Wave Speed, Frequency and Wavelength: What Changes at a Boundary

Q: How do I get the wavelength ratio without knowing the frequency?

Divide the speeds. λ2/λ1 = (v2/f)/(v1/f) = v2/v1, and the frequency cancels because it is the same on both sides. For light that ratio is n1/n2, which is where the usual statement “the wavelength in a medium is λvacuum/n” comes from. It is also the quickest route through most boundary questions: you rarely need the frequency at all.

Send a wave into a second medium and one of the three quantities survives unchanged, one is set by the medium and one has no choice but to move. This page takes a frequency and two media and prints both wavelengths, the ratio between them, and the opposite case — helium, where the wavelength is the fixed one and the pitch is what jumps.

Last updated 11 Nov 2026

Wavelength across a boundary

Frequency plus two media, both wavelengths

Frequency of the sourceIn hertz. 440 Hz is the A that orchestras tune to. The frequency is set by whatever is shaking — a string, a loudspeaker cone, an atom — and it is the one quantity on this page that a boundary cannot touch, because the far side of the boundary is driven by the near side and the two must stay in step. That is the whole content of the page and it is worth being suspicious of: nothing else about the wave is preserved.

First mediumTwelve media, each option carrying its own wave speed, because the engine here has no lookup table and because a student has the speed printed in the question anyway. The three air speeds are not quoted from a table: they are v = √(γRT/M) evaluated at 0, 20 and 30 °C with γ = 7/5, the exact SI gas constant and a dry-air molar mass of 28.96 g/mol, which is why they read 331.32, 343.24 and 349.04 rather than the rounded 331, 343 and 349 you will see elsewhere. The five light speeds are the exact defined c divided by the refractive index in NCERT’s Table 9.1. The sound speeds for water, sea water and steel are the figures in common tabulation and are the three numbers on this page with the weakest provenance — sea water in particular moves with salinity, temperature and depth, and 1500 m/s is a round nominal rather than a measurement of anything.

Second medium, across the boundaryThe medium the wave crosses into. Keep both selections in the same family: a sound speed on one side and a light speed on the other is not a boundary, it is two unrelated problems, and the page says so rather than quietly dividing one by the other. The default pair is air into water, which is the case every textbook uses because the wavelength changes by a factor of more than four.

3.368182mExample

A 440 Hz sound in air at 20 °C, crossing into fresh water at 20 °C

One equation, and the three different jobs the three symbols do

v = f λ · λ₂/λ₁ = v₂/v₁ · f₂ = f₁ · v_air(T) = √(γRT/M) ≈ 331.3 √(1 + T/273.15)

f: the frequency, in hertz. Set by the source and by nothing else. It is the quantity that crosses a boundary unchanged, and the reason is mechanical rather than mathematical: the molecules on the far side are pushed by the molecules on the near side, so they must go up and down the same number of times per second. Nothing in the second medium gets a vote
v: the wave speed, in metres per second. Set by the medium and by nothing else — for a gas by √(γRT/M), for a stretched string by √(T/μ), for light by c/n. Notice what is absent from all three: the frequency. A 20 Hz note and a 20 kHz note travel at the same speed through the same air, which is why an orchestra does not arrive out of order
λ: the wavelength, in metres. The dependent one. It is v/f, so it inherits whatever the medium does to v, and the ratio is exact: λ₂/λ₁ = v₂/v₁. Air into water multiplies it by 4.32; air into steel by 17.4; air into crown glass, for light, divides it by 1.52
T, the gas temperature: the speed of sound in a gas goes as the square root of the ABSOLUTE temperature, so the air figures on this page are three points on one curve rather than three independent constants. Near room temperature the square root is near enough linear and 331.3 + 0.606 T (T in °C) is good to about a tenth of a per cent up to 30 °C. This is the number that makes a nonsense of “the speed of sound is 340 m/s”: on a cold morning it is 331 and in a hot hall it is 349, a spread of 5 per cent
c: the speed of light in vacuum is 299 792 458 m/s exactly, because the metre is defined from it (SI Brochure, 9th edition). That is the one speed on this page with no uncertainty whatever. The speed of sound in air is not like that at all: it is a derived quantity that depends on temperature, on humidity and slightly on the carbon dioxide content

