In this chapter
NotesShala Class 11 · Physics
Chapter14

Waves

Waves and Their Types · Progressive Waves and Speed · Superposition and Standing Waves · Organ Pipes and Beats

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms
01
Part 1

Waves and Their Types

A wave carries energy from one place to another. The material of the medium does not travel along with it. All our communication depends on signals that travel as waves.

1.1Wave Motion

Definition

A wave is a disturbance that moves from one place to another through a medium or through vacuum. It is usually periodic. It carries energy and momentum, but there is no bulk motion of the medium.

A wave starts when a system is pushed away from its equilibrium (normal) state. This disturbance then spreads from one part of the system to the other parts.

System disturbed→Disturbance spreads→Energy and information travel→Matter stays in place
Remember

A wave carries information and energy, as signals, from one point to another. The matter of the medium does not move along with the wave as a whole.

1.2Types of Waves

We usually study three types of waves.

iMechanical Waves

Waves that need a material medium to travel are called mechanical waves.

They are also called elastic waves. The medium must have elasticity and inertia.

They obey Newton's laws of motion. They exist only in a medium like water, air, rock or a string.

e.g. Water waves, sound waves and seismic waves (energy waves through the Earth's layers).

iiElectromagnetic Waves

Waves made of oscillating electric and magnetic fields are called electromagnetic waves.

They need no medium to travel. So they are also called non-mechanical waves.

e.g. Visible light, ultraviolet light, X-rays, radio and TV waves, microwaves.

iiiMatter Waves

Waves linked with moving microscopic particles are called matter waves or de Broglie waves.

The particles can be electrons, protons, other fundamental particles, or even atoms and molecules.

e.g. An electron microscope uses the matter waves of moving electrons.

Speed of Electromagnetic Waves in Vacuum c=299 792 458 m s−1≈3×108 m s−1c = 299\,792\,458\ \text{m s}^{-1} \approx 3\times10^{8}\ \text{m s}^{-1}

All electromagnetic waves travel through vacuum with this same speed cc, the speed of light.

Note
Dual Nature of Light
  • Light is an electromagnetic wave. All the radiation we get from the Sun is electromagnetic.
  • Light shows a dual nature. It behaves both as a wave and as a particle.
  • In the same way, a moving particle can behave like a wave. This wave is its matter wave.

1.3Characteristics of Wave Motion

  1. The particles of the medium oscillate again and again about their mean positions. This passes the disturbance through the medium.
  2. Energy moves from one place to another. The particles themselves do not travel with the wave.
  3. Each particle gets the disturbance a little later than the particle before it. So there is a continuous phase difference between neighbouring particles.
  4. The wave velocity is not the same as the particle velocity. Particle velocity is the velocity of each particle in its SHM about the mean position.
  5. In a given medium, the wave velocity stays constant. But the particle velocity keeps changing during each oscillation.
Particle velocity at the mean positionMaximum
Particle velocity at an extreme positionZero

A medium can carry a mechanical wave only if it has these properties:

1Inertia 2Elasticity 3Very low friction

1.4Spring Model of Wave Propagation

Take a chain of springs joined end to end. One end is fixed to a rigid support. Pull the spring at the free end suddenly and let it go. The disturbance travels to the other end.

pull A free end rigid support
A chain of springs fixed at one end. A sudden pull at end AA travels along the chain
Why the Disturbance Travels
  1. The first spring is pulled, so it gets stretched.
  2. Due to elasticity, a restoring force develops in it.
  3. This force brings the first spring back to its mean position. At the same time, it stretches the second spring.
  4. The second spring then acts on the third, and so on.
Spring pulled and released→Restoring force→Next spring stretched→Disturbance travels

KeyThe disturbance reaches the far end. But every spring just oscillates about its own equilibrium position.

1.5Sound Waves in Air and in Solids

Sound Waves in Air

Think of a small region of air as a spring. It is joined to the regions around it, like springs in a chain. A sound wave compresses and expands each small region of air. This changes the density and the pressure of that region.

By Boyle's law (at constant temperature), pressure is inversely proportional to volume. So pressure is proportional to density:

p∝1V  ⇒  p∝ρ  ⇒  (Δp)∝(Δρ)p \propto \dfrac{1}{V} \;\Rightarrow\; p \propto \rho \;\Rightarrow\; (\Delta p) \propto (\Delta \rho)

Pressure is force per unit area. So a restoring force develops. It is proportional to the change in density. This is just like a stretched or compressed spring.

iA Region Is Compressed

Its molecules move out into the next region. The density there increases. So a compression forms in the next region. The first region now has lower density. It becomes a rarefaction.

iiA Region Is Rarefied

Air from the surroundings rushes in. So the rarefaction moves to the next region.

KeyCompressions and rarefactions pass from one region to the next. So a disturbance can travel through air.

Sound Waves in a Solid

In a crystalline solid, think of the atoms as end points. Springs join each pair of neighbouring atoms.

The forces from the surrounding atoms cancel out. So each atom is in equilibrium.

A sound wave displaces an atom from its equilibrium position. A restoring force develops. The disturbance is handed on to the neighbouring atom, then to the one after it.

Atom pushed from its place→Restoring force acts→Neighbour atom moves→Sound passes through the solid

1.6Transverse Waves

Definition

A transverse wave is a wave in which the particles of the medium do SHM about their mean positions perpendicular to the direction of travel of the wave.

e.g. Waves on a string, ripples on the surface of water, and the vibrating strings of a sitar or a guitar.

Illustration: Wave on a String

Take a long stretched string. Hold one end in your hand. The other end is fixed to a rigid support.

Give the free end a sudden upward jerk. An upward kink, or pulse, forms there. It travels along the string towards the fixed end.

Now move the free end up and down continuously. A train of sinusoidal waves travels along the string. These waves are transverse.

pulse harmonic wave C C′ T T′
A single pulse (top) and a harmonic wave (bottom) on a stretched string
iCrest

Points C,C′,…C, C', \ldots are crests. A crest is a point of maximum displacement in the positive (upward) direction.

iiTrough

Points T,T′,…T, T', \ldots are troughs. A trough is a point of maximum displacement in the negative (downward) direction.

Media for Transverse Waves

In a transverse wave, the particles move perpendicular to the wave. So each small part of the medium suffers a shearing strain. A transverse wave can travel only in a medium that can bear a shearing stress.

Solids and strings can bear shearing stressTransverse waves possible
Fluids cannot bear shearing stressTransverse waves not possible

1.7Longitudinal Waves

Definition

A longitudinal wave is a wave in which the particles of the medium do SHM about their mean positions along the direction of travel of the wave.

e.g. Waves in a spring, sound waves, and the vibrating air column in organ pipes.

