ARCHETYPE: Resonance-Tube & Pipe Problems

The resonance tube is the experiment that lets a school student measure the speed of sound with a tuning fork, a tall tube, and a jug of water — and get it right to within a per cent. It is also, for exactly that reason, one of the most heavily examined setups on the Indian boards: a prescribed practical on CBSE and ICSE, a reliable numerical on JEE and NEET. And it hides a genuinely clever trick, the kind that separates a student who memorised a formula from one who understands what a measurement is. Here is the apparatus. A long glass tube stands upright, its lower end connected to a water reservoir so you can raise or lower the water level, and with it the length of the air column above. A tuning fork of known frequency is struck and held over the open top. As you lower the water, at a certain column length the tube suddenly roars — the air column has hit resonance. Lower it further and it roars again. Those two roaring lengths are your data. The naive move is to take the first resonance, say the air column is a quarter of a wavelength, and compute the speed. It comes out about five per cent low, every time — and the reason is the end correction: the pressure antinode does not sit exactly at the rim, it floats a little above it, so the true quarter-wavelength is slightly longer than the column you measured. The clever move — the whole point of this archetype — is to take two resonances and subtract. The end correction is the same unknown amount in both, so subtracting the lengths cancels it exactly, and out drops a clean $\lambda = 2(L_2 - L_1)$ with no fudge factor at all. This article makes that subtraction, and everything you can be asked to do with it, completely routine.

Key takeaways

The resonance tube is a closed pipe of adjustable length: a fork of known frequency $f$ resonates the air column at lengths $L_1 = \lambda/4 - e$ and $L_2 = 3\lambda/4 - e$ .
Subtract to kill the end correction: $\lambda = 2(L_2 - L_1)$ , so the speed of sound is $v = f\lambda = 2f(L_2 - L_1)$ — no fudge factor.
The end correction falls out too: $e = (L_2 - 3L_1)/2$ , and the tube radius follows as $r = e/0.6$ .
Using only the first resonance, $v = 4fL_1$ , gives a value about $5\%$ too low — the end-correction trap the exam sets every year.
A measured speed can be corrected to a reference temperature with $v \propto \sqrt{T}$ (T in kelvin).
Key idea: two resonance lengths, subtracted, give $\lambda = 2(L_2 - L_1)$ and $v = 2f(L_2 - L_1)$ — the subtraction cancels the end correction exactly, which is why the two-length method is the right way and a single length is not.

The core insight

The resonance tube is a closed air column whose first two resonances occur at $L_1 = \lambda/4 - e$ and $L_2 = 3\lambda/4 - e$ ; subtracting them cancels the unknown end correction $e$ to give $\lambda = 2(L_2 - L_1)$ and $v = 2f(L_2 - L_1)$ , while their combination $e = (L_2 - 3L_1)/2$ recovers the correction itself.

Where this lives today

The resonance tube is a teaching apparatus, but "find a resonance by sweeping a length" is a working measurement technique. An acoustic engineer tuning a loudspeaker's ported enclosure adjusts the port length until the box resonates at the target bass frequency — the same node-and-antinode physics with a fixed fork replaced by the driver. A musician filling a bottle to a chosen pitch, an organ builder voicing a stopped pipe, and a physicist measuring the speed of sound in an unknown gas by resonance all use the identical idea: a known frequency, an adjustable column, and the length at which it roars. In research, cavity resonance measured by sweeping either the frequency or the geometry is how the speed of sound in exotic gases, the elastic moduli of materials, and even the dimensions of microwave and optical cavities are pinned down. The jug of water and glass tube on a school bench are the friendliest version of a measurement that, scaled up and refined, calibrates instruments across physics.

