Waves Class 11: Notes, Standing Waves, Beats and Solved Numericals

What Waves is about

Weightage: This chapter is part of Unit X, Oscillations and Waves, which the syllabus gives 10 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit X Oscillations and Waves

Key concepts

Mechanical waves: transverse and longitudinal

NCERT section 14.1 and 14.2

A wave is a disturbance that travels through a medium, carrying energy and momentum but not the medium itself. Each particle of the medium only oscillates about its mean position, passing the motion on to its neighbours through elastic forces.

• Transverse waves: particles oscillate perpendicular to the direction of travel; the wave has crests and troughs. They need a medium that resists shear, so they travel in solids and along strings but not through the bulk of liquids or gases.
• Longitudinal waves: particles oscillate along the direction of travel, forming compressions and rarefactions. They travel in solids, liquids and gases. Sound in air is longitudinal.

Water-surface waves are a mixture of both. Mechanical waves cannot pass through a vacuum; electromagnetic waves (studied in Class 12) can.

Worked example

Classify as transverse or longitudinal: (a) a wave on a guitar string, (b) sound in air, (c) a wave produced by pushing and pulling a long spring along its length. [1 mark]

Try it first, then show the solutionSolution

1. (a) The string moves sideways while the wave travels along it: transverse.
2. (b) and (c) The particles (or coils) move back and forth along the direction of travel: longitudinal.

Answer: (a) transverse (b) longitudinal (c) longitudinal

Displacement relation for a progressive wave

NCERT section 14.3

A harmonic wave travelling in the +x direction:
$y(x,t)=a\sin (kx-\omega t+\varphi )$
(a wave towards −x has $kx+\omega t$).

• a, amplitude: maximum displacement of a particle.
• (kx − ωt + φ), phase; φ is the initial phase.
• k, angular wave number: $k=\frac{2\pi }{\lambda }$ (rad m${}^{-1}$), where the wavelength λ is the distance between two nearest points in the same phase (crest to crest).
• ω, angular frequency: $\omega =\frac{2\pi }{T}=2\pi \nu$.

In one period T the wave moves forward by one wavelength, so the wave speed is
$v=\frac{\lambda }{T}=\nu \lambda =\frac{\omega }{k}$

Phase difference between two points Δx apart: $\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$.

Worked example

A wave is described by $y=0.03\sin (20x-600t)$ in SI units. Find its amplitude, wavelength, frequency and speed. [2 marks]

Try it first, then show the solutionSolution

1. a = 0.03 m; k = 20 rad m${}^{-1}$, so $\lambda =\frac{2\pi }{20}\approx 0.314m$
2. ω = 600 rad s${}^{-1}$, so $\nu =\frac{600}{2\pi }\approx 95.5Hz$; $v=\frac{\omega }{k}=\frac{600}{20}=30m{s}^{-1}$ (towards +x)

Answer: 3 cm; 0.314 m; 95.5 Hz; 30 m s${}^{-1}$

Speed of a transverse wave on a stretched string

NCERT section 14.4.1

The speed of a wave depends only on the properties of the medium: an inertial property (which resists motion) and an elastic property (which restores). For a string, these are the mass per unit length μ and the tension T. Dimensional analysis (and a full derivation) gives
$v=\sqrt{\frac{T}{\mu }}$

A tighter or lighter string carries waves faster. The frequency is set by the source; the speed by the medium; the wavelength then follows from $\lambda =\frac{v}{\nu }$.

Worked example

A string of mass per unit length 0.01 kg m${}^{-1}$ is under a tension of 100 N. Find the speed of transverse waves on it. [1 mark]

Try it first, then show the solutionSolution

1. $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{100}{0.01}}$
2. $v=100m{s}^{-1}$

Answer: 100 m s${}^{-1}$

Speed of a longitudinal wave: sound

NCERT section 14.4.2

For a longitudinal wave, the elastic property is the modulus for compression:

• In a fluid: $v=\sqrt{\frac{B}{\rho }}$ (B = bulk modulus).
• In a thin solid rod: $v=\sqrt{\frac{Y}{\rho }}$.

