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JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
8. Wave Motion
Content
1.
2.
3.
4.
Progressive Waves
Transverse and Longitudinal Waves
Polarisation
Determination of frequency and wavelength
Learning Outcomes:
Candidates should be able to:
(a)
show an understanding and use the terms displacement, amplitude, phase difference,
period, frequency, wavelength and speed.
(b)
deduce, from the definitions of speed, frequency and wavelength, the equation v = fλ.
(c)
recall and use the equation v = fλ.
(d)
show an understanding that energy is transferred due to a progressive wave.
(e)
recall and use the relationship, intensity ∝ (amplitude) .
(f)
analyse and interpret graphical representations of transverse and longitudinal waves.
(g)
show an understanding that polarisation is a phenomenon associated with transverse
waves.
(h)
determine the frequency of sound using a calibrated c.r.o.
(i)
determine the wavelength of sound using stationary waves.
2
Acknowledgements:
1.
Comprehensive Physics for ‘A’ level (3rd edition) Vol.2 – KF Chan, Charles Chew, SH
Chan (Federal Study Aids)
2.
Fundamentals of Physics (6th edition) – Halliday, Resnick, Walker (John Wiley & sons,
Inc.).
3.
Notes Wave Motion 2009 by FS Chin ©JJ
4.
2010 J2 H2 Wave Motion Notes by KW Chong ©JJ
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JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
Waves vs. Particles
Two ways to get in touch with a friend in a distant city are to write a letter and to use the
telephone. The first choice (the snail-mail) involves the concept of a “particle” : a material
object moves from one point to another, carrying with it information and energy.
The 2nd choice (the telephone) involves the concept of “waves”. The information and energy
move from one point to another but no material object makes that journey. In the telephone call,
a sound wave carries your message from your vocal cords to the telephone. There, an
electromagnetic wave takes over, passing along a copper wire or an optical fiber or through the
atmosphere (satellite). At the receiving end, there is another sound wave, from a telephone to
your friend’s ear. Although the message is passed, nothing that you have touched reaches
your friend.
“It often happens that the wave flees the place of its creation, while the water does not:
like the wave made in a field of grain by the wind, where we see the waves running across the field while
the grain remains in place.” ~ Leonardo da Vinci.
Particle and wave are the 2 great concepts in classical physics, although the concepts are very
different. The word particle suggests a tiny concentration of matter capable of transmitting
energy. The word wave suggests a broad distribution of energy, filling the space through which
it passes.
Wave Production
1.
The source of any wave is a vibration or oscillation.
2.
A wave is a mechanism for the transfer of energy from one point to another without the
physical transfer of any material between the points.
Types of Waves
There are 3 main types:
1.
Mechanical Waves. These waves are most familiar because we encounter them
almost constantly. Examples include water waves, sound waves and seismic waves.
The wave motion is transmitted by the particles of the medium oscillating to and fro.
Main features : (i) Newton’s laws govern them;
(ii) they can only exist within a material medium (e.g. water, air & rock)
2.
Electromagnetic (EM) waves. These are less familiar, but we use them constantly.
Common examples include visible and ultraviolet light, radio and television waves,
microwaves, x-rays and radar waves. The wave motion is in the form of varying electric
and magnetic fields.
Main features : (i) these waves require no material medium to exist
(ii) All electromagnetic waves travel through a vacuum at the
same speed, c (speed of light) ,
i.e. c = 299 792 458 m s-1 ≈ 3 x 108 m s-1.
3.
Matter waves. Although these waves are commonly used in modern technology, their
type is probably very unfamiliar to us. These waves are associated with electrons,
protons and other fundamental particles, even atoms and molecules. Because we
commonly think of these things as constituting matter, they are called matter waves.
(Not in your syllabus).
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JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
Visualising Waves
A typical way to visualize water waves is to use a ripple tank experiment. The set up is as
shown below.