Illustration: Piston in an Air-filled Pipe

Take a long pipe filled with air. A piston is fitted at one end, AA.

C R C R C R C R push · pull A B wave travels
Moving the piston back and forth sends compressions (C) and rarefactions (R) along the pipe
iPiston Pushed In

The layer of air next to the piston gets compressed. This layer moves right and compresses the next layer, and so on. So a pulse of compression reaches the other end. The air motion and the pressure change travel together as a pulse.

iiPiston Pulled Back

The layer next to the piston becomes rarefied, so its pressure falls. Air from the next layer moves in to make up the pressure. Now that layer becomes rarefied. So a pulse of rarefaction travels to the right.

Now push and pull the piston again and again in SHM. A sinusoidal wave travels along the pipe. It is a series of compressions (CC) and rarefactions (RR).

KeyThe air layers oscillate parallel to the direction of the wave. So sound waves in air are longitudinal.

Compression

A compression is a region where neighbouring particles are closer than normal.

  • Volume decreases
  • Density increases
Rarefaction

A rarefaction is a region where neighbouring particles are farther apart than normal.

  • Volume increases
  • Density decreases

Both changes are temporary. They last only for a short time, while the wave passes.

1.8Media for Longitudinal Waves

KeySolids and fluids can all bear a compressive strain. So longitudinal waves can travel in all elastic media.

Some facts about the medium and the waves in it:

A steel bar can carryBoth transverse and longitudinal waves
Air can carryOnly longitudinal waves
  • In the same medium, transverse and longitudinal waves travel with different speeds.
  • In wave motion, only the disturbance travels. The medium does not travel with it.
  • Waves can be one-, two- or three-dimensional. This depends on whether the energy spreads in one, two or three dimensions.
1-D Wave

Transverse waves along a string.

2-D Wave

Ripples on the surface of water.

3-D Wave

Sound waves from a point source.

02
Part 2

Progressive Waves and Speed

This part writes a travelling wave as an equation. It then finds the speed of waves on a string and the speed of sound in solids, liquids and gases.

2.1Equation of a Progressive Wave

To describe a travelling wave, we need the position xx and the time tt. Take a wave moving in the positive xx-direction. Let y(x,t)y(x,t) be the transverse displacement of the element at position xx at time tt. Then:

Displacement Relation (wave along +x) …(i) y(x,t)=a sin⁡(kx−ωt+ϕ)y(x,t) = a\,\sin(kx - \omega t + \phi)

The same wave can also be written as a sum of a sine term and a cosine term:

y(x,t)=A sin⁡(kx−ωt)+B cos⁡(kx−ωt)y(x,t) = A\,\sin(kx - \omega t) + B\,\cos(kx - \omega t)

Here,

a=A2+B2andϕ=tan⁡−1(BA)a = \sqrt{A^2 + B^2} \qquad \text{and} \qquad \phi = \tan^{-1}\left(\dfrac{B}{A}\right)

KeyThe equation shows that the particles of the medium at every position execute SHM.

The equation contains xx. So it gives the displacement of every element of the string at any time. It tells us the shape of the wave at any instant. It also tells how this shape changes as the wave moves along. In short, it shows how the wave progresses.

Wave along +x y(x,t)=a sin⁡(kx−ωt+ϕ)y(x,t) = a\,\sin(kx - \omega t + \phi)
Wave along −x …(ii) y(x,t)=a sin⁡(kx+ωt+ϕ)y(x,t) = a\,\sin(kx + \omega t + \phi)
Meaning of Each Symbol
SymbolMeaning
y(x,t)y(x,t)Displacement of the vibrating element, as a function of position xx and time tt
aaAmplitude of the vibrating element (particle)
ω\omegaAngular frequency of the wave
kkAngular wave number
xxPosition of the vibrating element
(kx−ωt+ϕ)(kx - \omega t + \phi)Phase of the element at position xx and time tt
ϕ\phiPhase constant: the phase at x=0x = 0, t=0t = 0

The figure plots Eq. (i) at equal time steps of a quarter period. A crest is a point of maximum positive displacement. A trough is a point of maximum negative displacement. The cross (×) marks one crest. It moves to the right as time passes. The dot (•) marks one element of the string. It only moves up and down at the same place.

x (a) t = 0 x (b) t = T/4 x (c) t = T/2 x (d) t = 3T/4 x (e) t = T λ
A harmonic wave moving along +x+x, drawn at t=0, T4, T2, 3T4, Tt = 0,\ \tfrac{T}{4},\ \tfrac{T}{2},\ \tfrac{3T}{4},\ T. In one period the crest (×) moves one wavelength λ\lambda

2.2Amplitude and Phase

Definition
Amplitude

The amplitude of a wave is the largest distance that a particle of the medium moves away from its equilibrium position while the wave passes.

In Eq. (i), y(x,t)y(x,t) varies between+a+a and −a-a
yy can be positive or negative, but aa isalways positive

So aa is the largest displacement of the particles of the medium from their mean position.

y x λ a Crest Trough O
Crest, trough, amplitude aa and wavelength λ\lambda on a wave
Definition
Phase

The phase of a wave is the quantity (kx−ωt+ϕ)(kx - \omega t + \phi). It is the argument of the sine (or cosine) function in the wave equation.

2.3Wavelength and Angular Wave Number

Definition
Wavelength (λ\lambda)

The wavelength is the smallest distance between two particles in the same phase.

It is also the distance the wave travels while a particle completes one oscillation.

Derivation
Wavelength in Terms of kk

Start from y(x,t)=asin⁡(kx−ωt+ϕ)y(x,t) = a\sin(kx - \omega t + \phi). Take ϕ=0\phi = 0 and look at the wave at t=0t = 0:

  1. y(x,0)=asin⁡kxy(x,0) = a\sin kx
  2. The sine function repeats after every 2π2\pi:
    sin⁡kx=sin⁡(kx+2nπ)=sin⁡k(x+2nπk)\sin kx = \sin(kx + 2n\pi) = \sin k\left(x + \dfrac{2n\pi}{k}\right)
  3. So the displacement is the same at xx and at x+2nπkx + \dfrac{2n\pi}{k}. Here n=1,2,3n = 1, 2, 3 and so on.
  4. The least such distance is for n=1n = 1:
    λ=(x+2πk)−x=2πk\lambda = \left(x + \dfrac{2\pi}{k}\right) - x = \dfrac{2\pi}{k}
Wavelength λ=2πk\lambda = \frac{2\pi}{k}
Angular Wave Number k=2πλ  rad m−1k = \frac{2\pi}{\lambda}\ \ \text{rad m}^{-1}
Definition
Angular Wave Number (kk)

The angular wave number kk of a wave is 2π2\pi times the number of waves that fit in a unit length. It is also called the propagation constant. Its SI unit is rad m−1\text{rad m}^{-1}.

