Resonance Tube Experiment
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Direct answer
A resonance tube is a vertical pipe partially filled with water, whose surface acts as a closed (reflecting) end; a tuning fork of frequency f held at the open end sounds markedly louder when the air column resonates. Because a closed pipe has a node at the water and an antinode at the open end, the first resonance occurs at column length l_1 ≈ λ/4 and the second at l_2 ≈ 3λ/4, so l_2 = 3l_1 ideally, and the end-corrected speed measurement gives v = 2f(l_2 − l_1) — end correction cancels in the difference, which is the experiment's elegance. The open end's effective length exceeds the physical length by e ≈ 0.6r (r = inner radius), and the fundamental of a closed pipe is f = v/4(l + e). Only odd harmonics (f, 3f, 5f...) exist in a closed pipe, against all harmonics in an open pipe.
What you must remember
- Boundary conditions: displacement node at the closed (water) end, antinode at the open end; resonance lengths l_1 = λ/4 − e and l_2 = 3λ/4 − e.
- The difference formula: v = 2f(l_2 − l_1), derived from l_2 − l_1 = λ/2; the end correction cancels, making this the measurement of choice.
- End correction: e ≈ 0.6r per open end; alternatively find e graphically by plotting l versus 1/f — the intercept on the l-axis equals −e.
- Harmonic content: closed pipe supports only odd multiples (f_1 : f_3 : f_5 = 1 : 3 : 5); open pipe supports all (f, 2f, 3f...) with f_1 = v/2L.
- Length ratio: l_2/l_1 = 3 ideally; measured ratios slightly below 3 reveal the end correction at work.
- Air temperature matters: v ≈ 331 + 0.6 T(°C) m/s, so v = 343 m/s at 20 °C — resonance-tube answers shift by about 0.6 m/s per degree, and NCERT uses the √T(kelvin) dependence.
- Loudness logic: at resonance the fork's energy transfers efficiently to the column (constructive reflection at the open end), the same phenomenon exploited in wind instruments like a flute's closed stops.
A worked measurement
A tuning fork of 512 Hz over a resonance tube gives the first resonance at 16.0 cm and the second at 50.4 cm. Find the speed of sound and the end correction. The difference: l_2 − l_1 = 34.4 cm = λ/2, so λ = 68.8 cm and v = fλ = 512 × 0.688 = 352 m/s — a plausible value for a warm laboratory (about 35 °C by v ≈ 331 + 0.6T). For the end correction: l_1 + e = λ/4 = 17.2 cm, so e = 17.2 − 16.0 = 1.2 cm; the tube's inner radius should then be about e/0.6 = 2.0 cm, a self-consistency check worth quoting in practical-based MCQs. Note what the arithmetic hid in plain sight: using the single-length formula v = 4f(l_1 + e) would have required knowing e beforehand, whereas the difference method delivered v without any knowledge of it — the entire reason laboratory manuals prescribe measuring both resonances.
How the exam frames it
NEET's resonance-tube questions come in three costumes. The first is direct computation: given f, l_1, l_2, find v (use the difference formula). The second is conceptual about harmonics: "a closed pipe of length L resonates at 3v/4L but never at v/2L" — testing that even multiples are missing, with an open pipe often offered for contrast. The third is the end-correction question itself: why does the measured l_1 fall short of λ/4, and why does the difference method survive the shortfall — because the same e afflicts both lengths and subtracts out. A recurring trap phrase is "the tube resonates with its open end at the water surface", inverting the geometry; remember the water is the closed end. And in instrument questions, a flute open at both ends sounds its fundamental at v/2L, an octave above a closed pipe of equal length — the number-pair 4L versus 2L settles such options instantly.
Frequently asked questions
Why does the resonance tube produce loud sound at specific lengths?
At those lengths the reflected wave returns in phase with the fork's, building a standing wave with an antinode at the open end, so energy transfer from fork to air column becomes efficient.
What is the relation between the first and second resonance lengths?
l_1 ≈ λ/4 and l_2 ≈ 3λ/4, so ideally l_2 = 3l_1 and l_2 − l_1 = λ/2, both measured from the open end to the water surface.
How is the speed of sound found without knowing the end correction?
Use v = 2f(l_2 − l_1); since the same end correction e shifts both lengths equally, it cancels in the difference.
Why does a closed pipe lack even harmonics?
Its boundary conditions (node at one end, antinode at the other) are satisfied only by odd quarter-wavelength multiples, so resonances occur at f, 3f, 5f and never at 2f, 4f.
What is end correction and how large is it?
The antinode sits slightly outside the open end, so the effective tube length is l + e with e ≈ 0.6r (r = tube radius); ignoring it underestimates v in single-length calculations.