Chapter 15: Waves

Introduction

Wave Definition and Properties

• Wave: disturbance that transfers energy without matter transfer.
• Mechanical waves need a medium; electromagnetic waves do not.
• Example: Water waves show energy propagation via ripples; particles vibrate about mean positions.

Wave Features

• Crest: highest point of wave.
• Trough: lowest point of wave.
• Wave direction: direction of energy propagation.

Solved Examples

Ex 1

• Identify crest and trough in a water wave profile.
• Answer: Crest is peak above normal level; trough is valley below normal level.

Practice Set

• Q1: What is transferred by a wave?
  • A) Matter
  • B) Energy
  • C) Both matter and energy
  • D) Neither
• Q2: Which wave can travel through vacuum?
  • A) Sound wave
  • B) Water wave
  • C) Electromagnetic wave
  • D) Seismic wave
• Q3: In water waves, particles move:
  • A) Along wave direction
  • B) Perpendicular to wave direction
  • C) In circular motion about mean position
  • D) Do not move

Answer Key

• Q1: B - Waves transfer energy, not matter.
• Q2: C - Electromagnetic waves travel through vacuum.
• Q3: C - Water particles vibrate in circular motion.

Memory Aids

• Wave = Energy transfer without matter movement.
• Crest = High point; Trough = Low point.
• EM waves = Vacuum; Mechanical waves = Medium.

Mechanical Waves

Definition and Characteristics

• Require material medium (solid, liquid, gas) for propagation.
• Obey Newton’s laws.
• Types: Transverse and Longitudinal.

Examples

• Water waves, sound waves, seismic waves.

Solved Examples

Ex 1

• Classify sound wave: Mechanical or electromagnetic?
• Answer: Mechanical wave (requires air medium).

Practice Set

• Q1: Mechanical waves require:
  • A) Vacuum
  • B) Medium
  • C) Electric field
  • D) Magnetic field
• Q2: Which is NOT a mechanical wave?
  • A) Sound wave
  • B) Water wave
  • C) Light wave
  • D) Seismic wave
• Q3: Mechanical waves obey:
  • A) Newton’s laws
  • B) Maxwell’s equations
  • C) Quantum mechanics
  • D) None

Answer Key

• Q1: B - Medium is essential for mechanical waves.
• Q2: C - Light is electromagnetic, not mechanical.
• Q3: A - Mechanical waves follow Newton’s laws.

Memory Aids

• Mechanical = Medium needed.
• Sound, water, seismic = Mechanical waves.
• Newton’s laws govern mechanical waves.

Electromagnetic Waves

Definition and Properties

• Oscillating electric and magnetic fields.
• Do not require material medium; can travel in vacuum.
• Examples: Light from sun, radio waves.

Solved Examples

Ex 1

• Can light travel through vacuum?
• Answer: Yes, electromagnetic waves travel through vacuum.

Practice Set

• Q1: Electromagnetic waves require:
  • A) Medium
  • B) Vacuum or medium
  • C) Only vacuum
  • D) None
• Q2: Which is an electromagnetic wave?
  • A) Sound
  • B) Water wave
  • C) Light
  • D) Seismic wave
• Q3: Electromagnetic waves are:
  • A) Mechanical
  • B) Non-mechanical
  • C) Both
  • D) None

Answer Key

• Q1: B - EM waves can travel in vacuum or medium.
• Q2: C - Light is an electromagnetic wave.
• Q3: B - EM waves are non-mechanical.

Memory Aids

• EM waves = Electric + Magnetic fields.
• Travel in vacuum, no medium needed.
• Light is EM wave.

Matter Waves

Definition and Importance

• Associated with microscopic particles in motion (electrons, protons, atoms).
• Important in quantum mechanics.
• Example: Electron microscopes use matter waves.

Atomic Model

• Electron orbits nucleus with positive charge.
• Electron has associated matter wave.

Solved Examples

Ex 1

• What wave is associated with an electron?
• Answer: Matter wave (de-Broglie wave).

Practice Set

• Q1: Matter waves are associated with:
  • A) Macroscopic objects
  • B) Microscopic particles
  • C) Electromagnetic fields
  • D) Sound waves
• Q2: Electron microscopes use:
  • A) Light waves
  • B) Matter waves
  • C) Sound waves
  • D) Seismic waves
• Q3: Matter waves are important in:
  • A) Classical mechanics
  • B) Quantum mechanics
  • C) Thermodynamics
  • D) Electromagnetism

Answer Key

• Q1: B - Matter waves relate to microscopic particles.
• Q2: B - Electron microscopes use matter waves.
• Q3: B - Matter waves are quantum mechanical concept.

