Amplitude and phase

Pluck a guitar string gently and it moves a whisker; pluck it hard and it swings through a wide arc — same note, same $\lambda$ , same $f$ , but a very different feel to the ear and a very different look on an oscilloscope screen. That "how big" is amplitude. Now watch two beads sitting on a long rope as a wave rolls through: at any instant one bead might be riding the crest while its neighbour, a little farther along, is still climbing up from the trough. They share the same wave, the same $A$ , the same $\omega$ — yet they are not doing the same thing at the same moment. What distinguishes them is phase, the "where exactly in the cycle" tag carried inside the sine. The wave equation $y(x,t) = A\sin(kx - \omega t)$ that you met while studying the displacement relation already contains both ideas in plain sight. $A$ sits outside the sine, fixing the ceiling and floor of every oscillation the wave will ever produce. Everything inside the sine — $kx - \omega t$ , together with any starting offset $\phi_0$ — is the phase, a running angle that ticks forward with $x$ and backward with $t$ . Two particles separated along the wave carry different phases at the same instant; the same particle carries different phases at different instants. Read that angle correctly and you can say, without any further calculation, whether two particles move in lock-step, in flat opposition, or somewhere in between. This article takes those two quantities apart and puts them back to work: amplitude as the single fixed number that never changes for an ideal progressive wave, and phase difference as the tool that compares any two points or any two moments on the wave. Along the way we derive the compact formula $\Delta\phi = \frac{2\pi}{\lambda}\Delta x$ that converts a distance you can measure with a ruler into an angle that tells you about timing — the single most exam-tested idea in this pocket of the Waves chapter — and we extend the two worked numbers you have already seen so that the same wave now yields a full family of answers.

Key takeaways

Amplitude $A$ is the maximum displacement of a particle from its mean position; it is a single fixed number for an ideal undamped progressive wave, independent of both $x$ and $t$ .
Phase is the argument of the sine/cosine in $y(x,t) = A\sin(kx - \omega t + \phi_0)$ — an angle, not a distance, that tells you exactly where a particle sits in its cycle at a given instant.
Initial phase $\phi_0$ is the phase at $x=0,\ t=0$ ; it is fixed by how the source starts and shifts the whole waveform sideways without changing $A$ , $k$ , or $\omega$ .
Phase difference between two points, at the same instant: $\Delta\phi = k\Delta x = \dfrac{2\pi}{\lambda}\Delta x$ — this is the workhorse formula linking a measured path difference to an angle.
Phase difference between two instants, at the same point: $\Delta\phi = \omega\Delta t = \dfrac{2\pi}{T}\Delta t$ .
In phase means $\Delta\phi = 2n\pi$ (equivalently $\Delta x = n\lambda$ ); exactly out of phase means $\Delta\phi = (2n+1)\pi$ (equivalently $\Delta x = (n+\tfrac12)\lambda$ ).
Key idea: amplitude A is the fixed height of every oscillation in the wave, while phase (kx − ωt + φ₀) is the angle that tells you where a particle stands in its cycle, and the phase difference Δφ = (2π/λ)Δx converts a measured spacing into that angle.

The core insight

Amplitude $A$ fixes how big every particle's oscillation is and never changes along an ideal progressive wave, while the phase $(kx-\omega t+\phi_0)$ is the angle that tells you exactly where a chosen particle sits in its cycle, so that the phase difference $\Delta\phi = \frac{2\pi}{\lambda}\Delta x$ between two points (or $\Delta\phi = \frac{2\pi}{T}\Delta t$ between two instants) tells you precisely how "out of step" they are.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Waves chapter (Unit: Oscillations and Waves), immediately after the displacement relation $y(x,t) = A\sin(kx-\omega t)$ that introduces $A$ , $k$ , and $\omega$ as pieces of one equation. Having met that equation, you are now expected to read $A$ physically as the amplitude and $(kx-\omega t+\phi_0)$ as the phase, and to use phase differences to compare two particles on the same wave or two instants for one particle. Everything here stays within algebra and the sine function — no wave-equation calculus, no energy/intensity formulas, and Doppler effect is a separate, currently out-of-syllabus topic studied later, so it is not touched here.

