6 WAVE BEHAVIOURBasic wave properties

• Review key wave properties

• Explain the meaning of the terms phase, phasor and superposition

Starter: Q1. Define the following wave terms: wavelength, frequency, amplitude, period, wave speed.

Q2. What is the wavelength of radio waves whose period is 9 ns?

Add on the labels: peak, trough, wavelength, amplitude.

Think about what frequency, wave speed, period are

Frequency is the number of wave cycles per second

Period is the time it takes for one complete oscillation

Waves have phase

Waves with 90° phase difference

Waves in phase

Waves out of phase

Using phasors to describe the phase of a wave

1. A phasor is a rotating arrow that represents the phase of wave without having to draw it out.2. The higher the frequency of the wave the more rapidly the phasor rotates.3. The larger the amplitude of the wave the longer the phasor.4. Different representations of the waves describe exactly the same thing.

Phase and angle

‘Clock arrow’ rotates at constant angular speed time

Phase angle

degrees 0 45 90 135 180 225 270 315 360

radians 0 /4 /2 3/4 5/4 3/2 7/4 2

phase angle radius aa sin

‘Clock arrow’ rotates 2 in periodic time T

angle = 2 (t /T )T = 1/f (f = frequency)angle = 2ftdisplacement = a sin = a sin 2ft

a

Adding waves with different phasesThe principle of superposition says that the resultant amplitude of two waves in the same point is the sum of their individual amplitudes at this point.

Phasors can also be used to describe what the resultant wave looks like when two waves superimpose. See page 126

Superposition and phase difference

Oscillations with 90 phase difference

For any phase difference,amplitude of resultant= arrow sum of components

A

B

C = A plus B

Rotating arrows add up:

+ =

arrows add tip to tail

0 5 10 15 20 25 30 35

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

Beats

time (s)

disp

lace

men

t (m

)

Superposition phenomena• Observe and explain some

phenomena involving wave superposition

• Explain how superposition is used in some practical situations

Starter: Q1. If two waves are to constructively interfere, what must be true about their frequency, wavelength and phase? Q2. Sound waves from two speakers driven from the same signal generator arrive at a point, 180o out of phase. What must be true about the path lengths taken by the two wave trains?

Path difference: the difference in distance (path) two waves have travelled

to reach the receiver

Maxima occur where the path difference is a whole number of wavelengths e.g. 1, 2, 3….. (the waves arrive in phase)

Minima occur when the path difference is an odd number of half wavelength e.g. 0.5, 1.5, 2.5…. (the waves arrive out of phase)

Now try questions 30S

In order for stable superposition effects it is necessary to have waves that are

coherent

Coherence means waves that have constant phase difference

If incoherent waves are used then the superposition effects will vary such as the beats observed with two slightly different frequencies.

All waves from the same source are inherently coherent

Radar guns

If the path to the car and back is a whole number of wavelengths then there is a maxima.If it is a whole number of half wavelengths then it is a minima.

When the car moves towards the detector maxima and minima are recorded sequentially.The frequency the signal varies at can be used to calculate the speed the car is approaching

Now try questions 70S

When you have finished; read textbook page 127 on oils and soap colours and take down

enough notes that you can explain the phenomenon to your partner

Look at the pretty bubbles!

Can you explain, in terms of wave superposition, why yousee a spectrum of colours in the surface of the bubble?

Colours of thin films

If light from the back is delayed byhalf a cycle, peaks coincide withtroughs in the waves from the twosurfaces.

Result: light of that colour is notreflected. Other colours are reflected.

For example, if red light is removedthe film looks bluish-green.

Light wave falling onthin filmtrough

peak

trough

peak

Note: there may also be achange of phase onreflection at a surface

thin film

Some is reflected atfront surface

Some passes throughand is reflected at backsurface

trough

peak

trough

peak

trough

peak

trough

peak

from front

from back

extra path travelled isodd number of halfwavelengths in film

Coherence

coherent waves with constant phase difference

Two waves will only show stable interference effects if they have a constantunchanging phase difference. If so they are said to be coherent.

Atoms emit bursts of light waves. A burst from one atom is not in phasewith a burst from another. So light waves from atoms are coherent onlyover quite short distances.

incoherent wave bursts with changing phase difference

Can you work out the superimposed wave from these phasors?

