ARCHETYPE: Reading & Building the Wave Equation

Somewhere in every waves paper — CBSE, ICSE, JEE-Main, NEET — sits a line of algebra pretending to be a physics question. It reads something like $y = 0.05 \sin(10x - 40t)$ , and then it asks for the wavelength, the frequency, the speed, the direction, the fastest the string ever moves. Students see six sub-questions and six chances to slip. There is only one skill. The skill runs in two directions, and they are the same skill wearing two coats. Reading: an equation is handed to you and you strip its physics off the argument — the amplitude sitting outside, the numbers multiplying $x$ and $t$ inside. Building: a wave is described in words — this tall, this long, this fast, going this way — and you assemble the equation that says it. Read enough of them and building becomes obvious; build a few and reading stops being a lottery. That is why this article treats the setup itself as the concept — an ARCHETYPE. The wave equation $y = A\sin(kx - \omega t + \varphi)$ is not a formula to memorise; it is a form with five slots, and every problem is either "read the slots" or "fill the slots." Master the slot machine and an entire family of exam questions collapses into bookkeeping. We will fix the master pattern, run it forwards and backwards, price the traps examiners bait it with — the deadly $\omega \cdot k$ , the centimetre-metre slip, the calculator stuck in degrees — and hand you a card you can run in ninety seconds. One honest word on scope up front. The algebraic wave function is an India-facing outcome — CBSE, ICSE and the JEE/NEET band live on it. A-Level, IB and AP treat the same wave graphically, from its snapshot, and never demand this algebra. If that is your board, read this for the ideas; the numbers are not on your paper.

Key takeaways

One equation runs the whole topic: $y = A\sin(kx - \omega t + \varphi)$ — a form with five slots (amplitude, wave number, angular frequency, sign, phase).
Reading strips the constants straight off the argument: $A$ sits outside, $k$ multiplies $x$ , $\omega$ multiplies $t$ . Then $\lambda = 2\pi/k$ , $f = \omega/2\pi$ , $T = 1/f$ .
Speed is $v = \omega/k$ — never $\omega \cdot k$ . The relative sign fixes direction: $(kx - \omega t)$ travels in $+x$ , $(kx + \omega t)$ in $-x$ .
Particle quantities come from $A$ and $\omega$ alone: max particle speed $v_{p,\max} = A\omega$ , max particle acceleration $a_{\max} = A\omega^2$ — nothing to do with the wave speed $v$ .
Building reverses the reading: turn $\lambda$ into $k = 2\pi/\lambda$ , turn $f$ into $\omega = 2\pi f$ , choose the sign from the direction, set $\varphi$ from the initial condition.
Key idea: the wave equation is a five-slot form — read the slots off the argument, or fill them from the words, and every wave-equation problem becomes fill-in-the-blanks.

The core insight

Every wave-equation problem is one of two moves on the single form $y = A\sin(kx - \omega t + \varphi)$ : read the five constants off a given equation, or assemble them from a description — and the only quantities you ever compute are $\lambda = 2\pi/k$ , $f = \omega/2\pi$ , $v = \omega/k$ , and the particle pair $A\omega$ , $A\omega^2$ .

Where this lives today

The habit of reading a signal off its equation is exactly what a phone does thousands of times a second. Every note, every syllable, every pixel of streamed video arrives as a sum of sinusoids, and the receiver's job is to strip the amplitude, frequency and phase off each one — the same three slots you read off $y = A\sin(kx - \omega t + \varphi)$ , only done in silicon and at radio frequencies. When your handset locks onto a $900\ \mathrm{MHz}$ carrier, it is solving $\omega = 2\pi f$ for a wave with a wavelength of about a third of a metre; when it corrects for the Doppler shift of a moving car, it is watching $\omega$ change and inferring a speed. Medical ultrasound builds an image by sending a wave equation into tissue and reading the reflected one, timing the echoes to place a boundary. Seismologists read the ground's motion as a superposition of travelling terms and solve for where the rock broke. In every case the underlying move is the one on this page: a wave is five numbers, and physics is the art of getting them out of, or into, the argument of a sine.

