Wave Motion

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

Wave motion is the transfer of energy and momentum through a medium (or vacuum, for electromagnetic waves) without bulk transfer of matter. NCERT Class 11 Physics Chapter 14, page 2 defines a wave as a disturbance that propagates through space and time, usually with a transfer of energy.

The trap NEET exploits: Students conflate the motion of the medium's particles with the motion of the wave itself. In a transverse wave on a string, particles move perpendicular to the wave's propagation direction. In a longitudinal wave (sound in air), particles oscillate parallel to propagation. The wave speed depends on medium properties — not on particle speed or amplitude.

Two key wave-speed relations:

For a transverse wave on a stretched string: v = √(T/μ), where T is tension and μ is linear mass density (NCERT Class 11 Physics Chapter 14, page 4). Speed depends on tension and mass distribution — not on frequency or amplitude.

For sound in a gas (longitudinal wave): v = √(γP/ρ), the Newton-Laplace formula (NCERT Class 11 Physics Chapter 14, page 5). Laplace's adiabatic correction fixed Newton's isothermal estimate, which underestimated the speed of sound by about 16%.

Common confusion: Changing frequency does NOT change wave speed in a given medium. The relation v = fλ means that if f changes (different source), λ adjusts — v stays fixed by the medium. Students who treat v as dependent on f lose marks on NEET problems that alter source frequency and ask for the new wavelength.

Watch-out: When tension on a string is quadrupled, wave speed doubles (v ∝ √T), not quadruples. The square-root relationship is a high-frequency distractor in NEET options.

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

In a transverse wave on a string, the particles of the medium move:

MCQ 2Easy RecallPractice

Which of the following correctly describes longitudinal waves?

MCQ 3Direct ApplicationPYQ Pattern

The speed of a transverse wave on a string is v. If the tension in the string is increased to four times its original value while keeping the linear mass density unchanged, the new wave speed is:

MCQ 4Direct ApplicationPractice

A wave of frequency 200 Hz travels through a medium with a speed of 400 m/s. If the frequency of the source is changed to 400 Hz, the wave speed in the same medium becomes:

MCQ 5Direct ApplicationPractice

A wave travelling along a string is described by y(x,t) = 0.005 sin(80.0x − 3.0t), where x is in metres and t in seconds. The wavelength of the wave is:

MCQ 6Easy RecallPractice

Newton's formula for the speed of sound in air gave a value about 16% lower than the experimental value. Laplace corrected this by assuming sound propagation is:

MCQ 7Direct ApplicationPractice

A transverse wave on a string has speed v₁ when the tension is T₁. A second string has twice the linear mass density but is under the same tension T₁. The wave speed on the second string is:

MCQ 8Concept TrapPractice

A wave pulse is travelling along a string. Which of the following statements is correct?

Quick recall before you leave

Worked Example

1
Given
A transverse wave travels on a wire of length 1.0 m with a linear mass density μ = 4.0 × 10⁻³ kg/m. The wire is under a tension of 64 N. The wire is then tightened so that the tension becomes 100 N.
2
Required
Find the wave speed (a) at the original tension and (b) at the new tension.
3
Concept
Wave speed on a string depends on tension and linear mass density: v = √(T/μ). Speed is proportional to √T when μ is constant (NCERT Class 11 Physics Chapter 14, page 4).
4
Formula
v = √(T/μ)
5
Substitution
(a) v₁ = √(64 / 4.0 × 10⁻³) = √(16000) m/s (b) v₂ = √(100 / 4.0 × 10⁻³) = √(25000) m/s
6
Calculation
(a) √(16000) = √(16 × 1000) = 4√1000 = 4 × 31.62 ≈ 126.5 m/s (b) √(25000) = √(25 × 1000) = 5√1000 = 5 × 31.62 ≈ 158.1 m/s **Note on exact values:** The tension values 64 N and 100 N, and the linear mass density 4.0 × 10⁻³ kg/m, are given data (treated as exact for this problem). The integer factors (4 and 5) extracted from the square roots are exact counting numbers and do not affect significant-figure counts.
7
Final answer
(a) v₁ ≈ 1.26 × 10² m/s (b) v₂ ≈ 1.58 × 10² m/s Ratio check: v₂/v₁ = √(100/64) = √(25/16) = 5/4 = 1.25. Indeed, 158.1/126.5 ≈ 1.25. ✓
8
Common trap
Treating v ∝ T (linear) instead of v ∝ √T. With linear scaling, a student would get v₂ = v₁ × (100/64) = 1.5625 × v₁ ≈ 197.7 m/s — significantly overshooting the correct value. Always check: is the relationship linear or square-root?
9
Similar NEET-style question
"A wave on a string has speed 50 m/s when the tension is T. If the tension is increased to 9T without changing the string, what is the new wave speed?" Answer: v' = √(9T/μ) = 3√(T/μ) = 3 × 50 = 150 m/s. ---

Before solving, remember these

Definition

Wave motion

A disturbance that propagates through a medium (or vacuum for EM waves) carrying energy and momentum without bulk transport of matter. Characterised by wavelength λ, frequency ν, period T, speed v.

-- NCERT, p. 2

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.