What is wave motion in physics?

Wave motion is a disturbance that propagates through a medium or space, transferring energy without the permanent displacement of matter. Waves can be mechanical (require medium) like sound waves or electromagnetic (don't require medium) like light waves.

What is the general equation of a progressive wave?

The general equation of a progressive wave traveling in +x direction is y = A sin(kx - ωt + φ) or y = A sin(ωt - kx + φ), where A is amplitude, k is wave number (2π/λ), ω is angular frequency (2π/T), and φ is initial phase.

What is the difference between transverse and longitudinal waves?

In transverse waves, particle vibration is perpendicular to wave propagation (like waves on a string). In longitudinal waves, particle vibration is parallel to wave propagation (like sound waves). Both carry energy without net particle displacement.

What is Doppler effect?

Doppler effect is the change in frequency (or wavelength) of a wave as observed when the source and observer are in relative motion. The apparent frequency increases when they approach each other and decreases when they move apart.

What are standing waves?

Standing waves (stationary waves) are formed by superposition of two identical waves traveling in opposite directions. They have fixed nodes (zero displacement) and antinodes (maximum displacement). Energy doesn't propagate in standing waves.

What is the formula for beat frequency?

Beat frequency equals the absolute difference of the two source frequencies: f_beat = |f₁ - f₂|. Beats are heard as periodic variations in loudness when two waves of slightly different frequencies interfere.

Wave Motion for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Wave Motion JEE notes, Formulas, PYQs — Wave Motion - Complete JEE Physics Notes

Introduction to Wave Motion

Wave motion is a fundamental concept in physics that describes how disturbances propagate through space and time. A wave transfers energy and momentum from one point to another without the permanent displacement of matter. Understanding waves is crucial for topics ranging from sound and light to quantum mechanics.

1.1 What is a Wave?

Definition of Wave

A wave is a disturbance that propagates through a medium (or vacuum) transferring energy from one point to another without the net movement of matter.

Key Characteristics:

Transfers energy and momentum
Particles oscillate about mean position
No net displacement of medium particles
Requires a source of disturbance

1.2 Wave Parameters

Parameter	Symbol	Definition	SI Unit
Amplitude	A	Maximum displacement from mean position	m
Wavelength	λ	Distance between two consecutive crests/troughs	m
Frequency	f or ν	Number of oscillations per second	Hz (s⁻¹)
Time Period	T	Time for one complete oscillation	s
Wave Velocity	v	Speed at which wave propagates	m/s
Wave Number	k	Spatial frequency (k = 2π/λ)	rad/m
Angular Frequency	ω	Angular frequency (ω = 2π/T = 2πf)	rad/s
Phase	φ	State of oscillation at a point	rad

1.3 Fundamental Wave Relations

Essential Formulas

Wave Velocity

v = f\lambda = \frac{\lambda}{T}

Frequency-Period Relation

f = \frac{1}{T}

Wave Number

k = \frac{2\pi}{\lambda}

Angular Frequency

\omega = 2\pi f = \frac{2\pi}{T}

Key Relationship:

v = \frac{\omega}{k}

🎯 JEE Tip: Quick Conversions

v = fλ - Most frequently used formula in wave problems
k = 2π/λ = ω/v - Wave number appears in wave equation
ω = 2πf = 2π/T - Convert frequency to angular frequency
Phase velocity = ω/k = λf - Always equal to wave velocity for non-dispersive medium

Types of Waves

Waves can be classified in multiple ways based on their properties, medium requirements, and particle motion. Understanding these classifications is essential for solving JEE problems correctly.

2.1 Based on Medium Requirement

Mechanical Waves

Require a material medium for propagation

Sound waves
Water waves
Seismic waves
Waves on a string

Cannot travel through vacuum

Electromagnetic Waves

Do NOT require a medium

Light waves
Radio waves
X-rays
Microwaves

Can travel through vacuum at speed c

2.2 Based on Particle Motion

Transverse Waves

Particle vibration ⊥ to wave propagation

Light waves
Waves on a string
S-waves (seismic)
All EM waves

Can be polarized

Longitudinal Waves

Particle vibration ∥ to wave propagation

Sound waves
P-waves (seismic)
Pressure waves in fluids
Waves in springs

Cannot be polarized

2.3 Based on Energy Propagation

Property	Progressive Waves	Stationary Waves
Energy Transfer	Energy propagates along with wave	No net energy transfer
Amplitude	Same for all particles	Varies from 0 (nodes) to 2A (antinodes)
Phase	Continuously changes along wave	Same between consecutive nodes
Formation	Single wave from source	Superposition of two identical waves
Examples	Sound from speaker, light from bulb	Vibrating string, organ pipe

2.4 Wave Velocity in Different Media

Important Velocity Formulas

String (Transverse Wave)

v = \sqrt{\frac{T}{\mu}}

T = Tension, μ = mass per unit length

Rod (Longitudinal Wave)

v = \sqrt{\frac{Y}{\rho}}

Y = Young's modulus, ρ = density

Fluid (Sound Wave)

v = \sqrt{\frac{B}{\rho}}

B = Bulk modulus, ρ = density

Ideal Gas (Sound Wave)

v = \sqrt{\frac{\gamma RT}{M}} = \sqrt{\frac{\gamma P}{\rho}}

γ = Cp/Cv, M = molar mass

📝 Solved Example 1

Question: A string has mass 10 g and length 2 m. It is stretched with a tension of 40 N. Find the velocity of transverse waves in the string.

