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 279 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 281). 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 287). 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.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
In a transverse wave on a string, the particles of the medium move:
Answer: B. By definition, in a transverse wave the particle displacement is perpendicular to the direction of wave propagation (NCERT Class 11 Physics Chapter 14, page 279).
Why A is wrong: A describes longitudinal wave motion, where particle oscillation is along the propagation direction — not transverse.
Why C is wrong: C describes surface water waves (approximately), which combine transverse and longitudinal components — not a purely transverse wave on a string.
Why D is wrong: D implies particles travel with the wave. In wave motion, energy propagates but particles oscillate about equilibrium — they do not travel along with the wave.
Which of the following correctly describes longitudinal waves?
Answer: D. Longitudinal waves propagate via compressions (high-pressure regions) and rarefactions (low-pressure regions) in the medium (NCERT Class 11 Physics Chapter 14, page 279).
Why A is wrong: A describes transverse waves. In longitudinal waves, particle displacement is parallel to wave velocity.
Why B is wrong: B is incorrect — sound is a longitudinal wave that propagates readily through gases.
Why C is wrong: C describes a requirement for transverse waves in a bulk medium. Longitudinal waves propagate through gases, liquids, and solids — shear modulus is not required.
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:
Answer: A. Wave speed on a string is v = √(T/μ). If T becomes 4T: v' = √(4T/μ) = 2√(T/μ) = 2v (NCERT Class 11 Physics Chapter 14, page 281).
Why B is wrong: B would result from halving the tension, not quadrupling it. v ∝ √T, so quadrupling T doubles v.
Why C is wrong: C applies a linear relationship v ∝ T. The correct relationship is v ∝ √T, so quadrupling tension gives 2v, not 4v. (Trap: linear-tension-scaling confusion.)
Why D is wrong: D would result from reducing tension to 1/16th. Quadrupling tension doubles speed via the square-root dependence.
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:
Answer: A. Wave speed in a given medium is determined by the medium's properties (tension and density for strings; γP/ρ for gases), not by the source frequency. Changing frequency changes wavelength, not speed. v remains 400 m/s (NCERT Class 11 Physics Chapter 14, pages 281–282).
Why B is wrong: B incorrectly assumes v ∝ f, doubling speed when frequency doubles. The relation v = fλ means λ adjusts, not v.
Why C is wrong: C incorrectly assumes v ∝ 1/f, halving speed when frequency doubles. Wave speed is independent of source frequency in a given medium.
Why D is wrong: D has no physical basis. Wave speed depends on medium properties, not on arithmetic combinations of the source frequency.
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:
Answer: D. D is correct. The wave number k = 80.0 rad/m is the coefficient of x. Wavelength λ = 2π/k = 2π/80.0 = π/40 m ≈ 0.0785 m (NCERT Class 11 Physics Chapter 14, page 281).
Why A is wrong: A is wrong because 0.0125 m is 1/k, which drops the factor 2π entirely. The relation is λ = 2π/k, not 1/k — the 2π is what converts radians per metre into metres per cycle.
Why B is wrong: B is wrong because 0.025 m is 2/k, which keeps the 2 but loses the π. The numerator of λ = 2π/k is the full 2π ≈ 6.28, not 2.
Why C is wrong: C mistakes the wave number k = 80.0 rad/m for the wavelength. The wave number is the spatial frequency (radians per metre), not the wavelength itself.
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:
Answer: B. Newton assumed isothermal compression (v = √(P/ρ)), giving ~280 m/s. Laplace corrected this by recognizing that sound compressions and rarefactions are rapid (adiabatic), giving v = √(γP/ρ) ≈ 332 m/s, matching experiment (NCERT Class 11 Physics Chapter 14, page 287).
Why A is wrong: A is Newton's original (incorrect) assumption. Isothermal compression underestimates the speed because it omits the adiabatic index γ.
Why C is wrong: C (constant pressure) does not apply to sound waves, which involve pressure oscillations by definition — compressions and rarefactions are pressure changes.
Why D is wrong: D (constant volume) does not describe the mechanics of sound propagation. Sound involves alternating compression and expansion of gas elements.
A fixed mass of gas is compressed to four times its original pressure while its temperature is held constant. Using v = √(γP/ρ) together with the ideal-gas relation PV = NkBT, the new speed of sound in the gas is:
Answer: A. From PV = NkBT at fixed T and fixed N, P/V is constant, so compressing the gas to 4× pressure also raises its density ρ (= Nm/V) to 4×. The ratio P/ρ is therefore unchanged, and so is v = √(γP/ρ). This is why the speed of sound in a gas depends on temperature, not on pressure alone (NCERT Class 11 Physics Chapter 14, page 286).
Why B is wrong: B assumes v ∝ √P alone, treating density as fixed. But density rises with pressure at constant temperature too, so it is the ratio P/ρ — not P by itself — that fixes v.
Why C is wrong: C assumes v ∝ P (linear), which ignores both the square root in the formula and the fact that ρ scales with P at constant T.
Why D is wrong: D would follow only if density quadrupled while pressure stayed fixed — the reverse of what the problem describes, and inconsistent with the stated pressure increase.
A wave pulse is travelling along a string. Which of the following statements is correct?
Answer: C. A wave (or pulse) transfers energy and momentum through the medium without bulk transport of matter. Each particle oscillates about its equilibrium position (NCERT Class 11 Physics Chapter 14, page 279).
Why A is wrong: A contradicts the fundamental definition of wave motion. Particles oscillate locally; they do not travel with the pulse.
Why B is wrong: B is incorrect for waves on a string: v = √(T/μ) depends on tension and linear mass density, not on amplitude. This is a common confusion — amplitude affects energy carried, not speed.
Why D is wrong: D is incorrect. Waves carry both energy and momentum. A wave pulse striking an obstacle exerts a force, demonstrating momentum transfer.
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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.
Required
Find the wave speed (a) at the original tension and (b) at the new tension.
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 281).
Formula
v = √(T/μ)
Substitution
(a) v₁ = √(64 / 4.0 × 10⁻³) = √(16000) m/s
(b) v₂ = √(100 / 4.0 × 10⁻³) = √(25000) m/s
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.
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. ✓
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?
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.
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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 Class 11 Physics, Ch. 14, p. 278More in Oscillations and Waves: 6 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
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