a normal voice.\n2. Play it back and notice how different, usually higher and thinner, it sounds compared with your own voice as you normally hear it.\n3. Now cover both ears firmly with your palms while you speak the same sentence again. Listen carefully.\n\nWith your ears covered, almost none of the air-conducted sound gets in, so what you hear is closer to pure bone conduction: often even deeper and stranger than usual. This shows the two paths clearly separated: air conduction (what the recording captured, and what everyone else hears) and bone conduction (the private, extra layer you are used to).",{"id":420,"type":263,"conceptId":421,"relation":265,"explanation":422},"e-conn-anatomy-bone","human-body-anatomy","Bone conduction depends on the skull's bones carrying vibration directly to the inner ear, a different route through the body's anatomy from the usual air-and-eardrum path.",{"id":424,"type":53,"title":425,"eyebrow":426,"navLabel":427},"e-ch05","Project: measure a Doppler shift yourself","Chapter 05","5 Project: Doppler",{"id":429,"type":430,"title":431,"items":432},"e-steps-doppler-project","steps","Recording and measuring a real Doppler shift",[433,437,441,445,449],{"title":434,"tag":435,"text":436},"Find a safe, steady sound source","safety first","Stand well back from any road. A passing train with a horn, a car with hazard-light clicking loud enough to hear, or (with an adult's help) a phone playing a steady tone from a bicycle rolling past are all safer than standing near traffic.",{"title":438,"tag":439,"text":440},"Record it approaching and receding","phone microphone","Record continuously as the source approaches, passes, and moves away, ideally standing to one side rather than directly in its path.",{"title":442,"tag":443,"text":444},"Look at the recording as a spectrogram","free software","Free audio-editing software (many are available for computers and some for phones) can display a spectrogram: a graph of frequency against time. The pitch trace should visibly step down as the source passes.",{"title":446,"tag":447,"text":448},"Estimate the frequency before and after","read the graph","Read off the approximate frequency just before and just after the moment of passing directly opposite you, which is where the shift is sharpest.",{"title":450,"tag":451,"text":452},"Compare with the formula","check your physics","If you can estimate the source's speed (a train timetable, a car's approximate speed, a measured cycling speed), use f' = f × v ÷ (v ∓ vₛ) with the frequency before passing to predict the frequency after, and compare with what you measured.",{"id":454,"type":455,"prompt":456},"e-reflect-doppler-project","reflection","If you tried the Doppler recording project, describe what you measured and how closely it matched the formula's prediction. If you could not try it, describe exactly what equipment and conditions you would need, and predict one source of error that might make a real measurement differ from the formula's clean prediction.",{"id":458,"type":53,"title":459,"eyebrow":460,"navLabel":461},"e-ch06","Project: build a resonance instrument","Chapter 06","6 Project: build one",{"id":463,"type":43,"markdown":464},"e-ch6-p1","Understand introduced the jal tarang, tuned bowls of water struck with a stick. Building and tuning your own version by ear, rather than just reading about one, is one of the best ways to feel resonance and mass-loading directly in your hands.",{"id":466,"type":430,"title":467,"items":468},"e-steps-jal-tarang-project","Building and tuning your own eight-note water xylophone",[469,473,477,481,485],{"title":470,"tag":471,"text":472},"Gather eight identical bowls or tumblers","same size","Identical containers matter: differences in glass thickness or shape change the note independently of the water level, making tuning confusing.",{"title":474,"tag":475,"text":476},"Set the two end bowls first","empty and full","The nearly empty bowl will ring highest; the nearly full one will ring lowest, giving you the top and bottom of your scale.",{"title":478,"tag":479,"text":480},"Tap each bowl and listen for a rough scale","tune by ear","Adjust the water in the six middle bowls, a small amount at a time, until each note sounds a clear step higher than the one before it, working from the fullest bowl to