Solutions Manual for
The Physics of Vibrations and Waves –
6th Edition
Compiled by
Dr Youfang Hu
Optoelectronics Research Centre (ORC), University of Southampton, UK
In association with the author
H. J. Pain
Formerly of Department of Physics, Imperial College of Science and Technology, London, UK
© 2008 John Wiley & Sons, Ltd
TABLE OF CONTENTS
1. Simple Harmonic Motion.
2. Damped Simple Harmonic Motion.
3. The Forced Oscillator.
4. Coupled Oscillations.
5. Transverse Wave Motion.
6. Longitudinal Waves.
7. Waves on Transmission Lines.
8. Electromagnetic Waves.
9. Waves in More than One Dimension.
10. Fourier Methods.
11. Waves in Optical Systems.
12. Interference and Diffraction.
13. Wave Mechanics.
14 Non-linear Oscillations and Chaos.
15 Non-linear Waves, Shocks and Solitons.
SOLUTIONS TO CHAPTER 1
1.1
In Figure 1.1(a), the restoring force is given by:
F = − mg sin θ
By substitution of relation sin θ = x l into the above equation, we have:
F = −mg x l
so the stiffness is given by:
s = − F x = mg l
so we have the frequency given by:
ω2 = s m = g l
Since
θ is a very small angle, i.e. θ = sin θ = x l , or x = lθ , we have the restoring force
given by:
F = − mgθ
Now, the equation of motion using angular displacement
second law:
θ can by derived from Newton’s
F = m&x&
i.e.
− mgθ = mlθ&&
i.e.
θ&& + θ = 0
g
l
which shows the frequency is given by:
ω2 = g l
In Figure 1.1(b), restoring couple is given by − Cθ , which has relation to moment of inertia I
given by:
− Cθ = Iθ&&
i.e.
θ&& +
C
θ =0
I
which shows the frequency is given by:
ω2 = C I
© 2008 John Wiley & Sons, Ltd
In Figure 1.1(d), the restoring force is given by:
F = −2T x l
so Newton’s second law gives:
F = m&x& = − 2Tx l
&x& + 2Tx lm = 0
i.e.
which shows the frequency is given by:
ω2 =
2T
lm
In Figure 1.1(e), the displacement for liquid with a height of x has a displacement of x 2 and
a mass of ρAx , so the stiffness is given by:
s=
G
2 ρAxg
=
= 2 ρAg
x2
x
Newton’s second law gives:
− G = m&x&
− 2 ρAxg = ρAl&x&
i.e.
&x& +
i.e.
2g
x=0
l
which show the frequency is given by:
ω 2 = 2g l
γ
In Figure 1.1(f), by taking logarithms of equation pV = constant , we have:
ln p + γ ln V = constant
so we have:
i.e.
dp
dV
+γ
=0
p
V
dp = −γp
dV
V
The change of volume is given by dV = Ax , so we have:
dp = −γp
Ax
V
The gas in the flask neck has a mass of ρAl , so Newton’s second law gives:
Adp = m&x&
© 2008 John Wiley & Sons, Ltd
− γp
i.e.
A2 x
= ρAl&x&
V
&x& +
i.e.
γpA
x=0
lρV
which show the frequency is given by:
ω2 =
γpA
lρV
In Figure 1.1 (g), the volume of liquid displaced is Ax , so the restoring force is − ρgAx . Then,
Newton’s second law gives:
F = − ρgAx = m&x&
&x& +
i.e.
gρA
x=0
m
which shows the frequency is given by:
ω 2 = gρA m
1.2
Write solution x = a cos(ωt + φ ) in form: x = a cos φ cos ωt − a sin φ sin ωt and
compare with equation (1.2) we find: A = a cos φ and B = −a sin φ . We can also
find, with the same analysis, that the values of A and B for solution
x = a sin(ωt − φ ) are given by: A = −a sin φ and B = a cos φ , and for solution
x = a cos(ωt − φ ) are given by: A = a cos φ and B = a sin φ .
Try solution x = a cos(ωt + φ ) in expression &x& + ω 2 x , we have:
&x& + ω 2 x = − aω 2 cos(ωt + φ ) + ω 2 a cos(ωt + φ ) = 0
Try solution x = a sin(ωt − φ ) in expression &x& + ω 2 x , we have:
&x& + ω 2 x = − aω 2 sin(ωt − φ ) + ω 2 a sin(ωt − φ ) = 0
Try solution x = a cos(ωt − φ ) in expression &x& + ω 2 x , we have:
&x& + ω 2 x = − aω 2 cos(ωt − φ ) + ω 2 a cos(ωt − φ ) = 0
© 2008 John Wiley & Sons, Ltd
1.3
(a) If the solution x = a sin(ωt + φ ) satisfies x = a at t = 0 , then, x = a sin φ = a
i.e. φ = π 2 . When the pendulum swings to the position x = + a
2 for the first
time after release, the value of ωt is the minimum solution of equation
a sin(ωt + π 2) = + a
2 , i.e. ωt = π 4 . Similarly, we can find: for x = a 2 ,
ωt = π 3 and for x = 0 , ωt = π 2 .
If the solution x = a cos(ωt + φ ) satisfies x = a at t = 0 , then, x = a cos φ = a
i.e. φ = 0 . When the pendulum swings to the position x = + a
2 for the first
time after release, the value of ωt is the minimum solution of equation
a cos ωt = + a
2 , i.e. ωt = π 4 . Similarly, we can find: for x = a 2 , ωt = π 3
and for x = 0 , ωt = π 2 .
If
the
solution
x = a sin(ωt − φ )
satisfies
x=a
at
t=0 ,
then,
x = a sin(−φ ) = a i.e. φ = − π 2 . When the pendulum swings to the position
x = +a
2 for the first time after release, the value of ωt is the minimum
solution of equation a sin(ωt + π 2) = + a
2 , i.e. ωt = π 4 . Similarly, we can
find: for x = a 2 , ωt = π 3 and for x = 0 , ωt = π 2 .
If
the
solution
x = a cos(ωt − φ )
satisfies
x=a
at
t=0 ,
then,
x = a cos(−φ ) = a i.e. φ = 0 . When the pendulum swings to the position
x = +a
2 for the first time after release, the value of ωt is the minimum
solution of equation a cos ωt = + a
2 , i.e. ωt = π 4 . Similarly, we can find: for
x = a 2 , ωt = π 3 and for x = 0 , ωt = π 2 .
(b) If
the
solution
x = a sin(ωt + φ )
satisfies
x = −a
at
t=0 ,
then,
x = a sin φ = −a i.e. φ = − π 2 . When the pendulum swings to the position
© 2008 John Wiley & Sons, Ltd
x = +a
2 for the first time after release, the value of ωt is the minimum
solution of equation a sin(ωt − π 2) = + a
2 , i.e. ωt = 3π 4 . Similarly, we can
find: for x = a 2 , ωt = 2π 3 and for x = 0 , ωt = π 2 .
If
the
solution
x = a cos(ωt + φ )
satisfies
x = −a
at
t=0 ,
then,
x = a cos φ = −a i.e. φ = π . When the pendulum swings to the position
x = +a
2 for the first time after release, the value of ωt is the minimum
solution of equation a cos(ωt + π ) = + a
2 , i.e. ωt = 3π 4 . Similarly, we can
find: for x = a 2 , ωt = 2π 3 and for x = 0 , ωt = π 2 .
If
the
solution
x = a sin(ωt − φ )
satisfies
x = −a
at
t=0 ,
then,
x = a sin(−φ ) = −a i.e. φ = π 2 . When the pendulum swings to the position
x = +a
2 for the first time after release, the value of ωt is the minimum
solution of equation a sin(ωt − π 2) = + a
2 , i.e. ωt = 3π 4 . Similarly, we can
find: for x = a 2 , ωt = 2π 3 and for x = 0 , ωt = π 2 .
If
the
solution
x = a cos(ωt − φ )
satisfies
x = −a
at
t=0 ,
then,
x = a cos(−φ ) = −a i.e. φ = π . When the pendulum swings to the position
x = +a
2 for the first time after release, the value of ωt is the minimum
solution of equation a cos(ωt − π ) = + a
2 , i.e. ωt = 3π 4 . Similarly, we can
find: for x = a 2 , ωt = 2π 3 and for x = 0 , ωt = π 2 .
1.4
The frequency of such a simple harmonic motion is given by:
ω0 =
s
=
me
e2
=
4πε 0 r 3me
(1.6 ×10 −19 ) 2
≈ 4.5 × 1016 [rad ⋅ s −1 ]
−12
−9 3
−31
4 × π × 8.85 ×10 × (0.05 ×10 ) × 9.1×10
Its radiation generates an electromagnetic wave with a wavelength λ given by:
© 2008 John Wiley & Sons, Ltd
λ=
2πc
ω0
=
2 × π × 3 ×108
≈ 4.2 ×10 −8 [m] = 42[nm]
4.5 ×1016
Therefore such a radiation is found in X-ray region of electromagnetic spectrum.
1.5
(a) If the mass m is displaced a distance of x from its equilibrium position, either
the upper or the lower string has an extension of x 2 . So, the restoring force of
the mass is given by: F = − sx 2 and the stiffness of the system is given by:
s′ = − F x = s 2 . Hence the frequency is given by ωa2 = s′ m = s 2m .
(b) The frequency of the system is given by: ωb2 = s m
(c) If the mass m is displaced a distance of x from its equilibrium position, the
restoring force of the mass is given by: F = − sx − sx = −2 sx and the stiffness of
the system is given by: s′ = − F x = 2 s . Hence the frequency is given by
ωc2 = s′ m = 2s m .
Therefore, we have the relation: ωa2 : ωb2 : ωc2 = s 2m : s m : 2s m = 1 : 2 : 4
1.6
At time t = 0 , x = x0 gives:
a sin φ = x0
(1.6.1)
aω cos φ = v0
(1.6.2)
x& = v0 gives:
From (1.6.1) and (1.6.2), we have
tan φ = ωx0 v0 and a = ( x02 + v02 ω 2 )1 2
1.7
The equation of this simple harmonic motion can be written as: x = a sin(ωt + φ ) .
The time spent in moving from x to x + dx is given by: dt = dx vt , where vt is
the velocity of the particle at time t and is given by: vt = x& = aω cos(ωt + φ ) .
© 2008 John Wiley & Sons, Ltd
Noting that the particle will appear twice between x and x + dx within one period
of oscillation. We have the probability η of finding it between x to x + dx given
by: η =
η=
2π
2dt
, so we have:
where the period is given by: T =
T
ω
2dt
2ωdx
dx
dx
dx
=
=
=
=
2
T
2πaω cos(ωt + φ ) πa cos(ωt + φ ) πa 1 − sin (ωt + φ ) π a 2 − x 2
1.8
Since the displacements of the equally spaced oscillators in y direction is a sine
curve, the phase difference δφ between two oscillators a distance x apart given is
proportional to the phase difference 2π between two oscillators a distance λ apart
by: δφ 2π = x λ , i.e. δφ = 2πx λ .