The physical picture

Start from the general form of a progressive wave travelling in the $+x$ direction, written with an explicit starting phase:

y(x,t) = A\sin(kx - \omega t + \phi_0)

where: $y$ is the displacement of the particle at position $x$ from its mean (equilibrium) position at time $t$ ; $A$ is the amplitude, the maximum value $y$ can ever take; $k = \dfrac{2\pi}{\lambda}$ is the angular wave number; $\omega = 2\pi f$ is the angular frequency; and $\phi_0$ is the initial phase, the value of the phase at $x = 0,\ t = 0$ , fixed once and for all by how the source began oscillating. The whole quantity inside the sine, $\theta = kx - \omega t + \phi_0$ , is called the phase of the wave at that particular $x$ and $t$ — it is an angle (radians, sometimes quoted in degrees), and it is what the sine function actually "reads" to decide the particle's displacement at that moment. The physical reading is this: $A$ answers "how far can this particle ever get from home?" — a single number true for the whole wave, forever, provided nothing damps it. The phase $\theta$ answers a completely different question: "at this exact instant, whereabouts in one full swing (0 to $2\pi$ ) is this particular particle?" A phase of $0$ or $2\pi$ means the particle is passing through the mean position moving one way; $\pi/2$ means it is exactly at a crest ( $y=+A$ ); $\pi$ means mean position again but moving the other way; $3\pi/2$ means the trough ( $y=-A$ ). Because $x$ and $t$ both sit inside $\theta$ , the phase changes as you move along the wave at a fixed instant, and it also changes at a fixed point as time ticks on — that dual dependence is exactly what lets you compare either two positions or two moments.

Figure

A snapshot of a sine wave at a fixed instant with dashed vertical lines dropped to two particles one wavelength apart showing equal displacement and labelled amplitude as the height from the axis to the crest

For a fixed instant of time, compare two particles at positions $x_1$ and $x_2$ on the same wave. Their phases are $kx_1 - \omega t + \phi_0$ and $kx_2 - \omega t + \phi_0$ ; the $\omega t$ and $\phi_0$ terms are common to both and cancel when you subtract, so only the spacing matters. For a fixed point in space, compare the same particle at two instants $t_1$ and $t_2$ ; now the $kx$ and $\phi_0$ terms cancel and only the elapsed time matters. Both comparisons reduce to the same idea — phase difference measures separation in units of angle rather than in units of length or time — and both are derived formally below.

Derivation APhase difference between two particles on a progressive wave

We want a formula that converts a measured distance between two particles on the same wave into the phase angle that separates their motion, so that we can immediately tell whether the two particles move together, move in exact opposition, or somewhere in between, purely from knowing how far apart they are compared to the wavelength.

Step 1. Write the phase of each particle at the same instant $t$ . For the progressive wave $y(x,t) = A\sin(kx - \omega t + \phi_0)$ , particle 1 at $x_1$ has phase $\theta_1 = kx_1 - \omega t + \phi_0$ , and particle 2 at $x_2$ has phase $\theta_2 = kx_2 - \omega t + \phi_0$ . Both are evaluated at the same $t$ because we are asking "what are these two particles doing right now."

Step 2. Subtract the two phases. $\Delta\phi = \theta_2 - \theta_1 = (kx_2 - \omega t + \phi_0) - (kx_1 - \omega t + \phi_0)$ . The $-\omega t$ terms are identical in both and cancel; the $\phi_0$ terms are identical and cancel too. What survives is $\Delta\phi = k(x_2 - x_1) = k\,\Delta x$ , where $\Delta x = x_2 - x_1$ is the path difference, the ordinary distance between the two particles measured along the direction of travel.

Step 3. Replace $k$ with $\dfrac{2\pi}{\lambda}$ , since $k$ is defined that way for any progressive wave. This gives the compact result

\Delta\phi = \frac{2\pi}{\lambda}\,\Delta x

Step 4. Read off the two landmark cases. If $\Delta x = n\lambda$ for a whole number $n$ , then $\Delta\phi = 2n\pi$ , an integer number of full turns — the two particles are in phase, moving identically (same displacement, same velocity direction) at every instant. If $\Delta x = \left(n+\tfrac12\right)\lambda$ , then $\Delta\phi = (2n+1)\pi$ , an odd multiple of $\pi$ — the two particles are exactly out of phase, always on opposite sides of the mean position with equal magnitude. Every other spacing gives an intermediate phase difference between $0$ and $\pi$ (taking the smaller angle), and the same three-step logic, run with $x$ fixed and $t$ varying instead, gives the companion time-domain result $\Delta\phi = \omega\,\Delta t = \dfrac{2\pi}{T}\Delta t$ for one particle compared across two instants.