the emptiest.",{"title":482,"tag":483,"text":484},"Check your scale against a known instrument","compare","Compare against a harmonium, keyboard app or singing voice you trust, adjusting water levels until the steps sound evenly spaced.",{"title":486,"tag":487,"text":488},"Play a simple, familiar tune","test it","A children's rhyme or a simple film tune is enough to prove the instrument actually works as a scale, not just eight separate notes.",{"id":490,"type":47,"variant":491,"title":492,"markdown":493},"e-careful-water-project","careful","Do this over a tray, not a good table","Water and tuning-by-trial-and-error go together with spills. Do this project over a large tray or outdoors, use plastic or china rather than anything fragile near the edge of a table, and mop up promptly so nobody slips.",{"id":495,"type":53,"title":496,"eyebrow":497,"navLabel":498},"e-ch07","Puzzles: combine what you know","Chapter 07","7 Combined puzzles",{"id":500,"type":77,"title":501,"problem":502,"steps":503,"help":508},"e-we-puzzle-rail","The old railway trick","A worker presses an ear to a steel rail 1,000 metres up the line while a hammer strikes the far end once. They also hear the same strike through the air. How much earlier does the sound arrive through the rail than through the air? (Speed in steel = 5,960 m\u002Fs; speed in air = 343 m\u002Fs.)",[504,505,506,507],"Find the time through air: t(air) = 1,000 ÷ 343 = **2.915 seconds**.","Find the time through steel: t(steel) = 1,000 ÷ 5,960 = **0.168 seconds**.","Difference: 2.915 − 0.168 = **2.75 seconds** earlier through the rail.","This is the real reason old films and stories show characters listening to railway lines: sound genuinely does arrive through solid steel dramatically sooner than through air, over a distance where the gap is easily long enough to notice without any instrument.",{"simplerExplanation":509},"Work out each travel time separately using time = distance ÷ speed, then subtract. The steel path is over seventeen times faster, so it wins by a clear margin.",{"id":511,"type":77,"title":512,"problem":513,"steps":514,"help":519},"e-we-puzzle-combined","A car horn: Doppler and echo together","A car horn sounding at 400 Hz approaches you at 72 km\u002Fh (20 m\u002Fs). At the same moment, someone claps once beside a wall 686 metres away. (a) What frequency do you hear from the approaching horn? (b) How long does the clap's echo take to return? Treat the two events separately.",[515,516,517,518],"(a) Convert to m\u002Fs: 72 km\u002Fh = 20 m\u002Fs exactly.","f' = 400 × 343 ÷ (343 − 20) = 400 × 343 ÷ 323 = **424.8 Hz**.","(b) Echo time = 2 × distance ÷ speed = 2 × 686 ÷ 343 = 1,372 ÷ 343 = **4.0 seconds**.","Two completely different calculations, both built from the same underlying fact (the speed of sound in air), used two different ways: one for a moving source's frequency shift, one for a stationary reflector's round trip.",{"simplerExplanation":520},"Treat the two parts as separate problems: Doppler's formula for the horn, and the round-trip echo formula for the clap. Both just need the 343 m\u002Fs speed of sound in air.",{"id":522,"type":77,"title":523,"problem":524,"steps":525,"help":531},"e-we-puzzle-firecracker","Stepping back from a firecracker is not enough on its own","A firecracker measures 140 dB at 2 metres. A bystander moves back to 16 metres away, eight times further. Using level(d₂) = level(d₁) − 20 × log₁₀(d₂ ÷ d₁), how loud does it still seem, and is that safe?",[526,527,528,529,530],"Distance ratio: 16 ÷ 2 = 8.","Drop = 20 × log₁₀(8) ≈ 20 × 0.903 ≈ 18.1 dB.","New level ≈ 140 − 18.1 = **about 121.9 dB**.","That is still above the roughly 120 dB level where even a single short exposure can cause damage.","The lesson: stepping back by a large factor still leaves a genuinely loud, risky sound; distance helps enormously but has limits, which is exactly why safety guidance also recommends covering ears and using proper hearing protection near very loud events, not distance alone.",{"simplerExplanation":532},"Eight times further away only costs about 18 dB. Starting from 140 dB, that still leaves nearly 122 dB: loud enough to hurt on its own.",{"id":534,"type":148,"itemId":535,"prompt":536,"check":537,"hints":541,"feedback":543},"e-prac-quake","sound.extend-quake-distance","A seismograph