Wave Speed: v = fλ

Sing a note underwater and it keeps its pitch. The pitch — the frequency — is set by your vocal cords, and they do not care whether the sound then travels through air or water. But the sound itself speeds up enormously on entering the water, from about 343 metres per second to about 1480. Something has to give, and it is the wavelength: the same note that spans 78 centimetres in air stretches to more than three metres in water. Frequency fixed, speed changed, wavelength adjusted — that single sentence is the whole of this article. It is captured in the most-used equation in all of wave physics:

v = f\lambda

Three symbols, and a strict division of labour between them. The source sets the frequency $f$ : a 440 Hz tuning fork emits 440 waves every second no matter what it is dipped into, and that number never changes as the wave travels or crosses into a new material. The medium sets the speed $v$ : a wave's speed is a property of the stuff it moves through — its stiffness and its density — and nothing the source does can alter it. That leaves the wavelength $\lambda = v/f$ as the quantity that adapts, stretching or shrinking to reconcile the two. Get that hierarchy right — source owns $f$ , medium owns $v$ , $\lambda$ is the adjustable slack — and a whole family of exam questions answers itself: why the frequency of light is unchanged but its wavelength shrinks entering glass, why making a sound louder does not make it faster, why a note keeps its pitch across a boundary. We will also line up the speeds of sound in air, water and steel against the speed of light, and meet a young Scottish engineer who chased a single wave down a canal on horseback to prove that it was the canal, not the boat, that set its speed.

Key takeaways

The one equation: $v = f\lambda$ — wave speed equals frequency times wavelength, equivalently $v = \lambda/T$ .
Division of labour: the source fixes the frequency $f$ ; the medium fixes the speed $v$ ; the wavelength $\lambda = v/f$ adapts to satisfy both.
Crossing into a new medium, the frequency stays the same (the boundary is driven at $f$ ), the speed changes, so the wavelength changes in the same proportion as the speed.
Speed is a property of the medium: $v = \sqrt{\text{elastic property}/\text{inertial property}}$ — string $\sqrt{T/\mu}$ , solid rod $\sqrt{Y/\rho}$ , fluid $\sqrt{B/\rho}$ .
Anchors: A4 (440 Hz) in air (343 m s⁻¹) has $\lambda = 0.78$ m; the same note in water (1480 m s⁻¹) has $\lambda = 3.36$ m — 4.3× longer. FM at 100 MHz has $\lambda = 3.0$ m. Sound in a steel rail runs at about 5.1 km s⁻¹.
Amplitude does not affect speed: a louder sound is not a faster one; a brighter light is not a quicker one.
Key idea: $v = f\lambda$ works because a wave carries exactly one wavelength past any point every period — and since the source owns $f$ and the medium owns $v$ , the wavelength is the quantity forced to adjust whenever a wave changes medium.

The core insight

Because one whole wavelength sweeps past a point in one period, $v = \lambda/T = f\lambda$ ; the source sets the frequency and the medium sets the speed, so on crossing into a new medium $f$ is unchanged, $v$ changes, and the wavelength $\lambda = v/f$ stretches or shrinks in exact proportion to $v$ .

Where this lives today

The division of labour in $v = f\lambda$ runs quietly through modern technology. Sonar and echo-sounding rely on the fixed speed of sound in seawater (about 1500 m s⁻¹): the frequency of the ping is the sonar operator's choice, but the speed is the ocean's, so the time for an echo to return reads directly as a depth. Medical ultrasound picks a frequency (a few megahertz) to trade resolution against penetration, knowing the speed in tissue is fixed near 1540 m s⁻¹, and reconstructs an image from echo timings. Every radio and phone band is chosen through $\lambda = c/f$ : a 100 MHz FM signal has a 3-metre wavelength that sets the length of the antenna on your car, while a 2.4 GHz Wi-Fi wave is 12.5 cm long and fits inside your router. And the ranging of light — from a surveyor's laser to the GPS constellation — turns on the same fixed medium speed $c$ that makes a light wave's frequency, not its speed, the carrier of a colour. Scott Russell's insight, that the channel sets the speed, now sends solitons of light down transoceanic fibres, each pulse a wave of translation carrying a piece of the internet.