2.4Period, Angular Frequency and Frequency

Now watch one element of the string: the one at x=0x = 0. Put ϕ=0\phi = 0 and x=0x = 0 in y(x,t)=asin⁡(kx−ωt+ϕ)y(x,t) = a\sin(kx - \omega t + \phi):

y(0,t)=asin⁡(−ωt)=−asin⁡ωty(0,t) = a\sin(-\omega t) = -a\sin \omega t

This graph is a sine curve in time. It shows the displacement of this element, its amplitude and its time period.

y t O t1 t1 + T T a
Element at x=0x = 0: y=−asin⁡ωty = -a\sin\omega t. The motion repeats after every period TT
Time Period
Definition

The time period (TT) of a wave is the time it takes to travel a distance of one wavelength.

Derivation
Angular Frequency
  1. Let the period be TT, counted from any time t1t_1. The displacement repeats after time TT:
    −asin⁡ωt1=−asin⁡ω(t1+T)=−asin⁡(ωt1+ωT)-a\sin \omega t_1 = -a\sin \omega(t_1 + T) = -a\sin(\omega t_1 + \omega T)
  2. The sine function repeats after every 2π2\pi. So:
    ωT=2π\omega T = 2\pi
Angular Frequency ω=2πT\omega = \frac{2\pi}{T}

SI unit of ω\omega: rad s−1\text{rad s}^{-1}

Frequency
Definition

The frequency (ν\nu) of a wave is the number of complete oscillations it makes in one second. Its SI unit is the hertz (Hz). It is related to the angular frequency ω\omega as shown below.

Frequency ν=1T=ω2π\nu = \frac{1}{T} = \frac{\omega}{2\pi}

The wavelength λ\lambda and the wave speed vv are related by:

λ=vν\lambda = \dfrac{v}{\nu}
Note

The frequency of a wave never changes, whatever interaction the wave goes through. It is fixed by the source alone.

2.5Speed of a Travelling Wave

To find the speed of a wave, follow any one point on it. The crest is the easiest point to follow.

The figure shows the wave at two instants, tt and t+Δtt + \Delta t. In the time Δt\Delta t, the whole pattern moves a distance Δx\Delta x.

y x Δx A time t time t + Δt
The wave moves to the right with velocity vv

KeyEach point of the moving waveform, like point AA on a crest, keeps its displacement yy. The points of the string do not keep it. Only the points of the waveform do.

Derivation
Speed of a Travelling Wave
  1. Point AA keeps its displacement as it moves. So its phase stays constant:
    kx−ωt=constantkx - \omega t = \text{constant}
  2. Let xx become x+Δxx + \Delta x and tt become t+Δtt + \Delta t. The phase is still the same:
    kx−ωt=k(x+Δx)−ω(t+Δt)kx - \omega t = k(x + \Delta x) - \omega(t + \Delta t)
  3. This gives k Δx=ω Δtk\,\Delta x = \omega\,\Delta t. So the wave speed is:
    ΔxΔt=ωk=v\dfrac{\Delta x}{\Delta t} = \dfrac{\omega}{k} = v
  4. Put ω=2πν\omega = 2\pi\nu and 2πk=λ\dfrac{2\pi}{k} = \lambda:
    v=2πνk=ν×(2πk)=νλv = \dfrac{2\pi\nu}{k} = \nu \times \left(\dfrac{2\pi}{k}\right) = \nu\lambda
Speed of a Travelling Wave v=νλ=λTv = \nu\lambda = \frac{\lambda}{T}

KeyThis relation holds for all progressive waves. In one time period, the wave pattern moves a distance of one wavelength.

2.6Speed of a Transverse Wave on a String

The speed of a mechanical wave depends on the restoring force in the medium. On a string, the tension TT gives this force. We find the speed by the method of dimensions.

Derivation
Speed on a String (Dimensional Method)
  1. The speed can depend on two quantities. The linear mass density (mass per unit length) is μ=[ML−1]\mu = [\text{ML}^{-1}]. The tension is T=[MLT−2]T = [\text{MLT}^{-2}].
  2. Divide the dimensions of TT by those of μ\mu:
    [MLT−2][ML−1]=[L2T−2]\dfrac{[\text{MLT}^{-2}]}{[\text{ML}^{-1}]} = [\text{L}^2\text{T}^{-2}]
  3. Speed has dimensions [LT−1][\text{LT}^{-1}]. This is the square root of [L2T−2][\text{L}^2\text{T}^{-2}]. So:
    v=c Tμv = c\,\sqrt{\dfrac{T}{\mu}}
  4. Here cc is a dimensionless constant. A full analysis shows that c=1c = 1.
Speed of a Transverse Wave on a String v=Tμv = \sqrt{\frac{T}{\mu}}

where, TT = tension in the string, μ\mu = mass per unit length of the string

Depends on the tension TT and the linear mass density μ\muYes
Depends on the frequency of the waveNo

The source of the wave fixes its frequency. The wavelength then follows from:

λ=vν\lambda = \dfrac{v}{\nu}

2.7Speed of Longitudinal Waves (Sound)

When a longitudinal wave passes, the particles of the medium move forward and backward along the direction of the wave. In a sound wave in air, they form compressions and rarefactions.

The restoring property of the medium is its bulk modulus BB. Its SI unit is the pascal (Pa). Here Δp\Delta p = change in pressure, ΔV\Delta V = change in volume and VV = initial volume:

B=−ΔpΔV/VB = -\dfrac{\Delta p}{\Delta V / V}
Derivation
Speed of Sound (Dimensional Method)
  1. For a longitudinal wave, the bulk modulus BB gives the restoring property. The density ρ\rho gives the inertia. Their dimensions are B=[ML−1T−2]B = [\text{ML}^{-1}\text{T}^{-2}] and ρ=[ML−3]\rho = [\text{ML}^{-3}].
  2. Divide the dimensions of BB by those of ρ\rho:
    [ML−1T−2][ML−3]=[L2T−2]\dfrac{[\text{ML}^{-1}\text{T}^{-2}]}{[\text{ML}^{-3}]} = [\text{L}^2\text{T}^{-2}]
  3. Match this with the dimensions of speed, [LT−1][\text{LT}^{-1}]:
    v=c Bρv = c\,\sqrt{\dfrac{B}{\rho}}
  4. Experiments show that c=1c = 1.
Speed of a Longitudinal Wave (Sound) in a Medium v=Bρv = \sqrt{\frac{B}{\rho}}
Solid Bar (a Linear Medium)

For a solid bar, the right modulus is Young's modulus YY. The same dimensional method gives the speed of longitudinal waves in the bar:

v=Yρv = \sqrt{\dfrac{Y}{\rho}}

where, YY = Young's modulus of the material of the bar, ρ\rho = density of the material

Speed of Sound in Some Media
StateMediumSpeed (m s⁻¹)
GasesAir (0 °C)331
Air (20 °C)343
Helium965
Hydrogen1284
LiquidsWater (0 °C)1402
Water (20 °C)1482
Seawater1522
SolidsAluminium6420
Copper3560
Steel5941
Granite6000
Vulcanised rubber54

2.8Speed of Mechanical Waves

The speed of a mechanical wave depends on the properties of the medium. For one type of wave in a given medium, the speed does not change when these properties of the wave change:

×Amplitude ×Wavelength ×Frequency

e.g. Infrasonic, audible and ultrasonic sound waves in a given medium all have the speed given by:

Speed of Mechanical Waves v=Eρv = \sqrt{\frac{E}{\rho}}

where, EE = the suitable elastic constant of the medium, ρ\rho = density of the medium

KeySo this speed is decided by the medium alone.