The setup

In an archetype the setup is the physics, so start by looking hard at the form itself. A single travelling sinusoid, written in SI units, is

y(x,t) = A\sin(kx - \omega t + \varphi) \qquad (1)

where: $y$ — the displacement of the particle at position $x$ at time $t$ (metres); $A$ — the amplitude, the largest displacement (metres); $k$ — the wave number, $2\pi$ divided by a wavelength (radians per metre); $\omega$ — the angular frequency, $2\pi$ times the frequency (radians per second); and $\varphi$ — the phase constant that sets where the wave is at $x = 0$ , $t = 0$ . Five numbers. Fix all five and the wave is completely specified; every question on this page either hands you those five and asks what they mean, or describes the wave and asks you to name them. The single most important habit is to see the argument $(kx - \omega t + \varphi)$ as a piece of bookkeeping with three separate jobs. The number multiplying $x$ is always $k$ ; the number multiplying $t$ is always $\omega$ ; the amplitude is always the factor sitting outside the sine, never inside. Confuse those three positions and every downstream number is wrong. Get them right and the rest is arithmetic.

Figure

Reading the five slots of a wave equation — where each quantity lives and what you compute from it

Two facts about the form protect more marks than anything else on the page, so pin them down before any numbers.

Wave speed is a ratio, particle speed is a product. The speed at which the pattern advances is $v = \omega/k$ — a division. The fastest any particle of the medium ever moves is $v_{p,\max} = A\omega$ — a multiplication involving the amplitude, which does not appear in $v$ at all. These are different physical quantities with different formulas, and the exam's favourite trap is to blur them. A string can carry a wave at $340\ \mathrm{m/s}$ while no particle of it moves faster than a few millimetres per second.

The relative sign, not either sign alone, sets the direction. A minus between the $x$ term and the $t$ term — $(kx - \omega t)$ — is a wave marching in $+x$ ; a plus — $(kx + \omega t)$ — marches in $-x$ . The reasoning is one line: to stay on a chosen crest you keep $kx - \omega t$ constant, so as $t$ grows $x$ must grow too. Reverse the inner sign and the crest runs backwards. Nothing else in the equation touches direction.

The master solution pattern

Every wave-equation problem, read or build, runs the same six steps. Learn them as a fixed sequence and you will never stare at a blank margin wondering where to start.

Step 1. Match the given equation to the standard form $y = A\sin(kx \pm \omega t + \varphi)$ , or — if you are building — write the standard form down and prepare to fill it. Get the units of $y$ , $x$ and $t$ straight now: metres and seconds, or centimetres and seconds. This is where the cm/m disasters are prevented.

Step 2. Strip the constants. The amplitude $A$ is the factor outside the sine. The wave number $k$ is whatever multiplies $x$ . The angular frequency $\omega$ is whatever multiplies $t$ . Write all three down with their units before doing anything with them.

Step 3. Convert to the everyday quantities. Wavelength $\lambda = 2\pi/k$ ; frequency $f = \omega/2\pi$ ; period $T = 1/f = 2\pi/\omega$ . These three are pure conversions — no physics, just $2\pi$ .

Step 4. Get the wave speed and direction. Speed $v = \omega/k$ — a division, and if a voice in your head says "multiply," silence it. Direction: minus sign inside means $+x$ , plus sign means $-x$ .

Step 5. Get the particle quantities, if asked, from $A$ and $\omega$ only. Maximum particle speed $v_{p,\max} = A\omega$ ; maximum particle acceleration $a_{\max} = A\omega^2$ . The wave speed plays no part here.

