Physics · Oscillations and waves

Wavelength, frequency and phase speed

The relation $c=\lambda f$ connects phase speed, wavelength and frequency of the same travelling periodic component.

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Formula

c = λ \cdot f

c in m/s · λ in metres [m] · f in hertz [Hz] = [1/s]

Variables & units

Variables & units
Symbol	Meaning	Unit
$c$	Phase speed of the periodic component	m/s
$λ$	Wavelength	m
$f$	Frequency	Hz

LaTeX

c = \lambda \cdot f

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Snapshot of a travelling harmonic component: λ is the separation of adjacent crests. The arrow marks phase speed vₚ = λf, not transverse particle motion.

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Wavelength and frequency calculator

Find the wavelength, frequency or wave speed of a periodic wave with c = λ · f, for example for sound, water waves or light.

Find

Wave speed c

How fast a wave crest moves along, e.g. about 340 m/s for sound in air.

Frequency f

Oscillations per second, e.g. 440 Hz for concert pitch A.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for λ, substitute, evaluate

1. Rearrange for λ
$\lambda=\frac cf$
2. Substitute values (in SI units)
$\lambda=\frac{340}{440}=0.772727\,\mathrm m$
3. Result
λ = 77.27 cm

Result: Wavelength λ

λ = 77.27 cm

With c = 340 m/s and f = 440 Hz, λ = 77.27 cm.

Quick tips

Formula triangle: c on top, λ and f below. Cover the quantity you need and the rest is the calculation, e.g. λ = c ÷ f.
For the same c: the higher the frequency, the shorter the wavelength. When the wave enters another medium, f stays the same and λ changes with c.

Assumptions and limits

Positive phase speed, wavelength and frequency of the same periodic wave. No particle speed, group velocity, direction or wave differential-equation solution.

Derivation & background

In a linear propagation and coupling model: Equal phase advances by λ in $T=\frac1f$ : $c=v_p=\frac\lambda T=\lambda f$ . In dispersive media phase speed can differ from envelope or signal speed. At a stationary time-independent interface f is conserved while λ follows the new medium. The problem value $\frac{343 m}{s}$ for air sound is approximate; vacuum light speed is exactly $\frac{299792458 m}{s}$ .

Exam blueprint

Validity range

In a linear propagation and coupling model: Applies to phase speed of a travelling periodic component in the same medium. In dispersive media distinguish group and signal speeds. Frequency is preserved at a stationary time-independent boundary.

Derivation steps

Follow equal phase of a travelling periodic component.

1λ is spacing between adjacent equal-phase points.
2T is repeat time at fixed position; $f=\frac1T$ .
3Write a harmonic component as $y=A\sin(kx-\omega t+\varphi)$ .
4For a travelling phase, $kx-\omega t+\varphi$ stays constant.
5Differentiation gives $k\frac{\mathrm dx}{\mathrm dt}=\omega$ and hence $v_p=\frac\omega k$ .
6Using $k=\frac{2\pi}{\lambda}$ and $\omega=2\pi f$ gives $v_p=\lambda f$ . Envelope speed can differ.

Rearrangements

Wavelength from speed and frequency

\lambda = \frac{c}{f}

High frequency means short wavelength, for a fixed propagation speed.

Frequency from speed and wavelength

f = \frac{c}{\lambda}

At a stationary time-independent linear interface frequency is preserved; wavelength changes with phase speed.

Task variant

Green light in vacuum has $\lambda=500\,\mathrm{nm}$ . Find frequency using the rounded problem value $c=3\times10^8\,\mathrm{m/s}$ .

Approach: $f=\frac c\lambda=\frac{3\times10^8}{5\times10^{-7}}\,\mathrm{Hz}=6\times10^{14}\,\mathrm{Hz}$ .

Check: $\lambda f=c$ .

A water wave has $f = 2 Hz$ and $\lambda{} = 1.5 m$ . How fast does it travel?

Approach: $v_p=\lambda f=1.5\,\mathrm m\cdot2\,\mathrm{Hz}=3\,\mathrm{m/s}$ .
This is the phase speed of that travelling component.

Common mistakes

Not converting nanometres to metres.

$1 nm = 10^{-9} m$ , otherwise the result is off by orders of magnitude.

Assuming the frequency changes when entering another medium.

At a stationary, time-independent linear interface, frequency is preserved while phase speed and wavelength adapt. A moving or time-modulated boundary requires a different model.

Mixing up the speeds of sound and light.

$\frac{343 m}{s}$ is a supplied approximate air-sound speed under suitable conditions. Vacuum light speed is exactly $\frac{299792458 m}{s}$ ; $\frac{3\times{}10^{8} m}{s}$ is a rounded problem value.

Exam context

A basic step in acoustics, optics and radio problems: $\frac{\lambda{}}{4}$ antenna length, string vibrations, colours of light.

Formula cluster

Wave physics

From oscillation to wave: the frequency comes from the source, c from the medium.

T = 2π·√(m/k)

Spring Pendulum (Oscillation Period)

n₁·sin(α₁) = n₂·sin(α₂)

Snell's Law of Refraction

λ = h/p

De Broglie Wavelength

f_B = f_Q·(c ± v_B)/(c ∓ v_Q)

Doppler Effect (Acoustics)

Worked example

For concert pitch $a^{1}$ at $f=440\,\mathrm{Hz}$ with supplied air-sound speed $c=343\,\mathrm{m/s}$ , $\lambda=\frac cf=\frac{343}{440}\,\mathrm m\approx0.779545\,\mathrm m$ , rounded to $0.78\,\mathrm m$ . Check: $\lambda f=c$ using unrounded wavelength.

Applications

Radio engineering (antenna length), musical instruments, ultrasound diagnostics, radar and optical fibres

How do you calculate with Wave Equation (c = λ·f)?

