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Class 11 Physics · NCERT Chapter 14

Waves Class 11 Physics Notes

Complete, exam-ready notes on waves: transverse and longitudinal wave motion, the wave equation and superposition, the speed of sound with Newton's formula and Laplace correction, reflection, standing waves in strings and organ pipes, beats and the Doppler effect — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.

ChapterWavesClassClass 11SubjectPhysicsBoardCBSEExamsCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is the Doppler effect in sound?

The apparent change in the frequency of a wave due to relative motion between the source and the observer: f′ = f(v ± v₀)/(v ∓ vₛ), using the upper sign when source or observer moves towards the other. The frequency rises on approach and falls on recession.

Wave Motion and Its Types

Wave

A wave is a disturbance that travels through a medium, transferring energy and momentum without the permanent transfer of matter — the particles of the medium oscillate about their mean positions while the disturbance moves on.

  • Transverse waves: particles vibrate perpendicular to the direction of propagation (e.g. waves on a string, light). Can travel through solids and the surface of liquids.
  • Longitudinal waves: particles vibrate parallel to the direction of propagation, forming compressions and rarefactions (e.g. sound). Can travel through solids, liquids and gases.
  • Mechanical waves need a medium; electromagnetic waves (light, radio) do not.
  • Categorised by wavelength: water ripples (tiny λ), sound (~15 m), light (~10⁻⁷ m).

Energy travels, matter does not

In a wave each particle merely oscillates about its mean position. What propagates is the disturbance — carrying energy and momentum — which is why the wave itself is the messenger.

MCQ Questions

1In a transverse wave, the particles of the medium vibrate —

2Sound waves are —

3A wave transfers —

4Which of the following requires a material medium to travel? —

1 Mark Questions

  1. Define a wave. Distinguish between mechanical and electromagnetic waves.
  2. Give one example each of a transverse and a longitudinal wave.
  3. In what direction do particles vibrate in a longitudinal wave?

2 Mark Questions

  1. Distinguish between transverse and longitudinal waves with respect to particle displacement.
  2. Explain why sound cannot travel in vacuum but light can.
  3. Why can transverse waves travel in solids but generally not in gases?

3 Mark Questions

  1. A wave travels the length of a 20 m string in 5 s. If consecutive crests are 0.4 m apart, find the frequency and period of the wave.
  2. Describe how a particle at the surface of water vibrates as a ripple passes, and how that differs from the motion of the ripple itself.

Characteristics of a Wave

Wave parameters

Amplitude A is the maximum displacement of a particle from equilibrium. Wavelength λ is the distance between two successive crests (or any two corresponding points). The period T is the time for one complete oscillation and the frequency f = 1/T is the number of oscillations per second.

v=fλ=λTv = f\lambda = \frac{\lambda}{T}
Wave speed relation

All particles of a wave oscillate with the same frequency and amplitude but with a continuously varying phase. The frequency is fixed by the source; the speed is fixed by the medium; the wavelength adjusts so that v = fλ. The unit of frequency is the hertz (Hz = s⁻¹).

  • Tuning a radio dial selects the frequency of the transmitted signal.
  • A 100 Hz sound in air at 20 °C (v ≈ 343 m/s) has wavelength about 3.43 m.
  • Speed does not change with frequency of the wave in a given medium (no dispersion for sound in air).

MCQ Questions

1The distance between two successive crests of a wave is called the —

2If the frequency of a wave is doubled while its speed stays constant, the wavelength —

3A sound wave of frequency 340 Hz travels with a speed of 680 m/s. Its wavelength is —

4The frequency of a mechanical wave is determined by the —

1 Mark Questions

  1. Define wavelength and frequency of a wave.
  2. Write the relation connecting wave speed, frequency and wavelength.
  3. Define the hertz.

2 Mark Questions

  1. Derive the relation v = fλ starting from the definition of wave speed.
  2. A tuning fork of frequency 512 Hz produces a wave in air with speed 345 m/s. Find the wavelength.
  3. Why is the frequency of a wave fixed by its source and not by the medium?

3 Mark Questions

  1. A source emits a wave of frequency 500 Hz. If the wave speed in the medium is 1500 m/s, find the wavelength and the time period, and state how far the wavefront travels in 2 s.
  2. Explain the meaning of amplitude, frequency, wavelength and speed of a wave, and give their SI units.

The Speed of Sound

Sound travels through a medium as a longitudinal wave, with a speed set by the elastic and inertial properties of the medium: the faster the medium returns to equilibrium and the lighter it is, the faster sound moves.