The physical picture

Imagine standing at a fixed point as a train of waves rolls past. In one period $T$ — the time for one full oscillation — exactly one complete wave goes by, so the wave pattern has advanced by exactly one wavelength $\lambda$ . Speed is distance over time, so the wave's speed is one wavelength per period: $v = \lambda/T$ . And since frequency is just periods-per-second, $f = 1/T$ , this is the same as $v = f\lambda$ . That is the entire content of the formula — it is a definition of speed, counted in wavelengths. The subtlety, and the reason this simple relation is examined every single year, is who controls which quantity. The frequency belongs to the source. A 440 Hz tuning fork vibrates 440 times a second because of its own stiffness and mass; it launches 440 waves per second into whatever touches it, and that rate is fixed at the moment of creation. The speed belongs to the medium. How fast the disturbance then propagates depends on the stuff it travels through — how stiffly the medium resists being deformed, and how much inertia it has — and the source has no say in it. So when frequency is set by one thing and speed by another, the wavelength is left with no freedom: it is forced to be whatever $v/f$ demands. Change the medium and the speed jumps; the frequency cannot change (the source is still driving at $f$ ); so the wavelength must absorb the entire change. That is why the same note has a short wavelength in air and a long one in water, and why light bends into glass — its wavelength shrinks while its colour (frequency) stays exactly the same.

Figure

A single wave train passing from air into water. In air the speed is 343 metres per second and the wavelength is short; crossing the boundary the frequency stays fixed but the speed jumps to 1480, so the wavelength stretches by the same factor, about 4.3 times. Frequency is owned by the source, speed by the medium, and the wavelength adapts.

The wave-speed relation

v = f\lambda = \frac{\lambda}{T} \qquad (1)

where: $v$ is the wave speed (m s⁻¹), $f$ the frequency (Hz = s⁻¹), $\lambda$ the wavelength (m), and $T = 1/f$ the period (s). Read it as bookkeeping on a single passing wave: one wavelength of distance is covered in one period of time, so multiplying "wavelengths per second" ( $f$ ) by "metres per wavelength" ( $\lambda$ ) gives "metres per second" ( $v$ ). The units confirm it: $\mathrm{s^{-1}} \times \mathrm{m} = \mathrm{m\,s^{-1}}$ . Equation (1) is deceptively even-handed — it looks as though any of the three could be the one that changes. In practice the physics assigns roles. Hold the medium fixed (so $v$ is constant) and raise the frequency, and the wavelength must fall: $\lambda = v/f$ . This is why a high-pitched note has a short wavelength and a bass note a long one, both travelling at the same 343 m s⁻¹ through the same air. Change the medium instead, and it is $v$ that jumps while $f$ holds — the case that fills exam papers.

Speed belongs to the medium

Where does a medium's wave speed come from? Always the same competition: a restoring stiffness that snaps the medium back, fighting an inertia that resists being moved. Every wave-speed formula in physics has the shape

v = \sqrt{\frac{\text{elastic (restoring) property}}{\text{inertial property}}} \qquad (2)

A stiffer medium restores faster, so waves run quicker; a denser, heavier medium is more sluggish, so they run slower. Filling in the specific properties for each case gives the standard family:

medium / wave	speed	restoring stiffness · inertia
stretched string	$v = \sqrt{T/\mu}$	tension $T$ · linear mass density $\mu$
solid rod (longitudinal)	$v = \sqrt{Y/\rho}$	Young's modulus $Y$ · density $\rho$
fluid — liquid or gas (sound)	$v = \sqrt{B/\rho}$	bulk modulus $B$ · density $\rho$
gas (sound, adiabatic)	$v = \sqrt{\gamma P/\rho}$	$\gamma P$ · density $\rho$

The string case, $v = \sqrt{T/\mu}$ , is the subject of the very next article, Speed of a Wave on a Stretched String, and derivation D2 below shows — without any of the dynamics — that it is the only formula the string's ingredients could possibly make. Two numbers worth carrying: sound in a steel rail travels at $v = \sqrt{Y/\rho} = \sqrt{2.0\times10^{11}/7800} = 5064$ m s⁻¹, about 5.1 km s⁻¹ — which is why you hear an approaching train through the rail long before through the air. And notice what is absent from every row: amplitude. Nowhere does how big the wave is enter the speed. A loud sound and a soft one travel at the identical 343 m s⁻¹.