Step 6. Sanity-check the magnitudes. Is $v$ physically sensible for the medium? Is $v_{p,\max}$ a small fraction of $v$ (it usually is)? Does the direction match the sign? A wave equation that survives all four checks is almost always right. Two conversions carry the whole build direction, so state them as numbered results. From a wavelength,

k = \frac{2\pi}{\lambda} \qquad (2)

where: $\lambda$ — the wavelength in metres, the distance between successive crests. Read it as "how many radians of the sine you turn through per metre of string" — a long wave turns slowly, so its $k$ is small. And from a frequency,

\omega = 2\pi f = \frac{2\pi}{T} \qquad (3)

where: $f$ — the frequency in hertz, the number of full oscillations a particle makes each second; $T$ — the period, the seconds per oscillation. Together with the direction sign and the initial condition that fixes $\varphi$ , equations (2) and (3) turn any spoken description into the form (1). Building is nothing more than reading run in reverse.

Derivation AThe master pattern in full — reading and building the anchor wave

The point of this derivation is to run the six-step pattern once, slowly and completely, on a single wave, so that the sequence becomes a reflex you can trust under exam pressure. We take the anchor wave $y = 0.05\sin(10x - 40t)$ in SI units, extract every quantity a paper could ask for, and then immediately turn round and rebuild the equation from those quantities — proving that reading and building are one skill, not two. Watch the units at every line: this is exactly where marks are won and lost.

Step 1. The equation is in SI ( $y$ , $x$ in metres, $t$ in seconds) and matches the standard form $y = A\sin(kx - \omega t)$ with $\varphi = 0$ . The inner sign is minus, so hold the thought: this wave travels in $+x$ .

Step 2. Strip the constants by position. Outside the sine: $A = 0.05\ \mathrm{m}$ , which is $5\ \mathrm{cm}$ . Multiplying $x$ : $k = 10\ \mathrm{rad/m}$ . Multiplying $t$ : $\omega = 40\ \mathrm{rad/s}$ .

Step 3. Convert. Wavelength $\lambda = 2\pi/k = 2\pi/10 = 0.6283\ \mathrm{m}$ . Frequency $f = \omega/2\pi = 40/2\pi = 6.366\ \mathrm{Hz}$ . Period $T = 1/f = 0.1571\ \mathrm{s}$ , which equals $2\pi/40$ as a check.

Step 4. Speed and direction. $v = \omega/k = 40/10 = 4.0\ \mathrm{m/s}$ , and because the inner sign is minus, the wave moves in $+x$ . Notice the trap sitting right here: $\omega \cdot k = 400$ , a plausible-looking number that is a hundredfold too big and physically meaningless.

Step 5. Particle quantities from $A$ and $\omega$ . Maximum particle speed $v_{p,\max} = A\omega = 0.05 \times 40 = 2.0\ \mathrm{m/s}$ . Maximum particle acceleration $a_{\max} = A\omega^2 = 0.05 \times 40^2 = 80\ \mathrm{m/s^2}$ .

Step 6. Rebuild, to prove the loop closes. Given " $A = 5\ \mathrm{cm}$ , $\lambda = 0.6283\ \mathrm{m}$ , $f = 6.366\ \mathrm{Hz}$ , moving in $+x$ , zero at the origin," we compute $k = 2\pi/\lambda = 10\ \mathrm{rad/m}$ , $\omega = 2\pi f = 40\ \mathrm{rad/s}$ , choose the minus sign for $+x$ , set $\varphi = 0$ , and write $y = 0.05\sin(10x - 40t)$ — the wave we started from. Verified: $\lambda = 0.6283$ m, $f = 6.366$ Hz, $T = 0.1571$ s, $v = 4.0$ m/s, $v_{p,\max} = 2.0$ m/s, $a_{\max} = 80$ m/s² are recomputed in verify_hs2-wave-equation-problems.py.