Solution:

Given: m = 10 g = 0.01 kg, L = 2 m, T = 40 N

Step 1: Calculate linear mass density

\mu = \frac{m}{L} = \frac{0.01}{2} = 0.005 \text{ kg/m}

Step 2: Apply velocity formula

v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.005}}

v = \sqrt{8000} = 89.44 \text{ m/s}

\boxed{v \approx 89.4 \text{ m/s}}

⚠️ Common Mistakes

Wrong: Sound requires air → Correct: Sound can travel through any medium (solid, liquid, gas)
Wrong: Longitudinal waves can be polarized → Correct: Only transverse waves can be polarized
Wrong: Wave velocity depends on amplitude → Correct: Wave velocity depends on medium properties only
Wrong: Frequency changes when wave enters new medium → Correct: Frequency remains constant, wavelength changes

Progressive Wave Equation

The mathematical representation of waves is crucial for solving JEE problems. A progressive wave equation describes how the displacement of particles varies with both position and time.

3.1 General Wave Equation

Standard Forms of Progressive Wave

Wave in +x direction

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

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

Wave in -x direction

y = A\sin(kx + \omega t + \phi)

y = A\sin(\omega t + kx + \phi)

Alternative Forms:

y = A\sin\left[\frac{2\pi}{\lambda}(x - vt)\right] = A\sin\left[2\pi\left(\frac{x}{\lambda} - \frac{t}{T}\right)\right]

3.2 Understanding the Wave Equation

Term	Symbol	Meaning	How to Find
Amplitude	A	Coefficient of sin/cos	Direct reading
Wave Number	k	Coefficient of x	λ = 2π/k
Angular Frequency	ω	Coefficient of t	f = ω/2π, T = 2π/ω
Initial Phase	φ	Constant term	Direct reading
Wave Velocity	v	v = ω/k = coefficient of t / coefficient of x	v = ω/k = λf
Direction	—	Sign between kx and ωt	Opposite signs → +x; Same signs → -x

3.3 Particle Velocity and Acceleration

Derivatives of Wave Equation

For y = A sin(kx - ωt):

Particle Velocity

v_p = \frac{\partial y}{\partial t}

v_p = -A\omega\cos(kx - \omega t)

(v_p)_max = Aω

Particle Acceleration

a_p = \frac{\partial^2 y}{\partial t^2}

a_p = -A\omega^2\sin(kx - \omega t)

(a_p)_max = Aω²

Slope

\frac{\partial y}{\partial x} = Ak\cos(kx - \omega t)

Slope_max = Ak

Important Relation:

\frac{\partial y}{\partial t} = -v \cdot \frac{\partial y}{\partial x}

Particle velocity = -(Wave velocity) × Slope

💡 Quick Tricks for Wave Equation

Direction check: If (kx - ωt) → wave moves in +x direction
Velocity = ω/k regardless of the form of equation
Max particle velocity = Aω (NOT wave velocity!)
Particle velocity ≠ Wave velocity: v_particle oscillates, v_wave is constant
At crest/trough: Particle velocity = 0, Particle acceleration = maximum
At mean position: Particle velocity = max, Particle acceleration = 0

📝 Solved Example 2 (JEE Main Pattern)

Question: A wave is described by y = 0.02 sin(3x - 2t) where x, y are in meters and t in seconds. Find (a) amplitude, (b) wavelength, (c) frequency, (d) wave velocity, (e) maximum particle velocity.

Solution:

Comparing with y = A sin(kx - ωt):

(a) Amplitude: A = 0.02 m = 2 cm

(b) Wavelength:

k = 3 \text{ rad/m} \quad \Rightarrow \quad \lambda = \frac{2\pi}{k} = \frac{2\pi}{3} \text{ m}

(c) Frequency:

\omega = 2 \text{ rad/s} \quad \Rightarrow \quad f = \frac{\omega}{2\pi} = \frac{2}{2\pi} = \frac{1}{\pi} \text{ Hz}

(d) Wave Velocity:

v = \frac{\omega}{k} = \frac{2}{3} \text{ m/s}

(e) Maximum Particle Velocity:

v_{p(max)} = A\omega = 0.02 \times 2 = 0.04 \text{ m/s} = 4 \text{ cm/s}

Direction: Wave travels in +x direction (opposite signs of kx and ωt)

3.4 Phase and Phase Difference

Phase Relations

Phase at point (x, t)

\phi = kx - \omega t + \phi_0

Path Difference to Phase Difference

\Delta\phi = \frac{2\pi}{\lambda} \times \Delta x = k \cdot \Delta x

Time Difference to Phase Difference

\Delta\phi = \frac{2\pi}{T} \times \Delta t = \omega \cdot \Delta t

Path Difference to Time Difference

\Delta t = \frac{\Delta x}{v} = \frac{T \cdot \Delta x}{\lambda}

Special Cases:

Δφ = 0, 2π, 4π... → In phase (constructive)
Δφ = π, 3π, 5π... → Out of phase (destructive)
Δφ = π/2 → Quadrature

Superposition of Waves

The principle of superposition states that when two or more waves meet at a point, the resultant displacement is the algebraic sum of individual displacements. This fundamental principle explains interference, beats, and standing waves.

4.1 Principle of Superposition

Mathematical Statement

y_{resultant} = y_1 + y_2 + y_3 + ... + y_n

The waves maintain their individual properties after superposition - they pass through each other without any permanent change.