2.9Speed of Electromagnetic Waves

The medium also decides how fast electromagnetic waves travel. These waves are made of electric and magnetic fields that oscillate in mutually perpendicular directions.

Two properties of the medium decide their speed: the permittivity ε\varepsilon and the permeability μ\mu.

In Vacuum (all EM waves) c=1μ0 ε0c = \frac{1}{\sqrt{\mu_0\,\varepsilon_0}}

μ0\mu_0 = permeability of vacuum, ε0\varepsilon_0 = permittivity of vacuum

In a Material Medium v=1μ εv = \frac{1}{\sqrt{\mu\,\varepsilon}}

μ\mu = permeability of the medium, ε\varepsilon = permittivity of the medium

Note

In a material medium, μ\mu and ε\varepsilon are not exactly the same for electromagnetic waves of different frequencies.

2.10Newton's Formula for Speed of Sound

Newton assumed that the pressure and volume of a gas change isothermally (at constant temperature) when sound passes through it.

Derivation
Newton's Formula
  1. For an ideal gas:
    pV=NkBTpV = N k_B T
    Here pp = pressure, VV = volume, NN = number of molecules in the volume VV, kBk_B = Boltzmann constant and TT = absolute temperature (in kelvin).
  2. For an isothermal change, ΔT=0\Delta T = 0. Take small changes on both sides:
    VΔp+pΔV=0V\Delta p + p\Delta V = 0
  3. Rearrange it into the form of the bulk modulus:
    −ΔpΔV/V=p-\dfrac{\Delta p}{\Delta V / V} = p
  4. The left side is the bulk modulus, B=−ΔpΔV/VB = -\dfrac{\Delta p}{\Delta V/V}. So the bulk modulus of the gas equals its pressure:
    B=pB = p
  5. Put B=pB = p in the formula for the speed of sound:
    v=Bρ=pρv = \sqrt{\dfrac{B}{\rho}} = \sqrt{\dfrac{p}{\rho}}
Newton's Formula v=pρv = \sqrt{\frac{p}{\rho}}

where, pp = pressure of the gas, ρ\rho = density of the gas

KeyThis is called Newton's formula for the speed of longitudinal waves (sound) in an ideal gas.

2.11Laplace Correction

Newton's formula for air at STP280 m s⁻¹ (about 15% too small)
Experimental value for air at STP331 m s⁻¹

Laplace found the mistake. When sound passes through a gas, its pressure and volume do not change isothermally. They change adiabatically.

Derivation
Laplace Correction
  1. For an adiabatic change:
    pVγ=constantpV^{\gamma} = \text{constant}
  2. Take small changes in pp and VV:
    pγVγ−1ΔV+VγΔp=0p\gamma V^{\gamma-1}\Delta V + V^{\gamma}\Delta p = 0
  3. So the adiabatic bulk modulus is:
    Bad=−ΔpΔV/V=γpB_{\text{ad}} = -\dfrac{\Delta p}{\Delta V / V} = \gamma p
    Here γ\gamma is the ratio of the two specific heats, Cp/CVC_p / C_V.
Speed of Sound in a Gas (Laplace) v=γpρv = \sqrt{\frac{\gamma p}{\rho}}

This change in Newton's formula is the Laplace correction. For air, γ=7/5\gamma = 7/5. The corrected formula then gives:

Speed of sound in air (Laplace's formula)331.3 m s⁻¹, matches experiment
Note
Speed of Sound and the Temperature of the Gas

The gas equation for one mole is pV=RTpV = RT. So p=RTVp = \dfrac{RT}{V}. Put this in v=γp/ρv = \sqrt{\gamma p / \rho}. Also, ρ×V=M\rho \times V = M, the molar mass:

v=γpρ=γRTρ×V=γRTMv = \sqrt{\dfrac{\gamma p}{\rho}} = \sqrt{\dfrac{\gamma R T}{\rho \times V}} = \sqrt{\dfrac{\gamma R T}{M}}

where, TT = temperature of the gas, MM = molar mass of the gas

03
Part 3

Superposition and Standing Waves

When two or more waves meet, their displacements add up. This one simple rule explains interference, standing waves and beats.

3.1Principle of Superposition of Waves

Principle

The principle of superposition states that when two or more waves (or pulses) overlap, the resultant displacement at any point is the algebraic sum of the displacements due to each wave.

Two pulses travel along the same stretched string in opposite directions. They meet, pass through each other and then move on independently, without any change.

(a) t = 0 s (b) t = 1 s pulses cancel: zero displacement (c) t = 2 s (d) t = 3 s (e) t = 4 s 0 1 2 3 4 5 6
Two equal and opposite pulses on one string. At t=2t = 2 s they overlap exactly and cancel. Then each moves on unchanged

KeyIn graph (c), the displacements due to the two pulses cancel exactly. The displacement is zero everywhere on the string.

Three important applications of superposition
1Standing waves

Two equal waves moving in opposite directions.

2Beats

Two waves of nearly equal frequencies.

3Interference

Waves meeting in the same medium.

3.2Interference of Waves

Definition

Interference is the effect seen when two or more waves pass through the same medium at the same time. The medium takes a shape that comes from the combined effect of all the waves on its particles.

Let two waves produce displacements y1(x,t)y_1(x,t) and y2(x,t)y_2(x,t) in a medium. The net displacement is:

y(x,t)=y1(x,t)+y2(x,t)y(x,t) = y_1(x,t) + y_2(x,t)

When nn waves overlap:

y=f1(x−vt)+f2(x−vt)+…+fn(x−vt)=∑i=1nfi(x−vt)y = f_1(x - vt) + f_2(x - vt) + \ldots + f_n(x - vt) = \displaystyle\sum_{i=1}^{n} f_i(x - vt)

Each function here stands for a moving wave. Superposition is the basis of interference. Take two harmonic waves with the same angular frequency ω\omega, the same wave number k=2π/λk = 2\pi/\lambda and the same amplitude aa.

