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. 291Fundamental Mode Harmonics
8 MCQs4 revision cards9-step worked example
Source: NCERT Oscillations and WavesOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Fundamental Mode Harmonics, explained for NEET
The trap that costs marks on harmonics questions: confusing which pipe supports which harmonics.
A string fixed at both ends or a pipe open at both ends supports all harmonics. The allowed frequencies are fₙ = nv/(2L), where n = 1, 2, 3, … The n = 1 mode is the fundamental (first harmonic). The n = 2 mode is the second harmonic (first overtone), and so on. Every integer multiple of the fundamental is present.
A pipe closed at one end behaves differently. The closed end forces a displacement node; the open end has an antinode. This boundary asymmetry permits only odd harmonics: fₙ = (2n−1)v/(4L), giving frequencies f, 3f, 5f, … No even harmonics exist. The fundamental frequency of a closed pipe is exactly half that of an open pipe of the same length.
For a string, the wave speed is v = √(T/μ), where T is tension and μ is linear mass density. So the fundamental of a string fixed at both ends is f₁ = (1/2L)√(T/μ). Changing tension or mass density shifts all harmonics proportionally.
The high-frequency trap (NCERT Class 11 Physics Chapter 14, pages 289–290): When a question describes a "resonance tube closed at one end" and asks for the next resonance after the fundamental, a common wrong answer picks 2f (the second harmonic). The correct answer is 3f — the next allowed mode in a closed pipe is the third harmonic. NEET has tested this distinction repeatedly (2023, 2025 papers).
Watch for the phrasing: "overtone" versus "harmonic." The first overtone of a closed pipe is the third harmonic (3f), not the second.
Can you answer these Fundamental Mode Harmonics MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
A string of length 1.0 m is fixed at both ends. If the wave speed on the string is 200 m/s, the frequency of the third harmonic is:
Show answer and why every option is right or wrong
Answer: A. For a string fixed at both ends, fₙ = nv/(2L). Third harmonic: f₃ = 3 × 200/(2 × 1.0) = 300 Hz (NCERT Class 11 Physics Chapter 14, page 289).
Why B is wrong: B (200 Hz) is the second harmonic f₂ = 2v/(2L) = 200 Hz, not the third.
Why C is wrong: C (100 Hz) is the fundamental f₁ = v/(2L) = 200/2 = 100 Hz, not the third harmonic.
Why D is wrong: D (400 Hz) would be the fourth harmonic. Possibly obtained by miscounting harmonics or confusing harmonic number with overtone number.
MCQ 2Direct ApplicationPractice
A pipe open at both ends has a fundamental frequency of 300 Hz. If one end is now closed, the new fundamental frequency is:
Show answer and why every option is right or wrong
Answer: B. Open pipe fundamental: f = v/(2L). Closed pipe fundamental: f' = v/(4L) = f/2 = 150 Hz. The closed-end boundary condition doubles the effective wavelength of the fundamental (NCERT Class 11 Physics Chapter 14, page 290).
Why A is wrong: A (600 Hz) doubles the frequency — this goes in the wrong direction. Closing one end lowers the fundamental, not raises it.
Why C is wrong: C (300 Hz) assumes closing an end doesn't change the fundamental. The boundary condition changes from antinode-antinode to node-antinode, halving the fundamental.
Why D is wrong: D (75 Hz) is f/4, which has no physical basis in the pipe formula. This likely results from applying the closed-pipe formula incorrectly as v/(8L).
MCQ 3Easy RecallPractice
Which of the following harmonics is NOT possible in a pipe closed at one end?
Show answer and why every option is right or wrong
Answer: C. A closed pipe supports only odd harmonics (1st, 3rd, 5th, 7th, …). The fourth harmonic is an even harmonic and cannot exist in a closed pipe (NCERT Class 11 Physics Chapter 14, page 290). Trap: confusing open-pipe and closed-pipe harmonic series.
Why A is wrong: A: The first harmonic (fundamental) exists in every resonating system — it is always the lowest allowed mode.
Why B is wrong: B: The third harmonic is the first overtone in a closed pipe and is certainly allowed (f₃ = 3f₁).
