SHOWING SOLUTIONS
Tuesday, 30 April 2019
09:30 – 11:00
EXAMINATION FOR THE DEGREES OF B.SC. (DESIGNATED AND
HONOURS), M.SCI. (HONOURS) AND M.SC.
[ PHYS4031 ]
Waves & Diffraction
Candidates should answer Question 1 (16 marks)
and either Question 2A or Question 2B (24 marks each)
Answer each question in a separate booklet
Candidates are reminded that devices able to store or display text or images
may not be used in examinations without prior arrangement.
Approximate marks are indicated in brackets as a guide for candidates.
PHYS4031 Waves & Diffraction
Fundamental constants
name symbol value
speed of light c 2.998× 108 m s 1
permeability of free space μ0 4pi × 10 7 H m 1
permittivity of free space 0 8.854× 10 12 F m 1
electronic charge e 1.602× 10 19 C
Avogadro’s number N0 6.022× 1023 mol 1
electron rest mass me 9.110× 10 31 kg
proton rest mass mp 1.673× 10 27 kg
neutron rest mass mn 1.675× 10 27 kg
Faraday’s constant F 9.649× 10 4 C mol 1
Planck’s constant h 6.626× 10 34 J s
fine structure constant α 7.297× 10 3
electron charge to mass ratio e/me 1.759× 1011 C kg 1
quantum/charge ratio h/e 4.136× 10 15 J s C 1
electron Compton wavelength λe 2.426× 10 12 m
proton Compton wavelength λp 1.321× 10 15 m
Rydberg constant R 1.097× 107 m 1
Bohr radius a0 5.292× 10 11 m
Bohr magneton μB 9.274× 10 24 J T 1
nuclear magneton μN 5.051× 10 27 J T 1
proton magnetic moment μp 1.411× 10 26 J T 1
universal gas constant R 8.314 J K 1 mol 1
normal volume of ideal gas – 2.241× 10 2 m3 mol 1
Boltzmann constant kB 1.381× 10 23 J K 1
First radiation constant 2pihc2 c1 3.742× 10 16 W m2
Second Radiation constant hc/kB c2 1.439× 10 2 m K
Wien displacement constant b 2.898× 10 3 m K
Stefan-Boltzmann constant σ 5.670× 10 8 W m 2 K 4
gravitational constant G 6.673× 10 11 m3 kg 1 s 2
impedance of free space Z0 3.767× 102 Ω
Derived units
quantity dimensions derived unit
energy ML2T 2 J
force MLT 2 N
frequency T 1 Hz
gravitational field strength LT 2 N kg 1
gravitational potential L2T 2 J kg 1
power ML2T 3 W
entropy ML2T 2 J K 1
heat ML2T 2 J
capacitance M 1L 2T 4I2 F
charge IT C
current I A
electric dipole moment LTI C m
electric displacement L 2TI C m 2
electric polarisation L 2TI C m 2
electric field strength MLT 3I 1 V m 1
electric (displacement) flux TI C
electric potential ML2T 3I 1 V
inductance ML2T 2I 2 H
magnetic dipole moment L2I A m2
magnetic field strength L 1I A m 1
magnetic flux ML2T 2I 1 Wb
magnetic induction MT 2I 1 T
magnetisation L 1I A m 1
permeability MLT 2I 2 H m 1
permittivity M 1L 3T 4I2 F m 1
resistance ML2T 3I 2 Ω
resistivity ML3T 3I 2 Ω m
M = mass, L = length, T = time, I = current
SHOWING SOLUTIONS
1 (a) A string of length 2 m and total mass 0.1 kg is placed under a tension of
30 N.
i. Calculate the propagation speed of transverse waves on the string. [2]
Solution: Unseen problem requiring recall of standard relations and
deduction of appropriate units.
The linear density ρ = 0.12 = 0.05 kg m
1 .
c =
√
T
ρ
=
√
30
0.05
=
√
600 = 10
√
6 = 24.5 m s 1 .
[. . . 2]
ii. Define the impedance of a stretched string. Calculate the impedance of
the string with parameters given above. Ensure you quote appropriate
units for your result. [3]
Solution: The impedance of a stretched string is the ratio of transverse
force to transverse velocity. The transverse force is T y x . For
non-dispersive propagation y x = 1c y t , so the transverse force is Tc y t . Thus
Z = Tc =
T√
T/ρ
=
√
Tρ . [. . . 2]
So for the problem given, Z =
√
30× 0.05 = √1.5 = 1.22 N s m 1 . [. . . 1]
iii. The ratio of reflected amplitude to incident amplitude for waves inci-
dent from a medium of impedance Z1 on a boundary with a medium of
impedance Z2 is
Z1 Z2
Z1 + Z2
.