The setup

The tube is a closed pipe with a movable bottom. The water surface is a rigid, reflecting boundary — the air cannot move there — so it is a displacement node. The open top is free, so it is a displacement antinode. Between a node and an antinode fits an odd number of quarter-wavelengths, exactly as for any closed pipe, which is why only certain column lengths resonate with a given fork. The first resonance (shortest roaring column) fits a single quarter-wavelength between the water and the top; the second fits three quarter-wavelengths. But there is the wrinkle that makes this an archetype rather than a plug-in: the antinode does not sit precisely at the rim. The vibrating air bulges a little way out of the open end, so the effective top of the air column is a small distance $e$ — the end correction — above the physical rim. Every measured length is therefore short of a true quarter-wavelength count by that same $e$ :

L_1 = \frac{\lambda}{4} - e, \qquad L_2 = \frac{3\lambda}{4} - e \qquad (1)

where: $L_1, L_2$ — the measured air-column lengths at the first and second resonances; $\lambda$ — the wavelength of the fork's sound in air; $e$ — the end correction, the same in both because the tube is the same. The minus signs say the true resonating length ( $\lambda/4$ or $3\lambda/4$ ) is a touch longer than what your metre scale reads, because the extra bit $e$ lives outside the tube.

Figure

A vertical resonance tube shown at its first and second resonance, water low for the longer column, with the displacement node at the water surface and the antinode floating a small distance e above the open rim

The two-length method

Now the trick. You do not know $e$ , and you do not need to. Subtract the two equations in (1): the $-e$ appears in both and cancels, leaving the wavelength in terms of quantities you measured:

L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{\lambda}{2} \quad\Rightarrow\quad \lambda = 2(L_2 - L_1) \qquad (2)

where: the gap between the two resonances is exactly half a wavelength, so twice that gap is the full wavelength. No end correction, no radius, no fudge — just the difference of two ruler readings. The speed of sound follows immediately from $v = f\lambda$ :

v = f\lambda = 2f(L_2 - L_1) \qquad (3)

This is the school measurement of the speed of sound, and it is honest precisely because the nuisance $e$ has been subtracted away. If, separately, the question wants the end correction itself, combine the two equations the other way. Multiply the first by three and subtract the second: $3L_1 - L_2 = 3(\lambda/4 - e) - (3\lambda/4 - e) = -2e$ , so

e = \frac{L_2 - 3L_1}{2}, \qquad r = \frac{e}{0.6} \qquad (4)

where: $r$ is the tube's internal radius, recovered from the textbook relation $e \approx 0.6r$ . So the same two lengths yield both the speed of sound (by subtracting) and the end correction with the tube radius (by this combination) — a favourite two-part exam question. Put the locked numbers through it. A $512\ \mathrm{Hz}$ fork gives a first resonance at $L_1 = 0.157\ \mathrm{m}$ and a second at $L_2 = 0.489\ \mathrm{m}$ . Wavelength: $\lambda = 2(0.489 - 0.157) = 2 \times 0.332 = 0.664\ \mathrm{m}$ . Speed: $v = 512 \times 0.664 = 340.0\ \mathrm{m\,s^{-1}}$ — the room-temperature value, recovered cleanly. End correction: $e = (0.489 - 3\times0.157)/2 = (0.489 - 0.471)/2 = 0.009\ \mathrm{m} = 0.9\ \mathrm{cm}$ . Radius: $r = 0.009/0.6 = 0.015\ \mathrm{m} = 1.5\ \mathrm{cm}$ . Four answers, two ruler readings.

The master recipe

Every resonance-tube question is these moves, in order:

Step 1. Read the two resonance lengths $L_1$ (first, shorter) and $L_2$ (second, longer). Step 2. Wavelength by subtraction: $\lambda = 2(L_2 - L_1)$ — never $4L_1$ . Step 3. Speed of sound: $v = f\lambda$ . Step 4. If asked, end correction $e = (L_2 - 3L_1)/2$ and radius $r = e/0.6$ . Step 5. If a reference temperature is given, correct the speed with $v \propto \sqrt{T}$ (kelvin).

Figure

The closed-pipe standing wave at the first resonance filling a quarter wavelength and at the second resonance filling three quarter wavelengths, showing why the two lengths differ by exactly half a wavelength

Derivation AThe two-length method — cancelling the end correction

The goal is to derive the working formulas $\lambda = 2(L_2 - L_1)$ , $v = 2f(L_2 - L_1)$ and $e = (L_2 - 3L_1)/2$ from the closed-pipe resonance condition, and to show why the subtraction is essential: it removes the one quantity — the end correction — you cannot measure directly. We run it on the locked $512\ \mathrm{Hz}$ data. The strategy is to write both resonance conditions with the same unknown $e$ , then combine them once to eliminate $e$ (for $v$ ) and once to isolate $e$ .

