Describing a Wave: y(x,t), Amplitude, Wavelength & Phase

Every signal you send or receive is a wave function with the numbers filled in. A radio station at 100 MHz is broadcasting $y = A\sin(kx - \omega t)$ with $\omega = 2\pi \times 10^8$ rad s⁻¹; your phone's modem recovers the message by tracking tiny, deliberate changes in the amplitude $A$, the frequency $f$ or the phase constant $\varphi$ — amplitude, frequency and phase modulation, the three ways to write information onto a carrier wave. The phase-difference equation (4) is the beating heart of every technology that locates things by waves: GPS compares the phase of signals from several satellites, noise-cancelling headphones generate a wave exactly $\pi$ out of phase with the noise ($\Delta x = \lambda/2$) to cancel it, and phased-array radar and ultrasound steer their beams purely by adjusting the phase constant across a row of emitters. And the split this article insists on — particle motion versus wave motion — is the same split a seismologist, an audio engineer and a fibre-optic technician all live by: the medium stays home while the pattern, and the information it carries, travels.

A wave lives in space and time, so a single graph can only ever show you a slice of it. There are two natural slices, and telling them apart is the whole game. Freeze time — imagine a camera flash at the instant $t = t_0$ — and plot the displacement $y$ of every particle against its position $x$. This snapshot graph is a photograph of the medium's shape: a sine curve laid out in space, and the distance over which it repeats is the wavelength $\lambda$. The horizontal axis is measured in metres. Now do the opposite. Fix on one particle, at position $x = x_0$, and plot its displacement $y$ against time $t$ as the wave rolls past. This history graph is the story of one particle's up-and-down life: again a sine curve, but the distance over which it repeats is the period $T$, and the horizontal axis is measured in seconds. Both graphs are sine curves, and to a careless eye they look identical — which is exactly the trap. One repeats in space (read off $\lambda$, in metres); the other repeats in time (read off $T$, in seconds). A wave function $y(x,t)$ is the single object that contains both slices at once, and everything below is bookkeeping on that object.

where: $y$ is the displacement of the particle at position $x$ at time $t$; $A$ is the amplitude, the greatest displacement (same units as $y$); $k$ is the wave number; $\omega$ is the angular frequency; and $\varphi$ is the phase constant, which sets what the wave is doing at $x = 0$, $t = 0$. The whole bracket, $kx - \omega t + \varphi$, is the phase — the stage the oscillation has reached at that place and moment, in radians. The two structural constants are fixed by the wave's repeats in space and time:

where: $\lambda$ is the wavelength (m), $f$ the frequency (Hz), $T$ the period (s). Read $k$ as "radians of phase per metre you move" and $\omega$ as "radians of phase per second you wait". A full cycle is $2\pi$ radians, so travelling one wavelength or waiting one period each advances the phase by exactly $2\pi$ — which is what equations (2) say. Note the units carefully: $k$ is radians per metre, not "the number of waves per metre" (that would be $1/\lambda$, smaller by a factor of $2\pi$). Combine the two and the wave's speed falls out. A crest is a point of fixed phase; for the phase $kx - \omega t$ to stay constant as time passes, $x$ must grow at the rate

where: $v$ is the wave (or phase) speed — the speed at which a crest moves along. Equation (3) is $v = f\lambda$, the subject of the next article, here dropping straight out of the wave function. Notice it is $\omega$ divided by $k$, never multiplied: a classic slip is to write $v = \omega k$, which is dimensionally wrong and can be off by orders of magnitude.

Here is the discipline, and it is worth drilling until it is automatic. When you are handed a graph, look at the label on the horizontal axis before you read anything off it. If the horizontal axis is position $x$ (metres), you have a snapshot — a frozen photograph of the wave in space. The repeat distance along it is the wavelength $\lambda$. You may read $\lambda$, and hence $k = 2\pi/\lambda$, but you cannot read the period or frequency from it, because time is frozen. If the horizontal axis is time $t$ (seconds), you have a history — the motion of one particle. The repeat interval along it is the period $T$. You may read $T$, and hence $f = 1/T$ and $\omega = 2\pi/T$, but you cannot read the wavelength from it, because you are standing at a single point. To get the wave speed you need one of each: $\lambda$ from a snapshot and $T$ (or $f$) from a history, combined as $v = \lambda/T = f\lambda$. A graph on its own gives you only half the wave. Examiners deliberately print a history graph and ask for the wavelength, or a snapshot and ask for the frequency; the correct answer is "not obtainable from this graph alone" unless the wave speed is also supplied.

The relative sign of the $x$ and $t$ terms tells you which way the wave travels, with no computation at all. Track a crest, a point of constant phase. For $y = A\sin(kx - \omega t)$, holding the phase $kx - \omega t$ constant as $t$ increases forces $x$ to increase: the crest moves toward larger $x$, so the wave travels in the +x direction. For $y = A\sin(kx + \omega t)$, holding $kx + \omega t$ constant as $t$ increases forces $x$ to decrease: the wave travels in the −x direction. The rule is blunt and worth memorising: opposite signs on $x$ and $t$ mean +x; the same sign means −x. (Some books write $y = A\sin(\omega t - kx)$, which is the same +x wave with an overall sign flip that does not move the crest anywhere — the relative sign is all that matters.)

