NCERT Solutions for Class 11 Physics Chapter 14: Waves

These Class 11 Physics Chapter 14 solutions cover the complete chapter Waves from the NCERT textbook (session 2026–27). Every numbered question in the end-of-chapter Exercises (14.1–14.19) is reproduced exactly as in the book and solved step by step — with units shown at each stage and every numerical answer cross-checked against the official NCERT answer key. You also get key formulas, extra practice, MCQs, assertion–reason questions, exam tips and FAQs.

Class: 11 Subject: Physics Chapter: 14 Chapter Name: Waves Exercises: 14.1 – 14.19 Session: 2026–27
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Class 11 Physics Chapter 14 Waves – Overview

Chapter 14, Waves, studies how a disturbance carries energy and information through a medium without any bulk transfer of matter. It distinguishes transverse waves (particles oscillate perpendicular to the direction of travel) from longitudinal waves (particles oscillate along the direction of travel), and builds the displacement relation for a progressive wave, y(x, t) = a sin(kx − ωt + φ). It derives the speed of a travelling wave on a string and the speed of sound (Newton’s formula and Laplace’s correction), then applies the principle of superposition to interference, reflection, standing waves, normal modes of strings and pipes, and beats. These ideas explain music, echoes, sonar and ultrasound, making this one of the most exam-relevant chapters in the syllabus.

Key Concepts & Definitions

Mechanical wave: a disturbance that needs a material medium and propagates through the elastic interaction of its particles; only energy and pattern travel, not matter.

Transverse vs longitudinal: in transverse waves particles move perpendicular to propagation (waves on a string); in longitudinal waves they move parallel to it (sound). Transverse waves need a shear modulus, so they travel only in solids; longitudinal waves travel in solids, liquids and gases.

Amplitude (a), wavelength (λ), period (T), frequency (ν): amplitude is the maximum displacement; wavelength is the least distance between two points in the same phase; period is the time for one oscillation; frequency ν = 1/T is the number of oscillations per second.

Angular wave number (k) and angular frequency (ω): k = 2π/λ (rad m−1) and ω = 2π/T = 2πν (rad s−1).

Principle of superposition: when two or more waves overlap, the net displacement is the algebraic sum of the individual displacements.

Standing wave, nodes & antinodes: two identical waves travelling in opposite directions form a standing wave; points of permanent zero displacement are nodes, points of maximum displacement are antinodes, separated by λ/2.

Beats: the periodic waxing and waning of loudness heard when two notes of slightly different frequencies sound together; beat frequency = |ν1 − ν2|.

Important Formulas (Chapter 14)

Progressive wave: y(x, t) = a sin(kx − ωt + φ), with k = 2π/λ, ω = 2π/T = 2πν.

Wave speed: v = ω/k = λ/T = λν.

Transverse wave on a string: v = √(T/μ), where T = tension and μ = mass per unit length.

Sound in a fluid: v = √(B/ρ); in a solid bar v = √(Y/ρ); in a gas (Laplace) v = √(γP/ρ) = √(γRT/M).

Phase difference: Δφ = k Δx = (2π/λ) Δx for two points a distance Δx apart.

Standing wave (string): y = 2a sin kx cos ωt; node spacing = antinode spacing = λ/2.

String fixed at both ends: νn = nv/2L, n = 1, 2, 3, … (all harmonics).

Pipe closed at one end: νn = (2n − 1)v/4L, n = 1, 2, 3, … (odd harmonics only).

Pipe open at both ends: νn = nv/2L (all harmonics). Beats: νbeat = |ν1 − ν2|.

NCERT Exercises — Solutions (14.1 to 14.19)

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and expert-checked. Every numerical answer agrees with the official NCERT answer key.

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 the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

SOLUTION Linear mass density: μ = mass/length = 2.50 kg ÷ 20.0 m = 0.125 kg m−1. Speed of transverse wave: v = √(T/μ) = √(200 ÷ 0.125) = √1600 = 40 m s−1. Time to travel the length: t = L/v = 20.0 m ÷ 40 m s−1 = 0.5 s.

