Frequencies of standing waves: ν_n = n v / (2L) for n = 1, 2, 3, ... ν_1 is the fundamental (first harmonic); ν_2 = 2ν_1 is the second harmonic, etc.
-- NCERT Class 11 Physics, Ch. 14, p. 291Standing Waves Strings Pipes
8 MCQs4 revision cards9-step worked example
Source: NCERT Oscillations and WavesPYQ coverage: NEET 2023, 2025, 2026Official key: NTA-verifiedLast updated: 24 Sep 2026
Standing Waves Strings Pipes, explained for NEET
The high-frequency trap in standing waves on strings and pipes is confusing which systems allow all harmonics and which allow only odd ones. A closed-end pipe supports only odd harmonics — yet students routinely plug in even harmonic numbers and lose marks.
Standing waves on a string fixed at both ends. When a transverse wave reflects at both fixed ends, superposition produces standing waves. Both ends are displacement nodes. The allowed frequencies are f_n = nv/(2L), where n = 1, 2, 3, … and v is the wave speed on the string (NCERT Class 11 Physics Chapter 14, page 289). All harmonics — fundamental, second, third, and so on — are present.
Standing waves in an open-open pipe. Both ends are displacement antinodes (pressure nodes). The formula is identical to the string: f_n = nv/(2L), with v now being the speed of sound. All harmonics are present (NCERT Class 11 Physics Chapter 14, page 290).
Standing waves in a closed-end pipe. One end is a displacement node (closed) and the other is a displacement antinode (open). The boundary asymmetry forces only odd harmonics: f_n = (2n−1)v/(4L), giving frequencies f, 3f, 5f, … (NCERT Class 11 Physics Chapter 14, page 290). Even harmonics (2f, 4f, …) are physically impossible in this geometry.
Why the closed pipe is different. The closed end demands a node and the open end demands an antinode. Fitting a standing wave between a node and the nearest antinode requires an odd number of quarter-wavelengths. That constraint eliminates every even harmonic.
The trap on paper. When a question says "pipe closed at one end," students who default to f_n = nv/(2L) — the string/open-pipe formula — will include forbidden even harmonics and pick a wrong option. Read the boundary condition first, then choose the formula.
Can you answer these Standing Waves Strings Pipes MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
A string fixed at both ends has a fundamental frequency of 200 Hz. What is the frequency of its third harmonic?
Show answer and why every option is right or wrong
Answer: B. For a string fixed at both ends, f_n = nf₁. Third harmonic: f₃ = 3 × 200 = 600 Hz (NCERT Class 11 Physics Chapter 14, page 289).
Why A is wrong: A. 300 Hz assumes the second harmonic (n = 1.5 is not valid) or confuses with 3/2 scaling — the harmonic number multiplies the fundamental directly.
Why C is wrong: C. 400 Hz likely doubles the fundamental, which gives the second harmonic (n = 2), not the third.
Why D is wrong: D. 800 Hz would be the fourth harmonic (n = 4). The third harmonic is n = 3, giving 600 Hz.
MCQ 2Easy RecallPractice
Which harmonics are present in a pipe closed at one end?
Show answer and why every option is right or wrong
Answer: D. A closed-end pipe has a displacement node at the closed end and antinode at the open end, forcing f_n = (2n−1)v/(4L) — only odd harmonics (NCERT Class 11 Physics Chapter 14, page 290).
Why A is wrong: A. All harmonics are present in an open-open pipe or a string fixed at both ends, not in a closed-end pipe (trap: confusing open-pipe formula with closed-pipe formula).
Why B is wrong: B. Even harmonics cannot exist in a closed pipe because fitting an even number of quarter-wavelengths between a node and an antinode is geometrically impossible.
Why C is wrong: C. The closed pipe supports infinitely many odd harmonics (f, 3f, 5f, …), not just the fundamental.