Memory Aids

• Matter waves = de-Broglie waves.
• Electron microscopes use matter waves.
• Quantum mechanics explains matter waves.

Transverse and Longitudinal Waves

Transverse Waves

• Particle vibration perpendicular to wave direction.
• Propagated as crests and troughs.
• Example: Waves on a string.

Longitudinal Waves

• Particle vibration parallel to wave direction.
• Propagated as compressions and rarefactions.
• Example: Sound waves in air.

Wave Combination and Speed

• Ocean waves combine transverse and longitudinal motions.
• Transverse and longitudinal waves travel at different speeds in same medium.

Solved Examples

Ex 1

• Identify wave type if particles vibrate perpendicular to wave motion.
• Answer: Transverse wave.

Practice Set

• Q1: In transverse waves, particle motion is:
  • A) Parallel to wave direction
  • B) Perpendicular to wave direction
  • C) Circular
  • D) Random
• Q2: Longitudinal waves consist of:
  • A) Crests and troughs
  • B) Compressions and rarefactions
  • C) Only compressions
  • D) Only rarefactions
• Q3: Ocean waves are:
  • A) Purely transverse
  • B) Purely longitudinal
  • C) Combination of transverse and longitudinal
  • D) Neither

Answer Key

• Q1: B - Transverse waves have perpendicular particle motion.
• Q2: B - Longitudinal waves have compressions and rarefactions.
• Q3: C - Ocean waves combine both wave types.

Memory Aids

• Transverse = perpendicular vibration.
• Longitudinal = parallel vibration.
• Ocean waves = mixed motion.

Water Surface Waves

Types and Restoring Forces

• Capillary waves: short wavelength, surface tension restores.
• Gravity waves: longer wavelength, gravity restores.

Solved Examples

Ex 1

• Identify restoring force for capillary waves.
• Answer: Surface tension.

Practice Set

• Q1: Capillary waves have restoring force due to:
  • A) Gravity
  • B) Surface tension
  • C) Elasticity
  • D) Pressure
• Q2: Gravity waves have wavelengths of:
  • A) Millimeters
  • B) Several meters to hundreds of meters
  • C) Micrometers
  • D) None
• Q3: Restoring force for gravity waves is:
  • A) Surface tension
  • B) Gravity
  • C) Elasticity
  • D) Magnetic force

Answer Key

• Q1: B - Surface tension restores capillary waves.
• Q2: B - Gravity waves have long wavelengths.
• Q3: B - Gravity is restoring force for gravity waves.

Memory Aids

• Capillary = surface tension, short waves.
• Gravity waves = gravity, long waves.

Displacement Relation in a Progressive Wave

Wave Displacement Equation

• Transverse displacement: \( y(x,t) = a \sin(kx - \omega t + \phi_0) \)
• \(a\): Amplitude (max displacement)
• \(k\): Angular wave number \(= \frac{2\pi}{\lambda}\)
• \(\omega\): Angular frequency \(= 2\pi f\)
• \(\phi_0\): Initial phase

Solved Examples

Ex 1

• Given \( \lambda = 0.5\,m \), find \( k \).
• \( k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.5} = 4\pi\,\text{rad/m} \)

Ex 2

• Given frequency \( f = 50\,Hz \), find angular frequency \( \omega \).
• \( \omega = 2\pi f = 2\pi \times 50 = 100\pi\,\text{rad/s} \)

Practice Set

Q1

What is the amplitude in the wave equation \( y = 0.02 \sin(10x - 500t) \)?

• A) 0.02 m
• B) 10 m
• C) 500 m
• D) 0.5 m

Q2

If the wavelength is 0.4 m, what is the angular wave number \( k \)?

• A) 5\pi rad/m
• B) 2.5\pi rad/m
• C) 10\pi rad/m
• D) 0.4 rad/m

Q3

A wave has frequency 60 Hz. Calculate its angular frequency \( \omega \).

• A) 120\pi rad/s
• B) 60 rad/s
• C) 30\pi rad/s
• D) 2\pi rad/s

Answer Key

• Q1: A. Amplitude is the coefficient before sine, 0.02 m.
• Q2: A. \( k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.4} = 5\pi \) rad/m.
• Q3: A. \( \omega = 2\pi f = 2\pi \times 60 = 120\pi \) rad/s.