Sound is fastest in solids, slower in liquids and slowest in gases.

Newton's formula assumed the compressions in air are isothermal, so B = P: $v=\sqrt{\frac{P}{\rho }}\approx 280$ m s${}^{-1}$ at STP, about 15% too low.

Laplace's correction: compressions and rarefactions are so fast that there is no time for heat to flow, so they are adiabatic, with $B=\gamma P$:
$v=\sqrt{\frac{\gamma P}{\rho }}\approx 331m{s}^{-1}$ at STP (γ = 1.4)

Consequences: for an ideal gas, v does not depend on pressure at constant temperature, but $v\propto \sqrt{T}$; it is greater in humid air (lower density) and in lighter gases.

Worked example

Find the speed of sound in hydrogen at STP using Laplace's formula. (P = $1.013\times 10^5$ Pa, ρ = 0.0899 kg m${}^{-3}$, γ = 1.4) [2 marks]

Try it first, then show the solutionSolution

1. $v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{1.4\times 1.013\times 10^5}{0.0899}}$
2. $v=\sqrt{1.58\times 10^6}\approx 1.26\times 10^{3}m{s}^{-1}$, nearly four times the speed in air (hydrogen is much less dense)

Answer: About 1260 m s${}^{-1}$

Principle of superposition and interference

NCERT section 14.5

When two or more waves overlap, the resultant displacement at any point is the algebraic sum of the displacements due to each wave:
$y=y_1+y_2+\dots$
After overlapping, each wave carries on unchanged.

For two waves of the same amplitude a and frequency travelling the same way with phase difference φ:
$y=\left[2a\cos \frac{\varphi }{2}\right]\sin \left(kx-\omega t+\frac{\varphi }{2}\right)$

• φ = 0 (in phase): amplitude 2a, constructive interference.
• φ = π (opposite phase): amplitude 0, destructive interference.

Superposition explains standing waves, beats and, in Class 12, interference of light.

Worked example

Two waves of amplitude 3 mm and the same frequency travel together along a string. Find the resultant amplitude if their phase difference is (a) 0, (b) π, (c) $\frac{2\pi }{3}$. [2 marks]

Try it first, then show the solutionSolution

1. Resultant amplitude $=2a\cos \frac{\varphi }{2}$
2. (a) 6 mm; (b) $2\times 3\times \cos 90°=0$; (c) $6\cos 60°=3$ mm

Answer: (a) 6 mm (b) 0 (c) 3 mm

Reflection of waves and standing waves on strings

NCERT section 14.6 and 14.6.1

Reflection: at a rigid boundary (fixed end), the reflected wave has a phase change of π (a crest returns as a trough). At an open boundary (free end), it reflects with no phase change.

Standing (stationary) waves: two identical waves travelling in opposite directions superpose:
$y=a\sin (kx-\omega t)+a\sin (kx+\omega t)=2a\sin kx\cos \omega t$
The pattern does not move. Nodes (zero amplitude) occur where sin kx = 0, at $x=0,\frac{\lambda }{2},\lambda ,\dots$; antinodes (amplitude 2a) halfway between. Adjacent nodes are $\frac{\lambda }{2}$ apart; a node and the next antinode are $\frac{\lambda }{4}$ apart. No energy flows along the string.

String fixed at both ends (length L): both ends are nodes, so $L=n\frac{\lambda }{2}$ and
${\nu }_n=\frac{nv}{2L}=\frac{n}{2L}\sqrt{\frac{T}{\mu }},n=1,2,3,\dots$
n = 1 is the fundamental mode (first harmonic); all integer harmonics are present.

Worked example

A string 1 m long, fixed at both ends, carries waves at 200 m s${}^{-1}$. Find its fundamental frequency and the next two harmonics. [1 mark]

Try it first, then show the solutionSolution

1. ${\nu }_1=\frac{v}{2L}=\frac{200}{2}=100Hz$
2. Harmonics: 200 Hz and 300 Hz

Answer: 100 Hz; 200 Hz and 300 Hz

Standing waves in organ pipes

NCERT section 14.6.1

Air columns also hold standing (longitudinal) waves. A closed end is a displacement node; an open end is a displacement antinode.

