When two or more waves overlap, the resultant displacement is the algebraic sum of the individual displacements: y = y₁ + y₂ + ... Foundation of interference, beats, standing waves.
-- NCERT Class 11 Physics, Ch. 14, p. 287Superposition Reflection Waves
8 MCQs3 revision cards9-step worked example
Source: NCERT Oscillations and WavesOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Superposition Reflection Waves, explained for NEET
The principle of superposition says: when two or more waves overlap at a point, the net displacement equals the algebraic sum of individual displacements. That is it — no interaction, no modification. Each wave continues as though the other does not exist. This linearity holds for all mechanical waves at small amplitudes (NCERT Class 11 Physics, Chapter 14, page 285).
Where NEET tests this: standing waves on strings and in pipes. Standing waves ARE superposition — two identical waves travelling in opposite directions produce nodes (zero displacement always) and antinodes (maximum displacement). The superposition principle is the reason standing waves exist.
The high-frequency trap: open pipe vs closed pipe harmonics. A pipe open at both ends supports all harmonics: f₁, 2f₁, 3f₁, … with fₙ = nv/(2L). A pipe closed at one end supports only odd harmonics: f₁, 3f₁, 5f₁, … with fₙ = (2n−1)v/(4L). Students routinely include even harmonics in a closed pipe — this is a common confusion in NEET.
Why closed pipes lose even harmonics: the closed end forces a displacement node; the open end has a displacement antinode. This boundary condition permits only quarter-wavelength fitting (λ/4, 3λ/4, 5λ/4, …), which maps to odd multiples of the fundamental.
Reflection at boundaries. A wave reflecting off a rigid (fixed) boundary inverts — phase shift of π. Reflection off a free (open) boundary preserves phase — no inversion. On a string fixed at both ends, the reflected wave superimposes with the incident wave to produce standing waves with nodes at the fixed ends.
For strings fixed at both ends: fₙ = nv/(2L), same harmonic series as open-open pipe. The wave speed on a string is v = √(T/μ), where T is tension and μ is linear mass density.
Watch-out: when a problem says "first overtone," that is the second harmonic (n = 2) for strings and open pipes, but the third harmonic (n = 3, i.e. 3f₁) for a closed pipe. Mislabelling overtone-to-harmonic mapping costs marks.
Can you answer these Superposition Reflection Waves 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
According to the principle of superposition of waves, the resultant displacement at a point where two waves overlap is:
Show answer and why every option is right or wrong
Answer: B. The principle of superposition states that the net displacement at any point is the algebraic sum of the displacements due to each individual wave (NCERT Class 11 Physics, Chapter 14, page 285).
Why A is wrong: A is wrong because superposition is additive (sum), not multiplicative. Product of displacements has no physical basis in linear wave theory.
Why C is wrong: C is wrong because the resultant is the sum, not the mean. Averaging would halve the combined effect, which does not match the superposition principle.
Why D is wrong: D is wrong because superposition adds instantaneous displacements at a point, not amplitudes. Amplitude difference applies only to beats, and even then the resultant displacement is still a sum.
MCQ 2Easy RecallPractice
When a transverse wave pulse on a string reflects from a rigid (fixed) boundary, the reflected pulse is:
Show answer and why every option is right or wrong
Answer: D. At a fixed boundary, the displacement must remain zero. The reflected pulse is inverted (phase reversal of π) to satisfy this boundary condition (NCERT Class 11 Physics, Chapter 14, page 285).
Why A is wrong: A is wrong because reflection at a fixed end inverts the pulse. No phase change occurs only at a free (open) boundary.
Why B is wrong: B is wrong because a rigid boundary reflects the wave; full transmission with zero reflection would require perfectly matched impedance, which a fixed end does not provide.
Why C is wrong: C is wrong because reflection does not double amplitude. Doubling occurs momentarily at a free end due to constructive overlap, not at a fixed boundary.