The phase $kx - \omega t + \varphi$ is the wave's internal clock. Two questions come up constantly: how does the phase differ between two places at one instant, and between two times at one place? Between two points separated by $\Delta x$, at the same instant, only the $kx$ term changes, so

where: $\Delta\varphi$ is the phase difference (rad) and $\Delta x$ the separation (m). Read it as a proportion: a full wavelength of separation is a full $2\pi$ out of step (which is the same as being in step); half a wavelength is $\pi$ (exactly out of phase); a quarter-wavelength is $\pi/2$, or 90°. This one equation is the engine of every interference and standing-wave problem later in the chapter, where "path difference" is exactly this $\Delta x$. Between two instants separated by $\Delta t$, at the same point, only the $\omega t$ term changes:

so waiting a quarter-period turns the same 90° of phase. Equations (4) and (5) are the space and time faces of the one idea: phase advances by $2\pi$ per wavelength travelled and per period elapsed.

This is the deepest point in the article, and the one boards probe hardest. As the wave passes, each particle is doing simple harmonic motion across the wave (for a transverse wave) — the SHM of SHM Kinematics: x(t), v(t), a(t) & Phase. Its displacement swings between $+A$ and $-A$, so, exactly as in SHM, its speed and acceleration peak at

where: $v_{p,\max}$ is the greatest speed of a medium particle (m s⁻¹) — reached as it flies through its rest position — and $a_{\max}$ its greatest acceleration (m s⁻²), reached at the extremes of the swing. These describe the particle's motion up and down. The wave speed $v = \omega/k$ is a completely separate quantity — the speed at which the crest pattern moves along. The two need not resemble each other at all; their ratio is a pure number,

fixed entirely by how big the amplitude is compared with the wavelength. Since $A$ is usually far smaller than $\lambda$, the particle typically crawls up and down far more slowly than the wave races along — but crank the amplitude up and the particle can, in principle, outrun the wave. They are different animals: the medium bobs; the pattern travels. One graphical result ties them together, and you may use it as a tool. The particle's velocity equals minus the wave speed times the slope of the snapshot graph at that point, $v_p = -v \times (\text{slope of } y\text{ vs } x)$. Where the snapshot rises steeply going forward, the particle just behind is about to drop; where the snapshot is flat (a crest or trough), the slope is zero and the particle is momentarily at rest. This is justified by the geometry of a rigid shape sliding sideways — no calculus required to use it, though a slope is a calculus idea underneath.

Board level — decoding a wave function (CBSE/ICSE/A-Level pattern)

A transverse wave on a string is $y = 0.05\sin(10x - 40t)$, with all quantities in SI units. Find the amplitude, wave number, angular frequency, wavelength, frequency, period, wave speed and direction, and the maximum speed and acceleration of a string particle.

Step 1. Read the three constants straight off the bracket and prefactor: $A = 0.05$ m ($= 5$ cm), $k = 10$ rad m⁻¹, $\omega = 40$ rad s⁻¹. The sign is $(kx - 40t)$ — opposite signs — so the wave travels in +x.

Step 2. Wavelength and the time quantities: $\lambda = 2\pi/k = 2\pi/10 = \boxed{0.6283\ \mathrm{m}}$; $f = \omega/2\pi = 40/6.2832 = \boxed{6.366\ \mathrm{Hz}}$; $T = 1/f = 2\pi/\omega = \boxed{0.1571\ \mathrm{s}}$.

Step 3. Wave speed — $\omega$ divided by $k$: $v = \omega/k = 40/10 = \boxed{4.0\ \mathrm{m\,s^{-1}}}$. Cross-check with $v = f\lambda = 6.366 \times 0.6283 = 4.0$ m s⁻¹. Agreed.

Step 4. Particle motion, a different pair of numbers: $v_{p,\max} = A\omega = 0.05 \times 40 = \boxed{2.0\ \mathrm{m\,s^{-1}}}$ and $a_{\max} = A\omega^2 = 0.05 \times 1600 = \boxed{80\ \mathrm{m\,s^{-2}}}$. Note the particle's top speed (2.0 m s⁻¹) is half the wave speed (4.0 m s⁻¹) — as it must be, since $v_{p,\max}/v = Ak = 0.05 \times 10 = 0.5$.

Competitive level — phase difference and the two speeds (JEE-Main/NEET pattern)

For the same wave $y = 0.05\sin(10x - 40t)$ (SI): (a) find the phase difference between two points 0.157 m apart at one instant; (b) confirm the ratio of a particle's maximum speed to the wave speed; (c) state where in its cycle a particle is moving fastest.

