Waves
Related chapter: Wave Phenomena
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4.1 Oscilla ons
Recognise simple harmonic mo on (SHM).
Describe the interchange of kine c and poten al energy involved within a cycle of SHM.
Define SHM in words and in graph form.
Interpret a–s graphs of SHM.
Understand the terms me period, frequency and amplitude as applied to SHM.
Use the equa on T = 1f , giving the frequency in Hertz.
Sketch s–t, v–t and a–t graphs of SHM.
Describe the phase difference between two oscillators with iden cal me periods.
4.2 Travelling waves
Wavelength, frequency, period, and wave speed
Solving problems involving wave speed, frequency, and wavelength.
Guidance
Students will be expected to derive c = f λ .
Transverse and longitudinal waves
Explaining the mo on of par cles of a medium when a wave passes through it for both
transverse
and longitudinal cases.
Sketching and interpre ng displacement–distance graphs and displacement– me graphs
for
transverse and longitudinal waves.
4.3 Wave characteris cs
Amplitude and intensity
Solving problems involving amplitude and intensity.
Superposi on
Sketching and interpre ng the superposi on of pulses and waves.
Students will be expected to calculate the resultant of two waves or pulses both
graphically and algebraically.
Guidance
Students will be expected to calculate the resultant of two waves or pulses both
graphically and algebraically.
Polariza on
Describing methods of polariza on.
4.4 Wave Behaviour
Reflec on and refrac on (light)
Sketching and interpre ng incident, reflected, and transmi ed waves at boundaries
between media.
Solving problems involving reflec on at a plane interface.
Snell’s law, cri cal angle, and total internal reflec on (light)
Solving problems involving Snell’s law, cri cal angle, and total internal reflec on.
Determining refrac ve index experimentally.
Guidance
Quan ta ve descrip ons of refrac ve index are limited to light rays passing between two or
more transparent media. If more than two media, only parallel interfaces will be considered.
4.5 Standing Waves
The nature of standing waves
Describing the nature and forma on of standing waves in terms of superposi on. Dis nguishing between standing and travelling waves.
Guidance
Students will be expected to consider the forma on of standing waves from the
superposi on of no more than two waves.
Boundary condi ons
Sketching and interpre ng standing wave pa erns in strings.
Guidance
Boundary condi ons for strings are: two fixed boundaries; fixed and free boundary; two
free boundaries.
Nodes and an nodes
Solving problems involving the frequency of a harmonic, length of the standing wave, and
the speed of the wave.
Guidance
The lowest frequency mode of a standing wave is known as the first harmonic - The terms
fundamental and overtone will not be used.
Definitions
1. A wave is a disturbance that travels in a medium, transferring energy and momentum from one place
to another. The direc on of propaga on of the wave is the direc on of energy transfer.
2. A transverse wave is one in which the displacement is perpendicular to the direc on of energy
transfer.
3. An oscilla on is a cyclical (repea ng) mo on of an object. If the mo on repeats in a fixed me
interval it is called isochronous, and the me interval of repe
on is called the period. An oscilla on
will have limits on its displacement from a central point.
4. Simple harmonic mo on is defined as mo on arising from accelera on of an object that is
propor onal to its displacement from a fixed equilibrium point (rest posi on) and directed toward
that point.
F , a ∝ −x
5. A restoring force (F ) is any force that is propor onal to the opposite of a displacement, which is
responsible for bringing an object in SHM back to its equilibrium.
6. The amplitude of a par cle in SHM is the maximum displacement from equilibrium.
7. A wavefront is a surface through crests and normal to the direc on of energy transfer of the wave.
8. Rays are lines at right angles to the wavefronts, and show the direc on of wave propaga on.
Formulae
F = −kx
f=
1
T
v = fλ
E = EP + EK
I=
P
a
I=
P
4πx2
I = I0 cos(θ)2
nm =
c
cm
sin(θ2 )
c2
=
sin(θ1 )
c1
sin(θ1 ) = n2 sin(θ2 )
λn =
2L
; n = 1, 2, 3, 4 … (both ends free/fixed)
n
λn =
4L
; n = 1, 3, 5, 7 … (one end free)
n
Symbol
What it represents
Unit
F
restoring force
N
k
spring constant
N m−1
x
displacement
m
E
total internal energy
J
EK
internal kine c energy
J
EP
internal poten al energy
J
P
power
W
a
area
m2
I
intensity/transmi ed intensity
W m−2
x
radius of the sphere in which a light source is emi ng light
m
I0
incident intensity on a polariser
W m−2
Theory
Simple Harmonic Motion
Phase difference must be an integral mul ple of π, π/2
Wave proper es
1. Time period (T )
2. Frequency
3. Phase
4. Phase difference
In SHM without damping, the sum of EK and EP is constant. In other words, energy is conserved.
