Class 11 · Physics · Chapter 14

Waves

This chapter builds the mathematics of wave motion — how a disturbance travels through a medium without carrying matter with it — and applies it to strings, sound in air, standing waves, beats and the Doppler effect. It is the foundation for optics and oscillations in Class 12 and comes back directly in JEE and NEET.

9 sections12 NCERT questions22 board questions15-question quiz

Before you startWhat the board asks from this chapter

Class 11 has no board exam, but this chapter is a major share of the school final/annual exam paper for the unit 'Oscillations and Waves' and reappears every year in JEE Main/Advanced and NEET as 1-3 questions.

Question typeMarksUsually asked
MCQ1Identifying transverse vs longitudinal, reading a, k, ω off a wave equation, or a quick Doppler/beats calculation.
Assertion–Reason1Tests whether you know WHY a fact is true — e.g. why sound needs a medium, why only transverse waves polarise.
Very short answer2State a formula and its terms, or a one-line conceptual difference (e.g. wave speed vs particle speed).
Short answer3A numerical on wave speed on a string, beats, or a standing-wave harmonic frequency.
Long answer5Full derivation — standing waves from superposition, normal modes of a string/pipe, or the Doppler effect.
Case-based4A passage on a musical instrument, a resonance tube experiment, or a moving-source scenario with sub-questions.

Hot topics, in order

  1. Displacement relation y = a sin(kx − ωt) and reading off a, λ, T, ν, v from it
  2. Speed of a transverse wave on a string, v = √(T/μ)
  3. Speed of sound in a gas — Newton's formula and Laplace's correction
  4. Standing waves: derivation, nodes and antinodes
  5. Normal modes of a stretched string and of open/closed organ pipes
  6. Beats and their use in tuning
  7. Doppler effect — all four cases (source moving, observer moving, both, and applications)

The whole chapter, in orderNotes

01 14.1 Introduction

A wave is a disturbance that travels through a medium, carrying energy and momentum from one point to another without the medium itself moving along with it. Drop a stone in a pond: the ripples spread outward, but a leaf floating on the surface only bobs up and down about its own position — it does not travel with the ripple. This is the central idea of wave motion: particles of the medium oscillate about their mean positions, and it is the pattern of oscillation, not the matter, that propagates.

Waves are broadly of two kinds. Mechanical waves — water waves, sound waves, waves on a string — need a material medium to travel through, because they work by one particle of the medium disturbing its neighbour through the medium's elasticity and inertia. Electromagnetic waves (light, radio waves) need no medium and can travel through vacuum; they are not part of this chapter but are worth knowing as the contrast. This chapter deals only with mechanical waves.

Two properties of the medium decide how fast a mechanical wave travels: its elasticity (how strongly it resists being deformed and springs back) and its inertia (its mass density, which decides how sluggishly it responds). A stiffer, lighter medium carries waves faster; a floppier, denser one carries them slower.

Exam tip A very common 1-mark trap: 'sound can travel through vacuum — true or false'. Sound is mechanical, so it needs a medium; the answer is false, unlike light.

02 14.2 Transverse and Longitudinal Waves PYQ CBSE 2020

In a transverse wave, the particles of the medium oscillate perpendicular to the direction in which the wave travels. A wave sent along a stretched string by moving one end up and down is transverse — the string moves up-down while the wave itself moves along the string's length. Transverse mechanical waves can only exist in a medium that can sustain shearing stress, which is why they travel through solids and along string/wire surfaces but not through the bulk of a liquid or a gas (liquids and gases cannot support a shear).

In a longitudinal wave, the particles oscillate along the same direction as the wave travels. Sound in air is the standard example: air molecules move back and forth along the direction the sound travels, creating alternate regions of compression (high pressure/density) and rarefaction (low pressure/density) that move forward. Longitudinal waves need only volume elasticity, so they travel through solids, liquids and gases alike.

A useful exam distinction: only transverse waves can be polarised — restricted to oscillate in a single plane — because their oscillation direction is perpendicular to propagation and can point anywhere in that perpendicular plane. Longitudinal waves oscillate only along the line of travel, so there is nothing left to restrict; they cannot be polarised. This is one way experiments proved that light is a transverse wave.

Exam tip If asked to justify why sound cannot be polarised but light can, say exactly this: sound is longitudinal (only one oscillation direction, along propagation), light is transverse (oscillation is confined to the plane perpendicular to propagation, which polarisation can further restrict).

03 14.3 Displacement Relation in a Progressive Wave PYQ CBSE 2019 · CBSE 2023

A sinusoidal wave travelling in the +x direction is written as y(x, t) = a sin(kx − ωt + φ), where y is the displacement of the particle at position x and time t. Here a is the amplitude (maximum displacement), k is the angular wave number (k = 2π/λ, unit rad m−1), ω is the angular frequency (ω = 2π/T = 2πν, unit rad s−1), and φ is the initial phase. The quantity (kx − ωt + φ) is the phase of the wave at that point and instant.

Wavelength (λ) is the distance between two consecutive points of the medium that are in the same phase, e.g. two consecutive crests. Time period (T) is the time for one particle to complete one full oscillation, and frequency (ν) = 1/T is the number of oscillations per second. The wave repeats itself over one wavelength in space and one time period in time.

The wave speed follows from the fact that in one time period T, the wave pattern advances exactly one wavelength λ: v = λ/T = νλ. In terms of the wave equation this is the same as v = ω/k, because both are the rate at which a point of constant phase moves. A wave written as a function of (kx + ωt) instead travels in the −x direction — the sign in front of ωt tells you the direction of travel.

