Displacement relation in a progressive wave

Pluck a long horizontal string once, gently, and watch a single hump race away from your hand. Nothing about that hump is mysterious to the eye — it moves left or right at a steady speed and keeps its shape. But describing it in one clean equation, an equation that tells you the height of the string at every point, at every instant, without you having to draw a fresh picture for each moment, is a different and much more useful problem. That equation is the displacement relation, and for a continuous sinusoidal wave it is $y(x,t) = A\sin(kx - \omega t)$ — deceptively short for something that packs an entire moving wave into four symbols. The power of this one line is that it secretly contains two different stories, and you get to choose which one to read. Freeze time and let position vary, and you see the shape of the string right now — a smooth sine curve stretching along the string, one full hump every wavelength. Freeze position instead — stand at one point on the string, say $x = 0$ , and watch only that one bead of string — and you see the history of that single particle bobbing up and down forever, tracing out simple harmonic motion with the same period as the wave. A wave is nothing but a huge number of particles doing SHM, each one slightly delayed compared to its neighbour, and the displacement relation is the compact bookkeeping device that keeps track of all of them at once. This chapter takes the transverse-versus-longitudinal picture you already have and gives it algebra. By the end you will be able to look at an equation like $y(x,t) = 0.02\sin(4x - 200t)$ , cold, and instantly read off the amplitude, wavelength, frequency, wave speed and direction of travel — and then go further, differentiating the same equation to find how fast and how hard any single particle of the string is accelerating at any chosen instant. That differentiation step is where this topic quietly reconnects with everything you already know about SHM, because it turns out $a_y = -\omega^2 y$ falls straight out of the wave equation, no new physics required.

Key takeaways

A progressive sinusoidal wave moving in the $+x$ direction is $y(x,t) = A\sin(kx - \omega t)$ ; moving in $-x$ it is $y(x,t) = A\sin(kx + \omega t)$ .
$k = \dfrac{2\pi}{\lambda}$ is the angular wave number and $\omega = 2\pi f$ is the angular frequency; both are read directly off the coefficients of $x$ and $t$ once the equation is in standard sine form.
The wave speed is $v = \dfrac{\omega}{k} = f\lambda$ , found by asking how fast the phase $(kx - \omega t)$ must move to stay constant.
Particle (transverse) velocity comes from $v_y = \partial y/\partial t$ at fixed $x$ ; particle acceleration comes from $a_y = \partial^2 y/\partial t^2 = -\omega^2 y$ — the SHM equation, particle by particle.
Wave speed and particle speed are entirely different quantities: the pattern glides at constant $v$ , while any one particle's speed oscillates between $0$ and $A\omega$ .
Fixing $t$ gives a spatial snapshot (a sine curve in $x$ ); fixing $x$ gives a temporal history (a sine oscillation in $t$ ) — switching between these two readings of the same equation is the core exam skill.
Key idea: the single expression $y(x,t) = A\sin(kx-\omega t)$ encodes an entire travelling wave, and differentiating it with respect to time at fixed position recovers the SHM of each individual particle, $a_y = -\omega^2 y$ .

The core insight

The displacement relation $y(x,t) = A\sin(kx-\omega t)$ describes a sinusoidal wave travelling in $+x$ at speed $v=\omega/k=f\lambda$ , and differentiating it with respect to time at a fixed point recovers the SHM of that one particle, $a_y = -\omega^2 y$ .

Where it fits in the syllabus

This sits at the algebraic heart of the CBSE Class 11 Waves chapter (Unit: Oscillations and Waves), directly after the transverse/longitudinal classification and directly before principle of superposition, reflection, and standing waves — all of which are built by adding two or more expressions of exactly this form. The prescribed 2025-26 syllabus asks you to write $y(x,t) = A\sin(kx-\omega t)$ for a given wave, identify $A$ , $k$ , $\omega$ , the wave speed, and the direction of travel from it, and to use it to find a particle's displacement, velocity and acceleration at a stated instant. The full wave equation as a partial differential equation and its general solution are beyond this course — you use the sinusoidal form directly, you do not derive it from a PDE.

The physical picture

A progressive sinusoidal wave moving in the positive x-direction is written as

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

and, changing only the sign in front of $\omega t$ , a wave moving in the negative x-direction is

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

where: $y(x,t)$ is the transverse (or longitudinal, for sound) displacement of the particle sitting at position $x$ , measured at time $t$ ; $A$ is the amplitude, the maximum displacement any particle reaches; $k = \dfrac{2\pi}{\lambda}$ is the angular wave number, with units radians per metre; $\omega = 2\pi f$ is the angular frequency, in radians per second; and the bracket $(kx - \omega t)$ is called the phase of the wave — it is what actually determines the value of the sine at any point and instant. Read the equation two ways. Freeze $t$ at some fixed instant and let $x$ run: you get a sine curve in space, a snapshot of the string's shape, repeating every wavelength $\lambda$ . Freeze $x$ at some chosen point and let $t$ run: you get a sine curve in time, the history of that one particle's up-and-down motion, repeating every period $T = 2\pi/\omega$ . The figure below shows both readings side by side.