What is the ratio of reflected energy to incident energy Hence deduce an
expression for the ratio of transmitted energy to incident energy. Finally,
deduce an expression for the ratio of the reflected wave amplitude to the
incident amplitude. [4]
Solution: Bookwork, with a small unseen element.
The ratio of reflected energy to incident energy = Z1Z1
[
Z1 Z2
Z1+Z2
]2
where
the first term is included here – though the numerator and denominator
cancel – as a reminder that the impedance of the string in which a wave
propagates must be taken into account when moving from wave amplitude to
energy. This will not be required in a student’s solution. [. . . 1]
Conservation of energy then dictates that the ratio of transmitted
energy to incident energy must be
1 Z1
Z1
[
Z1 Z2
Z1 + Z2
]2
PHYS4031 Waves & Diffraction 3/16 Q 1 continued over. . .
Q 1 continued
SHOWING SOLUTIONS
=
(Z1 + Z2)
2 (Z1 Z2)2
(Z1 + Z2)2
=
4Z1Z2
(Z1 + Z2)2
.
[. . . 1]
The ratio of transmitted energy to incident energy is Z2Z1
C2
A2
, where we
denote the incident wave amplitude by A and the transmitted wave
amplitude by C . [. . . 1]
The ratio of reflected amplitude to incident amplitude is
C
A
=
√
C2
A2
=
√
Z1
Z2
[
4Z1Z2
(Z1 + Z2)2
]
=
√
4Z1
2
(Z1 + Z2)2
=
2Z1
Z1 + Z2
.
[. . . 1]
(b) Waves propagating in a particular medium follow a dispersion relation
given by
ω = ak3 bk2 ,
where a and b are constants and k is the wavevector. Obtain expressions for
the phase and group velocities of the propagating waves and show that they
are equal for k = b/(2a) . [3]
Solution: Standard knowledge plus application to an
unseen problem.
Phase velocity = ωk = ak
2 bk . [. . . 1]
Group velocity = ω k = 3ak
2 2bk . [. . . 1]
Apart from the trivial solution when k = 0 these velocities are equal when
ak2 bk = 3k2 4bk ak b = 3ak 2b k = b2a . [. . . 1]
Sketch the 2-dimensional Fraunhofer diffraction pattern that would arise
from plane wave illumination of a circular aperture. Referring to your diagram,
give an expression for the angle subtended at the aperture by the first signif-
icant feature in the diffraction pattern. Without detailed calculation, sketch
the intensity profile along a straight line through the midpoint of the pattern. [4]
Solution:
Bookwork
2 marks for circular ring pattern plus formula, and 2 marks for a rough sketch
of the radial intensity distribution.
PHYS4031 Waves & Diffraction 4/16 Q 1 continued over. . .
Q 1 continued
SHOWING SOLUTIONS
[. . . 4]
PHYS4031 Waves & Diffraction 5/16 Paper continued over. . .
SHOWING SOLUTIONS
2A A string of length L and linear density ρ lies along the x-axis and is stretched
under tension T . The ends of the string are fixed at x = 0 and x = L .
(a) Sketch the first four normal modes of vibration of the string. [2]
Solution: Standard knowledge
Mode shapes for n = 1…4 are plotted below. Half a mark will be gained for
each of the four plots.
(In the plot, the horizontal axis extends from x = 0 to x = 1, corresponding to
the 0 to L points along the string length. The vertical axis is the transverse
displacement. All modes are illustrated with the same amplitude.) [. . . 2]
(b) The string is initially undisplaced and at time t = 0 is given a velocity
profile
v(x) =
4v0
L2
x(L x)
for a positive constant v0. Sketch this velocity profile. Without calculation,
discuss the expected excitation of normal modes of vibration of the string that
would result from this initial perturbation. [4]
Solution: Unseen problem; related problems have featured in examples sheet
and in past papers.
The initial velocity profile is a quadratic in x. The velocity is 0 at x = 0 and
x = L and has a maximum value of v0 at x = L/2.˙
PHYS4031 Waves & Diffraction 6/16 Q 2A continued over. . .