Do not confuse wave speed with particle speed. Wave speed v = νλ is how fast the pattern (crest, trough) moves along the medium — it is fixed by the medium. Particle speed is ∂y/∂t at a fixed x, which oscillates between 0 and aω as the particle goes through its cycle — it depends on where in the cycle the particle is, not on the medium alone.

y(x, t) = a sin(kx − ωt + φ)
aamplitudekangular wave number, k = 2π/λωangular frequency, ω = 2π/T = 2πνφinitial phase
v = νλ = ω/k
vwave speedνfrequencyλwavelength
Exam tip Given a wave equation with numbers plugged in, first match it term by term with y = a sin(kx − ωt + φ) — read off a, k and ω directly, then compute λ = 2π/k, T = 2π/ω, ν = 1/T and v = ω/k. Do not skip straight to a memorised shortcut; matching the form is what earns the marks.

04 14.4 The Speed of a Travelling Wave PYQ CBSE 2018 · CBSE 2022

Speed of a transverse wave on a stretched string. The speed depends only on the tension T in the string (the restoring force) and μ, the mass per unit length (the inertia). By dimensional analysis: let v = C Ta μb. Since [T] = MLT−2 and [μ] = ML−1, matching dimensions on both sides of [LT−1] = [MLT−2]a[ML−1]b gives a + b = 0, a − b = 1, and −2a = −1, so a = ½, b = −½. The constant C works out to 1 from the full mechanical treatment, giving v = √(T/μ). A heavier string (large μ) carries waves more slowly; a more tightly stretched string (large T) carries them faster.

Speed of a longitudinal wave. The same reasoning with elasticity in place of tension gives, for a wave in a fluid, v = √(B/ρ), where B is the bulk modulus of the medium and ρ is its density. For a longitudinal wave in a long solid rod, v = √(Y/ρ), using the Young's modulus Y in place of B. In general: the stiffer the medium (larger elastic modulus) and the lighter it is (smaller density), the faster sound travels through it — which is why sound is fastest in solids, slower in liquids, and slowest in gases.

Speed of sound in a gas. Newton first proposed that sound travels through a gas isothermally (temperature stays constant during compression and rarefaction), giving v = √(P/ρ), where P is the pressure. For air at STP this gives about 280 m/s — roughly 15% below the experimentally measured 332 m/s. Laplace corrected this: compressions and rarefactions happen too fast for heat to flow between them, so the process is adiabatic, not isothermal. Using PVγ = constant instead of PV = constant leads to v = √(γP/ρ), where γ = Cp/Cv is the ratio of specific heats (γ = 1.4 for air). This matches the measured speed of sound closely and is the accepted formula.

v = √(T/μ)
vspeed of transverse wave on a stringTtension in the stringμmass per unit length
v = √(B/ρ)
Bbulk modulus of the mediumρdensity of the medium
v = √(γP/ρ) (Laplace's formula)
γratio of specific heats, Cp/CvPpressure of the gas
Exam tip If asked 'why does Newton's formula fail', do not just say 'it's wrong' — say precisely that Newton assumed isothermal changes, but the rapid compressions in sound are adiabatic, so Laplace multiplied by γ inside the square root. This exact reasoning is what the marking scheme rewards.

05 14.5 The Principle of Superposition of Waves

When two or more waves travel through the same region of a medium at the same time, the resultant displacement at any point is simply the algebraic sum of the displacements each wave would have produced there on its own: y = y1 + y2 + y3 + ... This is the principle of superposition, and it holds because the wave equation is linear in y for the small oscillations considered in this chapter. The waves pass through each other completely unaffected — after they cross, each continues exactly as it would have if the other had never been there.

Superposition is the reason two waves can combine to reinforce each other (constructive interference, when crests meet crests) or cancel each other (destructive interference, when a crest meets a trough). It is also the tool used to build up every other result in this chapter: standing waves (superposition of a wave with its own reflection) and beats (superposition of two waves of slightly different frequency) are both direct applications of this one simple rule.

y(x, t) = y1(x, t) + y2(x, t) + ...
yresultant displacementy1, y2displacement due to each wave acting alone
Exam tip State the principle in words exactly this way for full marks: 'the resultant displacement is the algebraic sum of the displacements due to each wave separately' — examiners look for 'algebraic sum', not just 'sum'.

06 14.6 Reflection of Waves and Standing Waves PYQ CBSE 2020 · CBSE 2023

When a wave reaches the end of the medium — a rigid wall, a fixed end of a string — it reflects. At a rigid boundary the reflected wave undergoes a phase reversal (a crest returns as a trough); at a free/open boundary it reflects without phase reversal. A standing wave (or stationary wave) is what results when two identical waves of the same amplitude and frequency travel through the same medium in opposite directions — exactly the situation of an incident wave and its own reflection.

Derivation. Let y1 = a sin(kx − ωt) be the incident wave and y2 = a sin(kx + ωt) the wave travelling in the opposite direction. By superposition, y = y1 + y2 = a[sin(kx − ωt) + sin(kx + ωt)]. Using the identity sin A + sin B = 2 sin((A+B)/2) cos((A−B)/2), this simplifies to y(x, t) = 2a sin(kx) cos(ωt). Notice this is not a travelling wave at all — the space part sin(kx) and time part cos(ωt) are separate. Every particle at a given x oscillates with the same phase and amplitude 2a sin(kx), which depends on position.