Memory Aids

• Amplitude = max displacement (coefficient of sine)
• \( k = \frac{2\pi}{\lambda} \), remember \(2\pi\) radians per wavelength
• \( \omega = 2\pi f \), angular frequency relates to frequency

Wave Parameters

Key Definitions and Formulas

• Amplitude (a): Maximum displacement from equilibrium.
• Phase: Argument of sine/cosine in displacement equation.
• Wavelength (\(\lambda\)): Minimum distance between points of same phase.
• Time period (T): Time for one complete vibration.
• Frequency (f): Number of vibrations per second, \( f = \frac{1}{T} \), unit Hz.
• Angular frequency (\(\omega\)): Rate of phase change, \( \omega = \frac{2\pi}{T} = 2\pi f \), unit rad/s.
• Angular wave number (k): Number of waves per unit length, \( k = \frac{2\pi}{\lambda} \).

Solved Examples

Ex 1

• Find frequency if time period \( T = 0.01 s \).
• \( f = \frac{1}{T} = \frac{1}{0.01} = 100\,Hz \)

Ex 2

• Calculate angular frequency for \( f = 50 Hz \).
• \( \omega = 2\pi \times 50 = 100\pi\,rad/s \)

Practice Set

Q1

What is the frequency of a wave with period 0.02 s?

• A) 50 Hz
• B) 20 Hz
• C) 0.02 Hz
• D) 2 Hz

Q2

Angular frequency of a wave with frequency 25 Hz is:

• A) 50\pi rad/s
• B) 25 rad/s
• C) 100 rad/s
• D) 12.5\pi rad/s

Q3

Wavelength of a wave with wave number \( k = 4\pi \) rad/m is:

• A) 0.5 m
• B) 2 m
• C) 4 m
• D) 8 m

Answer Key

• Q1: A. \( f = 1/T = 1/0.02 = 50 Hz \)
• Q2: A. \( \omega = 2\pi f = 2\pi \times 25 = 50\pi \) rad/s.
• Q3: A. \( \lambda = \frac{2\pi}{k} = \frac{2\pi}{4\pi} = 0.5 m \)

Memory Aids

• Frequency and period are reciprocals: \( f = \frac{1}{T} \)
• Angular frequency = \( 2\pi \times \) frequency
• Wave number = \( \frac{2\pi}{\lambda} \)

Wave Progression Visualization

Wave Motion and Particle Oscillation

• Particles oscillate about mean position as wave passes.
• Crest moves in positive x-direction indicating wave travel.
• Graphs show sinusoidal waves with varying initial phases.
• Amplitude = peak height; wavelength = distance between peaks.
• Phase shifts change wave starting point along x-axis.

Solved Examples

Ex 1

• Identify wave direction from graph crest movement.
• Crest moves right → wave travels in positive x-direction.

Practice Set

Q1

What does the crest movement in the positive x-direction indicate?

• A) Wave travels left
• B) Wave travels right
• C) Particle displacement zero
• D) Wave is stationary

Q2

What does the distance between two crests represent?

• A) Amplitude
• B) Wavelength
• C) Frequency
• D) Phase

Q3

Changing initial phase shifts the wave along:

• A) Time axis
• B) Amplitude axis
• C) Space (x) axis
• D) Frequency axis

Answer Key

• Q1: B. Crest moving right means wave travels in positive x-direction.
• Q2: B. Distance between crests is wavelength.
• Q3: C. Initial phase shifts wave along x-axis (space).

Memory Aids

• Crest movement direction = wave travel direction
• Wavelength = distance between crests
• Initial phase shifts wave horizontally

Speed of a Travelling Wave

Wave Progression and Speed

• A harmonic wave shifts along the x-axis over time.
• Distance moved in time \( \Delta t \) is \( \Delta x \).
• Speed of wave \( v = \frac{\Delta x}{\Delta t} \).
• Phase condition: \( kx - \omega t = \text{constant} \).
• From phase condition, \( \frac{dx}{dt} = \frac{\omega}{k} = v \).
• General wave speed formula: \( v = \lambda \nu = \frac{\lambda}{T} \).

Solved Examples

Ex 1

• Given wavelength \( \lambda = 0.5\,m \), frequency \( \nu = 200\,Hz \).
• Calculate speed: \( v = \lambda \nu = 0.5 \times 200 = 100\,m/s \).

Practice Set

• Q1: If a wave travels 2 m in 0.01 s, its speed is?
  • A) 20 m/s
  • B) 200 m/s
  • C) 0.02 m/s
  • D) 2 m/s
• Q2: For a wave with \( \lambda = 1\,m \) and period \( T = 0.02\,s \), speed is?
  • A) 50 m/s
  • B) 0.02 m/s
  • C) 20 m/s
  • D) 2 m/s
• Q3: The phase velocity \( v = \frac{\omega}{k} \) implies?
  • A) Speed depends on amplitude
  • B) Speed depends on phase constant
  • C) Speed is ratio of angular frequency to wave number
  • D) Speed is product of frequency and wavelength

Answer Key

• Q1: B) 200 m/s — Speed = distance/time = 2/0.01 = 200 m/s.
• Q2: A) 50 m/s — Speed = \( \frac{\lambda}{T} = \frac{1}{0.02} = 50\,m/s \).
• Q3: C) Speed is ratio of angular frequency to wave number — By definition \( v = \frac{\omega}{k} \).

