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CBSE • Class XI • Physics • Ch 14
Estimated Time: 45 Mins
Study Progress: In Progress

Waves

In Class 11 Physics, "Waves" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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🌊 Have You Ever Wondered?

How do noise-canceling headphones silence roaring jet engine airplane noise using destructive wave interference, or why does a train horn shift from high pitch to low pitch as it speeds past a station platform? Wave kinematics and acoustic physics explain wave transmission.

Why This Chapter Matters

In Class 11 Physics, "Waves" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Oscillations from Chapter 13.
  • Frequency, wavelength, and speed.
  • Principle of superposition.

What You Will Learn (Core Objectives)

  • Distinguish between Transverse waves (crests/troughs, polarization) and Longitudinal waves (compressions/rarefactions).
  • Write equation of a Progressive Harmonic Wave: $y(x, t) = A\sin(kx - \omega t + \phi)$ and find wave speed $v = \frac{\omega}{k} = \nu\lambda$.
  • Apply Newton's formula for speed of sound and Laplace's correction ($v = \sqrt{\frac{\gamma P}{\rho}}$).
  • Apply the Principle of Superposition to analyze Standing (Stationary) Waves in stretched strings and organ pipes.
  • Analyze Beats: $f_{\text{beat}} = |f_1 - f_2|$.

Chapter Roadmap & Progression

1 1. Traveling Waves & Wave Velocity
2 2. Speed of Sound & Laplace's Corre...
3 3. Standing Waves & Beats

Complete Concept Guide (100% Curriculum Coverage)

1. Traveling Waves & Wave Velocity

A Wave is a traveling disturbance transferring energy and momentum through a medium without net transport of matter:
• Transverse: Particle oscillations perpendicular to wave propagation (Light, waves on a string).
• Longitudinal: Particle oscillations parallel to propagation (Sound waves in air).
• Displacement Equation: $$\mathbf{y(x, t) = A\sin(kx - \omega t + \phi)} \quad \left(k = \frac{2\pi}{\lambda}, \omega = 2\pi\nu\right)$$ Wave speed: $\mathbf{v = \frac{\omega}{k} = \nu\lambda}$.

2. Speed of Sound & Laplace's Correction

Newton assumed sound propagation in air is isothermal ($v = \sqrt{P/\rho} = 280\text{ m/s}$, failing vs experimental $332\text{ m/s}$). Laplace corrected this: compressions and rarefactions happen so rapidly that heat cannot exchange (Adiabatic!): $$\mathbf{v = \sqrt{\frac{\gamma P}{\rho}}} \quad (\gamma = 1.41 \text{ for air} \implies v \approx 332\text{ m/s}!)$$

3. Standing Waves & Beats

  • Standing Waves (Superposition): Incident and reflected waves interfere: $y = (2A\sin kx)\cos\omega t$. Generates motionless Nodes ($x = n\lambda/2$) and maximum vibration Antinodes.
    • Closed organ pipe: only odd harmonics ($1 : 3 : 5$).
    • Open organ pipe: all harmonics ($1 : 2 : 3 : 4$).
  • Beats: Waxing and waning of sound intensity from slightly different frequencies: $$\mathbf{f_{\text{beat}} = |f_1 - f_2|}$$

Waves - Key Conceptual & Analytical Model

Waves - Conceptual Architecture Physical Laws & Formulations Governing equations & conservation principles Calculus & Vector Foundations Differential models, limits & derivations Real-World Engineering & Competitive Edge CBSE board problem patterns, JEE/NEET diagnostic applications & lab experiments

Chapter Summary & 10 Key Takeaways

Takeaway 1
Wave Equation: $y = A\sin(kx - \omega t)$ with propagation velocity $v = \omega/k = \nu\lambda$.
Takeaway 2
Laplace Correction: Adiabatic bulk modulus $\gamma P$ yielding true sound velocity ($332\text{ m/s}$).
Takeaway 3
Nodes and Antinodes: Interference boundaries of zero motion vs maximum amplitude vibration.
Takeaway 4
Organ Pipe Harmonics: Open pipes produce all harmonics; closed pipes produce only odd frequencies.
Takeaway 5
Beat Frequency: $f_{beat} = |f_1 - f_2|$ used in precision musical instrument acoustic tuning.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Explain Laplace's correction to Newton's formula for the speed of sound in air.
Reveal Answer & Explanation
Answer: Newton assumed sound propagation is an isothermal process ($B = P$), which yielded $v = \sqrt{P/\rho} \approx 280\text{ m/s}$, 15% lower than experimental values. Laplace recognized that compressions and rarefactions occur so rapidly that heat cannot transfer, making it an adiabatic process ($B = \gamma P$). Hence, $v = \sqrt{\frac{\gamma P}{\rho}} \approx 332.5\text{ m/s}$, matching experiment perfectly.
Sound is adiabatic (not isothermal), giving v = √(γP/ρ).
2
A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Reveal Answer & Explanation
Answer: Mass per unit length $\mu = 2.5 / 20 = 0.125\text{ kg/m}$. Wave speed $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40\text{ m/s}$. Time taken $t = \frac{L}{v} = \frac{20}{40} = 0.5\text{ seconds}$.
0.5 seconds.
3
Why is the sound produced by an open organ pipe richer and more pleasing than that produced by a closed organ pipe?
Reveal Answer & Explanation
Answer: Because an open organ pipe produces all harmonics (both odd and even: $1f, 2f, 3f, 4f\dots$), whereas a closed organ pipe produces only odd harmonics ($1f, 3f, 5f\dots$). The presence of both even and odd harmonics creates richer musical timbre.
Open pipe contains both even and odd harmonics.
4
What are Beats? Two tuning forks A and B produce 5 beats per second. When fork A is filed slightly, the beat frequency increases to 7 beats/s. If frequency of B is 256 Hz, find the original frequency of A.
Reveal Answer & Explanation
Answer: Beats are periodic waxing and waning of sound intensity. Initially, $f_A = 256 \pm 5 = 261\text{ Hz}$ or $251\text{ Hz}$. Filing fork A increases its frequency. If $f_A = 261$, increasing it makes beat frequency $f_A - 256$ greater than 5 (which matches 7!). If $f_A = 251$, increasing it would reduce beat frequency. Hence, original $f_A = 261\text{ Hz}$.
f_A = 261 Hz.
5
What is the distance between: (i) two consecutive nodes, (ii) a node and adjacent antinode in a stationary wave?
Reveal Answer & Explanation
Answer: (i) Distance between two successive nodes is $\frac{\lambda}{2}$. (ii) Distance between a node and adjacent antinode is $\frac{\lambda}{4}$.
(i) λ/2, (ii) λ/4.
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