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Waves

CBSE Class 11 Physics � Chapter 14 � Detailed Notes

Chapter Overview

This chapter covers the detailed physics of Wave Motion, including mechanical waves, equation of progressive waves, reflection, superposition, standing waves, beats, and the Doppler effect.

14.1 Introduction to Waves

Wave Motion: It is a mode of energy transfer from one point to another without the permanent transport of matter. Patterns of disturbance move through the medium.

Mechanism: Particles of the medium oscillate about their mean positions. The disturbance is handed over from one particle to the next due to Elasticity and Inertia of the medium.

Elasticity: Brings the particle back to mean position.
Inertia: Allows the particle to overshoot the mean position and continue oscillation.

Types of Waves based on Medium:

Mechanical Waves: Require material medium (Solid, Liquid, Gas). Examples: Sound waves, Water waves, Seismic waves, Strings. governed by Newton's Laws.
Electromagnetic Waves: Do not require medium. Can travel in vacuum. Examples: Light, Radio, X-rays. Speed $c = 3 \times 10^8$ m/s.
Matter Waves: Associated with moving particles (Quantum nature). Example: Electron microscope.

14.2 Transverse and Longitudinal Waves

Comparison based on particle vibration direction relative to wave propagation.

Animation: Transverse vs Longitudinal Waves

Property	Transverse Wave	Longitudinal Wave
Particle Vibrations	Perpendicular ($\perp$) to wave propagation.	Parallel ($\parallel$) to wave propagation.
Formation	Travels in the form of Crests (Pos. Max) and Troughs (Neg. Max).	Travels in the form of Compressions (High Density) and Rarefactions (Low Density).
Medium Required	Requires Rigidity (Shear modulus). Solids and Surface of Liquids.	Requires Elasticity of Volume (Bulk modulus). Solids, Liquids, and Gases.
Pressure Variation	No pressure variation in the medium.	Pressure and density vary at every point.
Polarization	Can be polarized.	Cannot be polarized.

14.3 Progressive Wave (Travelling Wave)

Derivation of the Equation of a Plane Progressive Harmonic Wave:

Goal: To mathematically describe a wave where every particle performs Simple Harmonic Motion (SHM).

Step 1: Motion at the Source ($x=0$)

Assume the particle at the origin ($x=0$) oscillates in SHM starting from mean position:

$$ y(0, t) = A \sin(\omega t) $$

Step 2: Propagation & Time Lag

The disturbance travels with wave velocity $v$. To reach a particle at position $x$, the wave takes time $t_{lag} = \frac{x}{v}$.

Step 3: Motion at Position $x$

The particle at $x$ does exactly what the source particle did earlier at time $(t - t_{lag})$.

$$ y(x, t) = y(0, t - t_{lag}) $$

Substituting the SHM equation:

$$ y(x, t) = A \sin[\omega (t - \frac{x}{v})] $$

$$ y(x, t) = A \sin(\omega t - \frac{\omega}{v}x) $$

Step 4: Defining Wave Constants

We define Propagation Constant (or Angular Wave Number) as $k = \frac{\omega}{v} = \frac{2\pi \nu}{\nu \lambda} = \frac{2\pi}{\lambda}$.

Substituting $k$:

$$ y(x, t) = A \sin(\omega t - kx) $$

Note: Since $\sin(-\theta) = -\sin(\theta)$, and usually we start with $kx$, we often write it as:

$$ y(x, t) = A \sin(kx - \omega t + \phi) $$

This represents a wave traveling in the Positive X direction.

Why $(kx - \omega t)$?

As $t$ increases, the term $\omega t$ increases.
To keep the phase $(kx - \omega t)$ constant (tracking a specific crest), $x$ must also INCREASE.
Hence, wave moves in $+x$ direction.
For $(kx + \omega t)$, $x$ must DECREASE to keep phase constant $\rightarrow$ Wave moves in $-x$ direction.

General Sinusoidal Wave Equation:

$$ y(x, t) = A \sin(kx - \omega t + \phi) $$

Detailed Analysis of Terms:

$y(x,t)$: Instantaneous displacement of particle at $x$ at time $t$.
$A$ (Amplitude): Maximum displacement from mean positions.
$\omega$ (Angular Frequency): Rate of change of phase with time. $\omega = \frac{2\pi}{T}$.
$k$ (Angular Wave Number): Spatial Frequency (Phase change per unit length). $k = \frac{2\pi}{\lambda}$.
$(kx - \omega t + \phi)$: Total Phase. It determines the state of vibration of the particle.
Phase Difference ($\Delta \phi$): $\Delta \phi = k(x_2 - x_1) = \frac{2\pi}{\lambda}\Delta x$.