4.2 Interference of Two Waves

Resultant of Two Sinusoidal Waves

For two waves: y₁ = A₁sin(kx - ωt) and y₂ = A₂sin(kx - ωt + φ)

y = A_R \sin(kx - \omega t + \theta)

Resultant Amplitude

A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi}

Phase of Resultant

\tan\theta = \frac{A_2\sin\phi}{A_1 + A_2\cos\phi}

4.3 Special Cases of Interference

Condition	Phase Difference (φ)	Path Difference (Δx)	Resultant Amplitude
Constructive Interference (Maximum)	0, 2π, 4π, ... (2nπ)	0, λ, 2λ, ... (nλ)	A₁ + A₂
Destructive Interference (Minimum)	π, 3π, 5π, ... ((2n+1)π)	λ/2, 3λ/2, ... ((2n+1)λ/2)	\|A₁ - A₂\|
Equal Amplitudes (A₁ = A₂ = A)	Any φ	—	2A cos(φ/2)

💡 JEE Shortcut for Equal Amplitude Waves

When A₁ = A₂ = A, use this quick formula:

A_R = 2A\cos\left(\frac{\phi}{2}\right)

This comes from: A_R = √(A² + A² + 2A²cosφ) = A√(2 + 2cosφ) = 2A|cos(φ/2)|

4.4 Intensity and Interference

Intensity Relations

Since Intensity ∝ Amplitude²:

Resultant Intensity

I_R = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi

For Equal Intensities (I₁ = I₂ = I₀)

I_R = 4I_0\cos^2\left(\frac{\phi}{2}\right)

Maximum and Minimum Intensities:

I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2

I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2

📝 Solved Example 3

Question: Two waves of intensities I and 4I interfere. Find the ratio of I_max to I_min.

Solution:

Given: I₁ = I, I₂ = 4I

∴ √I₁ = √I, √I₂ = 2√I

Maximum Intensity:

I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I

Minimum Intensity:

I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I

\frac{I_{max}}{I_{min}} = \frac{9I}{I} = 9:1

Standing (Stationary) Waves

Standing waves are formed when two identical waves traveling in opposite directions superpose. They are crucial for understanding musical instruments, resonance in strings and pipes, and many JEE problems.

5.1 Formation of Standing Waves

Mathematical Formation

Incident wave: y₁ = A sin(kx - ωt)

Reflected wave: y₂ = A sin(kx + ωt)

Resultant:

y = y_1 + y_2 = 2A\sin(kx)\cos(\omega t)

Amplitude = 2A sin(kx) [varies with position] × cos(ωt) [oscillates with time]

5.2 Nodes and Antinodes

Nodes (Zero Amplitude)

Points where amplitude = 0 always

\sin(kx) = 0\] \[kx = 0, \pi, 2\pi, 3\pi...

Position of nodes:

x = 0, \frac{\lambda}{2}, \lambda, \frac{3\lambda}{2}...

Distance between consecutive nodes = λ/2

Antinodes (Maximum Amplitude)

Points where amplitude = 2A (maximum)

\sin(kx) = \pm 1\] \[kx = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}...

Position of antinodes:

x = \frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}...

Distance between consecutive antinodes = λ/2

Standing Wave Pattern

N = Node (zero amplitude), A = Antinode (maximum amplitude)

5.3 Properties of Standing Waves

Property	Description
Energy Transfer	No net energy transfer along the wave
Amplitude	Varies from 0 (node) to 2A (antinode)
Phase	All particles between two nodes vibrate in phase
Wavelength	Distance between alternate nodes = λ; consecutive nodes = λ/2
Node-Antinode Distance	λ/4

⚠️ Phase in Standing Waves

Particles between two consecutive nodes are in phase (same direction)
Particles on opposite sides of a node are 180° out of phase
At a node, phase changes by π
All particles pass through mean position simultaneously (except at nodes)

5.4 Boundary Conditions and Reflection

Fixed End (Rigid Boundary)

Reflection with phase change of π (180°)
Reflected wave: y = -A sin(kx + ωt)
Node forms at fixed end
Example: Fixed end of guitar string

Free End (Open Boundary)

Reflection with no phase change
Reflected wave: y = A sin(kx + ωt)
Antinode forms at free end
Example: Open end of organ pipe

Sound Waves

Sound waves are longitudinal mechanical waves that require a material medium for propagation. Understanding sound is crucial for JEE as it combines concepts of wave motion with thermodynamics and mechanics.

6.1 Nature of Sound Waves

Key Properties of Sound

Longitudinal wave
Requires material medium
Travels as compressions and rarefactions
Cannot be polarized

Speed: Solid > Liquid > Gas
Audible range: 20 Hz - 20 kHz
Infrasound: < 20 Hz
Ultrasound: > 20 kHz

6.2 Speed of Sound

Speed of Sound in Different Media

In Gases (Laplace Formula)

v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}}

γ = Cp/Cv, M = molar mass

In Liquids

v = \sqrt{\frac{B}{\rho}}

B = Bulk modulus

In Solids (Rod)

v = \sqrt{\frac{Y}{\rho}}

Y = Young's modulus

6.3 Effect of Various Factors on Speed of Sound

Factor	Effect on Speed	Explanation
Temperature	v ∝ √T	Speed increases with temperature
Pressure (at constant T)	No effect	P and ρ both change proportionally
Humidity	Increases with humidity	Moist air is less dense than dry air
Molecular Weight	v ∝ 1/√M	Lighter gases have higher speed

💡 Quick Formula: Temperature Effect

For small temperature changes near 0°C:

\[v_t = v_0 + 0.61t\]

where v₀ = 332 m/s (speed at 0°C), t = temperature in °C

Speed at 0°C ≈ 332 m/s, Speed at 20°C ≈ 344 m/s

6.4 Intensity and Loudness

Intensity of Sound

Definition

I = \frac{P_{avg}}{A} = \frac{E}{A \cdot t}

Power per unit area (W/m²)

In Terms of Wave Parameters

I = \frac{1}{2}\rho v \omega^2 A^2 = 2\pi^2 f^2 A^2 \rho v

Intensity Level (Loudness) in Decibels:

\beta = 10\log_{10}\left(\frac{I}{I_0}\right) \text{ dB}

where I₀ = 10⁻¹² W/m² (threshold of hearing)

📝 Solved Example 4

Question: The intensity of sound from a source is 10⁻⁶ W/m². What is the intensity level in decibels?