Here is how to work through a typical Wave Equation (c = λ·f) (c = λ·f) task step by step:

1
Task
Green light in vacuum has $\lambda=500\,\mathrm{nm}$ . Find frequency using the rounded problem value $c=3\times10^8\,\mathrm{m/s}$ .
Solution path
1. Approach: $f=\frac c\lambda=\frac{3\times10^8}{5\times10^{-7}}\,\mathrm{Hz}=6\times10^{14}\,\mathrm{Hz}$ .
Check: $\lambda f=c$ .
2
Task
A water wave has $f = 2 Hz$ and $\lambda{} = 1.5 m$ . How fast does it travel?
Solution path
1. Approach: $v_p=\lambda f=1.5\,\mathrm m\cdot2\,\mathrm{Hz}=3\,\mathrm{m/s}$ .
2. This is the phase speed of that travelling component.

Understand and apply the equation

Wavelength describes repetition in space; frequency describes repetition in time. Their product gives a harmonic wave’s phase velocity. Learn to extract these quantities from measurements or a wave function, determine direction, and distinguish the motion of a particle.

Two measurements describe the same wave

A photograph of a regular wave shows the spacing between neighbouring crests. A sensor at one fixed position instead measures the time between their passages. Together, these measurements determine how quickly a crest travels. This page relates spatial repetition to repetition in time.[1]

You need wavelength, full period, frequency, and prefix conversions. Here c denotes phase velocity of a chosen harmonic component; light in vacuum has the familiar special value. The differential equation also called the wave equation is a more detailed model and is not replaced by this relationship among quantities.[1]

What is the difference between wavelength and frequency?

Wavelength λ > 0 is the spacing of neighbouring equal-phase points, such as adjacent crests in a harmonic snapshot. Frequency f > 0 counts full oscillations per second at a fixed position. Period $T=\frac{1}{f}$ is a time and λ a length. A plot against x is therefore not automatically the same record as a plot against t.[1]

\begin{aligned}v_p&=\lambda f=\frac\lambda T=\frac\omega k\\k&=\frac{2\pi}\lambda\\\omega&=2\pi f\end{aligned}

The top symbol c denotes phase-speed magnitude vp here; it is not restricted to light.

Elsewhere c commonly denotes vacuum light speed. String waves and sound require speeds from their own medium models. Units m/s alone do not distinguish the cases. Frequency and wavelength must refer to the same component in the same medium.

How does constant phase give c = λ·f?

\begin{aligned}y(x,t)&=A\sin(kx-\omega t+\varphi)\\kx-\omega t+\varphi&=\text{constant}\end{aligned}

For positive k and ω, the minus sign before ωt describes a rightward harmonic wave.[1]

Follow a particular crest rather than a particular string element. Its phase stays constant. Differentiation gives $k\frac{dx}{dt}-\omega=0$ and hence $\frac{dx}{dt}=\frac{\omega}{k}$ . Definitions of k and ω produce λf. This is a kinematic relation for the periodic component; it does not yet determine the speed allowed by a medium.

\begin{aligned}k\frac{\mathrm dx}{\mathrm dt}-\omega&=0\\v_p&=\frac\omega k=\lambda f\end{aligned}

For $kx+\omega t$ , phase travels left with signed velocity $-\frac\omega k$ .

The statement “in T the wave advances by λ” follows the same phase of a travelling periodic component. A standing wave has fixed spatial nodes and requires a different description. An isolated nonperiodic pulse also has no single universal frequency or wavelength; it can contain many components.

Calculating wavelength, frequency and phase speed

\begin{aligned}\lambda&=\frac{v_p}f\\f&=\frac{v_p}\lambda\\{}[\lambda f]&=\mathrm{m\,s^{-1}}\end{aligned}

Assume positive frequency and wavelength; specify propagation direction separately.

Convert kilohertz to hertz by multiplying by 1000 and nanometres to metres by 10⁻⁹. ω is angular frequency in rad/s whereas f is in Hz. Do not apply 2π on only one side. Inserting ω as f makes the wavelength wrong by a factor of 2π.

The existing sound calculation remains useful: assumed air speed 343 m/s and 440 Hz give $\lambda=\frac{343\,\mathrm{m/s}}{440\,\mathrm{Hz}}\approx0.77955\,\mathrm m$ . This is a problem assumption near ordinary room-temperature conditions, not an exact universal air constant. Vacuum light instead uses the defined 299792458 m/s.[4]

Do medium particles move at the wave speed?

\begin{aligned}v_y&=\frac{\partial y}{\partial t}\\&=-A\omega\cos(kx-\omega t+\varphi)\\a_y&=-\omega^2y\end{aligned}

These quantities apply at fixed x in the harmonic transverse model.[1]

A taut string carries a pattern along x while a string element oscillates transversely in the linear model. Maximum particle speed is Aω, which depends on amplitude. Ideal-string phase speed instead follows tension and linear mass density. There is no general equality between these velocities.

Zero displacement does not imply rest: particle speed is maximal there. At its displacement extremum, particle speed is zero and inward acceleration maximal. Phase motion is independent; the crest may continue travelling.

A quarter period shifts the pattern

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For the third example,

\Delta t=12.5\,\mathrm{ms}=\frac T4

shifts the profile by

v_p\Delta t=0.15\,\mathrm m=\frac\lambda4

. These are snapshots, not a particle trajectory. Transverse displacement is visually enlarged.[1]

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How do you read wavelength, frequency and direction from a wave function?

The first example reads phase speed from the supplied wave function. To determine it from the string’s mechanical data instead, wave speed on a stretched string relates tension to linear mass density in the ideal model of small transverse displacements.[6][1]

Read a harmonic string function

Given: $y=0.003\sin(5\pi x-80\pi t)$ , with x in m, t in s and y in m; ideal travelling transverse component.
Find: Find λ, f, vp and particle motion at $x=0.10\,\mathrm m$ , t = 0.00625 s.

