Reflection of waves

Send a single sharp pulse down a long rope and watch it travel — a clean hump of displacement gliding steadily to the right. Now tie the far end to a wall and try it again. The pulse arrives, and something surprising happens: it does not just stop. It comes back, but flipped upside down, a crest returning as a trough. Untie that end and let it hang loose on a light ring instead, free to slide up and down a pole, and send the same pulse — this time it returns the right way up, crest still a crest. Nothing about the string material or the wave's speed has changed. The only thing that changed is what the wave met at the boundary, and that single fact — fixed or free — completely controls whether the reflected wave flips or not. This is reflection of waves, and it is one of those chapter topics that looks like a footnote but is secretly load-bearing for everything that follows. Every standing wave you will study next — on a guitar string, in a closed organ pipe, in a flute — is built from exactly this: a wave travelling one way, reflecting off a boundary, and superposing with its own reflection. Get the fixed-end/free-end phase rule wrong and every subsequent standing-wave diagram you draw will have nodes and antinodes in the wrong places. This article works entirely within CBSE Class 11 algebra and trigonometry. We track a single sinusoidal wave $y = A\sin(kx-\omega t)$ approaching a boundary, ask what boundary condition the end must obey, and read off the reflected wave's equation from that condition alone — no differential equations, no energy-flux calculus, just the physical requirement that a fixed end cannot move and a free end cannot support a sideways pulling force. Along the way we work through two numerical examples in full.

Key takeaways

A travelling wave reflects at a boundary; the type of boundary (fixed or free) decides whether the reflected wave is phase-flipped.
Fixed (rigid) end: displacement there must stay zero always, forcing a phase change of $\pi$ (180°) on reflection — a crest returns as a trough.
Free end: nothing forces zero displacement, and no sideways force can act at the very tip, so the wave reflects without any phase change — a crest returns as a crest.
Amplitude, wavelength, and frequency of the reflected wave are unchanged from the incident wave (ideal, lossless boundary) — only travel direction, and at a fixed end, phase, change.
For $y_i = A\sin(kx-\omega t)$ : fixed end gives $y_r = -A\sin(kx+\omega t)$ ; free end gives $y_r = A\sin(kx+\omega t)$ .
Sound obeys the same rule: a rigid/closed end behaves like a fixed end (phase reversal), an open end behaves like a free end (no reversal) — this is exactly what sets up organ-pipe standing waves.
Key idea: a boundary's constraint on displacement (fixed = zero always, free = no restoring force) fixes the phase of the reflected wave — π at a rigid end, zero at a free end — and this single rule is the seed from which every standing-wave pattern in the chapter grows.

The core insight

A wave reflecting off a rigid (fixed) end is phase-shifted by $\pi$ because the end cannot move, while a wave reflecting off a free end returns with no phase change because the end feels no restoring force, and both reflected waves keep the incident wave's amplitude, wavelength, and frequency.

Where it fits in the syllabus

This is part of the CBSE Class 11 Waves chapter, sitting right after the principle of superposition and right before standing waves. It is squarely inside the prescribed 2025-26 syllabus: reflection at fixed and free ends, together with the associated phase change, is explicitly listed scope. The topic is deliberately narrow — no wave-equation derivation, no energy-loss or partial-reflection calculations are asked at this level — but it is conceptually essential, because standing waves (the very next topic) are nothing but an incident wave permanently superposed with its own reflection.

The physical picture

Picture a long, taut string under tension, carrying a sinusoidal wave travelling toward positive $x$ . Before it meets any boundary at all, its shape is described completely by the standard progressive-wave equation from the previous topic:

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

where: $A$ is the amplitude, $k = \dfrac{2\pi}{\lambda}$ is the angular wave number, $\omega = 2\pi f$ is the angular frequency, and the minus sign in front of $\omega t$ is what makes this wave travel toward $+x$ . Now let this wave arrive at $x = 0$ , where the string meets a boundary of some kind. Whatever comes back must still be a valid wave on the same string, with the same $A$ , $k$ , and $\omega$ (no energy is assumed lost), but travelling toward $-x$ instead — which means its equation must have the form $\pm A\sin(kx+\omega t)$ . The only question physics has to settle is the sign, and that sign is decided entirely by what the boundary forces the displacement (or the force) at $x=0$ to do at every instant.