Solution:

Given: I = 10⁻⁶ W/m², I₀ = 10⁻¹² W/m²

\beta = 10\log_{10}\left(\frac{I}{I_0}\right)

\beta = 10\log_{10}\left(\frac{10^{-6}}{10^{-12}}\right)

\beta = 10\log_{10}(10^6) = 10 \times 6

\beta = 60 \text{ dB}

Doppler Effect

The Doppler Effect is the change in frequency of a wave as observed when the source and observer are in relative motion. This is one of the most important topics for JEE with guaranteed questions every year.

7.1 Master Formula for Doppler Effect

General Doppler Formula (Sound Waves)

f' = f_0 \left(\frac{v \pm v_o}{v \mp v_s}\right)

Where:

f' = Apparent/observed frequency
f₀ = Source/actual frequency
v = Speed of sound in medium
v_o = Speed of observer
v_s = Speed of source

🎯 Sign Convention (Easy Method)

Approaching: Use + for v_o (numerator), Use - for v_s (denominator)
Receding: Use - for v_o (numerator), Use + for v_s (denominator)

Remember: "+" when moving toward line of sight, "-" when moving away

7.2 Special Cases of Doppler Effect

Case	Condition	Formula
Source approaches stationary observer	v_o = 0, v_s ≠ 0 (approaching)	f' = f₀(v/(v - v_s))
Source recedes from stationary observer	v_o = 0, v_s ≠ 0 (receding)	f' = f₀(v/(v + v_s))
Observer approaches stationary source	v_s = 0, v_o ≠ 0 (approaching)	f' = f₀((v + v_o)/v)
Observer recedes from stationary source	v_s = 0, v_o ≠ 0 (receding)	f' = f₀((v - v_o)/v)
Both approaching	Both moving towards each other	f' = f₀((v + v_o)/(v - v_s))
Both receding	Both moving away from each other	f' = f₀((v - v_o)/(v + v_s))

💡 JEE Quick Trick: SOFA Rule

Source, Observer, Frequency increases when Approaching

Source approaching → f' > f₀ (higher pitch)
Source receding → f' < f₀ (lower pitch)
Observer approaching → f' > f₀
Observer receding → f' < f₀

7.3 Apparent Wavelength

Wavelength Change in Doppler Effect

Key Point:

Source moving: Wavelength changes (λ' ≠ λ)
Observer moving: Wavelength does NOT change (λ' = λ)

\text{When source moves: } \lambda' = \frac{v \mp v_s}{f_0} = \lambda \mp \frac{v_s}{f_0}

📝 Solved Example 5 (JEE Main Pattern)

Question: A train moving at 72 km/h emits a whistle of frequency 500 Hz. Find the apparent frequency heard by a stationary observer when (a) train approaches (b) train recedes. (Speed of sound = 340 m/s)

Solution:

Given: v_s = 72 km/h = 20 m/s, f₀ = 500 Hz, v = 340 m/s, v_o = 0

(a) Train approaches:

f' = f_0\left(\frac{v}{v - v_s}\right) = 500 \times \frac{340}{340 - 20}

f' = 500 \times \frac{340}{320} = 531.25 \text{ Hz}

(b) Train recedes:

f' = f_0\left(\frac{v}{v + v_s}\right) = 500 \times \frac{340}{340 + 20}

f' = 500 \times \frac{340}{360} = 472.22 \text{ Hz}

Answer: (a) 531.25 Hz (b) 472.22 Hz

Note: Approaching gives higher frequency, receding gives lower frequency

📝 Solved Example 6 (JEE Advanced Pattern)

Question: A car moving at 30 m/s honks (frequency 200 Hz) at a wall and hears the echo. What is the beat frequency between the original and reflected sound? (v = 330 m/s)

Solution:

Step 1: Find frequency received by wall (wall is stationary observer, car is approaching source)

f_{wall} = f_0\left(\frac{v}{v - v_s}\right) = 200 \times \frac{330}{330 - 30} = 220 \text{ Hz}

Step 2: Wall reflects at same frequency 220 Hz. Now wall is source (stationary), car is approaching observer

f_{echo} = f_{wall}\left(\frac{v + v_o}{v}\right) = 220 \times \frac{330 + 30}{330}

f_{echo} = 220 \times \frac{360}{330} = 240 \text{ Hz}

Step 3: Beat frequency

f_{beat} = |f_{echo} - f_{original}| = |240 - 200| = 40 \text{ Hz}

⚠️ Common JEE Mistakes in Doppler Effect

Wrong sign: Forgetting that approaching means "+" in numerator, "-" in denominator
Reflection problems: Not applying Doppler effect twice (source→wall + wall→observer)
Wavelength error: Assuming wavelength changes when observer moves (it doesn't)
Unit conversion: Forgetting to convert km/h to m/s
Wind effect: When wind blows from source to observer, add wind velocity to v

Beats & Resonance

8.1 Beats

Beats occur when two sound waves of slightly different frequencies interfere, resulting in periodic variations in loudness (or intensity).

