Reflection of Waves & Phase Change at Boundaries

Flick a rope tied to a wall and watch the pulse race away, hit the knot, and come back — upside down. Untie it, thread the end through a light ring that slides freely on a smooth pole, flick again, and the returning pulse comes back the right way up. Same rope, same flick; the only thing you changed was what happens at the end, and that alone decided whether the wave flips. That flip — a $\pi$ , or $180°$ , phase change on reflection — is one of the most quietly important facts in physics. It is why a guitar string held at both ends can only ring at certain pitches, why a closed organ pipe sounds an octave below an open one, and, dressed in light, why the centre of Newton's rings is dark. The rule has two clean halves. At a fixed (rigid) end the reflected pulse is inverted: it returns with a half-cycle phase change. At a free end it is erect: no phase change at all. And whatever crosses a boundary into a new medium — the transmitted part — is never inverted; only the reflection can flip. The physical reason is Newton's third law: the pulse hauls the wall upward, the wall hauls the rope back down with an equal and opposite jerk, and that downward kick is the inverted echo. This article makes the rule quantitative without leaving algebra. Every string carries a number called its impedance, $Z = \sqrt{T\mu}$ — its stubbornness against being wiggled — and reflection at any boundary is governed by the mismatch between the impedances on the two sides. From two boundary conditions we get the reflection and transmission coefficients $r = (Z_1 - Z_2)/(Z_1 + Z_2)$ and $t = 2Z_1/(Z_1 + Z_2)$ , watch a fixed end ( $Z_2 \to \infty$ ) force $r = -1$ and a free end ( $Z_2 = 0$ ) force $r = +1$ , and check that the reflected and transmitted energies always add back to the whole. A wave crossing from a light string into one four times heavier, we will find, sends back exactly one-third of its amplitude, inverted, and passes two-thirds on — with energies of one-ninth and eight-ninths that sum, exactly, to one.

Key takeaways

At a fixed end the reflected wave is inverted (a $\pi$ phase change); at a free end it is erect (no phase change). The transmitted wave is never inverted.
The physical cause is Newton's third law: the string pulls the wall, the wall pulls back, and that reaction is the flipped reflection. Equivalently, an inverted "image pulse" keeps the fixed end at rest.
Every string has a characteristic impedance $Z = \mu v = \sqrt{T\mu}$ ; reflection depends on the mismatch of impedances at the boundary.
Amplitude coefficients: $r = \dfrac{Z_1 - Z_2}{Z_1 + Z_2}$ , $t = \dfrac{2Z_1}{Z_1 + Z_2}$ . Reflection off a heavier string (higher $\mu$ , so higher $Z$ ) inverts; off a lighter one it does not.
Energy is conserved: power fractions $R = r^2$ and $T = (Z_2/Z_1)t^2$ satisfy $R + T = 1$ .
Key idea: a boundary reflects a wave inverted when it meets a stiffer, heavier medium and erect when it meets a lighter one, with the split of amplitude and energy fixed entirely by the impedance mismatch.

The core insight

A wave reaching a boundary is partly reflected and partly transmitted according to the mismatch of the two media's impedances $Z = \sqrt{T\mu}$ : the reflection inverts (a $\pi$ phase change) when the far medium is heavier or fixed and stays erect when it is lighter or free, while the transmitted wave never inverts and the reflected and transmitted energies always sum to the incident energy.

Where this lives today

The phase flip at a hard boundary is the working principle of medical ultrasound. A probe sends a pulse into the body; wherever it crosses from one tissue to another — muscle to bone, blood to vessel wall — the impedance mismatch reflects an echo, and the machine times that echo to place the boundary and builds an image from thousands of them. Radiographers even smear gel on the skin for exactly this reason: air and skin have wildly mismatched impedances, so without the gel almost the whole pulse would reflect at the surface (like a wave hitting a near-fixed end) and never reach the tissue beneath. The gel is an impedance match, nudging $r$ toward zero so the sound gets in. The same idea, in reverse, is a stealth aircraft's radar-absorbing skin, engineered so that reflected microwaves interfere away rather than return to the enemy dish. And in optics, Green's phase-change-on-reflection is why the coating on a camera lens can be tuned to make its reflection cancel, and why the exact centre of Newton's rings is dark rather than bright — the very puzzle that first convinced physicists a reflection can turn a wave upside down.