Newton's formula and Laplace correction

Newton assumed the compressions and rarefactions of a sound wave occur isothermally, giving v = √(P/ρ) ≈ 280 m s⁻¹ for air — well below the measured 332 m s⁻¹. Laplace showed the process is adiabatic, replacing P by γP: v = √(γP/ρ) with γ = Cp/Cv ≈ 1.4, giving ≈ 332 m s⁻¹.

v=Bρvgas=γPρv = \sqrt{\frac{B}{\rho}} \qquad v_{\text{gas}} = \sqrt{\frac{\gamma P}{\rho}}
Speed of sound in a medium
  • Speed order: solids > liquids > gases (bulk modulus divided by density).
  • In a gas, v ∝ √T — the speed rises with temperature (≈ 0.6 m/s per degree Celsius).
  • At constant temperature, speed is independent of pressure of the gas.
  • Moist air is denser-lighter: humidity increases the speed slightly because water-vapour molecules are light.

Air vs water vs steel

Sound is fastest in solids, then liquids, then gases: about 343 m/s in air, ~1500 m/s in water and ~5000 m/s in steel. This is why a railway track carries the rumble of a far-off train long before the air-borne sound arrives.

MCQ Questions

1Sound travels fastest in —

2Newton's formula for the speed of sound in air gave a value that was —

3The Laplace correction factor γ for air is approximately —

4At a fixed temperature, the speed of sound in a gas is —

1 Mark Questions

  1. State Newton's formula for the speed of sound in a gas.
  2. What correction did Laplace make and why?
  3. In which medium is the speed of sound maximum — solid, liquid or gas?

2 Mark Questions

  1. Explain the Laplace correction and derive v = √(γP/ρ).
  2. The speed of sound in air at 0 °C is 331 m/s. Estimate its speed at 20 °C.
  3. Why does the speed of sound increase with humidity?

3 Mark Questions

  1. Derive an expression for the speed of sound in a gas using Newton's formula and apply the Laplace correction to obtain the standard result.
  2. Compare the speed of sound in solids, liquids and gases, explaining the physical reasons for the order.
  3. Show that the speed of sound in a gas is proportional to the square root of its absolute temperature.

Progressive Waves and Superposition

Progressive (travelling) wave

A wave that travels through a medium, carrying energy away from the source, with a shape that does not change as it moves. A harmonic progressive wave is described by y = A sin(kx − ωt), where k = 2π/λ is the wave number and ω = 2πf is the angular frequency.

y=Asin⁡(kx−ωt)v=ωk=fλy = A\sin(kx - \omega t) \qquad v = \frac{\omega}{k} = f\lambda
Equation of a progressive wave and its wave speed

The sign between kx and ωt fixes the direction: (kx − ωt) is a wave moving in the +x direction, (kx + ωt) in the −x direction. The phase (kx − ωt) is what travels; its constancy, kx − ωt = constant, gives dx/dt = ω/k = v.

Superposition principle

When two or more waves overlap in a region, the resultant displacement at every point is the vector (algebraic) sum of the displacements each wave would produce alone. This is the basis of interference, beats and standing waves.

  • Phase difference Δφ and path difference Δx are related by Δφ = (2π/λ)Δx.
  • A path difference of λ gives a phase difference of 2π (one full cycle).
  • Phase and time difference: Δφ = ωΔt.

MCQ Questions

1In the wave equation y = A sin(kx − ωt), the quantity k is the —

2The equation y = A sin(kx + ωt) represents a wave travelling —

3A phase difference of 2π rad corresponds to a path difference of —

4According to the superposition principle, the resultant displacement of two overlapping waves at a point is the —

1 Mark Questions

  1. Define a progressive wave.
  2. State the superposition principle for waves.
  3. What is the wave number? Write its relation with wavelength.

2 Mark Questions

  1. Write the general equation of a one-dimensional progressive wave and identify each symbol.
  2. Show that the wave speed v = ω/k follows from the constancy of the phase.
  3. Derive the relation between phase difference and path difference.

3 Mark Questions

  1. A progressive wave is described by y = 0.05 sin π(200t − 2x) m. Find its (a) frequency, (b) wavelength and (c) speed of propagation.
  2. State and explain the principle of superposition of waves, and give three phenomena that it explains.

Reflection and Standing Waves

Reflection of a wave

When a wave hits a boundary it is reflected. At a rigid (fixed) boundary the reflected pulse is inverted — a phase change of π. At a free boundary it returns un-inverted with no phase change, because the boundary cannot exert a transverse force on the string.