A rigid, fixed end (the string end clamped to a wall) cannot move at all — its displacement is pinned at $y(0,t) = 0$ for every value of $t$ . If the incident wave alone tried to produce a nonzero displacement at $x=0$ , that would violate the clamp; the only way the total displacement (incident plus reflected, added by superposition) can stay zero at $x=0$ for all time is if the reflected wave exactly cancels the incident wave there at every instant. That forces the reflected wave to be the negative of what a simple, unflipped reflection would give — a phase change of $\pi$ :

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

A free end (the string end tied to a light ring that slides frictionlessly on a vertical rod) is not pinned to zero displacement — the ring is free to swing up and down through any value. What it cannot do is transmit a sideways pulling force past $x=0$ , since there is nothing beyond the ring for the string to pull against. The condition that the transverse force (and hence the slope $\partial y/\partial x$ ) vanish at the free end is satisfied without any inversion, so the reflected wave keeps the same sign as an ordinary, unflipped wave travelling the other way:

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

The physical reading of both results together: reflection never changes a wave's amplitude, wavelength, or frequency (those describe the wave's own energy and periodicity, untouched by a lossless boundary) — it only reverses the direction of travel, and, at a fixed end specifically, additionally reverses the sign of the displacement at every point, all the time. This same fixed-versus-free distinction transfers directly to sound: a rigid wall or a closed pipe end behaves like a fixed end (phase reversal on reflection, since the air there cannot be displaced along the pipe), while an open pipe end behaves like a free end (no phase reversal, since the air there is free to bulge in and out to atmosphere). That single analogy is exactly what will let you draw closed-pipe and open-pipe standing-wave diagrams correctly in the next topic.

Figure

A string pulse approaching a fixed wall returns inverted as a trough while the same pulse approaching a free ring returns upright as a crest, shown as two side by side panels

Derivation AWhy a fixed end forces a pi phase change and a free end forces none

The goal of this derivation is to show, using only the physical constraint each type of boundary places on the string and the principle of superposition, why a rigid fixed end must launch a reflected wave that is exactly inverted relative to the incident wave, why a free end must launch a reflected wave with no inversion at all, and how these two conclusions translate directly into the standard reflected-wave equations for an incident wave $y_i(x,t) = A\sin(kx-\omega t)$ .

Step 1. Set up what reflection must preserve. Since the boundary does no work on the wave and stores no energy (an idealised, lossless boundary), the reflected wave travelling back into the string must have the same amplitude $A$ , the same angular wave number $k$ , and the same angular frequency $\omega$ as the incident wave — a boundary cannot change how tightly the wave oscillates in space or in time, only which way it is heading and, possibly, its sign at the point of reflection. So the reflected wave must have the general shape $A\sin(kx+\omega t)$ , travelling toward $-x$ now instead of $+x$ , and the only thing left to decide is whether an extra minus sign belongs in front of it.

Step 2. Apply superposition right at the boundary. Once the reflected wave exists, both waves occupy the string at once, and the principle of superposition says the actual displacement anywhere near the boundary is $y = y_i + y_r$ , added algebraically point by point at every instant. Whatever constraint the boundary places on the total displacement (or on the total slope) must therefore be a constraint on this sum, not on the incident wave alone.

Step 3. Impose the fixed-end (rigid) condition. A rigidly clamped end is physically incapable of moving even slightly, so the total displacement at that point must equal zero at every single instant, with no exception. As the incident wave delivers a crest to the boundary, the only way the sum can stay pinned at zero at that instant is if the reflected disturbance leaving the boundary at that same instant is the exact negative of what just arrived — an inverted image, trough for crest. Because this cancellation must hold not just at one instant but continuously, the reflected wave must be the incident wave's mirror pattern turned upside down at every point in time, which is precisely a $\pi$ (180°) phase change relative to a plain, unflipped echo: $y_r(x,t) = -A\sin(kx+\omega t) = A\sin(kx+\omega t+\pi)$ .

