A mechanical wave needs a medium. How the medium's particles move relative to the wave's travel direction is the single distinction between the two fundamental wave types — and NEET regularly tests whether you can apply it under slight disguise.
Transverse waves: particle displacement is perpendicular to wave propagation. A pulse on a stretched string moves horizontally while each string element oscillates vertically. Transverse waves form crests and troughs. They travel through solids (and on liquid surfaces) but cannot propagate through the bulk of a fluid because fluids lack the shear restoring force needed for perpendicular displacement (NCERT Class 11 Physics, Chapter 14, page 280).
Longitudinal waves: particle displacement is parallel to wave propagation. Sound in air is the standard example — air molecules oscillate back and forth along the direction the sound travels, creating alternating compressions and rarefactions. Longitudinal waves can travel through solids, liquids, and gases because all three media support compressive restoring forces.
The speed connection. For a transverse wave on a string: v = √(T/μ), where T is tension and μ is linear mass density (NCERT Class 11 Physics, Chapter 14, page 281). Increasing tension increases wave speed — but the relationship is square-root, not linear. This is a high-frequency NEET distractor: if tension quadruples, speed doubles, not quadruples.
Watch out: NEET stems sometimes describe a wave scenario without labelling it "transverse" or "longitudinal." Identify the medium and the displacement direction. Sound in a metal rod? Longitudinal (compressions along the rod). Ripple on a water surface? Primarily transverse (surface particles move up and down). A wave on a slinky pushed end-to-end? Longitudinal. The classification follows from the physics, not from the exam's wording.
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, the particle displacement is:
Answer: D. By definition, in a transverse wave the medium's particles oscillate perpendicular to the wave's direction of travel (NCERT Class 11 Physics, Chapter 14, page 280).
Why A is wrong: A describes longitudinal waves, where particle motion is parallel to propagation — not transverse.
Why B is wrong: B would mean no wave at all — a wave requires particle displacement.
Why C is wrong: C is not a standard wave classification. Waves are classified as longitudinal (0°) or transverse (90°); no 45° type exists in classical mechanics.
Which of the following can transmit both longitudinal and transverse waves?
Answer: C. Solids possess both bulk modulus (supports compressions) and shear modulus (supports perpendicular displacement), so they transmit both longitudinal and transverse waves (NCERT Class 11 Physics, Chapter 14, page 280).
Why A is wrong: A is incorrect — air (a gas) has no shear modulus and cannot support transverse waves in its bulk. Only longitudinal waves propagate through air.
Why B is wrong: B is incorrect — bulk water lacks shear rigidity and cannot sustain transverse waves through its interior. (Surface waves are a separate phenomenon involving gravity, not shear restoring force.)
Why D is wrong: D is incorrect — a vacuum has no medium at all, so no mechanical wave of either type can propagate through it.
Sound waves in air are:
Answer: A. Sound in air propagates as compressions and rarefactions — particle displacement is along the direction of wave travel, making it longitudinal (NCERT Class 11 Physics, Chapter 14, page 280).
Why B is wrong: B is incorrect — sound requires a material medium; electromagnetic waves do not. Sound is a mechanical wave, not electromagnetic.
Why C is wrong: C is incorrect — air cannot support transverse mechanical waves because gases have zero shear modulus.
Why D is wrong: D is incorrect — in the bulk of a gas, only longitudinal modes exist. A combined mode requires a medium with shear rigidity (solids) or a surface boundary.
The speed of a transverse wave on a stretched string is given by v = √(T/μ). If the tension in the string is increased to 4 times its original value while the linear mass density remains the same, the new wave speed is:
Answer: C. v = √(T/μ). If T becomes 4T, v_new = √(4T/μ) = 2√(T/μ) = 2v. The speed doubles (NCERT Class 11 Physics, Chapter 14, page 281).
Why A is wrong: A results from treating v as directly proportional to T (v ∝ T), ignoring the square root. The correct relation is v ∝ √T, so quadrupling T doubles v, not quadruples it.
Why B is wrong: B would be correct if T were doubled (v ∝ √T → √2 · v), but T is quadrupled here, giving √4 = 2 times v.
Why D is wrong: D results from squaring 4 instead of taking its square root — an arithmetic direction error.
A transverse wave is sent along a string of linear mass density 5.0 × 10⁻² kg/m under a tension of 80 N. The speed of the wave is:
Answer: A. v = √(T/μ) = √(80 / 5.0 × 10⁻²) = √(1600) = 40 m/s = 4.0 × 10¹ m/s (NCERT Class 11 Physics, Chapter 14, page 281).
Why B is wrong: B (1600 m/s) results from computing T/μ = 1600 but forgetting to take the square root. The formula is v = √(T/μ), not T/μ.
Why C is wrong: C (4.0 m/s) results from an error in handling the exponent — dividing 80 by 5 to get 16, then taking √16 = 4, but dropping the 10⁻² correction. The correct denominator is 0.05, not 5.
Why D is wrong: D (160 m/s) likely comes from computing √(80 × 0.05 × some extra factor) or a unit-conversion slip; the correct value is √1600 = 40 m/s.
