In short

When a wave reaches a boundary, part (possibly all) of it reflects and part transmits into whatever medium lies beyond. The split is governed by the wave impedance Z = \mu v = \sqrt{\mu T} on each side of the boundary.

  • Fixed end (displacement forced to zero, like a rope knotted to a wall): the wave reflects with full amplitude and inverts (r = -1) — a crest comes back as a trough.
  • Free end (no transverse force, like a rope tied to a light ring that slides without friction on a smooth pole): the wave reflects with full amplitude and no inversion (r = +1) — a crest comes back as a crest.
  • Junction of two media (a thin string tied to a thick one, both same tension): partial reflection, partial transmission, with amplitudes
\boxed{\; r = \frac{Z_1 - Z_2}{Z_1 + Z_2}, \qquad \tau = \frac{2Z_1}{Z_1 + Z_2}. \;}

Here Z_1 is the impedance of the medium the wave arrives from, and Z_2 is the impedance it enters.

  • Impedance matching (Z_1 = Z_2): r = 0 and the wave transmits completely, with no reflection. This is how audio engineers kill echo, how stethoscopes collect heart sounds, and how ultrasound gel couples the probe to a patient's body.

Superposition of the incident wave with the reflected wave gives a standing wave (treated in the next article). The phase change on reflection — whether r is positive or negative — is what decides whether boundary conditions give nodes or antinodes.

Clip one end of a washing-line to the manja string wound on a charkhi, hold the other end, and flick a pulse. The pulse runs along the line, reaches the clip, and bounces back — but upside-down. A hump that set out above the line returns as a dip below it.

Now untie the far end from the clip and tie it instead to a light curtain ring that slides freely on a smooth vertical pole. Flick another pulse. This time the pulse returns right-side up — a hump going out comes back as a hump.

Two ends, two different echoes. The string did not change; the string's tension did not change; only the boundary changed. That boundary told the wave how to behave when it arrived — and the wave obeyed by inverting in one case and not in the other.

Every wave you will meet has to deal with boundaries. A sound wave from your voice in the corridor of a Delhi Metro station bounces off tiled walls and ceiling, producing the characteristic hollow echo. The ultrasound pulse from a pregnancy-scan probe hits the interface between different tissues and returns a fraction of its energy, which the machine interprets as a boundary. A light ray reaching the glass surface of a Vaishali apartment window reflects ~4% and transmits ~96%. A bat's echolocation call bounces off an insect and returns with timing information the bat's brain converts to range.

All of these are instances of the same physics: when a wave meets a boundary, it decomposes itself into a reflected piece and a transmitted piece, and the two pieces are set by matching the wave at the boundary. This article works out the details.

The three boundaries to understand are fixed end, free end, and impedance boundary between two media. The last generalises the first two — a fixed end is the limit Z_2 \to \infty, a free end is the limit Z_2 \to 0. You will need introduction to waves, the wave speed on a string, and the principle of superposition to make sense of everything below.

Reflection from a fixed end — the inverting echo

Start with the simplest, cleanest case: a wave on a long string, whose right end is rigidly fixed — tied immovably to a wall. The displacement of the string at the wall is always zero, no matter what the wave is doing. Formally, if the wall is at x = 0, then

y(x=0, t) = 0 \quad \text{for all } t. \tag{1}

This is a boundary condition. The wave that runs along the string has to satisfy it.

Send an incoming wave y_\text{inc}(x, t) = f(x - vt) travelling in the -x direction toward the wall at x = 0. (Written as a function of x + vt with v > 0; let us write the incoming wave as f(x + vt) going in the -x direction to keep signs clean.) At any point on the string away from the wall, there is also a reflected wave y_\text{ref}(x, t) = g(x - vt) going back out in the +x direction. The actual displacement of the string is the sum of the two waves — superposition:

y(x, t) = f(x + vt) + g(x - vt).

Evaluate at x = 0 to enforce the boundary condition:

y(0, t) = f(vt) + g(-vt) = 0 \quad \text{for all } t.

This must hold for every t. The only way for f(vt) + g(-vt) = 0 to hold for every t is

g(-u) = -f(u) \quad \text{for every } u, \tag{2}

or equivalently, g(s) = -f(-s) for every s.