Step 1. State the two resonances. A closed air column resonates when it holds an odd number of quarter-wavelengths, corrected by $e$ : first resonance $L_1 = \lambda/4 - e$ , second resonance $L_2 = 3\lambda/4 - e$ . The end correction $e$ is the same in both because it is the same tube and the same fork.

Step 2. Eliminate $e$ by subtracting. $L_2 - L_1 = (3\lambda/4 - e) - (\lambda/4 - e) = \lambda/2$ . The end corrections cancel exactly, so $\lambda = 2(L_2 - L_1)$ — a wavelength built only from measured lengths. With $L_1 = 0.157$ , $L_2 = 0.489\ \mathrm{m}$ : $\lambda = 2(0.489 - 0.157) = 0.664\ \mathrm{m}$ .

Step 3. Get the speed. $v = f\lambda = 512 \times 0.664 = 340.0\ \mathrm{m\,s^{-1}}$ , the correct room-temperature value — because $e$ never entered.

Step 4. Isolate $e$ by the other combination: $3L_1 - L_2 = 3(\lambda/4 - e) - (3\lambda/4 - e) = -2e$ , hence $e = (L_2 - 3L_1)/2 = (0.489 - 0.471)/2 = 0.009\ \mathrm{m}$ , and the radius $r = e/0.6 = 0.015\ \mathrm{m}$ . The same two readings deliver the speed of sound and the tube's geometry. Verified: λ = 2(0.489 − 0.157) = 0.664 m, v = 512·0.664 = 340.0 m/s, e = (0.489 − 3·0.157)/2 = 0.009 m, r = 0.009/0.6 = 0.015 m — recomputed in verify_hs2-resonance-tube-problems.py.

Go deeper · Derivation B · 1 minCorrecting the speed of sound to a reference temperature

Derivation BCorrecting the speed of sound to a reference temperature

The goal here is the second common variant: a speed measured on a warm lab day must often be reported at a standard temperature (usually $0\ \mathrm{^\circ C}$ ), and the tool is $v \propto \sqrt{T}$ with $T$ in kelvin. We derive the scaling and apply it to bring a measured $340\ \mathrm{m\,s^{-1}}$ at $15\ \mathrm{^\circ C}$ down to the textbook $0\ \mathrm{^\circ C}$ figure. The strategy is to take the ratio of the speed at two temperatures so the unknown constants cancel, exactly as the two-length method cancelled $e$ .

Step 1. State the temperature law. The speed of sound in an ideal gas is $v = \sqrt{\gamma RT/M}$ , so at fixed composition $v \propto \sqrt{T}$ with $T$ in kelvin. (This is quoted from the speed-of-sound article; here we only use the proportionality.)

Step 2. Form the ratio between the two temperatures. $v_0/v_{15} = \sqrt{T_0/T_{15}}$ , where $T_0 = 273\ \mathrm{K}$ ( $0\ \mathrm{^\circ C}$ ) and $T_{15} = 288\ \mathrm{K}$ ( $15\ \mathrm{^\circ C}$ ). Every constant — $\gamma$ , $R$ , $M$ — cancels, leaving pure kelvin temperatures.

Step 3. Substitute. $v_0 = v_{15}\sqrt{T_0/T_{15}} = 340\sqrt{273/288} = 340 \times 0.9736 = 331.0\ \mathrm{m\,s^{-1}}$ — recovering the standard $0\ \mathrm{^\circ C}$ value of the speed of sound.

Step 4. Note the direction as a sanity check: cooler air is denser and carries sound more slowly, so the $0\ \mathrm{^\circ C}$ value must be lower than the $15\ \mathrm{^\circ C}$ value, and $331 < 340$ confirms it. Using Celsius instead of kelvin here would be a catastrophe — $\sqrt{0/15}$ is nonsense — which is the standard trap. Verified: v₀ = 340·√(273/288) = 340·0.9736 = 331.0 m/s, with 15 °C = 288 K and 0 °C = 273 K — recomputed in verify_hs2-resonance-tube-problems.py.

