In short

A resonance tube is a vertical glass tube, open at the top and closed by a water surface at the bottom. The effective air column above the water behaves as a pipe closed at one end: pressure antinode at the water, pressure node (displacement antinode) at the open top. Strike a tuning fork of known frequency f above the mouth; as you lower the water, the air column lengthens. At certain lengths the column resonates — you hear the tube roar back at the fork's pitch. The first resonance sits at

L_1 + e \;=\; \tfrac{\lambda}{4},

the second at

L_2 + e \;=\; \tfrac{3\lambda}{4},

where e \approx 0.6\,R is the end correction (the antinode forms slightly above the physical mouth of a tube of inner radius R). Subtract:

L_2 - L_1 \;=\; \tfrac{\lambda}{2}, \qquad \boxed{\;v \;=\; f\lambda \;=\; 2f(L_2 - L_1).\;}

The end correction has cancelled. A 512 Hz fork and a 1-inch-bore tube give L_1 \approx 16 cm and L_2 \approx 50 cm; the subtraction returns v \approx 348 m/s at a lab temperature of 30 °C — within 1% of the theoretical v = \sqrt{\gamma RT/M} = 349 m/s for dry air. Raise the lab temperature on a May afternoon in Delhi and v rises: the same fork shows a slightly longer L_2 - L_1 because v \propto \sqrt{T}.

Walk into any well-stocked CBSE school laboratory in July and find the resonance-tube apparatus on the back bench: a tall glass cylinder, perhaps a metre long, clamped vertically on a stand, with a narrow rubber tube running from its lower end down to a reservoir of water clamped on a sliding platform. A hand-written label hangs from the metal: "RESONANCE COLUMN — HANDLE WITH CARE." On the shelf above sit half a dozen tuning forks stamped with their frequencies in neat black ink — 256 Hz, 288 Hz, 320 Hz, 384 Hz, 480 Hz, 512 Hz — and a small rubber mallet. There is a thermometer tucked into the stand's base. This apparatus is sixty years old in design and has not been improved since, for a reason: it measures the speed of sound in air to within about 1%, in a single lab period, using nothing but a tuning fork, a tube, some water, and your ears.

The physics is elegant. Strike the fork, hold it over the tube, and slide the reservoir. Each water level sets a different length of air column above the water surface. For most lengths, the fork's vibration couples weakly — you hear the fork, but the tube is silent. But at certain discrete lengths — L_1, L_2, L_3, \ldots — the column's own natural frequency matches the fork, and resonance kicks in: the air column amplifies the sound so strongly that the tube seems to hum on its own, loud enough to be heard across the room. The lengths at which this happens are spaced by exactly half a wavelength of the sound in air. Measure two consecutive lengths with a metre scale, subtract, multiply by 2f — and you have the speed of sound.

This article is the companion to the theoretical articles Sound Waves — Nature and Propagation and Vibrations of Strings and Pipes. Those derive the formulas v = \sqrt{\gamma P/\rho} and f_n = nv/(4L) for a closed pipe. This one is the lab counterpart: the apparatus, the derivation of why two consecutive resonances give v without any correction, the procedure as followed in a CBSE/ISC class-11 practical, the temperature dependence v \propto \sqrt{T} tested on a sweltering Indian afternoon, and the honest error analysis.

The apparatus

A "resonance tube" is really a resonance column — the column of air above the water surface inside the tube is what resonates, not the tube itself. The water's upper surface acts as a rigid wall (at a pressure antinode / displacement node); the open mouth at the top acts as a free pressure-release (at a pressure node / displacement antinode, approximately). A sliding water level lets you vary the column length continuously.

Typical apparatus in an Indian school/college lab:

Resonance tube apparatusA tall vertical glass tube mounted on a retort stand, with a water reservoir on the left connected to the lower end by rubber tubing. A tuning fork is held horizontally above the open top of the tube. The length L is labelled from the water surface inside the tube to the top edge of the tube. A thermometer is clipped to the side. Millimetre markings are shown along the tube.water surface0 cm40 cmL(air column length)reservoirrubber tubetuning fork (f)retort standthermometer
The resonance tube apparatus. The tuning fork is struck on a rubber mallet and held with its prongs vibrating horizontally just above the open mouth of the glass tube. Raising or lowering the water reservoir on the left changes the air-column length $L$ above the water surface. At the first resonance, $L = L_1$; at the second, $L = L_2$. A thermometer records the air temperature, which you will need to compare against $v = \sqrt{\gamma RT/M}$.