Why: the reflected-wave profile g is determined by the condition that the string's displacement at the wall is zero. The only reflected function that cancels f at every time is g(s) = -f(-s) — the incoming pulse, spatially reversed and sign-flipped.

What does that mean physically? Two operations have happened:

  1. Spatial reversal (f(u) \to f(-u)): the pulse's shape is mirrored in x. A pulse that was leading with its front now leads with its back.
  2. Sign flip (f \to -f): the pulse is inverted — crests become troughs and vice versa.

The net result: the reflected pulse is the incoming pulse turned upside down and running the other way. If you watch the moment when the incoming pulse and reflected pulse overlap at the wall, they cancel to produce zero displacement exactly at the boundary — which is what the wall demanded.

Animated: a pulse moving right hits the fixed wall and returns as an inverted pulse A red marker tracks the peak of a pulse initially at x = 2 m moving to the right at 2 m/s. When it reaches the wall at x = 10 m, it reflects back at 2 m/s. The marker's vertical position traces a hump going out (positive y) and a dip coming back (negative y), representing the sign flip on reflection from a fixed end. fixed wall (x = 10 m)
A pulse moves to the right at 2 m/s (first 4 s). At $t = 4$ s it reaches the fixed wall at $x = 10$ m and reflects. Notice the vertical position of the marker flips sign — a crest went out, a trough returns. This sign flip is the inversion phase change, sometimes written as a "π phase shift." Click replay to watch the crest transform into a trough as it bounces.

Equivalent statement: r = -1

In the language of amplitude coefficients, a sinusoidal incoming wave y_\text{inc} = A\sin(kx + \omega t) reflects from a fixed end as y_\text{ref} = -A\sin(kx - \omega t) = A\sin(-kx + \omega t + \pi) = A\sin(\pi - kx + \omega t). The amplitude reflection coefficient is

r_\text{fixed} = -1.

The minus sign encodes the inversion: the reflected amplitude has the same magnitude (|r| = 1, full reflection) but opposite sign (the "\pi phase shift on reflection from a denser medium," in the language of optics).

This inversion is the reason a stethoscope chestpiece, pressed to a patient's ribs, hears a much fainter sound than when pressed to soft tissue — the rib acts nearly like a fixed-end reflector, and the sound that bounces back from it cancels the outgoing wave over part of the cycle. More directly, on a string with a fixed end, the displacement at the end is pinned to zero, forming a node; when you set up a standing wave on a sitar string clamped at both nut and bridge, those ends are fixed, and the string's resonant frequencies are determined by fitting an integer number of half-wavelengths between them.

Reflection from a free end — the non-inverting echo

Now the other extreme. Imagine the string's right end is tied to a massless ring that slides without friction on a frictionless pole — so the ring can move freely in the transverse direction. There is nothing to resist the ring's motion, which means the tension in the string at the ring's position cannot have any component along the transverse direction. In other words: the string at the free end has zero slope.

Formally, if the free end is at x = 0, then

\left.\frac{\partial y}{\partial x}\right|_{x=0} = 0 \quad \text{for all } t. \tag{3}

Why: for a string under tension T, the transverse force on the end point is T \cdot \partial y/\partial x (see the derivation of the wave speed). A free end means no transverse force on the end, which means the slope \partial y/\partial x vanishes there.

Set up the same decomposition as before: incoming f(x + vt) plus reflected g(x - vt). Compute the slope at x = 0:

\left.\frac{\partial y}{\partial x}\right|_{x=0} = f'(vt) + g'(-vt) = 0.

For this to hold for every t: g'(-u) = -f'(u) for every u. Integrating (and choosing the constant of integration so the reflected pulse has no overall shift), this gives

g(s) = f(-s), \quad \text{so} \quad y_\text{ref}(x, t) = f(-(x - vt)) = f(vt - x). \tag{4}

Why: integrating g'(-u) = -f'(u) with respect to u gives -g(-u) = -f(u) + C, so g(-u) = f(u) - C. Setting C = 0 (no constant offset) gives g(s) = f(-s): the reflected profile is the mirror image of the incoming one — spatially reversed, but with no sign flip.

