Reflection of Sound: Echo, Reverberation & SONAR

Shout at a cliff face a hundred metres off and, about six-tenths of a second later, the cliff shouts back. That returning voice is an echo — sound reflected off a hard surface and heard as a separate sound — and it hides a small trap that sinks more exam answers than almost anything else in acoustics: the sound had to travel to the cliff and back, two hundred metres, not one. Forget that factor of two and every echo, depth-sounding and SONAR calculation you ever do will be exactly twice wrong. Reflection of sound is simply light's trick performed by pressure waves: a sound wave striking a wall bounces off it, angle in equal to angle out, and if the wall is far enough away the reflected wave arrives late enough to be heard on its own. "Far enough" turns out to be about seventeen metres, because the ear holds onto a sound for roughly a tenth of a second, and a reflection that returns within that window merges with the original instead of standing apart. Closer surfaces do not give a clean echo at all — they give reverberation, the blurred, decaying tail of many overlapping reflections that makes a cathedral boom and a padded studio fall dead. That same "send a pulse, time its echo, halve the distance" idea, pointed downward from a ship, measures the depth of the ocean; pointed at a shoal, it finds fish; pointed into the dark of the North Atlantic in 1912, weeks after the Titanic went down, it became SONAR. This article works through all of it — the echo condition and its seventeen metres, the round-trip formula $d = v t/2$ and its all-important factor of two, and Sabine's reverberation law that turned concert-hall design from guesswork into physics. Ultrasound and the Doppler shift, which push these ideas further, are the next chapter's business; here we master the plain reflection.

Key takeaways

Sound reflects off a hard surface with the angle of incidence equal to the angle of reflection, just like light.
An echo is a reflection heard as a separate sound, possible only if the delay exceeds the ear's persistence of about $0.1\ \mathrm{s}$ .
That sets a minimum reflector distance $d = v t/2 = 340 \times 0.1/2 = 17\ \mathrm{m}$ in air (at $v = 343\ \mathrm{m/s}$ , $17.15\ \mathrm{m}$ ).
Echo-ranging / SONAR: send a pulse, time its round trip $t$ , and the reflector is at $d = v t/2$ — the factor of two because the sound goes there and back.
Underwater use the speed of sound in water, $v \approx 1500\ \mathrm{m/s}$ : a $0.4\ \mathrm{s}$ round trip is a depth of $300\ \mathrm{m}$ ; a $2.0\ \mathrm{s}$ trip, $1500\ \mathrm{m}$ .
Reverberation is the smear of many overlapping reflections; Sabine's law gives its decay time, $T_{60} = 0.161\,V/A$ ( $0.805\ \mathrm{s}$ for a $1000\ \mathrm{m^3}$ room with $A = 200$ ).
Key idea: a reflected pulse tells you a distance through $d = v t/2$ — the factor of two, because sound travels to the reflector and back, is the whole point of echo-ranging.

The core insight

Sound reflects like light, and a reflection heard separately is an echo, which needs a delay of at least $0.1\ \mathrm{s}$ and hence a reflector at least $d = v t/2 \approx 17\ \mathrm{m}$ away; the same round-trip relation $d = v t/2$ — with its factor of two for the there-and-back path — underlies SONAR depth-sounding, while many overlapping reflections give reverberation, timed by Sabine's $T_{60} = 0.161\,V/A$ .

Where this lives today

Echo-ranging is one of the most widely used measurements on Earth. Every ship carries an echo-sounder that maps the seabed with $d = v t/2$ ; fishing fleets find shoals the same way, and the entire charted relief of the ocean floor was built pulse by pulse from reflected sound. Submarines hunt and hide by SONAR. A car's parking sensors and a bat's hunting cry both time reflected pulses to judge distance — biology discovered echo-ranging long before Richardson. Medical imaging sends ultrasonic pulses into the body and reconstructs organs and a beating fetal heart from the echoes, and engineers test bridges and aircraft wings for hidden cracks by the reflections of sound sent through the metal. Meanwhile Sabine's reverberation law is the working tool of every acoustic architect: concert halls are tuned to a rich two-second decay, recording studios and lecture theatres are damped to well under a second, and the difference is $0.161\,V/A$ made concrete in panels, drapes and shaped ceilings. The next chapter pushes reflected sound further still — into the megahertz world of ultrasound imaging and the frequency shift of the Doppler effect that reads a target's speed, not just its distance.