Why the water surface acts as a closed end

The water is essentially incompressible on the timescale of sound oscillations (milliseconds). Air molecules arriving at the surface bounce back almost without loss — the acoustic impedance of water is about 3500 times greater than that of air, so very little sound energy crosses into the water. The surface is therefore a near-perfect pressure-antinode / displacement-node boundary, exactly like the closed end of an organ pipe. The top of the tube, opening into the room, is a pressure-node / displacement-antinode boundary — to a good approximation. The resonance-tube system is, acoustically, a pipe closed at one end, with its length adjustable by moving the water up and down.

Why resonance happens — standing waves in a closed pipe

From Vibrations of Strings and Pipes, a pipe of length L closed at one end supports standing waves with a pressure antinode at the closed end and a pressure node at the open end. The allowed mode-shapes squeeze exactly (2n-1) quarter-wavelengths into the length L:

L \;=\; (2n-1)\,\tfrac{\lambda}{4}, \qquad n = 1, 2, 3, \ldots

The frequencies of these modes are f_n = (2n-1)\,v/(4L).

Turn this around: with the fork forcing a fixed frequency f (and hence fixing \lambda = v/f), resonance occurs for those lengths L at which the forced frequency matches one of the natural mode frequencies of the column. Solve for L:

L_n \;=\; (2n-1)\,\tfrac{\lambda}{4}.

So the resonance lengths are \lambda/4,\; 3\lambda/4,\; 5\lambda/4,\; 7\lambda/4,\;\ldots — equally spaced by \lambda/2.

Why: consecutive resonances correspond to adding one full half-wavelength of extra column. Geometrically, each time you lower the water by \lambda/2, you fit one more half-wavelength into the column while keeping the node/antinode pattern at the two ends intact.

First two resonance modes of a closed pipeTwo vertical tubes side by side, each with a closed bottom (the water surface) and an open top. The left tube shows the first mode, with L1 equal to a quarter wavelength — one node at the bottom and one antinode at the top. The right tube shows the second mode, with L2 equal to three-quarters of a wavelength — a node at the bottom, antinode above, node further up, and antinode at the top.node (water)antinodeL₁ = λ/4first resonanceL₂ = 3λ/4second resonance
The first two resonance modes of the closed air column. The water surface (shaded red at the bottom) enforces a displacement node; the open mouth (just above the tube's top edge) enforces a displacement antinode. The first resonance fits a single quarter-wavelength: $L_1 = \lambda/4$. The second fits three-quarters of a wavelength: $L_2 = 3\lambda/4$. Subtract: $L_2 - L_1 = \lambda/2$, independent of exactly where the antinode sits relative to the mouth.

The end correction — and why it does not matter

There is a small but important refinement. The pressure node at the open end is not exactly at the rim of the tube. It sits a short distance above it. The reason is physical: an oscillating parcel of air at the mouth of the tube pushes on, and is slightly loaded by, the cylinder of still air just outside the mouth. That outside cylinder — roughly of length 0.6\,R where R is the tube's inner radius — must also oscillate slightly as part of the standing wave. The effective acoustic length is therefore a little longer than the measured column length.

Let e be this end correction, with e \approx 0.6\,R for a uniform circular tube (derived by Rayleigh in 1894 for an unflanged pipe; the closely-related flanged-pipe correction is e \approx 0.82\,R). For the resonance condition, the effective length is L_n + e:

L_1 + e \;=\; \tfrac{\lambda}{4}, \tag{1}
L_2 + e \;=\; \tfrac{3\lambda}{4}. \tag{2}

Here is the clean move. Subtract (1) from (2):

L_2 - L_1 \;=\; \tfrac{3\lambda}{4} - \tfrac{\lambda}{4} \;=\; \tfrac{\lambda}{2}.

Why: the end correction e appears with the same sign in both equations and cancels on subtraction. You do not need to know e — you do not even need to know R — as long as the two resonances are of the same tube with the same mouth geometry. This is what makes the two-resonance method robust.

Multiply both sides by 2f, using v = f\lambda:

\boxed{\;v \;=\; f\lambda \;=\; 2f(L_2 - L_1).\;} \tag{3}

Equation (3) is the working formula of the experiment. Measure L_1 and L_2 with a metre scale, multiply their difference by twice the fork's frequency, and read off the speed of sound.

