About this calculator The wave speed calculator solves v = f λ v = f\lambda v = f λ : given a frequency f f f and a wavelength λ \lambda λ it finds the speed v v v at which the wave propagates; any two of the three quantities determine the third—how long the sound wave from a 440 Hz tuning fork is in air, how many metres a 100 MHz FM broadcast wavelength spans, how many centimetres Wi‑Fi's 2.4 GHz corresponds to, and how fast a 150 m ocean swell travels.
The equation applies to every periodic wave : sound, water waves, waves on a string, seismic waves, radio and visible light. It is not a law peculiar to one kind of wave but a kinematic description of “a periodic pattern on the move”: in one period the wave advances by one wavelength. The question with real physical content is what determines each of the three quantities : the speed is set by the medium, the frequency by the source, and the wavelength is the result of dividing the two. That is the main thread of this page and the key to refraction, dispersion, musical instruments and antenna dimensions.
What it does not do: it does not compute the period (T = 1 / f T = 1/f T = 1/ f , one hand-calculation); it does not derive the speed from medium parameters (temperature, tension, refractive index)—the speed must be entered as a known quantity; it does not distinguish phase velocity from group velocity (they differ in dispersive media; see Principles); it does not handle the Doppler effect; and it does not convert units (convert kHz, MHz, GHz and nm to Hz and m first).
Worked examples All four examples were recomputed by the engine's phys-wave compute routine; the display convention is at most 8 decimal places with trailing zeros removed, in units Hz, m and m/s.
Example 1: a tuning fork—the defaults are the speed of sound
Inputs: f = 440 f = 440 f = 440 , λ = 0.78 \lambda = 0.78 λ = 0.78 . v = 440 × 0.78 = 343.2 v = 440 \times 0.78 = 343.2 v = 440 × 0.78 = 343.2 . Interface shows: wave speed v 343.2 . This is the speed of sound in dry air at 20 °C (the empirical formula 331.3 1 + θ / 273.15 331.3\sqrt{1 + \theta/273.15} 331.3 1 + θ /273.15 gives 343.2 at 20 °C). The tuning fork vibrates 440 times per second, each time pushing out a 0.78 m compression pattern, and the front of the pattern advances 440 × 0.78 = 343.2 440 \times 0.78 = 343.2 440 × 0.78 = 343.2 m every second. The period is T = 1 / 440 = 2.27 ms T = 1/440 = 2.27\ \text{ms} T = 1/440 = 2.27 ms .
Figure 1: the sound wave of Example 1. In each period T the wave advances one wavelength λ, so v = λ/T = fλ
Example 2: how long is a sound wave the ear can hear?
Solving for wavelength. Clear λ \lambda λ and enter v = 343 v = 343 v = 343 , f = 440 f = 440 f = 440 : λ = 343 / 440 = 0.77954545 \lambda = 343/440 = 0.77954545 λ = 343/440 = 0.77954545 . Interface shows: wavelength λ 0.77954545 —differing from the default 0.78 in the third place because 0.78 is a rounded value.
The lower hearing limit f = 20 f = 20 f = 20 : λ = 343 / 20 = 17.15 \lambda = 343/20 = 17.15 λ = 343/20 = 17.15 . Interface shows: 17.15 —longer than a classroom. The upper limit f = 20000 f = 20000 f = 20000 : λ = 0.01715 \lambda = 0.01715 λ = 0.01715 . Interface shows: 0.01715 , i.e. 1.7 cm. Three orders of magnitude of frequency correspond to three orders of magnitude of wavelength, because v v v is fixed and λ ∝ 1 / f \lambda \propto 1/f λ ∝ 1/ f —an inverse proportion . Bass wavelengths are much larger than a doorway and bend around obstacles (diffraction), which is why through a wall you hear mainly drums and bass; treble wavelengths of a few centimetres travel in straight lines like light and are blocked. Large bass drivers and small tweeters have the same explanation.