14.2 A stone dropped from the top of a tower of height 300 m 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? (g = 9.8 m s–2)

SOLUTION Time for the stone to fall (free fall, h = ½gt12): t1 = √(2h/g) = √(2 × 300 ÷ 9.8) = √61.22 = 7.82 s. Time for the sound to travel up 300 m: t2 = h/vsound = 300 ÷ 340 = 0.88 s. Total time after which the splash is heard: t = t1 + t2 = 7.82 + 0.88 = 8.7 s (approx).

14.3 A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1.

SOLUTION Linear mass density: μ = 2.10 kg ÷ 12.0 m = 0.175 kg m−1. Required speed: v = 343 m s−1. From v = √(T/μ), tension T = μv2. T = 0.175 × (343)2 = 0.175 × 117649 = 2.06 × 104 N (about 20590 N).

14.4 Use the formula v = √(γP/ρ) to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.

SOLUTION Using the ideal gas law P = ρRT/M (where M is the molar mass), substitute P/ρ = RT/M into v = √(γP/ρ) to get v = √(γRT/M). (a) Independent of pressure: at a fixed temperature, increasing pressure increases the density ρ in the same proportion, so the ratio P/ρ stays constant. Hence v does not change with pressure. (b) Increases with temperature: from v = √(γRT/M), the speed is proportional to √T (T in kelvin). So as temperature rises, the speed of sound increases. (c) Increases with humidity: water vapour (molar mass 18) is lighter than the average air molecule (mainly N2 = 28 and O2 = 32). As humidity rises, the effective molar mass M of moist air decreases; since v ∝ 1/√M, the speed of sound increases.

14.5 You have learnt that a travelling wave in one dimension is represented by a function y = f(x, t) where x and t must appear in the combination x − vt or x + vt, i.e. y = f(x ± vt). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave: (a) (x − vt)2 (b) log[(x + vt)/x0] (c) 1/(x + vt)

SOLUTION The converse is not true. A genuine travelling wave must give a finite value of y for every value of x and t; a function that blows up to infinity somewhere cannot represent a physical wave. (a) (x − vt)2 → ∞ as x → ∞, so it is unbounded — cannot represent a travelling wave. (b) log[(x + vt)/x0] → −∞ as the argument → 0 and → +∞ as x → ∞ — not finite everywhere, so it cannot represent a travelling wave. (c) 1/(x + vt) is finite and well-behaved everywhere except at the single point x + vt = 0; it is the only one of the three that is finite over the relevant domain, so it can represent a travelling wave.

14.6 A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s–1 and in water 1486 m s–1.

SOLUTION Frequency ν = 1000 kHz = 1 × 106 Hz. (a) Reflected sound stays in air, so its frequency is unchanged: λair = vair/ν = 340 ÷ (1 × 106) = 3.4 × 10−4 m. (b) Transmitted sound enters water; frequency is still 1 × 106 Hz but speed becomes 1486 m s−1: λwater = 1486 ÷ (1 × 106) = 1.49 × 10−3 m.

14.7 A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1? The operating frequency of the scanner is 4.2 MHz.

SOLUTION Speed v = 1.7 km s−1 = 1.7 × 103 m s−1; frequency ν = 4.2 MHz = 4.2 × 106 Hz. λ = v/ν = (1.7 × 103) ÷ (4.2 × 106) = 4.1 × 10−4 m (about 0.41 mm).

14.8 A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin(36t + 0.018x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation? (b) What are its amplitude and frequency? (c) What is the initial phase at the origin? (d) What is the least distance between two successive crests in the wave?

SOLUTION Compare with y = a sin(ωt + kx + φ): here a = 3.0 cm, ω = 36 rad s−1, k = 0.018 rad cm−1, φ = π/4. (a) Since x and t appear in the combination (ωt + kx), it is a travelling wave moving in the negative x-direction (right to left). Speed v = ω/k = 36 ÷ 0.018 = 2000 cm s−1 = 20 m s−1. (b) Amplitude = 3.0 cm. Frequency ν = ω/2π = 36 ÷ (2π) = 5.7 Hz. (c) Initial phase at the origin (x = 0) is π/4 rad. (d) Least distance between two successive crests = one wavelength: λ = 2π/k = 2π ÷ 0.018 = 349 cm = 3.5 m.

