Speed of a travelling wave

Drop a pebble into a still pond and watch the ripple race outward. The water itself barely moves up and down at each point — yet the pattern, the ring of crests, sprints across the surface at a speed you could actually measure with a stopwatch and a metre stick. That travelling pattern, not any single molecule of water, is what "wave speed" means, and it is one of the most useful numbers in all of physics: it tells a doctor how fast an ultrasound pulse crosses tissue, tells a geophysicist how long an earthquake's tremor takes to reach the next city, and tells a musician why a thick guitar string sounds lower than a thin one under the same tension. For a plucked string, a stretched slinky, or a sound wave in air, the same question recurs: given the wave's frequency $f$ and its wavelength $\lambda$ , how fast does the disturbance itself travel? The answer is one of the shortest, most quoted formulas in the CBSE Class 11 syllabus, $v = f\lambda$ , and it is deceptively simple only until you have to use it in both directions — sometimes to find $\lambda$ from a given $v$ and $f$ , sometimes to find how $\lambda$ must change when a wave crosses into a new medium while its frequency stays locked to the source. This article builds that formula from first principles — from the displacement equation $y(x,t) = A\sin(kx - \omega t)$ up to $v = f\lambda$ — then shows exactly how it links to the two "real" wave-speed formulas of this chapter: $v = \sqrt{T/\mu}$ for a stretched string and the Newton–Laplace result for sound. By the end you should be able to move fluently between $v$ , $f$ and $\lambda$ , explain in one sentence why wave speed belongs to the medium and not the source, and never again confuse the speed of the wave pattern with the speed of an individual particle bobbing up and down as the wave passes through it.

Key takeaways

Wave speed is the speed of the pattern, not of any particle of the medium — it is how fast a crest, trough, or any fixed point of the wave's shape travels.
The master formula is $v = f\lambda$ , derived from tracking a point of constant phase in $y(x,t) = A\sin(kx-\omega t)$ : differentiate $kx - \omega t = \text{constant}$ with respect to time to get $v = \omega/k = f\lambda$ .
Wave speed is fixed by the medium, not by the source's frequency — tension and mass per length for a string ( $v=\sqrt{T/\mu}$ ), bulk modulus and density for sound (Newton–Laplace, qualitative here).
When a wave crosses into a new medium, $f$ stays fixed (set by the source) while $v$ and $\lambda$ both adjust so that $v=f\lambda$ still holds in the new medium.
$v=f\lambda$ is a kinematic bookkeeping relation — it links how fast the pattern moves, how long each repeat is, and how often it repeats, but it never by itself gives you the numerical value of $v$ for a real wave; that comes from a medium-specific formula.
Particle speed and wave speed are different things — particle speed is $\partial y/\partial t$ (how fast a bit of string oscillates up and down), wave speed is $\omega/k$ (how fast the whole pattern advances); one can be zero while the other is not.
Key idea: the speed of a travelling wave is the speed of its shape, $v = f\lambda = \omega/k$ , fixed entirely by the medium's own mechanical properties, so the source's frequency stays fixed across a boundary while wavelength and speed adjust together.

The core insight

Wave speed is how fast the wave's shape (a crest, a compression) advances through the medium — given by $v = f\lambda = \omega/k$ from the phase of $y(x,t)=A\sin(kx-\omega t)$ , and fixed by the medium's own elastic and inertial properties, never by the source's frequency.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Waves chapter (Unit: Oscillations and Waves), directly under the prescribed heading "speed of a travelling wave." You are expected to derive and apply $v = f\lambda$ starting from the general progressive-wave equation $y(x,t) = A\sin(kx-\omega t)$ , and to connect it with the two specific speed formulas the chapter also prescribes: $v=\sqrt{T/\mu}$ for a wave on a stretched string, and the Newton–Laplace expression for the speed of sound (treated qualitatively — its derivation is beyond this level). The topic is squarely in-syllabus and frequently examined both as a standalone numerical and as the opening step of a longer string- or sound-wave problem.