If you want to recover the end correction for its own sake — say, as an extra task in a practical exam — rearrange (1): e = \lambda/4 - L_1 = (L_2 - L_1)/2 - L_1. Compare this with the geometric prediction 0.6\,R for the given tube; the two should agree to about 10%. Discrepancies usually trace to slight non-circularity of the mouth, or to a fork held slightly to one side of the axis.

The procedure

The canonical CBSE / ISC / NCERT procedure runs like this. Treat it as a recipe you will actually follow in the lab next Monday.

Step 1: set up and level. Clamp the glass tube vertically using a spirit level or a plumb-bob against the tube. A tilt of 2–3° can shift L_1 and L_2 by a couple of millimetres each through parallax on the scale.

Step 2: fill the tube. Raise the reservoir so the water rises almost to the top of the glass tube. Watch for air bubbles inside the rubber tube; tap the rubber gently to clear them, because a trapped bubble makes the water level jumpy when you move the reservoir.

Step 3: pick a fork. Start with the highest-frequency fork available (say 512 Hz). Higher f means shorter \lambda, so L_1 and L_2 are shorter and both fit comfortably inside a 1 m tube. A 256 Hz fork gives L_2 \approx 100 cm, which can fall off the scale if the tube is not long enough.

Step 4: record the temperature. Note the air temperature inside the tube (or the laboratory's bulk temperature if they are the same — they usually are within 0.5 °C). You will need this to compare against theory.

Step 5: strike the fork. Hit the fork on a rubber mallet or a soft pad. Never on the corner of a desk or a metal edge — that produces audible overtones (modes other than the fundamental at the fork's stamped frequency) and makes resonance fuzzy.

Step 6: find the first resonance. Hold the vibrating fork horizontally with its prongs about 1 cm above the open mouth of the tube, prongs oriented so the air they push moves up and down the tube's axis. Starting from the water level high (short column), lower the reservoir slowly. The tube will suddenly roar at a certain length. That is L_1. Read the length from the water surface to the mouth of the tube using the scale marked on the glass. Record it.

Tip. The roar is unmistakable when you are close to it. As you approach from one side (water too high), the sound swells sharply. As you continue past it (water too low), it dies equally sharply. Bracket the maximum by going back and forth: mark the water level where the sound is loudest.

Step 7: find the second resonance. Keeping the fork struck (re-strike it as needed — it only rings for 5–10 seconds), continue lowering the reservoir past the first resonance. The sound will die, then swell again as you cross L_2 \approx 3L_1. Record L_2. A third resonance (at L_3 \approx 5L_1) may exist if the tube is long enough; finding it gives an independent check but is not required.

Step 8: repeat. Do the whole measurement five or six times with the same fork, using slightly different approach speeds and from both directions (water rising and water falling). Calculate the mean \bar{L}_1 and \bar{L}_2; these are your best estimates. Compute

v \;=\; 2f(\bar{L}_2 - \bar{L}_1).

Step 9: change the fork. Repeat with at least one other fork of different frequency (say, switch from 512 Hz to 384 Hz). You should get the same v within error — because the speed of sound is a property of the air, not the fork. If the two forks disagree by more than 1%, check the apparatus (verticality, clean tube, no trapped bubbles).

Interactive: find the resonance yourself

The figure below is a simplified idealised version of the experiment. Drag the slider to change the air column length L; the plot shows the amplitude of the air-column oscillation at the fork's frequency, as a function of L. Sharp peaks at L_1 = \lambda/4 - e and L_2 = 3\lambda/4 - e mark the two resonance positions. The tuning fork is set to 512 Hz and the simulation uses v = 348 m/s (a room at about 30 °C); the inner tube radius is R = 1.25 cm, giving an end correction e = 0.75 cm.