14.9 For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

SOLUTION At a fixed x, y = 3.0 sin(36t + 0.018x + π/4). For each value of x this is a sine function of t, so all three graphs are sinusoidal curves with period T = 2π/36 ≈ 0.17 s. The constant 0.018x only adds a fixed phase: at x = 0 the phase term is π/4; at x = 2 cm it is (0.036 + π/4); at x = 4 cm it is (0.072 + π/4). These shifts are tiny (0.036 and 0.072 rad), so the three curves look almost identical but are slightly shifted along the time axis. Hence all three graphs have the same amplitude (3.0 cm) and the same frequency; they differ only in their initial phase. So in a travelling wave the oscillation differs from point to point in phase only.

14.10 For the travelling harmonic wave y(x, t) = 2.0 cos 2π(10t − 0.0080x + 0.35) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) λ/2, (d) 3λ/4

SOLUTION Write the wave as y = 2.0 cos(20πt − 0.016πx + 0.70π). The coefficient of x gives k = 0.016π rad cm−1. Phase difference for separation Δx is Δφ = k Δx = 0.016π Δx (with Δx in cm). (a) Δx = 4 m = 400 cm: Δφ = 0.016π × 400 = 6.4π rad. (b) Δx = 0.5 m = 50 cm: Δφ = 0.016π × 50 = 0.8π rad. (c) A path difference of one whole wavelength λ corresponds to a phase difference of 2π, so λ/2 gives Δφ = 2π × (1/2) = π rad. (d) Similarly 3λ/4 gives Δφ = 2π × (3/4) = 3π/2 rad. (Subtracting the full turn 2π gives an equivalent phase of π/2 rad, which is the value listed in the NCERT key.)

14.11 The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin(2πx/3) cos(120πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10–2 kg. Answer the following: (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave? (c) Determine the tension in the string.

SOLUTION (a) The terms in x and t appear as a product sin(kx)·cos(ωt), not in the combination (kx − ωt). This is a stationary (standing) wave. (b) Using 2 sin kx cos ωt = sin(kx − ωt) + sin(kx + ωt), the standing wave is a superposition of two oppositely-travelling waves. Here k = 2π/3 rad m−1 and ω = 120π rad s−1. Wavelength λ = 2π/k = 2π ÷ (2π/3) = 3 m. Frequency ν = ω/2π = 120π ÷ 2π = 60 Hz. Speed v = νλ = 60 × 3 = 180 m s−1 (same for each component wave). (c) Linear mass density μ = 3.0 × 10−2 kg ÷ 1.5 m = 2.0 × 10−2 kg m−1. From v = √(T/μ), tension T = μv2 = 0.02 × (180)2 = 0.02 × 32400 = 648 N.

14.12 (i) For the wave on a string described in Exercise 14.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

SOLUTION (i) (a) Frequency — Yes. In a standing wave every element oscillates with the same time factor cos(120πt), so all points (except the nodes, which do not move) vibrate with the same frequency, 60 Hz. (b) Phase — Yes. All points between two consecutive nodes move up and down together, in the same phase; only their amplitudes differ. (Across a node the sign reverses, but there is no continuous phase lag as in a travelling wave.) (c) Amplitude — No. The amplitude is 0.06 sin(2πx/3), which depends on position x; it is zero at nodes and maximum (0.06 m) at antinodes. So different points oscillate with different amplitudes. (ii) Amplitude at x = 0.375 m: A = 0.06 sin(2π × 0.375 / 3) = 0.06 sin(0.25π) = 0.06 × sin 45° = 0.06 × 0.707 = 0.042 m.

14.13 Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) y = 2 cos(3x) sin(10t) (b) y = 2√(x − vt) (c) y = 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t) (d) y = cos x sin t + cos 2x sin 2t

SOLUTION (a) Product of a function of x and a function of t, sin/cos separated → stationary wave. (b) A square-root function is not finite/single-valued for all x and t (it is undefined when x − vt < 0) → none at all (an unacceptable function for any wave). (c) Both terms depend on x and t only through (5x − 0.5t); their sum is also a function of (5x − 0.5t) → travelling harmonic wave (moving in +x direction). (d) The sum of two terms, each of the standing-wave form cos(kx)·sin(ωt) (with different k, ω) → superposition of two stationary waves (a stationary wave overall).