When two identical harmonic waves travel in opposite directions along a string, their superposition gives a stationary (standing) wave. The pattern does not travel: energy is stored, not transported, and the medium divides into nodes (zero displacement) and antinodes (maximum displacement).

y(x,t)=2Acos⁡(kx) sin⁡(ωt)y(x,t) = 2A\cos(kx)\,\sin(\omega t)
Standing wave from two opposite travelling waves
  • Nodes occur where cos(kx) = 0; antinodes where cos(kx) = ±1.
  • Consecutive nodes (or antinodes) are λ/2 apart.
  • The standing wave does not transfer energy along the string.
  • A string fixed at both ends forms a standing wave with nodes at both ends.

Phase change on reflection

Fixed end — inverted pulse, phase change π. Free end — same orientation, no phase change. This one fact explains the node at a fixed end and the antinode at a free end.

MCQ Questions

1A wave pulse reflected from a rigid, fixed boundary undergoes a phase change of —

2In a standing wave, the distance between two consecutive nodes is —

3At an antinode of a stationary wave on a string, the amplitude of vibration is —

4A standing wave is formed by the superposition of two waves having —

1 Mark Questions

  1. Define a stationary or standing wave.
  2. What is a node? What is an antinode?
  3. What phase change does a wave suffer on reflection at a fixed boundary?

2 Mark Questions

  1. Show that consecutive nodes (or antinodes) of a stationary wave are λ/2 apart.
  2. Obtain the equation of a standing wave from two progressive waves of equal amplitude and frequency travelling in opposite directions.
  3. Explain why a stationary wave does not transport energy along the medium.

3 Mark Questions

  1. Describe the reflection of a travelling wave at a rigid boundary and at a free boundary, stating the phase change in each case with the reasoning.
  2. Two waves y₁ = A sin(kx − ωt) and y₂ = A sin(kx + ωt) superpose on a string. Obtain the resultant wave and identify the positions of nodes and antinodes.

Standing Waves in Strings and Organ Pipes

A stretched string fixed at both ends supports standing waves only at certain frequencies, called normal modes. The ends are nodes, so an integral number of half wavelengths must fit into the length: L = nλₙ/2, giving fₙ = nv/2L for n = 1, 2, 3, …

fn=nv2L(n=1,2,3,… )f_n = \frac{nv}{2L} \qquad (n = 1, 2, 3, \dots)
Frequencies of a string fixed at both ends
  • Fundamental (first harmonic): f₁ = v/2L.
  • All harmonics are present: f₂ = 2f₁, f₃ = 3f₁, … — the string can vibrate in any of these modes simultaneously.
  • The wave speed on a string is v = √(T/μ), where T is the tension and μ the mass per unit length.

Organ pipes

An open pipe has displacement antinodes at both ends and produces all harmonics: fₙ = nv/2L. A closed pipe has a node at the closed end and an antinode at the open end, so it produces only the odd harmonics: fₙ = (2n − 1)v/4L.

fn=(2n−1)v4Lclosed pipef_n = \frac{(2n-1)v}{4L} \qquad \text{closed pipe}
Frequencies of a closed organ pipe

Open vs closed pipes

Open pipe: all harmonics (f₁, 2f₁, 3f₁, …). Closed pipe: only odd harmonics (f₁, 3f₁, 5f₁, …) — its fundamental is one octave lower in timber and only about 7–8 dB quieter.

MCQ Questions

1The fundamental frequency of a string fixed at both ends of length L is —

2A closed organ pipe produces —

3To double the fundamental frequency of a stretched string without changing the tension, the length must be —

4A string of length L vibrates in its second harmonic. The number of nodes (excluding the fixed end points being nodes) visible is —

5The wave speed on a stretched string is proportional to —

1 Mark Questions

  1. Write the expression for the fundamental frequency of a stretched string.
  2. State the frequency formula for an open organ pipe and a closed organ pipe.
  3. Why does a closed organ pipe produce only odd harmonics?

2 Mark Questions

  1. Derive the expression for the frequencies of the normal modes of a string fixed at both ends.
  2. A string of length 1 m has a fundamental frequency of 110 Hz. Find its wavelength and the speed of waves on the string.
  3. Show that the fundamental frequency of a closed pipe is v/4L and that only odd harmonics are possible.

3 Mark Questions

  1. Compare the harmonics produced by an open organ pipe and a closed organ pipe of the same length.
  2. A stretched wire of length 0.90 m and mass 3.6 × 10⁻³ kg is under a tension of 90 N. Calculate its fundamental frequency and the frequencies of its first three harmonics.
  3. A closed organ pipe and an open organ pipe of the same length produce their fundamentals. Find the ratio of their frequencies.