Figure

Wave speed climbs across nine orders of magnitude on a logarithmic scale: sound in air about 343 metres per second, sound in water about 1480, sound in a steel rail about 5064, and light in vacuum three hundred million. Stiffer media carry mechanical waves faster, and light outruns them all.

Crossing a boundary: frequency stays, speed and wavelength change

Here is the rule that earns the most marks. When a wave passes from one medium into another, the boundary between them is driven up and down (or back and forth) by the arriving wave at its frequency, and it re-emits into the second medium at exactly that same frequency — a boundary cannot invent or swallow cycles. So the frequency is unchanged on crossing a boundary. The speed, however, is set by the new medium and is different. With $f$ fixed and $v$ changed, $\lambda = v/f$ must change in the same proportion as the speed:

\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} \qquad (f\ \text{unchanged}) \qquad (3)

A 440 Hz note entering water, where the speed is $1480/343 = 4.31$ times greater, comes out with a wavelength 4.31 times longer. Light entering glass does the reverse: it slows down, so its wavelength shrinks — while its frequency, and hence its colour, is untouched. The commonest mistake in the entire chapter is to "change the frequency" at a boundary. Refuse to: the source owns $f$ , and the boundary passes it along faithfully.

Phase velocity, and a whisper of dispersion

The speed $v = f\lambda = \omega/k$ is more precisely the phase velocity — the speed at which a single crest, a point of fixed phase, travels. In the simple media of this course every frequency travels at the same phase velocity, so a whole pulse (a mix of frequencies) travels at that speed too and keeps its shape. But in a dispersive medium different frequencies travel at slightly different speeds, so a pulse gradually spreads, and its overall envelope moves at a distinct group velocity. Deep-water waves are the classic example — long swells outrun short chop — which is exactly the kind of wave John Scott Russell was chasing. This article stays with non-dispersive media where phase velocity is the whole story; the richer picture is Phase and Group Velocity, a pointer, not a prerequisite.

Two derivations

Derivation AWhy v = fλ — the one-period argument

The relation $v = f\lambda$ is so familiar that it is often just asserted, but it deserves to be seen, because seeing it makes the roles of the three quantities obvious and immunises you against misusing it. The goal is to derive it from the bare meaning of the words "frequency", "wavelength" and "period", with no other physics at all — just the observation that a wave repeats itself in space every wavelength and in time every period. Once built this way, the formula reads not as a rule to memorise but as a statement about counting waves past a point.

Step 1. Fix your attention on a single point in the medium and watch the wave stream past it. By the definition of the period $T$ , one complete cycle of the wave passes that point in a time $T$ .

Step 2. By the definition of the wavelength $\lambda$ , one complete cycle of the wave occupies exactly one wavelength of distance along the direction of travel. So in the time $T$ that one cycle takes to pass, the wave pattern has moved forward by one wavelength $\lambda$ .

Step 3. Speed is distance travelled over time taken, so $v = \lambda/T$ . The wave advances one wavelength every period — nothing more is going on.

Step 4. Finally, frequency is the number of cycles per second, the reciprocal of the period: $f = 1/T$ . Substitute to get $v = \lambda/T = f\lambda$ , equation (1). The derivation lays the division of labour bare: $f$ comes from how often the source completes a cycle, $\lambda$ from how much space a cycle occupies in the medium, and $v$ is simply their product — so fixing any two fixes the third. Verified: v = fλ = λ/T for the anchor cases (343 = 440 × 0.7795) is recomputed in verify_hs2-wave-speed.py.

Go deeper · Derivation B · 1 minWhy the string speed must be √(T/μ) — dimensional analysis

Derivation BWhy the string speed must be √(T/μ) — dimensional analysis

Before the next article derives the string's wave speed from forces, we can discover its form almost for free, using only the requirement that both sides of a physics equation carry the same units. The goal is to show that out of a string's two governing quantities — its tension and its mass per unit length — there is exactly one combination with the units of a speed, namely $\sqrt{T/\mu}$ , so the answer is forced before any dynamics are done. This is a genuine problem-solving technique (dimensional analysis) and it is the reason the string formula could not have looked any other way.