Phase difference is the difference in synchronisa on of two waves, expressed in radians.
SHM can also be thought of as a one-dimensional projec on of uniform circular mo on:
Graphs of accelera on, velocity, and displacement
Rela onship between EK and EP
Waves
Each wave is simply a disturbance in a medium which transfers energy from one place to another.
This is done by oscilla ng par cles in the medium.
Waves have the following characteris cs:
Wavelength: the length of a complete oscilla on.
Frequency: the number of mes a complete oscilla on is done in unit me.
Time period: the me taken for one oscilla on.
Speed: how much distance each par cle in the medium covers in unit me.
Note: the speed of the wave depends only on the proper es of the medium and not on
how it is produced.
Deriving v = f λ
v=
distance
1
=λ⋅
= fλ
time
T
Transverse waves
In transverse waves, the direc on of propaga on is perpendicular to the movement of the par cles
in the medium.
Examples are water waves, light (electromagne c waves) , and waves on a guitar string.
enter image descrip on here
>
Each wave has a crest (posi on of maximum displacement) and a trough (posi on of minimum
displacement).
The amplitude of a transverse wave can be calculated by
A=
position of crest − position of trough
2
Electromagne c waves
These are transverse waves composed of perpendicularly oscilla ng magne c and electric fields.
These waves move at the speed of light (299, 792, 458ms−1 ) in a vacuum.
enter image descrip on here
Electromagne c waves are part of the electromagne c spectrum, which contains the following
categories of waves:
Longitudinal waves
In longitudinal waves, the direc on of propaga on is parallel to the direc on of energy transfer.
Examples are longitudinal waves along a spring, and sound.
Sound waves:
enter image descrip on here
Longitudinal waves in a spring:
enter image descrip on here
Just like transverse waves have crests and troughs, longitudinal waves have compressions and
rarefac ons.
Compressions are regions of the wave where par cles are closest to each other.
Rarefac ons are regions of the wave where par cles are furthest from each other.
Wavefronts and rays
A wavefront is a surface through crests and normal to the direc on of energy transfer of the wave.
Wavefronts can be
spherical
cylindrical
planar
A great explana on of how each work: h p://thescienceweek.blogspot.com/2016/07/wavefront-simplified-explana on.html
Rays are lines at right angles to the wavefronts, and show the direc on of wave propaga on.
Amplitude and intensity
Each wave carries energy, and the rate at which the energy carried is the power (P ) of the wave.
Intensity is the power of the light source incident on area a, expressed in W m−2 .
I=
P
a
If a light source is emi ng light in a spherical manner, then the light at a distance x away from the
light source has intensity:
I=
P
4πx2
The intensity at a par cular point is propor onal to the power P of the light source. Referring to
Wave Phenomena, we know that P ∝ Etotal ∝ A2 , so
I ∝ P ∝ Etotal ∝ A2
I ∝ A2
Superposition
The principle of superposi on states that when two or more waves of the same type arrive at a
given point in space at the same me, the displacement of the medium at that point is the algebraic
sum of the individual displacements.
Superposi on can happen construc vely or destruc vely.
Reflection of pulses
Reflec on from a fixed end
The pulse will be reflected by the fixed and become inverted.
This is because the fixed end exerts an equal and opposite force on the string/rope, so an upward
pulse now moves downward, and vice versa.
This corresponds to a phase change of 180∘ or π radians when reflected.
Reflec on from a free end
As the pulse arrives at the free end, the free end is pulled upwards. Once it falls back down, the
pulse is sent back not inverted, so there is no phase change.
Polarisation
An electromagne c wave is said to be plane polarised if the electric field oscillates on the same
plane.
Individual emi ers of light emit polarised light waves, but light from a large collec on of individual
emi ers is unpolarised.
A polariser is a sheet of material with a molecular structure that allows only a specific orienta on of
the electric field to pass through.
The ray of light consisted of a large number of differently polarised waves before entering the
polariser (the grey rectangle in the middle). A er coming through the polariser, light was polarised,
as the electric field oscillated on the same plane.
Malus’s Law
When an electromagne c field with electric field E0 passes through a polariser and makes an angle θ
with the transmission axis. the transmi ed electric field E will be E0 cos(θ).