MCQ 3Easy RecallPractice
A pipe closed at one end can produce:
Show answer and why every option is right or wrong
Answer: A. A closed-end pipe has a displacement node at the closed end and an antinode at the open end. This boundary condition allows only odd-harmonic standing waves: f, 3f, 5f, … (NCERT Class 11 Physics, Chapter 14, page 290).
Why B is wrong: B is wrong because even harmonics are specifically the ones excluded in a closed pipe. Only odd harmonics satisfy the node-at-closed-end, antinode-at-open-end requirement.
Why C is wrong: C is wrong because all harmonics occur in open-open pipes and strings fixed at both ends, not in a closed pipe. The closed-end boundary condition eliminates even harmonics.
Why D is wrong: D is wrong because a closed pipe does produce overtones — the 3rd, 5th, 7th harmonics are all allowed. It is not limited to the fundamental alone.
MCQ 4Easy RecallPractice
A transverse wave travelling along a string is reflected at a rigidly fixed end. The reflected wave undergoes a phase change of:
Show answer and why every option is right or wrong
Answer: B. B is correct. At a fixed end the displacement must stay zero at all times. The reflected wave can guarantee that only if it arrives inverted, so it cancels the incident wave there: a phase change of π, which is why a pulse on a string comes back upside down from a wall.
Why A is wrong: A is wrong because zero phase change is what happens at a FREE end, where the string can move and the pulse returns the same way up.
Why C is wrong: C is wrong because π/2 is not the reflection phase at either kind of end. The boundary condition at a fixed end demands complete cancellation there, which needs a full inversion.
Why D is wrong: D is wrong because 2π is the same as no phase change at all. It would leave the reflected wave upright, which cannot cancel the incident wave at the fixed point.
MCQ 5Direct ApplicationPractice
The third harmonic frequency of a closed pipe of length L is 900 Hz. What is its fundamental frequency?
Show answer and why every option is right or wrong
Answer: C. In a closed pipe, only odd harmonics exist. The third harmonic is 3f₁. So f₁ = 900/3 = 300 Hz (NCERT Class 11 Physics, Chapter 14, page 290).
Why A is wrong: A is wrong because 100 Hz would result from dividing by 9, treating the 'third harmonic' as the 9th multiple. In a closed pipe the third harmonic IS 3 times the fundamental, not 9 times.
Why B is wrong: B is wrong because 450 Hz comes from dividing by 2, as if the pipe had all harmonics (open pipe logic). In a closed pipe the harmonic numbers are 1, 3, 5, …, so the third harmonic is 3f₁, not 2f₁.
Why D is wrong: D is wrong because 900 Hz is the third harmonic itself, not the fundamental. The fundamental must be lower: f₁ = 900/3 = 300 Hz.
MCQ 6Direct ApplicationPractice
Two waves of the same frequency, each of amplitude A, travel in the same direction along a string with a phase difference of 60°. The amplitude of the resultant wave is:
Show answer and why every option is right or wrong
Answer: C. C is correct. For two equal waves the resultant amplitude is 2A cos(φ/2). With φ = 60°, that is 2A cos 30° = 2A × (√3/2) = √3 A. It falls between 2A (in phase) and 0 (out of phase).
Why A is wrong: A is wrong because A is 2A cos φ, using the full phase difference instead of half of it: 2A cos 60° = A. The formula halves the phase difference, 2A cos(φ/2).
Why B is wrong: B is wrong because 2A is the resultant only when the waves are exactly in phase, φ = 0. A 60° difference reduces it.
Why D is wrong: D is wrong because √2 A is the resultant for a 90° phase difference, 2A cos 45°. Here the waves are closer together in phase, so the resultant is larger.
MCQ 7Concept TrapPractice
In a closed organ pipe, the first overtone corresponds to which harmonic?
Show answer and why every option is right or wrong
Answer: A. A closed pipe supports only odd harmonics: 1st, 3rd, 5th, … The fundamental (1st harmonic) is the zeroth overtone. The first overtone is the next allowed harmonic, which is the 3rd (NCERT Class 11 Physics, Chapter 14, page 290). This is the overtone-to-harmonic mapping trap.