1.9
The mass loses contact with the platform when the system is moving downwards and
the acceleration of the platform equals the acceleration of gravity. The acceleration of
a simple harmonic vibration can be written as: a = Aω 2 sin(ωt + φ ) , where A is the
amplitude, ω is the angular frequency and φ is the initial phase. So we have:
Aω 2 sin(ωt + φ ) = g
A=
i.e.
g
ω sin(ωt + φ )
2
Therefore, the minimum amplitude, which makes the mass lose contact with the
platform, is given by:
Amin =
g
ω
2
=
g
4π f
2
2
=
9.8
≈ 0.01[m]
4 × π 2 × 52
1.10
The mass of the element dy is given by: m′ = mdy l . The velocity of an element
dy of its length is proportional to its distance y from the fixed end of the spring, and
is given by: v′ = yv l . where v is the velocity of the element at the other end of the
spring, i.e. the velocity of the suspended mass M . Hence we have the kinetic energy
© 2008 John Wiley & Sons, Ltd
of this element given by:
KEdy =
1
1 ⎛ m ⎞⎛ y ⎞
m′v′2 = ⎜ dy ⎟⎜ v ⎟
2
2⎝ l
⎠⎝ l ⎠
2
The total kinetic energy of the spring is given by:
KEspring
2
mv 2
1 ⎛ m ⎞⎛ y ⎞
= ∫ KEdy dy = ∫ ⎜ dy ⎟⎜ v ⎟ = 3
0
0 2
2l
⎝l
⎠⎝ l ⎠
l
l
l
1
∫ y dy = 6 mv
2
2
0
The total kinetic energy of the system is the sum of kinetic energies of the spring and
the suspended mass, and is given by:
1
1
1
KEtot = mv 2 + Mv 2 = (M + m 3)v 2
6
2
2
which shows the system is equivalent to a spring with zero mass with a mass of
M + m 3 suspended at the end. Therefore, the frequency of the oscillation system is
given by:
ω2 =
s
M +m 3
1.11
In Figure 1.1(a), the restoring force of the simple pendulum is − mg sin θ , then, the
stiffness is given by: s = mg sin θ x = mg l . So the energy is given by:
E=
1 2 1 2 1 2 1 mg 2
mv + sx = mx& +
x
2
2
2
2 l
The equation of motion is by setting dE dt = 0 , i.e.:
d ⎛ 1 2 1 mg 2 ⎞
x ⎟=0
⎜ mx& +
dt ⎝ 2
2 l
⎠
i.e.
&x& +
g
x=0
l
In Figure 1.1(b), the displacement is the rotation angle θ , the mass is replaced by the
moment of inertia I of the disc and the stiffness by the restoring couple C of the
wire. So the energy is given by:
1
1
E = Iθ& 2 + Cθ 2
2
2
The equation of motion is by setting dE dt = 0 , i.e.:
d ⎛ 1 &2 1
2⎞
⎜ Iθ + Cθ ⎟ = 0
dt ⎝ 2
2
⎠
© 2008 John Wiley & Sons, Ltd
θ&& +
i.e.
C
θ =0
I
In Figure 1.1(c), the energy is directly given by:
1
1
E = mv 2 + sx 2
2
2
The equation of motion is by setting dE dt = 0 , i.e.:
d ⎛1 2 1 2⎞
⎜ mx& + sx ⎟ = 0
dt ⎝ 2
2
⎠
&x& +
i.e.
s
x=0
m
In Figure 1.1(c), the restoring force is given by: − 2 Tx l , then the stiffness is given
by: s = 2T l . So the energy is given by:
E=
1 2 1 2 1 2 1 2T 2 1 2 T 2
mv + sx = mx& +
x = mx& + x
2
2
2
2 l
2
l
The equation of motion is by setting dE dt = 0 , i.e.:
d ⎛1 2 T 2⎞
⎜ mx& + x ⎟ = 0
l ⎠
dt ⎝ 2
&x& +
i.e.
2T
x=0
lm
In Figure 1.1(e), the liquid of a volume of ρAl is displaced from equilibrium
position by a distance of l 2 , so the stiffness of the system is given by
s = 2 ρgAl l = 2 ρgA . So the energy is given by:
E=
1 2 1 2 1
1
1
mv + sx = ρAlx& 2 + 2 ρgAx 2 = ρAlx& 2 + ρgAx 2
2
2
2
2
2
The equation of motion is by setting dE dt = 0 , i.e.:
d ⎛1
2
2⎞
⎜ ρAlx& + ρgAx ⎟ = 0
dt ⎝ 2
⎠
i.e.
© 2008 John Wiley & Sons, Ltd
&x& +
2g
x=0
l
In Figure 1.1(f), the gas of a mass of ρAl is displaced from equilibrium position by
a distance of x and causes a pressure change of dp = − γpAx V , then, the stiffness
of the system is given by s = − Adp x = γpA2 V . So the energy is given by:
E=
1 2 1 2 1
1 γpA2 x 2
mv + sx = ρAlx& 2 +
2
2
2
2 V
The equation of motion is by setting dE dt = 0 , i.e.:
d ⎛1
1 γpA2 x 2 ⎞
⎜⎜ ρAlx& 2 +
⎟=0
dt ⎝ 2
2 V ⎟⎠
&x& +
i.e.
γpA
x=0
lρV
In Figure 1.1(g), the restoring force of the hydrometer is − ρgAx , then the stiffness
of the system is given by s = ρgAx x = ρgA . So the energy is given by:
E=
1 2 1 2 1 2 1
mv + sx = mx& + ρgAx 2
2
2
2
2
The equation of motion is by setting dE dt = 0 , i.e.:
d ⎛1 2 1
2⎞
⎜ mx& + ρgAx ⎟ = 0
dt ⎝ 2
2
⎠
i.e.
&x& +
Aρg
x=0
m
1.12
The displacement of the simple harmonic oscillator is given by:
x = a sin ωt
so the velocity is given by:
x& = aω cos ωt
From (1.12.1) and (1.12.2), we can eliminate t and get:
x2
x& 2
+ 2 2 = sin 2 ωt + cos 2 ωt = 1
2
a
aω
which is an ellipse equation of points ( x, x& ) .
The energy of the simple harmonic oscillator is given by:
© 2008 John Wiley & Sons, Ltd
(1.12.1)
(1.12.2)
(1.12.3)
E=
1 2 1 2
mx& + sx
2
2
(1.12.4)
Write (1.12.3) in form x& 2 = ω 2 (a 2 − x 2 ) and substitute into (1.12.4), then we have:
E=
1 2 1 2 1
1
mx& + sx = mω 2 (a 2 − x 2 ) + sx 2
2
2
2
2
Noting that the frequency ω is given by: ω 2 = s m , we have:
E=
1
1
1
s (a 2 − x 2 ) + sx 2 = sa 2
2
2
2
which is a constant value.
1.13
The equations of the two simple harmonic oscillations can be written as:
y1 = a sin(ωt + φ )
and
y2 = a sin(ωt + φ + δ )
The resulting superposition amplitude is given by:
R = y1 + y2 = a[sin(ωt + φ ) + sin(ωt + φ + δ )] = 2a sin(ωt + φ + δ 2) cos(δ 2)
and the intensity is given by:
I = R 2 = 4a 2 cos 2 (δ 2) sin 2 (ωt + φ + δ 2)
I ∝ 4a 2 cos 2 (δ 2)
i.e.
Noting that sin 2 (ωt + φ + δ 2) varies between 0 and 1, we have:
0 ≤ I ≤ 4a 2 cos 2 (δ 2)
1.14
2
2
⎛x
⎞ ⎛ y
⎞
y
x
⎜⎜ sin φ2 − sin φ1 ⎟⎟ + ⎜⎜ cos φ1 − cos φ2 ⎟⎟
a2
a1
⎝ a1
⎠ ⎝ a2
⎠
2 xy
2 xy
x2
y2
y2
x2
= 2 sin 2 φ2 + 2 sin 2 φ1 −
sin φ1 sin φ2 + 2 cos 2 φ1 + 2 cos 2 φ2 −
cos φ1 cos φ2
a1
a2
a1a2
a2
a1
a1a2
=
2 xy
x2
y2
2
2
(sin
φ
+
cos
φ
)
+
(sin 2 φ1 + cos 2 φ1 ) −
(sin φ1 sin φ2 + cos φ1 cos φ2 )
2
2
2
2
a1
a2
a1a2
=
x 2 y 2 2 xy
+
−
cos(φ1 − φ2 )
a12 a22 a1a2
On the other hand, by substitution of :
x
= sin ωt cos φ1 + cos ωt sin φ1
a1
© 2008 John Wiley & Sons, Ltd
y
= sin ωt cos φ2 + cos ωt sin φ2
a2
2
2
⎛x
⎞ ⎛ y
⎞
y
x
into expression ⎜⎜ sin φ2 − sin φ1 ⎟⎟ + ⎜⎜ cos φ1 − cos φ2 ⎟⎟ , we have:
a2
a1
⎝ a1
⎠ ⎝ a2
⎠
2
2
⎛x
⎞ ⎛ y
⎞
y
x
⎜⎜ sin φ2 − sin φ1 ⎟⎟ + ⎜⎜ cos φ1 − cos φ2 ⎟⎟
a2
a1
⎝ a1
⎠ ⎝ a2
⎠
2
2
2
= sin ωt (sin φ2 cos φ1 − sin φ1 cos φ2 ) + cos ωt (cos φ1 sin φ2 − cos φ2 sin φ1 ) 2
= (sin 2 ωt + cos 2 ωt ) sin 2 (φ2 − φ1 )
= sin 2 (φ2 − φ1 )
From the above derivation, we have:
x 2 y 2 2 xy
+ 2−
cos(φ1 − φ2 ) = sin 2 (φ2 − φ1 )
2
a1 a2 a1a2
1.15
By elimination of t from equation x = a sin ωt and y = b cos ωt , we have:
x2 y 2
+
=1
a 2 b2
which shows the particle follows an elliptical path. The energy at any position of x ,
y on the ellipse is given by:
1 2 1 2 1 2 1 2
mx& + sx + my& + sy
2
2
2
2
1 2 2
1
1
1
= ma ω cos 2 ωt + ma 2ω 2 sin 2 ωt + mb 2ω 2 sin 2 ωt + mb 2ω 2 cos 2 ωt
2
2
2
2
1 2 2 1 2 2
= ma ω + mb ω
2
2
1
= mω 2 (a 2 + b 2 )
2
E=
The value of the energy shows it is a constant and equal to the sum of the separate
1
energies of the simple harmonic vibrations in x direction given by mω 2 a 2 and in
2
1
y direction given by mω 2b 2 .
2
At any position of x , y on the ellipse, the expression of m( xy& − yx& ) can be
written as:
© 2008 John Wiley & Sons, Ltd
m( xy& − yx& ) = m(−abω sin 2 ωt − abω cos 2 ωt ) = − abmω (sin 2 ωt + cos 2 ωt ) = − abmω
which is a constant. The quantity abmω is the angular momentum of the particle.
1.16
All possible paths described by equation 1.3 fall within a rectangle of 2a1 wide and
2a2 high, where a1 = xmax and a2 = ymax , see Figure 1.8.
When x = 0 in equation (1.3) the positive value of y = a2 sin(φ2 − φ1 ) . The value of
ymax = a2 . So y x=0 ymax = sin(φ2 − φ1 ) which defines φ2 − φ1 .