The physical picture

Send a single upward pulse down a string toward its far end. What comes back depends entirely on how that end is held, and there are two extreme cases worth burning into memory. A fixed end is tied to something immovable — a wall, a clamp. The end cannot move, so its displacement must stay zero at every instant. As the pulse arrives it tries to lift the end; by Newton's third law the wall pulls back on the string with an equal and opposite downward force, and that downward yank launches an inverted pulse back along the string. The reflected pulse is flipped: a crest returns as a trough. In the language of phase, it has suffered a $\pi$ (180°) phase change. A free end is attached to something that offers no resistance to transverse motion — a massless ring sliding on a frictionless rod. Now the end is free to fly up as the pulse arrives, and it overshoots to twice the pulse height before being pulled back. There is no reaction force to flip anything, so the reflected pulse returns erect — the right way up, with no phase change. The two cases are opposites, and everything in between is a partial mixture of them.

Figure

A pulse reflecting from a fixed end returns inverted, from a free end erect — four frames of each

The picture generalises the instant you replace the wall with a second string. A boundary between two strings is not all-or-nothing: part of the wave reflects and part transmits into the new string. The deciding quantity is each string's characteristic impedance — how hard it pushes back when you try to wiggle it:

Z = \mu v = \sqrt{T\mu} \qquad (1)

where: $\mu$ — the linear mass density; $v = \sqrt{T/\mu}$ — the wave speed; $T$ — the tension, common to both strings when they are joined. The two forms are equal because $\mu v = \mu\sqrt{T/\mu} = \sqrt{T\mu}$ . A heavy, taut string has a large $Z$ ; a light one a small $Z$ . And a rigid wall is the limit of an infinitely heavy string, $Z \to \infty$ , while a free end is the limit of an infinitely light one, $Z \to 0$ . Fixed and free, it turns out, are just the two ends of the impedance scale.

Reflection and transmission coefficients

At a junction two things must hold. The string is unbroken, so its displacement matches on both sides; and the junction point, having no mass, feels zero net transverse force. Turning those two conditions into algebra (done in the second derivation) gives the amplitude coefficients — how big the reflected and transmitted waves are, as fractions of the incident:

r = \frac{Z_1 - Z_2}{Z_1 + Z_2}, \qquad t = \frac{2Z_1}{Z_1 + Z_2} \qquad (2)

where: $r$ — the reflection coefficient, the reflected amplitude divided by the incident (its sign carries the phase: negative means inverted); $t$ — the transmission coefficient, the transmitted amplitude divided by the incident. Read the sign of $r$ off the impedances. If the second medium is heavier, $Z_2 > Z_1$ , then $r < 0$ — the reflection is inverted. If the second is lighter, $Z_2 < Z_1$ , then $r > 0$ — erect. Notice $t$ is always positive: the transmitted pulse never flips. Push (2) to its limits and the two extreme cases fall straight out. A fixed end is $Z_2 \to \infty$ : then $r \to -1$ (a perfectly inverted reflection of the full amplitude) and $t \to 0$ (nothing gets into the wall). A free end is $Z_2 = 0$ : then $r = +1$ (a perfectly erect reflection) and $t = 2$ (the end itself swings to twice the incident amplitude, since incident and erect reflected add there). This is the whole "fixed inverts, free does not" rule, now derived rather than asserted. Energy must balance, and it does. The fraction of power reflected and transmitted are

R = r^2, \qquad T = \frac{Z_2}{Z_1}\,t^2, \qquad R + T = 1 \qquad (3)

where: $R$ — the reflected power fraction; $T$ — the transmitted power fraction (the extra $Z_2/Z_1$ appears because power depends on the medium, not on amplitude alone). Their sum is exactly one for any $Z_1$ , $Z_2$ : no energy is created or lost at a boundary, only divided. Two warnings live in these signs. First, $r$ and $t$ are amplitude ratios, and $t$ can exceed $1$ (as at a free end) without violating anything, because transmitted power still obeys $T \le 1$ . Second, the frequency is untouched by any of this — reflection and transmission change amplitude and phase, never $f$ .