Points where sin(kx) = 0, i.e. x = 0, λ/2, λ, 3λ/2, ... never move — these are nodes. Points where sin(kx) = ±1, i.e. x = λ/4, 3λ/4, ... oscillate with the maximum amplitude 2a — these are antinodes. Nodes and antinodes alternate, each λ/4 apart.

Normal modes of a string fixed at both ends (length L). Both ends must be nodes, so L must be an integer number of half-wavelengths: L = nλ/2, giving allowed frequencies νn = nv/2L for n = 1, 2, 3, ... The lowest, ν1 = v/2L, is the fundamental or first harmonic; ν2 = 2ν1 is the second harmonic (first overtone), and so on — a stretched string (and hence a sonometer or guitar string) supports all harmonics.

Normal modes of air columns. An open pipe (open at both ends) has an antinode at each end, giving the same condition L = nλ/2, νn = nv/2L — all harmonics present. A closed pipe (closed at one end, open at the other) has a node at the closed end and an antinode at the open end, giving L = (2n−1)λ/4, so νn = (2n−1)v/4L — only odd harmonics are present, and the fundamental ν1 = v/4L is half that of an equally long open pipe.

y(x, t) = 2a sin(kx) cos(ωt)
2a sin(kx)position-dependent amplitude of the standing wave
νn = nv/2L (string, both ends fixed; open pipe)
n1, 2, 3, ... — all harmonicsLlength of string or pipe
νn = (2n−1)v/4L (pipe closed at one end)
n1, 2, 3, ... — only odd harmonics
Exam tip Draw the harmonic diagrams from memory before the exam. A very common error is drawing a node at the open end of a pipe — the open end is always an antinode.

07 14.7 Beats PYQ CBSE 2019

When two sound waves of slightly different frequencies (ν1 and ν2, close to each other) reach the same point together, the ear hears a sound whose loudness rises and falls periodically — this waxing and waning is called beats. Each rise-and-fall cycle is one 'beat'.

Derivation. Take two waves of equal amplitude a at a fixed point: y1 = a cos(ω1t) and y2 = a cos(ω2t). By superposition, y = y1 + y2 = 2a cos((ω1−ω2)t/2) · cos((ω1+ω2)t/2). This is a wave of frequency (ν1+ν2)/2 — essentially the average, which is what we hear as the pitch — whose amplitude itself is not constant but slowly varies as 2a cos((ω1−ω2)t/2). Loudness is maximum whenever this envelope amplitude is at its largest magnitude, which happens twice in every cycle of the slow cosine term; working through the timing gives a beat frequency Δν = |ν1 − ν2| — simply the difference of the two frequencies.

Beats are used practically to tune musical instruments: a musician plays a note next to a reference tone and adjusts the string tension until the beats slow down and disappear, meaning the two frequencies have become equal. A related trick used in numericals: loading a tuning fork with wax lowers its frequency, and clipping (filing) a fork raises it — watching whether the beat rate increases or decreases after loading tells you whether the unknown frequency was originally above or below the known one.

Δν = |ν1 − ν2|
Δνbeat frequency (number of beats heard per second)ν1, ν2the two source frequencies
Exam tip The classic numerical: beats increase after loading one fork with wax, so its frequency (which decreased) must originally have been on the far side, moving away from the other. Work it out by testing both possible original frequencies (ν + Δν and ν − Δν) and keeping the one consistent with the new information — never just guess.

08 14.8 The Doppler Effect PYQ CBSE 2020 · CBSE 2022 · CBSE 2023

The Doppler effect is the apparent change in the frequency of a wave when there is relative motion between the source and the observer. You hear this every time a fast car or train passes by: the pitch of the horn is higher as it approaches and drops as it recedes, even though the source is emitting a constant frequency ν throughout. Let v be the speed of sound in air (relative to the medium, taken as still).

Source moving, observer at rest. If the source moves toward the observer with speed vs, each successive wave crest is emitted from a position slightly closer to the observer than the last, compressing the wavelength to λ' = (v − vs)/ν. The observed frequency is therefore ν' = v/λ' = νv/(v − vs) — higher than ν. If the source recedes, the same reasoning with a sign flip gives ν' = νv/(v + vs) — lower than ν.

Observer moving, source at rest. If the observer moves toward the source with speed vo, sound crests reach the observer faster because the observer effectively meets them at a relative speed (v + vo); the observed frequency is ν' = ν(v + vo)/v. If the observer recedes, ν' = ν(v − vo)/v.

General case (both moving). Combining both effects gives the master formula ν' = ν(v ± vo)/(v ∓ vs), where the upper sign is used when the motion is toward the other and the lower sign when it is away — take + for vo and − for vs when they move to close the gap, and flip the sign for whichever one is moving apart.

Note carefully: this classical Doppler formula for sound depends on the speed of source and observer relative to the medium (still air), which is why the source and observer formulas are not symmetric — unlike light, where only relative velocity matters. Applications include radar speed guns, weather radar (Doppler radar), and estimating the speed of stars and galaxies from the shift in the frequency of their light (this optical case is covered again in Class 12).

ν' = ν(v ± vo)/(v ∓ vs)
vspeed of sound in the mediumvospeed of observer (+ toward source, − away)vsspeed of source (− toward observer, + away)
Exam tip Do not memorise four separate formulas — memorise the one general formula and reason the sign from the physical picture each time: motion that closes the distance always raises the observed frequency, motion that increases the distance always lowers it.