records the faster P wave, then the slower S wave 12 seconds later. Using the approximation that distance (km) = 8.4 × the time gap in seconds, how far away, in kilometres, was the earthquake?",{"kind":152,"answer":538,"tolerance":539,"unit":540},100.8,2,"km",[542],"Multiply the 12-second gap by 8.4.",{"correct":544,"incorrect":545},"Right: 8.4 × 12 = **100.8 km**, using exactly the same reasoning as the thunder rule, but with P and S waves instead of light and sound.","Multiply the time gap by 8.4: 8.4 × 12 = 100.8 km.",{"id":547,"type":53,"title":548,"eyebrow":549,"navLabel":550},"e-ch08","Wider context: listening to the Earth itself","Chapter 08","8 Listening to Earth",{"id":552,"type":47,"variant":125,"title":553,"markdown":554},"e-nuance-scale-jump","From a bedroom clap to a whole planet","It can feel like a leap to go from clapping in a room to locating an earthquake hundreds of kilometres away, but the reasoning has not actually changed at all: two signals leave the same event, they travel at different, known speeds, and the gap between their arrivals reveals a distance.\n\nThunder and lightning use light (near-instant) against sound (343 m\u002Fs). Earthquakes use a fast wave against a slower wave, both moving through solid rock at kilometres per second. The scale is a million times bigger; the physics is the same three-line argument every time.",{"id":556,"type":43,"markdown":557},"e-ch8-p1","The thunder rule from Discover, count the seconds and divide, turns out to be one example of a much bigger idea: whenever two signals of **different, known speeds** set off from the same event at the same instant, the time gap between their arrivals tells you how far away that event happened. Earthquakes give the clearest large-scale example on the whole planet.",{"id":559,"type":47,"variant":62,"title":560,"markdown":561},"e-def-p-s-waves","P waves and S waves","An earthquake sends out two main kinds of vibration through the solid Earth. **P waves** (primary waves) are longitudinal, exactly like sound, and travel fastest, roughly **6 kilometres per second** in the upper crust. **S waves** (secondary waves) are transverse, more like a shaken rope, and travel slower, roughly **3.5 kilometres per second**.\n\nBecause P waves outrun S waves, the longer a seismograph station waits between the two arrivals, the further away the earthquake happened.",{"id":563,"type":77,"title":564,"problem":565,"steps":566,"help":571},"e-we-quake-distance","How far away was that earthquake?","A seismograph station records the P wave, then the S wave 20 seconds later. Using approximate crust speeds of 6 km\u002Fs (P) and 3.5 km\u002Fs (S), how far away was the earthquake?",[567,568,569,570],"The distance travelled is the same for both waves; only the time differs, so distance = P-speed × P-time = S-speed × S-time, with S-time = P-time + 20.","Solving this gives a constant multiplier for the time gap: distance = time gap ÷ (1\u002FS-speed − 1\u002FP-speed) = time gap × 8.4 (km per second of gap), found once and reused for every earthquake.","Distance = 8.4 × 20 = **168 km**.","A single station only gives a distance, drawn as a circle on a map around it; seismologists combine circles from at least three stations to pin down the exact location, called triangulation.",{"simplerExplanation":572},"Multiply the time gap between the P and S waves by about 8.4 to get the distance in kilometres, the same style of shortcut as dividing thunder's delay by 3.",{"id":574,"type":47,"variant":125,"title":575,"markdown":576},"e-nuance-one-station","One station finds a distance, not a place","A single seismograph's P-S time gap gives only the **distance** to an earthquake, drawn as a circle of that radius around the station on a map; the earthquake could be anywhere on that circle. A second station's circle crosses the first at (usually) two points, and a third station's circle picks out which one is correct. This triangulation is exactly how earthquake epicentres are located worldwide, using nothing but arrival-time differences, the same style of reasoning as this entire topic's thunder rule, scaled up to a planet.",{"id":578,"type":53,"title":579,"eyebrow":580,"navLabel":581},"e-ch09","Careers built on sound","Chapter 09","9 Careers in sound",{"id":583,"type":90,"caption":584,"columns":585,"rows":589},"e-table-careers","A