The physical picture

A sound wave meeting a hard, smooth surface obeys the same law of reflection as light: the angle of incidence equals the angle of reflection, measured from the normal to the surface. A wall, a cliff, a metal sheet, the seabed — all send the wave back. Soft, porous materials (curtains, foam, people) absorb much of the wave instead, which is why a furnished room is quieter than a bare one. What you hear from a reflection depends entirely on how long it takes to come back. The ear does not respond instantly and then forget; a sound persists in your hearing for roughly a tenth of a second. If a reflected sound arrives within that $0.1\ \mathrm{s}$ window, it overlaps the original and you hear one slightly prolonged sound. Only if the reflection is delayed by more than about $0.1\ \mathrm{s}$ does it stand apart as a distinct repeat — an echo. Since the sound must travel to the reflector and back, that delay corresponds to a there-and-back path of $v t$ , so the reflector's distance is $d = v t/2$ . This round-trip idea — measure a distance by timing a reflected pulse — is the single most useful thing sound reflection does, and the figure below is its purest form.

Figure

A ship sends a sound pulse straight down to the seabed and times how long the echo takes to return, and because the pulse travels down and back the depth is the speed times the time divided by two

The echo condition and its seventeen metres

Put numbers to the echo condition. Taking the persistence of hearing as $t = 0.1\ \mathrm{s}$ and the speed of sound in air as $v = 340\ \mathrm{m/s}$ , the smallest reflector distance that yields a distinct echo is

d_{\min} = \frac{v\,t}{2} = \frac{340 \times 0.1}{2} = 17\ \mathrm{m}. \qquad (1)

where: $v$ is the speed of sound in air, $t = 0.1\ \mathrm{s}$ the minimum resolvable delay, and the division by $2$ accounts for the round trip. (Using the $20\ \mathrm{^\circ C}$ value $v = 343\ \mathrm{m/s}$ gives $17.15\ \mathrm{m}$ ; the exact figure depends on the temperature and on the tenth-of-a-second rule of thumb.) So a wall closer than about seventeen metres cannot give you a clean echo — the reflection comes back too soon and blurs into the original. This is why you hear echoes across a valley or in front of a distant cliff, but not from the wall of a small room. For a reflector at a known distance, the echo simply arrives after the round-trip time $t = 2d/v$ . A person standing $100\ \mathrm{m}$ from a cliff, clapping once, hears the echo after $t = 2 \times 100/340 = 0.588\ \mathrm{s}$ — comfortably more than $0.1\ \mathrm{s}$ , so it is a crisp, separate clap. Rearranged, this is the ranging formula: time the echo, and the reflector is at $d = v t/2$ .

Echo-ranging and SONAR

Point the idea downward and you can sound the depth of the sea. A ship sends a short pulse of sound straight down; it reflects off the seabed and returns; the depth is $d = v t/2$ , using the speed of sound in water, about $1500\ \mathrm{m/s}$ — four times the speed in air, and the number every depth calculation needs. A round-trip time of $t = 0.4\ \mathrm{s}$ means a depth of $d = 1500 \times 0.4/2 = 300\ \mathrm{m}$ ; a longer trip of $t = 2.0\ \mathrm{s}$ means $d = 1500 \times 2.0/2 = 1500\ \mathrm{m}$ , a kilometre and a half down. The same method, aimed sideways, detects submarines, shoals of fish and shipwrecks; this is SONAR (sound navigation and ranging), whose direct ancestor is Lewis Fry Richardson's 1912 echo-ranging patent, filed within weeks of the Titanic disaster. The one rule you must never break is to use the correct speed for the medium — $1500\ \mathrm{m/s}$ in water, not the $340\ \mathrm{m/s}$ of air.