1.17
In the range 0 ≤ φ ≤ π , the values of cos φi are − 1 ≤ cos φi ≤ +1 . For n random
values of φi , statistically there will be n 2 values − 1 ≤ cos φi ≤ 0 and n 2 values
0 ≤ cos φi ≤ 1 . The positive and negative values will tend to cancel each other and the
n
sum of the n values:
n
∑ cos φi → 0 , similarly
∑ cos φ
j =1
i =1
i≠ j
n
n
∑ cos φ ∑ cos φ
i =1
i≠ j
i
j =1
j
j
→ 0 . i.e.
→0
1.18
The exponential form of the expression:
a sin ωt + a sin(ωt + δ ) + a sin(ωt + 2δ ) + L + a sin[ωt + ( n − 1)δ ]
is given by:
ae iωt + ae i (ωt +δ ) + ae i (ωt + 2δ ) + L + ae i[ωt +( n−1)δ ]
From the analysis in page 28, the above expression can be rearranged as:
ae
⎡
⎛ n−1 ⎞ ⎤
i ⎢ ωt + ⎜
⎟δ ⎥
⎝ 2 ⎠ ⎦
⎣
sin nδ 2
sin δ 2
with the imaginary part:
⎡
⎛ n − 1 ⎞ ⎤ sin nδ 2
a sin ⎢ωt + ⎜
⎟δ ⎥
⎝ 2 ⎠ ⎦ sin δ 2
⎣
which is the value of the original expression in sine term.
© 2008 John Wiley & Sons, Ltd
1.19
From the analysis in page 28, the expression of z can be rearranged as:
z = aeiωt (1 + eiδ + ei 2δ + L ei ( n−1)δ ) = aeiωt
sin nδ 2
sin δ 2
The conjugate of z is given by:
z * = ae −iωt
sin nδ 2
sin δ 2
so we have:
zz * = aeiωt
© 2008 John Wiley & Sons, Ltd
sin nδ 2
sin nδ 2
sin 2 nδ 2
⋅ ae −iωt
= a2
sin δ 2
sin δ 2
sin 2 δ 2
SOLUTIONS TO CHAPTER 2
2.1
The system is released from rest, so we know its initial velocity is zero, i.e.
dx
=0
dt t =0
(2.1.1)
Now, rearrange the expression for the displacement in the form:
x=
F + G ( − p + q ) t F − G ( − p −q ) t
e
+
e
2
2
(2.1.2)
Then, substitute (2.1.2) into (2.1.1), we have
dx
F + G ( − p+q ) t
F − G ( − p −q ) t ⎤
⎡
= ⎢(− p + q )
+ (− p − q )
e
e
⎥⎦ = 0
dt t =0 ⎣
2
2
t =0
i.e.
qG = pF
(2.1.3)
By substitution of the expressions of q and p into equation (2.1.3), we have the ratio given by:
G
r
=
12
F
r 2 − 4ms
(
)
2.2
The first and second derivatives of x are given by:
r
⎡
( A + Bt )⎤⎥e −rt 2m
x& = ⎢ B −
2m
⎣
⎦
⎡ rB r 2
⎤
&x& = ⎢−
( A + Bt )⎥e −rt 2m
+
2
⎣ m 4m
⎦
We can verify the solution by substitution of x , x& and &x& into equation:
m&x& + rx& + sx = 0
then we have equation:
⎛
r2 ⎞
⎜⎜ s −
⎟( A + Bt ) = 0
4m ⎟⎠
⎝
which is true for all t, provided the first bracketed term of the above equation is zero, i.e.
© 2008 John Wiley & Sons, Ltd
s−
r2
=0
4m
r 2 4m 2 = s m
i.e.
2.3
The initial displacement of the system is given by:
(
)
x = e − rt 2 m C1eiω′t + C2e − iω′t = A cos φ at t = 0
So:
C1 + C2 = A cos φ
(2.3.1)
Now let the initial velocity of the system to be:
⎛ r
⎞
⎛ r
⎞
+ iω ′ ⎟C1e (−r 2 m+iω′ )t + ⎜ −
− iω ′ ⎟C2e(−r 2 m−iω′ )t = −ω ′A sin φ at t = 0
x& = ⎜ −
⎝ 2m
⎠
⎝ 2m
⎠
−
i.e.
r
A cos φ + iω ′(C1 − C2 ) = −ω ′A sin φ
2m
If r m is very small or φ ≈ π 2 , the first term of the above equation approximately equals zero,
so we have:
C1 − C2 = iA sin φ
(2.3.2)
From (2.3.1) and (2.3.2), C1 and C2 are given by:
A(cos φ + i sin φ )
=
2
A(cos φ − i sin φ )
C2 =
=
2
C1 =
A iφ
e
2
A −iφ
e
2
2.4
Use the relation between current and charge, I = q& , and the voltage equation:
q C + IR = 0
we have the equation:
Rq& + q C = 0
solve the above equation, we get:
q = C1e − t RC
© 2008 John Wiley & Sons, Ltd
where C1 is arbitrary in value. Use initial condition, we get C1 = q0 ,
q = q0e −t RC
i.e.
which shows the relaxation time of the process is RC s.
2.5
ω02 - ω ′2 = 10-6 ω02 = r 2 4m 2 => ω0 m r = 500
(a)
The condition also shows
ω ′ ≈ ω0 , so:
Q = ω ′m r ≈ ω0 m r = 500
Use
τ′ =
2π
, we have:
ω′
δ=
πr π
π
r
τ′ =
= =
2m
mω ′ Q 500
(b) The stiffness of the system is given by:
s = ω02 m = 1012 × 10−10 = 100[ Nm −1 ]
and the resistive constant is given by:
r=
ω0 m
Q
=
106 × 10−10
= 2 × 10− 7 [ N ⋅ sm −1 ]
500
(c) At t = 0 and maximum displacement, x& = 0 , energy is given by:
E=
1 2 1 2 1 2
1
mx& + sx = sxmax = × 100 × 10− 4 = 5 × 10− 3[ J ]
2
2
2
2
−1
Time for energy to decay to e of initial value is given by:
t=
m
10−10
=
= 0.5[ms]
r 2 × 10− 7
(d) Use definition of Q factor:
E
− ΔE
where, E is energy stored in system, and − ΔE is energy lost per cycle, so energy loss in the
Q = 2π
first cycle, − ΔE1 , is given by:
− ΔE1 = −ΔE = 2π
2.6
© 2008 John Wiley & Sons, Ltd
E
5 × 10−3
= 2π ×
= 2π × 10− 5[ J ]
Q
500
The frequency of a damped simple harmonic oscillation is given by:
ω ′2 = ω02 −
Use
r2
r2
r2
2
2
′
′
Δ
ω
=
ω
ω
=
=>
ω
−
ω
=
=>
0
0
4m 2
4m 2
4m 2 (ω ′ + ω0 )
ω ′ ≈ ω and Q =
ω0 m
r
we find fractional change in the resonant frequency is given by:
Δω
ω0
=
−1
ω ′ − ω0
r2
≈
= (8Q 2 )
2 2
ω0
8m ω0
2.7
See page 71 of text. Analysis is the same as that in the text for the mechanical case except that
inductance L replaces mass m , resistance R replaces r and stiffness s is replaced by
1 C , where C is the capacitance. A large Q value requires a small R .
2.8
Electrons per unit area of the plasma slab is given by:
q = −nle
When all the electrons are displaced a distance x , giving a restoring electric field:
E = nex / ε 0 ，the restoring force per unit area is given by:
F = qE = −
xn 2e 2l
ε0
Newton’s second law gives:
restoring force per unit area = electrons mass per unit area × electrons acceleration
i.e.
F =−
xn 2e 2l
ε0
= nlme × &x&
&x& +
i.e.
ne 2
x=0
mε 0
From the above equation, we can see the displacement distance of electrons, x , oscillates with
angular frequency:
ωe2 =
ne 2
meε 0
2.9
As the string is shortened work is done against: (a) gravity (mg cosθ ) and (b) the centrifugal
force ( mv r = mlθ& ) along the time of shortening. Assume that during shortening there are
2
2
© 2008 John Wiley & Sons, Ltd
many swings of constant amplitude so work done is:
A = −(mg cosθ + mlθ& 2 )Δl
where the bar denotes the average value. For small
θ , cosθ = 1 − θ 2 2 so:
A = − mgΔl + (mg θ 2 2 − mlθ& 2 )
The term − mgΔl is the elevation of the equilibrium position and does not affect the energy of
motion so the energy change is:
ΔE = (mg θ 2 2 − mlθ& 2 )Δl
Now the pendulum motion has energy:
E=
m 2 &2
l θ + mgl (1 − cos θ ) ,
2
that is, kinetic energy plus the potential energy related to the rest position, for small
becomes:
E=
θ this
ml 2θ& 2 mglθ 2
+
2
2
which is that of a simple harmonic oscillation with linear amplitude lθ 0 .
Taking the solution
θ = θ 0 cos ωt which gives θ 2 = θ 02 2 and θ& 2 = ω 2 θ 02 2 with
ω = g l we may write:
E=
ml 2ω 2θ 02 mglθ 02
=
2
2
and
⎛ mlω 2θ 02 mlω 2θ 02 ⎞
mlω 2θ 02
⎟⎟Δl = −
⋅ Δl
ΔE = ⎜⎜
−
4
2 ⎠
4
⎝
so:
1 Δl
ΔE
=−
E
2 l
Now
ω = 2πν = g l so the frequency ν varies with l −1 2 and
Δν
=−
ν
1 Δl ΔE
=
2 l
E
so:
E
ν
© 2008 John Wiley & Sons, Ltd
= constant
SOLUTIONS TO CHAPTER 3
3.1
The solution of the vector form of the equation of motion for the forced oscillator:
m&x& + rx& + sx = F0eiωt
is given by:
x=
iF0ei (ωt −φ )
iF
F
=−
cos(ωt − φ ) +
sin(ωt − φ )
ωZ m
ωZ m
ωZ m
Since F0eiωt represents its imaginary part: F0 sin ωt , the value of x is given by the
imaginary part of the solution, i.e.:
x=−
F
cos(ωt − φ )
ωZ m
The velocity is given by:
v = x& =
F
sin(ωt − φ )
Zm
3.2
The transient term of a forced oscillator decay with e − rt 2 m to e- k at time t , i.e.:
− rt 2m = − k
so, we have the resistance of the system given by:
r = 2mk t
(3.2.1)
ω ≈ ω0 = s m
(3.2.2)
For small damping, we have
We also have steady state displacement given by:
x = x0 sin(ωt − φ )
where the maximum displacement is:
x0 =
F0
ω r + (ωm − s m) 2
2
(3.2.3)
By substitution of (3.2.1) and (3.2.2) into (3.2.3), we can find the average rate of
growth of the oscillations given by:
© 2008 John Wiley & Sons, Ltd
x0
F0
=
t
2kmω0
3.3
Write the equation of an undamped simple harmonic oscillator driven by a force of
frequency ω in the vector form, and use F0eiωt to represent its imaginary part
F0 sin ωt , we have:
m&x& + sx = F0eiωt
(3.3.1)
We try the steady state solution x = Aeiωt and the velocity is given by:
x& = iωAeiωt = iωx
so that:
&x& = i 2ω 2 x = −ω 2 x
and equation (3.3.1) becomes:
(− Aω 2 m + As )eiωt = F0eiωt
which is true for all t when
− Aω 2 m + As = F0
F0
s − ω 2m
F0
i.e.
x=
eiωt
2
s −ω m
The value of x is the imaginary part of vector x , given by:
F0
x=
sin ωt
s − ω 2m
A=
i.e.
i.e.
x=
F0 sin ωt
m(ω02 − ω 2 )
Hence, the amplitude of x is given by:
A=
© 2008 John Wiley & Sons, Ltd
F0
m(ω − ω 2 )
2
0
where ω02 =
s
m
and its behaviour as a function of frequency is shown in the following graph:
A
F0
mω02
0
ω0
ω
By solving the equation:
m&x& + sx = 0
we can easily find the transient term of the equation of the motion of an undamped
simple harmonic oscillator driven by a force of frequency ω is given by:
x = C cos ω0t + D sin ω0t
where, ω0 = s m , C and D are constant. Finally, we have the general solution
for the displacement given by the sum of steady term and transient term:
x=
F0 sin ωt
+ C cos ω0t + D sin ω0t
m(ω02 − ω 2 )
(3.3.2)
3.4
In equation (3.3.2), x = 0 at t = 0 gives:
⎡ F sin ωt
⎤
+ A cos ω0t + B sin ω0t ⎥ = A = 0
x t =0 = ⎢ 0 2
2
⎣ m(ω0 − ω )
⎦ t =0
(3.4.1)
In equation (3.3.2), x& = 0 at t = 0 gives:
⎡ ωF0 cos ωt
⎤
dx
ωF0
=⎢
− ω0 A sin ω0t + ω0 B cos ω0t ⎥ =
+ ω0 B = 0
2
2
2
2
dt t =0 ⎣ m(ω0 − ω )
⎦ t =0 m(ω0 − ω )
B=−
i.e.