Relation between Phase Diff ($\Delta \phi$), Path Diff ($\Delta x$), and Time Diff ($\Delta t$):

1. Phase difference for two particles separated by distance $\Delta x$:

We know: $\lambda \longleftrightarrow 2\pi$

This means phase per unit length is: $\frac{2\pi}{\lambda}$

So if distance is $\Delta x$, phase difference will be: $\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$

This gives:

$$ \Delta \phi = \frac{2\pi}{\lambda} \Delta x $$

2. Phase difference for the SAME particle after time interval $\Delta t$:

We know: $T \longleftrightarrow 2\pi$

This means phase per unit time is: $\frac{2\pi}{T}$

So if time interval is $\Delta t$, phase difference will be: $\Delta \phi = \frac{2\pi}{T} \times \Delta t$

This gives:

$$ \Delta \phi = \frac{2\pi}{T} \Delta t $$

Key Definitions: Wavelength, Period & Frequency

Wavelength ($\lambda$): The minimum distance between two particles of the medium that are in the same phase of oscillation (e.g., crest to crest, trough to trough). SI unit: metre (m). $$ \lambda = \frac{2\pi}{k} $$
Period ($T$): The time taken by a particle to complete one full oscillation. Equivalently, the time for the wave to advance by one wavelength. $$ T = \frac{2\pi}{\omega} $$
Frequency ($\nu$): The number of oscillations per second. $$ \nu = \frac{1}{T} = \frac{\omega}{2\pi} $$
Angular Frequency ($\omega$): $\omega = 2\pi\nu = \frac{2\pi}{T}$

Snapshot of wave at t=0 and t=?t showing wavelength and speed

Practice Problem 1

Q: A wave equation is $y = 0.05 \sin(80x - 3t)$. Find velocity and wavelength. (SI units)

Solution: Compare with standard $y = A \sin(kx - \omega t)$.

$k = 80$ rad/m $\Rightarrow \frac{2\pi}{\lambda} = 80 \Rightarrow \lambda = \frac{2\pi}{80} = \frac{\pi}{40}$ m.
$\omega = 3$ rad/s.
Wave Velocity $v = \frac{\omega}{k} = \frac{3}{80} = 0.0375$ m/s.

14.4 Speed of a Travelling Wave

$$ v = \frac{\omega}{k} = \frac{2\pi\nu}{2\pi/\lambda} = \nu \lambda $$

Wave velocity depends ONLY on the properties of the medium (Elasticity $E$ and Inertia $\rho$).

$$ v = \sqrt{\frac{E}{\rho}} $$

Case 1: Transverse Wave on Stretched String (Pulse Method)

Goal: To derive $v = \sqrt{T/\mu}$ using the concept of Centripetal Force.

Logic: Imagine moving with the pulse at speed $v$. In this frame, the pulse appears stationary, and the string moves backward with speed $v$.

Step 1: Dynamics of Element

Consider a small element of length $dl = R(2\theta)$ (where $\theta$ is small half-angle). Mass $dm = \mu (2R\theta)$.

The element moves on a curved path of radius $R$ with speed $v$. Required Centripetal Force:

$$ F_c = \frac{(dm)v^2}{R} = \frac{(2\mu R \theta)v^2}{R} = 2\mu v^2 \theta $$

Step 2: Restoring Force

The tension $T$ at both ends provides the downward radial force. Vertical component is $2T \sin\theta$.

For small $\theta$, $\sin\theta \approx \theta$. So, $F_{res} \approx 2T\theta$.

Step 3: Equating Forces

$$ 2T\theta = 2\mu v^2 \theta $$

$$ T = \mu v^2 \Rightarrow v = \sqrt{\frac{T}{\mu}} $$

Case 2: Longitudinal Wave (Sound) in Fluids

General Formula: $v = \sqrt{\frac{B}{\rho}}$ where $B$ is Bulk Modulus.

Definition of Bulk Modulus: $B = -V \frac{dP}{dV}$.

1. Newton's Formula (Isothermal):

Assumption: Sound travels slowly, temperature remains constant ($PV = \text{constant}$).

Differentiating: $P dV + V dP = 0 \Rightarrow P dV = -V dP \Rightarrow P = -V \frac{dP}{dV}$.

This implies $B_{iso} = P$. So, $v = \sqrt{\frac{P}{\rho}}$.

