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