Approach: Wave-function coefficients determine spatial and temporal repetition.
$\begin{aligned}\lambda&=\frac{2\pi}k,&f&=\frac\omega{2\pi}\\v_p&=\frac\omega k,&T&=\frac1f\end{aligned}$
Substitute the given values with matching units.
$\begin{aligned}\lambda&=\frac{2\pi}{5\pi}=0.4\,\mathrm m\\{} f&=\frac{80\pi}{2\pi}=40\,\mathrm{Hz}\end{aligned}$
Phase travels rightward.
$\begin{aligned}v_p&=\lambda f=16\,\mathrm{m/s}\\{} T&=0.025\,\mathrm s\end{aligned}$
The specified point has zero phase.
$\begin{aligned}5\pi\cdot0.10-80\pi\cdot0.00625&=0\\{} y&=0\end{aligned}$
Particle velocity points downward.
$v_y=-0.003\cdot80\pi\approx-0.753982\,\mathrm{m/s}$

Result: Phase speed is 16 m/s rightward; the selected particle crosses $y=0$ downward at about 0.754 m/s.

Check: $\frac{\omega}{k}=16\,\mathrm{m/s}$ confirms λf. A crest advances 0.4 m in 0.025 s. Maximum slope $Ak=0.003\cdot5\pi\approx0.0471$ is small, making linear string geometry plausible.

The same rhythm at a stationary boundary

Given: Linear propagation and coupling model; Stationary time-independent boundary; monochromatic source f = 60 Hz. Corresponding phase speeds vp1 = 120 m/s and vp2 = 72 m/s.
Find: Find both wavelengths and their ratio.

Approach: The interface is stationary, linear, and time invariant: frequency is preserved.
$\begin{aligned}f_1&=f_2=f\\\lambda_1&=\frac{v_1}f\\\lambda_2&=\frac{v_2}f\\\frac{\lambda_2}{\lambda_1}&=\frac{v_2}{v_1}\end{aligned}$
Substitute the given values with matching units.
$f_1=f_2=60\,\mathrm{Hz}$
Use each medium’s own phase speed.
$\begin{aligned}\lambda_1&=\frac{120}{60}=2\,\mathrm m\\{} \lambda_2&=\frac{72}{60}=1.2\,\mathrm m\end{aligned}$
Calculate relative shortening.
$\frac{\lambda_2}{\lambda_1}=\frac{72}{120}=0.60$

Result: Wavelengths are 2 m and 1.2 m; the second is 40% shorter at unchanged temporal rhythm.

Check: $\lambda_1f=120\,\mathrm{m/s}$ and $\lambda_2f=72\,\mathrm{m/s}$ . This does not determine transmitted amplitude; reflection and energy partition require further boundary conditions.

Explore further: Compare phase and group velocity

\begin{aligned}v_p&=\frac\omega k\\v_g&=\frac{\mathrm d\omega}{\mathrm dk}\end{aligned}

Under suitable narrow-band conditions, group velocity describes envelope motion.[2]

Superpose two nearby frequencies and wavenumbers. The fast carrier moves approximately with $\frac\omega k$ and the slowly changing envelope with $\frac{\Delta\omega}{\Delta k}$ . As their separation shrinks this becomes $\frac{\mathrm d\omega}{\mathrm dk}$ . A linear dispersion relation $\omega=vk$ with constant v makes them equal; a curved relation generally does not.

Strong dispersion, absorption or broad bandwidth can reshape an envelope. A single vg value is then not automatically a unique pulse or information speed. The relation $v_p=\lambda f$ remains valid for a harmonic component without answering these additional transport questions.[2]

Why does frequency stay unchanged at a stationary boundary?

In a linear propagation and interface model, incident and transmitted harmonic components must continuously match temporal phase at a stationary time-independent boundary. Their frequency is therefore conserved. The new wavevector and wavelength follow from the medium and spatial phase matching. A moving boundary or time-modulated medium may change frequency. Snell’s law of refraction describes the associated direction change in the appropriate isotropic two-medium model.[5][3]

Relative motion between source and observer can also produce a Doppler shift. State the reference frame of a frequency measurement. “Frequency always stays the same” is too strong; conservation applies under the stated stationary conditions.

Use wavelengths without predicting an entire device

A quarter wavelength is λ/4, giving an initial geometric antenna scale. Actual resonance and matching also depend on geometry, surroundings and material. Likewise acoustic wavelength alone does not determine ultrasound resolution; bandwidth, aperture and measurement method remain relevant. For a standing wave on a string fixed at both ends, fixed nodes and string length instead determine the allowed modes; the pattern does not travel along the string like an individual crest.[7]

A travel time gives $d=vt$ only in the matching propagation model. An echo includes outbound and return paths, giving $d=\frac{vt}{2}$ for the same stationary path. A λf calculation alone cannot decide whether a measured envelope needs group rather than phase speed.

How do two sensors determine propagation direction?

One snapshot alone does not reveal direction. Two timed sensor readings can trace an identified crest. The following wave function also checks transverse motion at the second sensor.[1]

Measurements at two positions determine direction

Given: A small harmonic transverse wave travelling right has A=2.0 mm, λ=0.60 m, and f=20 Hz. At t=0 a crest is at x=0. Sensor B is 0.15 m to the right of sensor A at x=0. No reflection or dispersion occurs in the region considered.
Find: Write a suitable wave function and determine the second sensor’s delay and initial motion.