(Result: ~280 m/s for air. Incorrect.)

2. Laplace's Correction (Adiabatic):

Assumption: Compressions are rapid, heat cannot escape ($PV^\gamma = \text{constant}$).

Differentiating: $\gamma P V^{\gamma-1} dV + V^\gamma dP = 0$.

Divide by $V^{\gamma-1}$: $\gamma P dV + V dP = 0 \Rightarrow \gamma P = -V \frac{dP}{dV}$.

This implies $B_{adia} = \gamma P$. So, $v = \sqrt{\frac{\gamma P}{\rho}}$.

(Result: ~331 m/s. Correct!)

Factors Affecting Speed of Sound in Gas:

Temperature: $v \propto \sqrt{T}$. Speed increases by 0.61 m/s for every $1^\circ$C rise.
Molecular Weight: $v \propto \frac{1}{\sqrt{M}}$. (Hydrogen transmits sound faster than Oxygen).
Humidity: Moist air is lighter ($\rho$ decreases) than dry air. Hence speed increases with humidity.
Pressure: NO EFFECT (at constant temperature), because $P/\rho$ remains constant.

14.5 Principle of Superposition of Waves

Principle of Superposition (NCERT): When two or more waves traverse the same medium, the net displacement of any element of the medium at a given time is the algebraic sum of the displacements due to each individual wave.

In simple words: Imagine throwing two stones into a pond. The ripples from both will eventually cross each other. What happens to the water there? If Wave A tries to push a water particle UP, and Wave B tries to push it UP, the water particle will go up by the simple addition of both pushes.

$$ y(x,t) = y_1(x,t) + y_2(x,t) + \dots + y_n(x,t) $$

Mathematical Proof of Interference (from NCERT):

Let's take two identical waves traveling in the same direction, but one is slightly ahead of the other by a phase angle $\phi$:

Wave 1: $$ y_1(x,t) = a \sin(kx - \omega t) $$

Wave 2: $$ y_2(x,t) = a \sin(kx - \omega t + \phi) $$

By superposition, the new combined wave is $y = y_1 + y_2$. Using the math identity $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$, we get:

$$ y(x,t) = \left[ 2a \cos \left(\frac{\phi}{2}\right) \right] \sin \left(kx - \omega t + \frac{\phi}{2}\right) $$

Notice that the new wave has the same frequency, but a new Amplitude $A_{res} = 2a \cos(\phi/2)$. The intensity (loudness or brightness) of a wave is proportional to its Amplitude squared, so the resulting intensity is $I = 4I_0 \cos^2(\phi/2)$ (where $I_0$ is the intensity of a single wave).

Two Special Cases (Constructive & Destructive Interference):

1. Constructive Interference (Helping each other):

When the waves are exactly "in step" (in phase), meaning the phase difference is $\phi = 0, \pm 2\pi, \pm 4\pi \dots$
This happens when the path difference $\Delta x = n\lambda$ (where n = 0, 1, 2...). One wave traveled exactly a full wavelength further, so they still match perfectly.
The amplitude becomes maximum: $A_{max} = 2a$. The intensity reaches its peak: $I_{max} = 4I_0$.

2. Destructive Interference (Canceling out):

When the waves are completely "out of step" (out of phase), meaning the phase difference is $\phi = \pm\pi, \pm 3\pi, \pm 5\pi \dots$
This happens when the path difference $\Delta x = (n + \frac{1}{2})\lambda$ (where n = 0, 1, 2...). One wave traveled an extra half-wavelength, meaning a crest meets a trough.
The amplitude becomes zero: $A_{min} = 0$. The intensity drops to $I_{min} = 0$. They perfectly cancel each other out!

Transverse Wave Superposition

14.6 Reflection of Waves

When a wave hits a boundary, it bounces back. According to NCERT, the type of reflection depends completely on what it hits:

Rigid/Fixed Boundary: The wave cannot move the tied end. It pushes the wall, and by Newton's Third Law, the wall pushes back equally hard. The wave flips upside down (a crest returns as a trough).
Mathematical representation: If incident wave is $y_i = a \sin(kx - \omega t)$, the reflected wave is inverted, so $y_r = a \sin(kx + \omega t + \pi) = -a \sin(kx + \omega t)$. (This is a phase shift of $\pi$ radians).
Open/Free Boundary: The end is free to slide up and down. The wave whips it up and snaps back in the exact same shape. A crest returns as a crest.
Mathematical representation: The reflected wave is unchanged in orientation, so $y_r = a \sin(kx + \omega t)$. (This is a phase shift of 0).