Derivation AWhy a fixed end inverts — Newton's third law and the image pulse

The goal is to explain, from mechanics alone, why the reflection at a rigid end is upside down, and to give the elegant bookkeeping trick — the image pulse — that makes the fixed-end condition automatic. The physics is Newton's third law applied to the last bit of string touching the wall; the trick is to pretend an inverted twin pulse is travelling toward the boundary from the far side, so that the two always cancel exactly at the wall. Both routes deliver the same conclusion: a crest returns as a trough, a $\pi$ phase change.

Step 1. Hold the end fixed: its displacement is zero at all times, because the wall does not move. This is the boundary condition the reflected wave must arrange.

Step 2. Apply Newton's third law at the end. As the incident crest arrives it pulls the clamped end upward with a transverse force; the wall, being rigid, pulls back on the string with an equal and opposite downward force. That downward reaction is what accelerates the string below the crest, and it launches a pulse that is displaced downward — an inverted reflection.

Step 3. Get the same answer with the image trick. Imagine the string continued beyond the wall, carrying a mirror-image pulse that is both flipped left-to-right and turned upside down, travelling toward the boundary. At the wall the real crest and the imaginary trough always coincide and cancel, so the sum is zero there — exactly the fixed-end condition. What emerges on the real side after the meeting is that inverted image pulse, now travelling back: the reflected wave, inverted.

Step 4. Contrast the free end. There the boundary condition is not "zero displacement" but "zero slope" (no transverse force on the free ring), and the image pulse needed to satisfy it is flipped left-to-right but not turned over. So the free-end reflection is erect — no phase change — and the end swings to twice the incident height. Verified: the fixed-end limit $Z_2 \to \infty$ gives $r = -1$ and the free-end limit $Z_2 = 0$ gives $r = +1$ , recomputed in verify_hs2-wave-reflection-boundaries.py.

Go deeper · Derivation B · 1 minThe coefficients from two boundary conditions, and the energy audit

Derivation BThe coefficients from two boundary conditions, and the energy audit

The goal is to derive the reflection and transmission coefficients of equation (2) from two physical requirements at the junction, and then to prove the power fractions of equation (3) really do add to one, using the specific numbers of a light string joined to one four times heavier. The two conditions are that the string does not break (displacement is continuous) and that a massless point cannot feel a net force (the transverse pushes balance). The impedance $Z$ enters as the measure of how hard each side pushes back for a given wiggle. No calculus is needed; the whole thing is two linear equations.

Step 1. Continuity of displacement. Just left of the junction the string carries incident plus reflected, amplitude $A_i + A_r$ ; just right it carries the transmitted, $A_t$ . The string is unbroken, so these match: $A_i + A_r = A_t$ . Dividing by $A_i$ and writing $r = A_r/A_i$ , $t = A_t/A_i$ gives $1 + r = t$ .

Step 2. Balance of transverse force. The junction point has no mass, so the sideways pushes from the two strings must cancel. Each string resists transverse motion in proportion to its impedance, and working this through gives $Z_1(1 - r) = Z_2\,t$ .

Step 3. Solve the pair. Substitute $t = 1 + r$ into the force equation: $Z_1(1 - r) = Z_2(1 + r)$ , so $Z_1 - Z_1 r = Z_2 + Z_2 r$ , giving $r = (Z_1 - Z_2)/(Z_1 + Z_2)$ and then $t = 1 + r = 2Z_1/(Z_1 + Z_2)$ — equation (2).