Approach: a cosine places the initial crest; the minus sign describes rightward propagation.
$\begin{aligned}y(x,t)&=A\cos(kx-\omega t)\\k&=\frac{2\pi}{\lambda},&\omega&=2\pi f\\v_p&=\lambda f,&\Delta t&=\frac{\Delta x}{v_p}\\v_y(x,t)&=A\omega\sin(kx-\omega t)\end{aligned}$
Initial phase at the second sensor is π/2.
$\begin{aligned}k&=\frac{10\pi}{3}\,\mathrm{m^{-1}}\\\omega&=40\pi\,\mathrm{s^{-1}}\\v_p&=12\,\mathrm{m/s}\\\Delta t&=\frac{0.15}{12}\,\mathrm s=0.0125\,\mathrm s=\frac T4\\y(0.15,0)&=0\\v_y(0.15,0)&=0.002\cdot40\pi\,\mathrm{m/s}\\&\approx+0.251327\,\mathrm{m/s}\end{aligned}$

Result: Sensor B reaches the crest 12.5 ms after A. Initially it crosses zero in the positive transverse direction while the pattern travels right.

Check: At $t=0.0125\,\mathrm s$ , the second sensor has zero phase and hence $y=A$ . Also $v_p\Delta t=0.15\,\mathrm m$ and $Ak\approx0.0209$ , consistent with small slope.

Five exercises with complete checks

Calculate · Foundation

λ = 0.25 m and f = 48 Hz describe the same harmonic wave. Find vp and T.

Show hint

Multiply for vp and take the reciprocal for T.

Learning goal: Calculate phase speed and period.

Show solution and calculation

Approach: wavelength and frequency refer to the same component.
$v_p=12\,\mathrm{m/s}$ , $T=\frac1{48}\,\mathrm s\approx0.0208333\,\mathrm s$ .

Check: $v_pT=0.25\,\mathrm m$ , one wavelength per period.

\begin{aligned}v_p&=\lambda f=0.25\cdot48\,\mathrm{m/s}=12\,\mathrm{m/s}\\T&=\frac1f=\frac1{48}\,\mathrm s\approx0.0208333\,\mathrm s\end{aligned}

Calculate · Application

A component has $k=4\pi\,\mathrm{m^{-1}}$ and $\omega=24\pi\,\mathrm{s^{-1}}$ . Find λ, f and vp.

Show hint

Normalize both angular quantities using 2π.

Learning goal: Convert angular quantities to frequency and wavelength.

Show solution and calculation

Approach: convert wave number and angular frequency into spatial and temporal repetition.
$\lambda=0.50\,\mathrm m$ , $f=12\,\mathrm{Hz}$ , $v_p=6\,\mathrm{m/s}$ .
The direct ratio $\frac\omega k$ matches $\lambda f$ .

\begin{aligned}\lambda&=\frac{2\pi}k=\frac{2\pi}{4\pi}\,\mathrm m=0.50\,\mathrm m\\f&=\frac\omega{2\pi}=\frac{24\pi}{2\pi}\,\mathrm{Hz}=12\,\mathrm{Hz}\\v_p&=\frac\omega k=6\,\mathrm{m/s}\end{aligned}

Calculate · Application

In an ideal vacuum model, light has f = 2.00 GHz. Find λ using exact vacuum light speed.

Show hint

GHz = 10⁹ Hz; do not use an acoustic speed.

Learning goal: Find vacuum light wavelength.

Show solution and calculation

Approach: convert gigahertz to hertz and rearrange the wave relation for wavelength.
The calculation check is $0.149896229\,\mathrm m$ , reasonably rounded to $0.150\,\mathrm m$ .
Multiplying back by frequency reproduces the exact vacuum speed.
Finite precision of the supplied frequency does not establish matching measurement precision.

\begin{aligned}\lambda&=\frac cf\\&=\frac{299792458\,\mathrm{m/s}}{2.00\times10^9\,\mathrm{Hz}}\\&=0.149896229\,\mathrm m\approx0.150\,\mathrm m\end{aligned}

Interpret · Application

Two waves are $y_1=A\cos(kx-\omega t)$ and $y_2=A\cos(kx+\omega t)$ , with positive k and ω. Which propagates right and which left? Justify using a fixed phase point rather than a snapshot.

Show hint

Keep each phase argument constant and solve for x(t).

Learning goal: Identify propagation direction from the time-phase sign.

Show solution and calculation

Approach: A constant-phase point of the first wave moves with positive slope ω divided by k; one in the second moves with negative slope.
Thus y1 travels right and y2 left, at equal speed magnitude.
At time zero the profiles look identical, so one snapshot cannot establish direction.

\begin{aligned}kx-\omega t&=\phi\Rightarrow x(t)=\frac\phi k+\frac\omega k t\\{}kx+\omega t&=\phi\Rightarrow x(t)=\frac\phi k-\frac\omega k t\end{aligned}

[1]

Error analysis · Challenge

A model medium has $\omega=ak^2$ with a>0 for positive k. Someone calls $\frac{\omega}{k}$ the speed of a narrowband packet centre. Find phase and group velocities and their ratio; neglect absorption.

Show hint

For the packet use the slope of the dispersion relation.

Learning goal: Avoid equating phase and group velocity without a dispersion check.

Show solution and calculation

Approach: Phase ratio is ak whereas the derivative with respect to k is 2ak, twice as large.
A narrowband packet uses the local slope.
Wavelength times frequency remains valid for phase velocity, but does not alone establish a signal-front velocity.

\begin{aligned}v_p&=\frac\omega k=ak\\{}v_g&=\frac{\mathrm d\omega}{\mathrm dk}=2ak\\{}\frac{v_g}{v_p}&=2\end{aligned}

[2]

In brief

Assign wavelength and frequency to the same component. Their product describes a fixed phase point’s motion. A wave function also gives direction and particle motion. Frequency is preserved at stationary linear interfaces; dispersive media require a separate treatment of packet speed.[1]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made wavelength, frequency, phase derivation, propagation direction and group/particle motion accessible through learning questions; retained calculation and measurement examples and linked matching interface/string models in context.
October 1, 2026Connected spatial and temporal repetition more accessibly; added a two-sensor example and time-shifted profiles. Preserved existing component, interface, and dispersion exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