Interactive resonance-tube response curve A plot of the amplitude of air-column oscillation as a function of tube length L. Two sharp peaks are shown: a tall peak near L equals 16 cm (the first resonance) and another near L equals 50 cm (the second resonance). A draggable red dot on the horizontal axis lets the reader select a length and read off the corresponding amplitude; at the peaks, the tube is at maximum resonance. air column length L (cm) amplitude (arb.) 0 20 40 60 80 100 L₁ ≈ 16.3 cm L₂ ≈ 50.2 cm drag the red dot
Drag the red dot along the axis. The amplitude curve shows the air column's response to a 512 Hz fork as you vary $L$. Two sharp peaks — at $L_1 \approx 16.3$ cm and $L_2 \approx 50.2$ cm — mark the first and second resonances. Their difference $L_2 - L_1 \approx 33.98$ cm is exactly $\lambda/2$; the readout shows $v = 2f(L_2 - L_1) \approx 348$ m/s, in line with air at 30 °C.

Notice how sharp the peaks are. A mistuning of just 1 cm from L_1 already drops the amplitude to about a third of its peak value — this is why the resonance is easy to locate by ear, and why careful measurement can pin L_1 to about \pm 2 mm on a good day.

Worked examples

Example 1: A 512 Hz fork in a Delhi lab at 30 °C

A CBSE class-11 student in Delhi performs the experiment in May, when the laboratory is at 30 °C. Using a 512 Hz fork, the mean resonance lengths over five readings are \bar{L}_1 = 16.2 cm and \bar{L}_2 = 50.1 cm. Find (a) the speed of sound in the laboratory's air, (b) the end correction of the tube, and (c) the theoretical speed of sound at 30 °C, and compare.

Resonance-tube readings at 30 °CHorizontal number line showing the air column length in cm, with two marked positions: L1 at 16.2 cm and L2 at 50.1 cm. The distance between them is labelled L2 minus L1 equals 33.9 cm, and below this is written lambda over 2 equals 33.9 cm so lambda equals 67.8 cm.020406080100 cmL₁ = 16.2L₂ = 50.1L₂ − L₁ = 33.9 cm = λ/2
The two resonance positions on the tube scale. The distance between them is half a wavelength.

Step 1. Convert lengths to metres. \bar{L}_1 = 0.162 m, \bar{L}_2 = 0.501 m.

Why: SI units throughout keeps the formula v = 2f(L_2 - L_1) in m/s without any hidden centimetre factors.

Step 2. Compute the wavelength.

\lambda \;=\; 2(\bar{L}_2 - \bar{L}_1) \;=\; 2(0.501 - 0.162) \;=\; 2 \times 0.339 \;=\; 0.678\ \text{m}.

Why: the two consecutive resonances of a closed pipe are spaced by \lambda/2. Doubling their difference gives the wavelength.

Step 3. Compute the speed of sound.

v \;=\; f\lambda \;=\; 512 \times 0.678 \;=\; 347.1\ \text{m/s}.

Why: v = f\lambda is the universal relation between speed, frequency, and wavelength of a travelling wave. Here f is fixed by the fork; \lambda was measured via resonance.

Step 4. Compute the end correction from equation (1): L_1 + e = \lambda/4, so e = \lambda/4 - L_1 = 0.678/4 - 0.162 = 0.1695 - 0.162 = 0.0075 m, i.e. e \approx 0.75 cm.

Why: once \lambda is known, equation (1) is a direct read-off of e. Compare with the predicted 0.6\,R — if the tube's inner radius is 1.2 cm, the prediction is e \approx 0.72 cm, very close.

Step 5. Theoretical prediction at 30 °C. Using v = \sqrt{\gamma R T / M} with \gamma = 1.40 (diatomic air), R = 8.314 J/(mol·K), T = 303.15 K, M = 0.02897 kg/mol (dry air):

v_{\text{th}} \;=\; \sqrt{\frac{1.40 \times 8.314 \times 303.15}{0.02897}} \;=\; \sqrt{1.217 \times 10^5} \;\approx\; 349.0\ \text{m/s}.

Why: kinetic theory combined with the adiabatic process gives the closed-form expression. At 30 °C, this evaluates to just under 349 m/s for dry air.

Step 6. Compare. The experiment returned 347.1 m/s; theory gives 349.0 m/s. The measured value is lower by 1.9 m/s, about 0.5\% — well within the experiment's expected precision. Possible explanations for the small shortfall: slight humidity correction (ignored here), a fork whose stamped 512 Hz is really 510–511 Hz after years of use, or a small error in the mean column lengths.

Result: v_{\text{exp}} = 347.1 m/s, v_{\text{theory}} = 349.0 m/s, agreement within 0.5%. End correction e \approx 0.75 cm.