F0ω
mω0 (ω02 − ω 2 )
(3.4.2)
By substitution of (3.4.1) and (3.4.2) into (3.3.2), we have:
x=
⎞
⎛
1
ω
F0
⎟⎟
⎜
sin
ω
sin
ω
t
−
t
0
ω0
m (ω02 − ω 2 ) ⎜⎝
⎠
By substitution of ω = ω0 + Δω into (3.4.3), we have:
© 2008 John Wiley & Sons, Ltd
(3.4.3)
x=−
⎞
⎛
ω
F0
⎜⎜ sin ω0t cos Δωt + sin Δωt cos ω0t − sin ω0t ⎟⎟
ω0
m(ω0 + ω )Δω ⎝
⎠
(3.4.4)
Since Δω ω0 << 1 and Δωt << 1 , we have:
ω ≈ ω0 , sin Δωt ≈ Δωt , and cos Δωt ≈ 1
Then, equation (3.4.4) becomes:
x=−
⎞
⎛
ω
⎜⎜ sin ω0t + Δωt cos ω0t − sin ω0t ⎟⎟
2mω0 Δω ⎝
ω0
⎠
F0
x=−
i.e.
⎞
⎛ Δω
⎜⎜ −
sin ω0t + Δωt cos ω0t ⎟⎟
2mω0 Δω ⎝ ω0
⎠
F0
i.e.
x=
⎞
F0 ⎛ sin ω0t
⎜⎜
− t cos ω0t ⎟⎟
2mω0 ⎝ ω0
⎠
i.e.
x=
F0
(sin ω0t − ω0t cos ω0t )
2mω02
The behaviour of displacement x as a function of ω0t is shown in the following
x
F0π
2mω02
3F0π
2mω02
5F0π
2mω02
7 F0π
2mω02
F0t
2mω0
0
ω0 t
F0π
mω02
−
2 F0π
mω02
F0t
2mω0
3F0π
mω02
graph:
3.5
The general expression of displacement of a simple damped mechanical oscillator
driven by a force F0 cos ωt is the sum of transient term and steady state term, given
© 2008 John Wiley & Sons, Ltd
by:
x = Ce
−
rt
+ iωi t
2m
−
iF0 e i (ωt −φ )
ωZ m
where, C is constant, Z m = r 2 + (ωm − s ω ) 2 , ωt = s m − r 2 4m 2 and
⎛ ωm − s ω ⎞
⎟ , so the general expression of velocity is given by:
r
⎝
⎠
φ = tan −1 ⎜
⎛ r
⎞ − +iωit F0 i (ωt −φ )
v = x& = C ⎜ −
+ i ωt ⎟ e 2 m
+
e
Zm
⎝ 2m
⎠
rt
and the general expression of acceleration is given by:
rt
⎛ r2
iωt r ⎞ − 2 m +iωit iωF0 i (ωt −φ )
2
⎟e
v& = C ⎜⎜ 2 − ωt −
+
e
m ⎟⎠
Zm
⎝ 4m
⎛ r 2 − 2ms iωt r ⎞ − 2rtm +iωit iωF0 i (ωt −φ )
⎟⎟e
v& = C ⎜⎜
+
e
−
2
m
m
Z
2
⎠
⎝
m
i.e.
(3.5.1)
From (3.5.1), we find the amplitude of acceleration at steady state is given by:
v& =
ωF0
Zm
=
ωF0
r + (ωm − s ω ) 2
2
At the frequency of maximum acceleration:
dv&
=0
dω
⎤
d ⎡
ωF0
⎢
⎥=0
dω ⎢ r 2 + (ωm − s ω ) 2 ⎥
⎣
⎦
i.e.
r − 2ms +
2
i.e.
ω2 =
i.e.
2s 2
ω2
=0
2s 2
2sm − r 2
Hence, we find the expression of the frequency of maximum acceleration given by:
ω=
2s 2
2sm − r 2
The frequency of velocity resonance is given by: ω = s m , so if r = sm , the
acceleration amplitude at the frequency of velocity resonance is given by:
v& r =
sm
© 2008 John Wiley & Sons, Ltd
=
s m F0
ωF0
F
=
= 0
2
r + (ωm − s ω )
sm + ( sm − sm ) m
2
The limit of the acceleration amplitude at high frequencies is given by:
lim v& = lim
ω →∞
ω →∞
ωF0
= lim
r 2 + (ωm − s ω ) 2 ω →∞
F0
s ⎞
⎛
+ ⎜m − 2 ⎟
2
ω ⎝
ω ⎠
r2
2
=
F0
m
So we have:
v& r =
sm
= lim v&
ω →∞
3.6
The displacement amplitude of a driven mechanical oscillator is given by:
x=
x=
i.e.
F0
ω r 2 + (ωm − s ω ) 2
F0
ω 2 r 2 + (ω 2 m − s) 2
(3.6.1)
The displacement resonance frequency is given by:
ω=
s
r2
−
m 2m 2
(3.6.2)
By substitution of (3.6.2) into (3.6.1), we have:
F0
x=
2
r2 ⎞ ⎛ r2 ⎞
2⎛ s
⎟⎟
⎟+⎜
r ⎜⎜ −
2 ⎟ ⎜
⎝ m 2m ⎠ ⎝ 2m ⎠
x=
i.e.
F0
s
r2
r
−
m 4m 2
which proves the exact amplitude at the displacement resonance of a driven
mechanical oscillator may be written as:
F
x= 0
ω ′r
where,
ω ′2 =
s
r2
−
m 4m 2
3.7
(a) The displacement amplitude is given by:
x=
© 2008 John Wiley & Sons, Ltd
F0
ω r 2 + (ωm − s ω ) 2
At low frequencies, we have:
lim x = lim
ω →0
ω →0
F0
ω r + (ωm − s ω )
2
2
F0
= lim
ω r + (ω m − s)
ω →0
2 2
2
2
=
F0
s
(b) The velocity amplitude is given by:
F0
v=
r + (ωm − s ω ) 2
2
At velocity resonance: ω = s m , so we have:
vr =
F0
r + (ωm − s ω )
2
F0
=
2
r + ( sm − sm )
2
ω= s m
2
=
F0
r
(c) From problem 3.5, we have the acceleration amplitude given by:
v& =
ωF0
r + (ωm − s ω ) 2
2
At high frequency, we have:
lim v& = lim
ω →∞
ω →∞
ωF0
r + (ωm − s ω )
2
2
F0
= lim
ω →∞
r ω + (m − s ω )
2
2
2 2
=
F0
m
From (a), (b) and (c), we find lim x , vr and lim v& are all constants, i.e. they are all
ω →0
ω →∞
frequency independent.
3.8
The expression of curve (a) in Figure 3.9 is given by:
xa = −
where
F0 X m
F0 m(ω02 − ω 2 )
=
ωZ m2
m 2 (ω02 − ω 2 ) 2 + ω 2 r 2
(3.8.1)
ω0 = s m
xa has either maximum or minimum value when
dxa
=0
dω
i.e.
⎤
d ⎡
F0 m(ω02 − ω 2 )
=0
⎢ 2 2
2 2
2 2⎥
dω ⎣ m (ω0 − ω ) + ω r ⎦
i.e.
m 2 (ω02 − ω 2 ) 2 − ω02 r 2 = 0
Then, we have two solutions of ω given by:
ω1 = ω02 −
© 2008 John Wiley & Sons, Ltd
ω0 r
m
(3.8.2)
and
ω2 = ω02 +
ω0 r
m
(3.8.3)
Since r is very small, rearrange the expressions of ω1 and ω2 , we have:
ω0 r
2
ω0 r
2
r ⎞
r2
r
⎛
ω1 = ω −
= ⎜ ω0 −
≈ ω0 −
⎟ −
2
m
2m ⎠ 4m
2m
⎝
2
0
r ⎞
r2
r
⎛
ω2 = ω +
≈ ω0 +
= ⎜ ω0 +
⎟ −
2
m
2m ⎠ 4m
2m
⎝
2
0
The maximum and the minimum values of xa can found by substitution of (3.8.2)
and (3.8.3) into (3.8.1), so we have:
when ω = ω1 :
xa =
F0
F
≈ 0
2
2ω0 r − ω0 r m 2ω0 r
which is the maximum value of xa , and
when ω = ω2 :
xa = −
F0
F
≈− 0
2
2ω0 r + ω0 r m
2ω0 r
which is the minimum value of xa .
3.9
The undamped oscillatory equation for a bound electron is given by:
&x& + ω02 x = (− eE0 m) cos ωt
(3.9.1)
Try solution x = A cos ωt in equation (3.9.1), we have:
(−ω 2 + ω02 ) A cos ωt = (− eE0 m) cos ωt
which is true for all t provided:
(−ω 2 + ω02 ) A = − eE0 m
i.e.
