Superposition Reflection Waves

8 MCQs3 revision cards9-step worked example

Source: NCERT Oscillations and WavesPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 15, page 8).

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.

Practice 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:

MCQ 2Easy RecallPractice

When a transverse wave pulse on a string reflects from a rigid (fixed) boundary, the reflected pulse is:

MCQ 3Easy RecallPractice

A pipe closed at one end can produce:

MCQ 4Direct 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:

MCQ 5Direct ApplicationPractice

The third harmonic frequency of a closed pipe of length L is 900 Hz. What is its fundamental frequency?

MCQ 6Direct ApplicationPractice

A string fixed at both ends has a fundamental frequency of 200 Hz. The frequency of its third harmonic is:

MCQ 7Concept TrapPractice

In a closed organ pipe, the first overtone corresponds to which 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?

Quick recall before you leave

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.) ---

Before solving, remember these

Law

Principle of superposition

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, p. 8

Formulas

10 formulas — click to collapse

Beat frequency

f_{beat} = ∣ f_{1} - f_{2} ∣

When two waves of nearly equal frequencies superpose, amplitude oscillates at the difference frequency.

Symbol	Quantity	SI Unit
f_beat	beat frequency	Hz
f1, f2	superposed frequencies	Hz

Valid when

Linear superposition
f1, f2 close in value

Period of simple pendulum (small angle)

T = 2 π \frac{L}{g}

Period of simple pendulum of length L. Holds for small amplitudes (sin theta ~ theta).

Symbol	Quantity	SI Unit
T	period	s
L	pendulum length	m
g	gravity	m/s^2

Valid when

Small angular amplitude (typically <15°)
Massless string
Point bob

SHM displacement

x (t) = A cos (ω t + ϕ); ω = 2 π f

Displacement in simple harmonic motion. Velocity = -A*omega*sin(omega*t+phi); a = -omega^2 * x.

Symbol	Quantity	SI Unit
A	amplitude	m
omega	angular frequency	rad/s
phi	phase	rad
T	period	s
f	frequency	Hz

Valid when

Restoring force linear (F = -kx)
No damping

Total energy in SHM

E = \frac{1}{2} k A^{2} = \frac{1}{2} m ω^{2} A^{2}

Total mechanical energy is constant. Oscillates between KE (max at x=0) and PE (max at x=±A).

Symbol	Quantity	SI Unit
E	total energy	J
k	spring constant	N/m
A	amplitude	m
m	mass	kg
omega	angular frequency	rad/s

Valid when

Conservative SHM (no damping)
Elastic regime

Period of mass-spring oscillator

T = 2 π \frac{m}{k}

Period of horizontal spring with mass m, spring constant k. Independent of amplitude.

Symbol	Quantity	SI Unit
T	period	s
m	mass	kg
k	spring constant	N/m

Valid when

Hooke's law spring
No damping
Small enough amplitude to stay in elastic regime

Standing wave in closed-end pipe

f_{n} = \frac{( 2 n - 1 ) v}{4 L}

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

f_{n} = \frac{n v}{2 L}

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

f_{n} = \frac{n v}{2 L}

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

Speed of sound in gas (Newton-Laplace)

v = \frac{γ P}{ρ}

Speed of sound in gas. Adiabatic index gamma, pressure P, density rho. Increases with sqrt(T).

Symbol	Quantity	SI Unit
v	speed of sound	m/s
gamma	adiabatic index	-
P	pressure	Pa
rho	density	kg/m^3

Valid when

Ideal gas
Adiabatic compression/expansion of sound waves

Wave speed on string

v = \frac{T}{μ}

Speed of transverse wave on string under tension T, linear mass density mu.

Symbol	Quantity	SI Unit
v	wave speed	m/s
T	tension	N
mu	linear mass density	kg/m

Valid when

Stretched uniform string
Small amplitude

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

6 items — click to collapse

Trap

Pendulum: mass-dependence claim

Category: Overthinking

Student writes T as depending on bob mass. Simple pendulum T = 2π√(L/g); independent of m.

When it triggers

Question changes pendulum bob mass and asks for new period.

How to avoid

Mass cancels in derivation (gravitational mass = inertial mass). Mass changes the bob's KE and PE proportionally; period unaffected.

Trap

SHM period: amplitude dependence claim

Category: Overthinking

Student claims SHM period depends on amplitude. For ideal SHM (Hooke's law spring or simple pendulum at small angle), period is INDEPENDENT of amplitude.

When it triggers

Question gives changes in amplitude and asks for new period.

How to avoid

T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — neither depends on A. Only at large pendulum angles does T pick up a small amplitude correction.

Trap

Pipe harmonics: open vs closed end

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.

Mistake

Includes bob mass in simple pendulum period formula.

Root cause: concept gap

Correction

Simple pendulum T = 2π√(L/g) — independent of mass. Equivalence of inertial and gravitational mass cancels m.

Mistake

Claims SHM period depends on amplitude.

Root cause: concept gap

Correction

Ideal SHM: T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — no amplitude dependence. Doubling amplitude does not change period.

Mistake

Includes even harmonics in a closed-end pipe.

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.