Air (a gas) has bulk modulus B = 1.4 × 10⁵ Pa and density ρ = 1.2 kg/m³. Given that transverse waves need a shear modulus of elasticity while longitudinal waves need only a bulk modulus, which wave type can air support, and what is its speed using v = √(B/ρ)?
Answer: B. Gases have zero shear modulus, so they cannot support transverse waves — only longitudinal waves, which need just a bulk modulus, can propagate through them ('Transverse waves can propagate only in a medium with shear modulus of elasticity, Longitudinal waves need bulk modulus of elasticity and are therefore possible in all media, solids, liquids and gases', NCERT Class 11 Physics, Chapter 14, page 297). Speed: v = √(B/ρ) = √(1.4 × 10⁵ / 1.2) = √(1.17 × 10⁵) ≈ 342 m/s (NCERT Class 11 Physics, Chapter 14, pages 286, 297).
Why A is wrong: Air has zero shear modulus — it cannot support transverse waves at all; this option wrongly assigns a shear-dependent wave type to a shear-free gas, even though the numeric speed shown is correct for the longitudinal case.
Why C is wrong: Air (a gas) has no shear modulus, so it cannot support transverse waves — only longitudinal waves propagate in gases; 'both types' is wrong even though the speed value matches the longitudinal calculation.
Why D is wrong: This uses v = B/ρ instead of v = √(B/ρ) — omitting the square root gives 1.17 × 10⁵ m/s, far larger than the actual speed of sound in air.
A string of length 2.0 m has a mass of 1.0 × 10⁻² kg and is stretched with a tension of 40 N. If the string is plucked, the speed of the transverse wave produced is:
Answer: B. First, find μ = mass/length = 1.0 × 10⁻² / 2.0 = 5.0 × 10⁻³ kg/m. Then v = √(T/μ) = √(40 / 5.0 × 10⁻³) = √(8000) ≈ 89.4 m/s ≈ 89 m/s (NCERT Class 11 Physics, Chapter 14, page 281).
Why A is wrong: A (8.0 × 10³ m/s) results from computing T/μ = 8000 and forgetting the square root; v = √(T/μ).
Why C is wrong: C (63 m/s) could arise from using μ = 1.0 × 10⁻² directly (forgetting to divide mass by length), giving √(40/0.01) = √4000 ≈ 63.2 m/s. The correct μ requires dividing the string's total mass by its length.
Why D is wrong: D (about 45 m/s) results from multiplying the mass by the length instead of dividing, μ = 2.0 × 10⁻² kg/m, giving √(40/0.02) = √2000 ≈ 45 m/s.
A wave on a string has a speed of 50 m/s under tension T. If the tension is reduced to T/4 without changing the string, the new speed is:
Answer: B. v ∝ √T (μ unchanged). v_new = √(T/4)/√(T) × 50 = (1/2) × 50 = 25 m/s (NCERT Class 11 Physics, Chapter 14, page 281).
Why A is wrong: A (12.5 m/s) results from treating v ∝ T (linear) instead of v ∝ √T, dividing 50 by 4 instead of by 2.
Why C is wrong: C assumes speed is independent of tension — incorrect. v = √(T/μ) explicitly depends on T.
Why D is wrong: D reverses the effect, increasing speed instead of decreasing it. Reducing tension reduces speed, not increases it.
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Pattern: Given tension change, find new wave speed on string (PYQ pattern: wave speed–tension relationship, observed 2022).
Given
A transverse wave travels along a string with speed v₁ = 20 m/s when the string is under tension T₁ = 50 N. The tension is then increased to T₂ = 200 N. The string itself does not change.
Required
Find the new wave speed v₂.
Concept
The speed of a transverse wave on a stretched string depends on the tension and the linear mass density of the string: v = √(T/μ). Since the string does not change, μ is constant.
Formula
v = √(T/μ)
Taking the ratio: v₂/v₁ = √(T₂/T₁)
Substitution
v₂/v₁ = √(200/50) = √4
Calculation
v₂/v₁ = 2
v₂ = 2 × 20 = 40 m/s
Note on exact values: The tension values 50 N and 200 N, and the initial speed 20 m/s, are problem-defined exact values. They do not limit significant figures.
Final answer
v₂ = 40 m/s
Common trap
Treating v as directly proportional to T (v ∝ T) instead of v ∝ √T. That error gives v₂ = 4 × 20 = 80 m/s — exactly double the correct answer. Always check: the formula has a square root over T.
Similar NEET-style question
A string carries a transverse wave at 30 m/s under tension T. If the tension is reduced to T/9, what is the new wave speed?
Answer: v_new = 30/3 = 10 m/s (since v ∝ √T, factor = √(1/9) = 1/3).
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Transverse: particle displacement perpendicular to wave direction (e.g. waves on string, EM waves). Longitudinal: particle displacement along wave direction (e.g. sound in air). Both share v = ν λ.
-- NCERT Class 11 Physics, Ch. 14, p. 280Speed 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 |
More in Oscillations and Waves: 6 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
uses linear tension scaling
Treats v ∝ T not sqrt(T)
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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