Step 4. Audit the energy for a real case. Take a light string $\mu_1 = 1.0\times10^{-3}\ \mathrm{kg/m}$ joined to $\mu_2 = 4\mu_1$ under $T = 100\ \mathrm{N}$ . Then $Z_1 = \sqrt{100\cdot10^{-3}} = 0.3162$ and $Z_2 = \sqrt{100\cdot4\times10^{-3}} = 0.6325$ , so $r = (0.3162 - 0.6325)/(0.3162 + 0.6325) = -1/3$ and $t = 2/3$ . The power fractions are $R = r^2 = 1/9 = 0.1111$ and $T = (Z_2/Z_1)t^2 = 2\cdot(4/9) = 8/9 = 0.8889$ , and $R + T = 1/9 + 8/9 = 1.0000$ exactly. Verified: $Z_1 = 0.3162$ , $Z_2 = 0.6325$ , $r = -1/3 = -0.3333$ , $t = 2/3 = 0.6667$ , $R = 0.1111$ , $T = 0.8889$ , $R + T = 1.0000$ , recomputed in verify_hs2-wave-reflection-boundaries.py.

Limiting cases

boundary	$r$	$t$	what happens
fixed end, $Z_2 \to \infty$	$-1$	$0$	full inversion; nothing transmitted
free end, $Z_2 = 0$	$+1$	$2$	erect; end swings to twice the amplitude
heavier string, $Z_2 > Z_1$	$< 0$	$0 < t < 1$	reflected inverted, part transmitted
lighter string, $Z_2 < Z_1$	$> 0$	$1 < t < 2$	reflected erect, part transmitted
matched, $Z_2 = Z_1$	$0$	$1$	no reflection — perfect transmission
our junction, $\mu_2 = 4\mu_1$	$-1/3$	$2/3$	⅓ reflected inverted; $R = 1/9$ , $T = 8/9$

Common trapThe reflected pulse flips, so the transmitted one must flip too

It does not — the transmitted pulse is never inverted, whatever the boundary. Only the reflection can suffer a phase change, and it does so only when the wave meets a heavier or fixed medium. The clean way to see it: the transmission coefficient $t = 2Z_1/(Z_1 + Z_2)$ is always positive, because $Z_1$ and $Z_2$ are both positive, so $A_t$ always has the same sign as $A_i$ — the transmitted pulse comes through the right way up. The reflection coefficient $r = (Z_1 - Z_2)/(Z_1 + Z_2)$ can go negative, and that minus sign is the whole inversion. Two companion slips ride alongside this one. First, forgetting that the reflected amplitude is reduced as well as flipped at a two-string junction: at our junction the reflected pulse is inverted and only one-third as tall, not a full-height mirror image (that would be a fixed end). Second, imagining the frequency changes on reflection. It does not — reflection and transmission alter amplitude and phase, but the boundary is driven at the incident frequency, so $f$ is carried through unchanged on both the reflected and the transmitted waves, exactly as it is at any junction.

Worked examples

1 · Board level — fixed end, free end, and the transmitted pulse (CBSE/ICSE pattern, 3 marks)

A single upward (crest) pulse travels along a string toward its far end. State, with reasons, the nature of the reflected pulse when the end is (a) tied rigidly to a wall, (b) attached to a light ring that slides freely on a smooth vertical rod. (c) The string is instead joined to a heavier string; is the pulse transmitted into the heavy string inverted?

Step 1. Part (a), fixed end. The end cannot move, so its displacement stays zero. By Newton's third law the wall pushes back down on the string as the crest arrives, launching an inverted pulse — a crest returns as a trough, a phase change of $\pi$ (180°).

Step 2. Part (b), free end. The ring offers no transverse resistance, so the end is free to fly up; it overshoots to twice the pulse height, and the reflected pulse returns erect — no phase change.

Step 3. Part (c), transmission. The transmitted pulse is never inverted, regardless of the boundary: only the reflected pulse can flip. So the pulse entering the heavier string stays a crest (erect), although it is reduced in amplitude and slowed. (The reflected pulse at that junction, meeting the heavier string, would be inverted.)

Step 4. One-line summary to write in the margin: fixed inverts, free does not, transmission never inverts.

2 · Board level — the denser–rarer rule and the impedance limits (CBSE/JEE pattern, 3 marks)

Two strings under the same tension are joined; string 2 is denser than string 1. (a) When a wave travels from string 1 into string 2, is the reflected wave inverted? (b) Show that a fixed end and a free end are the two extremes of this rule, using $r = (Z_1 - Z_2)/(Z_1 + Z_2)$ .