Go deeper · Derivation B · 1 minA variant — the mixed-units wave, where the trap is the whole question

Derivation BA variant — the mixed-units wave, where the trap is the whole question

This variant exists because the deadliest wave-equation problems are not the ones with hard physics but the ones with mixed units, and the master pattern must survive them unchanged. We take $y = 3.0\sin(0.02\pi x - 4\pi t)$ with $y$ and $x$ measured in centimetres and $t$ in seconds — a genuine JEE-style trap — and run exactly the same six steps, converting to SI only at the very end, when a number needs to leave the problem. The lesson is that $k = 2\pi/\lambda$ and $v = \omega/k$ do not care what units you use, provided you keep them consistent and label every result.

Step 1. Units are centimetres for $y$ and $x$ , seconds for $t$ . So $k$ will come out in $\mathrm{rad/cm}$ and any speed in $\mathrm{cm/s}$ until we choose to convert. The inner sign is minus: the wave travels in $+x$ .

Step 2. Strip: $A = 3.0\ \mathrm{cm}$ ; $k = 0.02\pi\ \mathrm{rad/cm}$ ; $\omega = 4\pi\ \mathrm{rad/s}$ .

Step 3. Wavelength $\lambda = 2\pi/k = 2\pi/(0.02\pi) = 100\ \mathrm{cm} = 1.0\ \mathrm{m}$ — the $\pi$ cancels, which is the whole reason the setter chose $0.02\pi$ . Frequency $f = \omega/2\pi = 4\pi/2\pi = 2.0\ \mathrm{Hz}$ .

Step 4. Speed $v = \omega/k = 4\pi/(0.02\pi) = 200\ \mathrm{cm/s} = 2.0\ \mathrm{m/s}$ . Cross-check by the independent road $v = f\lambda = 2.0 \times 1.0 = 2.0\ \mathrm{m/s}$ — they agree, so the unit bookkeeping held.

Step 5. Particle speed $v_{p,\max} = A\omega = 3.0 \times 4\pi = 12\pi = 37.70\ \mathrm{cm/s} = 0.377\ \mathrm{m/s}$ . Keep $A$ in the same length unit as you want the answer; mixing a centimetre amplitude into a metre answer is the second trap hiding in this line. Verified: $\lambda = 100$ cm $= 1.0$ m, $f = 2.0$ Hz, $v = 200$ cm/s $= 2.0$ m/s, $v_{p,\max} = 37.70$ cm/s $= 0.377$ m/s are recomputed in verify_hs2-wave-equation-problems.py.

Common trapThe wave speed and the particle speed are the same kind of thing

They are not, and treating them as interchangeable is the single most expensive habit on this page. The wave speed $v = \omega/k$ is how fast the pattern — the crests, the shape — sweeps down the medium. The particle speed is how fast a single bit of the medium jiggles up and down about its rest position, and its maximum is $v_{p,\max} = A\omega$ . One is a property of the wave and the medium; the other depends on how hard you shook the source, through the amplitude $A$ . They are not even close in size, and their ratio is worth memorising: $v_{p,\max}/v = A\omega / (\omega/k) = Ak = 2\pi A/\lambda$ . For the anchor wave that ratio is $A k = 0.05 \times 10 = 0.5$ , so the fastest the string ever moves ( $2.0\ \mathrm{m/s}$ ) is half the speed at which the wave travels ( $4.0\ \mathrm{m/s}$ ). Because $A$ is almost always tiny next to $\lambda$ , this ratio is normally far below one — the particles barely twitch while the wave races past. The moment a problem asks for "the maximum speed of a point on the string," reach for $A\omega$ , and never, ever quote the wave speed.

Worked examples

1 · Board level — reading a wave equation completely (CBSE/ICSE pattern, 3 marks)

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

Step 1. SI units, standard form, $\varphi = 0$ , inner sign minus (so $+x$ ). Strip the constants: $A = 0.05\ \mathrm{m} = 5\ \mathrm{cm}$ , $k = 10\ \mathrm{rad/m}$ , $\omega = 40\ \mathrm{rad/s}$ .

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

Step 3. Speed and direction: $v = \omega/k = 40/10 = \boxed{4.0\ \mathrm{m/s},\ +x}$ . Guard: $\omega \cdot k = 400$ is the wrong answer the paper is fishing for — a wave on a string does not move at $400\ \mathrm{m/s}$ here.