Beat Frequency Formula

f_{beat} = |f_1 - f_2|

Beat frequency = Absolute difference of the two frequencies

Mathematical Derivation

Two waves: y₁ = A sin(2πf₁t) and y₂ = A sin(2πf₂t)

y = y_1 + y_2 = 2A\cos\left[2\pi\left(\frac{f_1 - f_2}{2}\right)t\right]\sin\left[2\pi\left(\frac{f_1 + f_2}{2}\right)t\right]

Resultant Frequency (Heard)

f_{resultant} = \frac{f_1 + f_2}{2}

Beat Frequency

f_{beat} = |f_1 - f_2|

💡 Important Points About Beats

Beat frequency must be < 10 Hz to be distinguishable by human ear
If beat frequency > 10 Hz, we hear a "rough" sound
Beats are used for tuning musical instruments
Number of beats per second = |f₁ - f₂|
Maximum amplitude = 2A (when in phase)
Minimum amplitude = 0 (when out of phase)

8.2 Resonance

Resonance is the phenomenon when an object is made to vibrate at its natural frequency by receiving energy from another vibrating object at that same frequency, resulting in maximum energy transfer.

Conditions for Resonance

Driving frequency = Natural frequency of the system
Energy is transferred from driver to driven system
Amplitude becomes maximum at resonance

Examples of Resonance:

Tuning fork resonating with air column
Radio/TV tuning (electrical resonance)
Swing pushed at natural frequency
Bridge collapse due to soldiers marching in step
Glass breaking with sound wave

📝 Solved Example 7

Question: Two tuning forks A and B give 5 beats per second. Fork A has frequency 256 Hz. On loading fork B with wax, beat frequency becomes 3 beats/s. Find original frequency of B.

Solution:

Step 1: Original beat frequency = 5 Hz

∴ f_B = 256 ± 5 = 251 Hz or 261 Hz

Step 2: After loading with wax, frequency of B decreases

New beat frequency = 3 Hz

Step 3: Analysis

If f_B = 261 Hz (originally), after loading f_B decreases → approaches 256 Hz → beat frequency should decrease. Beat becomes 3 Hz (decreased from 5). ✓ This works!
If f_B = 251 Hz (originally), after loading f_B decreases → moves away from 256 Hz → beat frequency should increase. But beat decreased to 3 Hz. ✗

\boxed{f_B = 261 \text{ Hz}}

Vibration in Strings & Air Columns

9.1 Vibrating Strings (Both Ends Fixed)

Harmonics in a Fixed String

Condition: Nodes at both ends \[L = n\frac{\lambda_n}{2} \quad \Rightarrow \quad \lambda_n = \frac{2L}{n}\] \[f_n = n \cdot f_1 = \frac{n}{2L}\sqrt{\frac{T}{\mu}} = \frac{nv}{2L}\]

Harmonic	n	Frequency	Wavelength	Loops
Fundamental (1st harmonic)	1	f₁ = v/2L	λ₁ = 2L	1
2nd harmonic (1st overtone)	2	f₂ = 2f₁ = v/L	λ₂ = L	2
3rd harmonic (2nd overtone)	3	f₃ = 3f₁ = 3v/2L	λ₃ = 2L/3	3
nth harmonic	n	fₙ = nf₁ = nv/2L	λₙ = 2L/n	n

Key Points:

All harmonics (odd + even) are present
Frequency ratio → 1:2:3:4:5...
Number of loops = harmonic number
Number of nodes = n + 1, Number of antinodes = n

Harmonics in a String (Both Ends Fixed)

Red dots = Nodes (fixed points), Colored lines = Antinodes (maximum displacement)

9.2 Laws of Vibrating Strings

Three Laws of Transverse Vibration

Law of Length

f \propto \frac{1}{L}

(T and μ constant)

Law of Tension

f \propto \sqrt{T}

(L and μ constant)

Law of Mass

f \propto \frac{1}{\sqrt{\mu}}

(L and T constant)

Combined Formula (Mersenne's Law):

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} = \frac{1}{2L}\sqrt{\frac{T}{\pi r^2 \rho}}

where μ = mass per unit length = πr²ρ (for wire of radius r and density ρ)

9.3 Open Organ Pipe (Both Ends Open)

Harmonics in Open Pipe

Condition: Antinodes at both ends \[L = n\frac{\lambda_n}{2} \quad \Rightarrow \quad \lambda_n = \frac{2L}{n}\] \[f_n = n \cdot f_1 = \frac{nv}{2L}\]

Harmonic	n	Frequency	Wavelength
Fundamental	1	f₁ = v/2L	λ₁ = 2L
1st Overtone	2	f₂ = 2f₁	λ₂ = L
2nd Overtone	3	f₃ = 3f₁	λ₃ = 2L/3

All harmonics (odd + even) are present

9.4 Closed Organ Pipe (One End Closed)

Harmonics in Closed Pipe

Condition: Node at closed end, Antinode at open end \[L = (2n-1)\frac{\lambda_n}{4} \quad \Rightarrow \quad \lambda_n = \frac{4L}{2n-1}\] \[f_n = (2n-1) \cdot f_1 = \frac{(2n-1)v}{4L}\]

Harmonic	n	Frequency	Wavelength
Fundamental (1st harmonic)	1	f₁ = v/4L	λ₁ = 4L
1st Overtone (3rd harmonic)	2	f₃ = 3f₁	λ₃ = 4L/3
2nd Overtone (5th harmonic)	3	f₅ = 5f₁	λ₅ = 4L/5

Only ODD harmonics are present (1, 3, 5, 7...)