Solving for frequency: a tube open at both ends and 0.65 m long has a fundamental wavelength of twice the tube length, λ = 1.3 \lambda = 1.3 λ = 1.3 . Clear f f f and enter v = 343 v = 343 v = 343 , λ = 1.3 \lambda = 1.3 λ = 1.3 : f = 343 / 1.3 = 263.84615385 f = 343/1.3 = 263.84615385 f = 343/1.3 = 263.84615385 . Interface shows: frequency f 263.84615385 —close to middle C (261.6 Hz). Flutes, pan pipes and pipe organs all choose pitch by length in this way.
Example 3: radio, Wi‑Fi and visible light
The speed of an electromagnetic wave in vacuum is c = 299 792 458 m/s c = 299\,792\,458\ \text{m/s} c = 299 792 458 m/s (a defined value). FM broadcast at 100 MHz: f = 100000000 f = 100000000 f = 100000000 , λ = 2.99792458 \lambda = 2.99792458 λ = 2.99792458 : v = 299792458 v = 299792458 v = 299792458 . Interface shows: wave speed v 299792458 . Conversely, the wavelength at 100 MHz is exactly 3 m, so an FM half-wave dipole is about 1.5 m.
Wi‑Fi at 2.4 GHz: clear λ \lambda λ and enter v = 299792458 v = 299792458 v = 299792458 , f = 2400000000 f = 2400000000 f = 2400000000 : λ = 0.12491352 \lambda = 0.12491352 λ = 0.12491352 . Interface shows: wavelength λ 0.12491352 —12.5 cm, almost identical to a microwave oven (2.45 GHz, 12.2 cm), which is why microwave ovens interfere with Wi‑Fi and why router antennas are only a few centimetres long.
Visible light: f = 500000000000000 f = 500000000000000 f = 500000000000000 (5 × 10 14 5 \times 10^{14} 5 × 1 0 14 Hz): λ = 299792458 / 5 × 10 14 = 6 × 10 − 7 \lambda = 299792458 / 5 \times 10^{14} = 6 \times 10^{-7} λ = 299792458/5 × 1 0 14 = 6 × 1 0 − 7 . Interface shows: 6e-7 , i.e. 600 nm, orange light. The frequency range of visible light is about 4 × 10 14 4 \times 10^{14} 4 × 1 0 14 –7.5 × 10 14 7.5 \times 10^{14} 7.5 × 1 0 14 Hz, with wavelengths of 750–400 nm.
Example 4: ocean swell and sonar
Ocean waves : a swell with a 10 s period (f = 0.1 f = 0.1 f = 0.1 ) and a 150 m wavelength. f = 0.1 f = 0.1 f = 0.1 , λ = 150 \lambda = 150 λ = 150 : v = 15 v = 15 v = 15 . Interface shows: wave speed v 15 (54 km/h). The speed of a deep-water wave is set by its wavelength (v = g λ / 2 π v = \sqrt{g\lambda / 2\pi} v = g λ /2 π , giving 15.3 m/s for 150 m), consistent with the example. Note that this is the speed of the waveform ; the water itself only circles in place. Surfers chase the waveform.
Sonar : the speed of sound in water is about 1500 m/s and the transducer works at a wavelength of 3 cm. Clear f f f and enter v = 1500 v = 1500 v = 1500 , λ = 0.03 \lambda = 0.03 λ = 0.03 : f = 50000 f = 50000 f = 50000 . Interface shows: frequency f 50000 —50 kHz, a common fish-finder band. The wavelength sets the smallest target that can be resolved, so seeing small fish requires a higher frequency and a shorter wavelength, at the cost of stronger absorption by the water and a shorter range.
Crossing into a new medium changes v and λ, not f
The 440 Hz sound wave of Example 2 entering water from air keeps its frequency of 440 (the source vibrates 440 times per second and the water surface does not “swallow” the vibration), while the speed changes from 343 to 1500 and the wavelength from 0.78 m to 1500 / 440 = 3.41 1500/440 = 3.41 1500/440 = 3.41 m. Light entering water from air (refractive index 1.333) behaves the same way: the frequency is unchanged and the speed and wavelength shrink to three quarters. The colour we see is set by the frequency, so red stays red under water.