Worked examples

Board level - phase difference between two particles

A wave is given by $y(x,t) = 0.05\sin(2\pi x - 100\pi t)$ (SI units, so $x$ in metres, $t$ in seconds, $y$ in metres). Two particles are located at $x_1 = 0.5$ m and $x_2 = 1.5$ m. Find the phase difference between them and state whether they oscillate in phase.

Step 1. Read $k$ off the equation by comparing with $y = A\sin(kx-\omega t)$ : here $k = 2\pi$ rad/m, so the wavelength is $\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{2\pi} = 1$ m.

Step 2. Find the path difference: $\Delta x = x_2 - x_1 = 1.5 - 0.5 = 1.0$ m.

Step 3. Apply the derived formula: $\Delta\phi = k\,\Delta x = 2\pi \times 1.0 = 2\pi$ rad.

Step 4. Interpret. Since $\Delta\phi = 2\pi$ , an integer multiple of $2\pi$ (equivalently $\Delta x = 1.0$ m $= 1\lambda$ , exactly one full wavelength), the two particles are in phase — at every instant they carry exactly the same displacement and the same velocity.

Competitive level - amplitude, time-domain phase, and crest identification

For the same wave $y(x,t) = 0.05\sin(2\pi x - 100\pi t)$ , find (a) the amplitude of oscillation, (b) the phase difference for the particle at $x = 0.25$ m between the times $t_1 = 0$ s and $t_2 = 0.005$ s, and (c) whether that particle is at a crest, trough, or zero displacement at $t_1 = 0$ .

Step 1. Amplitude is read directly off the coefficient outside the sine: $A = 0.05$ m $= 5$ cm. This value is fixed for the whole wave — it does not depend on which $x$ or which $t$ we pick, which is exactly why it can be quoted once and used everywhere.

Step 2. Get the period from $\omega$ . Comparing with $y = A\sin(kx-\omega t)$ , $\omega = 100\pi$ rad/s, so $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{100\pi} = 0.02$ s.

Step 3. Apply the time-domain phase-difference formula for a single fixed particle: $\Delta t = t_2 - t_1 = 0.005$ s, and $\Delta\phi = \omega\,\Delta t = 100\pi \times 0.005 = 0.5\pi$ rad $= 90°$ . That is a quarter of a full cycle, consistent with $\Delta t = 0.005$ s being exactly $T/4 = 0.02/4 = 0.005$ s.

Step 4. Evaluate the displacement at $t_1 = 0$ , $x = 0.25$ m directly: $y = 0.05\sin(2\pi \times 0.25 - 100\pi \times 0) = 0.05\sin(\pi/2) = 0.05 \times 1 = 0.05$ m. Since this equals $A$ itself, the particle sits exactly at a crest (maximum positive displacement) at $t_1 = 0$ . Answers: $A = 0.05$ m; $\Delta\phi = 90°$ (a quarter cycle) between $t_1$ and $t_2$ ; the particle is at a crest at $t_1 = 0$ . Timing note: in an exam, do not re-derive $\Delta\phi = \omega\Delta t$ from scratch under time pressure — quote it directly (it is the mirror image of $\Delta\phi = \frac{2\pi}{\lambda}\Delta x$ , with $T$ playing the role $\lambda$ played before) and spend your seconds on correctly reading $k$ , $\omega$ , and $A$ off the given equation, which is where marks are actually lost.