Memory Aids

• Wave speed \( v = \lambda \nu \) (wavelength × frequency).
• Phase velocity \( v = \frac{\omega}{k} \).
• Speed = distance/time for wave crest movement.

Speed of Transverse Wave on Stretched String

Factors Affecting Speed

• Depends on tension \( T \) and linear mass density \( \mu \) of string.
• Tension \( T \) units: Newtons (N) = \( [MLT^{-2}] \).
• Linear mass density \( \mu = \frac{m}{L} \), units: kg/m.
• Speed formula: \( v = \sqrt{\frac{T}{\mu}} \).
• Speed independent of frequency.

Solved Examples

Ex 1

• Given \( T = 100\,N \), \( \mu = 0.01\,kg/m \).
• Calculate speed: \( v = \sqrt{\frac{100}{0.01}} = \sqrt{10000} = 100\,m/s \).

Practice Set

• Q1: If tension doubles and \( \mu \) remains constant, speed changes by?
  • A) Doubles
  • B) Halves
  • C) Increases by \( \sqrt{2} \)
  • D) No change
• Q2: For \( T = 50\,N \), \( \mu = 0.02\,kg/m \), speed is?
  • A) 50 m/s
  • B) 35.36 m/s
  • C) 25 m/s
  • D) 100 m/s
• Q3: Speed of wave on string depends on?
  • A) Frequency and tension
  • B) Tension and linear mass density
  • C) Frequency and linear mass density
  • D) Amplitude and tension

Answer Key

• Q1: C) Increases by \( \sqrt{2} \) — Speed \( v \propto \sqrt{T} \).
• Q2: B) 35.36 m/s — \( v = \sqrt{\frac{50}{0.02}} = \sqrt{2500} = 50\,m/s \) (Check calculation: actually 50 m/s, so correct option is A). Correction: A) 50 m/s is correct.
• Q3: B) Tension and linear mass density — Speed depends only on these two.

Memory Aids

• Speed \( v = \sqrt{\frac{T}{\mu}} \).
• More tension → faster wave.
• More mass density → slower wave.

Speed of Longitudinal Wave

Dependence on Medium Properties

• Speed depends on bulk modulus \( B \) and density \( \rho \) of medium.
• Formula: \( v = \sqrt{\frac{B}{\rho}} \).
• For solids: \( v = \sqrt{\frac{Y}{\rho}} \), where \( Y \) is Young's modulus.
• Newton's formula for ideal gas: \( v = \sqrt{\frac{P}{\rho}} \), \( P \) = pressure.
• Laplace correction: \( v = \sqrt{\frac{\gamma P}{\rho}} \), \( \gamma = \frac{C_p}{C_v} \).
• At STP, speed of sound in air \( \approx 331.3\,m/s \).

Solved Examples

Ex 1

• Given air at STP: \( P = 1.01 \times 10^5\,Pa \), \( \rho = 1.29\,kg/m^3 \), \( \gamma = 1.4 \).
• Calculate speed: \( v = \sqrt{\frac{1.4 \times 1.01 \times 10^5}{1.29}} \approx 331.3\,m/s \).

Practice Set

• Q1: Speed of sound increases if?
  • A) Density increases
  • B) Bulk modulus decreases
  • C) Temperature increases (affects \( P \) and \( \rho \))
  • D) Pressure decreases
• Q2: For a solid rod with \( Y = 2 \times 10^{11} \) Pa and \( \rho = 8000\,kg/m^3 \), speed is?
  • A) 5000 m/s
  • B) 15811 m/s
  • C) 500 m/s
  • D) 1581 m/s
• Q3: Laplace correction factor \( \gamma \) for air is?
  • A) 1
  • B) 7/5
  • C) 5/7
  • D) 2

Answer Key

• Q1: C) Temperature increases — Speed increases as temperature affects pressure and density.
• Q2: D) 1581 m/s — \( v = \sqrt{\frac{2 \times 10^{11}}{8000}} \approx 15811\,m/s \) (Check calculation: actually \( \sqrt{2.5 \times 10^7} \approx 5000 \) m/s, so correct option is A) 5000 m/s.
• Q3: B) 7/5 — Ratio of specific heats for air.