Step 1. List the ingredients a taut string's wave speed could depend on. The restoring agency is the tension $T$ , a force, with dimensions $[T] = \mathrm{M\,L\,T^{-2}}$ (mass × acceleration). The inertia is the linear mass density $\mu =$ mass per unit length, with dimensions $[\mu] = \mathrm{M\,L^{-1}}$ . Amplitude and frequency are not on the list — the speed is a property of the medium alone.

Step 2. Seek a combination $T^{a}\mu^{b}$ whose dimensions are those of a speed, $\mathrm{L\,T^{-1}}$ . Its dimensions are $(\mathrm{M\,L\,T^{-2}})^{a}(\mathrm{M\,L^{-1}})^{b} = \mathrm{M}^{a+b}\,\mathrm{L}^{a-b}\,\mathrm{T}^{-2a}$ .

Step 3. Match the powers of M, L and T to those of a speed ( $\mathrm{M^0\,L^1\,T^{-1}}$ ). Mass: $a + b = 0$ . Time: $-2a = -1$ , so $a = \tfrac12$ . Then $b = -\tfrac12$ . Length check: $a - b = \tfrac12 - (-\tfrac12) = 1$ . ✓

Step 4. Therefore the only possible form is $v \propto T^{1/2}\mu^{-1/2} = \sqrt{T/\mu}$ . Dimensional analysis cannot supply the dimensionless prefactor, but the honest force derivation of the next article shows it is exactly 1, giving $v = \sqrt{T/\mu}$ . Notice the payoff already: doubling the tension multiplies the speed by $\sqrt 2$ , not by 2 — the square root was inevitable the moment tension carried one power of length that density had to cancel. Verified: the exponents a = 1/2, b = −1/2 solving a + b = 0 and −2a = −1 are recomputed in verify_hs2-wave-speed.py.

Limiting cases

Case	Result	Meaning
medium fixed, $f$ doubled	$\lambda$ halves	$v = f\lambda$ with $v$ constant — high notes have short wavelengths
cross into a faster medium ( $v$ up)	$\lambda$ up in proportion, $f$ fixed	air → water: $\lambda$ ×4.3
cross into a slower medium ( $v$ down)	$\lambda$ down, $f$ fixed	light into glass: colour unchanged, wavelength shrinks
amplitude changed	$v$ unchanged	louder ≠ faster; the speed is the medium's, not the source's
stiffer medium ( $Y$ , $B$ or $T$ up)	$v$ up	$v = \sqrt{\text{stiffness}/\text{inertia}}$
denser medium ( $\rho$ or $\mu$ up)	$v$ down	more inertia, slower wave
dispersive medium	each $f$ has its own $v$ ; pulse spreads	phase velocity ≠ group velocity

Common trapLouder is not faster, and the boundary does not change the pitch

Two beliefs must be unlearned. First, amplitude does not affect wave speed. A shout and a whisper reach you at the same 343 m s⁻¹; a bright light and a dim one travel at the same $c$ . Wave speed is fixed by the medium's stiffness and density — $v = \sqrt{\text{stiffness}/\text{inertia}}$ — and the size of the wave appears nowhere in it. So "the sound is louder, therefore faster" and "higher pitch, therefore faster" are both wrong: in a given medium every sound travels at one speed, regardless of loudness or pitch. Second, frequency does not change at a boundary. When a wave enters a new medium, the boundary is driven at the incoming frequency and re-radiates at exactly that frequency — cycles are conserved. It is the wavelength that changes (in proportion to the new speed), never the frequency. And a housekeeping trap: state which speed of sound you are using — $343$ m s⁻¹ at $20\,°\mathrm{C}$ or $331$ m s⁻¹ at $0\,°\mathrm{C}$ — and never mix the two in one problem.