The physical picture

A travelling wave on a string or in a medium is described by the displacement of each point of the medium as a function of position $x$ and time $t$ . For a sinusoidal wave moving in the positive $x$ -direction, this displacement relation is

y(x,t) = A\sin(kx - \omega t)

where: $y$ is the displacement of the medium at position $x$ and time $t$ , $A$ is the amplitude (maximum displacement), $k = 2\pi/\lambda$ is the angular wave number (radians per metre, packing in how many radians of phase fit per metre), and $\omega = 2\pi f$ is the angular frequency (radians per second, how fast the phase advances at a fixed point). The physical reading: at any instant, this equation is a snapshot of the wave's shape along $x$ ; at any fixed $x$ , it is a description of that one point oscillating in time. The whole wave pattern — the sequence of crests and troughs — slides along the $x$ -axis as time goes on, and that sliding speed is what we call the wave speed. To find how fast the pattern moves, follow one fixed feature of the shape — say a crest — and ask how its position $x$ must change with $t$ to keep the same value of the sine, i.e. to keep the phase constant:

kx - \omega t = \text{constant}

where: holding the phase fixed is exactly the statement "this is the same crest, just later in time, so it must have moved to a new $x$ ." Differentiating both sides with respect to time gives the wave speed directly:

v = \frac{dx}{dt} = \frac{\omega}{k} = f\lambda

where: the last step substitutes $\omega = 2\pi f$ and $k = 2\pi/\lambda$ , so $\dfrac{\omega}{k} = \dfrac{2\pi f}{2\pi/\lambda} = f\lambda$ . This is the single most-used relation in the whole chapter — it holds for any periodic travelling wave, transverse or longitudinal, mechanical or (as a kinematic relation) electromagnetic.

Figure

A snapshot of a sine wave along the x axis at two nearby times, the later curve shifted to the right by one small distance delta x in one small time delta t, with the ratio delta x over delta t labelled as the wave speed v

A crucial conceptual point follows immediately: wave speed is set by the properties of the medium, not by the source. For a string, the relevant properties are its tension $T$ and mass per unit length $\mu$ , giving $v=\sqrt{T/\mu}$ (derived in the next article of this chapter). For sound, the relevant properties are the medium's bulk modulus $B$ and density $\rho$ , giving the Newton–Laplace result $v=\sqrt{B/\rho}$ — quoted here without derivation, as its full development is beyond the Class 11 ceiling. When a wave crosses from one medium into another, $v$ changes because the medium's properties change; the frequency $f$ is fixed by whatever is driving the wave (the hand shaking the string, the tuning fork driving the air) and cannot change at a boundary, so the wavelength $\lambda$ must adjust to keep $v = f\lambda$ true in the new medium.

Derivation ADeriving v = f lambda from the phase of a travelling wave

The goal is to obtain the wave-speed formula $v = f\lambda$ directly from the mathematical description of a travelling sinusoidal wave $y(x,t) = A\sin(kx-\omega t)$ , using nothing beyond the idea that "the same point on the wave's shape" corresponds to "the same value of the sine function," i.e. the same phase, and that differentiating a constant with respect to time gives zero — no wave-equation PDE, no Fourier analysis, just algebra and one derivative.

Step 1. Identify what "the wave moves" actually means. The function $y(x,t)=A\sin(kx-\omega t)$ assigns a displacement to every $(x,t)$ pair. A specific feature of the wave — say the crest where $\sin(\cdot) = 1$ — is not tied to one fixed $x$ ; it is tied to whichever $x$ makes the argument of the sine equal to $\pi/2$ (or, more generally, to whatever constant value defines that feature) at the corresponding $t$ . So "the crest moves" means: as $t$ increases, the $x$ that keeps the argument at that same constant value must also change.

Step 2. Write the condition for tracking one fixed feature of the wave. Call the phase $\phi = kx - \omega t$ . To follow the same feature (same phase) as time passes, we require $\phi$ to stay constant: $kx - \omega t = \text{constant}$ .

Step 3. Differentiate this condition with respect to time. Since the right-hand side is a constant, its time derivative is zero: $k\dfrac{dx}{dt} - \omega = 0$ . This single line is the entire physical content of the derivation — it says the rate of change of $x$ (needed to keep phase fixed) is fixed by $k$ and $\omega$ alone.

Step 4. Solve for $dx/dt$ , which is by definition the speed at which this feature — and hence the whole periodic pattern, since every feature obeys the same equation — travels: $v = \dfrac{dx}{dt} = \dfrac{\omega}{k}$ .

Step 5. Convert to frequency and wavelength using the standard definitions $\omega = 2\pi f$ (angular frequency from frequency) and $k = 2\pi/\lambda$ (wave number from wavelength). Substituting, $v = \dfrac{2\pi f}{2\pi/\lambda} = f\lambda$ . The factors of $2\pi$ cancel exactly, leaving the compact final result $\boxed{v = f\lambda}$ , valid for any sinusoidal travelling wave and, by extension (via Fourier synthesis, which is beyond this level), for any periodic travelling wave shape at all.

