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Class 11 Physics Chapter 14 Waves — Formulas & Key Points

Chapter 14 Waves in NCERT Class 11 Physics introduces mechanical wave motion—transverse, longitudinal, progressive and standing waves—along with phenomena like reflection, refraction, Doppler effect, beats and resonance. Mastering the formulas is essential for CBSE board exams and competitive tests like NEET and JEE. This sheet organises every equation by topic, clarifies sign conventions and highlights common pitfalls to help you revise efficiently.

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Key takeaways

✓Wave speed v = νλ connects frequency, wavelength and medium properties for all mechanical waves.
✓Transverse waves oscillate perpendicular to propagation; longitudinal waves oscillate parallel to direction of travel.
✓Speed of sound in air: v = 331 + 0.6t m/s where t is temperature in °C; in solids v = √(E/ρ), in liquids v = √(B/ρ).
✓Doppler effect formula: ν' = ν[(v ± v₀)/(v ∓ vₛ)] where upper signs apply when source and observer approach each other.
✓Beat frequency f_beat = |f₁ - f₂| when two waves of nearly equal frequencies superpose.
✓Fundamental frequency of closed pipe f₀ = v/4L produces only odd harmonics; open pipe f₀ = v/2L produces all harmonics.
✓Principle of superposition: resultant displacement equals algebraic sum of individual displacements at any point.

Core Wave Formulas — Progressive Waves

Progressive waves transfer energy through a medium without net transport of matter. The general wave equation y = A sin(kx − ωt + φ) describes displacement y at position x and time t. Here A is amplitude, k = 2π/λ is the wave number, ω = 2πν is angular frequency, and φ is the initial phase. Speed v relates wavelength λ and frequency ν through v = νλ. Understanding these parameters is fundamental to solving any wave problem in CBSE Class 11 Physics Chapter 14. The table below consolidates the relationships you must recall instantly during exams and numerical problem-solving sessions.

Wave equation: y = A sin(kx − ωt + φ) where k = 2π/λ and ω = 2πν
Wave speed: v = νλ = ω/k = λ/T where T is the time period
For a wave travelling in +x direction, use (kx − ωt); for −x direction, use (kx + ωt)
Maximum particle velocity v_max = Aω; maximum particle acceleration a_max = Aω²

Speed of Waves in Different Media — Quick Reference Table

Wave speed depends on the medium's elastic and inertial properties. For transverse waves on strings, tension and linear mass density govern speed. In solids, Young's modulus E and density ρ determine longitudinal wave speed. In fluids, bulk modulus B replaces E because fluids cannot sustain shear stress. For sound in gases, the speed formula incorporates the adiabatic index γ, pressure P and density ρ. Temperature also affects speed in air linearly. The NCERT Class 11 Physics textbook emphasises dimensional correctness: all speed formulas yield units m/s when SI base units are substituted. Memorise these relationships for quick recall during numerical problems in Class 11 Physics solutions.

String: v = √(T/μ) where T is tension and μ is mass per unit length
Solid rod: v = √(E/ρ) where E is Young's modulus
Liquid: v = √(B/ρ) where B is bulk modulus
Gas (ideal): v = √(γP/ρ) = √(γRT/M) where γ is adiabatic index, R is gas constant, M is molar mass
Air at temperature t °C: v ≈ (331 + 0.6t) m/s

Doppler Effect — Complete Formula Set

The Doppler effect describes the change in observed frequency when source or observer is in motion relative to the medium. The general formula is ν' = ν[(v ± v₀)/(v ∓ vₛ)] where ν is the actual frequency emitted, ν' is observed frequency, v is wave speed in the medium, v₀ is observer speed and vₛ is source speed. Sign convention: use the upper sign in the numerator when observer moves toward the source; use the lower sign in the denominator when source moves toward the observer. Both motions increase observed frequency. This formula appears frequently in CBSE Class 11 Physics board exams and NEET. Practice applying the correct sign by visualising relative motion carefully before substitution.