Figure

Two panels side by side, the left panel a sine curve of displacement against position at one fixed instant showing one full wavelength, the right panel a sine curve of displacement against time at one fixed position showing one full period, with dashed guide lines marking the amplitude on both

The wave speed follows from a simple requirement: for a fixed feature of the waveform, such as a crest, to "travel" at all, the phase at that crest must stay constant even as both $x$ and $t$ increase.

kx - \omega t = \text{constant} \quad\Longrightarrow\quad \frac{dx}{dt} = \frac{\omega}{k} = v

where: the left side is just differentiating the constant-phase condition with respect to time, treating $x$ as the unknown function $x(t)$ that keeps the phase fixed; the result $v = \omega/k$ is the speed at which the crest — and hence the whole waveform — glides along. Since $\omega = 2\pi f$ and $k = 2\pi/\lambda$ , this is identically $v = f\lambda$ , tying the displacement relation back to the wave-speed formula from earlier in the chapter. Once $y(x,t)$ is known, two more quantities follow by straightforward differentiation with respect to time, holding $x$ fixed — that is, tracking one particular particle of the medium as time passes. The transverse velocity of that particle is

v_y = \frac{\partial y}{\partial t} = -A\omega\cos(kx-\omega t)

and its transverse acceleration is

a_y = \frac{\partial^2 y}{\partial t^2} = -A\omega^2\sin(kx-\omega t) = -\omega^2 y

where: $v_y$ and $a_y$ are the ordinary rate-of-change and rate-of-rate-of-change of $y$ with $t$ , both evaluated at one fixed $x$ — they describe how that one particle moves, not how the wave pattern moves. The last line, $a_y = -\omega^2 y$ , is exactly the defining equation of SHM: acceleration proportional to displacement and directed back toward equilibrium. It confirms, without any extra assumption, that every single particle a travelling wave passes through performs simple harmonic motion of angular frequency $\omega$ — the wave is nothing more than a chain of such oscillators, each slightly out of step with its neighbours.

Derivation AParticle velocity and acceleration from the displacement relation

The goal is to start from nothing but the displacement relation $y(x,t) = A\sin(kx-\omega t)$ and, using only ordinary differentiation with $x$ held fixed, obtain the transverse velocity and transverse acceleration of a single particle of the medium, then show that the acceleration obeys the SHM condition $a_y = -\omega^2 y$ — so that "wave motion" and "SHM of every particle" are seen to be the same statement viewed from two different angles, with no calculus beyond the chain rule for $\sin$ and $\cos$ .

Step 1. Write the displacement relation and fix attention on one particle. Choose a particular value of $x$ , call it $x_0$ — this is the coordinate of one bead of the string, a point that never moves left or right, only up and down. Its displacement as a function of time alone is $y(t) = A\sin(kx_0 - \omega t)$ , the same formula, now with $x_0$ treated as a constant.

Step 2. Differentiate once with respect to $t$ to get the transverse velocity. Using $\dfrac{d}{dt}\sin(u) = \cos(u)\cdot\dfrac{du}{dt}$ with $u = kx_0 - \omega t$ , so $du/dt = -\omega$ , gives $v_y = \dfrac{dy}{dt} = A\cos(kx_0-\omega t)\times(-\omega) = -A\omega\cos(kx_0-\omega t)$ . This is the up-down speed of that one particle at time $t$ ; it swings between $-A\omega$ and $+A\omega$ as the wave passes, and it is unrelated in size to the wave speed $v=\omega/k$ .

Step 3. Differentiate a second time to get the transverse acceleration. Differentiating $v_y = -A\omega\cos(kx_0-\omega t)$ with respect to $t$ , and again using $du/dt=-\omega$ with $\dfrac{d}{dt}\cos(u) = -\sin(u)\cdot du/dt$ , gives $a_y = \dfrac{dv_y}{dt} = -A\omega\times(-\sin(kx_0-\omega t))\times(-\omega) = -A\omega^2\sin(kx_0-\omega t)$ .

Step 4. Recognise the SHM form. The original displacement was $y = A\sin(kx_0-\omega t)$ , so the acceleration found in Step 3 is exactly $a_y = -\omega^2\left[A\sin(kx_0-\omega t)\right] = -\omega^2 y$ . Acceleration is proportional to displacement and points back toward $y=0$ — the defining test of simple harmonic motion — with angular frequency $\omega$ , the very same $\omega$ that appears in the wave equation. No new physics was introduced between Step 1 and Step 4; the SHM of the particle is an automatic consequence of the sinusoidal displacement relation.

