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 3).
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 4). 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:
Which of the following can transmit both longitudinal and transverse waves?
Sound waves in air are:
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:
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:
Which statement about longitudinal waves is correct?
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:
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:
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).* ---
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, p. 3When 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 | - |
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 |
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 |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student writes T as depending on bob mass. Simple pendulum T = 2π√(L/g); independent of m.
Question changes pendulum bob mass and asks for new period.
Mass cancels in derivation (gravitational mass = inertial mass). Mass changes the bob's KE and PE proportionally; period unaffected.
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.
Question gives changes in amplitude and asks for new period.
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.
Category: Similar Terms
Student includes even harmonics in a closed-end pipe. Closed pipe has only ODD harmonics (f, 3f, 5f, ...).
Question describes pipe closed at one end (e.g. resonance tube).
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.
Root cause: concept gap
Simple pendulum T = 2π√(L/g) — independent of mass. Equivalence of inertial and gravitational mass cancels m.
Root cause: concept gap
Ideal SHM: T = 2π√(m/k) (spring) or 2π√(L/g) (pendulum, small angle) — no amplitude dependence. Doubling amplitude does not change period.
Root cause: concept gap
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.
treats beat as sum
Adds frequencies instead of subtracting
expects mass dependence
Assumes T depends on bob mass
treats closed pipe like open
Includes even harmonics in closed pipe
uses pi 2 instead of pi
Confuses v-x phase with a-x phase
forgets 2pi factor
Drops 2*pi
uses amplitude in period
Believes T depends on amplitude
uses linear tension scaling
Treats v ∝ T not sqrt(T)
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