Memory Aids

• Speed \( v = \sqrt{\frac{B}{\rho}} \) for fluids.
• Speed \( v = \sqrt{\frac{Y}{\rho}} \) for solids.
• Laplace correction: multiply by \( \sqrt{\gamma} \).
• Speed of sound in air at STP \( \approx 331.3\,m/s \).

Principle of Superposition

Explanation

• When two or more waves travel in the same medium, the resultant displacement is the algebraic sum of individual displacements.
• Mathematically, \( y = \sum_{i=1}^n f_i(x - vt) \), where \( f_i \) are individual wave functions.
• Each wave function depends on position \( x \), time \( t \), and wave speed \( v \).
• This principle applies to all linear wave phenomena.

Solved Examples

• Ex 1: Two waves \( y_1 = a \sin(kx - \omega t) \) and \( y_2 = a \sin(kx - \omega t + \Phi) \) travel along a string. Find the resultant wave.
• Step 1: Apply superposition: \( y = y_1 + y_2 = a \sin(kx - \omega t) + a \sin(kx - \omega t + \Phi) \).
• Step 2: Use sine addition formula: \( y = 2a \cos(\frac{\Phi}{2}) \sin(kx - \omega t + \frac{\Phi}{2}) \).

Practice Set

• Q1: What is the resultant displacement when two waves of equal amplitude and zero phase difference interfere?
  • A) Zero
  • B) Double amplitude
  • C) Half amplitude
  • D) Same as individual amplitude
• Q2: Two waves \( y_1 = 3 \sin(kx - \omega t) \) and \( y_2 = 3 \sin(kx - \omega t + \pi) \) interfere. What is the amplitude of the resultant wave?
  • A) 6
  • B) 0
  • C) 3
  • D) 1.5
• Q3: Two waves of amplitude 5 cm and phase difference \( \frac{\pi}{3} \) interfere. Find the resultant amplitude.
  • A) 10 cm
  • B) 5 cm
  • C) 8.66 cm
  • D) 2.5 cm

Answer Key

• Q1: B) Double amplitude. Zero phase difference means constructive interference, amplitudes add up.
• Q2: B) 0. Phase difference \( \pi \) causes destructive interference, resulting in zero amplitude.
• Q3: C) 8.66 cm. Resultant amplitude = \( 2a \cos(\frac{\Phi}{2}) = 2 \times 5 \times \cos(\frac{\pi}{6}) = 10 \times 0.866 = 8.66 \) cm.

Memory Aids

• Superposition: Resultant = sum of displacements.
• Constructive interference: \( \Phi = 0, 2\pi, 4\pi, ... \)
• Destructive interference: \( \Phi = \pi, 3\pi, 5\pi, ... \)
• Resultant amplitude formula: \( 2a \cos(\frac{\Phi}{2}) \)

Interference of Harmonic Waves

Explanation

• Two harmonic waves of equal amplitude \( a \) and wavelength interfere according to phase difference \( \Phi \).
• Individual waves: \( y_1 = a \sin(kx - \omega t) \), \( y_2 = a \sin(kx - \omega t + \Phi) \).
• Resultant wave: \( y = 2a \cos(\frac{\Phi}{2}) \sin(kx - \omega t + \frac{\Phi}{2}) \).
• Amplitude depends on \( \cos(\frac{\Phi}{2}) \): maximum for \( \Phi = 0 \), zero for \( \Phi = \pi \).

Solved Examples

• Ex 1: Find the resultant amplitude when two waves of amplitude 4 cm interfere with phase difference \( \pi \).
• Step 1: Use formula \( A = 2a \cos(\frac{\Phi}{2}) = 2 \times 4 \times \cos(\frac{\pi}{2}) = 0 \).
• Result: Complete destructive interference, amplitude zero.

Practice Set

• Q1: For two waves with phase difference \( 0 \), the resultant amplitude is:
  • A) 0
  • B) \( 2a \)
  • C) \( a \)
  • D) \( a/2 \)
• Q2: Two waves of amplitude 2 cm interfere with phase difference \( \pi/2 \). Resultant amplitude is:
  • A) 4 cm
  • B) 2.83 cm
  • C) 2 cm
  • D) 1.41 cm
• Q3: Two waves of amplitude 3 cm interfere with phase difference \( \frac{3\pi}{2} \). Resultant amplitude is:
  • A) 6 cm
  • B) 0 cm
  • C) 2.12 cm
  • D) 4.24 cm

Answer Key

• Q1: B) \( 2a \). Zero phase difference means constructive interference, amplitude doubles.
• Q2: B) 2.83 cm. \( A = 2 \times 2 \times \cos(\pi/4) = 4 \times 0.707 = 2.83 \) cm.
• Q3: C) 2.12 cm. \( A = 2 \times 3 \times \cos(3\pi/4) = 6 \times (-0.707) = -4.24 \) amplitude magnitude is 4.24 cm but negative sign indicates phase; actual amplitude is 4.24 cm (absolute value). Correct option closest is 4.24 cm (D). Correction: Since options differ, correct is D) 4.24 cm.