Physically: the pulse bounces back from the free end right-side up — a hump going out returns as a hump. No inversion, no sign flip. The amplitude reflection coefficient is

r_\text{free} = +1.
Animated: a pulse moving right hits the free end and returns as a non-inverted pulse A red marker tracks the peak of a pulse initially at x = 2 m moving to the right at 2 m/s. When it reaches the free end at x = 10 m, it reflects back at 2 m/s, staying on the same side of the equilibrium line — no inversion. free end (x = 10 m)
At $t = 4$ s the pulse reaches the free end at $x = 10$ m and reflects. The marker's vertical position stays on the same side of the equilibrium line — a crest returns as a crest. At the exact moment of reflection, the incoming and reflected pulses overlap, and the total displacement at the boundary is *twice* the pulse amplitude, not zero. Click replay.

Two consequences:

A real-world intuition: imagine a rope tied to a small, light ring that slides freely up a vertical pole at the far end. Give the rope a flick. The pulse runs out, reaches the ring, and the ring itself is free to accelerate upward as the pulse arrives. The ring rides up and down, peaking when the pulse arrives, and as it comes back down it sends a matching pulse back along the rope — in the same sign (both "up" phases). That matching pulse is the reflected wave, and it inherits the positive sign of the incoming wave.

Reflection at a boundary between two media

The real interesting case — and the one most relevant for stethoscopes, ultrasound, and audio engineering — is when the wave meets not an end, but a junction: the string you are holding splices, at x = 0, to a different string of different mass density. Both strings are under the same tension T (if the junction is a knot, the tension is continuous through it). The string to the left (x < 0) has mass density \mu_1 and wave speed v_1 = \sqrt{T/\mu_1}. The string to the right (x > 0) has \mu_2 and v_2 = \sqrt{T/\mu_2}.

A wave comes in from the left, moving in +x: y_\text{inc}(x, t) = A_\text{inc}\sin(k_1 x - \omega t), with k_1 = \omega/v_1. At the junction, part of it reflects as y_\text{ref}(x, t) = A_\text{ref}\sin(-k_1 x - \omega t), going back in -x; and part transmits as y_\text{trans}(x, t) = A_\text{trans}\sin(k_2 x - \omega t), continuing into the second string with k_2 = \omega/v_2.

Two boundary conditions link the three amplitudes:

(i) The string is continuous across the junction. Displacement on the left at x = 0 equals displacement on the right at x = 0:

A_\text{inc}\sin(-\omega t) + A_\text{ref}\sin(-\omega t) = A_\text{trans}\sin(-\omega t),

which gives

A_\text{inc} + A_\text{ref} = A_\text{trans}. \tag{5}

Why: the two strings are knotted together at the junction — they must have the same displacement there. If they did not, the knot would have to stretch or break, which it cannot.

(ii) The slopes match. On the left at the junction, the slope is \partial y/\partial x |_{\text{left},\,x=0}; on the right, it is \partial y/\partial x |_{\text{right},\,x=0}. The tensions on each side of the junction pull with magnitudes T\,\partial y/\partial x; if these did not balance, the massless junction knot would have an infinite acceleration. So the slopes must be equal:

k_1(A_\text{inc} - A_\text{ref})\cos(-\omega t) = k_2 A_\text{trans}\cos(-\omega t),

which simplifies to

k_1(A_\text{inc} - A_\text{ref}) = k_2 A_\text{trans}. \tag{6}

Why: the incoming wave has slope +k_1 A_\text{inc}\cos(\cdot) at x=0. The reflected wave has spatial derivative -k_1 A_\text{ref}\cos(\cdot) because it runs in the opposite direction. Their sum — the slope on the left at the junction — must equal the slope on the right, k_2 A_\text{trans}\cos(\cdot), so that the tensions on each side pull equally and the junction is in mechanical equilibrium.

Solving for r and \tau

The amplitude reflection coefficient r and amplitude transmission coefficient \tau are the ratios A_\text{ref}/A_\text{inc} and A_\text{trans}/A_\text{inc}.