Step 1. Part (a). Denser means larger $\mu$ , hence larger impedance $Z = \sqrt{T\mu}$ , so $Z_2 > Z_1$ . In $r = (Z_1 - Z_2)/(Z_1 + Z_2)$ the numerator is then negative, so $r < 0$ : the reflected wave is inverted. Reflection off a heavier string flips; off a lighter string it does not.

Step 2. Part (b), the fixed-end extreme. A rigid wall behaves like an infinitely heavy string, $Z_2 \to \infty$ . Then $r = (Z_1 - Z_2)/(Z_1 + Z_2) \to -Z_2/Z_2 = \boxed{-1}$ — a full-amplitude inverted reflection, the strongest possible flip.

Step 3. The free-end extreme. A free end behaves like an infinitely light string, $Z_2 = 0$ . Then $r = (Z_1 - 0)/(Z_1 + 0) = \boxed{+1}$ — a full-amplitude erect reflection, no flip.

Step 4. So the whole "fixed inverts, free does not" rule is just the two ends of the impedance-mismatch scale, with every real two-string junction sitting somewhere between $-1$ and $+1$ .

3 · Competitive level — the two-string junction, in full (JEE-Advanced pattern)

A light string of linear density $\mu_1 = 1.0\times10^{-3}\ \mathrm{kg/m}$ is joined to a heavier string $\mu_2 = 4.0\times10^{-3}\ \mathrm{kg/m}$ , both under tension $T = 100\ \mathrm{N}$ . A wave travels from the light string toward the junction. Find (a) the wave speeds and impedances, (b) the reflection and transmission coefficients, (c) the reflected and transmitted power fractions, and check they sum to one. (d) What changes if the wave instead travels from the heavy string into the light one?

Step 1. Speeds and impedances. $v_1 = \sqrt{T/\mu_1} = \sqrt{100/10^{-3}} = \boxed{316.2\ \mathrm{m/s}}$ and $v_2 = \sqrt{100/(4\times10^{-3})} = \boxed{158.1\ \mathrm{m/s}}$ . Impedances $Z_1 = \sqrt{T\mu_1} = \sqrt{0.1} = \boxed{0.3162}$ and $Z_2 = \sqrt{T\mu_2} = \sqrt{0.4} = \boxed{0.6325}$ (units $\mathrm{kg/s}$ ).

Step 2. Coefficients, equation (2). $r = (Z_1 - Z_2)/(Z_1 + Z_2) = (0.3162 - 0.6325)/(0.9487) = \boxed{-0.3333} = -\tfrac13$ — inverted, and only one-third the amplitude. $t = 2Z_1/(Z_1 + Z_2) = 0.6325/0.9487 = \boxed{0.6667} = \tfrac23$ — erect.

Step 3. Power fractions, equation (3). $R = r^2 = (1/3)^2 = \boxed{0.1111} = \tfrac19$ . $T = (Z_2/Z_1)t^2 = 2 \times (2/3)^2 = 2 \times \tfrac49 = \boxed{0.8889} = \tfrac89$ . Sum: $R + T = \tfrac19 + \tfrac89 = \boxed{1.0000}$ — energy conserved exactly.

Step 4. Part (d), the reversed journey. Going from heavy into light swaps $Z_1$ and $Z_2$ , so $r = (Z_2 - Z_1)/(Z_2 + Z_1) = +\tfrac13 = \boxed{+0.3333}$ — now erect, because the wave meets a lighter string. The magnitudes of $R$ and $T$ are unchanged ( $\tfrac19$ and $\tfrac89$ ); only the phase of the reflection flips. Timing note: about three minutes. Compute $Z_1$ and $Z_2$ first — everything else is a ratio of them — and keep them as $\sqrt{0.1}$ and $\sqrt{0.4}$ so that $Z_2/Z_1 = 2$ falls out cleanly. The examiner's bait is to ask for the sign of $r$ and reward "inverted" only if you noticed the wave was entering the heavier string; the reversed part (d) checks you understood the mismatch, not a memorised sign.