Beats

Beats

When two sound waves of slightly different frequencies f₁ and f₂ reach a point together, the loudness alternately rises and falls. This periodic waxing and waning of intensity is called beats, and the number of intensity maxima heard per second equals the difference of the two frequencies.

fb=∣f1−f2∣f_b = |f_1 - f_2|
Beat frequency

Physically, the two waves fall alternately in phase (constructive — loud, sound reinforced) and out of phase (destructive — faint). The resultant amplitude varies slowly at the beat frequency while the wave oscillates rapidly at the average frequency. Beats are heard clearly only when the two frequencies are close (difference ≲ 10 Hz).

  • Beats are a demonstration of the superposition principle.
  • They are used to tune musical instruments: adjust a string until the beeps against a standard tuning fork vanish.
  • A slightly loaded tuning fork shifts its frequency; a fork of known frequency can be calibrated by counting beats with another.

MCQ Questions

1Two tuning forks of 256 Hz and 260 Hz are sounded together. The beat frequency is —

2Beats are produced when two waves of —

3For beats to be distinctly heard, the frequency difference between the two sources should be —

4When a musician tunes a string by listening to beats against a tuning fork, she adjusts the tension until —

1 Mark Questions

  1. Define beats.
  2. Write the formula for the beat frequency when two waves of frequencies f₁ and f₂ superpose.
  3. Why are beats heard only when the two frequencies are close together?

2 Mark Questions

  1. Explain how beats arise from the principle of superposition.
  2. Two tuning forks A (300 Hz) and B produce 4 beats per second. What are the possible frequencies of B?
  3. How are beats used to tune musical instruments?

3 Mark Questions

  1. Derive an expression for the resultant intensity of two waves of slightly different frequencies and show that the intensity oscillates at the beat frequency.
  2. A fork of unknown frequency produces 5 beats/s with a 256 Hz fork. When the unknown fork is slightly loaded with wax, the beats drop to 3 beats/s. Find the unknown frequency and justify your answer.

The Doppler Effect

Doppler effect

The apparent change in the frequency of a wave when the source and the observer are in relative motion. When they approach each other the observed frequency increases; when they recede it decreases. For sound, v₀ and vₛ are measured relative to the medium.

f′=f v±vov∓vsf' = f\,\frac{v \pm v_o}{v \mp v_s}
Apparent frequency for sound (upper signs on approach)
  • Observer moving towards a stationary source: f′ = f(v + v₀)/v.
  • Source moving towards a stationary observer: f′ = fv/(v − vₛ).
  • Both moving towards each other: f′ = f(v + v₀)/(v − vₛ).
  • No shift when the relative velocity is zero, or when the motion is perpendicular to the line of sight (no radial component).

Sound vs light Doppler

Sound needs a medium, so the speeds are measured relative to the air. For light (in vacuum) there is no medium, and only the relative velocity matters. The Doppler shift of light is what makes receding galaxies appear redder.

Applications: radar and speed guns, Doppler echocardiography, tracking weather and storms, and measuring the recession of stars and galaxies in astronomy.

MCQ Questions

1When a sound source moves towards a stationary observer, the apparent frequency is —

2The Doppler effect in sound is observed when —

3A stationary source of frequency f is heard by an observer moving towards it with speed v₀. If v is the speed of sound, the apparent frequency observed is —

4The apparent frequency of a whistle of a train approaching a stationary observer —

1 Mark Questions

  1. State the Doppler effect for sound.
  2. Write the expression for apparent frequency when the observer moves towards a stationary source.
  3. What happens to the apparent frequency as a source recedes from a stationary observer?

2 Mark Questions

  1. Derive the expression f′ = fv/(v − vₛ) for a source moving towards a stationary observer.
  2. A whistle of frequency 400 Hz is sounded by a train moving at 30 m/s. For a stationary observer towards whom the train moves, find the apparent frequency. (Speed of sound = 330 m/s.)
  3. Give two applications of the Doppler effect in technology.

3 Mark Questions

  1. Derive a general expression for the apparent frequency of sound when both the source and the observer are moving, and discuss the sign conventions.
  2. A source of sound of frequency 500 Hz moves towards a stationary observer at 20 m/s while the observer walks towards the source at 5 m/s. Find the apparent frequency heard. (Speed of sound = 340 m/s.)

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Wave speed

v=fλ=λTv = f\lambda = \frac{\lambda}{T}

frequency by source, speed by medium.