few careers where understanding sound is central",[586,587,588],"Career","What they do","Sound ideas they use daily",[590,594,598,602,606,610],[591,592,593],"Acoustic (architectural) engineer","Design concert halls, cinemas, offices and classrooms for the right amount of reverberation","Sabine's formula, absorption, reflection",[595,596,597],"Audiologist","Test hearing, fit hearing aids, diagnose hearing loss","Decibels, frequency range, how the ear works",[599,600,601],"Sound engineer \u002F Foley artist","Record, mix and create sound for films, music and games","Microphones, digital sampling, timbre and harmonics",[603,604,605],"Seismologist","Study earthquakes and the Earth's interior using vibration data","P and S waves, wave speed, triangulation",[607,608,609],"Marine bioacoustician","Study how whales, dolphins and fish use sound underwater","Speed of sound in water, echolocation, infrasound",[611,612,613],"Ultrasound (sonography) technician","Operate medical ultrasound scanners safely and accurately","Frequency-versus-detail trade-off, echo timing",{"id":615,"type":455,"prompt":616},"e-reflect-careers","Pick one career from the table above. Write three sentences: what a typical working day might involve, one piece of sound physics from this topic they would use constantly, and one question you would want to ask someone who actually does that job.",{"id":618,"type":47,"variant":203,"title":619,"markdown":620},"e-example-instrument-maker","A career older than any of these titles: the instrument maker","Long before \"acoustics\" was a word, sitar-makers, tabla-makers and the builders of temples with musical pillars were solving exactly these problems by ear and by trial, generation after generation. Skilled Indian instrument-makers today, tuning a tabla's syahi or choosing wood for a veena, are doing genuine applied acoustics, whether or not they use that word for it, carrying forward a craft that modern careers in acoustics grew directly out of.",{"id":622,"type":53,"title":623,"eyebrow":624,"navLabel":625},"e-ch10","Open questions","Chapter 10","10 Open questions",{"id":627,"type":47,"variant":628,"title":629,"markdown":630},"e-open-questions","question","Nobody has fully solved these yet. Pick one and dig in.","- Could a phone app measure a passing vehicle's speed accurately just from the Doppler shift in its recorded horn, without any other sensor? What would make this hard in a noisy street?\n- Musical pillars exist in several South Indian temples. Could similar acoustic engineering be deliberately designed into a modern building today, and would anyone want it?\n- As lossy digital compression (used to make music files smaller) keeps improving, is there a point at which nobody could ever hear the difference from an uncompressed recording? How would you test that fairly?\n- Whales' infrasound calls can apparently travel across whole ocean basins. How does noise from ships affect how far those calls actually carry today?\n- If a hearing aid could shift ultrasound or infrasound down into the range a person can hear, what new things might they notice about their everyday surroundings?",{"id":632,"type":455,"prompt":633},"e-reflect-open","Pick one open question above, or a sound-related open question of your own. Write down one small experiment or piece of research, doable by a school student, that would move it forward even slightly.",{"id":635,"type":53,"title":636,"eyebrow":637,"navLabel":638},"e-ch11","Check what you know","Chapter 11","11 Check yourself",{"id":640,"type":641,"title":642,"terms":643},"e-glossary","glossary","Extend words to keep",[644,647,651,655,658,661,664,667,670,673,677],{"term":171,"meaning":645,"example":646},"The change in frequency an observer hears because of relative motion between a source and the observer.","A siren sounding higher as it approaches, lower as it recedes.",{"term":648,"meaning":649,"example":650},"Redshift \u002F blueshift","Light's Doppler effect: shifted towards red when a source recedes, towards blue when it approaches.","Evidence that most distant galaxies are moving away from Earth.",{"term":652,"meaning":653,"example":654},"Sampling","Measuring an audio signal's height at many evenly spaced instants and storing each as a number.","CD audio samples 