09 Energy Transported by a Progressive Wave

A travelling wave carries energy — this is exactly how energy from an earthquake, a loudspeaker or a struck string reaches a distant point without any net transport of matter. For a wave on a string of mass per unit length μ, amplitude a and angular frequency ω, each element of the string undergoes simple harmonic motion, and the energy per unit length of the string works out to ½μω²a². As the wave moves at speed v, the average power transmitted across any point is Pav = ½μω²a²v. This shows the important scaling result you should remember for concept-based MCQs: the energy (and power) carried by a wave is proportional to the square of the amplitude and to the square of the frequency — doubling the amplitude quadruples the energy transported.

Pav = ½ μω²a²v
μmass per unit lengthωangular frequencyaamplitudevwave speed
Exam tip A quick MCQ trap: 'amplitude is doubled, frequency unchanged — energy transported becomes how many times?' Answer: 4 times, since P ∝ a².

VocabularyKey terms

Wave motionThe propagation of a disturbance through a medium, transferring energy without any net transport of the medium's particles.
Transverse waveA wave in which the particles of the medium oscillate perpendicular to the direction of propagation, e.g. a wave on a stretched string.
Longitudinal waveA wave in which the particles of the medium oscillate parallel to the direction of propagation, e.g. sound in air, made of compressions and rarefactions.
Amplitude (a)The maximum displacement of a particle of the medium from its mean position as the wave passes.
Wavelength (λ)The distance between two consecutive points of the medium that are in exactly the same phase, such as two successive crests.
Time period (T)The time taken by a particle of the medium to complete one full oscillation as the wave passes.
Frequency (ν)The number of complete oscillations a particle makes per second; ν = 1/T, measured in hertz.
Angular wave number (k)k = 2π/λ; describes how rapidly the phase of the wave changes with position, unit rad m−1.
Angular frequency (ω)ω = 2π/T = 2πν; describes how rapidly the phase of the wave changes with time, unit rad s−1.
Wave speed (v)The speed at which a point of constant phase (e.g. a crest) moves through the medium; v = νλ, fixed by the medium's properties, not the source.
Principle of superpositionWhen two or more waves overlap, the resultant displacement at any point and instant is the algebraic sum of the displacements each wave would produce alone.
Standing (stationary) waveThe pattern formed when two identical waves travel through the same medium in opposite directions; it does not travel, and has fixed nodes and antinodes.
NodeA point in a standing wave that always has zero displacement.
AntinodeA point in a standing wave that oscillates with maximum amplitude.
Fundamental frequencyThe lowest frequency at which a system (string or air column) can vibrate in a normal mode; also called the first harmonic.
Harmonics and overtonesHarmonics are integer multiples of the fundamental frequency; overtones are the modes above the fundamental (first overtone = second harmonic, and so on).
BeatsThe periodic rise and fall in loudness heard when two sounds of slightly different, close frequencies are heard together; beat frequency equals the difference of the two frequencies.
Doppler effectThe apparent change in the observed frequency of a wave caused by relative motion between the source and the observer.

Keep this on one pageFormula sheet

NameExpressionNotes
Progressive wave equationy(x, t) = a sin(kx − ωt + φ)+ωt for a wave travelling in −x direction; a, k, ω read off directly by matching terms
Wave speedv = νλ = ω/kspeed of the wave pattern, not the particle speed
Speed on a stretched stringv = √(T/μ)T = tension, μ = mass per unit length
Speed of longitudinal wave in fluidv = √(B/ρ)B = bulk modulus, ρ = density
Speed of sound in a gas (Laplace)v = √(γP/ρ)γ = Cp/Cv; adiabatic, matches measured value; Newton's isothermal v = √(P/ρ) underestimates by ~15%
Standing wave equationy = 2a sin(kx) cos(ωt)from superposition of two identical oppositely-travelling waves; nodes at kx = nπ
String / open pipe normal modesνn = nv/2L, n = 1, 2, 3, ...all harmonics present, both ends are antinodes (pipe) or nodes are fixed ends (string)
Closed pipe normal modesνn = (2n−1)v/4L, n = 1, 2, 3, ...only odd harmonics; fundamental is half that of an equal-length open pipe
Beat frequencyΔν = |ν1 − ν2|audible only when Δν is small (a few Hz)
Doppler effect (general)ν' = ν (v ± vo)/(v ∓ vs)+ toward, − away for each of source and observer
Average power of a wave on a stringPav = ½μω²a²vpower/energy transported ∝ (amplitude)² and ∝ (frequency)²

Practise with a pencilDiagrams that earn marks

A transverse progressive wave — snapshot at one instant

Draw a smooth sine curve on x-y axes. Mark two consecutive crests and label the horizontal distance between them as λ (wavelength). Mark the vertical distance from the mean line (x-axis) to a crest and label it a (amplitude). Draw an arrow along the x-axis marked v to show the direction of propagation, and a small vertical double-arrow at one particle to show its up-down oscillation direction.

Standing waves on a string fixed at both ends — first three harmonics

Draw three separate horizontal strings of the same length L, fixed (dots) at both ends. First: one loop (n=1) — label nodes N at both ends, one antinode A in the middle; write ν1 = v/2L. Second: two loops (n=2) with 3 nodes and 2 antinodes; write ν2 = 2ν1. Third: three loops (n=3) with 4 nodes and 3 antinodes; write ν3 = 3ν1. Clearly mark every N and A.