Step 1. Rewrite equations (5) and (6) in terms of r and \tau, dividing through by A_\text{inc}:

1 + r = \tau, \tag{5'}
k_1(1 - r) = k_2 \tau. \tag{6'}

Step 2. Substitute \tau = 1 + r into equation (6'):

k_1(1 - r) = k_2(1 + r).
k_1 - k_1 r = k_2 + k_2 r.
k_1 - k_2 = r(k_1 + k_2).

Why: collect terms involving r on one side, terms without r on the other. Standard algebra.

Step 3. Solve for r:

r = \frac{k_1 - k_2}{k_1 + k_2}. \tag{7}

Step 4. Substitute back to find \tau:

\tau = 1 + r = 1 + \frac{k_1 - k_2}{k_1 + k_2} = \frac{(k_1 + k_2) + (k_1 - k_2)}{k_1 + k_2} = \frac{2k_1}{k_1 + k_2}. \tag{8}

Step 5. Rewrite in terms of impedance Z = \mu v.

Using k_i = \omega/v_i = \omega \sqrt{\mu_i / T}, the ratio k_1 / k_2 = \sqrt{\mu_1 / \mu_2} = v_2/v_1. A cleaner combination emerges with the impedance Z_i = \mu_i v_i = \sqrt{\mu_i T}: we have Z_i \propto \sqrt{\mu_i} at fixed T, so k_i \propto Z_i. Equations (7) and (8) become

\boxed{\; r = \frac{Z_1 - Z_2}{Z_1 + Z_2}, \qquad \tau = \frac{2Z_1}{Z_1 + Z_2}. \;} \tag{9}

Why: Z_1/Z_2 = \sqrt{\mu_1/\mu_2} = k_1/k_2 (at common T). Substituting into (7) and (8) and cancelling the common factor gives the impedance form.

Interpretation

Equation (9) is one of the most useful pairs of formulas in wave physics. Three limits decode the whole story.

Limit 1: fixed end (Z_2 \to \infty). The second medium is so heavy or so stiff that the junction can barely move. Then

r = \frac{Z_1 - \infty}{Z_1 + \infty} = -1, \quad \tau = \frac{2Z_1}{Z_1 + \infty} = 0.

Full inverting reflection; zero transmission. A fixed end is the infinite-impedance limit.

Limit 2: free end (Z_2 \to 0). The second medium has no inertia. Then

r = \frac{Z_1 - 0}{Z_1 + 0} = +1, \quad \tau = \frac{2Z_1}{Z_1 + 0} = 2.

Full non-inverting reflection; the amplitude on the other side is twice the incoming (because A_\text{inc} + A_\text{ref} = A_\text{trans} becomes A + A = 2A). A free end is the zero-impedance limit.

Limit 3: matched media (Z_1 = Z_2). Then

r = 0, \quad \tau = 1.

No reflection; full transmission. The wave passes through the junction as if it were not there. This is called impedance matching, and it is the single most important principle in the design of any device that needs to couple wave energy from one medium to another — audio speakers to rooms, stethoscope diaphragms to skin, ultrasound probes to body tissue, microwave transmitters to antennas.

Does it matter which direction the wave came from?

Yes — partially. The magnitude of r is symmetric: |r| is the same whether the wave goes from medium 1 to medium 2 or the other way. But the sign of r flips: r_{1\to 2} = (Z_1 - Z_2)/(Z_1+Z_2) has the opposite sign to r_{2\to 1} = (Z_2 - Z_1)/(Z_1+Z_2).

The transmission coefficient \tau is not symmetric: \tau_{1\to2} = 2Z_1/(Z_1+Z_2) and \tau_{2\to 1} = 2Z_2/(Z_1+Z_2) are different (unless Z_1 = Z_2). Going from a heavy medium to a light one transmits a large amplitude; going the other way transmits a small amplitude. The power carried across the boundary, however, is the same both ways — which is a consequence of energy conservation and is worked out in the going-deeper section.

Explore the impedance ratio

Drag the impedance ratio below and watch how the reflection coefficient r = (1 - Z_2/Z_1)/(1 + Z_2/Z_1) and the transmission coefficient \tau = 2/(1 + Z_2/Z_1) change.

