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Physics · Waves

Reflection of Waves

Every wave considered so far has travelled through an unbounded medium. NCERT Section 14.6 asks the next question: what happens when a pulse meets a boundary? The answer turns on a single distinction — whether that boundary is rigid or free — and on the phase change the wave suffers there. This short topic supplies the boundary conditions on which standing waves, organ pipes and the entire later half of the chapter rest.

What Reflection at a Boundary Means

So far we considered waves propagating in an unbounded medium. When a pulse or a wave meets a boundary, it gets reflected. The phenomenon of echo — sound returning from a cliff or wall — is the everyday example of reflection by a rigid boundary. The character of the reflection, however, is not the same for every kind of boundary. Two ideal cases bracket the behaviour: a perfectly rigid boundary, where the end cannot move, and a perfectly free (open) boundary, where the end is completely free to move.

If the boundary is not completely rigid, or is an interface between two different elastic media, the situation is more involved. Part of the incident wave is reflected and part is transmitted into the second medium. When a wave is incident obliquely on the boundary between two media, the transmitted wave is called the refracted wave: the incident and refracted waves obey Snell's law of refraction, while the incident and reflected waves obey the usual laws of reflection. For NEET the two limiting cases — rigid and free — carry almost all the weight, so we treat each in turn.

Figure 1 — Two Limiting Boundaries

A single transverse pulse approaches a wall. At a rigid end it returns inverted; at a free end it returns upright. The two diagrams below make the contrast concrete.

rigid wall incident crest → ← reflected trough (inverted, phase π)

Reflection at a Rigid (Fixed) Boundary

Picture a pulse travelling along a stretched string toward an end clamped firmly to a wall. Assuming no absorption of energy by the boundary, the reflected wave has the same shape as the incident pulse, but it suffers a phase change of π (180°) on reflection. A crest returns as a trough.

NCERT gives two complementary reasons for this inversion. The first is a boundary-condition argument: the boundary is rigid, so the disturbance must have zero displacement at all times at that point. By the principle of superposition, the net displacement at the wall can be zero only if the reflected and incident waves differ in phase by π, so that they cancel there. The second is a dynamical argument from Newton's third law: as the pulse arrives at the wall it exerts an upward force on the wall; the wall exerts an equal and opposite (downward) force on the string, and this generates a reflected pulse that differs by a phase of π. Because the displacement is permanently zero at the fixed end, that point is a node.

NEET Trap

Rigid end → π phase change → node

Students routinely mix up which boundary inverts the pulse. Tie the three ideas together as one packet: a rigid / fixed / denser boundary forces zero displacement, so the reflected wave is phase-shifted by π and a crest comes back as a trough. The fixed point itself is a node.

Fixed end ⇒ phase reversal of π ⇒ node (zero displacement always).

Reflection at a Free (Open) Boundary

Now suppose the boundary point is not rigid but completely free to move — the textbook model is a string tied to a freely moving ring that slides without friction on a rod. With no constraint forcing the displacement to zero, the reflected pulse has the same phase and amplitude (assuming no energy dissipation) as the incident pulse. A crest returns as a crest, with no inversion.

At the instant the incident and reflected pulses overlap at the free end, their displacements add. The net maximum displacement at the boundary is therefore twice the amplitude of each pulse, so the free end is a point of maximum displacement — an antinode. An example of a non-rigid boundary is the open end of an organ pipe.

Figure 2 — Reflection at a Free End

At a free boundary there is no inversion. The reflected crest comes back as a crest, and momentarily the displacement at the end reaches 2a.

free ring incident crest → ← reflected crest (upright, no phase change)

Rigid vs Free — The Decisive Table

NCERT summarises the whole topic in a single sentence: a travelling wave or pulse suffers a phase change of π on reflection at a rigid boundary and no phase change on reflection at an open boundary. The table fixes every consequence side by side.

Feature Rigid / Fixed / Denser end Free / Open / Rarer end
Phase change on reflectionπ (180°)0 (none)
A crest returns asa trough (inverted)a crest (upright)
Displacement at boundaryzero at all timesmaximum (twice amplitude)
Point formed therenodeantinode
Reflected-wave equationy_r = − a sin(kx + ωt)y_r = a sin(kx + ωt)
Physical exampleecho from a cliff; closed end of a pipeopen end of an organ pipe

The Reflected-Wave Equations

To put the summary mathematically, NCERT writes the incident travelling wave as $y(x,t) = a\sin(kx - \omega t)$. The reflected wave that meets a boundary is obtained by adding the phase shift the boundary imposes.