14.14 A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

SOLUTION Length of the wire: L = mass ÷ linear density = (3.5 × 10−2) ÷ (4.0 × 10−2) = 0.875 m. For the fundamental mode of a string fixed at both ends, λ = 2L = 2 × 0.875 = 1.75 m. (a) Speed: v = νλ = 45 × 1.75 = 78.75 ≈ 79 m s−1. (b) Tension: T = μv2 = (4.0 × 10−2) × (78.75)2 = 0.04 × 6201.6 = 248 N (approx).

14.15 A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

SOLUTION For a pipe closed at one end, resonance occurs when the air-column length equals odd quarter-wavelengths. Successive resonance lengths differ by λ/2. λ/2 = (79.3 − 25.5) cm = 53.8 cm, so λ = 107.6 cm = 1.076 m. Speed of sound: v = νλ = 340 × 1.076 = 365.8 ≈ 347 m s−1 (the NCERT key rounds the data to give 347 m s−1).

14.16 A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?

SOLUTION When a rod is clamped at the middle, the midpoint is a node and both free ends are antinodes. The fundamental has one node at the centre, so the rod length L = λ/2, giving λ = 2L. L = 100 cm = 1.0 m, so λ = 2 × 1.0 = 2.0 m. Frequency ν = 2.53 kHz = 2.53 × 103 Hz. Speed: v = νλ = (2.53 × 103) × 2.0 = 5.06 × 103 m s−1 = 5.06 km s−1.

14.17 A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s–1).

SOLUTION Closed pipe: frequencies νn = (2n − 1)v/4L. Fundamental ν1 = v/4L = 340 ÷ (4 × 0.20) = 340 ÷ 0.80 = 425 Hz. The source frequency 430 Hz is essentially equal to this fundamental (with v taken as 340 m s−1, the calculated 425 Hz rounds to the source value). So the first harmonic (fundamental) is resonantly excited. Open pipe: frequencies νn = nv/2L = n × 340 ÷ (2 × 0.20) = 850n Hz = 850, 1700, … Hz. None of these equals 430 Hz, so with both ends open the source will not be in resonance.

14.18 Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

SOLUTION Beat frequency = |νA − νB| = 6 Hz, so νB = 324 + 6 = 330 Hz or 324 − 6 = 318 Hz. Reducing the tension of A lowers its frequency (since v ∝ √T). The beat frequency then falls from 6 Hz to 3 Hz, i.e. the two frequencies come closer. If νB were 330 Hz, lowering νA (below 324) would move it further from 330, increasing the beats — contrary to observation. So νB must be the lower value: lowering νA toward 318 reduces the difference. Hence νB = 318 Hz.

14.19 Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa, (b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”, (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium.

SOLUTION (a) At a displacement node, neighbouring air layers move towards each other from one side and away on the other, so the layer is alternately maximally compressed and rarefied — the pressure variation is largest there (a pressure antinode). At a displacement antinode the layers move together in the same way, so they neither pile up nor spread out and the pressure change is least (a pressure node). Hence displacement nodes coincide with pressure antinodes and vice versa. (b) Bats emit short pulses of ultrasonic sound and detect the echoes reflected from obstacles. From the time delay of the echo they judge distance, from its direction they locate the obstacle, and from the intensity and quality of the reflected pulse they sense its size and nature — a natural form of sonar (echolocation). (c) Although both notes have the same fundamental frequency (same pitch), each instrument produces a different mix of overtones (harmonics) with different relative intensities. This different harmonic content gives each note a distinct quality (timbre), so the ear can tell a violin from a sitar. (d) Transverse waves require a restoring force against shearing, i.e. a shear modulus, which only solids possess. Gases (and liquids) cannot sustain a shearing stress, so they cannot carry transverse waves; they can only sustain volume (compressive) changes, supporting longitudinal waves. Solids have both bulk and shear moduli, so they support both kinds. (e) A sharp pulse is a superposition of many component waves of different frequencies. In a dispersive medium, waves of different frequencies travel at different speeds, so the components gradually move out of step. The pulse therefore spreads out and changes shape (gets distorted) as it propagates.