This can be visualised as follows:
Only the ver cal component of E0 is transmi ed, calculated by E0 cos θ .
Since I ∝ E 2 , we can assume I0 = kE02 for some constant k . The new intensity would be
I = kE 2 = k(E0 cos(θ))2 = kE02 cos(θ)2 = I0 cos(θ)2
.
Therefore:
I = I0 cos(θ)2
where I represents transmi ed intensity and I0 represents incident intensity.
Polarisa on by reflec on
When unpolarised light reflects off a non-metallic surface, the reflected ray is par ally polarised.
The reflected light has various components of electric field of unequal magnitude.
The component with the greatest magnitude is found in the plane parallel to the surface.
Reflection
When a ray reflects off a surface, the angle of incidence with the normal to the surface is equal to
the angle of reflec on.
Refraction
Refrac on is the travel of light from one medium into another.
The refrac ve index of a par cular medium with respect to another medium is the ra o of the
speeds in the media.
In refrac on, the following laws are obeyed:
nm =
1.
c
cm
2.
sin(θ2 )
c2
=
sin(θ1 )
c1
3.
n1 sin(θ1 ) = n2 sin(θ2 )
Where:
c is the speed of light in vacuum
nm is the refrac ve index of medium m
θ1 is the angle at the normal to the surface for the ray passing through medium 1
θ2 is the angle at the normal to the surface for the ray passing through medium 2
The medium in which light is rela vely faster is called the op cally rarer medium.
The medium in which light is rela vely slower is called the op cally denser medium.
Whenever refrac on occurs, a small component of the incident ray is reflected.
Dispersion
Rays with the same angle of incidence but differerent wavelengths are refracted at different angles.
This is called dispersion.
This means that n for water with respect to a vacuum for red light is different than n for water with
respect to a vacuum for blue light.
Light with shorter wavelength refract more (bend more), and light with longer wavelength refract
less. This can be seen in the dispersion of white light in a prism.
Total Internal Reflec on
When light moves from a denser to a rarer medium, it diverges from the normal at a greater angle
than the angle of incidence.
The angle of incidence at which the angle of refrac on is 90∘ is called the cri cal angle, θc .
At any angle of incidence greater than θc, there is no refracted ray. The ray is completely reflected
back into the medium. This is called total internal reflec on.
sin θc =
1
n
Diffraction
Diffrac on takes place when a wave with wavelength comparable to or larger than the size of an
aperture or an obstacle moves through or past it and spreads out.
The larger the wavelength is rela ve to the slit width , the greater the effect.
If the aperture size equals the wavelength,the diffracted wavefronts are perfectly circular.
Interference
See Wave Phenomena for an explana on of single-slit and double-slit interference.
Standing Waves
When two waves of the same speed, wavelength and amplitude meet, a standing wave is formed.
Features of standing waves not seen in travelling waves:
The crests and troughs stay at the same place.
The points at which displacement is zero, nodes, do not move.
Half-way points between nodes are where displacement is maximum. These are called
an nodes.
Points between consecu ve nodes are in phase.
Points in between the next pair of consecu ve nodes have a velocity in the opposite direc on
of the points between a given pair.
The amplitude of oscilla on is different at different points on the string.
A standing wave does not transfer energy.
The ends of a standing wave are either nodes or an nodes. They determine the possible shape
of the wave.
Standing Waves on Strings
Harmonic
Number of nodes
Number of An nodes
Wavelength
1
2
1
2L
2
3
2
2L
2
3
4
3
2L
3
4
5
4
2L
4
n
n+1
n
2L
n
The first harmonic has the lowest frequency, as it has the highest wavelength.
All harmonics have frequencies that are integral mul ples of the first harmonic.
Standing Waves in Pipes
Pipe with both ends open
The end condi ons for this pipe are an node-an node.
2L
, n ∈ {1, 2, 3, 4, 5, 6, … }
n
λn =
Pipe with one end open
The end condi ons for this are an node-node.
There are only “odd” harmonics present here.
λn =
4L
, n ∈ {1, 3, 5, 7, 9, 11 … }
n
Pipe with both ends closed
The end condi ons for this are node-node.