Why D is wrong: D: The fifth harmonic is an odd harmonic and is allowed in a closed pipe (f₅ = 5f₁).
MCQ 4CalculationPractice
The first overtone of a closed pipe has the same frequency as the third harmonic of an open pipe. If the length of the open pipe is 30 cm, the length of the closed pipe is:
Show answer and why every option is right or wrong
Answer: B. B is correct. The first overtone of a closed pipe is its third harmonic, 3v/(4L_c). The third harmonic of an open pipe is 3v/(2L_o). Setting them equal, the 3v cancels on both sides and leaves 4L_c = 2L_o, so L_c = L_o/2 = 30/2 = 15 cm. A closed pipe is always half the length of the open pipe that shares a harmonic with it in this way (NCERT Class 11 Physics Chapter 14, pages 289–290).
Why A is wrong: A (10 cm) results from taking the closed pipe's first overtone as its second harmonic, 2v/(4L_c), as for an open pipe. A closed pipe has only odd harmonics, so its first overtone is 3v/(4L_c).
Why C is wrong: C (25 cm) has no clean algebraic origin from these formulas — likely an arithmetic slip.
Why D is wrong: D (45 cm) results from matching the closed pipe's first overtone to the open pipe's fundamental, v/(2L_o), instead of its third harmonic: 3v/(4L_c) = v/(2L_o) gives L_c = 3L_o/2 = 45 cm.
MCQ 5Easy RecallPractice
A string fixed at both ends vibrates in its second harmonic. The number of nodes (including the endpoints) is:
Show answer and why every option is right or wrong
Answer: B. For the nth harmonic on a fixed-fixed string, the number of nodes = n + 1. Second harmonic: 2 + 1 = 3 nodes (two fixed endpoints plus one interior node). Reference: NCERT Class 11 Physics Chapter 14, page 289.
Why A is wrong: A (1 node) ignores the fixed endpoints, which are always nodes. Even the fundamental has 2 nodes.
Why C is wrong: C (2 nodes) counts only the endpoint nodes and misses the interior node that forms in the second harmonic.
Why D is wrong: D (4 nodes) would correspond to the third harmonic (n + 1 = 4). This likely comes from confusing harmonic number with node count.
MCQ 6Easy RecallPractice
The ratio of the fundamental frequency of a pipe open at both ends to that of a pipe closed at one end, both of the same length, is:
Show answer and why every option is right or wrong
Answer: A. Open pipe fundamental: f_open = v/(2L). Closed pipe fundamental: f_closed = v/(4L). Ratio: f_open/f_closed = [v/(2L)] / [v/(4L)] = 2. So 2 : 1 (NCERT Class 11 Physics Chapter 14, page 290).
Why B is wrong: B (1:2) inverts the ratio. The open pipe has the higher fundamental, not the closed pipe.
Why C is wrong: C (1:1) incorrectly assumes both pipes have the same fundamental. The different boundary conditions (antinode-antinode vs node-antinode) change the allowed wavelengths.
Why D is wrong: D (4:1) likely results from squaring the ratio or confusing frequency ratio with wavelength ratio.
MCQ 7CalculationPractice
A wire of length 1.0 m is stretched between two fixed supports with tension 100 N. If the mass per unit length of the wire is 0.01 kg/m, the fundamental frequency of vibration is:
Show answer and why every option is right or wrong
Answer: D. Wave speed v = √(T/μ) = √(100/0.01) = √10000 = 100 m/s. Fundamental: f₁ = v/(2L) = 100/(2 × 1.0) = 50 Hz (NCERT Class 11 Physics Chapter 14, pages 281, 12).
Why A is wrong: A (25 Hz) results from using f = v/(4L) — the closed-pipe formula — instead of f = v/(2L) for a string fixed at both ends.
Why B is wrong: B (200 Hz) results from taking the fundamental wavelength as L/2 instead of 2L: f = v/(L/2) = 200 Hz. In the fundamental the string is half a wavelength long, so λ = 2L.
Why C is wrong: C (100 Hz) uses f = v/L, forgetting the factor of 2 in the denominator. The fundamental wavelength is 2L, not L.