Limiting cases and traps in the geometry

situation	result	meaning
subtract the two lengths	$\lambda = 2(L_2 - L_1)$	the end correction cancels — the honest wavelength
use only the first length	$v = 4fL_1$ , about $5\%$ low	ignores $e$ : the classic trap
combine $3L_1 - L_2$	$e = (L_2 - 3L_1)/2$	the end correction, recovered
$e \approx 0.6r$	$r = e/0.6$	the tube radius from the end correction
warmer air	$v$ larger ( $\propto \sqrt{T}$ , kelvin)	cooler air is slower; correct to reference $T$
lower the water	air column lengthens	resonances appear at longer columns

Common trapNever find the speed from a single resonance length

The tempting one-step answer is "the first resonance is a quarter-wavelength, so $\lambda = 4L_1$ and $v = 4fL_1$ ." It is wrong, and always in the same direction — about $5\%$ too low — because the true quarter-wavelength is $L_1 + e$ , not $L_1$ ; you have thrown away the end correction, which is a real length sitting outside the tube. With the locked data, $v = 4 \times 512 \times 0.157 = 321.5\ \mathrm{m\,s^{-1}}$ against the correct $340\ \mathrm{m\,s^{-1}}$ . The only reliable route is two resonances subtracted, $v = 2f(L_2 - L_1)$ , because the subtraction cancels $e$ exactly. If a question gives you two lengths, it is signalling that it wants the subtraction; if it gives you one length and the end correction separately, add them ( $\lambda/4 = L_1 + e$ ) before finding $v$ . Reporting $4fL_1$ as the speed is the single most common mark-losing move in this experiment.

Worked examples

1 · Board level — speed of sound, end correction and radius (CBSE and ICSE practical)

A resonance-tube experiment with a $512\ \mathrm{Hz}$ fork gives the first resonance at an air-column length of $0.157\ \mathrm{m}$ and the second at $0.489\ \mathrm{m}$ . Find the wavelength, the speed of sound, the end correction, and the tube's radius.

Step 1. Wavelength by subtraction: $\lambda = 2(L_2 - L_1) = 2(0.489 - 0.157) = 2 \times 0.332 = \boxed{0.664\ \mathrm{m}}$ .

Step 2. Speed of sound: $v = f\lambda = 512 \times 0.664 = \boxed{340.0\ \mathrm{m\,s^{-1}}}$ — the room-temperature value, with no end-correction fudge.

Step 3. End correction: $e = (L_2 - 3L_1)/2 = (0.489 - 3\times0.157)/2 = (0.489 - 0.471)/2 = \boxed{0.009\ \mathrm{m}} = 0.9\ \mathrm{cm}$ .

Step 4. Radius: $r = e/0.6 = 0.009/0.6 = \boxed{0.015\ \mathrm{m}} = 1.5\ \mathrm{cm}$ . Sense-check: a tube of radius $1.5\ \mathrm{cm}$ (diameter $3\ \mathrm{cm}$ ) is a realistic resonance tube.

2 · Board level — speed from a different fork (JEE-Main and NEET pattern)

A $480\ \mathrm{Hz}$ fork resonates a tube first at $0.17\ \mathrm{m}$ and again at $0.52\ \mathrm{m}$ . Find the speed of sound. Then state, without further data, whether the room was warmer or cooler than a room giving exactly $340\ \mathrm{m\,s^{-1}}$ .

Step 1. Subtract for the wavelength: $\lambda = 2(0.52 - 0.17) = 2 \times 0.35 = 0.70\ \mathrm{m}$ .

Step 2. Speed: $v = f\lambda = 480 \times 0.70 = \boxed{336.0\ \mathrm{m\,s^{-1}}}$ . The two-length method again does all the work in one subtraction and one product.

Step 3. Compare to $340\ \mathrm{m\,s^{-1}}$ . The measured speed is lower, and since $v \propto \sqrt{T}$ , a lower speed means a lower temperature — this room was cooler than the one giving $340\ \mathrm{m\,s^{-1}}$ .

Step 4. Note the habit: the subtraction is identical whatever the fork — only $f$ and the two lengths change. Once you trust $v = 2f(L_2 - L_1)$ , every board version of this question is the same two operations.

