Physics · Waves · NEET
No, and this is the number-one mistake in organ pipe PYQs. Overtone counting starts AFTER the fundamental. For an open pipe (all harmonics), 1st overtone = 2nd harmonic, 2nd overtone = 3rd harmonic, so nth overtone = (n+1)th harmonic. For a closed pipe (odd harmonics only), 1st overtone = 3rd harmonic, 2nd overtone = 5th harmonic, so nth overtone = (2n+1)th harmonic. Always convert to the harmonic number first before writing any frequency formula.
Closed pipes skip all even harmonics, so the allowed harmonics are 1, 3, 5, 7, 9, ... Just list them: fundamental = 1st harmonic, 1st overtone = 3rd, 2nd overtone = 5th, 3rd overtone = 7th, 4th overtone = 9th. The formula is: nth overtone = (2n+1)th harmonic. So the 4th overtone of a closed pipe = (2x4+1) = 9th harmonic, with frequency f = 9v/(4L).
Speed of sound v is the same air in both pipes, so it cancels. Write both frequencies fully, set them equal, then cross-cancel v. Example: closed 9v/(4Lc) = open 4v/(2Lo). Cancel v: 9/(4Lc) = 4/(2Lo) = 2/Lo. Cross multiply: 9 Lo = 8 Lc, so Lc/Lo = 9/8. Never plug in a number for v unless the question gives it (like speed of sound problems).
A closed pipe has a node (no motion) at the closed end and an antinode (max motion) at the open end. The distance node-to-antinode must be an odd number of quarter wavelengths (lambda/4, 3lambda/4, 5lambda/4...). This forces the harmonics to be odd multiples only: f, 3f, 5f, 7f. An open pipe has antinodes at both ends and allows all multiples: f, 2f, 3f, 4f.
Set the two frequencies equal and solve for the unknown length. If open pipe fundamental = closed pipe 3rd harmonic and closed length Lc = 20 cm: v/(2Lo) = 3v/(4Lc). Cancel v: 1/(2Lo) = 3/(4x20). So 2Lo = (4x20)/3 = 80/3, giving Lo = 40/3 = 13.3 cm. Keep the formula symbolic until the last step to avoid arithmetic errors.
The 4th overtone of a closed organ pipe has the same frequency as the 3rd overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
The fundamental frequency of an open organ pipe equals the third harmonic of a closed organ pipe. If the length of the closed pipe is 20 cm, the length of the open pipe is:
The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe of length L. The length of the open pipe is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Three steps: (1) Convert every overtone to a harmonic number (open: nth overtone = n+1; closed: nth overtone = 2n+1). (2) Write frequency as harmonic x fundamental: open uses v/(2L), closed uses v/(4L). (3) Equate frequencies, cancel v, and solve for the length or ratio.
Only if the question asks for a real frequency in Hz or gives you resonance lengths to find v. In pure ratio problems v cancels out because both pipes carry the same air, so you never need its value.
Open fundamental = v/(2L), closed fundamental = v/(4L). The ratio is 2 : 1 (the open pipe's fundamental is twice as high). This exact question appeared in NEET 2023 Phase 1.
Two successive resonance lengths always differ by lambda/2. So lambda = 2(L2 - L1), then use v = f x lambda. This links organ pipe theory to the resonance column experiment and is a very common NEET numerical.
For an open pipe the number of nodes equals the harmonic number. For a closed pipe, count only odd harmonics: 1st harmonic has 1 node, 3rd has 2, 5th has 3, and so on. ReNEET 2026 tested exactly this node-counting idea.