Normal modes of closed and open organ pipes

Draw two vertical tubes side by side. Left tube: closed at bottom (draw a node N there), open at top (antinode A); sketch the fundamental as half a loop, L = λ/4, and label ν1 = v/4L; note only odd harmonics exist. Right tube: open at both ends (antinode A at top and bottom); sketch the fundamental as a full loop between two antinodes, L = λ/2, label ν1 = v/2L; note all harmonics exist.

Formation of beats

Draw two sine waves of slightly different frequency, one above the other, both starting in phase at t = 0 and slowly drifting out of phase. Below both, draw the resultant (sum) waveform showing an amplitude that swells and shrinks periodically — a fast oscillation inside a slowly varying envelope. Mark two successive points of maximum loudness and label the time between them as 1/Δν.

TextbookNCERT questions the board keeps asking

14.1 A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If a transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?3 m

Mass per unit length μ = mass/length = 2.50/20.0 = 0.125 kg m−1.

Wave speed v = √(T/μ) = √(200/0.125) = √1600 = 40 m s−1.

Time to reach the other end t = length/v = 20.0/40 = 0.5 s.

14.2 A steel wire 12.0 m long has a mass of 2.10 kg. What tension should the wire be under so that the speed of a transverse wave on it equals the speed of sound in dry air at 20°C, taken as 343 m s−1?3 m

μ = 2.10/12.0 = 0.175 kg m−1.

v = √(T/μ) ⇒ T = v²μ = (343)² × 0.175 = 117649 × 0.175 ≈ 2.06 × 104 N.

14.3 A stone is dropped from the top of a tower 300 m high, and it splashes into the water of a pond near the base of the tower. When is the splash heard at the top, given that the speed of sound in air is 340 m s−1? Take g = 9.8 m s−2.3 m

Time for the stone to fall: h = ½gt1² ⇒ t1 = √(2h/g) = √(2×300/9.8) = √61.2 ≈ 7.82 s.

Time for sound to travel back up: t2 = h/v = 300/340 ≈ 0.88 s.

Total time = t1 + t2 ≈ 7.82 + 0.88 = 8.71 s.

14.4 A wave travelling along a string is described by y(x, t) = 0.005 cos(80x − 3t), where all quantities are in SI units. Find (i) the amplitude, wavelength and period, and (ii) the speed of the wave.3 m

Comparing with y = a cos(kx − ωt): a = 0.005 m = 5 mm; k = 80 rad m−1 ⇒ λ = 2π/k = 2π/80 ≈ 0.0785 m; ω = 3 rad s−1 ⇒ T = 2π/ω = 2π/3 ≈ 2.09 s.

Speed v = ω/k = 3/80 = 0.0375 m s−1 (= 3.75 cm s−1).

14.5 Derive the expression for the standing wave produced by two identical waves travelling in opposite directions, and use it to find the positions of the nodes and antinodes.5 m

Let y1 = a sin(kx − ωt) and y2 = a sin(kx + ωt). By superposition, y = y1 + y2 = a[sin(kx−ωt) + sin(kx+ωt)] = 2a sin(kx) cos(ωt), using sin A + sin B = 2 sin((A+B)/2)cos((A−B)/2).

Nodes: sin(kx) = 0 ⇒ kx = nπ ⇒ x = nλ/2 (n = 0, 1, 2, ...).

Antinodes: sin(kx) = ±1 ⇒ kx = (2n+1)π/2 ⇒ x = (2n+1)λ/4.

14.6 Two tuning forks A and B, when sounded together, produce 4 beats per second. Fork A has a known frequency of 256 Hz. When fork B is loaded with a little wax, the number of beats per second increases to 6. Find the original frequency of fork B.3 m

Before loading: |256 − νB| = 4 ⇒ νB = 252 Hz or 260 Hz.

Loading with wax always lowers the frequency of B. If νB were 260 Hz, loading would move it closer to 256 Hz, so beats should decrease — but beats increased, so this case is rejected.

If νB were 252 Hz, loading lowers it further away from 256 Hz, so beats increase — consistent with the data.

Original frequency of B = 252 Hz.

14.7 A pipe closed at one end resonates with a tuning fork of frequency 500 Hz at successive air-column lengths of 17.0 cm and 51.0 cm. Find the speed of sound in air.3 m

Consecutive resonance lengths of a closed pipe differ by λ/2 (successive odd-harmonic lengths L = λ/4, 3λ/4, 5λ/4, ...).

λ/2 = 51.0 − 17.0 = 34.0 cm ⇒ λ = 68.0 cm = 0.68 m.

v = νλ = 500 × 0.68 = 340 m s−1.

14.8 A train, standing at a station, blows a whistle of frequency 400 Hz in still air. What is the frequency of the whistle heard by a platform observer as the train (i) approaches and (ii) recedes from the platform with a speed of 10 m s−1? Take speed of sound in air as 340 m s−1.3 m

(i) Approaching: ν' = νv/(v−vs) = 400×340/(340−10) = 136000/330 ≈ 412.1 Hz.

(ii) Receding: ν' = νv/(v+vs) = 400×340/(340+10) = 136000/350 ≈ 388.6 Hz.

14.9 Using dimensional analysis, derive the expression for the speed of a transverse wave on a stretched string in terms of the tension T and the mass per unit length μ.3 m

Assume v = C Taμb, where C is a dimensionless constant. Dimensions: [v] = LT−1, [T] = MLT−2, [μ] = ML−1.