Derivation
Resultant of Two Interfering Waves
  1. Let the two waves differ in phase by a phase constant ϕ\phi:
    y1(x,t)=asin⁡(kx−ωt)y_1(x,t) = a\sin(kx - \omega t)
    y2(x,t)=asin⁡(kx−ωt+ϕ)y_2(x,t) = a\sin(kx - \omega t + \phi)
  2. By the principle of superposition, the net displacement is:
    y(x,t)=asin⁡(kx−ωt)+asin⁡(kx−ωt+ϕ)y(x,t) = a\sin(kx - \omega t) + a\sin(kx - \omega t + \phi)
  3. Use sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}:
    y=a[2sin⁡{(kx−ωt)+(kx−ωt+ϕ)2}cos⁡ϕ2]y = a\left[2\sin\left\{\dfrac{(kx - \omega t) + (kx - \omega t + \phi)}{2}\right\}\cos\dfrac{\phi}{2}\right]
Resultant Wave y(x,t)=2acos⁡ϕ2 sin⁡ ⁣(kx−ωt+ϕ2)y(x,t) = 2a\cos\frac{\phi}{2}\,\sin\!\left(kx - \omega t + \frac{\phi}{2}\right)

The resultant is also a harmonic wave moving in the positive xx-direction. It has the same frequency and wavelength. Its phase angle is ϕ/2\phi/2. Its amplitude is:

A(ϕ)=2acos⁡ϕ2A(\phi) = 2a\cos\dfrac{\phi}{2}
Iϕ=0\phi = 0 (waves in phase)
y(x,t)=2asin⁡(kx−ωt)y(x,t) = 2a\sin(kx - \omega t)

The amplitude is 2a2a. This is the maximum amplitude.

IIϕ=π\phi = \pi (out of phase by 180°)
y(x,t)=0y(x,t) = 0

The displacement is zero everywhere, at all times.

3.3Constructive and Destructive Interference

Constructive Interference

When the waves meet in phase, it is called constructive interference.

Phase differenceϕ=2nπ\phi = 2n\pi
Path differenceΔx=nλ\Delta x = n\lambda
Destructive Interference

When the waves meet in opposite phase, it is called destructive interference.

Phase differenceϕ=(2n+1)π\phi = (2n+1)\pi
Path differenceΔx=(2n+1)λ2\Delta x = (2n+1)\dfrac{\lambda}{2}

Here n=0,1,2,3n = 0, 1, 2, 3 and so on, and λ\lambda is the wavelength.

y y (a) 2a in phase: φ = 0 resultant: amplitude 2a y y (b) opposite phase: φ = π resultant: y = 0
(a) Constructive interference: waves in phase add up. (b) Destructive interference: waves with phase difference π\pi cancel
  • Constructive: a crest of one wave meets a crest of another wave of the same frequency at a point. The resultant displacement is the sum of the two magnitudes. See Fig. (a).
  • Destructive: a crest of one wave meets a trough of the other wave. The resultant displacement is the difference of the two magnitudes. See Fig. (b).
  • Two sine waves of the same amplitude and wavelength may travel in the same direction on a stretched string. They combine into one sine wave that moves in that same direction.
Intensity in Interference

Intensity is proportional to the square of the amplitude, I∝A2I \propto A^2. Take two sources with amplitudes A1A_1, A2A_2 and intensities I1I_1, I2I_2:

Ratio of Maximum to Minimum Intensity ImaxImin=(I1+I2I1−I2)2\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^{2}

When I1=I2=I0I_1 = I_2 = I_0, the resultant intensity is:

Iresultant=4I0cos⁡2(ϕ/2)I_{\text{resultant}} = 4I_0\cos^2(\phi/2)

where, ϕ\phi = phase difference between the two waves

Note

These rules also hold for sound waves and light waves. But the sources must be coherent. This means their phase difference must not change with time.

3.4Reflection of Waves

A pulse or wave that meets a rigid boundary is reflected. An echo is an example. If the boundary is not fully rigid, some of the wave comes back (reflected). The rest goes on into the second medium (transmitted).

A wave may fall obliquely on the boundary between two media. Then the transmitted wave is called the refracted wave.

Incident and refracted waves obeySnell's law of refraction
Incident and reflected waves obeyThe usual laws of reflection

Look at a pulse moving towards a rigid wall that absorbs no energy. The point at the wall cannot move, so its displacement must stay zero. By the principle of superposition, the reflected wave must have the same amplitude and a phase difference of π\pi.

Rigid boundary incident reflected: inverted phase change = π Open (free) boundary incident reflected: upright no phase change
Reflection of a pulse. At a rigid boundary it comes back inverted (phase change π\pi). At a free boundary it comes back upright

Let the incident wave be:

y(x,t)=asin⁡(kx−ωt)y(x,t) = a\sin(kx - \omega t)
iAt a Rigid Boundary

The reflected wave has a phase change of π\pi, so it is inverted. At the boundary:

yr(x,t)=asin⁡(kx−ωt+π)y_r(x,t) = a\sin(kx - \omega t + \pi)
=− asin⁡(kx−ωt)= -\,a\sin(kx - \omega t)

Net displacement at the boundary: y′=y+yr=0y' = y + y_r = 0 at all times.

iiAt an Open Boundary

The boundary point is completely free to move. The reflected pulse has the same phase and amplitude:

yr(x,t)=asin⁡(kx−ωt+0)y_r(x,t) = a\sin(kx - \omega t + 0)
=asin⁡(kx−ωt)= a\sin(kx - \omega t)

Net maximum displacement at the boundary = twice the amplitude.

3.5Standing Waves

Definition

Standing waves, or stationary waves, are formed when two progressive waves of the same type overlap while moving along the same straight line in opposite directions. The two waves have the same amplitude, the same frequency (time period and wavelength) and the same speed.

"Same type" means both are transverse, or both are longitudinal.

KeyA standing wave does not move in any direction. It carries no energy through the medium.

