Waves | Class 11 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
Waves – this guide gives you a concise, exam-ready overview of Waves from Class 11 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
14.4 The speed of a travelling wave
This section derives the speed of a traveling wave by examining the motion of a point of fixed phase (such as a crest) on the wave. The wave function phase is constant for such a point, leading to the relation kx - ωt = constant. Differentiating this with respect to time gives the wave speed v = ω/k.
Using the relations between ω, k, frequency ν, and wavelength λ, the wave speed can also be expressed as v = λν = λ/T.
The speed of a mechanical wave depends on the medium's elastic and inertial properties, not on the wave's frequency or wavelength.
The section then discusses the speed of transverse waves on a stretched string. The restoring force is due to the string's tension T, and the inertial property is the linear mass density μ (mass per unit length). Dimensional analysis yields the speed formula:
v = √(T/μ)
This formula shows that wave speed increases with tension and decreases with mass density.
Example 14.3 calculates the speed of transverse waves on a steel wire given its length, mass, and tension.
Next, the speed of longitudinal waves (sound waves) is discussed. The elastic property is the bulk modulus B, and the inertial property is the mass density ρ. Dimensional analysis gives:
v = √(B/ρ)
For solids under longitudinal strain, Young's modulus Y replaces B:
v = √(Y/ρ)
The section provides a table of sound speeds in various media, showing that sound travels faster in solids and liquids than in gases due to higher elastic moduli.
Newton's formula for sound speed in an ideal gas is v = √(P/ρ), assuming isothermal conditions. However, this underestimates the speed because sound propagation is adiabatic, not isothermal. Laplace corrected this by introducing the adiabatic bulk modulus B = γP, where γ is the ratio of specific heats, leading to:
v = √(γP/ρ)
This formula matches experimental values more closely.
Example 14.4 estimates the speed of sound in air at standard temperature and pressure using Laplace's correction.
📊 Diagram: Fig. 14.8 shows the wave profile at two times differing by Δt, illustrating the crest moving a distance Δx, defining wave speed v = Δx/Δt.
🧪 Activity: Calculate wave speed on a string and speed of sound in air using given parameters and formulas.
🔗 Connection: Prepares for the principle of superposition of waves and interference phenomena discussed in the next section.
Frequently asked questions
A string of mass $2.50\,\mathrm{kg}$ is under a tension of $200\,\mathrm{N}$. The length of the stretched string is $20.0\,\mathrm{m}$. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Given: Mass of string, m = 2.50 kg Tension, T = 200 N Length, L = 20.0 m
Step 1: Calculate the linear mass density (mass per unit length) \( \mu = \frac{m}{L} = \frac{2.50}{20.0} = 0.125\, \mathrm{kg/m} \).
Step 2: Calculate the speed of the transverse wave on the string using \( v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40\, \mathrm{m/s} \).
Step 3: Time taken for disturbance to travel length L is \( t = \frac{L}{v} = \frac{20.0}{40} = 0.5\, \mathrm{s} \).
Answer:
A stone dropped from the top of a tower of height $300\,\mathrm{m}$ 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 $340\,\mathrm{m\,s^{-1}}$? ($g = 9.8\,\mathrm{m\,s^{-2}}$)
Given: Height, h = 300 m Speed of sound, v = 340 m/s Acceleration due to gravity, g = 9.8 m/s²
Step 1: Calculate time taken by stone to fall: Using \( h = \frac{1}{2}gt^2 \), \( t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 300}{9.8}} = \sqrt{61.22} \approx 7.82 \text{ s} \).
Step 2: Time taken by sound to travel up: \( t_2 = \frac{h}{v} = \frac{300}{340} \approx 0.882 \text{ s} \).
Step 3: Total time when splash is heard: \( t = t_1 + t_2 = 7.82 + 0.882 = 8.70 \text{ s} \).
Answer: The s
A steel wire has a length of $12.0\,\mathrm{m}$ and a mass of $2.10\,\mathrm{kg}$. 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 $20^{\circ}\mathrm{C} = 343\,\mathrm{m\,s^{-1}}$.
Given: Length, L = 12.0 m Mass, m = 2.10 kg Speed of wave, v = 343 m/s
Step 1: Calculate linear mass density \( \mu = \frac{m}{L} = \frac{2.10}{12.0} = 0.175 \mathrm{kg/m} \).
Step 2: Use formula for wave speed on string: \( v = \sqrt{\frac{T}{\mu}} \Rightarrow T = \mu v^2 = 0.175 \times (343)^2 \).
Step 3: Calculate tension: \( T = 0.175 \times 117, 649 = 20,588.6 \mathrm{N} \).
Answer: The tension required is approximately 20,589 N.
Use the formula $v = \sqrt{\frac{\gamma P}{\rho}}$ to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.
Given formula for speed of sound in air: \( v = \sqrt{\frac{\gamma P}{\rho}} \)
(a) Independence of pressure: Using ideal gas law, \( P = \rho R T / M \), so \( \rho = \frac{P M}{R T} \). Substitute \( \rho \) in speed formula: \( v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma P}{\frac{P M}{R T}}} = \sqrt{\frac{\gamma R T}{M}} \). Pressure \( P \) cancels out, so speed is independent of pressure.
(b) Increase with temperature: From above, \( v = \sqrt{\frac{\gamma R T}{M}} \), so speed
Ready to ace this chapter?
Get the full Waves chapter — interactive notes, diagrams, worked solutions, polls and a free practice quiz — in the ConceptScroll app.
Study smarter with ConceptScroll
Daily NCERT-aligned reels, AI doubt solving and chapter quizzes — all free.
Start learning freeContinue reading
- Waves | Class 11 Physics Notes
Clear NCERT-aligned notes on Waves for Class 11 Physics.
- Waves | Class 11 Physics Notes
Clear NCERT-aligned notes on Waves for Class 11 Physics.
- Oscillations | Class 11 Physics Notes
Clear NCERT-aligned notes on Oscillations for Class 11 Physics.