Worked examples

Board level - reading off amplitude, wave number, frequency, speed and direction

A transverse wave travelling on a string is described by $y(x,t) = 0.02\sin(4x - 200t)$ , with $x$ and $y$ in metres and $t$ in seconds. Find the amplitude, the angular wave number and wavelength, the angular frequency and frequency, the wave speed, and the direction of travel.

Step 1. Match the given equation to the standard form $y = A\sin(kx-\omega t)$ term by term. The coefficient in front of the sine is $A = 0.02$ m $= 2$ cm — this is the amplitude, read off directly, no algebra needed.

Step 2. The coefficient of $x$ inside the sine is the angular wave number: $k = 4$ rad/m. The wavelength follows from $k = 2\pi/\lambda$ , so $\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{4} = 1.57$ m.

Step 3. The coefficient of $t$ inside the sine is the angular frequency: $\omega = 200$ rad/s. The ordinary frequency follows from $\omega = 2\pi f$ , so $f = \dfrac{\omega}{2\pi} = \dfrac{200}{2\pi} = 31.8$ Hz.

Step 4. The wave speed is $v = \dfrac{\omega}{k} = \dfrac{200}{4} = \mathbf{50\ m/s}$ . As a check, $v = f\lambda = 31.8 \times 1.57 \approx 50$ m/s, which agrees.

Step 5. Read the direction from the sign inside the sine. Since the equation is of the form $\sin(kx-\omega t)$ , with a minus sign between the two terms, the wave travels in the positive x-direction. Had the sign been a plus, it would be travelling in $-x$ . Final answers: $A = \mathbf{2\ cm}$ , $k=\mathbf{4\ rad/m}$ , $\lambda=\mathbf{1.57\ m}$ , $\omega=\mathbf{200\ rad/s}$ , $f=\mathbf{31.8\ Hz}$ , $v=\mathbf{50\ m/s}$ , travelling in $\mathbf{+x}$ .

Competitive level - particle displacement, velocity and acceleration at a given instant

For the same wave $y(x,t) = 0.02\sin(4x - 200t)$ , find the transverse displacement, velocity, and acceleration of the particle at $x = 0$ at time $t = 0.005$ s, and state whether the particle is moving toward or away from equilibrium at that instant.

Step 1. Substitute $x=0$ into the displacement relation to isolate the motion of that one particle: $y(0,t) = 0.02\sin(0 - 200t) = -0.02\sin(200t)$ .

Step 2. Evaluate the phase at $t = 0.005$ s: phase $= 200 \times 0.005 = 1$ rad. All trigonometric evaluations use radians here, never degrees.

Step 3. Compute the displacement: $y = -0.02\sin(1\ \text{rad}) = -0.02\times 0.8415 = \mathbf{-1.68\ cm}$ (that is, $-0.0168$ m — the particle is 1.68 cm below equilibrium).

Step 4. Compute the transverse velocity using $v_y = -A\omega\cos(kx-\omega t)$ evaluated at $x=0$ : $v_y = -(0.02)(200)\cos(1\ \text{rad}) = -4\times 0.5403 = \mathbf{-2.16\ m/s}$ . The negative sign means the particle is moving in the $-y$ direction, i.e. further downward, at this instant.

Step 5. Compute the transverse acceleration using the SHM shortcut $a_y = -\omega^2 y$ , avoiding a fresh trig evaluation: $a_y = -(200)^2\times(-0.0168) = \mathbf{672\ m/s^2}$ , directed in $+y$ — a restoring acceleration pulling the particle back up toward equilibrium, exactly as SHM demands even though the particle's velocity is still carrying it downward. So at $t=0.005$ s the particle at $x=0$ sits 1.68 cm below the line, still sinking at 2.16 m/s, but already being decelerated by a 672 m/s² restoring acceleration pointed back up — the same pattern any SHM oscillator shows past its equilibrium point on the way to an extreme. Timing note: In a 3-mark JEE-style sub-question, budget under a minute for Steps 1-3 (direct substitution) and use the $a_y=-\omega^2y$ shortcut in Step 5 rather than differentiating $v_y$ afresh — recomputing $-A\omega^2\sin(200t)$ from scratch costs an extra thirty seconds for the same number and is a common source of sign slips under time pressure.

