PhysicsClass 11Waves

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.

14.1 Introduction

This section introduces the concept of waves by contrasting the motion of isolated oscillating objects with that of a system consisting of many such objects coupled together. A material medium, such as water or air, consists of many particles bound by elastic forces, so the disturbance of one particle affects its neighbors, allowing the disturbance to propagate through the medium. For example, when a pebble is dropped into still water, circular ripples spread outward from the point of disturbance. However, the water itself does not flow outward; rather, the disturbance moves outward. This is demonstrated by cork pieces on the water surface that move up and down but do not drift away from the center. Similarly, sound waves propagate through air as pressure disturbances without bulk movement of air molecules from the source to the listener.

Waves are patterns of disturbance that transport energy and information without the physical transfer of matter. Communication systems rely on waves to transmit signals, such as sound waves for speech or electromagnetic waves for wireless communication. Mechanical waves, which include waves on strings, water waves, sound waves, and seismic waves, require a material medium for propagation and depend on the medium's elastic properties. In contrast, electromagnetic waves (like light, radio waves, and X-rays) can travel through vacuum at the speed of light, c = 299,792,458 m/s.

A third category, matter waves, arises from quantum mechanics and are associated with particles like electrons and atoms. These waves are more abstract but have practical applications, such as electron microscopes.

The section also discusses the historical development of wave physics, mentioning scientists like Huygens, Hooke, and Newton, and connects wave motion to harmonic oscillations in elastic media. An example with springs connected in series illustrates how a disturbance travels through the system as each spring oscillates about its equilibrium position. Similarly, sound waves in air involve compressions and rarefactions that propagate due to restoring forces analogous to spring forces.

In solids, atoms arranged in a lattice behave like masses connected by springs, allowing wave propagation through elastic restoring forces. This foundational understanding sets the stage for studying mechanical waves in detail.

📊 Diagram: Fig. 14.1 shows a series of springs connected end to end. When the spring at one end (point A) is pulled and released, the disturbance propagates through the springs as each spring oscillates about its equilibrium length.

🧪 Activity: Illustration of wave propagation using a series of connected springs to show how disturbance travels through oscillations.

🔗 Connection: Leads to the next section on types of mechanical waves, namely transverse and longitudinal waves, by establishing the nature of wave propagation in elastic media.

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

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