Q 2A continued
SHOWING SOLUTIONS
(In the plot, the horizontal axis extends from x = 0 to x = 1, corresponding to
the 0 to L points along the string length. The vertical v(x) axis is scaled so that 1
corresponds to the maximum initial transverse velocity v0 .) [. . . 2]
The shape of the velocity profile is similar to that of the n = 1 mode. It is not
precisely the same, however, so the expectation would be that the fundamental
mode would be the dominant one excited, with smaller amplitudes of higher order
modes. The initial profile has a large value at the midpoint of the string, so even
order modes – which would have a node at the midpoint – cannot be excited. [. . . 2]
(c) The motion of such a string can, in general, be described by a Fourier
decomposition of the form
y(x, t) =
∞∑
n=1
sin
(npix
L
)[
An cos
(npict
L
)
+Bn sin
(npict
L
)]
.
Explaining carefully each step of your calculation, show that for the given
initial condition the subsequent string motion can be expressed as
y(x, t) =
∞∑
n=1,3,5,…
32v0L
n4pi4c
sin
(npix
L
)
sin
(npict
L
)
,
where c is the propagation speed of transverse waves on the string. [10]
PHYS4031 Waves & Diffraction 7/16 Q 2A continued over. . .
Q 2A continued
SHOWING SOLUTIONS
Solution: Standard type of problem, but not seen before in this form.
Note that for educational purposes the solution given below includes even
relatively minor steps in the argument. A typical student-presented solution could
be expected to be significantly more concise.
y(x, t) =
∞∑
n=1
sin
(npix
L
)[
An cos
(
npict
L
)
+Bn sin
(
npict
L
)]
.
The string is initially undisplaced, so y(x, 0) = 0 for all x along the string.
y(x, 0) =
∞∑
n=1
sin
(npix
L
)
[An] = 0 An = 0
for all n . [. . . 2]
y(x, t) =
∞∑
n=1
Bn sin
(npix
L
)
sin
(
npict
L
)
.
We are given an initial velocity profile so we need
y˙(x, t) =
∞∑
n=1
Bn
(npic
L
)
sin
(npix
L
)
cos
(
npict
L
)
y˙(x, 0) =
∞∑
n=1
Bn
(npic
L
)
sin
(npix
L
)
.
Thus,
4v0
L2
x(L x) =
∞∑
n=1
Bn
(npic
L
)
sin
(npix
L
)
.
[. . . 2]
We solve this by multiplying both sides by sin mpixL and then integrating both
sides from 0 to L , making use of the orthogonality of the sine function to evaluate
the right hand side. Here are the steps:
4v0
L2
x(L x) sin
(mpix
L
)
=
∞∑
n=1
Bn
(npic
L
)
sin
(npix
L
)
sin
(mpix
L
)
∫ L
0
4v0
L2
x(L x) sin
(mpix
L
)
dx =
∫ L
0
∞∑
n=1
Bn
(npic
L
)
sin
(npix
L
)
sin
(mpix
L
)
dx
∫ L
0
4v0
L2
x(L x) sin
(mpix
L
)
dx =
∞∑
n=1
∫ L
0
Bn
(npic
L
)
sin
(npix
L
)
sin
(mpix
L
)
dx
4v0
L2
∫ L
0
x(L x) sin
(mpix
L
)
dx =
∞∑
n=1
Bn
(npic
L
)∫ L
0
sin
(npix
L
)
sin
(mpix
L
)
dx
PHYS4031 Waves & Diffraction 8/16 Q 2A continued over. . .
Q 2A continued
SHOWING SOLUTIONS
The set of integrals under the sum each evaluate to 0 for m 6= n ; for m = n the
integral evaluates to L/2 . [. . . 1]
To find the Bm coefficients we then have to solve
Bm
(mpic
L
) L
2
L2
4v0
= Bm
(
mpicL2
8v0
)
=
∫ L
0
x(L x) sin
(mpix
L
)
dx.
[. . . 1]
Using the given integrals (which are readily obtained by integration by parts)
the integral on the right hand side becomes
L
∫ L
0
x sin
mpi
L
xdx
∫ L
0
x2 sin
mpi
L
xdx
= L
[
Lx
mpi
cos
(mpix
L
)
+
L2
m2pi2
sin
(mpix
L
)]L
0