Reverberation and Sabine's law

When a room is small, the reflections do not stand apart. A sound in a hall bounces off every wall, floor and ceiling, again and again, each bounce a little weaker as some energy is absorbed, until the sound decays into silence. This lingering wash of overlapping reflections is reverberation — not a single echo but a smear of thousands, heard as a decaying tail after the source stops. Too much of it and speech turns to mush; too little and music sounds thin and dead. Wallace Clement Sabine, measuring lecture halls at Harvard around 1900, found the law that governs it. The reverberation time $T_{60}$ — the time for the sound to fall to a millionth of its energy, i.e. by $60\ \mathrm{dB}$ — is

T_{60} = \frac{0.161\,V}{A}, \qquad (2)

where: $V$ is the room's volume in $\mathrm{m^3}$ , $A$ is its total absorption in $\mathrm{m^2}$ of open-window equivalent (sabins), and $0.161$ is Sabine's constant in SI units. A bare, hard room ( $A$ small) rings for a long time; adding soft, absorbing surfaces raises $A$ and cuts $T_{60}$ . A classroom of volume $V = 1000\ \mathrm{m^3}$ with absorption $A = 200$ has $T_{60} = 0.161 \times 1000/200 = 0.805\ \mathrm{s}$ — good for speech. A concert hall of $V = 12000\ \mathrm{m^3}$ with $A = 1000$ has $T_{60} = 0.161 \times 12000/1000 = 1.93\ \mathrm{s}$ — a lush, singing acoustic. Sabine's single equation founded the whole field of architectural acoustics.

Figure

On the left a single clean reflection returns from a distant cliff and is heard as one separate echo, while on the right many overlapping reflections in a room pile up into reverberation, a decaying tail rather than a distinct repeat

Derivation AThe echo condition — why a reflector must be about 17 metres away

The goal is to turn the ear's tenth-of-a-second persistence into the minimum distance for a distinct echo, and to see why a nearer surface gives reverberation instead. The strategy is to demand that the reflected sound arrive at least $0.1\ \mathrm{s}$ after the original and to account, carefully, for the round-trip path — the factor of two that this whole topic turns on.

Step 1. State the perceptual rule. The ear holds a sound for about $t = 0.1\ \mathrm{s}$ . A reflection arriving within that time overlaps the original and is heard as one prolonged sound; a reflection arriving later is heard as a separate echo. So a distinct echo needs a delay of at least $0.1\ \mathrm{s}$ .

Step 2. Relate the delay to the distance. The sound travels from source to reflector and back — a total path of $2d$ for a reflector at distance $d$ — so the delay is $t = 2d/v$ . This is where the factor of two enters: the sound covers the gap twice.

Step 3. Impose the condition $t \ge 0.1\ \mathrm{s}$ : $2d/v \ge 0.1$ , so $d \ge v \times 0.1/2$ . With $v = 340\ \mathrm{m/s}$ , $d_{\min} = 340 \times 0.1/2 = 17\ \mathrm{m}$ (and $17.15\ \mathrm{m}$ at $v = 343$ ).

Step 4. Interpret the two regimes. Beyond $17\ \mathrm{m}$ a single reflection returns late enough to be a clean echo. Inside $17\ \mathrm{m}$ , and especially in a room where many surfaces reflect, the returns arrive too soon and too thickly to separate — they overlap into the decaying smear of reverberation rather than a distinct repeat. The same physics, near or far, gives an echo or a reverberation depending only on the timing. Verified: $d_{\min} = 340 \times 0.1/2 = 17.0\ \mathrm{m}$ and $343 \times 0.1/2 = 17.15\ \mathrm{m}$ are recomputed in verify_hs2-reflection-of-sound.py.

Go deeper · Derivation B · 1 minEcho-ranging: d = v t/2 and the depth of the sea

Derivation BEcho-ranging: d = v t/2 and the depth of the sea

Here we derive the ranging formula and apply it to SONAR, making the factor of two do useful work. The goal is $d = v t/2$ for a reflector found by timing a pulse's round trip, and the SONAR depths that follow. The strategy is simply to note that the measured time is for the whole journey out and back, so the one-way distance is half of $v t$ .