LT−1 = (MLT−2)a(ML−1)b = Ma+bLa−bT−2a.

Equating powers: a + b = 0; a − b = 1; −2a = −1 ⇒ a = ½, b = −½.

So v = C√(T/μ); the full mechanical derivation shows C = 1, giving v = √(T/μ).

14.10 Explain why Newton's formula for the speed of sound in air needed Laplace's correction, and derive the corrected expression.5 m

Newton assumed sound propagates isothermally, so PV = constant, giving the bulk modulus Bisothermal = P, and v = √(P/ρ). This underestimates the measured speed of sound by about 15%.

Laplace pointed out that compressions and rarefactions in sound happen very fast, so there is no time for heat exchange between them — the process is adiabatic, PVγ = constant, giving the adiabatic bulk modulus Badiabatic = γP.

Substituting: v = √(γP/ρ), with γ = Cp/Cv = 1.4 for air, which matches the experimental value of about 332 m s−1 closely.

14.11 State the differences between a progressive (travelling) wave and a stationary (standing) wave.2 m

A progressive wave transports energy and its waveform moves through the medium with speed v; every particle has the same amplitude but a phase that depends on its position. A stationary wave does not transport energy or move; it has fixed nodes (zero amplitude) and antinodes (maximum amplitude), and all particles between two nodes vibrate in the same phase but with different, position-dependent amplitudes.

14.12 Distinguish between wave speed and particle speed for a transverse wave on a string.2 m

Wave speed v = νλ = ω/k is the constant rate at which the waveform (a crest or trough) advances through the medium; it depends only on the properties of the string (T and μ), not on the wave's amplitude or frequency.

Particle speed is ∂y/∂t at a fixed x — the speed of a single particle of the string as it oscillates up and down. It varies continuously between 0 (at maximum displacement) and aω (while crossing the mean position), and it does depend on the amplitude and frequency of the wave.