+

+

=

=

Standing waves

• Describe how standing (stationary) waves are created by superposition of travelling waves

• Describe and explain the pattern of nodes and antinodes formed by standing waves on strings and in pipes

nodes andantinodesboth half awavelength apart

Standing waves

wave moving to leftwave moving to right

Waves travelling in opposite directions combine to make standing waves

nodes

antinodes

antinode: herethe wavescombine tomake a largeoscillation

node: here thewaves alwaysadd up to zero

harmonic n = 3

harmonic n = 2

at rest

fundamental n = 1

Strings of a guitar

Standing waves in air

Investigate and explain standing waves patterns in air columns

Starter:Write down two similarities and two differences between transverse waves and longitudinal waves

Standing waves in pipes

Closed pipes

The fundamental: The lowest frequency whichcan form a standing wave has wavelength equalto twice the length of the tube

A loudspeaker sends a sound into a long tube.Dust in the tube can show nodes and antinodes.Nodes are half a wavelength apart. So areantinodes. Maximum amplitude showsmaximum pressure variation and minimummotion of air (pressure antinode). Minimumamplitude shows minimum pressure variationand maximum motion of air (pressure node).

/2

/2 /2

Pipes open at both ends

Sound can be reflected from an open end aswell as from a closed end.

This is how open organ pipes and flutes work.

At a lower frequency, the wavelength is longer

/2

Pipes closed at one end

/2

Pipes closed at one end are shorter, for thesame note.

/4

3/4

A clarinet is like this. An oboe is too, but witha tapered tube.

Some organ pipes are stopped at one end.

loudspeaker

Frequencies of standing waves

pipes open or closed at both endsstrings fixed at both ends pipes open at one end

length L

fundamental

harmonics

L = n/2

f = v/2L

2f3f...nf

L = (2n–1) /4

f = v/4L

3f5f...

(2n–1)f

Apply your understanding of standing waves

Q1. What is the frequency of thesecond harmonic note produced byan organ pipe 1.3 m long, which is closedat one end? (Speed of sound in air = 340 ms-1.)

Q2. Can you explain, in termsof standing waves, why a flute produces higher pitch notes thana clarinet?

Generate questions from these answers

• Half a wavelength

• A node at each end

• A node at one end and an antinode at the other

• The harmonics follow the pattern f, 2f, 3f, 4f,.....

• The speed of sound will increase and the frequency will increase but the wavelength will remain the same

Fiendish problem• Suppose while an orchestra was

playing you pumped out all of the air from the concert hall and replaced it with helium, through which sound travels much faster than through air. What would happen to the pitch of the different instruments? Would all types be affected equally?

Hints: The pitch is an indication of the frequency. All of the sounds are produced by standing waves, either on strings under tension or in columns of air.

Ignore the fact that all of the members of the orchestra would be asphyxiated!

The nature of light• Explain Romer’s method

for estimating the speed of light

• Discuss differences between wave and particle models of light

• Review evidence for light behaving as either wave or particle

• What evidence from every day life is there to suggest that the speed of light is much higher than the speed of sound?

• Sketch diagrams to illustrate reflection, refraction, diffraction and interference.

• Which of the 4 wave phenomena above can be explained ONLY by a wave model of light? Why?

• “Looking at the night sky is looking back in time.” Explain this statement.

Light interference

• Observe light interference and explain it in terms of superposition

• Measure the wavelength of light of a laser using Young’s slits

Starter: Use the transparencies to create 2-source interference patterns. What happens to the spacing between the regions of constructive and destructive interference when the sources are moved apart?

Can you describe the interference pattern of sound and water waves?

RIPPLETank

If we were to try the same experiment with light. What problems might we face?

lightfromsource

Young’s two-slit interference experiment

narrowsource

two slits:1 mm spacing or less

bright anddark fringes

Geometry

several metres several metres

d

path difference d sin between light from slits

length L of lightpath from slits

angle

light combines atdistant screen

Approximations: angle very small; paths effectively parallel; distance L equal to slit–screen distance.Error less than 1 in 1000

path difference = d sin sin = x/Lpath difference = d(x/L)

x

Young’s slits experiment

Screen

d P

R

Q

L

x

Young’s two-slit interference experiment

Two simple cases

to brightfringe onscreen

Wavelength can be measured from the fringe spacing

d sin

d sin /2

to darkfringe onscreen

d

waves in phase: = d sin = d(x/L)

In general:for a bright fringe n = d sin spacing between fringes = (L/d)

waves in antiphase:/2 = d sin /2 = d(x/L)

d

sin = n/dn= dsin

Bright fringes occur when this length is a whole number of wavelengths

Screen

d P

R

Q

L

x

sin can also be calculated by looking at the big triangle that the fringes create

tan = x/Lsin = x/L

Predicting where bright fringes will be:

sin = n/dsin = x/L

x/L = n/d

x = nL/d= xd/L

Write down an estimation of the uncertainty in the measurements

of x, d and L

Look at the equations we have derived and make a prediction about what effect changing the slit spacing have on this

experiment.