Cite this reference page

Quanta. Wave Equation (c = λ·f). Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/wellengleichung

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Frequently asked questions about Wave Equation (c = λ·f)

How do you calculate with the wave equation

c = \lambda{}\cdot{}f

Use wavelength and frequency of the same travelling periodic component in the same medium. Their product $c=\lambda f$ gives its phase speed. Rearrangement gives $\lambda=\frac cf$ or $f=\frac c\lambda$ . For specified air sound speed $\frac{343 m}{s}$ and 440 Hz, $\lambda{} \approx{} 0.77955 m$ . Convert kilohertz to hertz and nanometres to metres. Angular frequency ω equals 2πf and must not be inserted unchanged as f. Check by multiplying λf back to the assumed speed. For a dispersive wave packet, this product does not automatically describe envelope or information speed; additional dispersion data are needed.

What stays the same when a wave passes into another medium?+

In a linear propagation and coupling model, a transmitted harmonic component retains the incident frequency at a stationary time-independent boundary. Phase speed follows the new medium, so $\lambda=\frac{v_p}f$ changes. For example, 60 Hz at $\frac{120 m}{s}$ gives wavelength 2 m, while $\frac{72 m}{s}$ gives 1.2 m. This does not determine transmitted amplitude, reflection or energy partition. A moving boundary or a time-modulated medium can change frequency. A Doppler measurement also requires the reference frame of source and observer. Frequency conservation is therefore a statement with specific conditions, rather than an unconditional rule.

Why do you not see lightning and hear thunder at the same time?+

Light and sound reach an observer after very different travel times. A simple model gives delay $\Delta t=\frac d{v_{\mathrm{sound}}}-\frac d{c_{\mathrm{vac}}}$ . With assumed sound speed $\frac{343 m}{s}$ , light time at ordinary distances is very small compared with sound time, so $d\approx(343\,\mathrm{m/s})\Delta t$ . A delay of 6 s therefore suggests about 2058 m, roughly 2 km. This is an estimate because temperature, wind, sound speed and the lightning’s spatial extent matter. An echo instead measures outbound and return travel and requires halving the travelled distance. A precise distance needs a matching propagation and measurement model.

Why are FM antennas shorter than long-wave antennas?+

At fixed propagation speed, higher frequency gives shorter wavelength through $\lambda=\frac{v_p}f$ . Using rounded vacuum speed $\frac{3\cdot{}10^{8} m}{s}$ , 100 MHz gives about 3 m, whereas 150 kHz gives about 2000 m. Quarter-wave lengths are therefore about 0.75 m and 500 m. These provide initial geometric scales for certain antenna designs, rather than mandatory lengths for every real antenna. Geometry, electrical shortening, material, surroundings and impedance matching affect resonance and efficiency. A real antenna may be much shorter and require an appropriate circuit and radiation model. The λf calculation alone cannot establish whether a particular device transmits efficiently.

Does

c = \lambda{}\cdot{}f

hold for all types of waves?+

The relation gives phase speed of a travelling periodic component, whether mechanical or electromagnetic. Wavelength and frequency must refer to that same component. A nonperiodic pulse generally contains many frequencies and has no single universal wavelength. In a dispersive medium, components may have different phase speeds. Under appropriate narrow-band conditions, envelope speed is $v_g=\frac{\mathrm d\omega}{\mathrm dk}$ , while $v_p=\frac\omega k=\lambda f$ describes phase. A standing wave has spatially fixed nodes; its v refers to the propagation speed of its components. The relation therefore does not provide a general speed for every medium particle or arbitrary signal.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §16.2 Mathematics of Waves

Related formulas

\lambda_n=\frac{2L}{n}, \quad f_n=\frac{n v}{2L}

Standing wave on a string fixed at both ends

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T = 2\pi \sqrt{\frac{l}{g}}

Simple Pendulum (Period)

x(t)=A_0 e^{-\beta t}\cos(\omega_d t+\varphi)

Damped harmonic oscillation

I=\frac{P}{A},\quad\beta=10\log_{10}\!\left(\frac{I}{I_0}\right)

Sound intensity and level

v=\sqrt{\frac{F_T}{\mu}}

Wave speed on a stretched string

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Wavelength, frequency and phase speed

The relation

c=\lambda f

connects phase speed, wavelength and frequency of the same travelling periodic component.

BasicExam-relevantLower secondaryUpper secondary

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Symbol

Meaning

Unit

c

Phase speed of the periodic component

m/s

λ

Wavelength

f

Frequency

Derivation & background

Exam blueprint

Validity range

Derivation steps

Follow equal phase of a travelling periodic component.

1λ is spacing between adjacent equal-phase points.
2T is repeat time at fixed position; $f=\frac1T$ .
3Write a harmonic component as $y=A\sin(kx-\omega t+\varphi)$ .
4For a travelling phase, $kx-\omega t+\varphi$ stays constant.
5Differentiation gives $k\frac{\mathrm dx}{\mathrm dt}=\omega$ and hence $v_p=\frac\omega k$ .
6Using $k=\frac{2\pi}{\lambda}$ and $\omega=2\pi f$ gives $v_p=\lambda f$ . Envelope speed can differ.

How do you calculate with Wave Equation (c = λ·f)?

Here is how to work through a typical Wave Equation (c = λ·f) (c = λ·f) task step by step:

Task

Green light in vacuum has $\lambda=500\,\mathrm{nm}$ . Find frequency using the rounded problem value $c=3\times10^8\,\mathrm{m/s}$ .

Solution path

Approach: $f=\frac c\lambda=\frac{3\times10^8}{5\times10^{-7}}\,\mathrm{Hz}=6\times10^{14}\,\mathrm{Hz}$ .

Check: $\lambda f=c$ .