General: ν' = ν[(v ± v₀)/(v ∓ vₛ)]
Observer approaching stationary source: ν' = ν[(v + v₀)/v]
Source approaching stationary observer: ν' = ν[v/(v − vₛ)]
Both approaching: ν' = ν[(v + v₀)/(v − vₛ)]
Both receding: ν' = ν[(v − v₀)/(v + vₛ)]

Beats and Resonance Formulas

When two waves of slightly different frequencies f₁ and f₂ superpose, periodic waxing and waning of intensity called beats occurs. Beat frequency f_beat = |f₁ − f₂| equals the absolute difference. Beats are used in tuning musical instruments. Resonance happens when driving frequency matches a natural frequency of a system, causing large-amplitude oscillations. In air columns (pipes), resonance conditions depend on boundary conditions: closed pipes support only odd harmonics because one end is a displacement node, while open pipes support all harmonics because both ends are displacement antinodes. The NCERT Class 11 Physics Chapter 14 Waves notes emphasise calculating resonant lengths for given frequencies.

Beat frequency: f_beat = |f₁ − f₂|
Closed organ pipe (one end closed): f_n = (2n − 1)v/4L where n = 1,2,3,... (only odd harmonics)
Open organ pipe (both ends open): f_n = nv/2L where n = 1,2,3,... (all harmonics)
Fundamental frequency closed pipe: f₀ = v/4L
Fundamental frequency open pipe: f₀ = v/2L

Standing Waves and Harmonics — Key Equations

Standing waves form when two identical progressive waves travelling in opposite directions interfere. The resultant wave has nodes (zero amplitude) and antinodes (maximum amplitude) at fixed positions. For a string fixed at both ends, length L = nλ/2 where n = 1,2,3,... gives the allowed wavelengths. Frequency of nth harmonic is f_n = nv/2L. The fundamental (n=1) has the lowest frequency. For closed pipes, only odd harmonics exist because the closed end is always a displacement node. Understanding node-antinode patterns is crucial for solving Class 11 Physics Chapter 14 numerical problems involving resonance and string vibrations. CBSE examiners frequently ask for positions of nodes or antinodes given the harmonic number.

String fixed at both ends: L = nλ/2, f_n = nv/2L (n = 1,2,3,...)
Distance between consecutive nodes or antinodes = λ/2
Amplitude at antinode = 2A where A is amplitude of each progressive wave
Standing wave equation: y = 2A sin(kx) cos(ωt)

Key Terms and Definitions — NCERT Terminology

Chapter 14 introduces precise definitions that form the conceptual foundation. A wave is a disturbance that transfers energy without net transport of matter. Transverse waves have particle oscillations perpendicular to propagation direction—examples include light waves and waves on strings. Longitudinal waves have particle oscillations parallel to propagation—sound in air is the classic example. Wavelength λ is the spatial period, the distance between consecutive crests or compressions. Frequency ν is the number of oscillations per second. Amplitude A is the maximum displacement from equilibrium. Phase describes the state of oscillation at a given instant. These definitions appear verbatim in CBSE Class 11 Physics board exams, so memorise them exactly as stated in NCERT.

Wave: disturbance propagating through a medium, transferring energy without net mass transport
Transverse wave: particle motion perpendicular to wave propagation (e.g. string waves, electromagnetic waves)
Longitudinal wave: particle motion parallel to wave propagation (e.g. sound waves in fluids)
Wavelength (λ): shortest distance between two points in phase (crest to crest or compression to compression)
Frequency (ν): number of complete oscillations per unit time, measured in hertz (Hz)
Amplitude (A): maximum displacement of a particle from its equilibrium position
Phase (φ): argument of the sine or cosine function specifying the state of oscillation
Wave speed (v): distance travelled per unit time by a wavefront, v = νλ

Important Constants and Standard Values for Quick Recall

Certain numerical values recur in Class 11 Physics Chapter 14 problems. Speed of sound in air at 0 °C is approximately 331 m/s; at room temperature (20 °C) it is about 343 m/s. The temperature coefficient is roughly 0.6 m/s per °C. For water, speed of sound is about 1480 m/s; in steel it is around 5000 m/s. Adiabatic index γ for air (diatomic) is 1.4. Bulk modulus of water B ≈ 2.2 × 10⁹ Pa. Young's modulus for steel E ≈ 2 × 10¹¹ Pa. Density of air ρ ≈ 1.29 kg/m³ at STP. Knowing these helps you estimate answers and check if your calculated result is physically reasonable during CBSE 11 Physics exams or NEET preparation.

