Figure 26: 2-dimensional water wave.
• Wave fronts are spaced exactly one wavelength λ apart – i.e. distance
between concentric circles is one wavelength.
• Wave fronts travel outwards from the source with a velocity v in a direction perpendicular to the tangent of the wave front.
• Every point on the same wave front is the same distance r from the
source. So mathematically, we can write the displacement as
u (r, t) = A (r) sin (kr − ωt + φ0 ) .
i.e. the amplitude A = A(r) (see later notes).
48
(2.19)
2.3.2
Three-dimensional Waves
For example, light waves, sound waves, shock wave from supernova explosion
(figure 27
19
).
Figure 27: Shock-heated remnant of Supernova 1987A.
We can generalise the idea of wave fronts to 3-dimensions:
• In this case, the crests of the waves form a series of concentric spherical
shells, separated by one wavelength.
• This can be visualised by looking at a cross section through the shell –
we get a cross section that looks like the 2-dimensional example shown
above (figure 26).
19
image from http://chandra.harvard.edu/press/00 releases/press 051100sn1987a.html
49
• Since every point on the same shell is at the same distance from the wave
source, we find that the displacement can be written mathematically as
(2.19).
2.4
Plane Waves
2-dimensions: consider a very, very small section of a wave front that is
very, very far away from the source.
• The wave fronts would appear as parallel lines, still spaced a wavelength apart and traveling at speed v (figure 28
20
).
Figure 28: 2-d plane waves.
• For example, a wave in the ocean:
– Wave starts out circular far out in the ocean.
– When it reaches land, the wave front is almost parallel.
3-dimensions: similarly, considering a very, very small section of a wave
front very, very far away from the source.
20
Knight, Figure 20.21(b), page 627
50
• The wave fronts will appear as a flat, rectangular area or plane – for
example, in the same way that to a person on the surface, the Earth
appears to be locally flat, even though it is globally spherical.
• Under these circumstances, we may model the spherical wave as a plane
wave (figure 29
21
) of speed v, the ”plane” being perpendicular to the
direction of motion of the wave front.
Figure 29: 3-dimensional plane waves.
• Observer encounters successive plane waves positioned one wavelength
apart.
• As shown in figure 29, the displacement of the plane wave depends only
upon x, i.e. the displacement u(x, t) describes a plane wave in the same
way as it describes a 1-dimensional wave, namely
u (x, t) = A sin (kx − ωt + φ0 ) .
21
Knight, Figure 20.22, page 627
51
(2.20)
Note: in reality, there are no perfect plane waves – but mathematically, they
are very good models for many waves of practical interest.
2.5
Phase and Phase Difference
Recall that in SHM, we call the argument of the sine or cosine function the
phase, φ, of the oscillation/wave. This is found to be very important when
adding various waves together (see later notes).
With respect to (2.20), if
φ = kx − ωt + φ0 ,
(2.21)
u (x, t) = A sin φ .
(2.22)
then
Note: in figures 26 and 29, since u(x, t) is given in the form of (2.22) – i.e.
each point on the wave front has the same displacement – the wave fronts are
surfaces of constant phase.
Consider the phase difference, ∆φ, between two points on a sinusoidal wave.
Figure 30
22
shows two points on a sinusoidal wave at time t. We have
∆φ = φ2 − φ1
= (kx2 − ωt + φ0 ) − (kx1 − ωt + φ0 )
= k (x2 − x1 )
(2.23)
= k∆x .
Recalling (2.12), the wave vector k = 2π/λ, and so in (2.23),
22
Knight, Figure 20.23, page 628
52
Figure 30: Leading edge.
∆φ = 2π
∆x
;
λ
(2.24)
i.e. the phase difference between two points on a wave depends upon the ratio
of their separation, ∆x, to the wavelength, λ. For example,
• Two points on a wave front, separated by ∆x = λ/2, have ∆φ = π.
• Two adjacent wave fronts, separated by ∆x = λ, have a phase difference
∆φ = 2π
λ
∆x
= 2π = 2π ,
λ
λ
(2.25)
i.e. moving from the crest of one wave to the next corresponds to changing the distance by λ, and the phase by 2π.
Example: a circular wave travels outwards from the origin. At one instant
53
of time, the phase at r1 = 30 cm is 0 rad, and the phase at r2 = 90 cm is 3π
rad. Calculate the wavelength of the wave.
Solution: using (2.24), the phase difference between the two points is
∆φ = φ2 − φ1 = 2π
∆r
λ
λ = 2π
−→
∆r
,
∆φ
i.e.
λ=
2.6
2π (90 − 30)
= 40 cm .
(3π − 0)
(2.26)
Sound Waves
Sound waves are longitudinal waves:
• Particles of the medium are displaced parallel to the direction in which
the wave is traveling.
• Consider the longitudinal wave in figure 31
23
showing the displacement
∆x against x at a time t1 = 0 s.
• Figure 32
24
shows what occurs at the particle level.
• Particles of the medium are represented by dots.
• First row of dots shows the positions of the particles at the equilibrium
position (corresponding to ∆x(x) = 0 in figure 31.
• Second row of dots corresponds to displacement at t1 shown in figure 31.
Particles in medium are displaced to the left or right of their equilibrium
positions: for example,
– Particle at x = 1 cm is displaced ∆x = 0.5 cm to the right.
– Particle at x = 2 cm has been displaced ∆x = 1 cm to the right.
23
24
Knight, Figure 20.10(a), page 618
Knight, Figure 20.10(b), page 618
54
Figure 31: Longitudinal wave.
55
Figure 32: Compressions and rarefactions.
56
– ...and so on.