Step 1. Send a pulse and time its return. The pulse travels to the reflector (distance $d$ ) and back (another $d$ ), a total of $2d$ , at the medium's sound speed $v$ . So the round-trip time is $t = 2d/v$ .

Step 2. Solve for the distance: $d = v t/2$ . The factor of two is not optional — it is the difference between the reflector's distance and the total path the sound covered.

Step 3. Apply to depth-sounding in water, where $v \approx 1500\ \mathrm{m/s}$ . A round-trip time $t = 0.4\ \mathrm{s}$ gives $d = 1500 \times 0.4/2 = 300\ \mathrm{m}$ . A longer $t = 2.0\ \mathrm{s}$ gives $d = 1500 \times 2.0/2 = 1500\ \mathrm{m}$ .

Step 4. Note the medium trap. These depths used the speed of sound in water; using the air value $340\ \mathrm{m/s}$ would underestimate the depth more than fourfold. Echo-ranging always demands the sound speed of the medium the pulse actually travels through — the formula is the easy part, choosing $v$ is where the error lives. Verified: $d = 1500 \times 0.4/2 = 300\ \mathrm{m}$ and $1500 \times 2.0/2 = 1500\ \mathrm{m}$ , and the cliff echo time $2 \times 100/340 = 0.588\ \mathrm{s}$ , are recomputed in verify_hs2-reflection-of-sound.py.

Limiting cases

case	result	meaning
reflector at $d < 17\ \mathrm{m}$ (air)	no distinct echo	reflection returns within $0.1\ \mathrm{s}$ ; blurs into the original
reflector at exactly $17\ \mathrm{m}$	echo just resolvable	the threshold $d = v t/2$ at $v = 340$ , $t = 0.1\ \mathrm{s}$
many surfaces, small room	reverberation	overlapping reflections, a decaying tail (Sabine's $T_{60}$ )
round trip in water, $t = 0.4\ \mathrm{s}$	depth $300\ \mathrm{m}$	SONAR with $v \approx 1500\ \mathrm{m/s}$
absorption $A$ increased	$T_{60}$ falls	soft surfaces deaden a room ( $T_{60} = 0.161V/A$ )
perfectly absorbing walls	no reflection at all	an anechoic chamber: no echo, no reverberation

Common trapThe factor of two, and the speed of the right medium

Two mistakes wreck echo problems, and both are about being careless with the round trip. First, the factor of two: the sound travels to the reflector and back, so the reflector is at $d = v t/2$ , never $d = v t$ . If you compute a cliff distance or a sea depth as "speed times time," you have doubled it. Always ask whether the time you were given is the round-trip time (halve it) or a one-way time (do not). Second, the speed of the medium: SONAR in the sea uses $v \approx 1500\ \mathrm{m/s}$ , the speed of sound in water, not the $340\ \mathrm{m/s}$ of air — get this wrong and every depth is out by more than a factor of four. And keep an echo distinct from reverberation: an echo is one clean, delayed repeat from a far surface; reverberation is the overlapping smear from many near ones. They are the same reflection physics separated only by timing and geometry.

Worked examples

Board level — the minimum echo distance and a cliff (CBSE, ICSE and IGCSE pattern)

The ear can just resolve two sounds separated by $0.1\ \mathrm{s}$ . Taking the speed of sound in air as $340\ \mathrm{m/s}$ , (a) find the minimum distance to a wall for a distinct echo. (b) A person stands $100\ \mathrm{m}$ from a cliff and claps once; after how long do they hear the echo?

Step 1. Minimum echo distance, equation (1): the sound must make a round trip in at least $0.1\ \mathrm{s}$ , so $d_{\min} = v t/2 = 340 \times 0.1/2 = \boxed{17\ \mathrm{m}}$ . (At the warmer $v = 343\ \mathrm{m/s}$ this is $17.15\ \mathrm{m}$ .) A wall nearer than this gives no clean echo.

Step 2. For the cliff, the sound travels $100\ \mathrm{m}$ out and $100\ \mathrm{m}$ back — $200\ \mathrm{m}$ in total — so the time is $t = 2d/v = 2 \times 100/340 = \boxed{0.588\ \mathrm{s}}$ .