Worked example

A 440 Hz sound in air at 20 °C, crossing into fresh water at 20 °C
THE TWO SPEEDS. Air at 20 °C: v = √(γRT/M) with γ = 1.4, R = 8.314 462 618 J mol⁻¹ K⁻¹ exactly, T = 293.15 K and M = 0.028 96 kg/mol, which gives 343.24 m/s. Fresh water at 20 °C is taken as 1482 m/s from the standard tabulation. The water figure is the weaker of the two: the air one is derived here from exact SI constants, the water one is quoted.
THE WAVELENGTH IN AIR. λ₁ = v₁/f = 343.24/440 = 0.7801 m, or 78.0 cm. Worth holding on to as a sanity anchor: a middle-A sound wave in air is about the length of your arm.
THE FREQUENCY IN WATER. Still 440 Hz. Not 440 times anything, not 440 divided by anything. The water is driven by the air at the interface and has to keep up. This is the step that is wrong on most of the answers you will find to this question, and the whole reason the page prints the far-side frequency as a row of its own.
THE WAVELENGTH IN WATER. λ₂ = v₂/f = 1482/440 = 3.368 182 m. You can get the ratio without touching the frequency at all: 1482/343.24 = 4.3177, and whatever f shortcut worth learning — the wavelength ratio IS the speed ratio.
AND THE CASE THAT GOES THE OTHER WAY. Breathe helium and your voice jumps about an octave and a half. Nothing about your vocal folds changed and nothing about the wavelength changed either: the resonant wavelengths of your vocal tract are set by its GEOMETRY, which is still the same shape. The speed went from 343.24 to 1007.43 m/s, so f = v/λ rose by 2.935×. Here the wavelength is pinned and the frequency moves — the exact opposite of the boundary case above, and the reason it is worth being precise about which quantity is being held fixed.

The same 440 Hz in twelve media — and what it is really telling you

Read the last column and the point of the page appears: the wavelength at a fixed frequency is just the speed, rescaled. Nothing else in the table does any work. The three air rows are the temperature dependence of the speed of sound made concrete — 331.32 on a freezing morning, 349.04 in a hot hall, a 5 per cent spread that a single quoted figure of “340 m/s” hides. The helium row is the one worth arguing with: 2.935 times the speed, so 2.935 times the wavelength at the same pitch — and yet breathing helium RAISES your pitch rather than lengthening anything, because what is actually fixed in your vocal tract is the wavelength and not the frequency. The light rows are all exact c divided by the index in NCERT’s Table 9.1, and they show why nobody speaks of the wavelength of a 440 Hz radio wave: it is longer than a marathon.
Medium	Wave	Speed, m/s	Wavelength at 440 Hz	Against air at 20 °C
Dry air, 0 °C	sound	331.32	0.7530 m	0.965×
Dry air, 20 °C	sound	343.24	0.7801 m	1.000×
Dry air, 30 °C	sound	349.04	0.7933 m	1.017×
Helium, 20 °C	sound	1007.43	2.2896 m	2.935×
Fresh water, 20 °C	sound	1482	3.3682 m	4.318×
Sea water, nominal	sound	1500	3.4091 m	4.370×
Steel, longitudinal	sound	5960	13.5455 m	17.36×
Vacuum	light	299 792 458 exactly	681.35 km	873 429×
Air, n = 1.000293	light	299 704 644	681.15 km	873 173×
Water, n = 1.33	light	225 407 863	512.29 km	656 714×
Crown glass, n = 1.52	light	197 231 880	448.25 km	574 624×
Diamond, n = 2.42	light	123 881 181	281.55 km	360 921×

Which of the three survives a boundary, and why the answer is never the wavelength

v = fλ is one multiplication, and a page whose whole content is one multiplication is worth nobody’s time. What is worth knowing is the asymmetry between the three symbols, because it is the thing that decides every boundary question on the syllabus. The frequency belongs to the SOURCE. The speed belongs to the MEDIUM. The wavelength belongs to neither and is simply the quotient, which is why it is the one that moves.