Step 4. Particle quantities from $A$ and $\omega$ : $v_{p,\max} = A\omega = 0.05 \times 40 = \boxed{2.0\ \mathrm{m/s}}$ ; $a_{\max} = A\omega^2 = 0.05 \times 1600 = \boxed{80\ \mathrm{m/s^2}}$ . Sanity: $v_{p,\max}$ is half of $v$ , comfortably below it, as it should be.

2 · Board level — building a wave equation from a description (CBSE/NEET pattern, 3 marks)

Write the equation of a transverse wave of amplitude $2.0\ \mathrm{cm}$ and wavelength $0.50\ \mathrm{m}$ , oscillating at $20\ \mathrm{Hz}$ , travelling in the $-x$ direction, with the particle at $x = 0$ at its rest position ( $y = 0$ ) at $t = 0$ . State the wave speed.

Step 1. Write the target form and fill it slot by slot. $A = 2.0\ \mathrm{cm} = 0.02\ \mathrm{m}$ .

Step 2. Wave number from the wavelength: $k = 2\pi/\lambda = 2\pi/0.50 = 4\pi = \boxed{12.566\ \mathrm{rad/m}}$ .

Step 3. Angular frequency from the frequency: $\omega = 2\pi f = 2\pi \times 20 = 40\pi = \boxed{125.66\ \mathrm{rad/s}}$ .

Step 4. Direction and phase: $-x$ travel needs the plus inner sign, $(kx + \omega t)$ ; $y = 0$ at the origin at $t = 0$ means $\varphi = 0$ . Assemble: $y = 0.02\sin(12.566x + 125.66t)\ \mathrm{m}$ . Speed $v = \omega/k = 125.66/12.566 = \boxed{10\ \mathrm{m/s}}$ , which the independent road $v = f\lambda = 20 \times 0.50 = 10\ \mathrm{m/s}$ confirms.

3 · Competitive level — the mixed-units wave (JEE-Main/NEET pattern)

A wave is given by $y = 3.0\sin(0.02\pi x - 4\pi t)$ , where $y$ and $x$ are in centimetres and $t$ in seconds. Find the wavelength (in metres), the frequency, the wave speed (in $\mathrm{m/s}$ ) and the maximum particle speed (in $\mathrm{m/s}$ ).

Step 1. The trap is announced in the units line: centimetres, not metres. Keep everything in cm and s, convert only the final answers. Strip: $A = 3.0\ \mathrm{cm}$ , $k = 0.02\pi\ \mathrm{rad/cm}$ , $\omega = 4\pi\ \mathrm{rad/s}$ .

Step 2. $\lambda = 2\pi/k = 2\pi/(0.02\pi) = 100\ \mathrm{cm} = \boxed{1.0\ \mathrm{m}}$ ; the $\pi$ cancels by design. $f = \omega/2\pi = 4\pi/2\pi = \boxed{2.0\ \mathrm{Hz}}$ .

Step 3. $v = \omega/k = 4\pi/(0.02\pi) = 200\ \mathrm{cm/s} = \boxed{2.0\ \mathrm{m/s}}$ ; check $v = f\lambda = 2.0 \times 1.0 = 2.0\ \mathrm{m/s}$ ✓.

Step 4. $v_{p,\max} = A\omega = 3.0 \times 4\pi = 12\pi = 37.70\ \mathrm{cm/s} = \boxed{0.377\ \mathrm{m/s}}$ . Timing note: ninety seconds, and every one of the four sub-answers hangs on the same decision made in Step 1 — do not convert $0.02\pi\ \mathrm{rad/cm}$ into $\mathrm{rad/m}$ midway; keep the units you were given and convert once at the end. Candidates who "tidy up" the units first are the ones who lose this mark.