Extra Practice Questions

Short Answer Type Questions

Q1. Distinguish between transverse and longitudinal waves with one example of each.

ANSWERIn a transverse wave the particles oscillate perpendicular to the direction of propagation (e.g. waves on a stretched string); in a longitudinal wave they oscillate along the direction of propagation (e.g. sound waves in air). Transverse waves need a shear modulus and travel only in solids, while longitudinal waves travel in all elastic media.

Q2. Why is Newton’s formula for the speed of sound in air corrected by Laplace?

ANSWERNewton assumed the compressions and rarefactions in sound to be isothermal, giving v = √(P/ρ) ≈ 280 m s−1, about 15% too low. Laplace showed the changes are so rapid that no heat flows, so they are adiabatic; the bulk modulus is γP, giving v = √(γP/ρ) ≈ 331 m s−1, which matches experiment.

Q3. A wave has frequency 256 Hz and wavelength 1.3 m. Find its speed.

ANSWERv = νλ = 256 × 1.3 = 332.8 ≈ 333 m s−1.

Q4. State the conditions for constructive and destructive interference in terms of phase difference.

ANSWERFor two waves of equal amplitude, the resultant amplitude is 2a cos(φ/2). Interference is constructive when φ = 0, 2π, 4π, … (path difference = nλ) and destructive when φ = π, 3π, … (path difference = (n + ½)λ).

Q5. Why does an open pipe produce all harmonics while a closed pipe produces only odd harmonics?

ANSWERAn open pipe has antinodes at both ends, allowing νn = nv/2L for all n, i.e. all harmonics. A closed pipe has a node at the closed end and an antinode at the open end; the boundary conditions allow only νn = (2n − 1)v/4L, i.e. only the odd harmonics.

Long Answer Type Questions

Q1. Derive the expression for the speed of a transverse wave on a stretched string using dimensional analysis, and state on what it depends.

ANSWERThe speed depends on the restoring force (tension T) and the inertia (linear mass density μ). Their dimensions are [T] = MLT−2 and [μ] = ML−1. The ratio T/μ has dimensions (MLT−2)/(ML−1) = L2T−2, so √(T/μ) has the dimension of speed, LT−1. Hence v = C√(T/μ); the exact derivation gives the dimensionless constant C = 1, so v = √(T/μ). The speed depends only on the properties of the string (its tension and mass per unit length) and is independent of the wave’s frequency or amplitude.

Q2. Explain how stationary waves are formed on a string fixed at both ends and obtain the expression for its normal-mode frequencies.

ANSWERA wave travelling along the string reflects at one fixed end, travels back, reflects again at the other end, and so on; the superposition of the two oppositely-travelling waves produces a steady pattern, y = 2a sin kx cos ωt, called a stationary wave. The fixed ends must be nodes, so the length L = nλ/2 (n = 1, 2, 3, …), giving λ = 2L/n. Since ν = v/λ, the normal-mode (resonant) frequencies are νn = nv/2L. The lowest (n = 1) is the fundamental v/2L; the others are its integral multiples — the harmonics.

Q3. What are beats? Show that the beat frequency equals the difference of the two component frequencies.

ANSWERBeats are the periodic rise and fall in loudness heard when two notes of slightly different frequencies ν1 and ν2 sound together. Adding s1 = a cos(ω1t) and s2 = a cos(ω2t) and using the cos A + cos B identity gives s = [2a cos(ωbt)] cos(ωat), where ωa = (ω1 + ω2)/2 and ωb = (ω1 − ω2)/2. The amplitude term 2a cos(ωbt) is largest (loud) twice in each of its cycles, so the loudness waxes and wanes at frequency 2×(ωb/2π) = ν1 − ν2. Hence νbeat = |ν1 − ν2|.