Why B is wrong: B is wrong because the second harmonic does not exist in a closed pipe — even harmonics are absent. The first overtone must be an odd harmonic.
Why C is wrong: C is wrong because the fourth harmonic is even and therefore absent in a closed pipe. The allowed sequence is 1st, 3rd, 5th harmonics.
Why D is wrong: D is wrong because the fifth harmonic is the second overtone in a closed pipe, not the first. First overtone = 3rd harmonic; second overtone = 5th harmonic.
MCQ 8CalculationPractice
An organ pipe closed at one end has a length of 0.50 m. The speed of sound in air is 340 m/s. What is the frequency of the first overtone of this pipe?
Show answer and why every option is right or wrong
Answer: D. Fundamental of closed pipe: f₁ = v/(4L) = 340/(4 × 0.50) = 170 Hz. First overtone in a closed pipe = 3rd harmonic = 3 × 170 = 510 Hz (NCERT Class 11 Physics, Chapter 14, page 290).
Why A is wrong: A is wrong because 170 Hz is the fundamental frequency. The problem asks for the first overtone, which is 3f₁ = 510 Hz in a closed pipe.
Why B is wrong: B is wrong because 340 Hz equals 2f₁, the second harmonic. But even harmonics do not exist in a closed pipe. The first overtone is 3f₁ = 510 Hz.
Why C is wrong: C is wrong because 680 Hz equals 4f₁, the fourth harmonic. Even harmonics are absent in a closed pipe. The next allowed frequency after the fundamental is 3f₁ = 510 Hz.
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Superposition Reflection Waves: quick recall before you leave
How do you solve a Superposition Reflection Waves question? A worked example
Pattern: Pipe harmonics — open vs closed comparison (based on PYQ pattern NEET pattern: pipe harmonics, observed 2023, 2025)
- 1
Given
An organ pipe of length L = 0.85 m is open at both ends. The speed of sound is v = 340 m/s.
- 2
Required
(a) Fundamental frequency when the pipe is open at both ends.
(b) If one end is closed, find the frequency of the second overtone. - 3
Concept
Open-open pipe: all harmonics, fₙ = nv/(2L). Closed pipe: only odd harmonics, fₙ = (2n−1)v/(4L). The second overtone of a closed pipe is the 5th harmonic.
- 4
Formula
Open pipe: f₁ = v/(2L)
Closed pipe second overtone: f₅ = 5v/(4L) - 5
Substitution
(a) f₁ = 340/(2 × 0.85) = 340/1.70
(b) f₅ = 5 × 340/(4 × 0.85) = 1700/3.40 - 6
Calculation
(a) f₁ = 200 Hz
(b) f₅ = 500 Hz
Note: L = 0.85 m and v = 340 m/s are given values treated as exact for this problem. The integer 5 is a counting number (harmonic index). These exact quantities do not limit significant figures. - 7
Final answer
(a) Fundamental frequency (open pipe) = 200 Hz
(b) Second overtone (closed pipe) = 500 Hz - 8
Common trap
The trap: treating the second overtone of a closed pipe as the 3rd harmonic. In a closed pipe, the overtone sequence maps as: 1st overtone → 3rd harmonic, 2nd overtone → 5th harmonic, 3rd overtone → 7th harmonic. Using 3f₁ instead of 5f₁ for the second overtone gives the wrong answer.
A second common error: using the open-pipe formula f = nv/(2L) for a closed pipe. The closed-pipe formula f = (2n−1)v/(4L) has a factor of 4L in the denominator, not 2L. - 9
Similar NEET-style question
A pipe closed at one end and 0.50 m long is in air where the speed of sound is 350 m/s. What is the frequency of the third overtone? (Answer: 7th harmonic = 7 × 350/(4 × 0.50) = 1225 Hz.)
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What to remember before solving Superposition Reflection Waves questions
More in Oscillations and Waves: 6 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Superposition Reflection Waves questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 14, p.285
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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