Worked examples

Board level — the wavelength of a musical note (CBSE/ICSE/IGCSE pattern)

The note A4 has frequency 440 Hz. Take the speed of sound in air at 20 °C to be 343 m s⁻¹. (a) Find its wavelength in air. (b) The note an octave higher, A5, has frequency 880 Hz. Find its wavelength in the same air, and comment.

Step 1. Rearrange $v = f\lambda$ for the wavelength: $\lambda = v/f$ . For A4, $\lambda = 343/440 = \boxed{0.7795\ \mathrm{m}} \approx 78$ cm — about the width of a doorway, which is why bass notes bend around corners so easily.

Step 2. For A5, the medium is unchanged so $v$ is still 343 m s⁻¹; only the frequency has changed: $\lambda = 343/880 = \boxed{0.390\ \mathrm{m}}$ .

Step 3. Comment: doubling the frequency (up one octave) halved the wavelength, because $v$ stayed fixed at the air's value and $\lambda = v/f$ . The medium set the speed; the source set the frequency; the wavelength did the adjusting — exactly the hierarchy of this article. Sense-check: $0.7795/0.390 = 2.0$ , the octave ratio, as it must be.

Competitive level — one note in two media, a rail, and a radio wave (JEE-Main/NEET pattern)

A 440 Hz tuning fork is sounded. (a) Find its wavelength in air (343 m s⁻¹) and in water (1480 m s⁻¹), and the ratio of the two. (b) Sound travels along a steel rail as a longitudinal wave; find its speed from $Y = 2.0\times10^{11}$ Pa and $\rho = 7800$ kg m⁻³. (c) An FM station broadcasts at 100 MHz; find the wavelength of its radio wave. In each part, say which quantity the source fixed and which the medium fixed.

Step 1. Part (a). In air: $\lambda_{\text{air}} = 343/440 = 0.7795$ m. In water the frequency is unchanged (the fork still drives at 440 Hz), but the speed is the water's: $\lambda_{\text{water}} = 1480/440 = 3.364$ m. The ratio is $\lambda_{\text{water}}/\lambda_{\text{air}} = 1480/343 = \boxed{4.31}$ — the wavelength is 4.31× longer in water, exactly the ratio of the speeds. Source fixed $f = 440$ Hz; each medium fixed its own $v$ .

Step 2. Part (b). The rail's longitudinal wave speed is $v = \sqrt{Y/\rho} = \sqrt{2.0\times10^{11}/7800} = \sqrt{2.564\times10^{7}} = \boxed{5064\ \mathrm{m\,s^{-1}}} \approx 5.1\ \mathrm{km\,s^{-1}}$ . Here the medium alone fixes $v$ ; no source frequency is involved. This is why an ear to the rail hears the train through the steel seconds before through the air.

Step 3. Part (c). A radio wave travels at the speed of light, $c = 3.0\times10^{8}$ m s⁻¹, and this speed is fixed by the vacuum (the medium): $\lambda = c/f = 3.0\times10^{8}/(100\times10^{6}) = \boxed{3.0\ \mathrm{m}}$ . Source fixed $f = 100$ MHz; the medium (vacuum) fixed $v = c$ . A useful reflex for radio: wavelength in metres $\approx 300/f_{\text{MHz}}$ . Timing note: about 2 minutes. The single trap that sinks people is "changing the frequency" in water — the fork's 440 Hz is fixed, and only $v$ and $\lambda$ move. Write $f$ once at the top and never touch it again; compute each $\lambda$ from that one $f$ and the local $v$ .

Exam traps

"Louder means faster." Amplitude has no effect on wave speed. In a given medium every wave — loud or soft, bright or dim — travels at the one speed set by the medium's stiffness and density. If a question offers "the sound is louder, so it arrives sooner", reject it.

"Higher pitch means faster." Also false, and for the same reason: $v$ is the medium's, not the source's. A piccolo and a tuba, played in the same room, send their sound to you at the identical 343 m s⁻¹. Pitch changes the wavelength, not the speed.

Changing the frequency at a boundary. The single biggest mark-loser. Frequency is conserved across a boundary; speed and wavelength change. When a wave enters water or glass, keep $f$ fixed and let $\lambda = v/f$ absorb the change.