Figure

A wave from a light string into a heavy one — one-third of the amplitude reflects inverted, two-thirds transmits erect

A word about sound, before the next chapter

The same rule governs sound in a pipe, with one twist in vocabulary. A compression reflecting off a closed (rigid) end returns as a compression — like the fixed end, it is a hard boundary — while a compression reaching an open end returns as a rarefaction, the phase flipping there instead. This swap is why a pipe closed at one end resonates an octave below an open one of the same length, and it is the reason wind instruments sound as they do. The full story of pipes and their harmonics is the next chapter's; here it is enough to know that "reflection with a phase change at a hard boundary" reappears, note for note, in the air of an organ.

Exam traps

Inverting the transmitted pulse. Only the reflection can flip. The transmission coefficient $t$ is always positive, so the transmitted pulse keeps the incident's sign — a crest stays a crest. Reserve inversion for reflections off heavier or fixed boundaries.

Treating a free end as "nothing happens." A free end reflects a full-amplitude erect pulse, and its endpoint swings to twice the incident height — an antinode. It is every bit as active as a fixed end; it simply does not flip the phase.

Forgetting the reflected amplitude is reduced at a junction. A fixed end gives $r = -1$ : a full-height inverted mirror. A two-string junction gives a fraction: our example reflects only one-third of the amplitude. Do not draw a full-height reflected pulse at a soft junction.

Confusing a phase change with a frequency change. Reflection changes amplitude and can add a $\pi$ phase, but it never changes $f$ . The boundary oscillates at the incident frequency, so reflected and transmitted waves share it. Any answer with a new frequency after reflection is wrong.

Board scope. This topic belongs to CBSE, ICSE, the JEE/NEET band, IB (SL and HL) and A-Level (Cambridge 9702, which explicitly examines the phase change on reflection). It is not on AP Physics 2 or IGCSE — if that is your board, the fixed-versus-free inversion is useful background for standing waves but the impedance coefficients are beyond your syllabus.

Origins · 1 minGeorge Green and how this was discovered

George Green

1793–1841 · England

George Green was a Nottingham miller with barely a year of schooling who taught himself mathematics from library books and, in 1828, privately printed an Essay introducing the potential function and what we now call Green's functions — work so far ahead of its time that it was nearly lost. Admitted to Cambridge at forty, he turned to waves, and in 1838 published "On the Reflexion and Refraction of Sound," showing that when a wave meets the boundary between two media it divides according to the mismatch of their impedances — precisely the algebra of this article. His analysis of how much reflects and how much transmits, and of the phase change that accompanies reflection off a denser medium, underlies everything from ultrasound imaging to anti-reflection coatings. Green died three years later, his genius recognised only after William Thomson (later Lord Kelvin) rescued the Essay in 1845.

History

1747

— Jean le Rond d'Alembert's wave equation lets a reflected wave be written as a second travelling solution, making the fixed-end image pulse exact.

1801

— Thomas Young explains the colours of thin films by a phase change when light reflects from a denser medium — the optical cousin of the inverted pulse.

1828

— George Green privately prints his Essay introducing the potential and the functions that bear his name, a different work from the one below.

1838

— Green's On the Reflexion and Refraction of Sound shows a wave splits at a boundary according to the impedance mismatch of the two media.

1849

— George Gabriel Stokes derives the relations fixing the phases of the reflected and transmitted waves, proving reflection off the denser medium inverts.

Going further

Texts. NCERT Physics Class 11, Ch. 15 "Waves", §15.6 (reflection of waves) · H. C. Verma, Concepts of Physics Vol. 1, "Waves on a String" — reflection and the two boundary conditions · Halliday, Resnick and Walker, "Waves — I" · Cambridge A-Level 9702 topic 8, which examines the phase change on reflection.

