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.
14.2 Transverse and longitudinal waves
This section classifies mechanical waves based on the direction of oscillation of the medium's particles relative to the direction of wave propagation. If the particles oscillate perpendicular to the wave's direction, the wave is transverse; if they oscillate parallel, it is longitudinal.
For transverse waves, an example is a pulse traveling along a stretched string. When the string is given an up-and-down jerk, the disturbance moves along the string in the x-direction, while the string elements oscillate in the y-direction, perpendicular to propagation. If the jerk is continuous and sinusoidal, the resulting wave is a sinusoidal transverse wave. The displacement of the string elements varies sinusoidally with time at a fixed position and sinusoidally with position at a fixed time.
Longitudinal waves are illustrated by sound waves in air. A piston moving back and forth in a pipe compresses and rarefies air particles along the direction of wave propagation. This creates regions of higher and lower density that travel through the air. The oscillations of air particles are parallel to the wave direction.
The section emphasizes that in both wave types, the medium as a whole does not move with the wave; only the disturbance travels. For example, water waves involve particle motion that is both up and down and back and forth, combining transverse and longitudinal components.
Transverse waves require a medium that can sustain shear stress, such as solids, and therefore do not propagate in fluids. Longitudinal waves require only compressive elasticity and can propagate in solids, liquids, and gases. For example, steel can support both wave types, while air supports only longitudinal waves.
The section also introduces capillary and gravity waves on water surfaces, distinguished by wavelength and restoring forces (surface tension for capillary waves and gravity for gravity waves).
An example problem classifies various wave motions as transverse, longitudinal, or a combination, reinforcing the conceptual understanding.
📊 Diagram: Fig. 14.2 shows a pulse traveling along a stretched string in the x-direction with string elements oscillating up and down (y-direction). Fig. 14.3 shows a sinusoidal transverse wave on a string with elements oscillating perpendicular to wave direction. Fig. 14.4 shows longitudinal waves in air inside a pipe generated by piston motion, with air particles oscillating along the pipe (wave) direction.
🧪 Activity: Classify given wave motions as transverse, longitudinal, or combination based on particle displacement direction.
🔗 Connection: Prepares for the mathematical description of wave displacement relations in progressive waves 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
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