44,100 times a second.",{"term":177,"meaning":656,"example":657},"How many samples are taken every second, in hertz.","44,100 Hz for CD-quality audio.",{"term":180,"meaning":659,"example":660},"The rule that a sampling rate must exceed twice the highest frequency to be captured faithfully.","40,000 Hz minimum for 20,000 Hz hearing.",{"term":183,"meaning":662,"example":663},"Sound reaching the inner ear through the skull's bones rather than through the air and ear canal.","Why your recorded voice sounds different from how you hear yourself.",{"term":186,"meaning":665,"example":666},"A fast, longitudinal earthquake wave, like sound, travelling through the solid Earth.","About 6 km\u002Fs in the upper crust.",{"term":189,"meaning":668,"example":669},"A slower, transverse earthquake wave, more like a shaken rope.","About 3.5 km\u002Fs in the upper crust.",{"term":192,"meaning":671,"example":672},"Locating an event by combining distance circles from at least three separate measuring stations.","How seismologists pinpoint an earthquake's epicentre.",{"term":674,"meaning":675,"example":676},"Sonic boom","The single strong crack heard when a source moves faster than sound, breaking the ordinary Doppler picture.","The sound of a supersonic aircraft passing overhead.",{"term":678,"meaning":679,"example":680},"Compression (digital audio)","Storing an almost-identical-sounding file in far less space, usually by discarding sound details hearing barely notices.","Why two files at the same sampling rate can differ greatly in size.",{"id":682,"type":683,"title":684,"questions":685},"e-quiz","quiz","Fourteen questions on Doppler shift, recording and wider contexts",[686,698,711,724,737,750,763,776,789,802,815,828,841,854],{"itemId":687,"prompt":688,"options":689,"correct":140,"why":697},"sound.extend-q-bat-sampling","To record a bat's calls up to 120,000 Hz without losing information, a recorder needs a sampling rate of at least",[690,691,693,695],{"id":134,"label":338},{"id":137,"label":692},"120,000 Hz",{"id":140,"label":694},"240,000 Hz",{"id":143,"label":696},"20,000 Hz","The Nyquist-Shannon theorem needs more than twice the highest frequency: 2 x 120,000 = 240,000 Hz.",{"itemId":699,"prompt":700,"options":701,"correct":137,"why":710},"sound.extend-q-sonic-boom","A sonic boom is best described as",[702,704,706,708],{"id":134,"label":703},"an extremely high-pitched Doppler shift",{"id":137,"label":705},"a single strong pressure wave from a source moving faster than sound itself",{"id":140,"label":707},"a kind of echo",{"id":143,"label":709},"the same as thunder","When a source outruns sound, compressions pile up into one strong wave trailing it in a cone, heard as a single crack rather than a smoothly shifting note.",{"itemId":712,"prompt":713,"options":714,"correct":134,"why":723},"sound.extend-q-doppler-approach","A source approaching a listener is heard at",[715,717,719,721],{"id":134,"label":716},"a higher frequency than its true note",{"id":137,"label":718},"a lower frequency than its true note",{"id":140,"label":720},"exactly its true note",{"id":143,"label":722},"zero frequency","An approaching source compresses its waves ahead of it, so more crests arrive each second: a higher heard frequency.",{"itemId":725,"prompt":726,"options":727,"correct":137,"why":736},"sound.extend-q-doppler-rider","A rider next to their own horn, while the bike moves, hears",[728,730,732,734],{"id":134,"label":729},"the same shift a bystander hears",{"id":137,"label":731},"the horn's true, unshifted frequency",{"id":140,"label":733},"no sound at all",{"id":143,"label":735},"only the receding shift","There is no relative motion between the rider and the horn, so there is no Doppler shift for the rider, whatever speed the bike reaches.",{"itemId":738,"prompt":739,"options":740,"correct":137,"why":749},"sound.extend-q-redshift","Astronomers use redshift to conclude that most distant galaxies are",[741,743,745,747],{"id":134,"label":742},"moving towards Earth",{"id":137,"label":744},"moving away from Earth",{"id":140,"label":746},"not moving at all",{"id":143,"label":748},"closer than nearby stars","Redshift is light's version of a receding source's lower pitch: light shifted towards red means the source is moving away.",{"itemId":751,"prompt":752,"options":753,"correct":137,"why":762},"sound.extend-q-nyquist","CD