3 · Competitive level — the end-correction trap (JEE-Advanced pattern)

A student, given only the first resonance at $0.157\ \mathrm{m}$ for the $512\ \mathrm{Hz}$ fork, computes the speed of sound as $v = 4fL_1$ . What value do they get, by how much is it wrong, and why?

Step 1. Compute the student's value: $v = 4fL_1 = 4 \times 512 \times 0.157 = \boxed{321.5\ \mathrm{m\,s^{-1}}}$ .

Step 2. Compare with the correct $340.0\ \mathrm{m\,s^{-1}}$ from the two-length method. The error is $(340.0 - 321.5)/340.0 = 0.054$ , about $5\%$ (precisely $5.4\%$ ) too low.

Step 3. Diagnose it. The single-length method assumes the first resonance is exactly a quarter-wavelength, $L_1 = \lambda/4$ . But truly $L_1 = \lambda/4 - e$ , so the real quarter-wavelength is $L_1 + e = 0.157 + 0.009 = 0.166\ \mathrm{m}$ , giving $\lambda = 0.664\ \mathrm{m}$ and $v = 340\ \mathrm{m\,s^{-1}}$ . Dropping $e$ shortened the wavelength and lowered the speed.

Step 4. The lesson, in one line: a single resonance always under-reports the speed of sound by the fractional amount $e/(L_1 + e)$ , here $0.009/0.166 = 5.4\%$ . Two resonances subtracted remove the error entirely. Timing note: under 2 minutes. The trap is doing exactly what the question describes and reporting $321.5\ \mathrm{m\,s^{-1}}$ as if it were correct. Whenever you see one resonance length and no end correction, expect the answer to be biased low, and reach for the two-length method or add $e$ first.

4 · Competitive level — correcting a measured speed to 0 degrees (NEET and JEE pattern)

A resonance-tube experiment done at $15\ \mathrm{^\circ C}$ yields a speed of sound of $340\ \mathrm{m\,s^{-1}}$ . Report the speed of sound at $0\ \mathrm{^\circ C}$ . Take $v \propto \sqrt{T}$ with $T$ in kelvin.

Step 1. Convert both temperatures to kelvin: $T_{15} = 15 + 273 = 288\ \mathrm{K}$ and $T_0 = 0 + 273 = 273\ \mathrm{K}$ . This conversion is the whole difficulty of the question.

Step 2. Form the ratio: $v_0 = v_{15}\sqrt{T_0/T_{15}} = 340\sqrt{273/288}$ .

Step 3. Evaluate: $\sqrt{273/288} = \sqrt{0.9479} = 0.9736$ , so $v_0 = 340 \times 0.9736 = \boxed{331.0\ \mathrm{m\,s^{-1}}}$ — exactly the standard textbook value of the speed of sound at $0\ \mathrm{^\circ C}$ .

Step 4. Sanity: colder air carries sound more slowly, so the $0\ \mathrm{^\circ C}$ value must be below the $15\ \mathrm{^\circ C}$ value, and $331 < 340$ ✓. Timing note: about 90 seconds. The only trap is using Celsius in the square root — $\sqrt{0/15}$ is meaningless; the law is $v \propto \sqrt{T}$ with $T$ in kelvin, always.

Exam traps

Never use one resonance for the speed. $v = 4fL_1$ ignores the end correction and comes out about $5\%$ low. Use $v = 2f(L_2 - L_1)$ ; the subtraction cancels $e$ . If forced to use one length, add the end correction first: the true quarter-wavelength is $L_1 + e$ .

It is a closed pipe — odd quarters only. The resonances are at $\lambda/4$ and $3\lambda/4$ , not $\lambda/4$ and $2\lambda/4$ . The gap between successive resonances is $\lambda/2$ , so $\lambda = 2(L_2 - L_1)$ . Writing $L_2 = \lambda/2$ (as if the second resonance were a half-wavelength) is the error to avoid.

Kelvin in $v \propto \sqrt{T}$ , never Celsius. A temperature correction uses absolute temperature. $15\ \mathrm{^\circ C}$ is $288\ \mathrm{K}$ ; using $15$ in the square root gives nonsense. This single slip is the most common way to lose the temperature-correction mark.