The inputs of Example 1 can be entered directly into the panel to reproduce it; change the frequency to 880 (an octave up) and see that the wavelength must halve while the speed is unchanged (clear the wavelength, enter 343.2, then change the frequency); change the wavelength to 3.41 to see the speed of 440 Hz in water.
Principle and derivation One wavelength advanced per period
Take a periodic wave travelling to the right and watch one crest. After one period T T T the source completes a full vibration and the next crest takes the place of the first—the whole pattern has shifted to the right by exactly one wavelength λ \lambda λ . The speed of the pattern is therefore v = λ / T v = \lambda / T v = λ / T . The frequency is the reciprocal of the period, f = 1 / T f = 1/T f = 1/ T ; substituting gives v = f λ v = f\lambda v = f λ .
The derivation uses no assumption about the medium or the type of wave, only periodicity and a translating pattern. That is why it holds for sound, light, water and string waves alike—and also why, being purely kinematic, it cannot tell you what v v v is .
What determines what
The medium sets the speed. For a string wave v = F / μ v = \sqrt{F/\mu} v = F / μ (tension over linear density); for sound in air v = γ R T / M v = \sqrt{\gamma R T / M} v = γ R T / M , depending only on temperature and gas composition, not on pitch or loudness; for electromagnetic waves in vacuum v v v is always c c c , and in a medium of refractive index n n n it is c / n c/n c / n . Nothing the source does can change any of these.
The source sets the frequency. The 440 Hz of a tuning fork is its own natural frequency; the 100 MHz of a broadcast station is the oscillator frequency of the transmitter. When a wave enters a new medium the vibrations on the two sides of the boundary must stay in step, so the frequency cannot jump.
The wavelength is a result. λ = v / f \lambda = v/f λ = v / f : the medium supplies v v v , the source supplies f f f , and the wavelength follows. The same tuning fork produces sound of three different wavelengths in air, water and steel.
These three rules explain why antenna dimensions change with band (Example 3), why instruments choose pitch by length (Example 2), and why the frequency is unchanged on refraction (the note).
Units: the hertz and the second
The hertz is an SI derived unit, 1 Hz = 1 s − 1 1\ \text{Hz} = 1\ \text{s}^{-1} 1 Hz = 1 s − 1 , adopted by the 11th General Conference on Weights and Measures in 1960 to replace “cycles per second”. The second itself has been defined since 1967 by the caesium-133 ground-state hyperfine transition at exactly 9 192 631 770 Hz—frequency is today's most precisely measured physical quantity—and the speed of light c c c was in turn defined as exactly 299 792 458 m/s in 1983, with the metre derived from c c c and the second. That f λ f\lambda f λ in Example 3 equals c c c exactly is not a coincidence but the definition of the metre.
Dispersion: when the wave speed depends on frequency
The statement that “the medium sets the speed” implicitly assumes it is independent of frequency. This is almost exactly true for sound in air and light in vacuum, but not in dispersive media : violet light is slower than red in glass (hence prism spectra), and long waves are faster than short ones in deep water (the v = g λ / 2 π v = \sqrt{g\lambda/2\pi} v = g λ /2 π of Example 4, which is why long swells from a distant storm arrive first). Under dispersion v = f λ v = f\lambda v = f λ still holds point by point, but the v v v computed is the phase velocity —the speed of a single-frequency pattern; a wave packet made of many frequencies travels as a whole at the group velocity , and the two can differ greatly (for deep-water waves the group velocity is only half the phase velocity). This calculator computes phase velocity only.
A little history
In 1636 Mersenne summarised the relations between a string's frequency, length, tension and linear density from experiments with vibrating strings—the earliest quantitative law of waves. In 1687 Newton computed the speed of sound in air as about 979 feet per second (298 m/s) under an “isothermal” assumption, about 15% below measurement; in 1816 Laplace pointed out that the compression of sound is adiabatic, and only after including the γ \sqrt{\gamma} γ factor did theory match experiment—the first precise confirmation that the speed is set by the properties of the medium. In 1887 Hertz measured the wavelength of radio waves with standing waves and multiplied by the known oscillator frequency to obtain the speed of light, confirming Maxwell's prediction that light is an electromagnetic wave: the most famous application of v = f λ v = f\lambda v = f λ .