Step 3. Check the factor of two both times. In part (a) we divided by two to get the one-way distance from the round-trip time; in part (b) we multiplied the one-way distance by two to get the round-trip path. Whenever a problem mentions a reflection, decide first whether you are given a one-way or a round-trip quantity — that decision is the whole difficulty.

Competitive level — SONAR depth and concert-hall reverberation (NEET and JEE pattern)

(a) A ship's SONAR sends a pulse to the seabed and receives the echo $0.4\ \mathrm{s}$ later; the speed of sound in seawater is $1500\ \mathrm{m/s}$ . Find the depth. Repeat for a $2.0\ \mathrm{s}$ echo. (b) Using Sabine's law $T_{60} = 0.161\,V/A$ , find the reverberation time of a classroom with $V = 1000\ \mathrm{m^3}$ , $A = 200$ , and of a concert hall with $V = 12000\ \mathrm{m^3}$ , $A = 1000$ .

Step 1. SONAR depth, $d = v t/2$ with the water speed: $d = 1500 \times 0.4/2 = \boxed{300\ \mathrm{m}}$ . The $0.4\ \mathrm{s}$ is the round-trip time, so it is halved.

Step 2. For the $2.0\ \mathrm{s}$ echo: $d = 1500 \times 2.0/2 = \boxed{1500\ \mathrm{m}}$ , a depth of a kilometre and a half. Note that using the air speed $340\ \mathrm{m/s}$ would have given a laughable $68\ \mathrm{m}$ — always the medium's own speed.

Step 3. Classroom reverberation, equation (2): $T_{60} = 0.161 \times 1000/200 = \boxed{0.805\ \mathrm{s}}$ — short enough for crisp speech.

Step 4. Concert hall: $T_{60} = 0.161 \times 12000/1000 = \boxed{1.93\ \mathrm{s}}$ — a long, rich decay that flatters music. The hall reverberates more than twice as long as the classroom, exactly what its bigger volume and lower absorption predict. Timing note: about two minutes for both parts. For SONAR, write $d = vt/2$ and check you have used the water speed before anything else. For Sabine, the arithmetic is trivial — $0.161 \times V/A$ — so the only slip is mixing up which number is $V$ and which is $A$ ; the volume is always the big one.

Exam traps

Forgetting the factor of two. The sound goes to the reflector and back, so $d = v t/2$ for a round-trip time, never $d = v t$ . Half of all wrong echo, cliff and SONAR answers are exactly double the right one. Decide up front whether a stated time is one-way or round-trip.

Using the air speed underwater (or vice versa). SONAR in water uses $v \approx 1500\ \mathrm{m/s}$ ; sound in air is $340\ \mathrm{m/s}$ . Plugging the wrong speed misjudges the distance by more than fourfold. The medium the pulse travels through fixes the speed.

Confusing an echo with reverberation. An echo is one distinct, delayed reflection from a far surface (beyond $\sim 17\ \mathrm{m}$ ). Reverberation is the overlapping smear of many reflections in an enclosed space, heard as a decaying tail, not a separate repeat. A question about a "hall's acoustics" wants reverberation and Sabine's law, not the echo formula.

Misreading the echo condition as a distance rule alone. The real requirement is a time gap of at least about $0.1\ \mathrm{s}$ ; the $17\ \mathrm{m}$ figure is what that time becomes in air. In water, where sound is far faster, the same $0.1\ \mathrm{s}$ corresponds to a much larger distance — so "an echo needs $17\ \mathrm{m}$ " is an air-only statement.

Swapping $V$ and $A$ in Sabine's law. $T_{60} = 0.161\,V/A$ : volume on top, absorption on the bottom. A bigger room reverberates longer (more $V$ ); more soft absorbing surface reverberates less (more $A$ ). Getting the ratio upside down inverts the physics.

Board scope. Echo, reflection, reverberation and echo-method distance-finding are CBSE, ICSE, NEET and IGCSE material; IGCSE 0625 in particular owns the echo-method distance calculation and sonar-by-echo. This topic is not on A-Level (CIE 9702), which does not examine echo, reverberation or SONAR. Ultrasound proper and the Doppler effect belong to the next chapter (the following lane) — they extend echo-ranging and add the moving-source frequency shift, and are only pointed to here, not developed.