🎯 JEE Quick Comparison: Open vs Closed Pipe

Property	Open Pipe	Closed Pipe
Boundary Condition	A-A (Antinode at both ends)	N-A (Node at closed, Antinode at open)
Fundamental Frequency	f₁ = v/2L	f₁ = v/4L
Harmonics Present	All (1, 2, 3, 4...)	Only odd (1, 3, 5...)
For Same L	Higher fundamental	Lower fundamental (f = f_open/2)
Sound Quality	Richer (more harmonics)	Less rich

9.5 End Correction

End Correction Formula

The antinode doesn't form exactly at the open end but slightly outside. This extra length is called end correction (e).

e \approx 0.6r

where r = radius of the pipe

Open Pipe (effective length)

L_{eff} = L + 2e = L + 1.2r

Closed Pipe (effective length)

L_{eff} = L + e = L + 0.6r

9.6 Resonance Tube Experiment

Finding Speed of Sound

In resonance tube experiment, first resonance occurs at length L₁ and second at L₂:

First Resonance

L_1 + e = \frac{\lambda}{4}

Second Resonance

L_2 + e = \frac{3\lambda}{4}

Subtracting to eliminate end correction:

L_2 - L_1 = \frac{\lambda}{2} \quad \Rightarrow \quad \lambda = 2(L_2 - L_1)

v = f\lambda = 2f(L_2 - L_1)

End Correction from experiment:

e = \frac{L_2 - 3L_1}{2}

📝 Solved Example 8

Question: In a resonance tube experiment, first and second resonance are obtained at lengths 20 cm and 62 cm respectively. Find (a) wavelength, (b) speed of sound if frequency = 512 Hz, (c) end correction.

Solution:

Given: L₁ = 20 cm, L₂ = 62 cm, f = 512 Hz

(a) Wavelength:

\lambda = 2(L_2 - L_1) = 2(62 - 20) = 2 \times 42 = 84 \text{ cm} = 0.84 \text{ m}

(b) Speed of Sound:

v = f\lambda = 512 \times 0.84 = 430.08 \text{ m/s}

(c) End Correction:

e = \frac{L_2 - 3L_1}{2} = \frac{62 - 3 \times 20}{2} = \frac{62 - 60}{2} = 1 \text{ cm}

Answers:

(a) λ = 84 cm (b) v = 430 m/s (c) e = 1 cm

Reflection & Transmission of Waves

When a wave encounters a boundary or interface between two media, part of it is reflected and part is transmitted. Understanding this is crucial for solving JEE problems on wave behavior at boundaries.

10.1 Reflection at Boundaries

Reflection from Rigid (Fixed) End

Phase change of π (180°)
Reflected pulse is inverted
Node forms at rigid boundary
Amplitude sign reverses

Example: Wave on string reflecting from fixed end

Reflection from Free (Open) End

No phase change
Reflected pulse is upright
Antinode forms at free boundary
Amplitude sign unchanged

Example: Wave on string with ring sliding on rod

10.2 Reflection at Interface of Two Media

Wave Traveling from Medium 1 to Medium 2

Rarer to Denser (v₁ > v₂)

Reflected wave: Phase change π
Transmitted wave: No phase change
Wavelength decreases in denser medium
Frequency remains same

Denser to Rarer (v₁ < v₂)

Reflected wave: No phase change
Transmitted wave: No phase change
Wavelength increases in rarer medium
Frequency remains same

10.3 Amplitude of Reflected and Transmitted Waves

Amplitude Relations (Strings)

For wave traveling from string of linear density μ₁ to string of linear density μ₂:

Reflected Amplitude

A_r = \frac{v_1 - v_2}{v_1 + v_2} A_i = \frac{\sqrt{\mu_2} - \sqrt{\mu_1}}{\sqrt{\mu_2} + \sqrt{\mu_1}} A_i

Transmitted Amplitude

A_t = \frac{2v_1}{v_1 + v_2} A_i = \frac{2\sqrt{\mu_2}}{\sqrt{\mu_2} + \sqrt{\mu_1}} A_i

Special Cases:

If v₂ → ∞ (free end): A_r = A_i, A_t = 2A_i
If v₂ → 0 (fixed end): A_r = -A_i, A_t = 0
If v₁ = v₂ (same medium): A_r = 0, A_t = A_i (no reflection)

10.4 Energy Relations

Power/Energy Coefficients

Reflection Coefficient (R)

R = \left(\frac{A_r}{A_i}\right)^2 = \left(\frac{v_1 - v_2}{v_1 + v_2}\right)^2

Transmission Coefficient (T)

T = \frac{v_1}{v_2}\left(\frac{A_t}{A_i}\right)^2 = \frac{4v_1v_2}{(v_1 + v_2)^2}

Energy Conservation:

\[R + T = 1\]

Total energy = Reflected energy + Transmitted energy

⚠️ Key Points for JEE

Frequency never changes when wave changes medium
Wavelength changes as velocity changes (λ = v/f)
Phase change of π occurs when wave reflects from a denser medium
A_r can be negative (indicating phase reversal)
A_t is always positive (no phase change in transmission)
For intensity: I ∝ A²ρv (depends on medium properties)

📝 Solved Example 9

Question: A wave traveling in a string of linear density 2 g/m enters another string of linear density 8 g/m. If incident amplitude is 4 cm, find the amplitudes of reflected and transmitted waves.