MCQ 3Direct ApplicationPractice
A pipe open at both ends has a fundamental frequency f. If one end is now closed (pipe length unchanged), the new fundamental frequency is:
Show answer and why every option is right or wrong
Answer: B. Open-open fundamental: f = v/(2L). Closed-end fundamental: f₁ = v/(4L) = f/2. Closing one end halves the fundamental frequency (NCERT Class 11 Physics Chapter 14, page 290).
Why A is wrong: A. Keeping the frequency unchanged would require the same formula v/(2L), but a closed pipe uses v/(4L) — the fundamental drops (trap: using the open-pipe formula for a closed pipe).
Why C is wrong: C. Doubling the frequency would require shortening the pipe or increasing v. Closing one end actually lowers the fundamental because the effective wavelength doubles.
Why D is wrong: D. f/4 has no physical basis from either formula. The closed-end fundamental is v/(4L) = f/2, not f/4.
MCQ 4Direct ApplicationPractice
The second overtone of an open-open pipe has a frequency of 900 Hz. What is the fundamental frequency of this pipe?
Show answer and why every option is right or wrong
Answer: C. For an open-open pipe, overtones follow: 1st overtone = 2f₁, 2nd overtone = 3f₁. So 3f₁ = 900 Hz, giving f₁ = 300 Hz (NCERT Class 11 Physics Chapter 14, page 290).
Why A is wrong: A. 150 Hz implies the 2nd overtone is 6f₁ (i.e. dividing 900 by 6), which confuses the overtone numbering — the 2nd overtone of an open pipe is the 3rd harmonic, not the 6th.
Why B is wrong: B. 450 Hz implies dividing by 2 (treating the 2nd overtone as 2f₁), but that is the 1st overtone. The 2nd overtone is the 3rd harmonic.
Why D is wrong: D. 225 Hz implies dividing by 4, which doesn't correspond to any harmonic relationship in an open-open pipe.
MCQ 5Direct ApplicationPractice
The second overtone of a pipe closed at one end has a frequency of 1250 Hz. What is the fundamental frequency?
Show answer and why every option is right or wrong
Answer: A. Closed pipe harmonics: f, 3f, 5f, … The 1st overtone is 3f₁ and the 2nd overtone is 5f₁. So 5f₁ = 1250 Hz, f₁ = 250 Hz (NCERT Class 11 Physics Chapter 14, page 290).
Why B is wrong: B. 500 Hz assumes the 2nd overtone is 2.5f₁, which would be the case if one naively divided by 2.5 — there is no such harmonic. In a closed pipe the 2nd overtone is 5f₁ (trap: using the open-pipe overtone numbering).
Why C is wrong: C. 625 Hz divides by 2, treating the 2nd overtone as 2f₁. That is the open-pipe 1st overtone formula, not the closed-pipe 2nd overtone.
Why D is wrong: D. 416.7 Hz divides by 3, treating the 2nd overtone as 3f₁. That would be the 1st overtone of a closed pipe, not the 2nd.
MCQ 6Easy RecallPractice
A string of length 1.0 m is fixed at both ends and vibrates in its fundamental mode. The positions of nodes are at:
Show answer and why every option is right or wrong
Answer: C. In the fundamental mode (n = 1), the string has nodes only at the two fixed ends: x = 0 and x = L = 1.0 m, with one antinode at the centre (NCERT Class 11 Physics Chapter 14, page 289).
Why A is wrong: A. 0.5 m is the position of the antinode in the fundamental, not a node. Fixed ends are always nodes.
Why B is wrong: B. Three nodes (at 0, 0.5, 1.0 m) describes the second harmonic (n = 2), which has one additional node at the midpoint. The fundamental has no interior nodes.
Why D is wrong: D. Nodes at 0.25 m and 0.75 m (without the ends) corresponds to the third harmonic's interior nodes. The fundamental has zero interior nodes.