Lowering the water lengthens the column. As the water level drops, the air column above it grows, so resonances appear at longer columns. Students sometimes track the water level and the air column in opposite senses — be clear that $L$ is the air length, measured from the water surface up to the open top.

End correction is per open end, radius from $0.6r$ . The resonance tube has one open end, so $L_{\text{eff}} = L + e$ with a single $e$ , and the radius comes from $e = 0.6r$ , i.e. $r = e/0.6$ . Do not double it (that is the both-ends-open pipe) and do not confuse $e$ with the radius itself.

Board scope. The resonance tube is an India-facing prescribed practical: CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. It is not part of AP Physics 2, IB or A-Level, which teach open and closed pipes but not this specific two-length lab drill. If you sit one of those boards, learn the closed-pipe physics but do not spend time on end-correction numericals.

Origins · 1 minGeorg Duffing and how this was discovered

Georg Duffing

1861–1944 · Berlin

Georg Duffing was a German engineer who, in a slim 1918 monograph, studied what happens when you drive an oscillator with a periodic force and sweep the driving frequency — the birth of what is now called the Duffing equation and the resonance curve it produces. A resonance tube is exactly a driven acoustic resonator: the tuning fork is the periodic drive, and you sweep the resonator's own natural frequency by sliding the water level until the air column roars at the fork's pitch. That roar is resonance — the driven system matched to a natural mode — which is Duffing's whole subject. His oscillator was nonlinear (a spring that stiffens as it stretches, giving a bent, hysteretic resonance peak), while the air column is its clean, linear cousin with a sharp symmetric peak. Duffing did not study pipes; he mapped the mathematics of driven resonance that the resonance tube demonstrates in its simplest form.

History

1636

— Marin Mersenne relates the pitch of an organ pipe to its length, the first quantitative link between a resonating air column and frequency.

1701

— Joseph Sauveur coins the terms node and antinode and describes the resonances of pipes and columns.

1866

— August Kundt measures the speed of sound in a tube from the dust patterns of a standing wave, a close cousin of the resonance-tube method.

1877

— Lord Rayleigh's Theory of Sound quantifies the end correction, roughly 0.6 times the tube radius, that the two-length method is built to cancel.

1918

— Georg Duffing publishes his study of forced oscillation and resonance, the framework in which the tuned air column is a driven resonator.

Going further

Texts. NCERT Physics Class 11, Ch. 15 ("Waves"), the resonance-column method for the speed of sound · H. C. Verma, Concepts of Physics Vol. 1, "Sound Waves", resonance-tube problems · the CBSE and ICSE physics practical manuals (resonance-column experiment) · previous years' JEE-Main and NEET waves questions.

Provenance. Standard results. Recompute: $\lambda = 2(0.489 - 0.157) = 0.664\ \mathrm{m}$ , $v = 512\times0.664 = 340.0\ \mathrm{m\,s^{-1}}$ , $e = (0.489 - 3\times0.157)/2 = 0.009\ \mathrm{m}$ , $r = 0.009/0.6 = 0.015\ \mathrm{m}$ ; second fork $\lambda = 2(0.52 - 0.17) = 0.70\ \mathrm{m}$ , $v = 480\times0.70 = 336.0\ \mathrm{m\,s^{-1}}$ ; single-length trap $v = 4\times512\times0.157 = 321.5\ \mathrm{m\,s^{-1}}$ , error $(340-321.5)/340 = 5.4\%$ ; temperature correction $v_0 = 340\sqrt{273/288} = 331.0\ \mathrm{m\,s^{-1}}$ ; consistency $L_1 + e = 0.166\ \mathrm{m}$ , $\lambda = 0.664\ \mathrm{m}$ .

Prerequisites. Standing Waves in Air Columns: Open and Closed Pipes — the closed-pipe physics (node at the wall, antinode at the opening, odd harmonics) this archetype measures. Leads to. Speed of Sound: Newton–Laplace and Factors — where the measured $v$ and its temperature dependence come from · Resonance (university mechanics), the general theory of a driven resonator.

Go deeper (UG). The general theory of a driven, damped oscillator and its resonance curve — of which the tuned air column is the acoustic case — is developed in Resonance in the university mechanics track. Enrichment only.

Why is the resonance tube a closed pipe and not an open one?