Interactive: reflection and transmission coefficients versus impedance ratio Two curves drawn against the impedance ratio Z2 over Z1, which ranges from 0 (free end) to 10 (heavy far medium). The r curve crosses from +1 at ratio 0 down through 0 at ratio 1 to -0.82 at ratio 10. The tau curve starts at 2 for ratio 0, drops to 1 at ratio 1, and approaches 0 at large ratios. A draggable red point sets the ratio; a readout panel shows the current r and tau. impedance ratio Z₂ / Z₁ r and τ (dimensionless) −1 0 1 2 0 1 5 10 r (reflection) τ (transmission) drag the red point along the x-axis
Drag the red point to vary the impedance ratio $Z_2/Z_1$. At ratio 1 (matched impedances), $r = 0$ and the wave transmits fully. At ratio 0 (free end), $r = +1$ and $\tau = 2$. As the ratio grows ($Z_2 \gg Z_1$, approaching a fixed end), $r \to -1$ and $\tau \to 0$. The two curves together capture every boundary condition on a string.

Power reflection and transmission

The amplitude coefficients r and \tau tell you how the displacements split at the boundary. For energy conservation, you want power coefficients — what fraction of the incoming power is reflected back and what fraction is transmitted onward.

From P = \tfrac12 \mu \omega^2 A^2 v = \tfrac12 Z \omega^2 A^2 (since Z = \mu v), the fraction of power reflected is

R = \frac{P_\text{ref}}{P_\text{inc}} = \frac{Z_1 A_\text{ref}^2}{Z_1 A_\text{inc}^2} = r^2,

and the fraction transmitted is

\mathcal{T} = \frac{P_\text{trans}}{P_\text{inc}} = \frac{Z_2 A_\text{trans}^2}{Z_1 A_\text{inc}^2} = \frac{Z_2}{Z_1}\tau^2 = \frac{4 Z_1 Z_2}{(Z_1 + Z_2)^2}.

Check: R + \mathcal{T} = 1.

R + \mathcal{T} = \frac{(Z_1 - Z_2)^2}{(Z_1 + Z_2)^2} + \frac{4Z_1 Z_2}{(Z_1 + Z_2)^2} = \frac{Z_1^2 - 2Z_1 Z_2 + Z_2^2 + 4Z_1 Z_2}{(Z_1 + Z_2)^2} = \frac{(Z_1 + Z_2)^2}{(Z_1 + Z_2)^2} = 1. \checkmark

Why: (Z_1 - Z_2)^2 + 4Z_1 Z_2 = Z_1^2 + 2Z_1Z_2 + Z_2^2 = (Z_1 + Z_2)^2. Algebra confirms that no power is lost at the junction — every joule of incoming energy either reflects or transmits.

Impedance matching (Z_1 = Z_2) gives R = 0 and \mathcal{T} = 1 — all the power transmits. Going from water to air, where Z_\text{water} is about 3500 times Z_\text{air} for sound, gives \mathcal{T} \approx 0.0011 — only 0.1% of the sound power in water gets into the air above it. (This is why fish below the surface of the Ganga do not hear the noise of a boat motor much from above — the sound largely reflects off the water surface.)

Worked examples

Example 1: A knot between two strings

A light string (mass density \mu_1 = 1.0 \times 10^{-3} kg/m) is joined at a knot to a heavier string (mass density \mu_2 = 4.0 \times 10^{-3} kg/m). Both strings are under the same tension T = 16 N. A wave of amplitude A_\text{inc} = 5 mm travels along the light string toward the knot. Find (a) the wave speeds v_1, v_2 on the two strings, (b) the reflection and transmission coefficients r and \tau, and (c) the reflected and transmitted amplitudes.

Two strings of different density joined at a knot A horizontal string: on the left, a thin line labelled mu1 = 0.001 kg/m; at x = 0, a knot (small filled circle); on the right, a thicker line labelled mu2 = 0.004 kg/m. Incoming wave arrow points right on the thin side; reflected wave arrow points left on the thin side (dashed); transmitted wave arrow points right on the thick side. The tension T = 16 N is labelled throughout. μ₁ = 1.0 × 10⁻³ kg/m μ₂ = 4.0 × 10⁻³ kg/m knot (x = 0) incoming reflected transmitted
Two strings of different densities, joined at a knot. A wave from the left partially reflects and partially transmits at the knot.