A=−
eE0
m(ω02 − ω 2 )
So, we find a solution to equation (3.9.1) given by:
x=−
eE0
cos ωt
m(ω02 − ω 2 )
(3.9.2)
For an electron number density n , the induced polarizability per unit volume of a
© 2008 John Wiley & Sons, Ltd
medium is given by:
χe = −
nex
ε0E
(3.9.3)
By substitution of (3.9.2) and E = E0 cos ωt into (3.9.3), we have
χe = −
nex
ne 2
=
ε 0 E ε 0 m(ω02 − ω 2 )
3.10
The forced mechanical oscillator equation is given by:
m&x& + rx& + sx = F0 cos ωt
which can be written as:
m&x& + rx& + mω02 x = F0 cos ωt
(3.10.1)
where, ω0 = s m . Its solution can be written as:
x=
F0 r
FX
sin ωt − 0 2m cos ωt
2
ωZ m
ωZ m
(3.10.2)
⎛ ωm − s ω ⎞
where, X m = ωm − s ω , Z m = r 2 + (ωm − s ω ) 2 , φ = tan −1 ⎜
⎟
r
⎝
⎠
By taking the displacement x as the component represented by curve (a) in Figure
3.9, i.e. by taking the second term of equation (3.10.2) as the expression of x , we
have:
x=−
F0 X m
F0 m(ω02 − ω 2 )
cos
ω
t
cos ωt
=
ωZ m2
m 2 (ω02 − ω 2 ) 2 + ω 2 r 2
(3.10.3)
The damped oscillatory electron equation can be written as:
m&x& + rx& + mω02 x = −eE0 cos ωt
(3.10.4)
Comparing (3.10.1) with (3.10.4), we can immediately find the displacement x for a
damped oscillatory electron by substituting F0 = −eE0 into (3.10.3), i.e.:
x=−
eE0 m(ω02 − ω 2 )
cos ωt
m 2 (ω02 − ω 2 ) 2 + ω 2 r 2
(3.10.5)
By substitution of (3.10.5) into (3.9.3), we can find the expression of χ for a
damped oscillatory electron is given by:
© 2008 John Wiley & Sons, Ltd
χ =−
nex
ne 2 m(ω02 − ω 2 )
=
ε 0 E ε 0 [m 2 (ω02 − ω 2 ) 2 + ω 2 r 2 ]
So we have:
ne 2 m(ω02 − ω 2 )
εr = 1+ χ = 1+
cos ωt
ε 0 [m 2 (ω02 − ω 2 ) 2 + ω 2 r 2 ]
3.11
The instantaneous power dissipated is equal to the product of frictional force and the
instantaneous velocity, i.e.:
P = (rx& ) x& = r
F02
cos 2 (ωt − φ )
Z m2
The period for a given frequency ω is given by:
2π
T=
ω
Therefore, the energy dissipated per cycle is given by:
T
2π ω
0
0
E = ∫ Pdt = ∫
=∫
r
2π ω
0
F02
cos 2 (ωt − φ )dt
2
Zm
rF02
[1 − cos 2(ωt − φ )]dt
2 Z m2
2π rF02
=
ω 2Z m2
(3.11.1)
πrF02
=
ωZ m2
The displacement is given by:
x=
F0
sin(ωt − φ )
ωZ m
so we have:
xmax =
F0
ωZ m
(3.11.2)
By substitution of (3.11.2) into (3.11.1), we have:
2
E = πrωxmax
3.12
The low frequency limit of the bandwidth of the resonance absorption curve ω1
satisfies the equation:
© 2008 John Wiley & Sons, Ltd
ω1m − s ω1 = −r
which defines the phase angle given by:
tan φ1 =
ω1m − s ω1
r
= −1
The high frequency limit of the bandwidth of the resonance absorption curve ω2
satisfies the equation:
ω2 m − s ω2 = r
which defines the phase angle given by:
tan φ2 =
ω2 m − s ω2
r
=1
3.13
For a LCR series circuit, the current through the circuit is given by
I = I 0eiωt
The voltage across the inductance is given by:
dI
d
L
= L I 0eiωt = iωLI 0 eiωt = iωLI
dt
dt
i.e. the amplitude of voltage across the inductance is:
VL = ωLI 0
(3.13.1)
The voltage across the condenser is given by:
1
1
q 1
iI
I 0eiωt = −
= ∫ Idt = ∫ eiωt dt =
C C
C
iωC
ωC
i.e. the amplitude of the voltage across the condenser is:
I
VC = 0
ωC
(3.13.2)
When an alternating voltage, amplitude V0 is applied across LCR series circuit,
current amplitude I 0 is given by: I 0 = V0 Z e , where the impedance Z e is given
by:
1 ⎞
⎛
Z m = R + ⎜ ωL −
⎟
ωC ⎠
⎝
2
2
At current resonance, I 0 has the maximum value:
I0 =
and the resonant frequency ω0 is given by:
© 2008 John Wiley & Sons, Ltd
V0
R
(3.13.3)
ω0 L −
1
1
= 0 or ω 0 =
ω0 C
LC
(3.13.4)
By substitution of (3.12.3) and (3.12.4) into (3.12.1), we have:
ωL
VL = 0 V0
R
By substitution of (3.12.3) and (3.12.4) into (3.12.2), we have:
VC =
L V0 ω0 L
=
V0
R
LC R
V0
LC
L V0
=
V0 =
=
ω0 RC
RC
C R
Noting that the quality factor of an LCR series circuit is given by:
ωL
Q= 0
R
so we have:
VL = VC = QV0
3.14
In a resonant LCR series circuit, the potential across the condenser is given by:
I
(3.14.1)
VC =
ωC
where, I is the current through the whole LCR series circuit, and is given by:
I = I 0eiωt
(3.14.2)
The current amplitude I 0 is given by:
I0 =
V0
Ze
(3.14.3)
where, V0 is the voltage amplitude applied across the whole LCR series circuit and is
a constant. Z e is the impedance of the whole circuit, given by:
1 ⎞
⎛
Z e = R 2 + ⎜ ωL −
⎟
ωC ⎠
⎝
2
From (3.14.1), (3.14.2), (3.14.3), and (3.14.4) we have:
VC =
V0
1 ⎞
⎛
Cω R 2 + ⎜ ωL −
⎟
ωC ⎠
⎝
which has the maximum value when
© 2008 John Wiley & Sons, Ltd
2
eiωt = VC 0eiωt
dVC 0
= 0 , i.e.:
dω
(3.14.4)
d
dω
V0
1 ⎞
⎛
Cω R 2 + ⎜ ω L −
⎟
ωC ⎠
⎝
2
=0
2
1 ⎞
1
⎛
2 2
R + ⎜ ωL −
⎟ +ω L − 2 2 = 0
ωC ⎠
ωC
⎝
2
i.e.
L
=0
C
i.e.
R 2 + 2ω 2 L2 − 2
i.e.
1
R2
1
− 2 = ω0 1 − Q02
LC 2 L
2
where ω02 =
ω=
ωL
1
, Q0 = 0
LC
R
3.15
In a resonant LCR series circuit, the potential across the inductance is given by:
VL = ωLI
(3.15.1)
where, I is the current through the whole LCR series circuit, and is given by:
I = I 0eiωt
(3.15.2)
The current amplitude I 0 is given by:
I0 =
V0
Ze
(3.15.3)
where, V0 is the voltage amplitude applied across the whole LCR series circuit and is
a constant. Z e is the impedance of the whole circuit, given by:
1 ⎞
⎛
Z e = R + ⎜ ωL −
⎟
ωC ⎠
⎝
2
2
From (3.15.1), (3.15.2), (3.15.3), and (3.15.4) we have:
VL =
ωLV0
1 ⎞
⎛
R 2 + ⎜ ωL −
⎟
ωC ⎠
⎝
which has the maximum value when
© 2008 John Wiley & Sons, Ltd
2
eiωt = VL 0eiωt
dVL 0
= 0 , i.e.:
dω
(3.15.4)
ωLV0
d
dω
1 ⎞
⎛
R 2 + ⎜ ωL −
⎟
ωC ⎠
⎝
2
=0
2
1 ⎞
1
⎛
2 2
R + ⎜ ωL −
⎟ −ω L + 2 2 = 0
ωC ⎠
ωC
⎝
2
i.e.
R2 +
i.e.
i.e.
where ω02 =
ω=
1
=
R 2C 2
LC −
2
1
LC
2
L
−2 = 0
2
ωC
C
2
1
1
ω0
= ω0
=
2
2
RC
R
1
1−
1− 2 2
1 − Q02
2L
2 L ω0
2
ωL
1
, Q0 = 0
LC
R
3.16
Considering an electron in an atom as a lightly damped simple harmonic oscillator,
we know its resonance absorption bandwidth is given by:
r
(3.16.1)
δω =
m
On the other hand, the relation between frequency and wavelength of light is given
by:
c
(3.16.2)
f =
λ
where, c is speed of light in vacuum. From (3.16.2) we find at frequency resonance:
δf = −
c
λ20
δλ
where λ0 is the wavelength at frequency resonance. Then, the relation between
angular frequency bandwidth δω and the width of spectral line δλ is given by:
δω = 2π δf =
2πc
λ20
δλ
From (3.16.1) and (3.16.3) we have:
δλ =
λ20 r
λr λ
= 0 = 0
2πcm ω0 m Q
So the width of the spectral line from such an atom is given by:
δλ =
© 2008 John Wiley & Sons, Ltd
λ0
Q
=
0.6 × 10 −6
= 1.2 × 10 −14 [m]
7
5 × 10
(3.16.3)
3.17
According to problem 3.6, the displacement resonance frequency ωr and the
xmax are given by:
corresponding displacement amplitude
ωr = ω02 −
r2
2m 2
F0
ωZ m
=
xmax =
ω =ωr
where, Z m = r 2 + (ωm − s ω ) 2 , ω ′ = ω02 −
F0
ω ′r
r2
, ω0 =
4m 2
s
m
Now, at half maximum displacement:
F0
x
F
= max = 0
2
2ω ′r
ωZ m
ω r 2 + (ωm − s ω ) 2 = 2ω ′r
i.e.
2
⎡ 2 ⎛
⎛s
s⎞ ⎤
r2 ⎞
ω ⎢r + ⎜ ωm − ⎟ ⎥ = 4⎜⎜ − 2 ⎟⎟r 2
ω ⎠ ⎥⎦ ⎝ m 4m ⎠
⎝
⎢⎣
2
i.e.
ω4 +
i.e.
(r 2 − 2sm) 2 s 2 4sr 2 r 4
ω + 2 − 3 + 4 =0
m2
m
m
m
2
i.e.
2
⎛s
r2 ⎞ 2 ⎛ s
r 2 ⎞ 3r 2 ⎛ s
r2 ⎞
ω − 2⎜⎜ − 2 ⎟⎟ω + ⎜⎜ − 2 ⎟⎟ − 2 ⎜⎜ − 2 ⎟⎟ = 0
⎝ m 2m ⎠ m ⎝ m 4m ⎠
⎝ m 2m ⎠
4
2
i.e.
i.e.
⎛ 3r 2 ⎞
(ω − ω ) − ⎜⎜
ω ′ ⎟⎟ = 0
m
⎝
⎠
2
2 2
r
ω 2 − ωr2 = ±
3r 2
ω′
m
(3.17.1)
If ω1 and ω2 are the two solutions of equation (3.17.1), and ω2 > ω1 , then:
ω22 − ωr2 =
ω12 − ω r2 = −
© 2008 John Wiley & Sons, Ltd
3r 2
ω′
m
3r 2
ω′
m
(3.17.2)
(3.17.3)
Since the Q-value is high, we have:
Q=
ω0 m
r
ω 02 >>
i.e.
>> 1
r2
m2
ωr ≈ ω ′ ≈ ω0
i.e.