At a rigid boundary the phase shift is π:

$$y_r(x,t) = a\sin(kx - \omega t + \pi) = -\,a\sin(kx - \omega t)$$

At an open boundary the phase shift is zero:

$$y_r(x,t) = a\sin(kx - \omega t + 0) = a\sin(kx - \omega t)$$

Two cautions are worth holding. First, the reflected wave actually travels in the opposite direction to the incident wave, so when it is used to build a standing wave it is written with the argument $kx + \omega t$; at the rigid end this reads $y_r = -\,a\sin(kx + \omega t)$, exactly as the NCERT summary states. Second, at the rigid boundary the incident and reflected displacements satisfy $y = y_i + y_r = 0$ at the wall at all times — the algebraic statement of the node.

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Builds on

Reflection is a direct application of the principle of superposition of waves: the resultant at the boundary is the algebraic sum of incident and reflected displacements.

Transmission at a Junction of Two Strings

A genuinely rigid or free end is an idealisation. More commonly a wave reaches a junction where the medium changes — for example a light string knotted to a heavy one. Here part of the wave is reflected and part is transmitted into the second medium, and the rule for the reflected part follows the same denser/rarer logic.

When a wave travelling on a string reflects from a denser medium (a heavier string ahead, which behaves like a fixed support in the limit), the reflected part is inverted — a π phase reversal, exactly as at a rigid wall. When it reflects from a rarer medium (a lighter string ahead, tending toward a free end), the reflected part is upright — no phase change. The transmitted wave, whichever way the density steps, always continues in phase with the incident wave; it is never inverted.

NEET Trap

Longitudinal waves: "change of type" at a rarer end

For sound (longitudinal waves) the NIOS treatment adds a subtlety. On reflection from a denser medium a compression returns as a compression (no change of type) but with change of sign. On reflection from a rarer medium a compression returns as a rarefaction and a rarefaction as a compression — a "change of type" but no change of sign. Do not blindly apply the transverse crest-to-trough rule to compressions.

Sound at a rarer (open) end ⇒ compression ↔ rarefaction (type flips), phase unchanged.

Reflection at a single boundary is only the first step. The familiar situations — a string fixed at both ends, or an air column closed at one end — involve reflection at two or more boundaries. A wave travelling one way reflects at one end, travels back, reflects at the other end, and so on, until a steady pattern is established. Superposing the incident $y_1 = a\sin(kx - \omega t)$ and reflected $y_2 = a\sin(kx + \omega t)$ waves gives

$$y(x,t) = 2a\,\sin kx\,\cos \omega t$$

in which $kx$ and $\omega t$ appear separately rather than in the combination $kx - \omega t$. This is a standing (stationary) wave, with permanently fixed nodes and antinodes set by the boundary conditions. The full development of normal modes, harmonics and organ-pipe frequencies follows from exactly the rigid-end and free-end conditions established here.

Quick Recap

Reflection of Waves in one screen

  • A pulse meeting a boundary is reflected; an echo is reflection at a rigid boundary.
  • Rigid / fixed / denser end: phase change of π, crest returns as trough, displacement zero, a node forms.
  • Free / open / rarer end: no phase change, crest returns as crest, displacement reaches 2a, an antinode forms.
  • Reflected equations for incident $a\sin(kx-\omega t)$: rigid → $-a\sin(kx+\omega t)$; open → $a\sin(kx+\omega t)$.
  • At a junction: part reflects (denser ⇒ inverted, rarer ⇒ upright), part transmits in phase. For sound, a rarer end flips compression ↔ rarefaction.
  • Superposing incident and reflected waves gives the standing wave $y = 2a\sin kx\,\cos\omega t$.

NEET PYQ Snapshot — Reflection of Waves

Reflection itself is usually tested through standing waves and pulses on strings; the cards below pair a real NEET item with the core concept.