Common traps

Confusing wave speed $v=\omega/k$ with particle (transverse) speed $v_y=\partial y/\partial t$ . These are entirely different quantities with entirely different values in general — the waveform itself glides at the constant speed $v$ , while any one particle only oscillates up and down (or back and forth, for sound) with a speed that continuously changes between $0$ and $A\omega$ . Mixing the two on an exam is the single most common error on this topic.
Sign confusion on direction. $\sin(kx-\omega t)$ travels in $+x$ ; $\sin(kx+\omega t)$ travels in $-x$ . It is easy to misremember this under pressure — the safest check is the phase-constant argument: as $t$ increases, $x$ must increase too to keep $kx-\omega t$ fixed, which is only possible if the wave moves toward $+x$ .
Forgetting that $k$ and $\omega$ are read straight off the coefficients of $x$ and $t$ once the equation is written in the standard $\sin(kx\pm\omega t)$ form — no extra algebra, no need to first find $\lambda$ or $T$ separately.
Evaluating trig functions in degrees instead of radians. The arguments $kx$ and $\omega t$ come out in radians because $k$ and $\omega$ are themselves defined through $2\pi$ ; switching a calculator to degree mode mid-problem silently produces a wrong final number that still looks plausible.
Differentiating $y$ with respect to $x$ instead of $t$ when asked for particle velocity. $\partial y/\partial x$ describes the slope of the waveform at an instant (useful for other purposes), not the particle's own speed — only $\partial y/\partial t$ , at fixed $x$ , gives the transverse velocity.

You should be able to…

Write the standard displacement relation $y(x,t)=A\sin(kx-\omega t)$ for a wave travelling in $+x$ , and correctly flip the sign for a wave travelling in $-x$ .
Extract $A$ , $k$ , $\omega$ directly from a given wave equation in standard form, and compute $\lambda$ , $f$ , and $v$ from them without extra derivation.
Differentiate $y(x,t)$ with respect to $t$ , at fixed $x$ , to obtain particle transverse velocity, and differentiate again to obtain particle transverse acceleration.
Clearly and consistently distinguish "wave speed" $v=\omega/k$ from "particle transverse speed" $v_y=\partial y/\partial t$ in both concept and calculation.
Verify, for any sinusoidal wave, that $a_y=-\omega^2 y$ , and explain in words why this ties wave motion back to SHM.
Work confidently in radians when evaluating $\sin$ or $\cos$ of a wave's phase at a specific $x$ and $t$ .

Ask the tutor: If you are given a wave equation with an added phase constant, like $y=A\sin(kx-\omega t+\phi)$ , ask the AI tutor to walk through how that constant shifts the waveform in time without touching $A$ , $k$ , or $\omega$ .

Why does the wave move at a constant speed v while individual particles keep changing speed

Because "wave speed" and "particle speed" answer two different questions. The wave speed $v=\omega/k$ describes how fast a fixed feature of the pattern, such as a crest, glides along $x$ — that is set once by the medium's properties (tension and mass per length, for a string) and the source frequency, and does not change as the wave travels. The particle speed $v_y=\partial y/\partial t$ describes how fast one bead of the string moves up and down, which is SHM, so it naturally swings between $0$ at the turning points and $A\omega$ at equilibrium. Nothing is inconsistent — you are simply asking about the motion of the pattern versus the motion of the matter the pattern moves through.

If I am given y at only one instant, can I still find the particle velocity

Not from that one snapshot alone — a single instant only gives you the spatial slope $\partial y/\partial x$ , which is a different quantity. To get the particle's transverse velocity $v_y=\partial y/\partial t$ you need the full displacement relation $y(x,t)$ , or at least the value of $\omega$ and enough information to evaluate the phase at your chosen $x$ and $t$ , so you can differentiate with respect to time. This is exactly why exam questions always give you the complete equation with both $x$ and $t$ in it, rather than a single frozen picture.

Does the direction rule change for a longitudinal wave like sound

No — the same displacement relation and the same sign rule apply, only the physical meaning of $y$ changes. For a transverse wave on a string, $y$ is a sideways displacement; for a longitudinal wave such as sound, $y$ represents the displacement of a small element of air along the direction of travel (or equivalently a pressure variation). Either way, $y(x,t)=A\sin(kx-\omega t)$ still travels in $+x$ and $y(x,t)=A\sin(kx+\omega t)$ still travels in $-x$ , and the same differentiation gives particle velocity and acceleration along the wave's own axis instead of sideways.

Why is a phase constant sometimes added, and does it change the answers to this kind of problem

A term like $y=A\sin(kx-\omega t+\phi)$ is used when the wave does not happen to pass through $y=0$ at $x=0,t=0$ — the constant $\phi$ just shifts the whole waveform earlier or later without changing its amplitude, wavelength, frequency, or speed, since none of $A$ , $k$ , $\omega$ appear inside $\phi$ . All the extraction rules in this article — reading $A$ , $k$ , $\omega$ off the equation, computing $v=\omega/k$ , and differentiating for particle velocity and acceleration — go through exactly as before; only the numerical value of $y$ , $v_y$ , or $a_y$ at a specific $x$ and $t$ shifts because the phase you plug in now includes the extra $\phi$ .