Think of a string whose two ends are both fixed. Or think of air in a pipe that is closed at both ends. The wave reflects at both boundaries. The incident and reflected waves interfere. So a steady wave pattern sets up on the string.

t = 0 t = T/8 t = T/4 t = 3T/8 N N N N N N N wave going right wave going left resultant λ/2
Two harmonic waves moving in opposite directions give a stationary wave. Its nodes N stay at fixed places, λ/2\lambda/2 apart
Derivation
Equation of a Stationary Wave
  1. Waves along the +x and the −x directions:
    y1(x,t)=asin⁡(kx−ωt)y2(x,t)=asin⁡(kx+ωt)y_1(x,t) = a\sin(kx - \omega t) \qquad y_2(x,t) = a\sin(kx + \omega t)
  2. The resultant wave:
    y(x,t)=y1(x,t)+y2(x,t)=a[sin⁡(kx−ωt)+sin⁡(kx+ωt)]y(x,t) = y_1(x,t) + y_2(x,t) = a\left[\sin(kx - \omega t) + \sin(kx + \omega t)\right]
  3. Use sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}. This gives the result below.
Stationary Wave y(x,t)=2asin⁡kx cos⁡ωty(x,t) = 2a\sin kx\,\cos \omega t

The amplitude of the stationary wave is 2asin⁡kx2a\sin kx.

  • The amplitude changes from point to point.
  • Every element oscillates with the same angular frequency ω\omega (same time period). There is no phase difference between the oscillations of different elements.
  • The pattern stays in place. The amplitude is fixed at a given point, but different at different points.

3.6Nodes, Antinodes and Normal Modes

Node (N)

A node is a point of a standing wave where the amplitude is zero. There is no motion at a node.

Antinode (A)

An antinode is a point of a standing wave where the amplitude is the largest.

Concept
Normal Modes

A system with standing waves cannot oscillate at any frequency it likes. It has a fixed set of natural frequencies. These are called its normal modes of oscillation.

We now find them for a stretched string with both ends fixed.

Derivation
Position of Nodes
  1. At a node, the amplitude 2asin⁡kx2a\sin kx is zero:
    sin⁡kx=0  ⇒  kx=nπ; n=0,1,2,3,4,…\sin kx = 0 \;\Rightarrow\; kx = n\pi;\ n = 0, 1, 2, 3, 4, \ldots
  2. Put k=2πλk = \dfrac{2\pi}{\lambda}:
    x=nπk=nπ2π/λx = \dfrac{n\pi}{k} = \dfrac{n\pi}{2\pi/\lambda}
Position of Nodes x=nλ2; n=0,1,2,3,…x = \frac{n\lambda}{2};\ n = 0, 1, 2, 3, \ldots
First node at x=0x = 0, second node atx=λ/2x = \lambda/2
Distance between two successive nodesλ/2\lambda/2
Derivation
Position of Antinodes
  1. At an antinode, the amplitude has its largest value, 2a2a. This happens when ∣sin⁡kx∣=1|\sin kx| = 1:
    kx=(n+12)π; n=0,1,2,3,…kx = \left(n + \dfrac{1}{2}\right)\pi;\ n = 0, 1, 2, 3, \ldots
  2. Put k=2πλk = \dfrac{2\pi}{\lambda} to get the positions below.
Position of Antinodes x=(2n+1)λ4; n=0,1,2,…x = (2n+1)\frac{\lambda}{4};\ n = 0, 1, 2, \ldots

At these points of the string, the amplitude is maximum.

First antinode at x=λ/4x = \lambda/4, second atx=3λ/4x = 3\lambda/4
Distance between two neighbouring antinodesλ/2\lambda/2
Node to node = λ/2·Antinode to antinode = λ/2

3.7String Fixed at Both Ends

Take a stretched string of length LL, fixed at both ends. Its boundary conditions are:

x=0x = 0one fixed end x=Lx = Lother fixed end

Both ends must be nodes. So the length must hold a whole number of half-wavelengths:

L=nλ2; n=1,2,3…⇒possible wavelengths λ=2Ln; n=1,2,3…L = \dfrac{n\lambda}{2};\ n = 1, 2, 3\ldots \quad\Rightarrow\quad \text{possible wavelengths } \lambda = \dfrac{2L}{n};\ n = 1, 2, 3\ldots
Natural Frequencies of the String ν=nv2Lfor n=1,2,3…\nu = \frac{n v}{2L} \quad \text{for } n = 1, 2, 3 \ldots

where, vv = speed of the wave on the string, LL = length of the string

Modes of Vibration
Definition

The mode of vibration of a string is the way it vibrates to form a standing wave.

First Mode (n=1n = 1)

The string vibrates in one loop. This is the lowest natural frequency. It is called the fundamental mode or first harmonic.

L=λ12,ν=v2LL = \dfrac{\lambda_1}{2}, \quad \nu = \dfrac{v}{2L}
Second Mode (n=2n = 2)

The string vibrates in two loops. It is called the second harmonic or first overtone.

L=λ2,ν2=2v2L=2ν1L = \lambda_2, \quad \nu_2 = \dfrac{2v}{2L} = 2\nu_1
Third Mode (n=3n = 3)

The string vibrates in three loops. It is called the third harmonic. Higher modes follow in the same way.

ν=3v2L=3ν1\nu = \dfrac{3v}{2L} = 3\nu_1

The figure shows the first four harmonics. A real string vibrates in a mixture of several modes at the same time. Some modes are strongly excited, others only weakly.

N N A (a) n = 1 · fundamental: L = λ/2 N N N A A (b) n = 2 · 2nd harmonic: L = λ N N N N A A A (c) n = 3 · 3rd harmonic: L = 3λ/2 N N N N N A A A A (d) n = 4 · 4th harmonic: L = 2λ
First four harmonics of a string fixed at both ends. The fixed ends are always nodes
Note

Musical instruments like the sitar and the violin work on this principle. Where the string is plucked or bowed decides which modes are stronger than the others.

3.8String Fixed at One End

A standing wave can also form on a string fixed at one end. Its other end can move freely sideways. Then:

The free end is alwaysan antinode (A)
The fixed end is alwaysa node (N)
Derivation
Frequencies of a String Fixed at One End
  1. Take the standing wave with a node at the fixed end, x=0x = 0:
    y(x,t)=2asin⁡kx cos⁡ωty(x,t) = 2a\sin kx\,\cos\omega t
  2. For an antinode at x=Lx = L, where n=0,1,2n = 0, 1, 2 and so on:
    sin⁡kL=±1  ⇒  kL=(2n+1)π2\sin kL = \pm 1 \;\Rightarrow\; kL = (2n+1)\dfrac{\pi}{2}
  3. Put k=2πλk = \dfrac{2\pi}{\lambda}:
    2πλ×L=(2n+1)π2  ⇒  2Lλ=(2n+1)×12\dfrac{2\pi}{\lambda}\times L = (2n+1)\dfrac{\pi}{2} \;\Rightarrow\; \dfrac{2L}{\lambda} = (2n+1)\times\dfrac{1}{2}
  4. Use λ=vν\lambda = \dfrac{v}{\nu}:
    2Lνv=(2n+1) 12\dfrac{2L\nu}{v} = (2n+1)\,\dfrac{1}{2}
Frequency of Vibration ν=(2n+1)4L v\nu = \frac{(2n+1)}{4L}\,v

where, n=0,1,2n = 0, 1, 2 and so on, ν\nu = frequency, vv = speed of the wave

These are the normal frequencies of the string. The fundamental frequency is for n=0n = 0. The other frequencies are:

