# Resonance Tube Experiment > Resonance tube for NEET Physics: closed pipe harmonics, first and second resonance lengths, end correction and speed of sound measurement. - Canonical URL: https://prepelephant.com/topics/neet-ug/physics/resonance-tube-neet - Exam / course: NEET-UG · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Resonance Tube Experiment", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/resonance-tube-neet ## 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.