Step 1. Part (a): use $\Delta\varphi = k\,\Delta x = 10 \times 0.157 = \boxed{1.57\ \mathrm{rad} = 90°}$. Since $0.157 \approx \lambda/4 = 0.6283/4 = 0.1571$ m, the two points are a quarter-wavelength apart, so a 90° phase difference is exactly right — a useful landmark to sanity-check against.

Step 2. Part (b): $v_{p,\max}/v = A k = 0.05 \times 10 = \boxed{0.5}$. So the particle's fastest motion (2.0 m s⁻¹) is half the wave speed (4.0 m s⁻¹). This ratio is a pure number set by $2\pi A/\lambda$; it says nothing mystical — just that this amplitude is small compared with this wavelength.

Step 3. Part (c): a particle in SHM moves fastest as it passes through its rest position ($y = 0$), where all its energy is kinetic, and is instantaneously at rest at the crest and trough ($y = \pm A$), where the snapshot slope is zero. The maximum-speed value there is the $v_{p,\max} = 2.0$ m s⁻¹ of part (b). Timing note: about 90 seconds. The examiner's trap is to slip a unit — leaving Δx in centimetres, or reading the wave speed as $\omega k = 400$ instead of $\omega/k = 4$. Write $v = \omega/k$ every time, and check that a quarter-wavelength gives 90° before trusting your phase answer.

Reading the wrong quantity off a graph. A $y$-vs-$x$ snapshot gives the wavelength; a $y$-vs-$t$ history gives the period. Check the horizontal-axis label first. If asked for the wave speed from a single graph, the honest answer is "need the other quantity too" — you cannot get $v$ from one graph alone.

$v = \omega k$ instead of $\omega/k$. The wave speed is angular frequency divided by wave number. Multiplying gives a number with the wrong units and often a wildly wrong size (here 400 instead of 4). Memorise the division.

$k$ as "waves per metre". The wave number is $2\pi/\lambda$ in radians per metre, not $1/\lambda$. The two differ by a factor of $2\pi \approx 6.28$ — a favourite distractor.

Setting particle speed equal to wave speed. $v_{p,\max} = A\omega$ (the particle bobbing) and $v = \omega/k$ (the wave travelling) are different quantities. Their ratio is $2\pi A/\lambda$, rarely one. Keep them in separate boxes.

Calculator in degree mode, or dropping $\varphi$. The phase is in radians; degree mode corrupts every displacement value. And if the wave is given with a phase constant, carry it — it fixes the wave's state at the origin and cannot be silently dropped.

Board scope — IGCSE. The wave function $y(x,t)$ is not on the IGCSE 0625 syllabus; IGCSE stops at the wave model and $v = f\lambda$. If you are an IGCSE candidate, read this article for understanding, but you will only be examined on $v = f\lambda$ and the shapes of waves, not on $A\sin(kx - \omega t + \varphi)$ or phase difference.

Texts. NCERT Physics Class 11, Ch. 15 "Waves", §§15.3–15.4 (the wave function, wavelength, wave number, phase) · H. C. Verma, Concepts of Physics Vol. 1, "Wave Motion and Waves on a String" · Halliday, Resnick & Walker, "Waves — I", the travelling-wave sections · Cambridge A-Level 9702 topic 7, "Waves" (graphical treatment of phase difference).

Provenance. Standard results. Recompute in `verify_hs2-describing-a-wave.py` for the anchor wave $y = 0.05\sin(10x - 40t)$: $A = 0.05$ m; $\lambda = 2\pi/10 = 0.6283$ m; $f = 40/2\pi = 6.366$ Hz; $T = 2\pi/40 = 0.1571$ s; $v = \omega/k = 40/10 = 4.0$ m s⁻¹, cross-checked as $f\lambda = 4.0$; $v_{p,\max} = A\omega = 2.0$ m s⁻¹; $a_{\max} = A\omega^2 = 80$ m s⁻²; the ratio $v_{p,\max}/v = Ak = 0.5$; the delay identity $\omega/v = k = 10$; and the phase difference $\Delta\varphi = k\,\Delta x = 10 \times 0.157 = 1.57$ rad $= 90°$ for $\Delta x = 0.157 \approx \lambda/4$ m.

Prerequisites. Wave Motion: Transverse & Longitudinal — the crest/trough picture this article puts to an equation · SHM Kinematics: x(t), v(t), a(t) & Phase — the single-particle motion the wave is built from, and the source of $v_{p,\max} = A\omega$. Leads to. Wave Speed: v = fλ — the medium's grip on $v = \omega/k$ · the wave-equation numericals archetype, which drills reading and building $y(x,t)$.

Go deeper (UG). Where the form $y = f(x - vt)$ comes from, and why $\partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2$ has exactly d'Alembert's two-shape solution, is developed in The One-Dimensional Wave Equation in the university mechanics track. It is enrichment; nothing here depends on it.