Speed of sound in air at 0 °C: v ≈ 331 m/s
Speed of sound in air at 20 °C: v ≈ 343 m/s
Temperature coefficient: Δv ≈ 0.6 m/s per °C
Speed of sound in water: v ≈ 1480 m/s
Speed of sound in steel: v ≈ 5000 m/s
Adiabatic index for air (γ): 1.4
Bulk modulus of water (B): ≈ 2.2 × 10⁹ Pa
Young's modulus of steel (E): ≈ 2 × 10¹¹ Pa
Density of air at STP (ρ): ≈ 1.29 kg/m³

Memory Tricks and Mnemonics for Wave Formulas

Remembering sign conventions in Doppler effect is easier with the mnemonic: 'Approach adds, recede subtracts'—when observer approaches source, add v₀ in numerator; when source approaches observer, subtract vₓ in denominator. For pipe harmonics, recall 'Closed is Odd' (closed pipes have only odd harmonics 1,3,5,...) and 'Open is All' (open pipes have all harmonics 1,2,3,...). The wave equation y = A sin(kx − ωt) has the minus sign for waves moving in +x direction; think 'k-x-minus-omega-t' as a phrase. To remember v = √(T/μ) for strings, note that higher tension T means faster wave, heavier string (larger μ) means slower—so T on top, μ below. These tricks save time during the CBSE Class 11 Physics board exam and reduce silly errors in Class 11 Physics solutions.

Doppler signs: 'Approach adds frequency, recede subtracts'—visualise relative motion direction
Pipe harmonics: 'Closed has Odd, Open has All'
Wave direction: minus sign (kx − ωt) for +x travel, plus sign (kx + ωt) for −x travel
String speed: 'Tension up, speed up; mass up, speed down' ⇒ v = √(T/μ)
Beat frequency: 'Difference in frequencies' ⇒ f_beat = |f₁ − f₂|
Node spacing: 'Half a wavelength apart' ⇒ λ/2

Common Mistakes — Sign Conventions, Units and Notation Pitfalls

Students frequently mix up Doppler signs: always draw a diagram showing source and observer motion relative to the medium before applying the formula. Another common error is confusing particle velocity (v_particle = Aω) with wave velocity (v_wave = νλ). Particle velocity oscillates in time, wave velocity is constant for a given medium. In standing wave problems, forgetting that closed-pipe length L relates to only odd multiples of λ/4 leads to wrong harmonic identification. Units must be consistent: convert all lengths to metres, frequencies to hertz and speeds to m/s before substitution. Sign errors in phase (φ) can flip the waveform; always check initial conditions. Writing k = 2π/T instead of k = 2π/λ is a notation blunder that costs marks in CBSE Class 11 Physics exams. Double-check dimensional homogeneity in every formula to catch algebraic slips early.

Doppler effect: draw motion diagram first; use correct signs for approach vs. recession
Distinguish particle speed (Aω) from wave speed (νλ)—they are different physical quantities
Closed pipe: harmonics are odd multiples of fundamental, L = (2n−1)λ/4 not nλ/2
Unit consistency: always convert to SI base units (m, s, kg) before calculation
Wave number k = 2π/λ, NOT 2π/T; angular frequency ω = 2π/ν
Phase constant φ: check initial conditions to determine the correct sign and value
Standing wave amplitude at antinode is 2A, not A

Solved Mini-Examples Applying Chapter 14 Formulas

Working through numerical examples cements formula recall and builds confidence for CBSE board exams. Each example below illustrates a different concept from Class 11 Physics Chapter 14 Waves. Example 1 applies the wave equation to extract parameters. Example 2 uses the Doppler formula with moving observer. Example 3 calculates beat frequency from two tuning forks. Practising similar problems from NCERT Class 11 Physics textbook exercises and Class 11 Physics solutions PDFs reinforces method and accuracy. For step-by-step video walkthroughs and instant doubt clearing with photo uploads, students can try CBSETUTOR.ai—a 24×7 AI tutor platform offering personalised help for Classes 6-12 at just ₹999/month, with a 3-day free trial to explore before committing.