λn =
2L
, n ∈ {1, 2, 3, 4, 5, 6, … }
n
References
K. A. Tsokos - Physics for the IB Diploma, Sixth Edi on
Class notes
h ps://en.wikipedia.org/wiki/Electromagne c_radia on
h ps://sites.ualberta.ca/~pogosyan/teaching/PHYS_130/FALL_2010/lectures/lect23/lecture23.html
Simplified Wave Front Explana on
Penn State: Superposi on of Waves
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Wave Phenomena
Related chapter: Waves
Home IB IGCSE ContributeYouTube
Syllabus
About
IB
Us Physics
9.1 Simple harmonic mo on
State the defining equa on for SHM and associate it with the graph of a against x.
Understand that x refers to the displacement from the equilibrium posi on.
Associate ω with the hardware of the system genera ng the SHM.
From ω calculate T and f , understanding that they do not change with x0, the
amplitude.
Iden fy (in the data booklet) and apply the equa ons rela ng to:
x as a func on of t.
v as a func on of t.
v as a func on of x.
kine c energy as a func on of x.
total energy as a func on of amplitude.
Sketch the graphs for the rela onships above.
Show graphically how the kine c and poten al energy changes with displacement and
with me.
Discuss the parameters of a mass–spring system that affect its frequency and apply the
correct equa on.
Discuss the parameters of a pendulum that affect its frequency and apply the correct
equa on.
Formulae
f=
1
T
v = fλ
a = −ω 2 x
Tf =
Etotal =
2π
ω
1
1
2
mvmax
= mA2 ω 2
2
2
EK =
1
1
mv 2 = mω 2 (A2 − x2 )
2
2
EP =
1
mω 2 x2
2
θ=
λ
b
b sin θ = nλ
s=
λD
d
1
2dn = (m + )λ
2
θA ≥
λ
b
λavg
= mN
Δλ
us
λ′ = λ(1 − )
v
R=
f′ = f(
v
)
v − us
Δf
v
≈
f
c
Δλ
v
≈
λ
c
Spring
F = −kx
k
m
ω2 =
T = 2π
m
k
T = 2π
l
g
Pendulum
Symbol
What it represents
Unit/Value
F
restoring force
N
k
spring constant
N m−1
x
displacement
m
ω
angular frequency
rad ⋅ s−1
l
length of pendulum
m
g
accelera on due to gravity
9.81ms−2
Single slit diffrac on
First loca on of destruc ve interference:
b sin θ = λ
For small angles of θ :
θ=
λ
b
s=λ
D
d
Double slit diffrac on
Separa on between maximums:
Mul ple slit diffrac on
With N slits, there are N − 2 maxima.
Intensity of the central maximum: N 2 × I0 , where I0 is the intensity of the central maximum in
single slit diffrac on.
Theory
Simple harmonic mo on can be modelled by the following equa ons:
x = A sin(ωt + ϕ)
v = ω ⋅ A cos(ωt + ϕ)
a = −ω 2 ⋅ A sin(ωt + ϕ) = −ω 2 x = −
k
x
m
where:
A: amplitude
ω : angular frequency (how many radians on the graph correspond to 1 second)
ϕ: the le ward phase shi
t: me
x: displacement from equilibrium posi on.
v : velocity.
a: accelera on.
In addi on:
vmax = ωA
amax = −ω 2 A
Velocity and accelera on as a func on of displacement
Energy
In an ideal system with no drag forces, energy is conserved:
EK + EP = Etotal remains constant.
The total energy of a system can be expressed in terms of the maximum kine c energy of the
system:
Etotal =
1
1
2
mvmax
= mA2 ω 2
2
2
So kine c and poten al energy are:
EK =
=
1
1
mv 2 = mω 2 (A2 cos2 (ωt + ϕ))
2
2
1
mω 2 (A2 (1 − sin2 (ωt + ϕ)))
2
=
1
mω 2 (A2 − x2 )
2
EP = Etotal − EK =
1
1
mA2 ω 2 − mω 2 (A2 − x2 )
2
2
=
1
mω 2 x2
2
note that v 2 = A2 cos2 (ωt + ϕ) and x2 = A2 sin2 (ωt + ϕ).
Single slit diffraction
For destruc ve interference, the path difference must be (n + 12 )λ, and the first minimum is
obtained where path difference = 12 λ.
With the aid of the diagram, we know that AB is the path difference between the two waves,
the slit width is b, and λ is the wavelength:
AB = path difference
1
λ (for the first minimum)
2
1
AB = λ
2
b
AB = sin θ (using trignometry)
2
1
b
λ = sin θ
2
2
b sin θ = λ
λ
sin θ =
b
Since sin θ ≈ θ for very small angles:
λ
θ=
b
path difference =
So, we can say that the first minimum for a single slit diffrac on is observed at an angle θ
where θ = bλ .