4 · Competitive level — particle speed versus wave speed (JEE-Advanced pattern)

For the wave $y = 0.05\sin(10x - 40t)$ (SI), (a) show that a particle of the string can never move faster than the wave; (b) find the particle's displacement when its speed is exactly half of its maximum; (c) rewrite the wave in the form $y = A\sin 2\pi(x/\lambda - t/T)$ .

Step 1. From Example 1, $A = 0.05\ \mathrm{m}$ , $k = 10\ \mathrm{rad/m}$ , $\omega = 40\ \mathrm{rad/s}$ , so $v = 4.0\ \mathrm{m/s}$ and $v_{p,\max} = A\omega = 2.0\ \mathrm{m/s}$ .

Step 2. Part (a). The ratio of the two speeds is $v_{p,\max}/v = A\omega/(\omega/k) = Ak = 0.05 \times 10 = \boxed{0.5}$ . Because this is less than $1$ , the fastest the string ever moves ( $2.0\ \mathrm{m/s}$ ) is only half the wave speed ( $4.0\ \mathrm{m/s}$ ): the particle never outruns the wave here.

Step 3. Part (b). Each particle executes SHM (this is the content of SHM Kinematics), so its speed is $A\omega$ at the centre and zero at the extremes, and it takes half its maximum speed when its displacement is $y = \pm\frac{\sqrt{3}}{2}A$ — the standard SHM result $v = \omega\sqrt{A^2 - y^2}$ with $v = \tfrac{1}{2}A\omega$ . Numerically $y = \pm 0.866 \times 0.05 = \boxed{\pm 0.0433\ \mathrm{m}} = \pm 4.33\ \mathrm{cm}$ .

Step 4. Part (c). Since $k = 2\pi/\lambda$ and $\omega = 2\pi/T$ , the argument factorises exactly: $kx - \omega t = 2\pi(x/\lambda - t/T)$ . With $\lambda = 0.6283\ \mathrm{m}$ and $T = 0.1571\ \mathrm{s}$ , $y = 0.05\sin 2\pi(x/0.6283 - t/0.1571)\ \mathrm{m}$ . This is the same wave; the $(t/T - x/\lambda)$ ordering some books use is the identical wave shifted by half a cycle. Timing note: about two minutes. Part (a) is a five-second ratio once you know $A k$ ; the time sink is part (b), so recall the SHM fact $v = \omega\sqrt{A^2 - y^2}$ rather than trying to reconstruct it — an archetype rewards recognised sub-results, not fresh derivations.

Figure

The two graphs of one wave — a snapshot in space gives the wavelength, a history in time gives the period

Exam traps

The limits of this topic live in this table too — the extreme readings are exactly where the one-mark distractors are bred.

the trap	why it is wrong	the fix
$v = \omega \cdot k$	speed is a ratio, not a product; $\omega k = 400$ for the anchor wave is a hundredfold nonsense	$v = \omega/k$ always — divide the $t$ -coefficient by the $x$ -coefficient
reading $\lambda$ off the history graph or $T$ off the snapshot	the snapshot is $y$ vs $x$ (space → $\lambda$ ); the history is $y$ vs $t$ (time → $T$ )	label the horizontal axis first, then read only what lives on it
$k$ read as "waves per metre"	$k = 2\pi/\lambda$ is in radians per metre; waves per metre is $1/\lambda$ , smaller by $2\pi$	always $\lambda = 2\pi/k$ , never $\lambda = 1/k$
cm left inside an SI answer	$A = 3.0\ \mathrm{cm}$ fed into $A\omega$ gives cm/s, not m/s — a factor of 100	keep one length unit throughout; convert only the final number
the direction sign	$(kx + \omega t)$ is $-x$ , $(kx - \omega t)$ is $+x$ ; the relative sign, not the sign of $\omega$ alone	to hold a crest, keep $kx \mp \omega t$ constant and see which way $x$ must move
calculator in degrees	$\sin(10x - 40t)$ has a radian argument; degree mode silently wrecks any numeric value	set radians before evaluating any sine in this chapter
the limits	$A \to 0$ kills the wave but not $v$ (speed is set by $k$ , $\omega$ , i.e. the medium); $\omega \to 0$ freezes it ( $v \to 0$ , $T \to \infty$ )	amplitude changes never change $v$ — a louder wave is not a faster one