Origins · 1 minLewis Fry Richardson and how this was discovered

Lewis Fry Richardson

1881–1953 · England

Lewis Fry Richardson was an English mathematician and physicist best remembered for the first attempt at numerical weather forecasting and for his mathematical studies of the causes of war, but his tie to this article is direct and early. In 1912, within weeks of the Titanic striking an iceberg, Richardson filed British patents for detecting distant objects by the echo of sound — in air and, crucially, in water — timing a reflected pulse to find the range of an obstacle. That is precisely the $d = v t/2$ of echo-ranging, and it is the conceptual ancestor of depth-sounding, fish-finding and submarine-detecting SONAR, which Paul Langevin and Constantin Chilowsky turned into working ultrasonic devices during the First World War. A Quaker and lifelong pacifist, Richardson was uneasy that his ideas served naval warfare; but the simple, powerful notion — send a pulse, time its echo, halve the distance — is his, and it is exactly this article.

History

1490

— Leonardo da Vinci notes that a tube placed in the water lets a listener hear distant ships, an early hint of underwater sound detection.

1826

— Jean-Daniel Colladon and Charles Sturm measure the speed of sound in water across Lake Geneva, the number every echo-depth calculation needs.

1900

— Wallace Clement Sabine publishes the first quantitative reverberation law, T60 = 0.161 V/A, founding architectural acoustics.

1912

— Lewis Fry Richardson files British patents for echo-ranging in air and water, weeks after the Titanic sinks.

1917

— Paul Langevin obtains the first echoes from a working piezoelectric ultrasonic device, realising Constantin Chilowsky's 1915 proposal — the origin of sonar.

Going further

Texts. NCERT Physics Class 9, Ch. 12 and Class 11, Ch. 15 ("Waves"), reflection of sound, echo and reverberation · H. C. Verma, Concepts of Physics Vol. 1, "Sound Waves," reflection and echo · Cambridge IGCSE 0625 (reflection of sound, echo method, sonar by echo).

Provenance. Standard results. Recompute: $d_{\min} = 340 \times 0.1/2 = 17.0\ \mathrm{m}$ (and $343 \times 0.1/2 = 17.15\ \mathrm{m}$ ); cliff echo $2 \times 100/340 = 0.588\ \mathrm{s}$ ; SONAR depths $1500 \times 0.4/2 = 300\ \mathrm{m}$ and $1500 \times 2.0/2 = 1500\ \mathrm{m}$ (and the air-speed blunder $340 \times 0.4/2 = 68\ \mathrm{m}$ ); Sabine $0.161 \times 1000/200 = 0.805\ \mathrm{s}$ and $0.161 \times 12000/1000 = 1.93\ \mathrm{s}$ . All are checked in verify_hs2-reflection-of-sound.py.

Prerequisites. Sound as a Longitudinal Mechanical Wave — the wave that reflects, and the speed $v$ that turns a delay into a distance. Leads to. the university-track Elastic Waves in Solids, where reflection at a boundary is treated through acoustic impedance.

Go deeper (UG). How much of a sound reflects versus transmits at a boundary is governed by the acoustic impedance mismatch between the two media, developed in Elastic Waves in Solids in the university mechanics track — the quantitative story behind "hard surfaces reflect, soft ones absorb." Enrichment only; nothing here depends on it.

Why does an echo need the reflector to be at least about 17 metres away?

Because the ear cannot separate two sounds that arrive less than about a tenth of a second apart, and $0.1\ \mathrm{s}$ is how long a round trip to a $17\ \mathrm{m}$ wall takes. The reflected sound must travel to the reflector and back, a distance of $2d$ , at $340\ \mathrm{m/s}$ ; setting that time equal to $0.1\ \mathrm{s}$ gives $d = 340 \times 0.1/2 = 17\ \mathrm{m}$ . Nearer than that, the echo returns within the ear's persistence and merges with the original into one slightly prolonged sound; farther, it stands apart as a distinct repeat. The precise figure shifts a little with air temperature (it is $17.15\ \mathrm{m}$ at $343\ \mathrm{m/s}$ ) and with exactly how fine your hearing is, but seventeen metres is the standard textbook value.