Previous years and frequently askedBoard questions with model answers

MCQ Sound waves in air are:
  1. Transverse waves
  2. Longitudinal waves
  3. Electromagnetic waves
  4. Neither transverse nor longitudinal
Frequently asked1 m
Answer: (b)
Sound in air propagates as compressions and rarefactions, with particles oscillating along the direction of propagation — this is a longitudinal wave.
MCQ A transverse wave is described by y = a sin(kx − ωt). The wave is travelling in the:
  1. +x direction
  2. −x direction
  3. +y direction
  4. −y direction
Frequently asked1 m
Answer: (a)
A wave written with (kx − ωt) inside the sine travels in the +x direction; (kx + ωt) would indicate the −x direction.
MCQ The speed of sound in a gas, according to Laplace's correction, is given by:
  1. v = √(P/ρ)
  2. v = √(γP/ρ)
  3. v = √(ρ/γP)
  4. v = √(γρ/P)
Frequently asked1 m
Answer: (b)
Laplace treated sound propagation as adiabatic, giving v = √(γP/ρ), which matches the experimentally measured speed of sound.
MCQ In a stationary wave, the distance between a node and the next antinode is:
  1. λ/2
  2. λ/4
  3. λ
  4. 3λ/4
Frequently asked1 m
Answer: (b)
Nodes and antinodes in a standing wave occur alternately at intervals of λ/4.
MCQ A closed organ pipe of length L resonates only at frequencies that are:
  1. All integer multiples of v/2L
  2. Only odd multiples of v/4L
  3. Only even multiples of v/4L
  4. All integer multiples of v/4L
Frequently asked1 m
Answer: (b)
A closed pipe has a node at the closed end and an antinode at the open end, which restricts the allowed modes to L = (2n−1)λ/4, giving only odd harmonics ν = (2n−1)v/4L.
MCQ Two sound waves of frequencies 256 Hz and 260 Hz are sounded together. The number of beats heard per second is:
  1. 2
  2. 4
  3. 258
  4. 516
Frequently asked1 m
Answer: (b)
Beat frequency Δν = |ν1 − ν2| = |260 − 256| = 4 Hz.
Assertion–Reason Assertion (A): Sound waves cannot be polarised. Reason (R): Sound waves are longitudinal waves, and the oscillation is entirely along the direction of propagation.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
Polarisation restricts the oscillation direction, which is only meaningful when that direction is perpendicular to propagation (a transverse wave). Since sound oscillates only along the direction of travel, it cannot be polarised — R correctly explains A.
Assertion–Reason Assertion (A): The frequency of a wave does not change when it passes from one medium to another. Reason (R): Frequency is a property of the source producing the wave, not of the medium through which it travels.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
Frequency is fixed by the rate of oscillation of the source, so it stays constant as the wave crosses a boundary; the wave speed and wavelength change instead, since they depend on the medium's properties. R correctly explains A.
Assertion–Reason Assertion (A): A stretched string can produce all harmonics of its fundamental frequency. Reason (R): Both ends of a stretched string fixed at both ends are nodes, which allows the length to be an integer number of half-wavelengths.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
With both ends as nodes, L = nλ/2 for any integer n, so every integer multiple of the fundamental (ν = v/2L) is an allowed mode — all harmonics are present. R correctly explains A.
Very short answer Why can transverse mechanical waves not travel through a gas, while longitudinal waves can?Frequently asked2 m
Transverse waves require the medium to sustain a shearing stress, so that a displaced layer pulls its neighbouring layer sideways with it. Gases (and liquids) have no rigidity — they offer no resistance to shear — so a transverse disturbance cannot be transmitted. Longitudinal waves need only volume elasticity (the medium's resistance to compression), which every gas possesses, so sound travels through gases perfectly well.
Very short answer Write the differential relation between wave speed v, angular frequency ω and angular wave number k, and state what each symbol means.Frequently asked2 m
v = ω/k. Here v is the speed at which the waveform (a point of constant phase) advances through the medium, ω = 2πν is the angular frequency describing how fast the phase changes with time, and k = 2π/λ is the angular wave number describing how fast the phase changes with position.
Very short answer State the principle of superposition of waves.Frequently asked2 m
When two or more waves travel through the same region of a medium simultaneously, the resultant displacement of a particle of the medium at any point and instant is the algebraic sum of the displacements that each wave would have produced there acting alone: y = y1 + y2 + y3 + ...
Very short answer What is meant by the fundamental frequency and the first overtone of a vibrating string?Frequently asked2 m
The fundamental frequency (or first harmonic) is the lowest frequency at which a string fixed at both ends can vibrate in a standing-wave mode, ν1 = v/2L, with just one antinode between the two end nodes. The first overtone is the next allowed mode above the fundamental, ν2 = 2ν1 (the second harmonic), with two antinodes.
Short answer A wave travelling on a string is described by y(x, t) = 0.02 sin(1.5x − 400t), all quantities in SI units. Find the amplitude, wavelength, frequency and speed of the wave.Frequently asked3 m
Comparing with y = a sin(kx − ωt): a = 0.02 m; k = 1.5 rad/m ⇒ λ = 2π/k = 2π/1.5 ≈ 4.19 m; ω = 400 rad/s ⇒ ν = ω/2π = 400/6.28 ≈ 63.7 Hz. Speed v = ω/k = 400/1.5 ≈ 266.7 m/s (equivalently v = νλ ≈ 63.7 × 4.19 ≈ 266.9 m/s).
Short answer Two tuning forks of frequencies 320 Hz and 326 Hz are sounded together. (i) What is the beat frequency? (ii) What is the time interval between two successive maxima of loudness?Frequently asked3 m
(i) Beat frequency Δν = |326 − 320| = 6 Hz. (ii) Time between successive maxima of loudness = 1/Δν = 1/6 ≈ 0.167 s.
Short answer An open organ pipe has a length of 0.5 m. Taking the speed of sound in air as 340 m/s, find its fundamental frequency and the frequency of its first overtone.Frequently asked3 m
For an open pipe, ν(n) = nv/2L. Fundamental (n = 1): ν1 = 340/(2 × 0.5) = 340 Hz. First overtone is the second harmonic (n = 2): ν2 = 2ν1 = 680 Hz.
Short answer A source of sound emits a note of frequency 300 Hz. Find the frequency heard by an observer moving away from the stationary source at 20 m/s. Take the speed of sound as 340 m/s.Frequently asked3 m
Observer moving away from a stationary source: ν' = ν(v − vo)/v = 300 × (340 − 20)/340 = 300 × 320/340 ≈ 282.4 Hz.
Long answer Derive the expression for the standing wave set up on a string, and use it to obtain the frequencies of the normal modes of a string of length L fixed at both ends.Frequently asked5 m
Let two identical waves travelling in opposite directions be y1 = a sin(kx − ωt) and y2 = a sin(kx + ωt). By the principle of superposition, y = y1 + y2 = a[sin(kx − ωt) + sin(kx + ωt)]. Using sin A + sin B = 2 sin((A+B)/2) cos((A−B)/2), this gives y(x, t) = 2a sin(kx) cos(ωt) — a standing wave, since the space and time dependence have separated. Nodes occur where sin(kx) = 0, i.e. kx = nπ; antinodes where sin(kx) = ±1. For a string fixed at both ends (x = 0 and x = L), both ends must be nodes: sin(kL) = 0 ⇒ kL = nπ (n = 1, 2, 3, ...) ⇒ k = nπ/L ⇒ λ = 2L/n. Since v = νλ, the allowed frequencies are ν(n) = nv/2L, n = 1, 2, 3, ... The lowest, ν1 = v/2L, is the fundamental; ν2 = 2ν1, ν3 = 3ν1, ... are the higher harmonics, all of which the string supports.
Long answer Derive an expression for the apparent frequency heard by a stationary observer when a source of sound moves towards the observer with speed vs, and when it moves away. Take the speed of sound as v and the actual frequency of the source as ν.Frequently asked5 m
Let the source emit crests at intervals of the period T = 1/ν. When the source moves toward a stationary observer with speed vs, in the time T between the emission of two successive crests the source moves a distance vsT closer to the observer, so the second crest is emitted from a point vsT nearer than the first. The distance between successive crests (the apparent wavelength) is therefore reduced: λ' = vT − vsT = (v − vs)T = (v − vs)/ν. The observer hears frequency ν' = v/λ' = νv/(v − vs), which is greater than ν since the denominator is smaller. When the source instead moves away from the observer with speed vs, the same reasoning gives an apparent wavelength that is stretched: λ' = (v + vs)/ν, so the observed frequency is ν' = νv/(v + vs), which is less than ν.
Long answer What are beats? Derive an expression for the beat frequency produced when two sound waves of nearly equal frequencies ν1 and ν2 superpose.Frequently asked5 m
When two waves of slightly different frequency are heard together, the loudness of the resultant sound rises and falls periodically; each such cycle of rise and fall is called a beat. Consider two waves of equal amplitude a reaching a fixed point: y1 = a cos(ω1t) and y2 = a cos(ω2t), where ω1 = 2πν1 and ω2 = 2πν2. By superposition, y = y1 + y2 = 2a cos((ω1 − ω2)t/2) cos((ω1 + ω2)t/2), using the sum-to-product identity for cosines. This resultant behaves like a wave of frequency (ν1 + ν2)/2 (the average pitch heard) whose amplitude is modulated by the slowly varying envelope 2a cos((ω1 − ω2)t/2). The loudness is maximum whenever this envelope reaches its largest magnitude, and working through the timing of these maxima shows that they occur at a rate equal to the difference of the two frequencies: beat frequency Δν = |ν1 − ν2|.
Case-based Read the passage and answer the questions that follow.
A sonometer wire of length 1.0 m is stretched between two fixed points. The wire has a mass per unit length of 4.0 × 10−3 kg/m and is under a tension of 64 N. The wire is plucked and made to vibrate in its fundamental mode.
(i) Find the speed of the transverse wave on the wire.
(ii) Find the fundamental frequency of vibration of the wire.
(iii) If the wire vibrates in its first overtone instead, what would be the frequency?
(iv) How many nodes and antinodes are present when the wire vibrates in its first overtone?
Frequently asked4 m
(i) v = √(T/μ) = √(64/4.0×10⁻³) = √16000 ≈ 126.5 m/s.
(ii) Fundamental frequency ν1 = v/2L = 126.5/(2×1.0) ≈ 63.2 Hz.
(iii) First overtone is the second harmonic: ν2 = 2ν1 ≈ 126.5 Hz.
(iv) In the first overtone (two loops) there are 3 nodes (including the two fixed ends) and 2 antinodes.
Case-based Read the passage and answer the questions that follow.
A police car sounding a siren of frequency 500 Hz moves towards a stationary pedestrian at a speed of 20 m/s. The pedestrian is standing still on the pavement. Take the speed of sound in air as 340 m/s.
(i) State which case of the Doppler effect formula applies here.
(ii) Calculate the frequency of the siren as heard by the pedestrian while the car approaches.
(iii) After the car crosses the pedestrian and moves away at the same speed, calculate the frequency now heard.
(iv) Explain, without further calculation, why the pedestrian perceives a sudden drop in pitch as the car passes.
Frequently asked4 m
(i) Source moving, observer stationary — the case ν' = νv/(v ∓ vs).
(ii) Approaching: ν' = νv/(v − vs) = 500×340/(340−20) = 170000/320 ≈ 531.3 Hz.
(iii) Receding: ν' = νv/(v + vs) = 500×340/(340+20) = 170000/360 ≈ 472.2 Hz.
(iv) The observed frequency jumps from the higher value (531.3 Hz) to the lower value (472.2 Hz) essentially as the car passes, because the direction of the source's velocity relative to the observer reverses from 'approaching' to 'receding', flipping the sign in the denominator of the Doppler formula.