NEET 2016

A uniform rope of length L and mass m₁ hangs vertically from a rigid support. A block of mass m₂ is attached to the free end of the rope. A transverse pulse of wavelength λ₁ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ₂. The ratio λ₂/λ₁ is:

  1. $\sqrt{(m_1 + m_2)/m_2}$
  2. $\sqrt{m_2/m_1}$
  3. $\sqrt{(m_1 + m_2)/m_1}$
  4. $\sqrt{m_2/(m_1 + m_2)}$
Answer: (1)

Tension at the bottom is $T_1 = m_2 g$; at the top it carries the rope too, $T_2 = (m_1 + m_2)g$. With $v = \sqrt{T/\mu}$ and $v = f\lambda$ at fixed frequency, $\lambda \propto \sqrt{T}$, so $\lambda_2/\lambda_1 = \sqrt{T_2/T_1} = \sqrt{(m_1+m_2)/m_2}$. The rigid support at the top is the fixed boundary from which such a pulse would reflect with a π phase change.

Concept

A transverse wave pulse travelling along a string is reflected from a fixed (rigid) end. Which statement is correct?

  1. The pulse returns upright with no phase change.
  2. The pulse returns inverted, with a phase change of π, and the fixed end is a node.
  3. The pulse is fully transmitted with no reflection.
  4. The reflected pulse has twice the amplitude of the incident pulse.
Answer: (2)

At a rigid boundary the displacement must be zero at all times, so by superposition the reflected wave differs in phase by π — a crest returns as a trough — and the fixed point is a node. Option (4) describes the free-end (antinode) case, not the rigid end.

FAQs — Reflection of Waves

The six confusions NEET aspirants raise most often on boundary reflection.

Does a wave undergo a phase change on reflection at a rigid boundary?

Yes. A travelling wave or pulse suffers a phase change of π (180°) on reflection at a rigid boundary. A crest returns as a trough. This is required because the rigid boundary is fixed, so the displacement there must be zero at all times; only a π phase difference between the incident and reflected waves makes the resultant displacement zero at the wall, which forms a node.

Why does no phase change occur at a free (open) boundary?

At a free boundary the end is completely free to move, so there is no constraint forcing the displacement to zero. The reflected pulse has the same phase and amplitude as the incident pulse, so a crest returns as a crest. The incident and reflected displacements add at the boundary, giving a net maximum displacement of twice the amplitude of each pulse — an antinode.

What are the reflected-wave equations at a rigid and an open boundary?

For an incident wave y(x,t) = a sin(kx − ωt), the reflected wave at a rigid boundary is y_r(x,t) = a sin(kx − ωt + π) = − a sin(kx − ωt), and at an open boundary it is y_r(x,t) = a sin(kx − ωt + 0) = a sin(kx − ωt). The reflected wave actually travels in the opposite direction, so it is written with kx + ωt; at the rigid end this gives y_r = − a sin(kx + ωt).

What happens when a wave reaches a junction of two strings?

At an interface between two media a part of the incident wave is reflected and a part is transmitted into the second medium. Reflection from a denser medium (for example a heavier string, or a fixed support treated as infinitely dense) gives a π phase reversal of the reflected part. Reflection from a rarer medium (a lighter string, or a free end) gives no phase change in the reflected part. The transmitted wave always continues in phase with the incident wave.

How does reflection create standing waves?

When a string is fixed at both ends, a wave travelling one way is reflected at one end, travels back and is reflected at the other end. The incident and reflected waves of equal amplitude and wavelength move in opposite directions and superpose, giving y(x,t) = 2a sin kx cos ωt — a standing wave with fixed nodes and antinodes.

Is an echo an example of reflection at a rigid boundary?

Yes. The phenomenon of echo is an example of reflection of sound by a rigid boundary such as a cliff or a wall. The open end of an organ pipe, by contrast, behaves as a non-rigid (open) boundary, where reflection occurs without phase change.

Related Topics
Waves — Chapter Home All subtopics of the Waves chapter in one place. Standing Waves & Normal Modes Where reflection at two boundaries leads — nodes, antinodes, harmonics. Superposition of Waves The principle that fixes the displacement at a boundary. Progressive Wave Equation The y = a sin(kx − ωt) form the reflected wave is built from. Transverse & Longitudinal Waves Why crests invert but compressions change type. Speed of a Travelling Wave The v = √(T/μ) relation behind reflection at a junction.