Then, from (3.17.2) and (3.17.3) we have:
(ω 2 − ω1 )(ω 2 + ω1 ) ≈
2 3r 2
ω0
m
ω1 + ω 2 ≈ 2ω 0
and
Therefore, the width of displacement resonance curve is given by:
ω 2 − ω1 ≈
3r
m
3.18
In Figure 3.9, curve (b) corresponds to absorption, and is given by:
x=
F0 r
F ωr
sin ωt = 2 2 0 2 2
sin ωt
2
ωZ m
m (ω0 − ω ) + ω 2 r 2
and the velocity component corresponding to absorption is given by:
v = x& =
F0ω 2 r
cos ωt
m 2 (ω02 − ω 2 ) 2 + ω 2 r 2
For Problem 3.10, the velocity component corresponding to absorption can be given
by substituting F0 = −eE0 into the above equation, i.e.:
v = x& = −
eE0ω 2 r
cos ωt
m 2 (ω02 − ω 2 ) 2 + ω 2 r 2
(3.18.1)
For an electron density of n , the instantaneous power supplied equal to the product
of the instantaneous driving force − neE0 cos ωt and the instantaneous velocity, i.e.:
⎛
⎞
eE0ω 2 r
⎜
⎟⎟
cos
P = (−neE0 cos ωt ) × ⎜ − 2 2
ω
t
2 2
2 2
(
)
m
ω
−
ω
+
ω
r
0
⎝
⎠
2 2 2
ne E0 ω r
= 2 2
cos 2 ωt
m (ω0 − ω 2 ) 2 + ω 2 r 2
© 2008 John Wiley & Sons, Ltd
The average power supplied per unit volume is then given by:
ω 2π ω
Pdt
2π ∫0
ω 2π ω
ne 2 E02ω 2 r
=
cos 2 ωt
2
2
2 2
2 2
∫
0
2π
m (ω0 − ω ) + ω r
Pav =
=
ne 2 E02
ω 2r
2 m 2 (ω02 − ω 2 ) 2 + ω 2 r 2
which is also the mean rate of energy absorption per unit volume.
© 2008 John Wiley & Sons, Ltd
SOLUTIONS TO CHAPTER 4
4.1
The kinetic energy of the system is the sum of the separate kinetic energy of the two
masses, i.e.:
Ek =
1 2 1 2 1 ⎡1
1
1
⎤ 1
mx& + my& = m ⎢ ( x& + y& ) 2 + ( x& − y& ) 2 ⎥ = mX& 2 + mY& 2
2
2
2 ⎣2
2
4
⎦ 4
The potential energy of the system is the sum of the separate potential energy of the
two masses, i.e.:
1 mg 2 1
1 mg 2 1
Ep =
x + s( y − x)2 +
y + s( x − y)2
2 l
2
2 l
2
1 mg 2
( x + y 2 ) + s( x − y)2
=
2 l
1 mg ⎡ 1
1
⎤
2
x
y
(
)
( x − y)2 ⎥ + s( x − y)2
=
+
+
⎢
2 l ⎣2
2
⎦
=
1 mg 2 ⎛ 1 mg
⎞
X +⎜
+ s ⎟Y 2
4 l
⎝2 l
⎠
Comparing the expression of Ek and E p with the definition of E X and EY given
by (4.3a) and (4.3b), we have:
mg
1
1
mg
, c = m , and d =
a = m, b =
+s
4l
2l
2
4
Noting that:
12
12
12
12
⎛m⎞
⎛m⎞
X q = ⎜ ⎟ ( x + y) = ⎜ ⎟ X
⎝2⎠
⎝2⎠
⎛m⎞
⎛m⎞
Yq = ⎜ ⎟ ( x − y ) = ⎜ ⎟ Y
⎝2⎠
⎝c⎠
and
⎛m⎞
X =⎜ ⎟
⎝2⎠
i.e.
−1 2
Xq
⎛m⎞
and Y = ⎜ ⎟
⎝2⎠
−1 2
Yq
we have the kinetic energy of the system given by:
2
2
1 ⎡ 2 & ⎤
1 ⎡ 2 &⎤
1
1
E k = aX& + cY& = m ⎢
X q ⎥ + m⎢
Yq ⎥ = X& q2 + Y&q2
4 ⎣ m ⎦
4 ⎣ m ⎦
2
2
2
2
and
2
2
mg ⎡ 2 ⎤ ⎛ mg
1 g 2 ⎛ g 2s ⎞ 2
⎞⎡ 2 ⎤
E p = bX + dY =
+ s ⎟⎢
Xq⎥ + ⎜
Yq ⎥ =
X q + ⎜ + ⎟Yq
⎢
4l ⎣ m ⎦ ⎝ 2l
2 l
⎠⎣ m ⎦
⎝l m⎠
2
2
© 2008 John Wiley & Sons, Ltd
which are the expressions given by (4.4a) and (4.4b)
4.2
The total energy of Problem 4.1 can be written as:
1
1
1 mg 2
E = Ek + E p = mx& 2 + my& 2 +
( x + y 2 ) + s( x − y) 2
2
2
2 l
The above equation can be rearranged as the format:
E = ( Ekin + E pot ) x + ( Ekin + E pot ) y + ( E pot ) xy
where, ( Ekin + E pot ) x =
1 2 ⎛ mg
1
⎞
⎛ mg
⎞
+ s ⎟ x 2 , ( Ekin + E pot ) y = my& 2 + ⎜
+ s ⎟ y2 ,
mx& + ⎜
2
2
l
2
2
l
⎝
⎠
⎝
⎠
and E pot = −2 sxy
4.3
x = −2a , y = 0 :
-X
+
≡
-2a
y=0
Y
+
-a
-a
a
-a
x = 0 , y = −2 a :
-X
-
≡
x=0
-2a
+
-a
-a
4.4
For mass m1 , Newton’s second law gives:
m1&x&1 = sx
For mass m2 , Newton’s second law gives:
m2 &x&2 = − sx
© 2008 John Wiley & Sons, Ltd
Y
-a
a
Provided x is the extension of the spring and l is the natural length of the spring,
we have:
x2 − x1 = l + x
By elimination of x1 and x2 , we have:
−
s
s
x−
x = &x&
m2
m1
&x& +
i.e.
m1 + m2
sx = 0
m1m2
which shows the system oscillate at a frequency:
ω2 =
s
μ
where,
μ=
m1m2
m1 + m2
For a sodium chloride molecule the interatomic force constant s is given by:
s = ω 2μ =
(2πν ) 2 mNa mCl 4π 2 × (1.14 × 1013 ) 2 × (23 × 35) × (1.67 × 10−27 ) 2
=
≈ 120[ Nm −1 ]
mNa + mCl
(23 + 35) × 1.67 × 10 −27
4.5
If the upper mass oscillate with a displacement of x and the lower mass oscillate
with a displacement of y , the equations of motion of the two masses are given by
Newton’s second law as:
m&x& = s ( y − x) − sx
m&y& = s ( x − y )
i.e.
m&x& + s ( x − y ) − sx = 0
m&y& − s ( x − y ) = 0
Suppose the system starts from rest and oscillates in only one of its normal modes of
frequency ω , we may assume the solutions:
x = Aeiωt
y = Beiωt
where A and B are the displacement amplitude of x and y at frequency ω .
© 2008 John Wiley & Sons, Ltd
Using these solutions, the equations of motion become:
[−mω 2 A + s( A − B) + sA]eiωt = 0
[−mω 2 B − s( A − B)]eiωt = 0
We may, by dividing through by me iωt , rewrite the above equations in matrix form
as:
⎡2 s m − ω 2
⎢
⎣ −s m
− s m ⎤ ⎡ A⎤
⎥⎢ ⎥ = 0
s m − ω 2 ⎦ ⎣ B⎦
(4.5.1)
which has a non-zero solution if and only if the determinant of the matrix vanishes;
that is, if
(2s m − ω 2 )( s m − ω 2 ) − s 2 m 2 = 0
i.e.
ω 4 − (3s m) ω 2 + s 2 m 2 = 0
ω 2 = (3 ± 5 )
i.e.
s
2m
In the slower mode, ω 2 = (3 − 5 ) s 2m . By substitution of the value of frequency
into equation (4.5.1), we have:
A
s
s − mω 2
5 −1
=
=
=
2
B 2 s − mω
s
2
which is the ratio of the amplitude of the upper mass to that of the lower mass.
Similarly, in the slower mode, ω 2 = (3 + 5 ) s 2m . By substitution of the value of
frequency into equation (4.5.1), we have:
A
s
s − mω 2
5 +1
=
=
=−
2
B 2 s − mω
s
2
4.6
The motions of the two pendulums in Figure 4.3 are given by:
(ω − ω1 )t
(ω + ω2 )t
x = 2a cos 2
cos 1
= 2a cos ωmt cos ωat
2
2
(ω − ω1 )t
(ω + ω2 )t
y = 2a sin 2
sin 1
= 2a sin ωmt sin ωat
2
2
where, the amplitude of the two masses, 2a cos ωmt and 2a sin ωmt , are constants
over one cycle at the frequency ωa .
Supposing the spring is very weak, the stiffness of the spring is ignorable, i.e. s ≈ 0 .
© 2008 John Wiley & Sons, Ltd
Noting that ω12 = g l and ω22 = ( g l + 2s m) , we have:
g
⎛ ω + ω2 ⎞
2
= ω12 ≈ ω22 ≈ ⎜ 1
⎟ = ωa
l
⎝ 2 ⎠
2
Hence, the energies of the masses are given by:
1
1 mg
(2a cos ωmt )2 = 2ma 2ωa2 cos 2 ωmt
s x a x2 =
2
2 l
1
1 mg
(2a sin ωmt )2 = 2ma 2ωa2 sin 2 ωmt
E y = s y a y2 =
2
2 l
Ex =
The total energy is given by:
E = Ex + E y = 2ma 2ωa2 (cos2 ωmt + sin 2 ωmt ) = 2ma 2ωa2
Noting that ωm = (ω2 − ω1 ) 2 , we have:
E
[1 + cos(ω2 − ω1 )t ]
2
E
E y = 2ma 2ωa2 sin 2 (ω2 − ω1 )t = [1 − cos(ω2 − ω1 )t ]
2
which show that the constant energy E is completed exchanged between the two
E x = 2ma 2ωa2 cos 2 (ω2 − ω1 )t =
pendulums at the beat frequency (ω2 − ω1 ) .
4.7
By adding up the two equations of motion, we have:
m1&x& + m2 &y& = −(m1 x + m2 y )( g l )
By multiplying the equation by 1 (m1 + m2 ) on both sides, we have:
m1 &x& + m2 &y&
g m1 x + m2 y
=−
m1 + m2
l m1 + m2
m1 &x& + m2 &y& g m1 x + m2 y
+
=0
m1 + m2
l m1 + m2
i.e.
which can be written as:
X&& + ω12 X = 0
(4.7.1)
where,
X =
© 2008 John Wiley & Sons, Ltd
m1 x + m2 y
and
m1 + m2
ω12 = g l
On the other hand, the equations of motion can be written as:
&x& = −
g
s
x − ( x − y)
l
m1
&y& = −
g
s
y+
( x − y)
l
m2
By subtracting the above equations, we have:
⎛ s
g
s ⎞
&x& − &y& = − ( x − y ) − ⎜⎜ +
⎟⎟( x − y )
l
⎝ m1 m2 ⎠
⎡g ⎛ 1
1 ⎞⎤
&x& − &y& + ⎢ + s⎜⎜ +
⎟⎟⎥ ( x − y ) = 0
⎝ m1 m2 ⎠⎦
⎣l
i.e.
which can be written as:
Y&& + ω22Y = 0
(4.7.2)
where,
Y = x− y
and
ω22 =
g ⎛ 1
1 ⎞
⎟⎟
+ s⎜⎜ +
l
⎝ m1 m2 ⎠
Equations (4.7.1) and (4.7.2) take the form of linear differential equations with
constant coefficients and each equation contains only one dependant variable,
therefore X and Y are normal coordinates and their normal frequencies are given
by ω1 and ω2 respectively.