Task

A water wave has $f = 2 Hz$ and $\lambda{} = 1.5 m$ . How fast does it travel?

Solution path

Approach: $v_p=\lambda f=1.5\,\mathrm m\cdot2\,\mathrm{Hz}=3\,\mathrm{m/s}$ .
This is the phase speed of that travelling component.

Two measurements describe the same wave

What is the difference between wavelength and frequency?

\begin{aligned}v_p&=\lambda f=\frac\lambda T=\frac\omega k\\k&=\frac{2\pi}\lambda\\\omega&=2\pi f\end{aligned}

The top symbol c denotes phase-speed magnitude vp here; it is not restricted to light.

How does constant phase give c = λ·f?

\begin{aligned}y(x,t)&=A\sin(kx-\omega t+\varphi)\\kx-\omega t+\varphi&=\text{constant}\end{aligned}

For positive k and ω, the minus sign before ωt describes a rightward harmonic wave.[1]

\begin{aligned}k\frac{\mathrm dx}{\mathrm dt}-\omega&=0\\v_p&=\frac\omega k=\lambda f\end{aligned}

For $kx+\omega t$ , phase travels left with signed velocity $-\frac\omega k$ .

Calculating wavelength, frequency and phase speed

\begin{aligned}\lambda&=\frac{v_p}f\\f&=\frac{v_p}\lambda\\{}[\lambda f]&=\mathrm{m\,s^{-1}}\end{aligned}

Assume positive frequency and wavelength; specify propagation direction separately.

Do medium particles move at the wave speed?

\begin{aligned}v_y&=\frac{\partial y}{\partial t}\\&=-A\omega\cos(kx-\omega t+\varphi)\\a_y&=-\omega^2y\end{aligned}

These quantities apply at fixed x in the harmonic transverse model.[1]

A quarter period shifts the pattern

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For the third example,

\Delta t=12.5\,\mathrm{ms}=\frac T4

shifts the profile by

v_p\Delta t=0.15\,\mathrm m=\frac\lambda4

. These are snapshots, not a particle trajectory. Transverse displacement is visually enlarged.[1]

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How do you read wavelength, frequency and direction from a wave function?

Read a harmonic string function

Given: $y=0.003\sin(5\pi x-80\pi t)$ , with x in m, t in s and y in m; ideal travelling transverse component.
Find: Find λ, f, vp and particle motion at $x=0.10\,\mathrm m$ , t = 0.00625 s.

Approach: Wave-function coefficients determine spatial and temporal repetition.
$\begin{aligned}\lambda&=\frac{2\pi}k,&f&=\frac\omega{2\pi}\\v_p&=\frac\omega k,&T&=\frac1f\end{aligned}$
Substitute the given values with matching units.
$\begin{aligned}\lambda&=\frac{2\pi}{5\pi}=0.4\,\mathrm m\\{} f&=\frac{80\pi}{2\pi}=40\,\mathrm{Hz}\end{aligned}$
Phase travels rightward.
$\begin{aligned}v_p&=\lambda f=16\,\mathrm{m/s}\\{} T&=0.025\,\mathrm s\end{aligned}$
The specified point has zero phase.
$\begin{aligned}5\pi\cdot0.10-80\pi\cdot0.00625&=0\\{} y&=0\end{aligned}$
Particle velocity points downward.
$v_y=-0.003\cdot80\pi\approx-0.753982\,\mathrm{m/s}$

Result: Phase speed is 16 m/s rightward; the selected particle crosses $y=0$ downward at about 0.754 m/s.

Check: $\frac{\omega}{k}=16\,\mathrm{m/s}$ confirms λf. A crest advances 0.4 m in 0.025 s. Maximum slope $Ak=0.003\cdot5\pi\approx0.0471$ is small, making linear string geometry plausible.

The same rhythm at a stationary boundary

Given: Linear propagation and coupling model; Stationary time-independent boundary; monochromatic source f = 60 Hz. Corresponding phase speeds vp1 = 120 m/s and vp2 = 72 m/s.
Find: Find both wavelengths and their ratio.

Approach: The interface is stationary, linear, and time invariant: frequency is preserved.
$\begin{aligned}f_1&=f_2=f\\\lambda_1&=\frac{v_1}f\\\lambda_2&=\frac{v_2}f\\\frac{\lambda_2}{\lambda_1}&=\frac{v_2}{v_1}\end{aligned}$
Substitute the given values with matching units.
$f_1=f_2=60\,\mathrm{Hz}$
Use each medium’s own phase speed.
$\begin{aligned}\lambda_1&=\frac{120}{60}=2\,\mathrm m\\{} \lambda_2&=\frac{72}{60}=1.2\,\mathrm m\end{aligned}$
Calculate relative shortening.
$\frac{\lambda_2}{\lambda_1}=\frac{72}{120}=0.60$

Result: Wavelengths are 2 m and 1.2 m; the second is 40% shorter at unchanged temporal rhythm.

Check: $\lambda_1f=120\,\mathrm{m/s}$ and $\lambda_2f=72\,\mathrm{m/s}$ . This does not determine transmitted amplitude; reflection and energy partition require further boundary conditions.

Explore further: Compare phase and group velocity

\begin{aligned}v_p&=\frac\omega k\\v_g&=\frac{\mathrm d\omega}{\mathrm dk}\end{aligned}

Under suitable narrow-band conditions, group velocity describes envelope motion.[2]

Why does frequency stay unchanged at a stationary boundary?

Use wavelengths without predicting an entire device

How do two sensors determine propagation direction?

One snapshot alone does not reveal direction. Two timed sensor readings can trace an identified crest. The following wave function also checks transverse motion at the second sensor.[1]

Measurements at two positions determine direction

Given: A small harmonic transverse wave travelling right has A=2.0 mm, λ=0.60 m, and f=20 Hz. At t=0 a crest is at x=0. Sensor B is 0.15 m to the right of sensor A at x=0. No reflection or dispersion occurs in the region considered.
Find: Write a suitable wave function and determine the second sensor’s delay and initial motion.