Fig 1(a)
Fig 1(b)
Learning points:
• Our usual representation of waves is typically sinusoidal curves, as seen in Fig 1(a), which
is the wave pattern as seen from the side view.
• Another common representation of waves is seen in Fig 1(b), which is the wave pattern
projected by a light source from the top (i.e. Top View). The dark lines indicate the crests of
the wave.
a)
Show an understanding and use the term displacement, amplitude, phase
difference, period, frequency, wavelength and speed.
1.
Displacement, x
Displacement of a particle is its distance in a given direction from its equilibrium
position.
(Similar as in SHM. It is NOT the displacement of the whole wave.)
2.
Amplitude, A
Amplitude is the magnitude of the maximum displacement of a particle from its
equilibrium position.
It is a scalar.
3.
Period, T
The period of oscillation of a wave is the time taken for a particle (element) in the
wave to complete one oscillation.
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4.
WAVE MOTION
Frequency, f
Frequency of a wave is the number of oscillations per unit time made by a particle
(element) in the wave.
The frequency of a wave is the same as the frequency of its source.
It is independent of the medium through which the wave propagates.
f =
It is related to period, T, by
5.
1 ω
=
T 2π
where ω = angular frequency.
Wavelength, λ
Wavelength of a wave is the shortest distance between two points which are in
phase.
6.
Speed, v
Speed of a wave is the distance travelled by the wave per unit time.
7.
Phase and Phase Difference, φ
Phase difference between two points in a wave is the difference between the stages
of oscillations, expressed in terms of an angle.
(e.g. Two points half a wavelength apart has a phase difference of π radians)
1.
Two points being in the same phase means that they are in the same state of
disturbance at the same time (e.g. the points x x’ and y y’ in the diagram above).
They are said to be in phase. Their phase difference is zero.
2.
Two crests or two troughs are in phase, whereas a crest and an adjacent trough
are out of phase by 180° or π rad. They are said to be anti-phase. Their phase
difference is 180° or π rad.
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JJ 2011 J2/H1 Physics (8866)
3.
WAVE MOTION
The phase difference, φ, between two particles in a wave (e.g. particle P and
another particle B, below) separated by distance x, is given by
x
λ
=
φ
2π
The L.H.S is a ratio of angles in radians The denominator is the angle for a
complete oscillation.
The R.H.S is a ratio of distance travelled by the wave. The denominator is the
distance travelled by the wave in the time for a complete oscillation, the
wavelength λ.
4.
The phase difference, φ, between two oscillations separated by a time t is
given by
φ
t
=
2π T
Now the R.H.S. is a ratio of time. The denominator is the time for a complete
oscillation, the period T.
(b)
Deduce, from the definitions of speed, frequency and wavelength, the
equation v = fλ.
Speed, v, is the distance travelled per unit time. In the time of one period, T, the wave
travels a distance of one wavelength, λ. So the speed of a wave motion can be
expressed as
distance travelled λ
=
- - - - - (1)
v=
time taken
T
Frequency f is related to period T by the relation
f=
1
T
- - - - - (2)
From the relations (1) and (2) above, the speed of a wave motion can hence be
expressed as
v = fλ
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JJ 2011 J2/H1 Physics (8866)
(c)
WAVE MOTION
Recall and use the equation v = fλ
Example 1
Visible light has wavelengths between 400 nm and 700 nm, and its speed in a vacuum is
3.0 × 108 m s-1.
What is the maximum frequency of visible light?
Solution:
From v = fλ, the frequency f =
v
λ
, i.e. f is inversely proportional to λ.
For maximum frequency, minimum wavelength should be used.
Hence,
fmax =
v
λmin
=
3 × 108 m s−1
= 7.5 × 1014 Hz.
400 × 10−9 m
Example 2
A sound wave of frequency 400 Hz is travelling in a gas at a speed of 320 m s-1.
What is the phase difference between two points 0.2 m apart in the direction of travel?
Solution:
Wavelength, λ =
v 320 m s−1
=
= 0.80 m
f
400 Hz
φ x 0.2 m 1
= =
=
2π λ 0.8 m 4
Æ
φ=
π
2
rad
Example 3
The speed of electromagnetic (EM) waves (which include visible light, radio and x-rays) in
vacuum is 3.0 x 108 m s-1.
a)
Wavelengths of visible light waves range from about 400 nm in the violet to about 700
nm in the red. What is the range of frequencies of these waves?
b)
The range of frequencies for short-wave radio (e.g. Class 95 FM) is 1.5 to 300 MHz.
What is the corresponding wavelength range?
c)
X-ray wavelengths range from about 5.0 nm to about 1.0 x 10-2 nm. What is the
frequency range for x-rays?
Solution
a)
Using v = fλ Æ f =
∴ f red =
v
λ
8
3.0 x10
= 4.29 x10 14 Hz
−9
700 x10
&
f violet =
3.0 x10 8
= 7.50 x1014 Hz
−9
400 x10
14
14
Hence, range of frequencies for visible light is from 4.29 x 10 Hz to 7.50 x 10 Hz.
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b)
From v = fλ Æ λ =
WAVE MOTION
v
f
3 x108
3 x108
200
m
=
&
λ
=
=1 m
2
1.5 x106
300 x106
Hence, wavelength range is from 1 m to 200 m
∴ λ1 =
c)
From v = fλ Æ f =
f1 =
v
λ
8
3.0 x10
= 6.0 x10 16 Hz
−9
5.0 x10
& f2 =
3.0 x10 8
= 3.0 x10 19 Hz
−11
1.0 x10
16
Hence, frequency range for x-ray is from 6.0 x 10
19
to 3.0 x 10
Hz.
Example 4
What happens to the speed, frequency and wavelength of light when it enters glass from air?
A
B
C
D
speed
frequency
wavelength
decreases
increases
unchanged
decreases
increases
unchanged
decreases
unchanged
unchanged
increases
decreases
decreases
Solution:
Frequency of a wave is the same as the frequency of its source, independent of the medium.
Speed of light is highest in vacuum (or air).
(d)
Ans: D
Show an understanding that energy is transferred due to a progressive
wave.
•
Oscillation (or oscillatory motion) refers to the to-and-fro motion of a particle about
an equilibrium position.
The oscillatory motion of the particle is a continuous exchange of potential and
kinetic energy of the particle. It is illustrated by a graph of displacement from that
equilibrium position, y, vs time, t.
•
Wave refers to the combined motion of a series of linked-particles, each of which
is originally at rest at its respective equilibrium position.
Starting from the oscillation of the first particle about its equilibrium position, the
energy of the oscillation is passed to the second particle, which in turn is passed to
the third particle and subsequent particles in the series of linked-particles.
So wave motion is the motion of energy passed from one particle to the next in a
series, through oscillatory motion of these particles, in sequence.
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JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
Examples:
(1) Sound wave:
When a sound wave is propagated from a tuning fork to an ear of a person some
distance away, the vibration of the fork sets the air layer next to it into vibration.
The second layer of air is then set into vibration by the transfer of energy from the
first layer.
This transfer of energy continues for subsequent layers until the layer of air next to
the ear is also set into vibration, which in turn vibrates the ear-drum of the ear,
enabling the person to hear the sound originated from the tuning fork.
There is no net transfer of air particles from the tuning fork to the ear. The ear-drum
in the ear can vibrate because energy has been transferred to it from the tuning fork,
through the sequential vibration of the layers of air between the tuning fork and the
ear. (Diagram above)
This sequential vibration of the layers of air forms regions of compression (where
air layers are closer to each other) and regions of rarefaction (where air layers are
further apart). The one-way movement of such regions from the tuning fork to the
ear signifies the propagation of sound wave energy.
(2)
Wave in a rope:
The wave travelling in a rope may originate from the vibration of the first particle at
one end of the rope.
The energy of the vibrating first particle is transferred to the second particle, setting
it into vibration.
This transfer of energy continues to subsequent particles in the rope until it reaches
the other end of the rope.
(3)
Water wave:
The energy of the vibrating water molecules is transferred to subsequent molecules
along the surface of water, causing these molecules further from the vibrating
source to be set into up-down motion.
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JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
The examples (1), (2) and (3) are examples of progressive waves, where energy is
transferred from one region to another region through sequential vibration of a series of
linked-particles.
The energy of a first vibrating particle is propagated along a series of linked-particles to
another region. Sound energy is propagated from the tuning fork to the ear, energy from
one end of a rope is propagated to the other end, and energy from one region of water
surface next to a vibrating source is propagated to another region in the ripple tank.
(e)
Recall and use the relationship, intensity ∝ (amplitude)2.
Intensity, I, is the rate of incidence of energy per unit area normal to the direction of
incidence.
The rate of incidence of energy can be regarded as power.
The plane of the area, which the wave energy is incident onto, has to be normal
(perpendicular) to the direction of the incidence of the wave energy.
The unit of intensity is W m-2.
Intensity on an area A can be expressed as
I=
P
A
where P is the power incident on the area normally.
Intensity ∝ (amplitude)2
To help to recall
For rationale:
From S.H.M., total energy can be expressed as
So energy ∝ (amplitude)2
Î
1
2
mω 2 xo2 , where xo represents amplitude.
Intensity ∝ (amplitude)2.
[Further information on Intensity can be found in the Extra Reading section]
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WAVE MOTION
Example 5
A sound wave of amplitude 0.20 mm has an intensity of 3.0 W m-2.
What will be the intensity of a sound wave of the same frequency which has an amplitude of
0.40 mm?
Solution:
The relation
I ∝ (amplitude)2
can be expressed as
I = k(amplitude)2
where k is the constant of proportionality.
Substituting,
3.0 W m-2 = k(0.20 mm)2 - - - - - (1)
I = k(0.40 mm)2 - - - - - (2)
New intensity,
I
=4
3.0 W m-2
(2)
:
(1)
Î I = 12.0 W m-2.
•
For sound waves, intensity is a measure of loudness.
•
For light waves, intensity is a measure of brightness.
Example
For a 40 W lamp, if the surface area of
the light bulb is 0.0010 m2, the intensity on the surface
of the light bulb is 40/0.0010 W m-2 = 40 kW m-2.
The bulb acts as a point source where light wave
energy is propagated uniformly in all directions.
r
•
•
If the lamp has a transparent spherical shell of radius
r enclosing the light bulb at its centre, the spherical
surface area of the shell is 4πr2.
bulb
transparent spherical
shell
The intensity on the surface of the shell is the light power per unit area incident on the
shell.
Since it is not easy to determine the amount of light power incident on a unit area of the
shell, it is not easy to determine the intensity on the shell surface by using the amount of
power incident on unit area of the shell.
For easy calculation, we consider all the power from the source to be incident on the
whole area of the shell.
The intensity at the surface of the shell, from a point source, can then be expressed as
I =
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power of source
1
∝ 2
2
r
4π r
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WAVE MOTION
Example 6
A point source of sound radiates energy uniformly in all directions. At a distance of 3.0 m from
the source, the amplitude of vibration of air molecules is 1.0 × 10-7 m.
Assuming that no sound energy is absorbed, calculate the amplitude of vibration 5.0 m from the
source.
Solution:
Using the relation above,
and the relation
we get the relation
Î
1
r2
I ∝ (amplitude)2
I∝
1
r
c
amplitude =
r
amplitude ∝
where c is the constant of proportionality.
Substituting,
(2)
:
(1)
- - - - - (1)
- - - - - (2)
new amplitude = 6.0 × 10-8 m.
Î
(f)
c
3.0 m
c
New amplitude =
5.0 m
New amplitude 3
=
5
1.0 × 10−7 m
1.0 × 10-7 m =
Analyse and interpret graphical representations of transverse and
longitudinal waves.
In a wave, there are two directions of motions:
(1) direction of propagation of energy (which is the direction of motion of the wave),
(2) direction of oscillation of the particles in the wave.
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WAVE MOTION
A transverse wave is one in which the direction of propagation of energy is
perpendicular to the direction of oscillation of the particles in the wave.
The string element’s motion is
perpendicular to the wave’s direction of
travel. This is a transverse wave.
In the example of a wave travelling along a string (or a wave travelling along a slinky
diagrams above, page 9), the direction of propagation of the wave is along the string. If
the wave is started from one end of the string by the oscillation of the first element in the
direction perpendicular to the string, then this wave travelling along the string is an
example of a transverse wave.
A longitudinal wave is one in which the direction of propagation of energy is parallel to
the direction of oscillation of the particles in the wave.
A visual demonstration of
a longitudinal wave.
When a sound wave set up by the vibrating piston propagates along the pipe of air, the
direction of propagation of sound energy is along the pipe to the right. The direction of
oscillation of the air layers is back and forth, parallel to this direction. Hence sound wave
is an example of a longitudinal wave, as explained in the diagram below.
A sound wave is set up in an air-filled pipe by moving a piston back
and forth.
The rightward motion of the piston moves the elements of air next to it
rightward, changing the air pressure there.
The increased air pressure then pushes rightward on the elements of
air further along the pipe.
Moving the piston leftward reduces the air pressure next to it.
Hence the elements move backward and so forth.
The motion of the air (and the change in air pressure) travel rightward
along the pipe as a pulse.
Since the motion of the elements of air is parallel to the direction of
the wave’s travel, it is a longitudinal wave.
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JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
Graphs used to represent transverse or longitudinal waves are the same.
Graph 1 : Displacement vs. Position graphs
These are plotted with displacement, y, against distance or position, x.
For a transverse wave moving from left to right along the x-axis, displacement of the
particles in the wave, y, may be given a +ve sign for displacement upwards, and a –ve
sign for displacement downwards.
For a longitudinal wave moving from left to right along the x-axis, displacement of the
particles in the wave, y, may be given a +ve sign for displacement to the right, and a –ve
sign for displacement to the left.
distance
In Displacement vs. Position graphs,
• the graph represents the actual wave
at an instant in time
• the distance between consecutive
crests or consecutive troughs is one
wavelength
• The maximum height of the vertical
axis = amplitude of wave
raph 2 : Displacement vs. Time graphs
In contrast, graphs used to represent an oscillation of a particle are plotted with
displacement, y, against time, t.
In the graph above, we are tracking the displacement of one particle only as time goes
by. This does NOT represent the wave.
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WAVE MOTION
one
period
time
In Displacement vs. Time graphs,
• The graph represents the oscillation
of one particle on the wave with time.
• the “distance” between consecutive
crests or consecutive troughs is one
period
• The maximum height of the vertical
axis = amplitude of oscillation
• Fig.2 shows 5 ‘snapshots’ of a sinusoidal
transverse wave in a string, travelling in the +ve
direction of an x-axis (left to right).
• The movement of the wave is indicated by the
right-ward progress of the short down-pointing
arrow, pointing at the middle ‘crest’ of the wave
in snapshot (a).
• From snapshots (a) to (e), the short arrow moves
to the right with the wave, but each particle in the
string moves parallel to the y-axis (up and down).
An example of such a particle is along the y-axis
(shown darkened).
• Each snapshot is taken at an interval of ¼ period.
One full oscillation takes place from (a) to (e).
(e)
Fig.2
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WAVE MOTION
In the 3 diagrams below (for longitudinal wave), diagram
(1)
shows the equilibrium positions of 15 particles and their displacements at a
particular instant,
(2)
shows the corresponding displacement-distance graph (+ve displacement to the
right),
(3)
shows the corresponding change in pressure for air layers in the atmosphere (for a
sound wave travelling through air).
Diagram (1) – Actual positions of layers within the longitudinal wave
Diagram (2) – Displacement vs. Distance
Diagram (3) – Air Pressure vs. Distance
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WAVE MOTION
Summary For Part (f):
(1)
Displacement-time graph is for the oscillation of a particle in the wave.
The graph above shows an element oscillating with an amplitude of 4 mm.
Its period of oscillation is about 5 ms.
(2)
Displacement-distance graph is for a snapshot of a wave motion at an instant.
The graph above shows an instant of a wave with an amplitude of 4 mm.
Its wavelength is about 1.8 m.
Example 7
The diagram below shows an instantaneous position of a string as a transverse progressive
wave travels along it from left to right.
Which one of the following correctly shows the directions of the velocities of the points 1, 2 and
3 on the string?
Solution
1
2
3
Knowing that the wave is traveling from LEFT to
A
→
→
→
RIGHT, sketch how the wave would look like just an
instant after:
B
→
←
→
C
↓
↓
↓
D
↓
↑
↓
E
↑
↓
↑
Wave
motion
Then look at the points concerned.
Since it is a TRANSVERSE wave, the particles only
oscillates perpendicular to the wave direction.
This eliminates Answers A & B.
Ans : [
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JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
Example 8
The graph shows the shape at a particular instant of part of a transverse wave travelling
along a string.
Which statement about the motion of elements of the string is correct?
A
B
C
D
The speed of the element at P is a maximum
The displacement of the element at Q is always zero
The energy of the element at R is entirely kinetic
The acceleration of the element at S is a maximum
Solution:
Although the graph represents the whole wave at an instant in time, the question requires you
to analyse the motion of the individual particles within the wave at this instant.
Element P: At extreme end of oscillation
Element Q: At equilibrium position
Element R: At extreme end of oscillation
Element S: At extreme end of oscillation
(i)
Æ
Æ
Æ
Æ
stationary
moving fastest
stationary, no kinetic energy
max displacement, max acceleration
Ans: D
Show an understanding that polarisation is a phenomenon associated with
transverse waves.
Polarisation is a phenomenon in a transverse wave where the vibrations of the elements
in the wave are restricted to a plane.
Consider a transverse wave in a rope travelling along the x-direction:
Transverse waves with initial oscillations in the y-direction
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WAVE MOTION
Transverse waves initially oscillating (a) in the z-direction, and (b) at an angle in the y-z plane.
In the cases above, the wave is plane-polarised (i.e. the oscillations of the elements in
the wave are in a plane).
The following is adapted from Comprehensive Physics for ‘A’ level (3rd edition) Vol.2
– KF Chan, Charles Chew, SH Chan (Federal Study Aids), page 32.
•
In Figure 1, in a horizontal rope WZ, a transverse wave is set up by vibrations in
many different planes by holding the end W, and moving it in all directions
perpendicular to WZ, as illustrated by the arrows in plane P.
•
2 parallel slits X and Y are placed
between W and Z. A wave then
emerges along XY. Unlike the wave
along WX (not shown) due to
vibrations in different planes, the
wave along XY is due only to
vibrations parallel to slit X. This
plane-polarised
wave
passes
through the parallel slit Y.
•
In Figure 2, the slit Y is turned such
that it is perpendicular to slit X. No
wave is observed beyond Y.
•
To show that polarisation cannot be
obtained with longitudinal waves,
WZ is replaced by a thick elastic
cord. Longitudinal waves can be produced along the cord, by vibrating the end W
parallel to WZ.
•
The longitudinal wave set up in cord WZ will pass through the slits X and Y when they
are parallel (as in Fig.1) and when they are perpendicular (as in Fig.2). This shows
that there is no polarisation.
A longitudinal waves cannot be polarised because the particles in the wave oscillate
parallel to the wave direction and cannot be restricted to vibrate in any plane.
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(h)
WAVE MOTION
Determine the frequency of sound using a calibrated c.r.o.
A calibrated c.r.o. (cathode-ray oscilloscope) implies that the time-base is set such that
the period, T, of oscillations of the air layers detected by the microphone may be read.
Using the relation
f=
1
T
the frequency, f, of sound produced by the vibrating loudspeaker may be determined.
Example 9
The trace shown appeared on an oscilloscope screen with the time-base set to 2.0 ms cm-1.
1 cm
What is the frequency of the signal?
A 40 Hz
B 125 Hz
C 250 Hz
D 500 Hz
Solution
Period, T = 2.0 ms cm-1 × 4 cm = 8.0 ms
1
1
Frequency, f = =
= 125 Hz
T 8 × 10−3 s
(i)
Determine the wavelength of sound using stationary waves.
Please refer to notes on Stationary Waves in the topic Superposition.
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Ans: B
JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
Extra Reading
Electromagnetic Waves
James Clerk Maxwell’s (1831 – 1879) crowning achievement was to show that a beam of light
is a travelling wave of electric and magnetic field, an electro-magnetic wave.
In Maxwell’s time, the visible, infrared and ultraviolet form of light were the only electromagnetic
waves known. Heinrich Hertz then discovered what we now call radio waves and verified that
they move through the laboratory at the same speed as visible light.
We now know a wide spectrum of electromagnetic waves. The Sun, being the dominant source
of these waves, continually bathes us with electromagnetic waves throughout this spectrum.
Type of EM wave
Gamma (γ) rays
x-rays
UV ultraviolet
Visible light
IR (infra-red)
Radio wave (includes
microwaves, UHF, VHF etc)
Typical Wavelengths λ and its
corresponding frequency, f.
λ
= 1 pm = 10-12 m
f
= 3 x 1020 Hz
λ
= 100 pm = 10-10 m
f
= 3 x 1018 Hz
λ
= 10 nm = 10-8 m
f
= 3 x 1016 Hz
λred
= 700 nm
λgreen = 600 nm = 0.6 μm
λviolet = 400 nm
fgreen = 5 x 1014 Hz
λ
= 100 μm = 10-4 m
f
= 3 x 1012 Hz
λ
= 3m
f
= 108 Hz
Orders of magnitude
for wavelength, λ / m
10-12
10-10
10-8
10-6
10-4
100 ~ 10-2
Properties of Electromagnetic Waves
1)
EM waves have the same speed, c, in vacuum (c ≈ 3 x 108 m s-1).
2)
EM waves consist of oscillating electric and magnetic fields that are perpendicular to
each other.
3)
EM waves are all transverse waves.
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Extra Reading
Energy (E) and Intensity (I) of a Progressive Wave
When we set up a wave on a stretched string, we provide energy for the motion of the string.
As the wave moves away from us, it transports that energy as both kinetic energy and elastic
potential energy.
Kinetic Energy
An element of the string of mass Δm, oscillating transversely in
simple harmonic motion as the wave passes through it, has KE
G
associated with its transverse velocity u .
• When the element is rushing through its y = 0 position
(element b in the diagram), its transverse velocity – and
thus its KE - is a maximum.
• When the element is at its extreme position y = A (element
a), its transverse velocity – and thus its KE – is zero.
Elastic Potential Energy
To send a sinusoidal wave along a previously straight string, the wave must necessarily stretch
the string. As a string element of length Δx oscillates transversely , its length must increase
and decrease in a periodic way if the string element is to fit the sinusoidal waveform. Elastic
potential energy is associated with these length changes, just as for a spring.
• When the string element is at y = A, its length has its normal undisturbed value Δx, so its
elastic potential energy is zero.
• However, when the element is rushing through its y = 0 position, it is stretched to its
maximum extent, and its elastic potential energy then is a maximum.
Energy Transmitted
As waves travel through a medium, energy is transmitted as vibrational energy from particle to
particle of the medium. The energy, E, is given by:
E ∝ f2A2,
Since E ∝ f2A2 ⇒
where f = frequency, and A = amplitude
Intensity, I ∝ f2A2.
For a given source of fixed vibrations, f is constant.
∴
where A = amplitude
The unit for Intensity, I, is W m-2.
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I ∝ A2
JJ 2011 J2/H1 Physics (8866)
WAVE MOTION
Extra Reading
Polarisation of light
The electromagnetic (E.M.) waves emitted by any common source of light (e.g. the Sun or a
lamp) are polarized randomly or unpolarised; i.e., the electric field at any given point is always
perpendicular to the direction of travel of the waves but changes direction randomly.
Since E.M. waves are transverse in nature, they can be polarised. We can produce and detect
polarised light by using polarising sheets, commercially known as Polaroids or Polaroid filters
(invented by Edwin Land in 1932 while he was an undergraduate student). A polarising sheet
consisits of certain long molecules embedded in plastic. When the sheet is manufactured, it is
stretched to align the molecules in parallel rows.
ELECTRIC
FIELD
MAGNETIC
FIELD
3-D representation of an EM wave
When light shines through a polariser, the electric field component parallel to the polarising
direction passes through (transmitted); a component perpendicular to it is absorbed. Hence, the
electric field of the light wave emerging from the sheet consists of only the componenets that
are parallel to the polarising direction.
Malus’ Law
Taken from : http://www.saburchill.com/physics/chapters2/0041.html
Light travelling parallel to polariser Æ the transmitted light has (almost) the same intensity as
the polarised light (i.e. the amplitude of the light wave is identical).
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WAVE MOTION
When the 2nd polariser, or the Analyser is perpendicular to polariser, no transmitted light is
observed. Hence, intensity is zero. (i.e. the amplitude of the light wave is zero).
With the polariser and analyser at some other angle, θ, the amplitude of the transmitted light
waves is equal to component of the amplitude of the polarised light parallel to the plane of
the analyser.
Therefore, amplitude of transmitted light is given by
(amplitude of polarised light) × cos θ
Since intensity of a wave is proportional to its amplitude squared, we conclude that if polarised
light is incident on a polarising filter, the intensity of the transmitted light is proportional to the
cos2 of the angle between the plane of polarisation and the plane of the filter:
Itransmitted = I0 cos2 θ
where I0 is the intensity of the light incident on the analyser and I is the intensity of the
transmitted light.
This is called Malus’ law.
Uses of Polarisers : LCD panels
In LCD panels, a bright white light is emitted behind the panels. This source of light is typically a
fluorescent backlight. In recent years, there has been a move to use LEDs to illuminate the
panels leading to better energy efficiency.
Liquid crystals are the 4th state of matter, with plasma being the 5th state. Liquid crystal is a
liquid substance that has solid-like properties, often rod-shaped and their orientation can be
changed using electric fields. The LCD panels is made from 2 polarizer with axis aligned 90°
apart. Between these polarizer are 2 layers of glass panel, with an “invisible electrode” etched
on each side. The electrodes are typically made of indium-tin-oxide, a transparent and
conductive material. The liquid crystal is placed between these 2 glass panels, but their
orientation gradually twists till they are 90° apart, each orientation corresponding to the
polarizer’s axis nearest to them.
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WAVE MOTION
Diagram taken from : http://qxwujoey.tripod.com/lcd.htm
White light passing through the 1st polarizer will be polarized in one direction, which then travels
through the liquid crystals. The crystals change the plane of polarization, allowing light to be
able to pass through the 2nd polarizer. An observer will hence see that the screen is “bright” or
“white”.
When an electric field is applied across the liquid crystal, their crystal orientation is now
straightened. Light passing the 1st polarizer will hence travel along the liquid crystal but cannot
pass through the 2nd polarizer as it is 90° to it. An observer will observe the screen as “dark” or
“black”. By controlling the electric field over very tiny areas (called pixels), light can be filtered
out or allowed to pass through, creating an image on the screen.
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