What is the real difference between an echo and reverberation?

Both are reflected sound, and the difference is timing and multiplicity. An echo is a single reflection from a distant surface, delayed enough (more than about $0.1\ \mathrm{s}$ , so more than about $17\ \mathrm{m}$ away in air) to be heard as a separate repeat of the original — the "hello… hello" off a cliff. Reverberation is the overlapping wash of many reflections in an enclosed space: the sound bounces off every surface repeatedly, each bounce weaker, and you hear not a distinct repeat but a smooth decaying tail after the source stops. A cathedral reverberates; a canyon echoes. Sabine's law $T_{60} = 0.161\,V/A$ quantifies reverberation by giving the time for that tail to fade by $60\ \mathrm{dB}$ , whereas the echo condition is about a single well-separated reflection.

Why must I use 1500 m/s for SONAR instead of 340 m/s?

Because SONAR pulses travel through water, and sound goes about four times faster in water than in air — roughly $1500\ \mathrm{m/s}$ versus $340\ \mathrm{m/s}$ . The echo-ranging formula $d = v t/2$ needs the speed in the medium the sound actually crosses, so a sea-depth calculation must use the water value. If you carelessly plug in the air speed, every depth comes out more than four times too small: a $0.4\ \mathrm{s}$ echo would look like $68\ \mathrm{m}$ instead of the true $300\ \mathrm{m}$ . The lesson generalises: whenever you time a reflected pulse, identify the medium first and use its own sound speed. This is exactly why Colladon and Sturm's 1826 measurement of the speed of sound in water mattered so much for later depth-sounding.

Where does the 0.161 in Sabine's law come from, and what are its units?

The constant $0.161$ (in SI, with $V$ in cubic metres and $A$ in square-metre sabins) bundles together the speed of sound and some geometry of how sound energy decays in a room after many reflections; Sabine found it empirically from measurements of real halls around 1900 and it was later derived from the physics of a diffuse sound field. Its job is to convert a room's volume-to-absorption ratio into a time. A larger volume $V$ means the sound travels farther between bounces, so it takes longer to decay — hence $V$ on top. A larger absorption $A$ (more soft, sound-eating surface) removes energy faster — hence $A$ on the bottom. The result $T_{60}$ is the time for the sound to fall by $60\ \mathrm{dB}$ , a factor of a million in energy, which is the standard measure of how "live" a room sounds.

Does sound get weaker each time it reflects?

Yes. No real surface is a perfect mirror for sound; every reflection absorbs some of the wave's energy (and some is scattered or transmitted through the surface), so each successive reflection is a little quieter. Hard, smooth materials like stone or glass reflect most of the sound and absorb little, which is why bare halls reverberate for a long time; soft, porous materials like curtains, carpet and acoustic foam absorb a large fraction on each hit, which is why furnished and padded rooms fall quiet quickly. This progressive weakening is exactly what makes reverberation a decaying tail rather than an endless one, and it is what Sabine's absorption term $A$ measures: the more absorbing surface a room has, the faster the reflected sound dies away.

How is this different from the ultrasound and Doppler topics coming next?

This article is about plain reflection: a pulse bounces off a surface, and by timing its round trip you find a distance, $d = v t/2$ . The next chapter builds on it in two ways. Ultrasound is simply sound above $20\ \mathrm{kHz}$ — too high to hear, but reflected and timed by exactly the same $d = v t/2$ ; its short wavelength lets it image fine detail, which is why medical scanners and industrial flaw-detectors use it. The Doppler effect adds something new: when the reflector is moving, the echo comes back at a shifted frequency, and that shift measures the target's speed rather than its distance — the principle of Doppler radar and of the fetal-heartbeat monitor. Both extend the reflection physics here, and both are developed properly in the following lane; this article deliberately stops at the stationary-reflector, distance-only case.