Pipe closed at one end (length L): node at the closed end, antinode at the open end, so $L=(2n-1)\frac{\lambda }{4}$:
$\nu =(2n-1)\frac{v}{4L},n=1,2,3,\dots$
Only odd harmonics are present: ${\nu }_1,3{\nu }_1,5{\nu }_1,\dots$ with ${\nu }_1=\frac{v}{4L}$.

Pipe open at both ends: antinodes at both ends, so $L=n\frac{\lambda }{2}$:
$\nu =\frac{nv}{2L}$
All harmonics are present, and the fundamental is twice that of a closed pipe of the same length.

This is the basis of flutes, organ pipes and resonance-tube experiments in the lab.

Worked example

A pipe 25 cm long is closed at one end. Find its fundamental frequency and the next two possible resonant frequencies. (v = 340 m s${}^{-1}$) [2 marks]

Try it first, then show the solutionSolution

1. ${\nu }_1=\frac{v}{4L}=\frac{340}{4\times 0.25}=340Hz$
2. Only odd harmonics: $3{\nu }_1=1020Hz$ and $5{\nu }_1=1700Hz$

Answer: 340 Hz; 1020 Hz and 1700 Hz

Beats

NCERT section 14.7

When two sound waves of slightly different frequencies ${\nu }_1$ and ${\nu }_2$ and similar amplitude are heard together, the loudness rises and falls periodically. These periodic changes are beats.

Adding $s_1=a\cos {\omega }_{1}t$ and $s_2=a\cos {\omega }_{2}t$:
$s=\left[2a\cos {\omega }_{b}t\right]\cos {\omega }_{a}t$, with ${\omega }_b=\frac{{\omega }_1-{\omega }_2}{2}$ and ${\omega }_a=\frac{{\omega }_1+{\omega }_2}{2}$.
The loudness peaks whenever $\cos {\omega }_{b}t=\pm 1$, which happens twice per cycle of ${\omega }_b$, so the
$\text{beat frequency}=|{\nu }_1-{\nu }_2|$

The ear can count beats up to about 10 per second. Use: musicians tune an instrument against a standard by adjusting it until the beats disappear. Loading a tuning fork with wax lowers its frequency; filing its prongs raises it, which tells you which of two forks is higher.

Worked example

A tuning fork of 256 Hz gives 4 beats per second with an unknown fork. When the unknown fork is loaded with a little wax, the beats decrease. Find its frequency. [2 marks]

Try it first, then show the solutionSolution

1. Unknown frequency is 256 ± 4: 252 Hz or 260 Hz
2. Wax lowers its frequency. If it were 252 Hz, beats would increase; they decrease, so it must be 260 Hz

Answer: 260 Hz

Physics in real life: Waves

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Progressive Wave Equation and Speed of Waves
• Standing Waves, Organ Pipes and Beats

Waves formula sheet

Progressive wave

$y=a\sin (kx-\omega t+\varphi ),k=\frac{2\pi }{\lambda },\omega =2\pi \nu$

Wave speed

$v=\nu \lambda =\frac{\omega }{k}$

Phase difference and path difference

$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$

Speed on a string

$v=\sqrt{\frac{T}{\mu }}$

Speed of sound

$v=\sqrt{\frac{B}{\rho }},v_{solid}=\sqrt{\frac{Y}{\rho }},v_{gas}=\sqrt{\frac{\gamma P}{\rho }}$

Superposition of two waves

$A_{res}=2a\cos \frac{\varphi }{2}$

Standing wave

$y=2a\sin kx\cos \omega t$

String fixed at both ends; pipe open at both ends

${\nu }_n=\frac{nv}{2L},n=1,2,3,\dots$

Pipe closed at one end

$\nu =(2n-1)\frac{v}{4L}$

Beat frequency

${\nu }_{beat}=|{\nu }_1-{\nu }_2|$

How the school exam asks Waves

The question patterns to expect, and the method that works for each.