Reflection at Fixed End (Wave Flips upside down, $\pi$ shift)

Reflection at Free End (Wave Stays the Same, 0 shift)

Standing Waves (Stationary Waves)

How are they formed? When two identical waves travel in opposite directions through the same medium (like a continually reflecting string wave), they superimpose to form a beautiful pattern that just bounces up and down in place instead of traveling forward.

Wave 1 (Moving Right): $y_1 = a \sin(kx - \omega t)$

Wave 2 (Moving Left): $y_2 = a \sin(kx + \omega t)$

Superposition: $y = y_1 + y_2 = a [\sin(kx - \omega t) + \sin(kx + \omega t)]$. Using math identities:

$$ y(x,t) = [2a \sin(kx)] \cos(\omega t) $$

Crucial NCERT Insights on the Math:

This is NOT a traveling wave $(kx \pm \omega t)$. The position $x$ and time $t$ are in separate sine/cosine terms!
The amplitude is not constant everywhere! The term $[2a \sin(kx)]$ acts as the Amplitude at any given position $x$.
Nodes (The "dead spots"): Points where the medium never moves. This means $\sin(kx) = 0$.
This happens when $kx = n\pi \implies \frac{2\pi}{\lambda}x = n\pi \implies x = n\frac{\lambda}{2}$. (n = 0, 1, 2...)
Antinodes (The "max spots"): Points that bounce up and down with maximum possible amplitude $2a$. This means $\sin(kx) = \pm 1$.
This happens when $kx = (n + \frac{1}{2})\pi \implies \frac{2\pi}{\lambda}x = (n + \frac{1}{2})\pi \implies x = (n + \frac{1}{2})\frac{\lambda}{2}$.

Normal Modes (Harmonics in Strings and Air Columns)

Musical instruments (like guitars and flutes) rely on bound systems. Because the boundaries force Nodes or Antinodes to always form at specific places, a string or pipe can only hold standing waves of certain "allowed" perfect lengths and frequencies. These are called Normal Modes or Harmonics.

System Type	NCERT Boundary Conditions	What frequencies can it play?
Stretched String (fixed at both ends)	Both walls are fixed, so we must have Nodes at both ends of length $L$. Allowed lengths: $L = n \frac{\lambda}{2}$ Base (Fundamental) Frequency: $\nu_1 = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$	ALL Harmonics Present! $\nu_n = n \nu_1$ Ratio: 1 : 2 : 3 : 4... Produces a very rich, full sound.
Open Organ Pipe (open at both ends)	Both ends are free to the air, so we must have Antinodes at both ends. Allowed lengths: $L = n \frac{\lambda}{2}$ Base (Fundamental) Frequency: $\nu_1 = \frac{v}{2L}$	ALL Harmonics Present! $\nu_n = n \nu_1$ Ratio: 1 : 2 : 3 : 4... It works symmetrically like the string.
Closed Organ Pipe (one end closed)	The closed wall forces a Node, the open end forces an Antinode. Allowed lengths: $L = (n + \frac{1}{2}) \frac{\lambda}{2} = (2n+1)\frac{\lambda}{4}$ Base (Fundamental) Frequency: $\nu_1 = \frac{v}{4L}$	ONLY ODD Harmonics Present! $\nu_n = (2n+1) \nu_1$ Ratio: 1 : 3 : 5 : 7... Missing evens gives a "hollow" sound.

System Type

NCERT Boundary Conditions

What frequencies can it play?

Stretched String
(fixed at both ends)

Both walls are fixed, so we must have Nodes at both ends of length $L$.

Allowed lengths: $L = n \frac{\lambda}{2}$

Base (Fundamental) Frequency: $\nu_1 = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

ALL Harmonics Present!
$\nu_n = n \nu_1$
Ratio: 1 : 2 : 3 : 4...
Produces a very rich, full sound.

Open Organ Pipe
(open at both ends)

Both ends are free to the air, so we must have Antinodes at both ends.

Allowed lengths: $L = n \frac{\lambda}{2}$

Base (Fundamental) Frequency: $\nu_1 = \frac{v}{2L}$

ALL Harmonics Present!
$\nu_n = n \nu_1$
Ratio: 1 : 2 : 3 : 4...
It works symmetrically like the string.

Closed Organ Pipe
(one end closed)

The closed wall forces a Node, the open end forces an Antinode.