The mechanical reason the frequency is preserved is worth saying out loud, because stated as a rule it sounds arbitrary. Imagine the boundary as a sheet of molecules shared by both media. Those molecules are being pushed up and down 440 times a second by the air behind them. They are the same molecules that push the water in front of them. A molecule cannot be shaken 440 times a second on one face and 300 times a second on the other, so the water is shaken 440 times a second too. There is no mechanism by which a boundary could change the count. Once that is clear, the usual exam trap stops working: a sound wave passes from air into water; what happens to its frequency? Nothing happens to its frequency. The wavelength multiplies by 4.32.

The speed, meanwhile, has nothing to do with how loud or how high the wave is. For a gas it is √(γRT/M) — temperature and the gas, nothing else. For a stretched string it is √(T/μ), the tension over the mass per unit length, which is why tuning a guitar is a matter of tension and why a thicker string is a lower note at the same tension. For light it is c/n. Not one of those expressions contains the frequency, and that independence is the reason an orchestra is audible as an orchestra: if the speed of sound depended on pitch, the double basses would arrive after the piccolos and the dispersion would grow with distance. It does happen a little for light in glass, which is exactly why a prism makes a spectrum, and it is called dispersion precisely because it is the exception.

Where this goes wrong, almost always, is the helium case run backwards. Breathe helium and your voice goes up. Students who have just learned that frequency survives a boundary conclude that it cannot, and get confused; students who have not learned it conclude that helium somehow raises frequencies. Neither is right. There is no boundary in the helium problem at all. Your vocal tract is a resonator whose resonant WAVELENGTHS are fixed by its length and shape, and the frequency it produces is v/λ for whatever gas is inside it. Fill it with a gas in which sound travels 2.935 times faster and the resonant frequencies rise 2.935 times. The quantity held fixed is the wavelength, so the frequency is the dependent one, and the page prints that row deliberately alongside the boundary row so the two cases cannot be confused.

A note on the numbers rather than the physics. The three air speeds here are derived, not quoted: √(γRT/M) with the exact SI gas constant. That matters because “the speed of sound” is quoted as 330, 331, 340, 343 and 346 m/s in different school papers and all five are defensible at some temperature. If a question gives you a speed, use the one it gives you. If it does not, say which temperature you assumed — examiners mark that. The speed of light needs no such hedging: it is 299 792 458 m/s by definition, exactly, because the metre is defined from it, and it is the only number on this page with no uncertainty at all.

Frequently asked questions

When a wave goes into a new medium, does the frequency or the wavelength change?

The wavelength. The frequency is set by the source and is preserved across the boundary, because the far side of the boundary is physically driven by the near side and the two cannot oscillate at different rates. The speed is set by the new medium. Wavelength is v/f, so it changes by exactly the ratio of the two speeds: air into water multiplies it by 4.32, air into crown glass (for light) divides it by 1.52. If an answer tells you the frequency changed at a boundary, it is wrong.

Why does breathing helium make your voice squeaky if the frequency cannot change?

Because there is no boundary in that problem. Your vocal tract is a resonator and what its geometry fixes is the WAVELENGTH, not the frequency. Sound travels at about 1007 m/s in helium against 343 m/s in air, so f = v/λ at the same fixed λ rises by about 2.94× — roughly an octave and a half. The pitch of your vocal FOLDS barely changes, which is why a helium voice sounds odd rather than simply high: the resonances move and the fundamental does not.

What speed of sound in air should I use, 330, 340 or 343 m/s?

Whichever the question gives you; and if it gives you none, say what you assumed. The speed of sound in a gas is √(γRT/M) and therefore depends on the absolute temperature: 331.32 m/s in dry air at 0 °C, 343.24 at 20 °C, 349.04 at 30 °C. Near room temperature 331.3 + 0.606 T with T in degrees Celsius is good to about a tenth of a per cent. Humidity raises it slightly, pressure alone does not change it at all for an ideal gas, and the frequency never enters.

Does the speed of sound depend on the frequency or the loudness?