One-Glance Last-Minute Revision Box

Before entering the CBSE Class 11 Physics exam hall, scan this box to refresh core formulas. Wave equation: y = A sin(kx − ωt + φ); wave speed v = νλ; string speed v = √(T/μ); sound in air v = 331 + 0.6t; Doppler ν' = ν[(v ± v₀)/(v ∓ vₛ)]; beats f_beat = |f₁ − f₂|; closed pipe f₀ = v/4L (odd harmonics); open pipe f₀ = v/2L (all harmonics); standing wave nodes separated by λ/2. Remember sign conventions, check units and draw diagrams for Doppler problems. Keep this revision checklist handy on your phone or print it on a card. Consistent practice with NCERT Class 11 Physics exercises and past board papers ensures you apply these formulas accurately under exam pressure.

y = A sin(kx − ωt + φ); v = νλ = ω/k
String: v = √(T/μ); Solid: v = √(E/ρ); Liquid: v = √(B/ρ); Gas: v = √(γP/ρ)
Doppler: ν' = ν[(v ± v₀)/(v ∓ vₛ)]—approach adds frequency
Beats: f_beat = |f₁ − f₂|
Closed pipe: f_n = (2n−1)v/4L; Open pipe: f_n = nv/2L
Standing wave: nodes at λ/2 intervals; antinode amplitude = 2A
Air sound speed ≈ 340 m/s at 20 °C; water ≈ 1480 m/s; steel ≈ 5000 m/s

Frequently asked questions

What is the difference between transverse and longitudinal waves in Chapter 14?+

In transverse waves, particles oscillate perpendicular to the direction of wave propagation—examples include waves on strings and electromagnetic waves. In longitudinal waves, particles oscillate parallel to the propagation direction—sound waves in air are the classic example from NCERT Class 11 Physics.

How do I apply the Doppler effect formula correctly in CBSE exams?+

Use ν' = ν[(v ± v₀)/(v ∓ vₛ)]. Draw a diagram showing source and observer motion. Use the upper sign in numerator when observer approaches source; use lower sign in denominator when source approaches observer. Both increase observed frequency. Practice with Class 11 Physics solutions to master sign conventions.

Why do closed organ pipes produce only odd harmonics?+

A closed pipe has one end closed (displacement node) and one open (displacement antinode). Only odd multiples of λ/4 fit this boundary condition: L = (2n−1)λ/4, giving frequencies f_n = (2n−1)v/4L where n = 1,2,3,... This is a key concept in CBSE Class 11 Physics Chapter 14 Waves.

What is beat frequency and how is it calculated?+

Beat frequency is the rate at which intensity waxes and wanes when two waves of slightly different frequencies superpose. It equals the absolute difference: f_beat = |f₁ − f₂|. Musicians use beats to tune instruments by listening for zero beats when frequencies match exactly.

How does temperature affect the speed of sound in air?+

Speed of sound in air increases linearly with temperature: v ≈ 331 + 0.6t m/s where t is in °C. At 0 °C, v ≈ 331 m/s; at 20 °C, v ≈ 343 m/s. This formula is derived from v = √(γRT/M) and appears in many NCERT Class 11 Physics numerical problems.

What is the difference between wave speed and particle speed?+

Wave speed v = νλ is the speed at which the wavefront propagates through the medium and remains constant for a given medium. Particle speed is the speed of individual medium particles oscillating about equilibrium, given by v_particle = Aω, and varies with time. Confusing these is a common mistake in Class 11 Physics Chapter 14.

How do I find the positions of nodes and antinodes in a standing wave?+

Nodes (zero amplitude) occur at positions where sin(kx) = 0, i.e. x = nλ/2 where n = 0,1,2,... Antinodes (maximum amplitude 2A) occur where sin(kx) = ±1, i.e. x = (2n+1)λ/4. Consecutive nodes or antinodes are separated by λ/2. CBSE examiners often ask for these positions given harmonic number.

Which online platform offers 24×7 doubt solving for Class 11 Physics formulas?+

CBSETUTOR.ai provides a 24×7 AI tutor for Classes 6-12 at ₹999/month, covering all subjects including NCERT Class 11 Physics. Students can upload photos of numericals or formulas and get instant step-by-step solutions. A 3-day free trial is available to test the service before subscribing.

What is the fundamental frequency of an open organ pipe of length L?+

For an open pipe (both ends open), the fundamental frequency is f₀ = v/2L where v is the speed of sound. The pipe supports all harmonics f_n = nv/2L (n = 1,2,3,...). This differs from closed pipes which have f₀ = v/4L and only odd harmonics, a key distinction in Class 11 Physics notes.

How do I remember the formula for speed of transverse waves on a string?+

Use the mnemonic: 'Tension up, speed up; mass up, speed down.' The formula v = √(T/μ) has tension T in the numerator (higher tension → faster wave) and linear mass density μ in the denominator (heavier string → slower wave). Dimensional analysis also confirms [√(N/(kg/m))] = m/s.

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