4.8
Since the initial condition gives x& = y& = 0 , we may write, in normal coordinate, the
solutions to the equations of motion of Problem 4.7 as:
X = X 0 cos ω1t
Y = Y0 cos ω2t
i.e.
m1 x + m2 y
= X 0 cos ω1t
m1 + m2
x − y = Y0 cos ω2t
By substitution of initial conditions: t = 0 , x = A and y = 0 into the above
equations, we have:
X 0 = ( m1 M ) A
Y0 = A
© 2008 John Wiley & Sons, Ltd
M = m1 + m2
where,
so the equations of motion in original coordinates x , y are given by:
m1 x + m2 y m1
=
A cos ω1t
m1 + m2
M
x − y = A cos ω2t
The solutions to the above equations are given by:
A
x=
(m1 cos ω1t + m2 cos ω2t )
M
m
y = A 1 (cos ω1t − cos ω2t )
M
Noting that
ω1 = ωa − ωm
and
ω2 = ωa + ωm , where ωm = (ω2 − ω1 ) 2
and
ωa = (ω1 + ω2 ) 2 , the above equations can be rearranged as:
A
[m1 cos(ωa − ωm )t + m2 cos(ωa + ωm )t ]
M
A
= [m1 (cos ωmt cos ωa t + sin ωmt sin ωat ) + m2 (cos ωmt cos ωa t − sin ωmt sin ωat )]
M
A
= [(m1 + m2 ) cos ωmt cos ωa t + (m1 − m2 ) sin ωmt sin ωa t ]
M
A
= A cos ωmt cos ωat + (m1 − m2 ) sin ωmt sin ωat
M
x=
and
m1
[cos(ωa − ωm )t − cos(ωa + ωm )t ]
M
m
= 2 A 1 sin ωmt sin ωat
M
y=A
4.9
From the analysis in Problem 4.6, we know, at weak coupling conditions, cos ωmt
and sin ωmt are constants over one cycle, and the relation: ωa ≈ g l , so the energy
of the mass m1 , E x , and the energy of the mass m2 , E y , are the sums of their
separate kinetic and potential energies, i.e.:
1
1
1
1 mg 2 1
1
m1 x& 2 + s x x 2 = m1 x& 2 +
x = m1 x& 2 + m1ωa2 x 2
2
2
2
2 l
2
2
1
1
1
1 mg 2 1
1
E y = m21 y& 2 + s y y 2 = m2 y& 2 +
y = m1 y& 2 + m1ωa2 y 2
2
2
2
2 l
2
2
Ex =
© 2008 John Wiley & Sons, Ltd
By substitution of the expressions of x and y in terms of cos ωa t and sin ωa t
given by Problem 4.8 into the above equations, we have:
2
Ex =
A
1 ⎡
⎤
m1 ⎢− Aωa cos ωmt sin ωa t + ωa (m1 − m2 ) sin ωmt cos ωa t ⎥ +
M
2 ⎣
⎦
A
1
⎡
⎤
m1ωa2 ⎢ A cos ωmt cos ωa t + (m1 − m2 ) sin ωmt sin ωa t ⎥
M
2
⎣
⎦
=
=
=
=
[
2
1
2 A
m1ωa 2 (m1 + m2 ) 2 cos 2 ωmt + (m1 − m2 ) 2 sin 2 ωmt
M
2
A2
1
m1ωa2 2 m12 + m22 + 2m1m2 (cos 2 ωmt − sin 2 ωmt )
M
2
A2
1
m1ωa2 2 m12 + m22 + 2m1m2 cos 2ωmt
M
2
E
m12 + m22 + 2m1m2 cos(ω2 − ω1 )t
2
M
[
2
]
]
[
]
[
]
and
2
1 ⎡ m
⎤ 1
⎡ m
⎤
E y = m2 ⎢2 A 1 ωa sin ωmt cos ωat ⎥ + m1ωa2 ⎢2 A 1 sin ωmt sin ωa t ⎥
2 ⎣ M
⎦ 2
⎣ M
⎦
[
A2
sin 2 ωmt (cos 2 ωat + sin 2 ωat )
M2
A2
= 2m12 m2ωa2 2 sin 2 ωmt
M
⎛1
⎞⎛ 2m m ⎞
= ⎜ m1ωa2 A2 ⎟⎜ 1 2 2 ⎟[1 − cos 2ωmt ]
⎝2
⎠⎝ M ⎠
= 2m12 m2ωa2
2
]
⎛ 2m m ⎞
= E ⎜ 1 2 2 ⎟[1 − cos(ω2 − ω1 )t ]
⎝ M ⎠
where,
E=
1
m1ωa2 A2
2
4.10
Add up the two equations and we have:
mg
m( &x& + &y&) +
( x + y ) + r ( x& + y& ) = F0 cos ωt
l
mg
i.e.
mX&& + rX& +
X = F0 cos ωt
l
Subtract the two equations and we have:
mg
m( &x& − &y&) +
( x − y ) + r ( x& − y& ) + 2s( x − y ) = F0 cos ωt
l
© 2008 John Wiley & Sons, Ltd
(4.10.1)
i.e.
⎛ mg
⎞
mY&& + rY& + ⎜
+ 2s ⎟Y = F0 cos ωt
⎝ l
⎠
(4.10.2)
Equations (4.10.1) and (4.10.2) shows that the normal coordinates X and Y are
those for damped oscillators driven by a force F0 cos ωt .
By neglecting the effect of r , equation of (4.10.1) and (4.10.2) become:
mg
X ≈ F0 cos ωt
l
⎞
⎛ mg
mY&& + ⎜
+ 2s ⎟Y ≈ F0 cos ωt
⎠
⎝ l
mX&& +
Suppose the above equations have solutions: X = X 0 cos ωt and Y = Y0 cos ωt , by
substitution of the solutions to the above equations, we have:
mg ⎞
⎛
2
⎜ − mω +
⎟ X 0 cos ωt ≈ F0 cos ωt
l ⎠
⎝
mg
⎛
⎞
2
+ 2 s ⎟Y0 cos ωt ≈ F0 cos ωt
⎜ − mω +
l
⎝
⎠
These equations satisfy any t if
mg ⎞
⎛
2
⎜ − mω +
⎟ X 0 ≈ F0
l ⎠
⎝
mg
⎛
⎞
2
+ 2 s ⎟Y0 ≈ F0
⎜ − mω +
l
⎝
⎠
X0 ≈
i.e.
Y0 ≈
F0
m( g l − ω 2 )
F0
m( g l + 2 s m − ω 2 )
so the expressions of X and Y are given by:
X = x+ y ≈
F0
cos ωt
m( g l − ω 2 )
Y = x− y ≈
F0
cos ωt
m( g l + 2 s m − ω 2 )
By solving the above equations, the expressions of x and y are given by:
© 2008 John Wiley & Sons, Ltd
⎡
⎤
F0
1
1
cos ω t ⎢ 2
+ 2
2
2 ⎥
2m
ω2 − ω ⎦
⎣ ω1 − ω
⎡
⎤
F
1
1
y ≈ 0 cos ω t ⎢ 2
− 2
2
2 ⎥
2m
ω2 − ω ⎦
⎣ ω1 − ω
x≈
where,
ω12 =
g
l
ω22 =
and
g 2s
+
l m
The ratio of y x is given by:
⎡ 1
F0
1 ⎤
1
1
+ 2
cos ωt ⎢ 2
+ 2
2
2⎥
2
2
2
y 2m
ω22 − ω12
⎣ ω1 − ω ω2 − ω ⎦ = ω1 − ω ω2 − ω =
≈
1
1
x
ω22 + ω12 − 2ω 2
⎡ 1
F0
1 ⎤
−
−
cos ωt ⎢ 2
2
2
2
2
2
2
2⎥
2m
⎣ ω1 − ω ω2 − ω ⎦ ω1 − ω ω2 − ω
y
x
1
ω22 + ω12
ω −ω
ω +ω
2
2
2
2
2
1
2
1
2
0
ω1
ω2
ω
-1
The behaviour of y x as a function of frequency ω is shown as the figure below:
The figure shows y x is less than 1 if ω < ω1 or ω > ω2 , i.e. outside frequency
range ω2 − ω1 the motion of y is attenuated.
4.11
Suppose the displacement of mass M is x , the displacement of mass m is y ,
and the tension of the spring is T . Equations of motion give:
M&x& + kx = F0 cos ωt + T
m&y& = −T
s ( y − x) = T
© 2008 John Wiley & Sons, Ltd
(4.11.1)
(4.11.2)
(4.11.3)
Eliminating T , we have:
M&x& + kx = F0 cos ωt + s ( y − x)
so for x = 0 at all times, we have
F0 cos ωt + sy = 0
that is
y=−
F0
cos ωt
s
Equation (4.11.2) and (4.11.3) now give:
m&y& + sy = 0
with ω 2 = s m , so M is stationary at ω 2 = s m .
This value of ω
satisfies all equations of motion for
x=0
including
T = − F0 cos ωt
4.12
Noting the relation: V = q C , the voltage equations can be written as:
q1 q2
dI
− =L a
C C
dt
q2 q3
dI
− =L b
C C
dt
so we have:
q&1 − q& 2 = LCI&&a
q& − q& = LCI&&
2
3
b
i.e.
q&1 − q&2 = LCI&&a
q& − q& = LCI&&
2
3
b
By substitution of q&1 = − I a , q& 2 = I a − I b and q&3 = I b into the above equations, we
have:
− I a − I a + I b = LCI&&a
I − I − I = LCI&&
a
b
b
b
i.e.
LCI&&a + 2 I a − I b = 0
LCI&&b − I a + 2 I b = 0
© 2008 John Wiley & Sons, Ltd
By adding up and subtracting the above equations, we have:
LC ( I&& + I&& ) + I + I = 0
a
b
a
b
LC ( I&&a − I&&b ) + 3( I a − I b ) = 0
Supposing the solutions to the above normal modes equations are given by:
I a + I b = A cos ωt
I a − I b = B cos ωt
so we have:
(− Aω 2 LC + A) cos ωt = 0
(− Bω 2 LC + 3B) cos ωt = 0
which are true for all t when
ω2 =
1
LC
and
B=0
or
3
and
A=0
LC
which show that the normal modes of oscillation are given by:
1
I a = I b at ω12 =
LC
and
3
I a = − I b at ω22 =
LC
ω2 =
4.13
From the given equations, we have the relation between I1 and I 2 given by:
I2 =
iω M
I1
Z 2 + iωLs
so:
⎛
ω 2M ⎞
⎟ I1
E = iωL p I1 − iωMI 2 = ⎜⎜ iωL p +
Z 2 + iωLs ⎟⎠
⎝
i.e.
E
ω 2M 2
= iωL p +
I1
Z 2 + iωLs
which shows that E I1 , the impedance of the whole system seen by the generator, is
the sum of the primary impedance, iωL p , and a ‘reflected impedance’ from the
© 2008 John Wiley & Sons, Ltd
secondary circuit of ω 2 M 2 Z s , where Z s = Z 2 + iωLs .