Step 1. Compute the wave speeds on each side.

v_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{16}{10^{-3}}} = \sqrt{16000} = 126.5 \text{ m/s}.
v_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{16}{4\times 10^{-3}}} = \sqrt{4000} = 63.2 \text{ m/s}.

Why: the wave speed formula v = \sqrt{T/\mu} comes from speed-of-waves-on-a-string. The thicker string — four times heavier per metre — carries waves at half the speed.

Step 2. Compute the impedances.

Z_1 = \sqrt{\mu_1 T} = \sqrt{10^{-3} \times 16} = \sqrt{0.016} = 0.1265 \text{ kg/s}.
Z_2 = \sqrt{\mu_2 T} = \sqrt{4\times 10^{-3} \times 16} = \sqrt{0.064} = 0.253 \text{ kg/s}.

The ratio Z_2/Z_1 = 2 (which could have been found faster as \sqrt{\mu_2/\mu_1} = \sqrt{4} = 2).

Step 3. Compute the amplitude reflection and transmission coefficients.

r = \frac{Z_1 - Z_2}{Z_1 + Z_2} = \frac{1 - 2}{1 + 2} = -\frac{1}{3}.
\tau = \frac{2Z_1}{Z_1 + Z_2} = \frac{2}{1 + 2} = \frac{2}{3}.

Why: the negative r means the reflected wave is inverted — a crest that went in returns as a trough. The transmitted wave keeps the same sign (positive \tau), and since \tau < 1, its amplitude is reduced.

Step 4. Compute the reflected and transmitted amplitudes.

A_\text{ref} = r \cdot A_\text{inc} = -(1/3) \times 5 \text{ mm} = -1.67 mm (the sign indicates inversion, magnitude 1.67 mm).

A_\text{trans} = \tau \cdot A_\text{inc} = (2/3) \times 5 \text{ mm} = 3.33 mm.

Step 5 (check). A_\text{inc} + A_\text{ref} = 5 + (-1.67) = 3.33 = A_\text{trans}. Continuity of displacement is satisfied.

Result: r = -1/3, \tau = 2/3. Reflected amplitude = 1.67 mm (inverted). Transmitted amplitude = 3.33 mm.

What this shows: Going from a light string to a heavy string, the wave partially inverts on reflection — the signature of reflecting into a "denser" medium. Power check: R = r^2 = 1/9 \approx 11\% reflected, \mathcal{T} = 4 Z_1 Z_2/(Z_1 + Z_2)^2 = 8/9 \approx 89\% transmitted. The incoming wave loses about 11% of its power to the echo, and 89% continues into the thicker string.

Example 2: Echolocation in a Delhi Metro tunnel

A train announcer's voice at a Delhi Metro station emits a sound pulse (frequency 500 Hz, pressure amplitude 0.5 Pa) travelling down the tunnel at v = 343 m/s in air. The pulse reaches the end of the tunnel, which is closed off by a thick concrete wall. The wave impedance of air at sea level is Z_\text{air} = \rho_\text{air}\, v_\text{air} = 1.2 \times 343 \approx 412 N·s/m³. The wave impedance of concrete is about Z_\text{conc} \approx 7.5 \times 10^6 N·s/m³. Find the fraction of the sound power that reflects back as an echo.

Sound pulse in a metro tunnel reaching a concrete wall A horizontal tunnel cross-section: on the left, a person (speaker) emits a sound wave rightward down the tunnel. The wave is depicted as alternating compression bands. On the right, a thick grey wall labelled concrete stops the wave. A reflected wave arrow points leftward. speaker incident sound concrete wall echo (reflected)
A sound wave travelling through air in a tunnel reaches a concrete wall with very high impedance. Almost all of the incident acoustic power reflects back as an echo.

Step 1. Identify the impedance ratio.

Z_1 = 412 N·s/m³ (air), Z_2 = 7.5\times 10^6 N·s/m³ (concrete). Z_2/Z_1 \approx 18{,}200.