FrequencyName
ν0=v4L\nu_0 = \dfrac{v}{4L}First harmonic (fundamental)
ν1=3v4L=3ν0\nu_1 = \dfrac{3v}{4L} = 3\nu_0Third harmonic (first overtone)
ν2=5v4L=5ν0\nu_2 = \dfrac{5v}{4L} = 5\nu_0Fifth harmonic (second overtone)
ν3=7v4L=7ν0\nu_3 = \dfrac{7v}{4L} = 7\nu_0Seventh harmonic (third overtone)
ν₀ : ν₁ : ν₂ : ν₃ = 1 : 3 : 5 : 7⇒Only odd harmonics are present
N A (a) fundamental: L = λ/4 N N A A (b) 1st overtone: L = 3λ/4 N N N A A A (c) 2nd overtone: L = 5λ/4
(a) Fundamental, (b) first overtone and (c) second overtone. Node N at the fixed end, antinode A at the free end
Law
Laws of Transverse Vibration of a String
Given string, fixed tension TT: frequency is inversely proportional to the vibrating length LLν∝1L\nu \propto \dfrac{1}{L}
Uniform string of fixed length: frequency is proportional to the square root of the tension TTν∝T\nu \propto \sqrt{T}
Fixed length and tension: frequency is inversely proportional to the square root of the mass per unit lengthν∝1μ\nu \propto \dfrac{1}{\sqrt{\mu}}

A sonometer measures the fundamental frequency of a vibrating string. It is also used to verify these laws.

04
Part 4

Organ Pipes and Beats

Air inside a pipe can also form standing waves. This part also explains beats, which come from two waves of nearly equal frequency.

4.1Vibrations of Air Columns

Air vibrating inside an organ pipe is a familiar case of stationary waves. The pipe is a tube shaped like a cylinder. There are two kinds:

iClosed Organ Pipe

Closed at one end and open at the other end.

iiOpen Organ Pipe

Open at both ends.

Make the air at the open end of the pipe vibrate. A longitudinal wave travels along the pipe to its far end. There it is reflected back. The incident and reflected waves overlap. So stationary waves form in the pipe.

Air vibrates at the open end→Longitudinal wave travels→Reflected at the far end→Stationary waves

4.2Closed Organ Pipe

Now find the normal modes of an air column that is closed at one end and open at the other. This is a closed organ pipe. A glass tube partly filled with water is a simple example.

The closed end (at the water surface) is a nodePressure changes the most
The open end is an antinodePressure changes the least
Derivation
Frequencies of a Closed Organ Pipe
  1. Take the end touching the water as x=0x = 0 (a node). The other end, x=Lx = L, is an antinode. So:
    ∣sin⁡kx∣=1  ⇒  ∣sin⁡kL∣=1|\sin kx| = 1 \;\Rightarrow\; |\sin kL| = 1
  2. The sine has magnitude 1 at odd multiples of π/2\pi/2:
    kL=(n+12)πkL = \left(n + \dfrac{1}{2}\right)\pi
  3. Put k=2πλk = \dfrac{2\pi}{\lambda}:
    L=(2n+1)λ4; n=0,1,2,3,…L = (2n+1)\dfrac{\lambda}{4};\ n = 0, 1, 2, 3, \ldots
Frequency of a Closed Organ Pipe ν=(2n+1)v4L; n=0,1,2,3,…\nu = (2n+1)\frac{v}{4L};\ n = 0, 1, 2, 3, \ldots

where, vv = speed of sound in air, LL = length of the air column

Fundamental frequency (n=0n = 0)ν=v4L\nu = \dfrac{v}{4L}
Higher frequencies: the odd harmonics3v4L, 5v4L,\dfrac{3v}{4L},\ \dfrac{5v}{4L}, etc.

KeyA pipe closed at one end gives only odd harmonics.

N A (a) 1st harmonic: L = λ/4 N N A (b) 3rd harmonic: L = 3λ/4 N N N A (c) 5th harmonic: L = 5λ/4 N N N N A (d) 7th harmonic: L = 7λ/4 N N N N N A (e) 9th harmonic: L = 9λ/4 N N N N N N A (f) 11th harmonic: L = 11λ/4
Odd harmonics of an air column closed at the left end and open at the right end. (a) is the fundamental. Dots mark the nodes
Note
  • A system is in resonance when the outside (driving) frequency is close to one of its natural frequencies.
  • Take a circular membrane clamped tightly all around its edge. Its normal modes follow from one boundary condition: no point on the edge vibrates.
  • Standing waves in a closed organ pipe are like those on a string fixed at one end.

4.3Open Organ Pipe

In a pipe open at both ends, each end is an antinode. Such an air column gives all the harmonics. Its frequency formula is the same as for a string fixed at both ends.

Frequency of an Open Organ Pipe ν=nv2Lfor n=1,2,3,…\nu = \frac{nv}{2L} \quad \text{for } n = 1, 2, 3, \ldots

KeyBoth ends are antinodes. Nodes and antinodes alternate between them.

N A A (a) fundamental (1st harmonic): L = λ/2 N N A A (b) 2nd harmonic: L = λ N N N A A (c) 3rd harmonic: L = 3λ/2 N N N N A A (d) 4th harmonic: L = 2λ
First four harmonics of an open organ pipe: antinode A at each open end, nodes N in between
nnFrequencyName
1ν1=v2L\nu_1 = \dfrac{v}{2L}Fundamental or first harmonic
2ν2=vL=2ν1\nu_2 = \dfrac{v}{L} = 2\nu_1Second harmonic (first overtone)
3ν3=3v2L=3ν1\nu_3 = \dfrac{3v}{2L} = 3\nu_1Third harmonic (second overtone)
4ν4=4v2L=4ν1\nu_4 = \dfrac{4v}{2L} = 4\nu_1Fourth harmonic (third overtone)
ν₁ : ν₂ : ν₃ = 1 : 2 : 3⇒An open pipe gives all harmonics
Note
  • An open pipe gives a shriller sound (higher pitch) than a closed pipe of the same length.
  • The quality of sound from an open pipe is better than from a closed pipe.
  • Standing waves in an open organ pipe are like those on a string fixed at both ends.

4.4Beats

Beats come from the interference of two waves with nearly equal frequencies.

Definition

Beats are the periodic rise and fall in the intensity of sound. They are heard when two sound waves of nearly equal frequencies and amplitudes, moving in the same direction, overlap.