Approach: a cosine places the initial crest; the minus sign describes rightward propagation.
$\begin{aligned}y(x,t)&=A\cos(kx-\omega t)\\k&=\frac{2\pi}{\lambda},&\omega&=2\pi f\\v_p&=\lambda f,&\Delta t&=\frac{\Delta x}{v_p}\\v_y(x,t)&=A\omega\sin(kx-\omega t)\end{aligned}$
Initial phase at the second sensor is π/2.
$\begin{aligned}k&=\frac{10\pi}{3}\,\mathrm{m^{-1}}\\\omega&=40\pi\,\mathrm{s^{-1}}\\v_p&=12\,\mathrm{m/s}\\\Delta t&=\frac{0.15}{12}\,\mathrm s=0.0125\,\mathrm s=\frac T4\\y(0.15,0)&=0\\v_y(0.15,0)&=0.002\cdot40\pi\,\mathrm{m/s}\\&\approx+0.251327\,\mathrm{m/s}\end{aligned}$

Result: Sensor B reaches the crest 12.5 ms after A. Initially it crosses zero in the positive transverse direction while the pattern travels right.

Check: At $t=0.0125\,\mathrm s$ , the second sensor has zero phase and hence $y=A$ . Also $v_p\Delta t=0.15\,\mathrm m$ and $Ak\approx0.0209$ , consistent with small slope.

Five exercises with complete checks

Calculate · Foundation

λ = 0.25 m and f = 48 Hz describe the same harmonic wave. Find vp and T.

Show hint

Multiply for vp and take the reciprocal for T.

Learning goal: Calculate phase speed and period.

Show solution and calculation

Approach: wavelength and frequency refer to the same component.
$v_p=12\,\mathrm{m/s}$ , $T=\frac1{48}\,\mathrm s\approx0.0208333\,\mathrm s$ .

Check: $v_pT=0.25\,\mathrm m$ , one wavelength per period.

\begin{aligned}v_p&=\lambda f=0.25\cdot48\,\mathrm{m/s}=12\,\mathrm{m/s}\\T&=\frac1f=\frac1{48}\,\mathrm s\approx0.0208333\,\mathrm s\end{aligned}

Calculate · Application

A component has $k=4\pi\,\mathrm{m^{-1}}$ and $\omega=24\pi\,\mathrm{s^{-1}}$ . Find λ, f and vp.

Show hint

Normalize both angular quantities using 2π.

Learning goal: Convert angular quantities to frequency and wavelength.

Show solution and calculation

Approach: convert wave number and angular frequency into spatial and temporal repetition.
$\lambda=0.50\,\mathrm m$ , $f=12\,\mathrm{Hz}$ , $v_p=6\,\mathrm{m/s}$ .
The direct ratio $\frac\omega k$ matches $\lambda f$ .

\begin{aligned}\lambda&=\frac{2\pi}k=\frac{2\pi}{4\pi}\,\mathrm m=0.50\,\mathrm m\\f&=\frac\omega{2\pi}=\frac{24\pi}{2\pi}\,\mathrm{Hz}=12\,\mathrm{Hz}\\v_p&=\frac\omega k=6\,\mathrm{m/s}\end{aligned}

Calculate · Application

In an ideal vacuum model, light has f = 2.00 GHz. Find λ using exact vacuum light speed.

Show hint

GHz = 10⁹ Hz; do not use an acoustic speed.

Learning goal: Find vacuum light wavelength.

Show solution and calculation

Approach: convert gigahertz to hertz and rearrange the wave relation for wavelength.
The calculation check is $0.149896229\,\mathrm m$ , reasonably rounded to $0.150\,\mathrm m$ .
Multiplying back by frequency reproduces the exact vacuum speed.
Finite precision of the supplied frequency does not establish matching measurement precision.

\begin{aligned}\lambda&=\frac cf\\&=\frac{299792458\,\mathrm{m/s}}{2.00\times10^9\,\mathrm{Hz}}\\&=0.149896229\,\mathrm m\approx0.150\,\mathrm m\end{aligned}

Interpret · Application

Two waves are $y_1=A\cos(kx-\omega t)$ and $y_2=A\cos(kx+\omega t)$ , with positive k and ω. Which propagates right and which left? Justify using a fixed phase point rather than a snapshot.

Show hint

Keep each phase argument constant and solve for x(t).

Learning goal: Identify propagation direction from the time-phase sign.

Show solution and calculation

Approach: A constant-phase point of the first wave moves with positive slope ω divided by k; one in the second moves with negative slope.
Thus y1 travels right and y2 left, at equal speed magnitude.
At time zero the profiles look identical, so one snapshot cannot establish direction.

\begin{aligned}kx-\omega t&=\phi\Rightarrow x(t)=\frac\phi k+\frac\omega k t\\{}kx+\omega t&=\phi\Rightarrow x(t)=\frac\phi k-\frac\omega k t\end{aligned}

[1]

Error analysis · Challenge

A model medium has $\omega=ak^2$ with a>0 for positive k. Someone calls $\frac{\omega}{k}$ the speed of a narrowband packet centre. Find phase and group velocities and their ratio; neglect absorption.

Show hint

For the packet use the slope of the dispersion relation.

Learning goal: Avoid equating phase and group velocity without a dispersion check.