audio samples at 44,100 Hz mainly because",[754,756,758,760],{"id":134,"label":755},"it is a round number",{"id":137,"label":757},"it exceeds twice the top of human hearing (about 20,000 Hz), satisfying the Nyquist-Shannon theorem",{"id":140,"label":759},"microphones cannot sample any faster",{"id":143,"label":761},"it matches the speed of sound","The theorem requires more than twice the highest frequency of interest: 2 x 20,000 = 40,000 Hz, and 44,100 Hz clears that with a safety margin.",{"itemId":764,"prompt":765,"options":766,"correct":137,"why":775},"sound.extend-q-bone","A recording of your voice sounds different from what you hear yourself speaking mainly because",[767,769,771,773],{"id":134,"label":768},"microphones distort all voices equally",{"id":137,"label":770},"you normally also hear yourself through bone conduction, which a microphone cannot pick up",{"id":140,"label":772},"recordings always play back too fast",{"id":143,"label":774},"your ears are different from a microphone's frequency range","Bone conduction adds extra low-frequency richness only you can hear from inside your own skull; a recording captures only the air-conducted sound everyone else already hears.",{"itemId":777,"prompt":778,"options":779,"correct":134,"why":788},"sound.extend-q-rail","Sound reaches a listener sooner through a long steel rail than through the air alongside it because",[780,782,784,786],{"id":134,"label":781},"steel carries sound much faster than air",{"id":137,"label":783},"air absorbs all sound over long distances",{"id":140,"label":785},"steel amplifies the sound",{"id":143,"label":787},"there is no real difference","Steel carries sound at about 5,960 m\u002Fs against air's 343 m\u002Fs, over seventeen times faster, so the rail path arrives first over a long enough distance.",{"itemId":790,"prompt":791,"options":792,"correct":137,"why":801},"sound.extend-q-p-wave","Compared with an S wave, a P wave from the same earthquake",[793,795,797,799],{"id":134,"label":794},"arrives later, since it is slower",{"id":137,"label":796},"arrives earlier, since it is faster",{"id":140,"label":798},"arrives at exactly the same time",{"id":143,"label":800},"does not reach seismographs at all","P waves travel faster than S waves (roughly 6 km\u002Fs against 3.5 km\u002Fs in the upper crust), so they always arrive first.",{"itemId":803,"prompt":804,"options":805,"correct":137,"why":814},"sound.extend-q-triangulation","A single seismograph station's P-S time gap tells scientists",[806,808,810,812],{"id":134,"label":807},"the exact location of the earthquake",{"id":137,"label":809},"only the distance to the earthquake, as a circle on a map",{"id":140,"label":811},"nothing useful on its own",{"id":143,"label":813},"the earthquake's magnitude directly","One station gives only a distance (a circle of possible locations); at least three stations' circles are combined by triangulation to find the exact location.",{"itemId":816,"prompt":817,"options":818,"correct":140,"why":827},"sound.extend-q-quake-calc","Using distance (km) = 8.4 × the P-S time gap in seconds, a 15-second gap corresponds to a distance of about",[819,821,823,825],{"id":134,"label":820},"15 km",{"id":137,"label":822},"56 km",{"id":140,"label":824},"126 km",{"id":143,"label":826},"840 km","8.4 x 15 = 126 km.",{"itemId":829,"prompt":830,"options":831,"correct":137,"why":840},"sound.extend-q-career","An audiologist's daily work mainly draws on understanding",[832,834,836,838],{"id":134,"label":833},"reverberation time in concert halls",{"id":137,"label":835},"decibels, frequency range and how the ear works",{"id":140,"label":837},"seismic wave speeds",{"id":143,"label":839},"digital sampling rates","Audiologists test and treat hearing, which relies on the decibel scale, the range of human hearing and the workings of the ear.",{"itemId":842,"prompt":843,"options":844,"correct":137,"why":853},"sound.extend-q-sitar-recording","A sitar recording sounds noticeably different played back through cheap earphones instead of good speakers mainly because",[845,847,849,851],{"id":134,"label":846},"the recording changes itself over time",{"id":137,"label":848},"different playback equipment reproduces the recorded frequencies, including harmonics, with different accuracy",{"id":140,"label":850},"sitars