Step 2. Compute the amplitude reflection coefficient.

r = \frac{Z_1 - Z_2}{Z_1 + Z_2} = \frac{412 - 7.5\times 10^6}{412 + 7.5\times 10^6} \approx \frac{-7.5\times 10^6}{7.5\times 10^6} = -1 + \frac{824}{7.5\times 10^6} \approx -0.99989.

Why: when Z_2 \gg Z_1, the formula r = (Z_1 - Z_2)/(Z_1+Z_2) is approximately -1 + 2Z_1/Z_2 — slightly above -1 but very close. The wall is acting very much like a fixed end.

Step 3. Compute the power reflection coefficient.

R = r^2 \approx (0.99989)^2 \approx 0.99978.

About 99.978% of the sound power reflects as echo. Only 0.022% transmits into the concrete as a structural vibration.

Step 4. Estimate the echo pressure amplitude.

A_\text{echo} = |r| \times A_\text{inc} \approx 0.9999 \times 0.5 \text{ Pa} = 0.4999 \text{ Pa}.

Result: R \approx 0.9998 (almost complete reflection); echo pressure amplitude \approx 0.5 Pa; the echo is essentially the full announcement returning from the wall, which is why metro tunnel announcements are audibly reverberant.

What this shows: Concrete walls and air are so impedance-mismatched for sound that they behave almost exactly like fixed-end boundaries. This is true for most situations where sound meets a building — tiled walls, stone pillars, glass facades. The trivial fraction that does transmit into the wall is why building-to-building sound isolation works so well (and why loud music in one flat barely penetrates an adjacent one through concrete). The echoes you hear in a metro station, a temple hall, or an empty Connaught Place arcade are this physics in action.

Impedance matching in the real world

Why do stethoscopes have a rubber diaphragm and a chestpiece that presses flat against the skin? Because the skin has high acoustic impedance (close to water), the stethoscope's air column inside has low impedance, and a direct air-to-skin boundary would reflect almost all the heart sound away. The rubber diaphragm is chosen to have an intermediate impedance, and when pressed firmly against the chest it acts as a matching layer — reducing the mismatch in two smaller steps rather than one big one, and letting more acoustic energy into the stethoscope tube.

Why does an ultrasound technician smear a pregnancy-scan probe with gel before scanning? Because air between the probe and the patient's skin would reflect nearly all of the ultrasound (air-to-tissue impedance mismatch is massive, almost as bad as air-to-concrete). The gel has impedance close to water (close to tissue), so the path probe-to-gel-to-skin is nearly matched, and most of the ultrasound energy enters the body.

Why is a speaker cone flared into a horn? A bare speaker driver has much higher mechanical impedance than air; an exponential horn gradually transitions from the high-impedance driver to the low-impedance free air, matching the two over many half-wavelengths and letting more acoustic power radiate into the room. The same principle is why bansuri flutes have flared ends and why church organs use massive resonant pipes.

Why does a bat's echolocation work? Because insect bodies are mostly water — high acoustic impedance — against the low impedance of air around them. A bat's ultrasonic call suffers a big impedance mismatch at the insect's body, reflecting a substantial fraction back. That echo, timed by the bat's brain, gives range. The same physics underlies naval sonar (water-steel mismatch) and fish-finders (water-air-bladder mismatch).

Common confusions

The fixed-end, free-end, and impedance-junction results in the main text are plenty for JEE. What follows extends them.

Deriving the phase change on reflection — the quantum-mechanical connection

In quantum mechanics, a particle's wave function \psi(x, t) satisfies the Schrödinger equation. When the particle meets a step potential (a sudden jump from potential V = 0 to V = V_0 > 0 at x = 0), the wave function splits into a reflected and transmitted piece with amplitude coefficients that look exactly like equation (9), with the wave number k playing the role of impedance. For an energy E above the barrier:

k_1 = \sqrt{2mE}/\hbar, \qquad k_2 = \sqrt{2m(E - V_0)}/\hbar,
r = \frac{k_1 - k_2}{k_1 + k_2}, \qquad \tau = \frac{2k_1}{k_1 + k_2}.

Same algebra, same limits. The R + \mathcal{T} = 1 check becomes probability conservation. This is not a coincidence: every wave phenomenon — classical or quantum — obeys matching conditions at boundaries, and the linear algebra of those conditions produces the same r and \tau expressions.