Solution:

Given: μ₁ = 2 g/m, μ₂ = 8 g/m, A_i = 4 cm

Step 1: Find velocity ratio

Since v ∝ 1/√μ (for same tension)

\frac{v_1}{v_2} = \sqrt{\frac{\mu_2}{\mu_1}} = \sqrt{\frac{8}{2}} = 2

Let v₁ = 2v and v₂ = v

Step 2: Reflected Amplitude

A_r = \frac{v_1 - v_2}{v_1 + v_2} A_i = \frac{2v - v}{2v + v} \times 4 = \frac{v}{3v} \times 4 = \frac{4}{3} \text{ cm}

Step 3: Transmitted Amplitude

A_t = \frac{2v_1}{v_1 + v_2} A_i = \frac{2 \times 2v}{2v + v} \times 4 = \frac{4v}{3v} \times 4 = \frac{16}{3} \text{ cm}

Answers:

Reflected amplitude = 4/3 cm ≈ 1.33 cm (same phase - going to denser)

Transmitted amplitude = 16/3 cm ≈ 5.33 cm

📝 Previous Year Questions Analysis (Wave Motion)

JEE Main (Last 5 Years)

✓ Standing Waves & Harmonics: 25%
✓ Doppler Effect: 25%
✓ Wave Equation & Parameters: 20%
✓ Beats & Resonance: 15%
✓ Sound Intensity & Speed: 10%
✓ Superposition: 5%

JEE Advanced (Last 5 Years)

✓ Standing Waves (Strings/Pipes): 30%
✓ Doppler Effect (Complex): 25%
✓ Wave Equation Analysis: 20%
✓ Reflection/Transmission: 10%
✓ Multi-concept Problems: 10%
✓ Phase & Superposition: 5%

Weightage Analysis

JEE Main: 12-16 marks (3-4 questions)

JEE Advanced: 15-20 marks (4-5 questions)

Difficulty Level: Medium to Hard

Time Required (prep): 8-10 hours complete study

Year-wise Question Distribution

Year	JEE Main	JEE Advanced	Hot Topics
2024	4 Questions (16 marks)	5 Questions (18 marks)	Doppler, Standing waves
2023	3 Questions (12 marks)	4 Questions (15 marks)	Resonance tube, Beats
2022	4 Questions (16 marks)	5 Questions (20 marks)	Wave equation, Organ pipes
2021	3 Questions (12 marks)	4 Questions (16 marks)	Superposition, Phase

Download Complete PYQ PDF (2015-2024) - Wave Motion

🎯 Practice Problem Set - Wave Motion

Level 1: Basic (JEE Main Standard)

A wave is represented by y = 5 sin(4πt - 0.02x) where y and x are in cm and t in seconds. Find amplitude, frequency, wavelength, and wave velocity.
Calculate the speed of sound in air at 27°C if it is 332 m/s at 0°C.
Two tuning forks A and B give 4 beats/s. A has frequency 480 Hz. Find possible frequencies of B.
A string 1 m long has fundamental frequency 400 Hz. Find frequency of 3rd harmonic.
A closed organ pipe has fundamental frequency 500 Hz. Find frequencies of first two overtones.
Convert intensity 10⁻⁵ W/m² to decibels.
A train moving at 36 km/h emits sound of 500 Hz. Find apparent frequency when it approaches a stationary observer. (v = 340 m/s)
What is the phase difference between two particles 0.25 m apart if wavelength is 0.5 m?

Level 2: Intermediate (JEE Main/Advanced)

The equation of a wave is y = 0.05 sin(10πx - 200πt) where x, y are in meters and t in seconds. Find (a) wavelength (b) frequency (c) wave velocity (d) max particle velocity (e) max particle acceleration.
Two identical strings have same tension. One is 60 cm and the other is 90 cm long. Find ratio of their fundamental frequencies.
An open pipe and a closed pipe have same fundamental frequency. If length of open pipe is 60 cm, find length of closed pipe.
A train moving at 54 km/h blows a horn of frequency 600 Hz. Find apparent frequency heard by a person moving at 18 km/h (a) towards the train (b) away from the train. (v = 340 m/s)
Two waves y₁ = 3 sin(ωt) and y₂ = 4 sin(ωt + π/2) superpose. Find resultant amplitude and phase.
In resonance tube experiment, first resonance is at 17 cm and second at 51 cm. Find (a) wavelength (b) end correction.
A tuning fork of frequency 340 Hz is vibrated just above a tube of length 100 cm. Find the height of water in tube for resonance. (v = 340 m/s)
A source of sound of frequency 256 Hz moves with velocity 25 m/s towards a wall. What beat frequency is heard by the observer at the source? (v = 330 m/s)
The intensity of sound from a point source at 4 m is 10 W/m². What is the intensity at 8 m?
A string vibrates in 3 loops. If tension is increased by 44%, in how many loops will it vibrate at same frequency?

Level 3: Advanced (JEE Advanced/Olympiad)

A wave pulse is traveling on a string with speed v. The shape of pulse is triangular with height h and base 2l. Find the total kinetic energy of the pulse. (linear density = μ)
Two coherent sources S₁ and S₂ are separated by 4λ. A circular wire of radius r (r >> λ) is placed with center midway between sources. Find number of maxima and minima on the wire.
A tube of length L₁ open at both ends and another tube of length L₂ closed at one end have same fundamental frequency. Find the ratio of 6th harmonic of open pipe to 5th harmonic of closed pipe.
A source of sound moves along a circle of radius R with constant angular velocity ω. An observer is at distance d from center. Find maximum and minimum frequencies heard. (d > R)
Prove that when a wave travels from string of linear density μ₁ to string of linear density μ₂, the reflection and transmission coefficients satisfy R + T = 1.
Two strings of same material and length have radii r and 2r. They are joined and stretched with tension T. A wave in thinner string has amplitude A. Find amplitude of transmitted wave in thicker string.
A cylindrical tube closed at one end contains air column of variable length. The minimum length for resonance with a fork of frequency f is 20 cm. Find the next two lengths for resonance with same fork and also the speed of sound.
A police car sounding a siren of frequency 1000 Hz moves towards a building at 20 m/s. What frequency does the driver hear from the reflected wave? (v = 340 m/s)
Two speakers facing each other are 10 m apart. Both emit sound of frequency 680 Hz. A person walks from one speaker to other. How many maxima will he hear? (v = 340 m/s)
Show that the displacement node is a pressure antinode and vice versa in a standing sound wave.