Because one end is stopped by water. The water surface is a rigid, reflecting boundary that the air cannot move through, so it behaves exactly like the closed end of a pipe — a displacement node. The top is open to the room, a displacement antinode. A column that is a node at one end and an antinode at the other is precisely a closed pipe, so it resonates at odd quarter-wavelengths: $\lambda/4$ , $3\lambda/4$ , $5\lambda/4$ , and so on. That is why the first two resonances are a quarter and three-quarters of a wavelength, why the gap between them is $\lambda/2$ , and why the wavelength is $2(L_2 - L_1)$ . If you tried to treat it as an open pipe (both ends antinodes) you would get the resonance lengths wrong by a factor related to the harmonics it does not actually have.

What exactly is the end correction, and why does subtracting cancel it?

The vibrating air at the open top does not stop precisely at the rim — it bulges a little way out into the room before turning around, so the pressure antinode floats a small distance $e$ (about $0.6$ times the tube radius) above the physical opening. Every measured column length is therefore short of a true quarter-wavelength count by that same $e$ . The reason subtraction cancels it is that $e$ is identical at both resonances: same tube, same opening, same fork. So when you write $L_1 = \lambda/4 - e$ and $L_2 = 3\lambda/4 - e$ and subtract, the two $-e$ terms are equal and vanish, leaving $L_2 - L_1 = \lambda/2$ with no trace of the correction. It is the cleanest way to measure around an unknown you cannot get rid of physically.

Can I find the end correction as well, or only the speed of sound?

Both, from the same two readings. Subtracting the lengths gives the speed (the $e$ cancels); combining them the other way isolates $e$ . Specifically, three times the first length minus the second, $3L_1 - L_2$ , equals $-2e$ , so $e = (L_2 - 3L_1)/2$ . With the locked data that is $e = (0.489 - 0.471)/2 = 0.009\ \mathrm{m}$ , and since $e \approx 0.6r$ , the tube radius is $r = e/0.6 = 0.015\ \mathrm{m}$ . So a single experiment with two resonance lengths yields the wavelength, the speed of sound, the end correction, and the tube's radius — which is exactly why examiners like it as a multi-part question.

Why must temperature go in kelvin for the v ∝ √T correction?

Because the law comes from the ideal-gas result $v = \sqrt{\gamma RT/M}$ , in which $T$ is the absolute temperature — the one that is zero when molecular motion stops, at $-273\ \mathrm{^\circ C}$ . Only on the kelvin scale is temperature proportional to the average molecular kinetic energy that sets the speed of sound; the Celsius scale has its zero in the wrong place. If you put Celsius into $\sqrt{T}$ you get absurdities — the speed at $0\ \mathrm{^\circ C}$ would come out zero, and the speed at negative Celsius temperatures would be imaginary. Always convert: $15\ \mathrm{^\circ C}$ is $288\ \mathrm{K}$ , $0\ \mathrm{^\circ C}$ is $273\ \mathrm{K}$ , and only then take the ratio and the square root.

What if I get a third resonance? Does it help?

Yes — a third resonance at $L_3 = 5\lambda/4 - e$ gives you more data and a better result. The gap between any two successive resonances is $\lambda/2$ , so you can use $\lambda = 2(L_2 - L_1)$ , or $\lambda = 2(L_3 - L_2)$ , or best of all average them (or plot resonance length against the odd numbers $1, 3, 5$ and take the slope). More resonances reduce the effect of random reading errors and let you check consistency: successive gaps should all be equal to $\lambda/2$ . In a careful experiment you take three or four resonances and fit a straight line, which is more accurate than any single pair. The end correction still cancels, because it is the same constant subtracted from every length.

Is the resonance tube on my board's syllabus?

It is a staple of the Indian boards and their entrance exams: CBSE and ICSE list it as a prescribed practical, and JEE-Main, JEE-Advanced and NEET all set numericals on it. It is not part of AP Physics 2, IB (either level) or CIE A-Level, which cover open and closed pipes and their harmonics but do not use the specific two-length end-correction lab method. If you are on one of those boards, understand the closed-pipe physics behind it — node at the water, antinode at the top, resonances at $\lambda/4$ and $3\lambda/4$ — but you will not be asked to compute end corrections or crunch two-length data.