Show solution and calculation

Approach: Phase ratio is ak whereas the derivative with respect to k is 2ak, twice as large.
A narrowband packet uses the local slope.
Wavelength times frequency remains valid for phase velocity, but does not alone establish a signal-front velocity.

\begin{aligned}v_p&=\frac\omega k=ak\\{}v_g&=\frac{\mathrm d\omega}{\mathrm dk}=2ak\\{}\frac{v_g}{v_p}&=2\end{aligned}

[2]

In brief

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made wavelength, frequency, phase derivation, propagation direction and group/particle motion accessible through learning questions; retained calculation and measurement examples and linked matching interface/string models in context.
October 1, 2026Connected spatial and temporal repetition more accessibly; added a two-sensor example and time-shifted profiles. Preserved existing component, interface, and dispersion exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

Cite this reference page

Quanta. Wave Equation (c = λ·f). Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/wellengleichung

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Frequently asked questions about Wave Equation (c = λ·f)

How do you calculate with the wave equation

c = \lambda{}\cdot{}f

What stays the same when a wave passes into another medium?+

Why do you not see lightning and hear thunder at the same time?+

Why are FM antennas shorter than long-wave antennas?+

Does

c = \lambda{}\cdot{}f

hold for all types of waves?+

Wavelength, frequency and phase speed

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Wavelength and frequency calculator

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Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Wave physics

Worked example

Applications

How do you calculate with Wave Equation (c = λ·f)?

Two measurements describe the same wave

What is the difference between wavelength and frequency?

How does constant phase give c = λ·f?

Calculating wavelength, frequency and phase speed

Do medium particles move at the wave speed?

A quarter period shifts the pattern

How do you read wavelength, frequency and direction from a wave function?

Read a harmonic string function

The same rhythm at a stationary boundary

Explore further: Compare phase and group velocity

Why does frequency stay unchanged at a stationary boundary?

Use wavelengths without predicting an entire device

How do two sensors determine propagation direction?

Measurements at two positions determine direction

Five exercises with complete checks

λ = 0.25 m and f = 48 Hz describe the same harmonic wave. Find vp and T.

A component has k=4π m−1k=4\pi\,\mathrm{m^{-1}}k=4πm−1 and ω=24π s−1\omega=24\pi\,\mathrm{s^{-1}}ω=24πs−1. Find λ, f and vp.

In an ideal vacuum model, light has f = 2.00 GHz. Find λ using exact vacuum light speed.

Two waves are y1=Acos⁡(kx−ωt)y_1=A\cos(kx-\omega t)y1​=Acos(kx−ωt) and y2=Acos⁡(kx+ωt)y_2=A\cos(kx+\omega t)y2​=Acos(kx+ωt), with positive k and ω. Which propagates right and which left? Justify using a fixed phase point rather than a snapshot.

A model medium has ω=ak2\omega=ak^2ω=ak2 with a>0 for positive k. Someone calls ωk\frac{\omega}{k}kω​ the speed of a narrowband packet centre. Find phase and group velocities and their ratio; neglect absorption.

In brief

Sources for this explanation

Frequently asked questions about Wave Equation (c = λ·f)

Subject references and further reading

Related formulas

More formulas: Oscillations and waves

Retain Wave Equation (c = λ·f) for exams

Wavelength, frequency and phase speed

Variables & units

Wavelength and frequency calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Wave physics

Worked example

Applications

How do you calculate with Wave Equation (c = λ·f)?

Two measurements describe the same wave

What is the difference between wavelength and frequency?

How does constant phase give c = λ·f?

Calculating wavelength, frequency and phase speed

Do medium particles move at the wave speed?

A quarter period shifts the pattern

How do you read wavelength, frequency and direction from a wave function?

Read a harmonic string function

The same rhythm at a stationary boundary

Explore further: Compare phase and group velocity

Why does frequency stay unchanged at a stationary boundary?

Use wavelengths without predicting an entire device

How do two sensors determine propagation direction?

Measurements at two positions determine direction

Five exercises with complete checks

λ = 0.25 m and f = 48 Hz describe the same harmonic wave. Find vp and T.

A component has k=4π m−1k=4\pi\,\mathrm{m^{-1}}k=4πm−1 and ω=24π s−1\omega=24\pi\,\mathrm{s^{-1}}ω=24πs−1. Find λ, f and vp.

In an ideal vacuum model, light has f = 2.00 GHz. Find λ using exact vacuum light speed.

Two waves are y1=Acos⁡(kx−ωt)y_1=A\cos(kx-\omega t)y1​=Acos(kx−ωt) and y2=Acos⁡(kx+ωt)y_2=A\cos(kx+\omega t)y2​=Acos(kx+ωt), with positive k and ω. Which propagates right and which left? Justify using a fixed phase point rather than a snapshot.

A model medium has ω=ak2\omega=ak^2ω=ak2 with a>0 for positive k. Someone calls ωk\frac{\omega}{k}kω​ the speed of a narrowband packet centre. Find phase and group velocities and their ratio; neglect absorption.

In brief

A component has $k=4\pi\,\mathrm{m^{-1}}$ and $\omega=24\pi\,\mathrm{s^{-1}}$ . Find λ, f and vp.

Two waves are $y_1=A\cos(kx-\omega t)$ and $y_2=A\cos(kx+\omega t)$ , with positive k and ω. Which propagates right and which left? Justify using a fixed phase point rather than a snapshot.

A model medium has $\omega=ak^2$ with a>0 for positive k. Someone calls $\frac{\omega}{k}$ the speed of a narrowband packet centre. Find phase and group velocities and their ratio; neglect absorption.

A component has $k=4\pi\,\mathrm{m^{-1}}$ and $\omega=24\pi\,\mathrm{s^{-1}}$ . Find λ, f and vp.

Two waves are $y_1=A\cos(kx-\omega t)$ and $y_2=A\cos(kx+\omega t)$ , with positive k and ω. Which propagates right and which left? Justify using a fixed phase point rather than a snapshot.

A model medium has $\omega=ak^2$ with a>0 for positive k. Someone calls $\frac{\omega}{k}$ the speed of a narrowband packet centre. Find phase and group velocities and their ratio; neglect absorption.