Essentially no, and that independence is why music works at a distance. √(γRT/M) contains no frequency and no amplitude term. At very high amplitudes sound does become mildly nonlinear and a shock front travels faster than the ordinary sound speed, which is the physics of a thunder crack rather than of a classroom problem. Light in a transparent medium is a different story: its index varies a little with wavelength, and that dispersion is what a prism exploits.

Is the speed of light really exact?

Yes, by definition rather than by measurement. The SI Brochure, 9th edition, defines the metre by fixing c at 299 792 458 m/s, so the number has no uncertainty and never will — a better experiment now refines the metre, not c. Everything derived from it on this page inherits that: the speed of light in water, 225 407 863 m/s, is exact c divided by an index of 1.33, so its only uncertainty is the index’s.

How do I get the wavelength ratio without knowing the frequency?

Divide the speeds. λ₂/λ₁ = (v₂/f)/(v₁/f) = v₂/v₁, and the frequency cancels because it is the same on both sides. For light that ratio is n₁/n₂, which is where the usual statement “the wavelength in a medium is λ_vacuum/n” comes from. It is also the quickest route through most boundary questions: you rarely need the frequency at all.

Related calculators

References

The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019. The metre is defined by fixing the speed of light in vacuum at c = 299 792 458 m/s exactly, so c carries no uncertainty at all: it is a definition, not a measurement, and every refractive index on these pages divides that exact number. The same brochure fixes the Boltzmann and Avogadro constants exactly, which makes the molar gas constant R = 8.314 462 618 153 24 J mol⁻¹ K⁻¹ exact as well — and that is what the speed-of-sound figures here are derived from rather than copied out of a table.
NCERT, Physics Textbook for Class XI, Part II, Chapter 14, Waves. The chapter number was checked against NCERT’s own answers appendix for Part II (keph2an.pdf, read 11 November 2026), which runs Chapters 8 to 14 and puts Oscillations at 13 and Waves at 14; older printings numbered Waves as 15 and much of the web still says 15. Cited for the syllabus treatment of wave speed, standing waves, beats and the Doppler effect. Citing the chapter is not a claim that any board has approved this page.
NCERT, Physics Textbook for Class XII, Part II, Chapter 9, Ray Optics and Optical Instruments (read 11 November 2026). Section 9.2.1 states the Cartesian sign convention: all distances measured from the pole of the mirror or the optical centre of the lens, distances measured in the same direction as the incident light taken as positive, heights above the axis positive. Equation 9.7 is the mirror equation 1/v + 1/u = 1/f and Equation 9.9 gives m = −v/u; Equation 9.23 is the thin lens formula 1/v − 1/u = 1/f and Equation 9.24 gives m = v/u. Table 9.1 lists refractive indices against air with their critical angles: water 1.33 and 48.75°, crown glass 1.52 and 41.14°, diamond 2.42 and 24.41°.
The speed of sound in air on this page is derived, not quoted: v = √(γRT/M) with γ = 7/5 for a diatomic ideal gas, R the exact SI molar gas constant above, and M = 28.96 g/mol for dry air. That returns 331.32, 343.24 and 349.04 m/s at 0, 20 and 30 °C, which agrees with the figures school texts tabulate to within about 0.01 m/s. The reason for deriving rather than quoting is the brief’s own standard of proof: the tabulated school figure is in a thousand places and citable in almost none of them, while every ingredient of the derivation has a published definition. The one ingredient that is a convention rather than a definition is the dry-air molar mass, which standard-atmosphere documents give as 28.9644 g/mol and the CIPM air-density work as 28.9655; the difference moves the answer by 0.02 m/s and nothing on this page turns on it.
The sound speeds for fresh water (1482 m/s at 20 °C), sea water (1500 m/s nominal) and steel (5960 m/s longitudinal) are the figures in common tabulation. This batch did not reach a primary standards publication for any of the three, so they are presented as tabulated values and not as sourced measurements. The usual reference for pure water is Marczak’s 1997 fit in the Journal of the Acoustical Society of America, which was not read here. Sea water is the weakest: its sound speed moves with temperature, salinity and depth across roughly 1450 to 1550 m/s, and 1500 is a round nominal rather than a property of anything. Steel depends on the alloy and on whether the wave is a bulk longitudinal wave or a thin-bar wave, which differ by about 15 per cent.