What this shows: A tuning fork, a tube of water, and a metre scale measure the speed of sound at your local temperature to half a per cent of the true value — and the end-correction correction cancels cleanly out of the subtraction. That is why the two-resonance method has been the CBSE practical-exam standard for two generations.

Example 2: The same lab in December — temperature dependence

The same student repeats the experiment in December, when the Delhi laboratory cools to 15 °C. Using the same 512 Hz fork and the same tube, she obtains \bar{L}_1 = 15.8 cm and \bar{L}_2 = 48.9 cm. Does the speed of sound change as predicted by theory?

Comparison of resonance lengths at 30 °C and 15 °CTwo horizontal bars one above the other, each representing an air column. The top bar (at 30 °C) shows L1 at 16.2 cm and L2 at 50.1 cm, with the spacing L2 minus L1 equal to 33.9 cm. The bottom bar (at 15 °C) shows L1 at 15.8 cm and L2 at 48.9 cm, with spacing 33.1 cm. The narrower spacing at the lower temperature is visually emphasised.30 °CL₂ − L₁ = 33.9 cm15 °CL₂ − L₁ = 33.1 cmlower T → smaller λ/2 → slower v
Summer readings above, December readings below. Lower temperature compresses the spacing $L_2 - L_1$, because sound travels slower in cold air.

Step 1. Measured speed in December:

v_{15} \;=\; 2 \times 512 \times (0.489 - 0.158) \;=\; 2 \times 512 \times 0.331 \;=\; 338.9\ \text{m/s}.

Why: the working formula (3) applies equally at any temperature. Only the measured L_2 - L_1 changes.

Step 2. Measured speed in May (from Example 1): v_{30} = 347.1 m/s.

Step 3. Theory predicts v \propto \sqrt{T} with T in kelvins. The ratio:

\frac{v_{30}}{v_{15}} \bigg|_{\text{theory}} \;=\; \sqrt{\frac{T_{30}}{T_{15}}} \;=\; \sqrt{\frac{303.15}{288.15}} \;=\; \sqrt{1.0520} \;=\; 1.0257.

Why: v = \sqrt{\gamma RT/M} has \gamma, R, and M all independent of T (within a couple of per cent — the ideal-gas approximation). Only the square root of temperature varies.

Step 4. Measured ratio:

\frac{v_{30}}{v_{15}} \bigg|_{\text{exp}} \;=\; \frac{347.1}{338.9} \;=\; 1.0242.

Why: a direct ratio of the two measured speeds, no theory inputs used.

Step 5. Compare: 1.0242 measured vs 1.0257 predicted. Agreement is within 0.15\% — well below the \sim 0.5\% precision floor of the experiment. The v \propto \sqrt{T} scaling is confirmed.

Step 6. Linearised form. For small temperature differences, \sqrt{T} \approx \sqrt{T_0}(1 + (T - T_0)/(2 T_0)), so v \approx v_0 + (v_0 / (2 T_0))(T - T_0) \approx 331 + 0.61\,\Delta T m/s at T_0 = 273 K, with \Delta T in °C. A rule of thumb worth remembering: the speed of sound in air rises by about 0.6 m/s for every 1 °C rise in temperature. Between 15 °C and 30 °C, that predicts \Delta v \approx 0.61 \times 15 \approx 9.1 m/s; the experiment gave 347.1 - 338.9 = 8.2 m/s — again within 1 m/s.

Result: Summer: v = 347 m/s. December: v = 339 m/s. The \sqrt{T} scaling holds.

What this shows: The resonance tube is not only a speed-of-sound measurement; by repeating it at different temperatures, it tests the kinetic-theory prediction v \propto \sqrt{T} directly. The same apparatus, used twice, checks a core result of thermodynamics.

Sources of error — and how to beat them

Honest error analysis is part of the experiment. Here are the dominant error sources in a CBSE/ISC setup, with their typical magnitudes and the standard fixes.

Adding in quadrature: a careful experiment with a calibrated fork and no humidity correction reaches about \pm 0.6\% precision — better than \pm 2 m/s out of \sim 345 m/s. That is impressive for a sub-₹500 apparatus.

Common confusions

If you came here to do the experiment, read the formula, and turn in your practical file, you already have what you need. What follows is for readers who want the formal derivation of the end correction, the alternative graphical methods used in competitive practicals, and the correction for humidity.