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Switzerland

Leonhard Euler, the most prolific mathematician in history, gave physics the very idea that a wave is a function of position and time. One of his earliest papers, in 1727, was on sound — Dissertatio physica de sono, written for a chair at Basel — and he returned to vibrating strings for decades. When d'Alembert wrote down the wave equation in 1747 and solved it as $f(x - vt) + g(x + vt)$ , a fierce argument broke out over what kinds of shape $f$ were allowed to be. Euler took the boldest line: any shape you can draw, even one with corners, is a legitimate travelling wave, because a wave is whatever the string is doing, not whatever a tidy formula permits. That insistence — that $y(x,t)$ can be read straight from the physical shape — is precisely the habit this article trains. It took Fourier, in 1822, to prove Euler and Bernoulli had both been right.

History

1727

— Leonhard Euler, aged twenty, publishes an early paper, Dissertatio physica de sono, on the nature and propagation of sound.

1747

— Jean le Rond d'Alembert writes the vibrating string as the wave equation and solves it in the travelling-wave form f(x − vt) + g(x + vt).

1748

— Euler argues that any shape at all, however drawn, is a legitimate solution — the "vibrating-string controversy" begins.

1753

— Daniel Bernoulli insists every vibration is a sum of simple sinusoidal modes, the seed of the modern picture.

1759

— Joseph-Louis Lagrange enters the dispute, treating the string as a limit of many small beads.

1822

— Joseph Fourier proves that Bernoulli's sums can represent Euler's arbitrary shapes, reconciling the whole quarrel.

Going further

Texts. NCERT Physics Class 11, Ch. 15 "Waves", §15.2–15.4 · H. C. Verma, Concepts of Physics Vol. 1, "Wave Motion and Waves on a String" — the reading/building drills live here · D. C. Pandey, Waves and Thermodynamics, the wave-equation problem sets.

Provenance. Every number is recomputed in `verify_hs2-wave-equation-problems.py`: the anchor wave's $\lambda = 0.6283$ m, $f = 6.366$ Hz, $T = 0.1571$ s, $v = 4.0$ m/s, $v_{p,\max} = 2.0$ m/s, $a_{\max} = 80$ m/s², and the distractor $\omega k = 400$ ; the build $k = 12.566$ rad/m, $\omega = 125.66$ rad/s, $v = 10$ m/s; the mixed-units wave's $\lambda = 100$ cm $= 1.0$ m, $f = 2.0$ Hz, $v = 200$ cm/s $= 2.0$ m/s, $v_{p,\max} = 37.70$ cm/s $= 0.377$ m/s; the speed ratio $Ak = 0.5$ and the half-speed displacement $\pm 0.0433$ m.

Prerequisites. Describing a Wave: y(x,t), Amplitude, Wavelength & Phase — where the form (1) is built from a source doing SHM · Wave Speed: v = fλ — the speed this article reads off. The particle-SHM facts used in Example 4 are from SHM Kinematics: x(t), v(t), a(t) & Phase. Leads to. ARCHETYPE: String Tension & Wave-Speed Numericals — where the same $v$ is pinned to a real wire's tension and mass. Go deeper. The university treatment, in which this form is a solution of a differential equation rather than a shape to memorise, is The One-Dimensional Wave Equation — enrichment, not a dependency.

How do I instantly tell which way the wave is going?

Look only at the relative sign between the $x$ term and the $t$ term inside the sine. Minus, as in $(kx - \omega t)$ , means the wave travels toward $+x$ ; plus, as in $(kx + \omega t)$ , means it travels toward $-x$ . The quick reasoning: to ride a chosen crest you must keep the whole argument constant, so if it is $kx - \omega t$ , then as $t$ increases $x$ must increase to compensate — the crest moves to larger $x$ . Do not be distracted by whether $\omega$ or $k$ looks positive on its own; it is the sign between the two terms that decides.