Memory Aids

• Resultant amplitude depends on \( \cos(\frac{\Phi}{2}) \).
• Constructive: \( \Phi = 0, 2\pi, ... \) amplitude = \( 2a \).
• Destructive: \( \Phi = \pi, 3\pi, ... \) amplitude = 0.
• Use sine addition formula for combining waves.

Reflection of Waves

Key Points

• Wave pulse on a stretched string reflects differently at fixed and free ends.
• At fixed end: reflected pulse is inverted; phase change of \(\pi\).
• At free end: reflected pulse is not inverted; no phase change.
• When wave crosses boundary between two media, part reflects, part refracts.

Solved Examples

• Ex 1: A pulse on a string fixed at one end reflects. What is the phase change?
  Answer: \(\pi\) (inversion)
• Ex 2: Pulse reflects from free end. Orientation?
  Answer: Same as incident pulse (no inversion)

Practice Set

• Q1: What happens to a wave pulse at a fixed end?
  A) No reflection B) Reflection without inversion C) Reflection with inversion D) Absorbed
  
• Q2: Phase change on reflection at free end is:
  A) \(\pi\) B) \(\pi/2\) C) 0 D) \(2\pi\)
  
• Q3: When a wave passes from medium 1 to medium 2, what occurs?
  A) Full reflection B) Full transmission C) Partial reflection and refraction D) No change
  

Answer Key

• Q1: C - Fixed end causes reflection with inversion (phase change \(\pi\)).
• Q2: C - Free end reflection has no phase change.
• Q3: C - Wave partly reflects and partly refracts at boundary.

Memory Aids

• Fixed end = Inverted reflection (phase \(\pi\)).
• Free end = Same orientation reflection (no phase change).
• Boundary crossing = Partial reflection + refraction.

Standing Waves and Normal Modes

Key Points

• Standing wave formed by superposition of two waves traveling opposite directions with same amplitude, frequency, wavelength.
• Equation: \(y(x,t) = 2a \sin kx \cos \omega t\)
• Nodes: points of zero amplitude at \(x = n \frac{\lambda}{2}\)
• Antinodes: points of maximum amplitude at \(x = (n + \frac{1}{2}) \frac{\lambda}{2}\)
• Normal modes: natural frequencies at which system vibrates.

Solved Examples

• Ex 1: Find node positions for \(\lambda = 0.4\,m\).
  Answer: \(x = 0, 0.2, 0.4, 0.6, ...\) m
• Ex 2: Write standing wave equation for amplitude \(a\).
  Answer: \(y = 2a \sin kx \cos \omega t\)

Practice Set

• Q1: Nodes occur at positions where amplitude is:
  A) Maximum B) Zero C) Half maximum D) Variable
  
• Q2: Standing wave formed by:
  A) One traveling wave B) Two traveling waves in same direction C) Two traveling waves in opposite directions D) No waves
  
• Q3: Antinode position for wavelength \(\lambda\) is:
  A) \(n \lambda\) B) \(n \frac{\lambda}{2}\) C) \((n + \frac{1}{2}) \frac{\lambda}{2}\) D) \(\frac{\lambda}{4}\)
  

Answer Key

• Q1: B - Nodes have zero amplitude.
• Q2: C - Standing wave is superposition of two opposite traveling waves.
• Q3: C - Antinodes at \(x = (n + 1/2) \lambda/2\).

Memory Aids

• Standing wave = two opposite waves superposed.
• Nodes = zero amplitude points.
• Antinodes = maximum amplitude points.

Vibrations of Stretched String

Key Points

• Wave velocity: \(v = \sqrt{\frac{T}{\mu}}\), \(\mu\) = mass per unit length (kg/m).
• Frequency: \(\nu = \frac{1}{\lambda} \sqrt{\frac{T}{\mu}}\)
• Wavelength: \(\lambda = \frac{2l}{n}\), \(n=1,2,3...\)
• Fundamental frequency (1st harmonic): \(\nu_1 = \frac{v}{2l}\)
• Higher harmonics: \(\nu_n = n \nu_1\)
• Mass per unit length: \(\mu = \frac{\pi D^2 \rho}{4}\)
• Frequency laws:
  - \(\nu \propto \frac{1}{L}\) (length)
  - \(\nu \propto \sqrt{T}\) (tension)
  - \(\nu \propto \frac{1}{\sqrt{\mu}}\) (mass per unit length)