Step 4. Impose the free-end condition. A free end (light ring on a frictionless rod) is not pinned to zero displacement — it is free to swing to whatever value the wave carries it to. What it cannot do is transmit a transverse pulling force past that point, because there is nothing beyond the ring for the string to pull against; equivalently, the string must meet the ring smoothly, with zero slope there at every instant. Unlike the fixed end, this condition does not require the departing disturbance to cancel the arriving one — it only requires the shape to flatten out momentarily at that point, which is satisfied when the reflected wave matches the incident wave's sign rather than opposing it. No inversion is needed, so no phase change occurs: $y_r(x,t) = A\sin(kx+\omega t)$ .

Step 5. State the results side by side. Fixed end: $y_r(x,t) = -A\sin(kx+\omega t)$ , a $\pi$ phase change, the boundary forced to stay at zero displacement (it will become a node once standing waves are studied). Free end: $y_r(x,t) = A\sin(kx+\omega t)$ , no phase change, the boundary free to swing through its largest displacement (it will become an antinode). In both cases $A$ , $k$ , and $\omega$ carry over unchanged from the incident wave — reflection only ever touches the direction of travel and, at a fixed end only, an overall sign.

Worked examples

Board level - identifying the reflection at a clamped wall

A pulse travelling along a stretched string reaches an end that is rigidly clamped to a wall. Describe, in words, what happens to a crest arriving at this end, and state the phase change involved.

Step 1. Identify the boundary type. The wall end is rigidly clamped, so it is a fixed end: the displacement there must remain exactly zero at every instant, no matter what pulse arrives.

Step 2. Apply the zero-displacement condition. As the crest (a positive-displacement pulse) arrives at the wall, the string right at the wall is forced to stay at $y=0$ . By superposition, this is only possible if an equal-and-opposite pulse is simultaneously travelling back into the string from that point, exactly cancelling the incident crest there at every instant.

Step 3. Read off the outcome. The equal-and-opposite pulse is an inverted pulse of the same size — so the crest is reflected back as a trough. This corresponds to a phase change of $\pi$ (180°) on reflection from a fixed end.

Competitive level - writing the reflected-wave equation at a free end

A wave $y_i(x,t) = 0.02\sin(4x-100t)$ (SI units) travels along a string toward a free end located at $x=0$ , where it reflects. Write the equation of the reflected wave, and state its amplitude, angular wave number, and angular frequency.

Step 1. Identify the boundary type and the rule. The end at $x=0$ is free, so reflection introduces no phase change, and the reflected wave must travel in the direction opposite to the incident wave.

Step 2. Read off the incident wave's parameters. Comparing $y_i = 0.02\sin(4x-100t)$ with the standard form $A\sin(kx-\omega t)$ gives $A = 0.02\ \text{m}$ , $k = 4\ \text{rad/m}$ , $\omega = 100\ \text{rad/s}$ — and since the incident wave has $-\omega t$ , it travels toward $+x$ , so the reflected wave, travelling toward $-x$ , must carry $+\omega t$ .

Step 3. Apply the free-end (no phase change) rule and write the equation. Keeping the same amplitude, wave number, and angular frequency, and flipping only the direction term (no additional sign flip, since there is no phase change at a free end):

y_r(x,t) = 0.02\sin(4x+100t)

Step 4. State the final unchanged parameters explicitly. Amplitude $A = 0.02\ \text{m}$ , angular wave number $k = 4\ \text{rad/m}$ , angular frequency $\omega = 100\ \text{rad/s}$ — all identical to the incident wave; only the sign of the $\omega t$ term has flipped, showing reversed direction of travel with no additional phase inversion. Timing note: this should take under a minute in an exam. The only two things to check, in order, are (1) which end — fixed flips the overall sign, free does not, and (2) which direction the incident wave already travels, so you flip the correct term ( $kx-\omega t \to kx+\omega t$ ) rather than guessing. A common slip under time pressure is flipping the wrong term or forgetting that a free end still reverses direction of travel even though it does not reverse phase — direction always reverses on reflection, phase reverses only at a fixed end.