MCQ 8Direct ApplicationPractice
An organ pipe closed at one end resonates at its fundamental frequency of 250 Hz. Its next higher resonant frequency is:
Show answer and why every option is right or wrong
Answer: C. A closed pipe supports only odd harmonics. Fundamental = f = 250 Hz. The next allowed mode is the third harmonic = 3f = 750 Hz. There is no second harmonic in a closed pipe (NCERT Class 11 Physics Chapter 14, page 290). Trap: picking 500 Hz (2f) by assuming all harmonics are present.
Why A is wrong: A (375 Hz = 1.5f) is not a harmonic of any pipe — harmonics are always integer multiples of the fundamental.
Why B is wrong: B (500 Hz = 2f) is the second harmonic, which does NOT exist in a closed pipe. Only odd multiples are allowed. This is the high-frequency trap for this topic.
Why D is wrong: D (1000 Hz = 4f) is the fourth harmonic — an even harmonic, not allowed in a closed pipe. The next mode after 3f is 5f = 1250 Hz.
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Fundamental Mode Harmonics: quick recall before you leave
How do you solve a Fundamental Mode Harmonics question? A worked example
Pattern: Open-open vs closed-end pipe harmonic comparison (based on NEET pattern: pipe harmonics).
- 1
Given
• L = 0.50 m• v = 340 m/s
- 2
Required
(a) First three harmonics of open-open pipe.
(b) First three allowed modes of closed-end pipe. - 3
Concept
Open pipe: all harmonics, fₙ = nv/(2L). Closed pipe: odd harmonics only, fₙ = (2n−1)v/(4L) (NCERT Class 11 Physics Chapter 14, pages 289–290).
- 4
Formula
• Open: fₙ = nv/(2L)• Closed: fₙ = (2n−1)v/(4L)
- 5
Substitution
(a) Open pipe:• f₁ = 1 × 340 / (2 × 0.50) = 340/1.0• f₂ = 2 × 340 / (2 × 0.50) = 680/1.0• f₃ = 3 × 340 / (2 × 0.50) = 1020/1.0
(b) Closed pipe:• n = 1: f = (2×1 − 1) × 340 / (4 × 0.50) = 1 × 340/2.0• n = 2: f = (2×2 − 1) × 340 / (4 × 0.50) = 3 × 340/2.0• n = 3: f = (2×3 − 1) × 340 / (4 × 0.50) = 5 × 340/2.0 - 6
Calculation
(a) Open pipe: f₁ = 340 Hz, f₂ = 680 Hz, f₃ = 1020 Hz (ratio 1 : 2 : 3).
(b) Closed pipe: f₁ = 170 Hz, f₃ = 510 Hz, f₅ = 850 Hz (ratio 1 : 3 : 5).
Note: The integers 1, 2, 3, 5 are exact counting numbers and do not limit significant figures. The given values L = 0.50 m (2 sig figs) and v = 340 m/s (3 sig figs as conventionally treated) govern precision. Results are reported to 2–3 significant figures accordingly. - 7
Final answer
| Mode | Open pipe | Closed pipe |
|------|-----------|-------------|
| Fundamental | 340 Hz | 170 Hz |
| Next allowed | 680 Hz (2nd harmonic) | 510 Hz (3rd harmonic) |
| Third allowed | 1020 Hz (3rd harmonic) | 850 Hz (5th harmonic) |
Key observation: The closed pipe's fundamental is exactly half the open pipe's. The closed pipe skips all even harmonics. - 8
Common trap
For part (b), a common error is writing the second mode of the closed pipe as 340 Hz (2 × 170), treating it like an open pipe. The second harmonic does not exist in a closed pipe — the next mode jumps to the third harmonic (3 × 170 = 510 Hz).
- 9
Similar NEET-style question
A resonance tube closed at one end has a fundamental frequency of 200 Hz. What are the frequencies of the next two resonances? *(Answer: 600 Hz and 1000 Hz — odd multiples 3f and 5f.)*
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What to remember before solving Fundamental Mode Harmonics 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 Fundamental Mode Harmonics formulas do you need for NEET?
1 formula — click to collapse
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
More in Oscillations and Waves: 6 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
Fundamental Mode Harmonics questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Fundamental Mode Harmonics?
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
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.
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