Sound gets louder once and softer onceOne beat
Beats heard in one secondBeat frequency
Remember
The 10 Hz Limit

The two frequencies should not differ by more than 10 Hz. For a bigger difference, the ear cannot tell the rises and falls apart. This is due to the persistence of hearing.

Good to know
Tuning Instruments

Musicians use beats to tune their instruments with each other. They keep tuning until their trained ears hear no beats. The frequencies of all the instruments in the band then match.

4.5Analytical Treatment of Beats

Take two harmonic sound waves with nearly equal angular frequencies ω1\omega_1 and ω2\omega_2. Look only at the point x=0x = 0. These are sound waves, so y1y_1 and y2y_2 stand for longitudinal displacements.

Derivation
Beat Frequency
  1. The two waves at x=0x = 0:
    y1=acos⁡ω1tandy2=acos⁡ω2ty_1 = a\cos\omega_1 t \quad\text{and}\quad y_2 = a\cos\omega_2 t
  2. By the principle of superposition, the resultant displacement is:
    y=y1+y2=a[cos⁡ω1t+cos⁡ω2t]y = y_1 + y_2 = a\left[\cos\omega_1 t + \cos\omega_2 t\right]
  3. Use cos⁡A+cos⁡B=2cos⁡A−B2cos⁡A+B2\cos A + \cos B = 2\cos\dfrac{A-B}{2}\cos\dfrac{A+B}{2}. Write ωb=ω1−ω22\omega_b = \dfrac{\omega_1 - \omega_2}{2} and ωa=ω1+ω22\omega_a = \dfrac{\omega_1 + \omega_2}{2}:
    y=2acos⁡(ω1−ω2)t2 cos⁡(ω1+ω2)t2y = 2a\cos\dfrac{(\omega_1 - \omega_2)t}{2}\,\cos\dfrac{(\omega_1 + \omega_2)t}{2}
    ⇒  y=2acos⁡ωbt cos⁡ωat\Rightarrow\; y = 2a\cos\omega_b t\,\cos\omega_a t
  4. Assume ∣ω1−ω2∣≪ω1,ω2|\omega_1 - \omega_2| \ll \omega_1, \omega_2. Then ωa≫ωb\omega_a \gg \omega_b. So the resultant oscillates with the average angular frequency ωa\omega_a. Its amplitude is maximum when:
    ∣cos⁡ωbt∣=1|\cos\omega_b t| = 1
  5. So the resultant wave waxes and wanes (grows louder and softer) with angular frequency:
    2ωb=ω1−ω22\omega_b = \omega_1 - \omega_2
Beat Frequency νbeat=ν1−ν2\nu_{\text{beat}} = \nu_1 - \nu_2

The figure shows beats from two harmonic waves of 11 Hz and 9 Hz. The amplitude of the resultant wave rises and falls twice every second. So the beat frequency is 2 Hz.

(a) 11 Hz (b) 9 Hz t (s) (c) resultant 2 beats/s 0 0.25 0.5 0.75 1.0 one beat = 0.5 s
(a) Wave of 11 Hz, (b) wave of 9 Hz, (c) their resultant, which shows beats at 2 Hz
Good to know
Musical Pillars

Some temple pillars of rock give basic musical notes when tapped. They are of three types:

  • Shruti pillars produce the basic notes, the 'swaras'.
  • Gana Thoongal produce the basic tunes that make up the 'ragas'.
  • Laya Thoongal produce 'taal' (beats) when tapped.

4.6Quick Revision

TermKey points
Wave motionA disturbance that travels through a medium because its particles oscillate again and again about their mean positions. Each particle hands the disturbance to the next. Three types: mechanical, electromagnetic and matter waves.
Transverse waveParticles oscillate perpendicular to the wave. Needs a medium with a shear modulus, e.g. a solid.
Longitudinal waveParticles oscillate along the wave. Needs a bulk modulus, so it travels in all media: solids, liquids and gases.
Progressive waveA wave that moves from one point of the medium to another. y(x,t)=asin⁡(kx−ωt+ϕ)y(x,t) = a\sin(kx - \omega t + \phi) along +x+x, and y(x,t)=asin⁡(kx+ωt+ϕ)y(x,t) = a\sin(kx + \omega t + \phi) along −x-x.
AmplitudeMaximum displacement from the mean position.
Time period TTTime for one complete oscillation.
Frequency ν\nuNumber of waves produced per second.
Wavelength λ\lambdaDistance between the two nearest particles oscillating in the same phase.
Basic relationsλ=2πk\lambda = \dfrac{2\pi}{k}, ω=2πT\omega = \dfrac{2\pi}{T}, ν=1T\nu = \dfrac{1}{T}, v=νλ=λTv = \nu\lambda = \dfrac{\lambda}{T}
Wave speedsString: v=Tμv = \sqrt{\dfrac{T}{\mu}} · Mechanical: v=Eρv = \sqrt{\dfrac{E}{\rho}}
Solid: v=Yρv = \sqrt{\dfrac{Y}{\rho}} · Liquid: v=Bρv = \sqrt{\dfrac{B}{\rho}}
Gas (Newton): v=pρv = \sqrt{\dfrac{p}{\rho}} · Laplace: v=γpρv = \sqrt{\dfrac{\gamma p}{\rho}}
Speed of soundvsolid>vliquid>vgasv_{\text{solid}} > v_{\text{liquid}} > v_{\text{gas}}
Superpositiony′(x,t)=y1(x,t)+y2(x,t)y'(x,t) = y_1(x,t) + y_2(x,t)
Interferencey=2acos⁡ϕ2sin⁡ ⁣(kx−ωt+ϕ2)y = 2a\cos\dfrac{\phi}{2}\sin\!\left(kx - \omega t + \dfrac{\phi}{2}\right)
In phase (ϕ=0\phi = 0): y=2asin⁡(kx−ωt)y = 2a\sin(kx - \omega t) · Out of phase (ϕ=π\phi = \pi): y=0y = 0
Constructive / destructiveConstructive: ϕ=2nπ\phi = 2n\pi, Δx=nλ\Delta x = n\lambda. Destructive: ϕ=(2n+1)π\phi = (2n+1)\pi, Δx=(2n+1)λ2\Delta x = (2n+1)\dfrac{\lambda}{2}.
Phase and path differenceϕ=2πλ×x  ⇒  x=ϕλ2π\phi = \dfrac{2\pi}{\lambda} \times x \;\Rightarrow\; x = \dfrac{\phi\lambda}{2\pi}
Organ pipesClosed pipe: ν=(2n+1)v4L\nu = (2n+1)\dfrac{v}{4L} · Open pipe: ν=nv2L\nu = \dfrac{nv}{2L}
Beatsνbeat=ν1−ν2\nu_{\text{beat}} = \nu_1 - \nu_2
End of Chapter 14 · notesshala.in
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