Why does tension not appear in r?

Look again at equation (9): r depends only on the ratio of impedances. For a string junction at common tension T, Z_1/Z_2 = \sqrt{\mu_1/\mu_2}, so r depends only on the mass density ratio, not on the tension at all. Double both strings' tension (while leaving the mass densities alone) and r is unchanged — the wave speeds both change by the same factor, but the reflection coefficient does not notice.

What if the junction has mass?

Real knots have mass. Mathematically, a massive junction adds an extra boundary condition: the knot's equation of motion, m_\text{knot}\,\ddot y_\text{knot} = F_\text{net}. For ordinary physical knots, m_\text{knot} is so small that this correction is tiny. In extreme cases — say, a small bead slid onto a string — it can be significant, and the reflection/transmission coefficients acquire a frequency-dependent piece, turning the massless-knot linear equations into a more complex impedance problem. The generalisation is called a mechanical-impedance network and is worked out in courses on transmission-line theory.

Multiple reflections and the Fabry-Perot interferometer

If you have two boundaries — a finite string between two different media, or a thin sheet of glass between air on both sides — a wave bounces back and forth between them. Each bounce picks up an r factor, so after many bounces, the net reflected and transmitted amplitudes are sums of geometric series:

A_\text{net, reflected} = r_1 A_\text{inc} + \tau_1 r_2 \tau_1' A_\text{inc} e^{2i\phi} + \tau_1 r_2 r_1' r_2 \tau_1' A_\text{inc} e^{4i\phi} + \ldots,

where \phi is the phase accumulated in one traversal of the middle layer and the primed coefficients are for waves going the other way. Summing the geometric series gives the Fabry-Perot interferometer formula, which is used in every high-precision laser instrument to filter wavelengths. It is also the mathematical basis for anti-reflection coatings on camera lenses — a single quarter-wavelength layer of magnesium fluoride on glass can reduce reflection from 4% per surface to well under 1% at a chosen wavelength.

Energy-conservation check for the general case

Starting from the amplitude coefficients r = (Z_1 - Z_2)/(Z_1+Z_2) and \tau = 2Z_1/(Z_1+Z_2), the power coefficients are:

R = r^2 = \left(\frac{Z_1 - Z_2}{Z_1 + Z_2}\right)^2, \qquad \mathcal{T} = \frac{Z_2}{Z_1}\tau^2 = \frac{Z_2}{Z_1}\cdot \frac{4 Z_1^2}{(Z_1+Z_2)^2} = \frac{4Z_1 Z_2}{(Z_1+Z_2)^2}.

Adding:

R + \mathcal{T} = \frac{(Z_1 - Z_2)^2 + 4Z_1 Z_2}{(Z_1+Z_2)^2} = \frac{Z_1^2 - 2Z_1Z_2 + Z_2^2 + 4Z_1Z_2}{(Z_1+Z_2)^2} = \frac{(Z_1 + Z_2)^2}{(Z_1+Z_2)^2} = 1.

No power is lost; the boundary is a perfect splitter.

A preview of standing waves

The next article, Standing Waves and Normal Modes, takes the reflection apparatus further. If you reflect a sinusoidal wave from a fixed end, the incident A\sin(kx + \omega t) and reflected -A\sin(kx - \omega t) superpose to give

y = A[\sin(kx + \omega t) - \sin(kx - \omega t)] = 2A\cos(\omega t)\sin(kx),

which is a standing wave — a product of a spatial pattern \sin(kx) and a temporal oscillation \cos(\omega t), with nodes at kx = n\pi (where \sin(kx) = 0) and antinodes at kx = (n+\tfrac12)\pi. The fixed-end boundary automatically creates a node at the boundary; the free-end boundary automatically creates an antinode. That is why a sitar string has nodes at both ends (both fixed) and a bansuri flute has an antinode at the open end (free).

The whole theory of musical instrument modes — pitches, overtones, harmonics — comes from applying the reflection rules of this article at two ends of a finite-length medium. The next article derives the resulting discrete spectrum of allowed frequencies.

Where this leads next