cannot be recorded properly at all",{"id":143,"label":852},"earphones always play at a lower sampling rate","Timbre depends on harmonics across a wide frequency range; equipment that reproduces that range less accurately changes how the recorded timbre is heard.",{"itemId":855,"prompt":856,"options":857,"correct":137,"why":866},"sound.extend-q-light-doppler-limit","The everyday sound Doppler formula cannot simply be reused, unmodified, for light travelling near the speed of light because",[858,860,862,864],{"id":134,"label":859},"light has no Doppler effect at all",{"id":137,"label":861},"at such enormous speeds, Einstein's special relativity changes the mathematics needed",{"id":140,"label":863},"light never shows redshift or blueshift",{"id":143,"label":865},"the sound formula already works perfectly for light","The directional idea (approaching shifts higher, receding shifts lower) carries over, but the precise arithmetic for speeds near light's own needs relativity, not the everyday sound formula.",{"id":868,"type":869,"title":870,"points":871},"e-cheat","summary","Cheat sheet",[872,873,874,875,876,877,878,879],"**Doppler effect:** f' = f × v ÷ (v − vₛ) approaching, f' = f × v ÷ (v + vₛ) receding. No relative motion, no shift, whatever the source's own speed.","**Faster sources shift more**: vₛ is a bigger share of v (343 m\u002Fs), changing the fraction more.","**Light shows the same directional idea**: redshift for a receding source, blueshift for an approaching one, used in astronomy and radar\u002Fspeed guns, though the precise maths differs from sound's.","**Digital audio samples a signal**, typically 44,100 times a second, chosen to exceed twice the 20,000 Hz top of human hearing (the Nyquist-Shannon theorem).","**Bone conduction** adds extra bass to the voice you hear from inside your own head; a recording captures only what everyone else already hears.","**A long steel rail carries sound to a listener's ear far sooner than air does**, over 17 times faster, a real, testable effect.","**P and S waves** from an earthquake travel at different speeds (about 6 and 3.5 km\u002Fs); their arrival-time gap gives distance, and three stations' distances triangulate the exact location.","**Careers built on sound** include acoustic engineering, audiology, sound engineering, seismology, marine bioacoustics and medical sonography.",{"id":881,"type":882,"sourceIds":883},"e-sources","sources",[884,885,886,887,888,889,890,891],"sound-britannica-doppler","sound-hyperphysics-speed","sound-wikipedia-44100hz","sound-wikipedia-bone-conduction","sound-usgs-earthquake-locate","sound-ncert-class9-sound","sound-physicsclassroom-sound","sound-hypertextbook-radar",[884,885,886,887,888,889,890,891],"needs_review",{"generatedBy":895,"notes":896},"claude-code","Draft generated locally; pending owner review. Every Doppler shift, sampling calculation and earthquake distance computed and asserted in scratchpad\u002Fsound\u002Fnumbers.py.","0361572d2cc434deb346d651524007fb0e2dd09fc72c256fc627a07087f1fbc5",{"logic:practice":899,"component:match-pairs@1":900,"component:sort-game@1":901,"component:process-steps@1":902,"source:sound-britannica-doppler":903,"source:sound-hyperphysics-speed":904,"source:sound-hypertextbook-radar":905,"source:sound-ncert-class9-sound":906,"source:sound-physicsclassroom-sound":907,"source:sound-usgs-earthquake-locate":908,"source:sound-wikipedia-44100hz":909,"source:sound-wikipedia-bone-conduction":910},"3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","c2f918c426383c50d52939054780add1488f3f282c9d1a965c3a38345ddcb265","838644206b59946eaed329bdfe25d227ac6dbb22279805052fee24175ed3ce31","4cb248d8beea032b112212f11d2a3d10751a3a68da1b26d2ef93b11835467932","a027fae0e45ccfe5b03e0a9336864300932afe79520cfcde96d9f92aed00a482","615e6ca7b7252ccb2523eea00746ef69215849397defe43c853a9fcb617799dd","967196792bc122ee73ed66cefbf9d62ac069b2c688af723b4f4eb52fcdd8174a","b2c9464b8e51c98308ebd63e76a062662ba23f7c7150ac64cc555216cf2ffab8","85166e3b268f56304a2070992c53f631f7e9fa6a0b1864bba5770b13c2471743","1c161fa64b415c06c85cc03facf17983594bcde1c6e94c2470c385ea3f36a7fc",{"state":912,"reviewer":913,"selfReview":914,"reviewedAt":915,"method":916},"approved","The library owner",true,"2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899598377]