Common traps

Treating amplitude as position- or time-dependent. For an ideal undamped progressive wave, $A$ is a single fixed number for the whole wave — it never varies with $x$ or $t$ . That only happens for damped waves or for the envelope of a standing wave, both beyond what this equation describes.
Confusing path difference with phase difference. $\Delta x$ is a distance, measured in metres; $\Delta\phi$ is an angle, measured in radians (or degrees). They are proportional, not equal — always route through $\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x$ , never write $\Delta\phi = \Delta x$ by mistake.
Mixing up the space and time formulas. $\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x$ compares two different points at the same instant; $\Delta\phi = \dfrac{2\pi}{T}\Delta t$ compares the same point at two different instants. Reaching for the wrong one when a question gives a spacing versus a time is a common slip.
Misreading "exactly out of phase." $\Delta\phi = \pi$ means opposite displacement sign at every instant (one particle at $+y$ while the other is at $-y$ of the same magnitude), not opposite direction of wave travel or unrelated motion.
Dropping the initial phase $\phi_0$ . If a problem states or implies the wave does not start at $y = 0$ at $x=0,t=0$ , an extra phase constant $\phi_0$ is present and must be included or solved for — it shifts every particle's phase by the same fixed amount without touching $A$ , $k$ , or $\omega$ .

You should be able to…

State that amplitude $A$ is the coefficient of the sine in $y = A\sin(kx-\omega t+\phi_0)$ and that it is a single constant for the whole wave.
Identify the phase $(kx-\omega t+\phi_0)$ of a wave at any given $x$ and $t$ , and evaluate it as an angle in radians or degrees.
Derive and apply $\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x$ to convert a path difference between two particles into a phase difference.
Derive and apply $\Delta\phi = \dfrac{2\pi}{T}\Delta t$ to convert an elapsed time for one particle into a phase difference.
Classify two particles as in phase ( $\Delta\phi = 2n\pi$ ) or exactly out of phase ( $\Delta\phi = (2n+1)\pi$ ) from either $\Delta x$ or $\Delta\phi$ .
Use the phase to determine whether a particle sits at a crest, a trough, or the mean position at a given instant.

Ask the tutor: Want a few quick-fire problems converting between path difference and phase difference, or working out when two particles on a wave will next line up in phase?

If amplitude never changes, why do real sounds and waves seem to die out with distance?

Good catch — that is exactly why this article calls it an "ideal, undamped" progressive wave. The formula $y = A\sin(kx-\omega t+\phi_0)$ is a simplified model that assumes no energy is lost as the wave travels, so $A$ stays constant everywhere. Real waves lose energy to friction, air resistance, or simply spread their energy over a larger area (a ripple on a pond gets weaker as its circle grows), so their true amplitude does shrink with distance or time. That richer behaviour, damping and energy spreading, sits beyond the Class 11 syllabus for this topic; here you can safely treat $A$ as a constant unless a question explicitly says otherwise.

Is phase the same thing as phase angle, and can it be negative?

Yes, "phase" and "phase angle" mean the same quantity here — the argument $kx - \omega t + \phi_0$ inside the sine, always expressed in radians (sometimes converted to degrees for convenience). It absolutely can come out negative, or larger than $2\pi$ , or any real number at all, because it just keeps accumulating as $x$ grows and $t$ ticks forward. What actually matters physically is the phase modulo $2\pi$ — that is, its remainder after removing whole multiples of $2\pi$ — since a phase of $-\pi/2$ and a phase of $3\pi/2$ describe the exact same point in the oscillation. When comparing two particles, always reduce $\Delta\phi$ into a sensible range like $0$ to $2\pi$ before deciding "in phase" or "out of phase."

How is phase difference different from what we mean by "path difference" in optics questions I've heard about?

They are closely related but you will formally study the optics version (interference of light) only in Class 12. Here, "path difference" $\Delta x$ is simply the physical separation between two particles measured along the wave, and "phase difference" $\Delta\phi = \frac{2\pi}{\lambda}\Delta x$ converts that separation into an angle using the same wavelength. The logic is identical to what you will later meet in light interference, but for now keep it grounded in mechanical waves on a string or in sound, exactly as this chapter frames it — no extra optics ideas are needed or expected at this stage.

Why does the initial phase phi-naught matter if it cancels out of Delta phi anyway?

It cancels out precisely because Delta phi compares two particles (or two instants) on the same wave — any constant offset common to both simply subtracts away, which is a useful shortcut. But $\phi_0$ still matters whenever a question asks about a single particle's actual displacement, velocity, or "is it at a crest right now" at a specific $x$ and $t$ — there $\phi_0$ directly shifts the answer. It also matters if you are asked to compare two different waves (say, from two separate sources) rather than two points on one wave, since their initial phases need not be equal or cancel. Treat $\phi_0$ as "don't need it for Delta phi within one wave, but never drop it when writing the full equation or evaluating a single point."