Conceptual Questions (Theory Based)

Why can't longitudinal waves be polarized?
Explain why sound travels faster in humid air than in dry air.
Why is Laplace's correction necessary for speed of sound formula?
What happens to wavelength when a wave enters from a rarer to denser medium?
Why do closed pipes produce only odd harmonics?
Explain the physical significance of phase velocity and group velocity.
Why does a siren sound higher when approaching and lower when receding?
What is the difference between interference and beats?
Why does resonance occur only at specific frequencies?
Explain end correction in organ pipes with diagram.

Get Complete Solutions PDF with Step-by-Step Working

📋 Wave Motion - Complete Formula Sheet

Basic Wave Relations

v = fλ = λ/T = ω/k

k = 2π/λ (wave number)

ω = 2πf = 2π/T

y = A sin(kx - ωt + φ)

Wave Velocity

String: v = √(T/μ)

Gas: v = √(γP/ρ) = √(γRT/M)

Solid: v = √(Y/ρ)

Liquid: v = √(B/ρ)

Particle Properties

v_particle = -Aω cos(kx - ωt)

v_max = Aω

a_max = Aω²

v_p = -v × (dy/dx)

Standing Waves

y = 2A sin(kx) cos(ωt)

Nodes: x = 0, λ/2, λ, ...

Antinodes: x = λ/4, 3λ/4, ...

Node-Antinode dist = λ/4

String (Both Ends Fixed)

f_n = nv/2L = (n/2L)√(T/μ)

λ_n = 2L/n

All harmonics present

n = 1, 2, 3, ...

Open Pipe

f_n = nv/2L

λ_n = 2L/n

All harmonics (1,2,3...)

Antinodes at both ends

Closed Pipe

f_n = (2n-1)v/4L

λ_n = 4L/(2n-1)

Only odd harmonics (1,3,5...)

Node at closed, antinode at open

Doppler Effect

f' = f₀[(v ± v_o)/(v ∓ v_s)]

Upper: approaching

Lower: receding

Beat: f_beat = |f₁ - f₂|

Sound & Intensity

I = P/A = ½ρvω²A²

β = 10 log₁₀(I/I₀) dB

I₀ = 10⁻¹² W/m²

v_t = 332 + 0.61t m/s

📑 Quick Navigation - Wave Motion

Wave Motion JEE Main & Advanced 2025-26

Introduction to Wave Motion

1.1 What is a Wave?

Definition of Wave

1.2 Wave Parameters

1.3 Fundamental Wave Relations

Essential Formulas

🎯 JEE Tip: Quick Conversions

Types of Waves

2.1 Based on Medium Requirement

Mechanical Waves

Electromagnetic Waves

2.2 Based on Particle Motion

Transverse Waves

Longitudinal Waves

2.3 Based on Energy Propagation

2.4 Wave Velocity in Different Media

Important Velocity Formulas

📝 Solved Example 1

⚠️ Common Mistakes

Progressive Wave Equation

3.1 General Wave Equation

Standard Forms of Progressive Wave

3.2 Understanding the Wave Equation

3.3 Particle Velocity and Acceleration

Derivatives of Wave Equation

💡 Quick Tricks for Wave Equation

📝 Solved Example 2 (JEE Main Pattern)

3.4 Phase and Phase Difference

Phase Relations

Superposition of Waves

4.1 Principle of Superposition

Mathematical Statement

4.2 Interference of Two Waves

Resultant of Two Sinusoidal Waves

4.3 Special Cases of Interference

💡 JEE Shortcut for Equal Amplitude Waves

4.4 Intensity and Interference

Intensity Relations

📝 Solved Example 3

Standing (Stationary) Waves

5.1 Formation of Standing Waves

Mathematical Formation

5.2 Nodes and Antinodes

Nodes (Zero Amplitude)

Antinodes (Maximum Amplitude)

5.3 Properties of Standing Waves

⚠️ Phase in Standing Waves

5.4 Boundary Conditions and Reflection

Fixed End (Rigid Boundary)

Free End (Open Boundary)

Sound Waves

6.1 Nature of Sound Waves

Key Properties of Sound

6.2 Speed of Sound

Speed of Sound in Different Media

6.3 Effect of Various Factors on Speed of Sound

💡 Quick Formula: Temperature Effect

6.4 Intensity and Loudness

Intensity of Sound

📝 Solved Example 4

Doppler Effect

7.1 Master Formula for Doppler Effect

General Doppler Formula (Sound Waves)

7.2 Special Cases of Doppler Effect

💡 JEE Quick Trick: SOFA Rule

7.3 Apparent Wavelength

Wavelength Change in Doppler Effect

📝 Solved Example 5 (JEE Main Pattern)

📝 Solved Example 6 (JEE Advanced Pattern)

⚠️ Common JEE Mistakes in Doppler Effect

Beats & Resonance

8.1 Beats

Beat Frequency Formula

Mathematical Derivation

💡 Important Points About Beats

8.2 Resonance

Conditions for Resonance

📝 Solved Example 7