Speed of sound in gas

v=γPρv = \sqrt{\frac{\gamma P}{\rho}}

γ = 1.4 for air; ~343 m/s at 20 °C.

Progressive wave

y=Asin⁡(kx−ωt)y = A\sin(kx - \omega t)

k = 2π/λ, ω = 2πf; v = ω/k.

Phase–path relation

Δϕ=2πλΔx\Delta\phi = \frac{2\pi}{\lambda}\Delta x

λ path difference ↔ 2π phase difference.

Standing wave

y=2Acos⁡(kx)sin⁡(ωt)y = 2A\cos(kx)\sin(\omega t)

nodes every λ/2.

String / open pipe frequencies

fn=nv2Lf_n = \frac{nv}{2L}

n = 1, 2, 3, … (all harmonics).

Closed pipe frequencies

fn=(2n−1)v4Lf_n = \frac{(2n-1)v}{4L}

only odd harmonics.

Beats

fb=∣f1−f2∣f_b = |f_1 - f_2|

intensity waxes and wanes at this rate.

Doppler effect

f′=f v±vov∓vsf' = f\,\frac{v \pm v_o}{v \mp v_s}

upper sign on approach; v₀, vₛ measured relative to the medium.

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • v = fλ; speed depends only on the medium, frequency only on the source.
  • Newton's isothermal value ≈ 280 m/s was corrected by Laplace to v = √(γP/ρ) ≈ 332 m/s.
  • Standing wave: nodes and antinodes λ/2 apart; string and open pipe give all harmonics, closed pipe only odd ones.
  • Beats occur only when the two frequencies are close; beat frequency = |f₁ − f₂|.
  • Doppler: frequency rises on approach, falls on recession; formula f′ = f(v ± v₀)/(v ∓ vₛ).
  • Phase difference Δφ = (2π/λ)Δx.

Solved problems

Worked examples

JEE / NEET-style numericals, solved step by step.

1

Wavelength of a tuning fork's sound

A tuning fork of frequency 512 Hz is sounded in air where sound travels at 345 m/s. Find the wavelength of the wave.

  1. λ = v/f = 345 ÷ 512 ≈ 0.674 m.
  2. The crests are about 67 cm apart in the air column.

Answer

≈ 0.674 m

2

Speed of sound using Laplace correction

Given P = 1.013 × 10⁵ N m⁻², ρ = 1.29 kg m⁻³ and γ = 1.4 for air, calculate the speed of sound.

  1. v = √(γP/ρ) = √(1.4 × 1.013 × 10⁵ ÷ 1.29).
  2. γP/ρ = 1.418 × 10⁵ / 1.29 ≈ 1.099 × 10⁵.
  3. v ≈ √(1.099 × 10⁵) ≈ 331.5 m s⁻¹.

Answer

≈ 332 m s⁻¹

3

Beats between two tuning forks

Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. Find the beat frequency and the time period of the beat.

  1. Beat frequency = |260 − 256| = 4 Hz.
  2. Time period of beats = 1/4 = 0.25 s.
  3. The intensity of the combined sound rises and falls four times each second.

Answer

4 beats per second; period 0.25 s

FAQ

Common questions

Why did Newton's formula underestimate the speed of sound and how was it corrected?

Newton assumed the compressions and rarefactions were isothermal, giving v = √(P/ρ) ≈ 280 m/s. Laplace pointed out the changes are adiabatic and rapid, replacing P by γP, so v = √(γP/ρ) ≈ 332 m/s — matching experiment.

What is the difference between a node and an antinode?

In a standing wave a node is a point of permanent zero displacement where waves meet in opposite phase, while an antinode is a point of maximum displacement. Nodal and antinodal points alternate, spaced λ/2 apart.

Why does a closed organ pipe produce only odd harmonics?

A closed pipe has a displacement node at the closed end and an antinode at the open end, so its length fits an odd number of quarter wavelengths, giving fₙ = (2n − 1)v/4L — only odd harmonics are possible.

What are beats, and how are they used?

Beats are the periodic waxing and waning of loudness heard when two sources of slightly different frequency (say 256 Hz and 260 Hz) sound together; the beat frequency is |f₁ − f₂|. Musicians use beats to tune instruments by minimising the beat rate.

When does the Doppler effect apply and what changes on approach?

It applies whenever source or observer moves relative to the medium (for sound) carrying waves. On approach the apparent frequency rises (f′ = f(v + v₀)/(v − vₛ)) and on recession it falls — the pitch you hear from a siren changes as it passes you.

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