Rayleigh's calculation of the end correction

Why is e \approx 0.6\,R? Physically, just outside the tube's mouth, the oscillating air couples to a small "radiating" region. Energy leaks outward (this is what lets you hear the tube from across the room). But most of the oscillation is still confined to a plug of air just above the mouth, moving in sympathy with the column below. This plug adds to the inertia of the oscillating column.

For an unflanged (cylindrical, no outside baffle) pipe of inner radius R, Rayleigh computed the load by solving for the velocity potential outside the mouth using spherical harmonic expansions. The result is

e_{\text{unflanged}} \;\approx\; 0.6133\,R.

For a flanged pipe (one where the tube's mouth sits flush in a large flat plate, like an organ flue), the plug is bigger because the air outside is forced to move in a hemisphere instead of a full sphere, and

e_{\text{flanged}} \;\approx\; 0.8216\,R.

A standard school resonance tube is unflanged (a free cylindrical mouth), so e \approx 0.6\,R is the right prediction. For an inner radius of 1.25 cm, that gives e \approx 0.75 cm, which is what Example 1 measured experimentally. The agreement is usually good to 10%, with the residual disagreement coming from slight asymmetries in the mouth and from the finite wavelength (Rayleigh's calculation is a long-wavelength limit).

The graphical method

A neat generalisation of the two-resonance method, often asked in ISC and engineering practical exams, is the graphical method. Repeat the experiment at several frequencies f_1, f_2, \ldots, f_k, always measuring just the first resonance L_1 for each. From equation (1), L_1 + e = v/(4f), so

L_1 \;=\; \frac{v}{4f} - e \;=\; \frac{v}{4} \cdot \frac{1}{f} - e.

Plot L_1 (vertical) against 1/f (horizontal). The points should lie on a straight line of slope v/4 and intercept -e on the L_1 axis. Read both off the graph: multiply the slope by 4 to get v, negate the y-intercept to get e. This method automatically averages over random errors and gives e without the subtraction trick.

The line-of-best-fit analysis follows the same pattern as in the simple-pendulum experiment: compute residuals, take the slope from the centroid-slope formula or a least-squares fit, and estimate the uncertainty on v from the scatter. This graphical treatment is often the "viva" question in ISC/CBSE class-12 practicals, even though the experiment itself is introduced in class-11.

The humidity correction

Moist air has an effective molar mass lower than dry air's. Let x be the mole fraction of water vapour in air (at 100% relative humidity at 30 °C, x \approx 0.042). The effective molar mass is

M_{\text{eff}} \;=\; (1-x)\,M_{\text{dry}} + x\,M_{\text{H}_2\text{O}}.

With M_{\text{dry}} = 28.97 g/mol and M_{\text{H}_2\text{O}} = 18.02 g/mol, a 4% water content drops M_{\text{eff}} by about 1.5\%. Since v \propto 1/\sqrt{M}, this raises v by about 0.75\% — roughly 2.6 m/s at 30 °C. A Delhi monsoon lab reading will therefore sit slightly above the dry-air theoretical value, and matching measurement to theory requires this correction. In modern CBSE practicals, the humidity correction is typically acknowledged in the error analysis rather than applied, because a wet-and-dry-bulb hygrometer is rarely at hand.

There is also a tiny correction to \gamma: water vapour is a polyatomic molecule with \gamma_{\text{H}_2\text{O}} \approx 1.33 versus \gamma_{\text{air}} = 1.40. For a 4% water content, this lowers \gamma_{\text{eff}} by about 0.2\% and hence v by about 0.1\%. It partially cancels the molar-mass effect, but not fully.

Kundt's tube — a faster cousin

An entirely different apparatus, Kundt's tube, measures the speed of sound in a gas using a horizontal glass tube sprinkled with fine cork dust or lycopodium powder. A loudspeaker at one end drives the air; the powder piles up at the displacement nodes of the standing wave, which mark spacings of \lambda/2. Measuring the pile-to-pile distance with a metre scale gives \lambda directly; multiplying by the known driving frequency gives v. Kundt's tube is faster and visually striking — you see the standing wave, not just hear it — and is used in many Indian university labs for the speed of sound in hydrogen, CO₂, and other gases. The underlying physics is identical to the resonance tube's; the experimental details differ.

Where this leads next