• The medium is compressed to a higher density (so-called compressions)
at the centre of the pulse, and to compensate, expanded to lower density
(so-called rarefactions) at the leading and trailing edges.
• Further lines at t2 = 1 s and t3 = 2 s correspond to graphs shown in
figures 33 and 34, respectively.
Figure 33: Longitudinal wave at t2 = 1 s.
Figure 34: Longitudinal wave at t3 = 2 s.
So this periodic sequence of compressions and rarefactions travels outwards
from the source as a longitudinal wave.
57
For example, a loudspeaker (figure 35
25
):
Figure 35: Example of a loudspeaker.
• Each time the cone moves forwards, it collides with the particles and
pushes them closer together, producing compressions.
• Half a cycle later, as the cone moves backwards, the fluid has room to
expand, leading to rarefactions.
Note: in fluids (liquids and gases), sound waves are always longitudinal,
whereas in solids, sound waves can be either longitudinal or transverse.
• For a transverse wave to propagate, a plane of molecules, oscillating
perpendicular to the direction of motion, has to be able to ”drag” neighbouring planes of molecules along with it.
– Fluids: neighbouring ”planes” of molecules slip – so media cannot
support transverse waves.
25
Knight, Figure 20.24, page 629
58
– Solids: stronger molecular bonds between planes allows transverse sound waves to propagate – these are sometimes called shear
waves. The speed of transverse sound waves is usually different
from that of longitudinal sound waves.
We will consider sound waves to be longitudinal.
2.6.1
Properties of Sound Waves
The speed of sound waves through a medium is found to depend upon properties of the medium such as
• Temperature.
• Molecular mass of medium.
For example, the speed of sound in air at 20◦ is vsound = 343 m s−1 ;
• At higher temperatures, vsound > 343 m s−1 (by small amount).
• At lower temperatures, vsound < 343 m s−1 (by small amount).
Liquids and solids are less compressible than gases, which leads to v sound being
greater in liquids and solids than in gases (see table 1.
For humans:
• 20 Hz ≤ νsound ≤ 20 kHz – typical range of human detection.
Low frequencies are perceived as a low pitch bass note.
High frequencies are perceived as a high pitch treble note.
• νsound > 20 Hz – so-called ultrasonic frquencies (undetectable to human
ear).
59
Medium
Speed of sound (m s−1 )
Air (0◦ C)
331
Air (20◦ C)
343
Helium (0◦ C)
970
Hydrogen (0◦ C)
1284
Ethyl alcohol
1170
Water (0◦ C)
1402
Water (0◦ C)
1482
Seawater (20◦ C)
1522
Steel
5941
Granite
6000
Aluminium
6420
Table 1: Speeds of sound in various materials (all values at 1 atm pressure,
seawater salinity is 3.5%).
Example: a 10.0 m-long metal bar at room temperature is tapped on at one
end by a hammer. A microphone at the other end of the bar picks up two
pulses of sound – one that travels through the metal, and another that travels
through the air. If the pulses are separated by 27.5 ms, calculate the speed of
sound in the metal bar.
Solution: the time interval for the sound pulse traveling through the air is
∆tair =
10.0
∆x
= 0.02915 s = 29.15 ms .
=
343
vair
Sound travels faster through solids than gases, so ∆tmet < ∆tair . If the pulses
are separated by ∆t = 27.5 ms, then
∆tmet = ∆tair − ∆t = 29.2 ms − 27.5 ms = 1.7 ms .
Therefore, the speed of sound in the metal is
60
vmet =
2.7
10.0
∆x
= 5880 m s−1 .
=
1.7 × 10−3
∆tmet
Light Waves
Recall that light is an electromagnetic wave – a self-sustaining oscillation
of the electromagnetic field, i.e. electromagnetic waves require no material
medium is required for it to propagate.
By the late 19th century, it was predicted theoretically, and subsequently confirmed experimentally, with outstanding precision, that all electromagnetic
waves travel through a vacuum with the same speed, the speed of
light, c. The modern accepted value is
c = 2.99792458 × 108 m s−1
in vacuum
(an appropriate value for calculations is 3.00 × 108 m s−1 ).
2.7.1
The Electromagnetic Spectrum
The entire wavelength/frequency range of electromagnetic waves is known as
the electromagnetic spectrum. This is shown in figure 36
26
Note: in figure 36, the frequency ν and wavelength λ are, of course, related
by the expression
c = νλ .
2.7.2
Refractive Index
• Light waves travel at a speed c in a vacuum.
26
see http://lasp.colorado.edu/cassini/education/Electromagnetic Spectrum.htm
61
(2.27)
Figure 36: Electromagnetic spectrum.
62
• However, they slow down slightly as they pass through transparent
materials such as glass or air.
• This reduction in speed is due to interactions between the electromagnetic field of the wave and the electrons of the material.
We characterise the speed of light v in a material by the refractive index,
n, of the material, defined by
n=
c
speed of light in a vacuum
= .
v
speed of light in material
(2.28)
• So for a material, since v < c, we have that n < 1.
• Values of n for various media are shown in table 2.
If electromagnetic waves slow down as they enter a transparent material, then
since the speed changes, and v = νλ, either ν or λ must change.
The frequency of a wave does not change as the wave moves from
one medium to another.
Thus, it must be the wavelength that changes.
Indeed, for a light wave entering a material the only way it can go slower,
while maintaining the same oscillation frequency, is to reduce its wavelength.
Consider figure 37,
27
showing light passing through a transparent material
of refractive index n.
• Before entering material: the light wave has a speed c, wavelength
λvac and frequency νvac , related via
c = νvac λvac .
27
Knight, Figure 20.26, page 632
63
(2.29)