The diffraction grating• Observe and explain

transmission and reflection diffraction grating effects

• Measure a laser wavelength using a transmission grating

Starter: Use the equation nλ= d sin θ to predict the effect on a 2-slit interference fringe pattern of changing to slits that are further apart. Verify using the transparencies.

Diffraction grating

narrowsource

grating: many finelyspaced slits

bright angledbeam

bright ‘straightthrough’beam

bright angledbeam

light combinesat distant screen

lightfromsource

path difference d sin between light fromadjacent slits

Geometry

Waves from many sources all in phase

Sharp bright spectral lines at angles where n = d sin

When = d sin wavesfrom all slits are in phase

Bright lines at = d sin and n = d sin = d sin

d

to bright lineon screen

d

d sin

d

Gratings d is the spacing between adjacent slits on the grating

The first maximum occurs when the length indicated is equal to

Gratings

d

The angle is increased until the length indicated becomes equal to 2

The order of the maxima corresponds to the number of difference in the path length

Note that by the same argument it can be shown that: n =d sin

Now calculate the wavelength of the maxima.

Can you explain why the maxima are more spread out than when we used a double slit?

Diffraction gratings can be used in:1. Separating different frequencies of

light for analysis2. Observing the spectrum of stars3. Selecting particular wavelength for

use

Now try Qs 3-6 p144

Reflection grating

Starter: Draw a diagram to show how plane waves diffract when they pass through an aperture:

(a) when the aperture is much larger than the wavelength of the waves

(b) when the aperture is comparable in size to the wavelength of the wave

Single aperture diffraction

• Observe and explain single slit diffraction

• Explain some practical consequences in astronomy and every day life

DiffractionPatterns

1. How does the diffraction spreading depend on the wavelength of light?2. Carefully adjust the width of the slit. How does the diffraction spreading depend upon slit width?

.

Diffraction at a single aperture

intensityacross

Single slit

distant screen

Diffraction at a single aperture

Useful approximation

beamwidth W

sin = W/L approximately= beam angle in radian

Beam angle in radian = /ddistance L

Diffraction at a single aperture

phasors all in same direction

large resultant

nodifferencebetweenpaths

Simplified case:screen distant, paths nearly parallel

constantdifferencebetweenadjacentpaths

angle

pathdifferenceacross wholeslit = d sin

d

d

Phasors add to zero if they make acomplete circle.

Path lag over slit must be one cycle.

Path difference across slit must beone wavelength.

each phasor at sameangle to the next

First zero intensity at angle = d sin

pairs ofphasorsadd to zero

zero resultant

Beam widthDiffraction at a single aperture

Useful approximation

beamwidth W

sin = W/L approximately= beam angle in radian

Beam angle in radian = /ddistance L

sin =W/L = /dso for small angles:beam angle in radian = /d

Link

Note that W is really referring to half the beam width

Angular resolutionHow close can two objects be and still be resolved?What is the beam half width?

separation of objects wavelength of radiationbeam ½ width

W = (sin)L d

distance to object size of aperture

Synthetic Aperture Radar

Problem:

Can you see evidence of Apollo landingon the Moon? Eagle module base = 4.3 m acrossEarth-Moon distance = 384 000 kmWavelength of light = 550 nmLargest telescope mirror diameter (La Palma) = 10.4 m.

From these data, angular resolution needed = 10-8 radians; angular resolution available = 5 x 10-8, so would not be able to resolve module from surroundings.

Starter: The diagram shows light rays from a star arriving at a concave telescope mirror.Q1. How can you tell that the star is very far away?Q2. What can you say about the phasors for each of the light rays as they pass the line XX’ ?Q3. Why must the mirror be shaped as shown in order to focus the light? Explain your answer in terms of path lengths and phasor rotations.

X

X’

Using phasors to explain superposition

• Use the phasor model to explain interference and diffraction

Superposition and phase difference

Oscillations in phase

If phase difference = 0then amplitude of resultant= sum of amplitudes ofcomponents

Rotating arrows add up:

+ =

arrows add tip to tail

A

B

C = A plus B