Provenance. Every number is recomputed in `verify_hs2-wave-reflection-boundaries.py`: the speeds $v_1 = 316.2$ , $v_2 = 158.1$ m/s; the impedances $Z_1 = 0.3162$ , $Z_2 = 0.6325$ ; the coefficients $r = -1/3 = -0.3333$ , $t = 2/3 = 0.6667$ ; the power fractions $R = 1/9 = 0.1111$ , $T = 8/9 = 0.8889$ , and $R + T = 1.0000$ ; and the limits $r = -1$ (fixed), $r = +1$ (free), and the reversed $r = +1/3$ .

Prerequisites. Superposition & Interference of Waves — reflection sets a wave interfering with itself · Speed of a Wave on a Stretched String — the source of $v = \sqrt{T/\mu}$ and hence of the impedance $Z = \sqrt{T\mu}$ . Leads to. Standing Waves & Stationary-Wave Formation — where a wave and its reflection lock into a fixed pattern. Go deeper. The optical version, in which reflection off a denser medium flips a light wave's phase, is Stokes Relations and Phase Change on Reflection — enrichment, not a dependency.

Why exactly does a fixed end flip the pulse but a free end does not?

It comes down to what the end is allowed to do. A fixed end cannot move, so its displacement must stay zero; the only way a returning wave can keep it at zero is to be the inverted mirror of the incoming one, so the two always cancel there — hence the flip. A free end, by contrast, feels no transverse force, so instead of "zero displacement" its condition is "zero slope," and satisfying that requires a reflection that is not turned over. Mechanically, the fixed end pulls back (Newton's third law) and inverts; the free end is dragged along and overshoots, staying erect.

What is impedance, physically? It sounds like an electrical word.

It is borrowed from electricity, but here it means a string's resistance to being wiggled sideways — how much transverse force it takes to give the string a certain transverse velocity. A heavy, tightly stretched string is hard to wiggle and has a large impedance $Z = \sqrt{T\mu}$ ; a light, slack one is easy and has a small $Z$ . Reflection at a boundary is governed by the mismatch between the impedances on the two sides, exactly as an electrical signal reflects off a mismatched cable. A rigid wall is the limit of infinite impedance, a free end the limit of zero — which is why they give the two extreme reflections.

Can the transmission coefficient really be bigger than 1?

Yes, and it does not break energy conservation. At a free end $t = 2$ , meaning the endpoint moves with twice the incident amplitude — but that is an amplitude ratio, not an energy ratio. The energy that flows on is measured by the power fraction $T = (Z_2/Z_1)t^2$ , which stays at or below $1$ because a larger $t$ occurs only when the far medium is lighter (smaller $Z_2$ ), and the $Z_2/Z_1$ factor pulls the power back down. Amplitude and energy are different quantities; $t > 1$ is allowed, $T > 1$ is not.

Does the wave's frequency change when it reflects or transmits?

No — the frequency is fixed by the source and survives every boundary. Reflection can change the amplitude and add a half-cycle phase shift, and transmission changes the amplitude and (through a new wave speed) the wavelength, but neither touches $f$ . The boundary point is driven up and down at the incident frequency, and it cannot pass on or send back any other rate. So on the reflected wave, the transmitted wave and the incident wave alike, the frequency is one and the same. Any solution that reports a new frequency after a boundary has made an error.

Is the reflected pulse always the same height as the incident one?

Only at a perfect fixed or free end, where $|r| = 1$ . At a junction between two real strings the reflected amplitude is a fraction of the incident, set by $r = (Z_1 - Z_2)/(Z_1 + Z_2)$ . In this article's example the reflection is one-third the incident height (and inverted); the other two-thirds of the amplitude carries on into the second string. So when you sketch a reflection at a two-string junction, draw it shorter than the incident pulse — a full-height reflection means a true fixed end, not a soft boundary.

How does this connect to standing waves?

Directly — it is the reason they exist. When a wave reflects off a boundary, the reflected wave travels back through the incident wave and the two superpose. If the geometry is right (a string fixed at both ends, say), the wave keeps reflecting back and forth, and the overlapping forward and backward waves lock into a fixed pattern of nodes and antinodes — a standing wave. The phase change at a fixed end is exactly what forces a node there. The next article builds standing waves from precisely this superposition of a wave with its own reflection.