MCQ 7CalculationPractice
An open-open pipe of length L and a closed-end pipe of length L have the same speed of sound. The ratio of the fundamental frequency of the open pipe to that of the closed pipe is:
Show answer and why every option is right or wrong
Answer: D. Open pipe: f_open = v/(2L). Closed pipe: f_closed = v/(4L). Ratio = [v/(2L)] / [v/(4L)] = 2. So f_open : f_closed = 2 : 1 (NCERT Class 11 Physics Chapter 14, pages 289–290).
Why A is wrong: A. 1:1 would mean both pipes have the same fundamental, but the denominators differ (2L vs 4L), so the open pipe's fundamental is twice the closed pipe's (trap: applying the same formula to both pipes).
Why B is wrong: B. 4:1 would require dividing the closed-pipe frequency by an extra factor of 2, which has no physical basis. The ratio of 2L to 4L is 1:2, giving a frequency ratio of 2:1.
Why C is wrong: C. 1:2 inverts the ratio. The open pipe has the higher fundamental (v/(2L) > v/(4L)).
MCQ 8CalculationPractice
A closed-end pipe resonates at frequencies 850 Hz and 1190 Hz with no resonant frequency between them. What is the fundamental frequency of the pipe?
Show answer and why every option is right or wrong
Answer: A. Consecutive resonant frequencies of a closed pipe are successive odd harmonics: (2n−1)f₁ and (2n+1)f₁. Their difference = 2f₁. So 2f₁ = 1190 − 850 = 340 Hz, giving f₁ = 170 Hz. Verify: 850/170 = 5 (5th harmonic) and 1190/170 = 7 (7th harmonic) — both odd, as required (NCERT Class 11 Physics Chapter 14, page 290).
Why B is wrong: B. 340 Hz is 2f₁ (the difference between consecutive odd harmonics), not f₁ itself. Dividing by 2 is still needed.
Why C is wrong: C. 425 Hz is half of 850 Hz — this treats 850 Hz as the 2nd harmonic (n = 2), which assumes all harmonics are present. A closed pipe has only odd harmonics (trap: using open-pipe harmonic numbering).
Why D is wrong: D. 119 Hz divides 1190 by 10, which doesn't correspond to any harmonic relationship. The method requires finding the difference between consecutive resonant frequencies.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Standing Waves Strings Pipes: quick recall before you leave
How do you solve a Standing Waves Strings Pipes question? A worked example
Pattern: Open-open vs closed-end pipe harmonic comparison (based on NEET pattern: compare frequency ratios across pipe types).
- 1
Given
An organ pipe open at both ends has a fundamental frequency of 480 Hz. The speed of sound is 340 m/s.
- 2
Required
(a) Find the length of the pipe.
(b) If one end is now closed, find the fundamental frequency and the frequency of the 2nd overtone. - 3
Concept
Open-open pipe: all harmonics, f_n = nv/(2L). Closed-end pipe: only odd harmonics, f_n = (2n−1)v/(4L). Closing one end changes the boundary condition from antinode-antinode to node-antinode.
- 4
Formula
Open pipe fundamental: f₁ = v/(2L) → L = v/(2f₁)
Closed pipe fundamental: f₁' = v/(4L)
Closed pipe 2nd overtone = 5th harmonic: f = 5v/(4L) - 5
Substitution
L = 340/(2 × 480) = 340/960
f₁' = 340/(4 × 0.354)
2nd overtone = 5 × 340/(4 × 0.354) - 6
Calculation
L = 340/960 ≈ 0.354 m
f₁' = 340/(4 × 0.354) = 340/1.417 ≈ 240 Hz
2nd overtone of closed pipe = 5 × 240 = 1200 Hz
Note on exact values: 480, 340, and the integer multipliers (2, 4, 5) are given or exact values — they do not limit significant figures. The division 340/960 is carried to three significant figures consistent with the input data. - 7
Final answer
(a) L ≈ 0.354 m (about 35.4 cm)
(b) Closed-pipe fundamental = 240 Hz; 2nd overtone = 1200 Hz
Cross-check: 240 Hz is exactly half of 480 Hz, consistent with closing one end halving the fundamental. The 2nd overtone (5th harmonic) of the closed pipe at 1200 Hz is an odd multiple of 240 Hz. Both checks confirm the result. - 8
Common trap
Using f_n = nv/(2L) for the closed pipe gives the 2nd overtone as 3 × 480 = 1440 Hz (the open pipe's 3rd harmonic), which is wrong. The closed pipe's 2nd overtone is the 5th harmonic at (2×3−1)v/(4L) = 5f₁', not 3f₁. Always check whether the pipe is open or closed before selecting the formula.