Tricks to remember Waves

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Source sets frequency, medium sets speed

  When a wave enters a new medium, ν stays the same; v changes, so λ changes.

  Use it when: Refraction-type and speed questions.

• Fixed end flips, free end doesn't

  Reflection at a rigid end: phase change π (crest returns as trough). At an open end: no change.

  Use it when: Reflection questions.

• Open-open: all; closed-open: odd

  Open pipe and string: ν, 2ν, 3ν… Closed pipe: ν, 3ν, 5ν… and its fundamental is v/4L.

  Use it when: Harmonic series.

• N to N is half lambda

  Node to node or antinode to antinode: λ/2. Node to antinode: λ/4.

  Use it when: Standing-wave spacing.

• Wax down, file up

  Loading a fork with wax lowers its frequency; filing it raises the frequency.

  Use it when: Beats problems.

Waves: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Waves

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using Newton's formula for the speed of sound

  Forgetting Laplace's correction.

  Wrong working

  $v=\sqrt{P/\rho }\approx 280$ m s${}^{-1}$

  Right working

  $v=\sqrt{\gamma P/\rho }\approx 331$ m s${}^{-1}$

  ✓Do this: Sound compressions are adiabatic, so use γP.

  Usually costs: 1-2 marks.

• ✗Mistake: Including even harmonics for a closed pipe

  Treating every pipe like a string.

  ✓Do this: A pipe closed at one end has only odd harmonics.

  Usually costs: 1 mark.

• ✗Mistake: Saying pressure affects the speed of sound in air

  Seeing P in the formula.

  ✓Do this: At constant T, P/ρ is constant for an ideal gas, so v does not change with pressure; it does change with temperature and humidity.

  Usually costs: 1 mark.

• ✗Mistake: Mixing up k and λ

  Reading the coefficient of x as the wavelength.

  ✓Do this: The coefficient of x is k = 2π/λ; compute λ = 2π/k.

  Usually costs: 1 mark.

• ✗Mistake: Writing Doppler-effect answers

  Old notes.

  ✓Do this: The Doppler effect was removed for 2026-27.

  Usually costs: Time.

NCERT exercise guide: Waves

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Waves

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The distance between a node and the nearest antinode in a standing wave is:
   (a) λ
   (b) λ/2
   (c) λ/4
   (d) 2λ

   Hint

   Half of node-to-node.

   Answer

   (c) λ/4

2. Q2 · 1 mark · MCQExam level

   Transverse mechanical waves cannot travel through:
   (a) steel
   (b) a stretched string
   (c) air
   (d) a solid rod

   Hint

   Need shear rigidity.

   Answer

   (c) air

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): An organ pipe open at both ends produces all harmonics.
   Reason (R): Both open ends are displacement antinodes, so a whole number of half-wavelengths fits in the pipe.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   L = nλ/2.

   Answer

   (a)

4. Q4 · 2 marks · Very short answerExam level

   What are beats? Write the expression for beat frequency and give one use of beats.

   Hint

   Slightly different frequencies.

   Answer

   Beats are the periodic rise and fall in loudness when two sounds of slightly different frequencies are heard together. Beat frequency = $|{\nu }_1-{\nu }_2|$. Musicians use beats to tune instruments: when the beats vanish, the frequencies match.

5. Q5 · 2 marks · Very short answerHarder exam level

   A wave of frequency 400 Hz travels at 340 m s${}^{-1}$. How far apart are two points with a phase difference of π/3?

   Hint

   Δx = λΔφ/2π.

   Answer

   λ = 0.85 m; $\Delta x=\frac{0.85}{2\pi }\times \frac{\pi }{3}\approx 0.14$ m (about 14 cm).

6. Q6 · 2 marks · Very short answerHarder exam level

   The wave $y=0.05\sin (4\pi x+200\pi t)$ (SI) travels along a string. Find its direction of travel, wavelength, frequency and speed.

   Hint

   Sign between kx and ωt.