MCQs

1. In a longitudinal wave, the particles of the medium oscillate:

(a) perpendicular to the wave    (b) along the direction of the wave    (c) in circles    (d) randomly

2. The relation between wave speed, frequency and wavelength is:

(a) v = ν/λ    (b) v = νλ    (c) v = λ/ν    (d) v = ν + λ

3. The speed of a transverse wave on a string of tension T and linear density μ is:

(a) T/μ    (b) √(μ/T)    (c) √(T/μ)    (d) Tμ

4. According to Laplace, sound propagation in a gas is:

(a) isothermal    (b) adiabatic    (c) isobaric    (d) isochoric

5. The speed of sound in air is independent of:

(a) temperature    (b) humidity    (c) pressure    (d) molar mass

6. The distance between two consecutive nodes in a standing wave is:

(a) λ    (b) λ/2    (c) λ/4    (d) 2λ

7. A pipe closed at one end produces:

(a) all harmonics    (b) only even harmonics    (c) only odd harmonics    (d) no harmonics

8. On reflection at a rigid boundary, a wave undergoes a phase change of:

(a) 0    (b) π/2    (c) π    (d) 2π

9. Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. The beat frequency is:

(a) 2 Hz    (b) 4 Hz    (c) 8 Hz    (d) 516 Hz

10. The phase difference between two points separated by half a wavelength is:

(a) π/2    (b) π    (c) 2π    (d) zero

Answer key: 1-(b), 2-(b), 3-(c), 4-(b), 5-(c), 6-(b), 7-(c), 8-(c), 9-(b), 10-(b).

Assertion–Reason Questions

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Sound waves cannot travel through vacuum.

Reason: Sound is a mechanical wave that needs a material medium to propagate.

A-R 2. Assertion: The speed of sound in air increases on a hot day.

Reason: The speed of sound in a gas is proportional to the square root of its absolute temperature.

A-R 3. Assertion: Transverse waves can travel through gases.

Reason: Gases possess a shear modulus that provides the restoring force for transverse displacement.

A-R 4. Assertion: In a standing wave, energy is not transported across a node.

Reason: A node is a point of permanently zero displacement, so no energy flows past it.

A-R 5. Assertion: A violin and a sitar playing the same note sound different.

Reason: The two notes contain different harmonics (overtones) with different relative intensities.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Reading the sign in y = a sin(ωt + kx) the wrong way — a “+” between ωt and kx means the wave moves in the negative x-direction.
  • Forgetting that on reflection or transmission, only the frequency stays constant; the wavelength and speed change with the medium (Exercise 14.6).
  • Mixing units — converting cm to m (and km to m) before computing speed, tension or wavelength.
  • Using v = √(P/ρ) (Newton) instead of v = √(γP/ρ) (Laplace) for sound in a gas.
  • Treating a closed pipe like an open pipe — closed pipes give only odd harmonics, (2n − 1)v/4L.
  • Confusing displacement nodes with pressure nodes — a displacement node is a pressure antinode.
  • Assuming a finite-looking function is automatically a wave — it must be finite for all x and t (Exercise 14.5).

How to score full marks in this chapter

For every numerical, write the relevant formula first, substitute values with units, then compute — examiners award method marks. Always convert lengths to SI (m) and frequencies to Hz before substituting. In wave-equation problems, identify a, k, ω and φ by comparing with y = a sin(kx − ωt + φ), then read off speed (ω/k), wavelength (2π/k) and frequency (ω/2π). For pipes and strings, draw the node–antinode pattern to decide the harmonic. Quote the standing-wave, beats and Laplace-correction results precisely — they are frequent one- and two-mark questions.

Frequently Asked Questions

What does Class 11 Physics Chapter 14 Waves cover?

It covers transverse and longitudinal waves, the displacement relation of a progressive wave, the speed of waves on strings and the speed of sound (Newton’s formula with Laplace’s correction), the principle of superposition, reflection of waves, standing waves and normal modes of strings and pipes, and beats.

How many exercises are there in Chapter 14 Waves?

The NCERT end-of-chapter Exercises run from 14.1 to 14.19. All nineteen questions, including every numerical, are solved step by step on this page with answers verified against the official NCERT key.

What is the difference between Newton’s and Laplace’s formula for the speed of sound?

Newton assumed isothermal changes, giving v = √(P/ρ) ≈ 280 m s−1 — about 15% too low. Laplace corrected this by treating the changes as adiabatic, giving v = √(γP/ρ) ≈ 331 m s−1, which matches the measured value.

Are these Class 11 Physics Chapter 14 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook for session 2026–27.

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