Allowed lengths: $L = (n + \frac{1}{2}) \frac{\lambda}{2} = (2n+1)\frac{\lambda}{4}$

Base (Fundamental) Frequency: $\nu_1 = \frac{v}{4L}$

ONLY ODD Harmonics Present!
$\nu_n = (2n+1) \nu_1$
Ratio: 1 : 3 : 5 : 7...
Missing evens gives a "hollow" sound.

NCERT Terminology: Harmonics vs Overtones

Harmonic: Any integer multiple of the fundamental frequency base ($2\nu_1$ is the 2nd harmonic, $3\nu_1$ is the 3rd).
Overtone: Simply the physical pitches that appear over the base tone. In a Closed Pipe, the first tone higher than $\nu_1$ is $3\nu_1$. Therefore, its First Overtone is the 3rd Harmonic!

14.7 Beats

What are Beats? If you listen to two sound waves with nearly equal frequencies together, the intensity of the sound repeatedly rises and falls (waxing and waning). This acoustic phenomenon is called beats.

NCERT Mathematical Proof:

Let two frequencies be $\omega_1 = 2\pi\nu_1$ and $\omega_2 = 2\pi\nu_2$ (very close to each other).

$y_1 = a \cos(\omega_1 t)$ and $y_2 = a \cos(\omega_2 t)$

By superposition: $y = y_1 + y_2 = a [\cos(\omega_1 t) + \cos(\omega_2 t)]$

Using the trigonometric identity $\cos A + \cos B = 2\cos(\frac{A-B}{2})\cos(\frac{A+B}{2})$:

$$ y(t) = \left[ 2a \cos \left( 2\pi\left(\frac{\nu_1 - \nu_2}{2}\right) t \right) \right] \cos \left( 2\pi\left(\frac{\nu_1 + \nu_2}{2}\right) t \right) $$

Simple Meaning behind the Math:

The first bracket is a slowly drifting Amplitude Envelope. It goes from max magnitude to zero repeatedly.
A loud sound (maximum intensity) occurs whenever the amplitude envelope is maximized ($\pm 1$). Because cosine peaks twice per complete cycle (at $0$ and $\pi$), the frequency of these loudness peaks is twice the envelope's frequency.
Therefore, the number of beats you hear per second (Beat Frequency) is absolutely simple: $\nu_{beat} = |\nu_1 - \nu_2|$.

14.8 Doppler Effect

What is the Doppler Effect? Have you ever noticed how an ambulance siren sounds high-pitched as it races towards you, and suddenly drops to a noticeably deeper pitch the moment it passes you by? This apparent change in the frequency (pitch) of sound due to relative motion is the Doppler Effect.

Source moves TOWARDS Observer: The wave crests are squeezed closer together. Wavelength decreases $\implies$ Pitch goes UP.
Source moves AWAY from Observer: The wave crests are stretched widely apart. Wavelength increases $\implies$ Pitch goes DOWN.

The General Formula from NCERT:

$$ \nu = \nu_0 \left( \frac{v + v_0}{v + v_s} \right) $$

How to use it realistically without getting confused by coordinate signs:

Instead of strict vector signs, use this foolproof logic formula:

$$ \nu_{apparent} = \nu_{actual} \left( \frac{v \pm v_{observer}}{v \mp v_{source}} \right) $$

The "Common Sense" Sign Rule:

If the motion tries to bring the Source and Observer Closer together $\implies$ Pitch must INCREASE. Choose the math signs that make the final fraction Bigger (Numerator $+$, Denominator $-$).
If the motion tries to pull them Further apart $\implies$ Pitch must DECREASE. Choose the math signs that make the final fraction Smaller (Numerator $-$, Denominator $+$).

?? Quick Formula Reference

NCERT Topic	Mathematical Formula
Superposition Eq.	$y(x,t) = a\sin(kx - \omega t + \phi)$
Wave Speed	$v = \nu\lambda = \frac{\omega}{k}$
Speed on Stretched String	$v = \sqrt{\frac{T}{\mu}}$
Standing Wave Equation	$y(x,t) = [2a \sin(kx)] \cos(\omega t)$
Fund. Freq (String / Open Pipe)	$\nu_1 = \frac{v}{2L}$
Fund. Freq (Closed Pipe)	$\nu_1 = \frac{v}{4L}$
Beat Frequency	$\nu_{beat} = \|\nu_1 - \nu_2\|$
Doppler Effect	$\nu' = \nu_0\left(\frac{v \pm v_o}{v \mp v_s}\right)$