$v = \omega/k$ , not $\omega k$ ; and $\lambda = v/f$ , not $vf$ . Rearrange $v = f\lambda$ carefully — dividing where you should divide. A dimension check ( $\mathrm{m\,s^{-1}} = \mathrm{s^{-1}} \times \mathrm{m}$ ) catches most slips.

Mixing 331 and 343 for the speed of sound. These are the values at $0\,°\mathrm{C}$ and $20\,°\mathrm{C}$ respectively. State which one the question intends and use it throughout — do not compute one wavelength with 343 and another with 331.

Origins · 1 minJohn Scott Russell and how this was discovered

John Scott Russell

1808–1882 · Scotland

In 1834, a young Scottish naval engineer was watching a barge being drawn along the Union Canal near Edinburgh when the boat stopped suddenly — and a smooth, solitary heap of water rolled on ahead of it, keeping its shape. Scott Russell gave chase on horseback, followed it for nearly two miles until it faded, and spent the next decade studying what he called the "wave of translation". His Report on Waves to the British Association (1844) gave its speed as $c = \sqrt{g(h+a)}$ , showing that the speed of such a shallow-water wave is set by the depth of the canal, not by the barge that launched it — a vivid demonstration that the medium, not the source, owns the speed. Long dismissed by contemporaries, he was vindicated in 1895 when Korteweg and de Vries found the equation whose solitary-wave solution — the soliton — is exactly his wave; solitons now carry data down the world's optical fibres.

History

1636

— Marin Mersenne, in Harmonie universelle, makes the first serious measurement of the speed of sound in air and states the three empirical laws of the vibrating string.

1687

— Isaac Newton, in the Principia, gives the first theoretical speed of sound — about 15 % too low, a gap unresolved until Laplace's adiabatic correction of 1816.

1834

— John Scott Russell, riding along the Union Canal near Edinburgh, chases a solitary wave for nearly two miles and finds its speed is set by the canal's depth, not the barge that made it.

1844

— Scott Russell presents his Report on Waves to the British Association, with the shallow-water speed law

c = \sqrt{g(h+a)}

1877

— Lord Rayleigh's The Theory of Sound places the speeds of waves in strings, air and water on one unified footing.

1895

— Diederik Korteweg and Gustav de Vries publish the equation whose solitary-wave solution finally explains Scott Russell's "wave of translation".

Going further

Texts. NCERT Physics Class 11, Ch. 15 "Waves", §15.4 (speed of a travelling wave) · H. C. Verma, Concepts of Physics Vol. 1, "Wave Motion and Waves on a String" · Halliday, Resnick & Walker, "Waves — I" · Cambridge IGCSE 0625 §3.2 and A-Level 9702 topic 7, which state $v = f\lambda$ for all boards.

Provenance. Standard results. Recompute in `verify_hs2-wave-speed.py`: $\lambda_{\text{air}} = 343/440 = 0.7795$ m ( $\approx 78$ cm) and A5 at $343/880 = 0.390$ m; $\lambda_{\text{water}} = 1480/440 = 3.364$ m with the speed ratio $1480/343 = 4.31$ ; the steel-rail speed $\sqrt{2.0\times10^{11}/7800} = 5064$ m s⁻¹ ( $\approx 5.1$ km s⁻¹); the FM wavelength $3.0\times10^{8}/(100\times10^{6}) = 3.0$ m; Scott Russell's canal wave $c = \sqrt{9.8(1.2 + 0.3)} = \sqrt{14.7} = 3.83$ m s⁻¹ $= 13.8$ km h⁻¹; and the dimensional-analysis exponents $a = \tfrac12$ , $b = -\tfrac12$ for $\sqrt{T/\mu}$ .

Prerequisites. Wave Motion: Transverse & Longitudinal — what is travelling · Describing a Wave: y(x,t), Amplitude, Wavelength & Phase — where $v = \omega/k = f\lambda$ comes from. Leads to. Speed of a Wave on a Stretched String — the honest derivation of $v = \sqrt{T/\mu}$ that D2 previews · Energy & Intensity of Waves — what the travelling wave carries.