Solved Examples

• Ex 1: Calculate fundamental frequency for string length 1 m, tension 100 N, \(\mu = 0.01\,kg/m\).
  Solution: \(v=\sqrt{100/0.01}=100\,m/s\), \(\nu_1=100/(2\times1)=50\,Hz\)
• Ex 2: Find wavelength for 3rd harmonic on string length 1 m.
  Answer: \(\lambda_3 = \frac{2l}{3} = \frac{2}{3} m\)

Practice Set

• Q1: Frequency is proportional to:
  A) Length B) \(\sqrt{T}\) C) Mass per unit length D) \(1/\sqrt{T}\)
  
• Q2: Wavelength of 2nd harmonic for string length \(l\) is:
  A) \(2l\) B) \(l\) C) \(l/2\) D) \(4l\)
  
• Q3: If diameter of string doubles, frequency:
  A) Doubles B) Halves C) Decreases by factor \(\sqrt{2}\) D) Increases by factor \(\sqrt{2}\)
  

Answer Key

• Q1: B - Frequency \(\propto \sqrt{T}\).
• Q2: B - \(\lambda_2 = l\).
• Q3: C - \(\mu \propto D^2\), frequency \(\propto 1/\sqrt{\mu}\), so frequency decreases by \(\sqrt{2}\).

Memory Aids

• \(v = \sqrt{T/\mu}\)
• \(\nu_1 = v/2l\)
• Frequency \(\propto \sqrt{T}\), inversely to length and mass density.

Organ Pipe Oscillations

Key Points

• Organ pipes produce sound by standing longitudinal waves.
• Closed pipe (one end closed):
  Frequency \(\nu = \frac{(2n+1)v}{4l}\), only odd harmonics.
• Open pipe (both ends open):
  Frequency \(\nu = \frac{nv}{2l}\), all harmonics present.
• Positions:
  Closed pipe nodes at closed end, antinodes at open end.
  Open pipe antinodes at both ends.
• Overtone ratios:
  Closed pipe: 3:5:7...
  Open pipe: 2:3:4...

Solved Examples

• Ex 1: Frequency of fundamental mode in closed pipe length 0.5 m, speed of sound 340 m/s.
  \(\nu = \frac{v}{4l} = \frac{340}{4 \times 0.5} = 170\,Hz\)
• Ex 2: Frequency of 2nd harmonic in open pipe length 1 m.
  \(\nu_2 = \frac{2v}{2l} = \frac{v}{l}\)

Practice Set

• Q1: Closed pipe fundamental frequency formula:
  A) \(\frac{v}{2l}\) B) \(\frac{v}{4l}\) C) \(\frac{3v}{4l}\) D) \(\frac{v}{l}\)
  
• Q2: Which harmonics are present in closed pipe?
  A) All B) Even only C) Odd only D) None
  
• Q3: Open pipe length 0.75 m, speed 340 m/s, frequency of 3rd harmonic:
  A) 226.7 Hz B) 340 Hz C) 453.3 Hz D) 113.3 Hz
  

Answer Key

• Q1: B - Closed pipe fundamental frequency \(v/4l\).
• Q2: C - Only odd harmonics present in closed pipe.
• Q3: A - \(\nu_3 = \frac{3v}{2l} = \frac{3 \times 340}{2 \times 0.75} = 226.7\,Hz\).

Memory Aids

• Closed pipe: odd harmonics only.
• Open pipe: all harmonics.
• Frequency formulas:
  Closed: \(\nu = \frac{(2n+1)v}{4l}\)
  Open: \(\nu = \frac{nv}{2l}\)

End Correction

Key Points

• Wave reflects slightly beyond open end; extra length called end correction \(e\).
• Effective length: \(l + e\) for closed pipe, \(l + 2e\) for open pipe.
• End correction depends on pipe radius \(r\): \(e = 0.6r\) or \(e = 0.3d\).
• Adjusted frequency formulas:
  Closed pipe: \(f = \frac{(2n+1)v}{4(l+e)}\)
  Open pipe: \(f = \frac{nv}{2(l+2e)}\)

Solved Examples

• Ex 1: Calculate fundamental frequency of closed pipe length 0.5 m, radius 0.01 m, speed 340 m/s.
  \(e=0.6 \times 0.01=0.006\,m\)
  \(f=\frac{v}{4(l+e)}=\frac{340}{4(0.5+0.006)}=168.3\,Hz\)

Practice Set

• Q1: End correction depends on:
  A) Length B) Radius C) Tension D) Density
  
• Q2: Effective length of open pipe with end correction \(e\):
  A) \(l+e\) B) \(l+2e\) C) \(l-e\) D) \(l-2e\)
  
• Q3: Frequency of closed pipe with end correction is:
  A) \(\frac{nv}{2l}\) B) \(\frac{(2n+1)v}{4(l+e)}\) C) \(\frac{nv}{2(l+2e)}\) D) \(\frac{v}{4l}\)
  

Answer Key

• Q1: B - End correction depends on pipe radius.
• Q2: B - Open pipe effective length is \(l + 2e\).
• Q3: B - Closed pipe frequency with end correction is \(\frac{(2n+1)v}{4(l+e)}\).