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Confusing wave speed (v = νλ, fixed by the medium) with particle speed (∂y/∂t, which varies over a cycle) — these answer different questions and are often mixed up in numericals.
  • Forgetting the sign convention in y = a sin(kx ∓ ωt): a minus sign before ωt means the wave travels in +x, a plus sign means −x. Many students get this backwards.
  • Assuming a closed pipe supports all harmonics like an open pipe or a string — a closed pipe supports only odd harmonics, and this is frequently tested.
  • Marking a node at the open end of a pipe or the free end of a string — the open/free end is always an antinode, not a node.
  • Using Newton's formula v = √(P/ρ) for the speed of sound instead of Laplace's corrected v = √(γP/ρ) — always include γ for a gas.
  • Applying the wrong sign in the Doppler formula — remember 'closing the distance always raises the observed frequency' and reason it out physically rather than guessing the sign.
  • Writing the beat frequency as the sum ν1 + ν2 instead of the difference |ν1 − ν2| — beats depend on the difference in frequency, not the sum.
  • Treating the amplitude of a standing wave, 2a sin(kx), as constant — it varies with position; only in a travelling wave is the amplitude the same at every point.

How to prepareStrategy for this chapter

This chapter rewards careful bookkeeping more than raw memorisation. Start by getting completely comfortable with the wave equation y = a sin(kx − ωt + φ): practise pulling a, k, ω and φ out of any numerical wave equation you are given, since half of every numerical in this chapter starts there. Learn the three speed formulas (string, fluid, gas with Laplace's correction) as a family — they all come from the same 'restoring force over inertia, inside a square root' pattern, which makes them much easier to recall under pressure than as three unrelated facts.

The two long-answer derivations that appear most often are: standing waves from superposition (leading to node/antinode positions and the normal-mode frequencies of a string or pipe), and the Doppler effect for a source moving towards a stationary observer. Write each of these out fully at least three times from a blank page before the exam — the marking scheme gives points for each individual step, not just the final formula.

For numericals, beats and Doppler-effect problems come up the most, so drill several of each with different combinations of 'source moving / observer moving / both moving, towards / away'. Keep the harmonic-series diagrams for a string, an open pipe and a closed pipe fresh in memory, since a diagram-based short question is common and quick marks are lost by mislabelling nodes and antinodes.

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