The formula to find the angle at which addi onal minima form becomes:
b sin θ = nλ
Where:
b : slit width.
n: the nth minima.
λ: wavelength.
We can also conclude that:
θ∝λ
As wavelength ↑, θ ↑, and the angular width of the central maxima ↑.
θ∝
1
b
As b ↑, θ ↓, and so the angular width of the central maxima ↓.
Young’s Double Slit Experiment
When light from two slits intefere in the following setup, a set of fringes are formed:
The following condi ons are required:
The light from both slits must be coherent, ie, constant or zero phase difference.
The distance of between the slits and the slit width must be negligible compared to the
distance between the screen and the slits.
The slit width should be comparable to the wavelength of the light for circular diffrac on.
Since D >> d, it can be assumed that the two rays are parallel.
For construc ve interference at point P , Δx = nλ, where n = 0, 1, 2, 3, 4, ….
In this case, n = 1.
Δx = λ = d sin θ
tan θ =
s
D
We can assume tan θ = sin θ because θ is extremely small.
λ = d(tan θ) =
ds
D
Finally, we can conclude that:
s=
λD
d
The graph of light intensity vs angle from center for both single and double slit diffrac on is as
follows:
The intensity of the double slit pa ern is modulated by the one-slit pa ern.
Mul -slit diffrac on
If the number of slits are increased (with the same length between each slit), the fringes are
more dis nctly pronounced:
For N slits, there are N − 2 secondary maxima between two primary maxima.
With an increase in the number of slits to N :
the primary maxima will become thinner and sharper
The N − 2 secondary maxima will become unimporant
The intensity of the central maximum is propor onal to N 2 .
Diffraction grating
used in spectroscopy to measure the wavelength of light.
Has rulings, which are slits, which help determine the slit sepera on.
x lines/rulings per millimetre corresponds to a x1 slit sepera on.
d = x1
Since we know that the condi on for construc ve interference is d sin θ = nλ, we can use
this to calculate the wavelength of light.
Thin-film interference
Upon reflec on on a surface with a refrac ve index greater than the medium a ray is already
in, the ray undergoes a phase change of π .
If d is the thickness of the film, the condi on for construc ve interference, when only one
phase change occurs (light is reflected off a medium with a greater refrac ve index only once):
1
2dn = (m + )λ
2
Where:
d : thickness of the film.
n : refrac ve index.
m : an integer.
Note: It is similar to the condi on for destruc ve interference (n + 12 λ), but it is the condi on for
construc ve interference in this case because of the phase shi of π , resul ng in crests becoming
troughs and troughs becoming crests.
The condi on for destruc ve interference, when only one phase change occurs(light is
reflected off a medium with a greater refrac ve index only once):
2dn = mλ
The condi on for construc ve interference when there is no or two phase changes (light is
reflected off a medium with a greater refrac ve index twice):
2dn = mλ
The condi on for destruc ve interference when there is no or two phase changes (light is
reflected off a medium with a greater refrac ve index twice):
1
2dn = (m + )λ
2
Resolution
The angular separa on of two objects is given by
θA =
s
D
where s is the distance between the objects, and D is the distance between the observer and
the objects.
According to the Rayleigh criterion, resolu on is possible when the angular separa on is
greater than the angle fo the first diffrac on minimum:
θA ≥
λ
b
For a circular slit, the following criteria is used:
θA ≥ 1.22
λ
b
In a diffrac on gra ng
R=
λavg
= mN
Δλ
where:
R: resolving power of the gra ng
λavg : The average of the two wavelengths to be resolved
Δλ: difference in the wavelengths that are to be resolved
m:
Doppler Effect
The doppler effect is the change in the observed frequency of a wave which happens
whenever there is rela ve mo on between the source and the observer.
If the wavelength of the light decreases, the source is moving towards the observer, which is
called a blueshi .
If the wavelength of the light increases, the source is moving away from the observer, which is
called a redshi .
If a source is moving towards an observer with velocity us :
λ′ = λ(1 −
f′ = f(
us
)
v
v
)
v − us
If an observer is moving towards a source with velocity uo :
f′ = f(
v + uo
)
v
For light
If the speed of the observer is small compared to the speed of light, then:
Δf
v
≈
f
c
Δλ
v
≈
λ
c
References
K. A. Tsokos - Physics for the IB Diploma, Sixth Edi on
Class Notes
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