Worked examples

Board level - frequency and wavelength of a sound wave

A sound wave in air has a wavelength of $0.75\ \text{m}$ and the speed of sound in air is $340\ \text{m/s}$ . Find its frequency, and state whether it lies in the audible range for humans (roughly $20\ \text{Hz}$ to $20{,}000\ \text{Hz}$ ).

Step 1. Write down what is given: $v = 340\ \text{m/s}$ , $\lambda = 0.75\ \text{m}$ . Both are already in SI units, so no conversion is needed.

Step 2. Apply $v = f\lambda$ , rearranged to solve for frequency: $f = \dfrac{v}{\lambda} = \dfrac{340}{0.75}$ .

Step 3. Compute: $f = 453.3\ \text{Hz}$ (to one decimal place).

Step 4. Check the range: since $20\ \text{Hz} < 453.3\ \text{Hz} < 20{,}000\ \text{Hz}$ , this frequency is well within the human audible range — comparable to a note near the piano's middle-A region. Answer: $f \approx 453.3\ \text{Hz}$ , which is audible.

Competitive level - a wave crossing from a light string into a heavy string

A transverse wave with frequency $100\ \text{Hz}$ and wavelength $0.5\ \text{m}$ travels along a light section of string, where its speed is $v_1$ . Find $v_1$ . The string then joins a heavier section where the wave speed drops to $v_2 = 25\ \text{m/s}$ . Find the new wavelength $\lambda_2$ , and state which quantity among $v$ , $f$ , $\lambda$ stayed fixed across the junction and why.

Step 1. Find the wave speed on the light section using $v = f\lambda$ : $v_1 = f\lambda_1 = 100 \times 0.5 = 50\ \text{m/s}$ . This matches the Pass-1 numbers for this exact wave — a light-to-heavy string junction always slows a transverse wave down, since $v=\sqrt{T/\mu}$ falls as $\mu$ rises for the same tension.

Step 2. Identify what is fixed across the junction. The driving source — whatever end of the string is being shaken — sets the frequency, and both sides of a continuous string must oscillate together at the junction point, so $f$ cannot change: $f = 100\ \text{Hz}$ on both sides.

Step 3. Use $v=f\lambda$ again, now on the heavy section, to find the new wavelength: $\lambda_2 = \dfrac{v_2}{f} = \dfrac{25}{100} = 0.25\ \text{m}$ — exactly half the original wavelength, because the speed also halved while frequency stayed fixed.

Step 4. State the conclusion explicitly: $f$ stayed fixed at $100\ \text{Hz}$ because it is set by the source, not the medium; both $v$ (from $50\ \text{m/s}$ to $25\ \text{m/s}$ ) and $\lambda$ (from $0.5\ \text{m}$ to $0.25\ \text{m}$ ) changed together, in the same proportion, precisely to keep $v=f\lambda$ satisfied on each side. Answer: $v_1 = 50\ \text{m/s}$ , $\lambda_2 = 0.25\ \text{m}$ , and $f = 100\ \text{Hz}$ is the invariant. Timing note: about ninety seconds. The whole problem is two applications of $v=f\lambda$ bracketing one conceptual sentence — do not be tempted to recompute $f$ on the heavy side; recognising that $f$ is source-fixed is the entire trick, and skipping straight to $\lambda_2 = v_2/f$ saves the most time.

Common traps

Believing wave speed depends on frequency or amplitude. For a fixed medium, $v$ is set only by the medium's mechanical properties; changing the source's frequency changes $\lambda$ (via $v=f\lambda$ ), never $v$ itself.
Confusing wave speed with particle (transverse) speed. Wave speed $v=\omega/k$ is how fast the pattern advances along $x$ ; particle speed $\partial y/\partial t$ is how fast one bit of the medium oscillates up and down at fixed $x$ — these have different formulas, different units in the same sense of "speed," and are not numerically equal in general.
Assuming frequency changes when a wave changes medium. It is $\lambda$ and $v$ that adjust at a boundary; $f$ is locked to the source and is unchanged on both sides.
Sloppy unit conversion, especially for frequencies quoted in kHz/MHz or wavelengths in cm/mm/nm — always convert everything to SI (Hz, m, m/s) before applying $v=f\lambda$ , or the arithmetic silently goes wrong by powers of ten.
Using $v=f\lambda$ as if it "gives" the wave speed from nothing. It is a relation between three quantities; you need an independent piece of information (either the medium's own speed formula, or two of the three quantities given directly) before it can be solved.