Go deeper (UG). When a medium is dispersive and different frequencies travel at different speeds, the single crest speed and the speed of a pulse's envelope split apart; this is Phase and Group Velocity in the university electromagnetism track. It is enrichment; every result on this page holds in the non-dispersive media assumed here.

If v = fλ, why can I not just change any one of the three whenever I like?

Because physics, not algebra, decides which two are pinned down. The source fixes the frequency $f$ — a 440 Hz fork makes 440 waves a second, full stop. The medium fixes the speed $v$ — it depends on the stiffness and density of the stuff the wave moves through. With two of the three quantities fixed by physical causes, the wavelength has no choice but to be $\lambda = v/f$ . The equation is true in the sense that any two determine the third; but in a real situation you must ask what is holding each quantity fixed. Almost every mistake in this topic comes from "freeing up" a quantity — usually the frequency — that the physics has actually nailed down.

Why does the frequency stay the same when a wave crosses into a new medium?

Think about the boundary itself, the last layer of the first medium touching the first layer of the second. That boundary is being shaken by the incoming wave, and it shakes at the incoming wave's frequency — say 440 times a second. It cannot do anything else: it is physically attached to the wave arriving. Whatever it re-radiates into the second medium therefore also cycles 440 times a second, because it is the boundary's own motion that launches the transmitted wave. Cycles are conserved — the boundary can neither create nor destroy them. The speed in the new medium is different, so with $f$ pinned the wavelength must change: $\lambda = v/f$ . Frequency is the passport that survives every border crossing.

Does a louder sound really travel at the same speed as a quiet one?

Yes — and it is worth being sure of, because intuition rebels. Loudness is amplitude, the size of the pressure swings; speed is set by how stiff and how dense the air is, which a bigger swing does not change. So a thunderclap and a whisper both cross a room at about 343 m s⁻¹; the thunderclap merely arrives bigger, not sooner. (There is a genuine exception at extreme amplitudes — a shock wave from an explosion or a supersonic jet does travel faster than ordinary sound — but that is a nonlinear regime far outside ordinary waves, and no normal loud sound comes close to it.) For every wave in this course, amplitude and speed are independent.

What is the difference between phase velocity and group velocity?

The phase velocity is the speed of a single crest — a point of constant phase — and it is what $v = f\lambda = \omega/k$ gives. In the simple, non-dispersive media of this course, every frequency travels at the same phase velocity, so a whole pulse (which is a bundle of frequencies) travels at that one speed and keeps its shape. In a dispersive medium, different frequencies travel at slightly different phase velocities, so the pulse gradually spreads out, and the speed of its overall envelope — the speed at which the energy and the information actually travel — is the group velocity, which can differ from the phase velocity. Deep-water waves disperse (long swells outrun short ripples); light in glass disperses (which is why a prism splits colours). It is developed in Phase and Group Velocity; here, phase velocity is all you need.

Why is sound so much faster in steel than in air?

Because steel is enormously stiffer than air, and stiffness is the numerator of the wave-speed formula. Sound speed goes as $\sqrt{\text{stiffness}/\text{density}}$ : steel is far denser than air (which would slow the wave), but it is vastly stiffer (Young's modulus around $2\times10^{11}$ Pa against air's bulk modulus of about $10^{5}$ Pa), and the stiffness wins overwhelmingly. The numbers give about 5064 m s⁻¹ in a steel rail against 343 m s⁻¹ in air — roughly fifteen times faster. That is why, in old films, a character presses an ear to the railway track: the sound of a distant train reaches them through the steel several seconds before it arrives through the air.

The formula gives one speed, but ocean waves of different sizes seem to move at different speeds. Why?

Because deep water is a dispersive medium — a case $v = f\lambda$ still describes at each instant, but where the speed itself depends on the wavelength. Long-wavelength swells travel faster than short-wavelength chop, which is why the smooth, long swell from a distant storm arrives at a beach before the choppy local waves. In such a medium there is no single wave speed; each wavelength has its own, and a mixed group of waves spreads out as it travels (this is dispersion). The simple media in most of this course — a taut string, still air, the vacuum for light — are non-dispersive, so a single $v$ describes them all. Scott Russell's canal wave was special precisely because it fought dispersion and held its shape.