Memory Aids

• End correction \(e = 0.6r\).
• Effective length longer than physical length.
• Adjust frequency formulas by replacing \(l\) with \(l + e\) or \(l + 2e\).

Difference Between Travelling and Standing Waves

Key Points

• Travelling wave: disturbance moves through space; standing wave: disturbance confined.
• Travelling wave: all particles move similarly; standing wave: particles have different amplitudes.
• Standing wave nodes are always at rest; no such points in travelling waves.
• Particles in standing wave cross mean position simultaneously; not so in travelling waves.
• Standing wave particles between nodes move in phase; travelling wave particles have phase differences.
• Energy transmitted in travelling waves; energy confined in standing waves.

Solved Examples

• Ex 1: Identify wave type where energy is confined.
  Answer: Standing wave.
• Ex 2: Wave with nodes at fixed points is:
  Answer: Standing wave.

Practice Set

• Q1: Which wave transmits energy?
  A) Standing B) Travelling C) Both D) None
  
• Q2: Nodes are present in:
  A) Travelling wave B) Standing wave C) Both D) None
  
• Q3: Particles move in phase in:
  A) Travelling wave B) Standing wave C) Both D) None
  

Answer Key

• Q1: B - Travelling waves transmit energy.
• Q2: B - Nodes exist only in standing waves.
• Q3: B - Particles between nodes move in phase in standing waves.

Memory Aids

• Travelling wave: energy moves, no nodes.
• Standing wave: energy confined, nodes present.
• Standing wave particles move in phase between nodes.

Beats

Definition and Basic Concept

• Beats: Regular variation in sound intensity due to superposition of two waves with slightly different frequencies.
• Wave equations: y₁ = a \sin \omega_1 t = a \sin 2\pi
  u_1 t, y₂ = a \sin \omega_2 t = a \sin 2\pi
  u_2 t
• Resultant wave: y = 2a \cos \pi (
  u_1 -
  u_2) t \cdot \sin \pi (
  u_1 +
  u_2) t
• Beat frequency:
  u_{beat} = |
  u_1 -
  u_2| (in Hz)
• Beat period: T = \frac{1}{
  u_{beat}} = \frac{1}{|
  u_1 -
  u_2|} (in seconds)

Solved Examples

• Ex 1: Two waves of frequencies 11 Hz and 9 Hz interfere. Find beat frequency.
  • Beat frequency = |11 - 9| = 2 Hz
• Ex 2: Calculate beat period for above waves.
  • Beat period = 1 / 2 = 0.5 s

Practice Set

• Q1: Two tuning forks produce beats of frequency 5 Hz. If one has frequency 256 Hz, what is the other frequency?
  • A) 251 Hz
  • B) 261 Hz
  • C) Either 251 Hz or 261 Hz
  • D) 256 Hz
• Q2: Two waves of frequencies 440 Hz and 444 Hz interfere. What is the beat frequency?
  • A) 4 Hz
  • B) 884 Hz
  • C) 440 Hz
  • D) 444 Hz
• Q3: A tuning fork of unknown frequency produces 3 beats per second with a 256 Hz fork. When wax is applied to the unknown fork, beat frequency increases to 5 Hz. What is the unknown frequency?
  • A) 259 Hz
  • B) 253 Hz
  • C) 261 Hz
  • D) 251 Hz

Answer Key

• Q1: C) Either 251 Hz or 261 Hz
  Beat frequency = |256 - ?| = 5 Hz, so unknown frequency = 256 ± 5
• Q2: A) 4 Hz
  Beat frequency = |444 - 440| = 4 Hz
• Q3: B) 253 Hz
  Wax loading lowers frequency; beat frequency increases means unknown frequency = 256 - 3 = 253 Hz

Memory Aids

• Beat frequency = difference of frequencies.
• Beat period = reciprocal of beat frequency.
• Loading wax lowers frequency; filing increases frequency.
• Use ± sign in unknown frequency: ν_B = ν_A ± ν_beat.