Why is it v = ω/k and not ω × k? I keep mixing them up.

Check the units, and the mistake becomes impossible. $\omega$ is in radians per second, $k$ is in radians per metre. Divide them and the radians cancel: $(\mathrm{rad/s})/(\mathrm{rad/m}) = \mathrm{m/s}$ , a speed. Multiply them and you get $\mathrm{rad^2\,m^{-1}\,s^{-1}}$ , which is not a speed and not anything. So $v = \omega/k$ is the only combination that can possibly be a velocity. A second anchor: $v = f\lambda$ , and substituting $f = \omega/2\pi$ , $\lambda = 2\pi/k$ gives $v = (\omega/2\pi)(2\pi/k) = \omega/k$ — the same result by a different road.

What exactly is the difference between the wave speed and the particle speed?

The wave speed $v = \omega/k$ is how fast the shape travels — how quickly a crest moves down the string. The particle speed is how fast one small piece of the string moves up and down as the wave passes it, and its maximum is $v_{p,\max} = A\omega$ . They are different in kind and usually very different in size: their ratio is $A k = 2\pi A/\lambda$ , which is tiny whenever the amplitude is small compared with the wavelength. A wave can race along at hundreds of metres per second while every particle only shivers a few millimetres per second. When a question asks for "the maximum speed of a point on the string," it wants $A\omega$ , never $v$ .

Does the amplitude change the wave speed?

No — and this is worth over-learning, because it feels wrong. The wave speed depends only on the medium (for a string, on its tension and mass per length, the subject of the next article) and shows up in the equation through $k$ and $\omega$ . The amplitude $A$ sets how far the particles swing, and therefore the particle speed $A\omega$ and the energy the wave carries, but not how fast the pattern moves. Double the amplitude and the wave is "louder," carries four times the energy, and travels at exactly the same speed. Distractors that offer a bigger speed for a bigger amplitude are baiting this misconception.

The equation had no φ. Does that mean the phase is zero, or that I can ignore it?

It means it was chosen to be zero — the source happened to be at $y = 0$ and moving in a definite sense at $x = 0$ , $t = 0$ . You cannot simply ignore $\varphi$ ; when a problem specifies a different initial condition (say the particle starts at its crest), you must include a non-zero $\varphi$ to match it. A particle starting at maximum displacement needs a cosine, i.e. $\varphi = \pi/2$ , since $\sin(\theta + \pi/2) = \cos\theta$ . In building problems, $\varphi$ is the last slot you fill, and you fill it from the stated position at the origin.

My calculator gives a weird value for the sine. What am I doing wrong?

Almost certainly it is in degree mode. The argument $kx - \omega t$ is measured in radians — that is what $k$ in $\mathrm{rad/m}$ and $\omega$ in $\mathrm{rad/s}$ guarantee — so your calculator must be set to radians before you evaluate any sine or find any phase. A calculator in degrees will turn a phase of $1.57$ radians (a quarter cycle) into the sine of $1.57$ degrees, which is nearly zero, and every numeric answer downstream will be wrong. Switch to radian mode at the start of the waves chapter and leave it there.

Is this algebra actually on my board's paper?

Depends on the board, and it is worth knowing. The full algebraic wave function $y = A\sin(kx - \omega t + \varphi)$ and the numericals built on it are core for CBSE, ICSE and the JEE-Main, JEE-Advanced and NEET band — expect at least one every year. A-Level, IB and AP Physics 2 teach the same wave but assess it graphically: they will give you a snapshot or a history graph and ask you to read $\lambda$ , $T$ and $v$ off the axes, not to manipulate the equation. If you sit one of those boards, the graph-reading half of this article is your syllabus and the pure-algebra manipulations are enrichment.