Common trapWave speed is not particle speed, and it is not set by the source's frequency

The single biggest source of lost marks in this topic is blurring two distinct speeds and two distinct causes. First: the wave speed $v = f\lambda = \omega/k$ describes how fast the shape of the disturbance travels through the medium; the particle (or transverse) speed $\partial y/\partial t$ describes how fast an individual point of the medium moves up and down as the wave passes it — these come from different derivatives of $y(x,t)$ and are, in general, different numbers. Second: wave speed is a property of the medium (tension and mass per length for a string, bulk modulus and density for a fluid) and stays fixed as long as the medium does not change, even if the source's frequency is dialled up or down — only the wavelength responds to a frequency change, via $\lambda = v/f$ with $v$ held fixed.

You should be able to…

Derive $v = f\lambda$ starting from the phase condition $kx - \omega t = \text{constant}$ for the displacement relation $y(x,t) = A\sin(kx-\omega t)$ .
Apply $v=f\lambda$ fluently in both directions — finding $\lambda$ given $v$ and $f$ , or $f$ given $v$ and $\lambda$ , with correct SI unit conversion.
Explain, in one sentence, why wave speed is fixed by the medium's mechanical properties and not by the source's frequency.
Predict what changes ( $v$ , $\lambda$ ) and what stays fixed ( $f$ ) when a wave crosses from one medium into another.
Distinguish wave speed ( $\omega/k$ , the speed of the pattern) from particle speed ( $\partial y/\partial t$ , the speed of a point of the medium) both conceptually and by their different formulas.
Recognise that $v=f\lambda$ is a bookkeeping relation, not a source of new physical information, and know when a medium-specific formula ( $v=\sqrt{T/\mu}$ or Newton–Laplace) is additionally needed.

Ask the tutor: Want to see $v=f\lambda$ combined with the string-wave formula $v=\sqrt{T/\mu}$ in one worked problem that starts from tension and mass per length and ends at wavelength?

Why does wavelength change but frequency stay the same when a wave enters a new medium?

Frequency is set by whatever is driving the wave — the hand shaking the string, the tuning fork vibrating the air — and at the boundary between two media, the point of contact has to move at exactly one rate that both sides agree on, so $f$ cannot suddenly jump. Wave speed, on the other hand, is a property of each medium separately (its tension and mass density, or its elastic and inertial properties), so $v$ genuinely does change at the boundary. Since $v=f\lambda$ must hold on both sides with the same $f$ , the wavelength $\lambda = v/f$ is forced to change in exact proportion to $v$ .

Is v = f lambda a law of nature, or does it just come from the definitions of f and lambda?

It is best thought of as a consequence of definitions plus the wave picture, not an independent law like Newton's second law. Once you define wavelength as the spatial repeat-distance and frequency as the temporal repeat-rate of a periodic wave, tracking a fixed point of the shape (a crest) automatically gives $v=f\lambda$ — it is baked into what "periodic travelling wave" means. The genuine physics — the part that requires real experimental input about a real medium — is in formulas like $v=\sqrt{T/\mu}$ , which tell you the actual numerical value of $v$ for a given physical system.

If wave speed does not depend on frequency, why do different musical notes on the same string sound different?

Different notes correspond to different frequencies of vibration, but the wave speed on that particular string (fixed tension, fixed mass per length) stays the same for every note you play on it — what changes between notes is either the wavelength (by changing where you press the string, i.e. the effective length) or you retune the string's tension, which does change $v=\sqrt{T/\mu}$ and hence the note. The frequency you hear is a consequence of the standing-wave condition on that length of string, not a free choice that alters the travelling-wave speed on its own.

How is wave speed different from the speed of sound I hear quoted, like 340 metres per second?

They are the same idea applied to one specific medium. "The speed of sound in air, 340 m/s" is exactly the wave speed $v$ for sound waves in air at typical room conditions — a number fixed by air's bulk modulus and density through the Newton–Laplace relation, which this chapter only asks you to know qualitatively. Once you have that one number, $v=f\lambda$ lets you convert between any audible frequency and its corresponding wavelength in air, which is exactly what the board-level worked example above does.