- 9
Similar NEET-style question
A pipe closed at one end has a fundamental frequency of 150 Hz. An open pipe of the same length is placed nearby. What is the fundamental frequency of the open pipe, and which is the lowest harmonic of the open pipe that matches a resonant frequency of the closed pipe?
Hint: The open pipe's fundamental is 2 × 150 = 300 Hz. Its harmonics are 300, 600, 900, 1200, … Hz. The closed pipe's harmonics are 150, 450, 750, 1050, … Hz. Find the first common frequency.
---
What to remember before solving Standing Waves Strings Pipes questions
Formula
Standing waves in pipes
Open at both ends: ν_n = n v/(2L), all harmonics present. Closed at one end: ν_n = (2n-1) v/(4L), only odd harmonics.
-- NCERT Class 11 Physics, Ch. 14, p. 292Which Standing Waves Strings Pipes formulas do you need for NEET?
3 formulas — click to collapse
Standing wave in closed-end pipe
Pipe closed at one end has only odd harmonics: f, 3f, 5f, ...
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_n | n-th harmonic | Hz |
| v | sound speed | m/s |
| L | pipe length | m |
Valid when
- Closed at one end (open at other)
- End correction neglected
Standing wave in open-open pipe
Pipe open at both ends has all harmonics. Same formula as string.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_n | n-th harmonic | Hz |
| v | sound speed | m/s |
| L | pipe length | m |
Valid when
- Open at both ends
- End correction neglected
Standing wave frequencies on fixed-fixed string
Allowed frequencies on string fixed at both ends. n=1 fundamental; harmonics 2f, 3f, ...
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_n | n-th harmonic | Hz |
| v | wave speed on string | m/s |
| L | string length | m |
| n | harmonic number | - |
Valid when
- String fixed at both ends
- Wave speed v as defined above
Where do students lose marks on Standing Waves Strings Pipes?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student includes even harmonics in a closed-end pipe. Closed pipe has only ODD harmonics (f, 3f, 5f, ...).
When it triggers
Question describes pipe closed at one end (e.g. resonance tube).
How to avoid
Open both ends: all harmonics, f_n = nv/(2L). Closed one end: odd only, f_n = (2n-1)v/(4L). Fundamental of closed pipe is HALF that of open pipe of same L.
Root cause: concept gap
Correction
Closed-end pipe has only ODD harmonics (f, 3f, 5f, ...). Open-both-ends pipe has all (f, 2f, 3f, ...). Reason: closed end has displacement node and pressure antinode.
More in Oscillations and Waves: 4 exam traps and mistakes · 7 formulas · 5 question patterns from its other lessons.
Standing Waves Strings Pipes questions from past NEET papers
3 questions from NEET 2023, 2025, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
15/9
29/5
31
43/5
NTA Answer: Option 3(final)
NEET 2025
12f
2f/2
3f
43f/2
NTA Answer: Option 3(final)
NEET 2023
12 : 1
21 : 3
33 : 1
41 : 2
NTA Answer: Option 1(final)
How does NEET ask about Standing Waves Strings Pipes?
1 recurring pattern from past papers — click to collapse
Open-open pipe (all harmonics) vs closed-end pipe (odd only). Compare frequency ratios.
InterpretationMedium
Common distractors
treats closed pipe like open
Includes even harmonics in closed pipe
Sources
NCERT refs: Class 11 Physics Chapter 14, p.289 | Class 11 Physics Chapter 14, p.290
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →