l 09 waves

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XMLECTURE

09 WAVES NO DEFINITIONS. JUST PHYSICS.

9.1 What are Waves? ....................................................................................................................... 2

9.2 Transverse vs Longitudinal Waves ............................................................................................. 3

9.3 Graphical Representation ........................................................................................................... 4

9.4 Wave speed ................................................................................................................................ 6

9.5.1 Phase Relationship .................................................................................................................. 8

9.5.2 Graphical Representation of Phase Relationship ................................................................ 10

9.6.1 Longitudinal Wave ................................................................................................................. 12

9.6.2 Interpretation of Longitudinal Wave Graphs ........................................................................ 13

9.7 Electromagnetic wave ............................................................................................................... 16

9.8 Intensity .................................................................................................................................... 18

9.9.1 Polarization ............................................................................................................................ 21

9.9.2 The Three Polarizer Problem .............................................................................................. 24

Online resources are provided at https://xmphysics.com/waves

https://xmphysics.com/waves


9.1 What are Waves?

We observe in nature that when we have an elastic medium, a disturbance originating at one point

can be propagated to another point in the medium. This mechanism for transfer of energy or

momentum is called wave.

For example, the water surface is kind of like an elastic sheet. A disturbance caused by a falling rain

drop will be propagated outward. The energy of the fallen rain drop is carried outward along water

surface by a circular wave.

It is important to note that each individual particle in the medium merely oscillates about its own

equilibrium position. The particles do not travel with the wave, only the energy is passed along.

see animation at xmphysics.com

http://www.xmphysics.com/9.1



9.2 Transverse vs Longitudinal Waves

A wave is basically a vibration that gets propagated along the medium.

If the direction of those vibrations are perpendicular to the direction of wave travel, we call it a

transverse wave.

Transverse Wave see animation at xmphysics.com

If the direction of those vibrations are parallel to the direction of wave travel, it’s called a longitudinal

wave.

Longitudinal Wave see animation at xmphysics.com

One thing that transverse waves can do but longitudinal waves cannot, is this phenomena called

polarization. We know that light can be polarized. So we can tell that light is a transverse wave.

On the other hand, longitudinal waves can produce regions of high and low pressures called

compressions and rarefactions. Transverse waves cannot. Sound is an example of a longitudinal

wave.

Both polarization and pressure waves will be discussed in detail in later sections.

direction of vibration

wave direction crest

trough

direction of vibration

wave direction

compression rarefaction





9.3 Graphical Representation

Graphs are used to depict waves. You will commonly encounter two types of graphs:

1. Displacement-position graph: which shows the wave profile at one particular instant.

2. Displacement-time graph: which describe the vibration at one particular position.

Suppose we have a sinusoidal wave1 travelling rightward. The displacement-position graph below

shows the “frozen” profile of the wave at 0t .

Graph of a Wave

At each position along the wave is a SHM. Bearing in mind that the wave is travelling rightward, we

can work out that displacement-time graphs of the oscillations at position A and B to be as follow.

Graph of Two Points of a Wave

1 In the H2 syllabus, we are concerned mainly with sinusoidal waves. The cute thing about sinusoidal waves is

that every point on a sinusoidal wave is a sinusoidal oscillations, i.e. simple harmonic motion. So all the things

you learn about SHM applies here.

A

B

displacement

position

direction of wave

λ

A B displacement

time

T



Worked Example 1

Deduce the direction of motion of point A (at this instant), if the graph is

a) a displacement-position graph of a wave traveling rightward.

b) a displacement-time graph.

Assume that positive displacement is upward, and negative displacement is downward.

Solution

a) downward

b) upward

A

displacement

displacement

position

direction of wave

A

displacement

time

A

a)

b)


9.4 Wave speed

The speed of a wave refers to the speed at which the energy or momentum is being passed along by

the wave. It is different from the speed of the individual particles in the medium as they perform their

oscillations (about their equilibrium positions).

You can think of wave speed as the speed at which a crest of the wave advances. Do you realize that

a crest always advances by one wavelength after every one period T? Yup, that’s why the speed of

the wave is given by the formula v fT

.

watch animation at xmphysics.com

v = fλ

Do you think high pitched sound travel faster or slower than low pitched sound? Interestingly, the

speed of a wave is largely dependent on the medium, and changing the frequency does not change

its speed. Based on the formula v f , it is obvious that a higher frequency wave must have a

shorter wavelength, in order for the wave speed to be constant.

Same medium

So what determines the speed of a wave? It is the medium, its elasticity and its inertia. Roughly

speaking, the stiffer the medium and lighter the medium particles, the faster the disturbance will be

passed along. For example, sound travels faster in solids than gas because solid is a stiffer medium.

On the other hand, sound travels faster in helium gas than normal air because helium molecules are

lighter.

displacement

distance

crest advances by one λ in one T

λ T

y

x

y

x

higher frequency




When a wave crosses from one medium to another, its speed will change (that’s why refraction

occurs). This time round, it is the frequency that remains unchanged. Based on the formulav f , it

is obvious that the wavelength must change accordingly. v=fλ

Different medium

watch video at xmphysics.com

y

x

y

x

higher speed



9.5.1 Phase Relationship

Phase relationship is a concept unique to periodic repetitive patterns.

Take the phases of the Moon as an example: It is a repetitive pattern with a period of one month, with

fanciful names for different phases such as the New Moon, the Crescent, the Gibbous and the Full

Moon.

In Mathematics, instead of words, we simply denote the phase with a number between 0° and 360°,

or between 0 radian and 2π radians. (As usual, 360° or 2π radians represent one full complete cycle.)

Now we can talk about phase relationships. Suppose we have two pendulums of the same period,

but released one after another so that one is always “ahead” of the other. The lead or lag of one

oscillation over the other is called the phase difference.

If the phase difference between two oscillations is zero, they are said to be in-phase.

In-phase

315° 270° 225° 180° 135° 90° 45° 0°

http://www.xmphysics.com/9.5.1


If the phase difference between two oscillations is 180° or π radian, they are said to be completely

out of phase, or in antiphase.

Antiphase

If the phase difference between two oscillations is 90° or π/2 radian, they are said to be a quarter-

cycle out of phase.

Quarter-cycle Lag

In the video below we line up eight oscillations, and arrange for each oscillation to lag the earlier one

on the left by 45° or π/4 radian. Do you see what I see? A travelling wave!

One-eighth-cycle Lag watch video at xmphysics.com



9.5.2 Graphical Representation of Phase Relationship

Sometimes you’re given a displacement-distance graph (like the one below) and asked to calculate

the phase difference between two oscillations (at two different positions of the wave).

Between 2 points on a x-x graph

First, let’s agree that the oscillation at A leads the oscillation at B. How can we tell? Because the

wave is coming from the left, meaning the wave will hit A first before B. The one further from the wave

source must be lagging since it is the delayed version.

Now back to the calculation. The phase difference between 2 points on a wave depends on how far

apart they are along the wave. The larger the separation x, the larger the phase difference θ. We

know that if 2 points are one wavelength apart, the phase difference between them would be 2π rad.

By simple proportion, if they are separated by a distance of x, the phase difference θ would be

2x

A

B

displacement

distance

direction of wave

λ

x



Other times, you’re given the displacement-time graphs of two oscillations (see below), and asked to

calculate the phase difference between them.

Between 2 x-t graphs

Firstly, let’s agree that A leads B. This is easy to tell since B is a delayed version of A. Whatever A

did, B will do at a later time.

As for the calculation, you must first obtain the misalignment in time t between the two oscillations.

Just look for the delay between two crests, or two troughs, or any two corresponding points from each

of the two graphs. Since a misalignment of one period corresponds to a phase difference of 2π rad,

a misalignment of t would correspond to phase difference of

2t

T

A B

displacement

time

T

t


9.6.1 Longitudinal Wave

Recall that in a longitudinal wave, the individual particles in the medium oscillate in the direction

parallel to wave propagation. These “parallel” oscillations change the spacing between the medium

particles, causing regions of higher density (compression) and lower density (rarefaction) in the

medium to propagate through the medium.

A longitudinal wave that is familiar to all of us, is sound. When we hear silence, it is because the air

pressure outside our ear drum is constant, at 1 atm. When the diaphragm of a loudspeaker vibrates,

it causes the air pressure in the immediate vicinity to alternate between high and low pressure, above

and below 1 atm. This disturbance is then propagated away from the loudspeaker. How come?

It may not look like it, but the air around us, consisting of air molecules, is actually “springy” and acts

like an elastic medium. The pressure fluctuations at the loudspeaker will be relayed by the air

molecules in the air, eventually arriving at our ear drum. Our ear drum reacts to the pressure

fluctuations, and our brain interprets that as sound. The amplitude of the pressure fluctuations

determines the loudness, and the frequency the pitch.

see animation at xmphysics.com

compression

rarefaction




9.6.2 Interpretation of Longitudinal Wave Graphs

Let’s have a sound wave travelling rightward. Figure (a) shows an imaginary snapshot of the air

molecules at one instant of time. You can see that there are regions of high density denoted by C

(compression) and regions of low density denoted by R (rarefaction).

Figure (b) is a displacement-position (x-x) graph showing the displacement of air molecules (from

their equilibrium positions) at different positions. By convention, a positive displacement represents a

rightward displacement, and a negative displacement represents a leftward displacement.

Figure (c) shows the pressure at each position along the wave. Note that the pressure at

compressions and rarefactions are only slightly above and below the atmospheric pressure. Even for

the noise level of a rock concert, we are talking about a pressure variation of a few pascals only.

Now let’s try to make sense of the x-x (displacement-position) graphs.

pressure

displacement

C C R C R R C

position

position

direction of wave

(a)

(b)

(c)

1 atm



First, if all the air molecules are undisturbed, they will all be sitting at their equilibrium positions and

they will be spaced out evenly. No compression, no rarefaction, just constant atmospheric pressure.

Average pressure

Now there is a sound wave passing through. Air molecules are displaced from their equilibrium

positions in such a way that alternating regions of compressions and rarefactions are produced.

Compressions and rarefactions

So why is there a compression at position 5? Because the molecule at position 5 has displacement

zero, but its neighbor on the left (position 4) has a positive (rightward) displacement, while its neighbor

on the right (position 6) has a negative (leftward) displacement. As a result, there is a higher

concentration of air molecules at position 5, resulting in high pressure.

1 2 3 4 5 6 7 8 9 10 11 12 13

x

x 1 2 3 4 5 6 7 8 9 10 11 12 13

air molecules at their equilibrium positions

1 2 3 4 5 6 7 8 9 10 11 12 13

x

x

air molecules displaced from their equilibrium positions

1 2 3 4 5 6 7 8 9 10 11 12 13


So why is there a rarefaction at position 9? Because the molecule at position 9 has displacement zero,

but its neighbor on the left (position 8) has a negative (leftward) displacement, while its neighbor on

the right (position 10) has a positive (rightward) displacement. As a result, there is a lower

concentration of air molecules at position 10, resulting in low pressure.

How about at position 3? Notice that even though the molecule at position 3 is displaced rightward,

so are its two neighbors on either side (position 2 and 4). The net result is there is no change in the

concentration of air molecules at position 3. The pressure at position 3 is still atmospheric pressure.

It’s the same thing at position 7. Even though the molecule at position 7 is displaced leftward, so are

its two neighbors on either side (position 6 and 8). The net result is there is no change in the

concentration of air molecules and the pressure at position 7 remains at atmospheric pressure.

From experience, students tend to take some time before they internalize the interplay between

displacement and pressure graphs. A few pointers may be useful. (1) Compression and rarefaction

can only happen at the zero points in the x-x graph, because that’s where the displacement is

opposite in signs on either side of the zero point, corresponding to air molecules either congregating

or dispersing from these positions. (2) No change in pressure occurs at the amplitude points in the

x-x graph, because at those turning points, everybody has the same displacement. Do you realize

that if everybody makes the same displacement, the spacing between them will remain unchanged?

1 2 3 4 5 6 7 8 9 10 11 12 13

x

x

air molecules displaced from their equilibrium positions

1 2 3 4 5 6 7 8 9 10 11 12 13


9.7 Electromagnetic wave

Having gone through your O-Level education, you are probably comfortable with the thought that light

is a wave. But has it occurred to you that there is something unique about light2?

Firstly, instead of oscillating particles, light has oscillating fields. Two fields in fact, an electric field and

a magnetic field (hence the name electromagnetic waves). And the two fields oscillate at right angle

to each other.

And since there is no oscillating particles, electromagnetic waves do not require any medium to

propagate. The next time you marvel at the beauty of a sunrise, remember that the light managed to

reach you, despite the vast vacuum between the Sun and the Earth.

Furthermore, electromagnetic waves propagates at incredible speed. In vacuum the speed of light is

8 -13.00 10 m sc . In glass, it is about one-third slower. But still incredibly fast.

The range of wavelengths for electromagnetic waves we encounter is truly mind boggling, from

picometers (gamma rays) to kilometers (radio waves). In comparison, the human eye is only sensitive

to light of wavelength between 400 nm (violet) to 700 nm (red).

2 Spoiler Alert: If you always had your suspicion of light as a wave, you are so right. In the Quantum

topic, you will learn that light is neither wave nor particles, but something that’s nothing like anything

you have seen before. Nevertheless, for most applications, e.g. optics and radio communications, it

is easier to work with the wave model of light.



listen to the Electromagnetic song on xmphysics.com

The syllabus DOES require you to “memorize” the wavelengths of the different type of EM waves.

You should find the following table and the above song useful.

Common Name Wavelength

Radio >m

Micro ~cm

IR > 750 nm

Visible (ROYGBIV)

400 nm to 750 nm

UV < 400 nm

X-ray ~10-10 m

-ray <10-12 m



9.8 Intensity

Which is worse? Staring into the light emitted by a 1 mW laser pointer or staring into the light emitted

by a 10 W filament light bulb? The answer depend on the distance between you and the light sources.

This has something to do with this quantity called the intensity.

The intensity of a wave (at a location) refers to the power per unit area (at that location). Take for

example a power source of 10 W. Now think of a point at a distance of 0.5 m away from the source.

Do we get all 10 W of power arriving at this point? If the wave propagates uniformly in all directions,

the total power of 10 W would have to be transmitted to each and every point that is 0.5 m away from

the source. So we are talking about 10 W being spread evenly onto a spherical surface (not volume)

of area 2 24 0 5 3 14( . ) . m . The intensity of the wave at any point on this spherical surface is thus 10

W ÷ 3.14 m2 = 3.18 W m-2.

Since the power P is propagated onto 24 r areas, the intensity of a point source at distance r away

is given by

24

P

rI .

In general, waves can propagate in one, two or three dimensions. Sound wave traveling down a

narrow tube is an example of a 1D wave. The circular ripples on a water surface is a 2D wave. The

light leaving the Sun is a 3D wave.

x

y

z

10 W Power

0.5 m

0.5 m

0.5 m

3.18 W m-2 Intensity



A 3D wave propagates in 3 dimensions onto spherical surfaces whose area varies with the square of

the distance (since surface area of sphere is 24 r ). As such we can deduce that for a 3D wave,

2

1

rI (and hence3

1A

r ).

3D wave propagation

A 2-D wave propagates onto circular surfaces that expand with r (since circumference of circle is

2 r ). So for a 2-D wave, 1

rI and

1A

r .

2D wave propagation

A 1-D wave propagates only in one direction along a line, so the power P is propagated within a

constant area. For a 1-D wave, both I and A are constant and do not diminish with distance.

1D wave propagation

3 In the topic of SHM we learnt that the energy of oscillation is proportional to the square of its

amplitude. Since a sinusoidal wave consists of SHMs, it means that 2AI , so A I .


Now back to the question of light bulb vs laser. A light bulb is 3-D source. Staring into the light bulb

1 m away and 100 m away is two completely different experiences because the intensity of the light

at those two distances are different by a factor of 10,000! A laser beam on the other hand is a 1-D

source. The intensity of the laser beam is the same whether 1 m or 100 m away. Ouch.


9.9.1 Polarization

Polarization

Imagine an attempt to transmit a rope wave through a vertical slit. If the rope is oscillating vertically,

then the wave would travel unimpeded through the slit. On the other hand, if the rope is vibrating

horizontally, then the wave would be absorbed by the vertical slit. Welcome to the magical world of

polarization, a phenomena unique to transverse waves.

Light is a transverse wave. If a light wave is propagating along the z-axis, then its E-field could be

oscillating in the x-axis, or y-axis, or any direction in the x-y plane. It turns out that light waves emitted

by most light sources are unpolarized light, which is to say the light beam contains a mixture of waves

with E-field oscillating in all possible directions in the x-y plane.4

4 Mathematically, we can model natural light as two arbitrary, incoherent, perpendicularly polarized waves of equal amplitude.

vertical oscillation

horizontal oscillation

vertical slit

vertical slit

unpolarized light

polarizer

polarization direction

polarized light

z

x

y



There are materials such as Polaroids which, due to their chemical structure, absorb only electric

fields oscillating in the direction of their polymer chains. Electric fields oscillating in the direction

perpendicular to the polymer chains are not absorbed. The Polaroid is an example of a polarizer, and

the direction of the polymer chains is called the polarization direction. When a beam of unpolarized

light passes through a polarizer, it becomes polarized. After polarization, the oscillation of electric field

will only be in the polarization direction of the polarizer (because all the components of electric field

in the perpendicular direction have been absorbed).

The fun part begins now. Suppose we pass a beam of polarized light with amplitude A0 and intensity

I0 through a second polarizer.

Parallel and crossed polarizers

If the 2nd polarizer’s polarization direction is parallel to the 1st polarizer’s, then we get 100%

transmission. On the other hand, if the two polarizers’ polarization directions are perpendicular to

each other, then we get 0% transmission.

unpolarized light

A=A0

I=I0

A0

I0

unpolarized light

A=0 I=0

A0

I0


Now, what happens if they are neither parallel nor perpendicular, but misaligned by some angle θ?

Easy. We can always resolve the amplitude of the electric field A0 (of the polarized light) into the

components parallel and perpendicular to the polarization direction (of the polarizer), A0cosθ and

A0sinθ.

Malu’s Law

The parallel component will be completely passed through, while the perpendicular component will

be completely absorbed. Just by simple logic, we have deduced the Malu’s Law.

0cosA A

2

0cos I I

watch videos at xmphysics.com

unpolarized light

θ



A0

I0

A=A0cosθ

I=I0cos

2θ

θ θ



9.9.2 The Three Polarizer Problem

watch videos at xmphysics.com

In the above demonstration, the bear became visible again when a third polarizer is inserted

between the original two polarizers.

We understand why there was only darkness with the original two polarizers X and Y. They were

aligned perpendicular to each other so no light was able to pass through Y. The fact that the bear

became visible again with the third polarizer Z means that some light is now able to pass through Y.

Adding another layer of polarization cuts out LESS light! How is that possible?

unpolarized light

A=0 I=0

A0

I0

X

Y

Z




This “mystery” can be solved once you realize that a polarizer is different from a filter. It does not just

attenuate the light passing through. It also re-aligns the polarization direction of the light passing

through. By choosing the polarization direction of the 3rd polarizer (Z) to be neither vertical nor

horizontal, we avoid having two consecutive polarizers which are perpendicular to each other. This

prevents the light from being totally cut off.

Let’s confirm this with some math. Since X and Z are misaligned by angle θ, the intensity of light after

passing through Z is

2

1 0cos I I

Since Z and Y are misaligned by angle 90 , the intensity of light after passing through Y is

2

2 1

2 2

0

2 2

0

20

90

90

24

cos ( )

cos cos ( )

cos sin ( )

sin

I I I

I

I

I

In fact, the maximum intensity of 0

4

I is achieved by arranging Z to have polarization angle midway

between that of X and Y, i.e. 45°,

unpolarized light

θ


A0

I0

θ

90°-θ

A1

I1

A2

I2

X

Y

Z

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