   Answer

   Towards −x (plus sign). λ = 2π/4π = 0.5 m; ν = 200π/2π = 100 Hz; v = νλ = 50 m s${}^{-1}$.

7. Q7 · 2 marks · Very short answerHarder exam level

   By what percentage must the tension in a string be increased to raise its fundamental frequency by 20%?

   Hint

   ν ∝ √T.

   Answer

   $T_2/T_1=1.2^2=1.44$: a 44% increase.

8. Q8 · 2 marks · Very short answerHarder exam level

   A metal rod 0.8 m long is clamped at its middle and set into longitudinal vibration in its fundamental mode at 3.2 kHz. Find the speed of sound in the metal.

   Hint

   Node at the clamp, antinodes at the free ends: L = λ/2.

   Answer

   λ = 2L = 1.6 m; v = νλ = 3200 × 1.6 ≈ 5.1 × 10${}^3$ m s${}^{-1}$.

9. Q9 · 3 marks · Short answerHarder exam level

   Explain why the speed of sound in air increases with humidity but does not change with pressure at constant temperature.

   Hint

   $v=\sqrt{\gamma P/\rho }$.

   Answer

   Water vapour (M = 18) is lighter than dry air (M ≈ 29), so humid air has a lower density at the same P and T, giving a higher $\sqrt{\gamma P/\rho }$. Changing the pressure at constant T changes ρ in the same proportion (P/ρ = RT/M), so the speed stays the same.

10. Q10 · 3 marks · Short answerHOTS / competency

    Two identical strings are each under a tension of 100 N and sound 200 Hz. The tension in one is raised slightly and 4 beats per second are heard. Find its new frequency and new tension.

    Hint

    Higher tension, higher frequency.

    Answer

    New frequency = 204 Hz (it must be higher). $T_2=100\times {\left(\frac{204}{200}\right)}^2\approx 104$ N.

11. Q11 · 3 marks · Short answerHOTS / competency

    A pipe closed at one end and a pipe open at both ends have the same fundamental frequency. Find the ratio of their lengths. If the open pipe is 60 cm long, what is the second resonant frequency of the closed pipe? (v = 330 m s${}^{-1}$)

    Hint

    v/4L₁ = v/2L₂.

    Answer

    $\frac{v}{4L_c}=\frac{v}{2L_o}\Rightarrow L_c=\frac{L_o}{2}$: ratio 1 : 2. Fundamental $=\frac{330}{2\times 0.6}=275$ Hz; the closed pipe's next resonance is $3\times 275=825$ Hz.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Distinguish transverse and longitudinal waves.Show answer

   Transverse: particles move perpendicular to the direction of travel. Longitudinal: particles move along it.

2. 2.Write the relation between wave speed, frequency and wavelength.Show answer

   $v=\nu \lambda$.

3. 3.Write the speed of a transverse wave on a string.Show answer

   $v=\sqrt{T/\mu }$.

4. 4.What was Laplace's correction?Show answer

   Sound compressions are adiabatic, so B = γP and $v=\sqrt{\gamma P/\rho }$.

5. 5.State the principle of superposition.Show answer

   The resultant displacement is the algebraic sum of the displacements due to each wave.

6. 6.What is the distance between two adjacent nodes?Show answer

   λ/2.

7. 7.Which harmonics does a pipe closed at one end produce?Show answer

   Only odd harmonics.

8. 8.What is the beat frequency of two sources of 400 Hz and 405 Hz?Show answer

   5 Hz.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why sound cannot be a transverse wave in air?
• ☐Can you read a, λ, ν and v from a wave equation?
• ☐Can you find the phase difference between two points on a wave?
• ☐Can you find the speed of a wave on a string?
• ☐Can you explain Laplace's correction?
• ☐Can you predict how temperature and humidity affect the speed of sound?
• ☐Can you derive the standing-wave equation and locate nodes and antinodes?
• ☐Can you write the harmonic series for a string and for open and closed pipes?
• ☐Can you solve a resonance-tube problem?
• ☐Can you find an unknown frequency from beats with waxing or filing?

Waves: frequently asked questions