Common traps

Mixing up which end gives the phase change. Fixed end gives a $\pi$ phase change (crest reflects as trough); free end gives no phase change (crest reflects as crest). This is one of the most commonly reversed facts in the chapter — a helpful memory hook is that a fixed end "fights back" and inverts, while a free end "goes along with it" and does not.
Thinking the reflected wave has a different amplitude, frequency, or wavelength. In ideal reflection with no energy loss, only the direction of travel changes, and possibly the phase at a fixed end — amplitude, frequency, and wavelength of the reflected wave equal those of the incident wave.
Forgetting that direction reverses even at a free end. Students sometimes think "no phase change" means "nothing changes," but the reflected wave always travels the opposite way — only the extra $\pi$ flip is absent at a free end, not the direction reversal itself.
Forgetting reflection is the physical basis of standing waves. The next topic, standing waves, arises precisely from superposing an incident wave with its own reflected wave — treating reflection and standing waves as unrelated ideas is a common gap that costs marks in longer derivation questions.
Assuming only strings show this fixed/free distinction. The same fixed-end (phase-reversed) versus free-end (no phase reversal) distinction applies to sound reflecting off closed versus open pipe ends, which becomes essential for organ-pipe standing-wave patterns in the next topic.

You should be able to…

State what happens to phase, amplitude, and direction of a wave reflecting off a fixed (rigid) end.
State what happens to phase, amplitude, and direction of a wave reflecting off a free end.
Explain physically why a fixed end forces a $\pi$ phase change, using the zero-displacement boundary condition.
Derive or write the equation of the reflected wave, $y_r(x,t)$ , given the incident wave's equation and the type of boundary.
Extend the fixed/free-end phase rule to sound reflecting off closed and open pipe ends.
Connect reflection, via superposition, to the formation of standing waves as the immediate next topic.

Ask the tutor: Still shaky on how a $\pi$ phase change at a fixed end turns into a node in the very next standing-wave diagram — ask the AI tutor to walk through the superposition step by step.

Why does a fixed end flip the wave but a free end does not — is there an intuitive way to see it, not just the boundary-condition algebra?

Yes. Think of the fixed end as a wall pushing back exactly as hard as the string pushes on it, at every instant — that reaction is what keeps the displacement pinned at zero, and a reaction that opposes the incoming disturbance is naturally an inverted copy of it, which is exactly a crest becoming a trough. A free end, like a light ring on a frictionless rod, has nothing to push back against sideways, so it simply follows the incoming disturbance and swings up to match it, sending back an upright copy instead. The algebra with the boundary condition ( $y=0$ always at a fixed end, zero slope at a free end) is the precise version of this same picture.

Does the reflected wave lose any energy or amplitude at the boundary?

In the ideal case assumed throughout this CBSE topic, no — the boundary is taken to be perfectly rigid or perfectly free, so no energy is absorbed and the reflected wave has exactly the same amplitude as the incident wave. In a real string tied to a real wall, a small amount of energy is always lost to the support (as sound or heat), so the reflected pulse is very slightly smaller in practice, but this partial-reflection and energy-loss detail is beyond the Class 11 syllabus — for every exam question here, treat reflection as complete and lossless.

How is this connected to the standing waves I am about to study next?

Very directly — reflection is literally the mechanism that creates a standing wave. Once a wave reflects off a boundary, the incident wave and the reflected wave both exist on the same string at the same time, travelling in opposite directions, and by the principle of superposition they add up everywhere. Because the reflected wave is exactly the incident wave with either a $\pi$ flip (fixed end) or no flip (free end), that sum works out to a wave pattern that does not travel at all — it just oscillates in place, with fixed nodes and antinodes. That fixed, non-travelling pattern is precisely the standing wave you will study in the next topic, and the position of its first node or antinode is decided entirely by which type of end you have.

Does sound really reflect the same way as a wave on a string — what does closed versus open even mean for air?

Yes, the same fixed/free logic carries over, just with different physical quantities. A closed pipe end (a rigid cap) stops the air molecules right there from moving along the pipe's length, so it behaves like a fixed end for the displacement wave — a displacement node forms there, analogous to the $\pi$ phase change on a string. An open pipe end lets air bulge freely in and out to the atmosphere, so it behaves like a free end — a displacement antinode forms there, analogous to the no-phase-change case on a string. This is exactly why closed and open organ pipes end up with different sets of harmonics in the next topic.