4.14
Problem 4.13 shows the impedance seen by the generator Z is given by:
ω 2M 2
Z = iωL p +
Z 2 + iωLs
Noting that M = L p Ls and L p Ls = n 2p ns2 , the impedance can be written as:
Z=
iωL p Z 2 − ω 2 L p Ls + ω 2 M 2
Z 2 + iωLs
=
iω L p Z 2 − ω 2 M 2 + ω 2 M 2
Z 2 + iωLs
=
iωL p Z 2
Z 2 + iωLs
so we have:
1 Z 2 + iωLs
1
1
1
1
=
==
+
=
+ 2
Z
iωL p Z 2
iωL p L p
iωL p n p
Z2
Z2
Ls
ns2
which shows the impedance Z is equivalent to the primary impedance iωL p
connected in parallel with an impedance (n p ns ) 2 Z 2 .
4.15
Suppose a generator with the internal impedance of Z1 is connected with a load with
an impedance of Z 2 via an ideal transformer with a primary inductance of L p and
the ratio of the number of primary and secondary transformer coil turns given by
n p ns , and the whole circuit oscillate at a frequency of ω . From the analysis in
Problem 4.13, the impedance of the load is given by:
1
1
1
=
+ 2
Z L iω L p n p
Z2
ns2
At the maximum output power: Z L = Z1 , i.e.:
1
1
1
1
=
+ 2
=
Z L iωL p n p
Z1
Z2
2
ns
which is the relation used for matching a load to a generator.
© 2008 John Wiley & Sons, Ltd
4.16
From the second equation, we have:
I1 = −
Z2
I2
ZM
By substitution into the first equation, we have:
−
Z1Z 2
I2 + ZM I2 = E
ZM
I2 =
i.e.
E
I1
⎛
Z1Z 2 ⎞
⎟
⎜⎜ Z M −
Z M ⎟⎠
⎝
Noting that Z M = iωM and I 2 has the maximum value when X 1 = X 2 = 0 , i.e.
Z1 = R1 and Z 2 = R2 , we have:
I2 =
E
iωM −
R1 R2
iω M
I1 =
E
ωM +
R1R2
ωM
E
I1 ≤
2 ωM
which shows I 2 has the maximum value of
R1 R2
ωM
I1 =
E
I1
2 R1R2
RR
E
I1 , when ωM = 1 2 , i.e.
ωM
2 R1R2
ωM = R1 R2
4.17
By substitution of j = 1 and n = 3 into equation (4.15), we have:
⎡
2⎤
π⎤
⎡
2
ω12 = 2ω02 ⎢1 − cos ⎥ = 2ω02 ⎢1 −
⎥ = (2 − 2 )ω0
4
2
⎣
⎦
⎣
⎦
By substitution of j = 2 and n = 3 into equation (4.15), we have:
⎡
⎣
ω12 = 2ω02 ⎢1 − cos
2π ⎤
= 2ω02
⎥
4 ⎦
By substitution of j = 3 and n = 3 into equation (4.15), we have:
⎡
⎣
ω12 = 2ω02 ⎢1 − cos
© 2008 John Wiley & Sons, Ltd
⎡
3π ⎤
2⎤
2
= 2ω02 ⎢1 +
⎥ = (2 + 2 )ω0
⎥
4⎦
2
⎣
⎦
In equation (4.14), we have A0 = A4 = 0 when n = 3 , and noting that ω02 = T ma ,
equation (4.14) gives:
⎛
ω2 ⎞
− A0 + ⎜⎜ 2 − 2 ⎟⎟ A1 − A2 = 0
ω0 ⎠
⎝
when r = 1 :
⎛
ω2 ⎞
⎜⎜ 2 − 2 ⎟⎟ A1 − A2 = 0
ω0 ⎠
⎝
(4.17.1)
⎛
ω2 ⎞
− A1 + ⎜⎜ 2 − 2 ⎟⎟ A2 − A3 = 0
ω0 ⎠
⎝
(4.17.2)
i.e.
when r = 2 :
⎛
ω2 ⎞
− A2 + ⎜⎜ 2 − 2 ⎟⎟ A3 − A4 = 0
ω0 ⎠
⎝
when r = 3 :
⎛
ω2 ⎞
⎜
− A2 + ⎜ 2 − 2 ⎟⎟ A3 = 0
ω0 ⎠
⎝
i.e.
(4.17.3)
Write the above equations in matrix format, we have:
⎛ 2 − ω 2 ω02
⎞⎛ A1 ⎞
−1
0
⎜
⎟⎜ ⎟
−1
−1
2 − ω 2 ω02
⎜
⎟⎜ A2 ⎟ = 0
⎜
2
2 ⎟⎜
−1
0
2 − ω ω0 ⎠⎝ A3 ⎟⎠
⎝
which has non zero solutions provided the determinant of the matrix is zero, i.e.:
(2 − ω 2 ω02 )3 − 2(2 − ω 2 ω02 ) = 0
The solutions to the above equations are given by:
ω12 = (2 − 2 )ω02 ,`
ω22 = 2ω02 , and
ω12 = (2 + 2 )ω02
4.18
By substitution of ω12 = (2 − 2 )ω02 into equation (4.17.1), we have:
2 A1 − A2 = 0
i.e.
A1 : A2 = 1 : 2
By substitution of ω12 = (2 − 2 )ω02 into equation (4.17.3), we have:
− A2 + 2 A3 = 0 i.e.
A2 : A3 = 2 : 1
Hence, when ω12 = (2 − 2 )ω02 , the relative displacements are given by:
A1 : A2 : A3 = 1 : 2 : 1
By substitution of ω22 = 2ω02 into equation (4.17.1), we have:
© 2008 John Wiley & Sons, Ltd
A2 = 0
By substitution of ω22 = 2ω02 into equation (4.17.2), we have:
− A1 + A3 = 0
i.e.
A1 : A3 = 1 : −1
Hence, when ω22 = 2ω02 , the relative displacements are given by:
A1 : A2 : A3 = 1 : 0 : −1
By substitution of ω22 = (2 + 2 )ω02 into equation (4.17.1), we have:
− 2 A1 − A2 = 0 i.e.
A1 : A2 = 1 : − 2
By substitution of ω12 = (2 + 2 )ω02 into equation (4.17.3), we have:
− A2 − 2 A3 = 0 i.e.
A2 : A3 = − 2 : 1
Hence, when ω12 = (2 + 2 )ω02 , the relative displacements are given by:
A1 : A2 : A3 = 1 : − 2 : 1
The relative displacements of the three masses at different normal frequencies are
shown below:
ω 2 = (2 − 2 )ω02
ω 2 = 2ω02
ω 2 = (2 + 2 )ω02
As we can see from the above figures that tighter coupling corresponds to higher
frequency.
4.19
Suppose the displacement of the left mass m is x , and that of the central mass M
is y , and that of the right mass m is z . The equations of motion are given by:
© 2008 John Wiley & Sons, Ltd
m&x& = s ( y − x)
M&y& = − s ( y − x) + s ( z − y )
m&z& = − s ( z − y )
If the system has a normal frequency of ω , and the displacements of the three masses
can by written as:
x = η1eiωt
y = η 2eiωt
z = η3eiωt
By substitution of the expressions of displacements into the above equations of
motion, we have:
− mω 2η1eiωt = s (η 2 − η1 )eiωt
− Mω 2η 2eiωt = − s (η 2 − η1 )eiωt + s (η3 − η 2 )eiωt
− mω 2η3eiωt = − s (η3 − η 2 )eiωt
i.e.
[( s − mω 2 )η1 − sη 2 ]eiωt = 0
[− sη1 + (2s − Mω 2 )η 2 − sη3 ]eiωt = 0
[− sη 2 + ( s − mω 2 )η3 ]eiωt = 0
which is true for all t if
( s − mω 2 )η1 − sη 2 = 0
− sη1 + (2s − Mω 2 )η 2 − sη3 = 0
− sη 2 + ( s − mω 2 )η3 = 0
The matrix format of these equations is given by:
⎛ s − mω 2
0 ⎞⎛ η1 ⎞
−s
⎟⎜ ⎟
⎜
2
2s − Mω
− s ⎟⎜η 2 ⎟ = 0
⎜ −s
⎜ 0
s − mω 2 ⎟⎠⎜⎝η3 ⎟⎠
−s
⎝
which has non zero solutions if and only if the determinant of the matrix is zero, i.e.:
( s − mω 2 ) 2 (2s − Mω 2 ) − 2s 2 ( s − mω 2 ) = 0
i.e.
( s − mω 2 )[(s − mω 2 )(2s − Mω 2 ) − 2s 2 ] = 0
i.e.
( s − mω 2 )[mMω 4 − s ( M + 2m)ω 2 ] = 0
i.e.
ω 2 ( s − mω 2 )[mMω 2 − s ( M + 2m)] = 0
The solutions to the above equation, i.e. the frequencies of the normal modes, are
given by:
s
s ( M + 2m)
ω12 = 0 , ω22 =
and ω32 =
m
mM
© 2008 John Wiley & Sons, Ltd
At the normal mode of ω 2 = 0 , all the atoms are stationary, η1 = η 2 = η3 , i.e. all the
masses has the same displacement;
s
At the normal mode of ω 2 = , η 2 = 0 and η1 = −η3 , i.e. the mass M is
m
stationary, and the two masses m have the same amplitude but are “anti-phase” with
respect to each other;
s ( M + 2m)
At the normal mode of ω 2 =
, η1 : η 2 : η3 = M : −2m : M , i.e. the two
mM
mass m have the same amplitude and are “in-phase” with respect to each other.
They are both “anti-phase” with respect to the mass M . The ratio of amplitude
between the mass m and M is M 2m .
4.20
In understanding the motion of the masses it is more instructive to consider the range
n 2 ≤ j ≤ n . For each value of the frequency ω j the amplitude of the r th mass is
Ar = C sin
rjπ
where C is a constant. For j = n 2 adjacent masses have a π 2
n +1
phase difference, so the ratios: Ar −1 : Ar : Ar +1 = −1 : 0 : 1 , with the r th masses
stationary and the amplitude Ar −1 anti-phase with respect to Ar +1 , so that:
Ar → 0
Ar +1
Ar
Ar −1
j=n 2
j→n
As n 2 → n , Ar begins to move, the coupling between masses tightens and when
j is close to n each mass is anti-phase with respect to its neighbour, the amplitude
of each mass decreases until in the limit j = n no motion is transmitted as the cut off
frequency ω 2j = 4T ma is reached. The end points are fixed and this restricts the
motion of the masses near the end points at all frequencies except the lowest.
4.21
By expansion of the expression of ω 2j , we have:
© 2008 John Wiley & Sons, Ltd
ω 2j =
⎤
jπ ⎞ 2T ⎡ ( jπ n + 1) 2 ( j π n + 1) 4 ( jπ n + 1) 6
2T ⎛
−
+
− L⎥
⎜1 − cos
⎟=
⎢
ma ⎝
n + 1 ⎠ ma ⎣
2!
4!
6!
⎦
If n >> 1 and j << n , jπ n + 1 has a very small value, so the high order terms of
the above equation can be neglected, so the above equation become:
2T ⎡ ( jπ n + 1) 2 ⎤ T ⎛ jπ ⎞
ω =
⎥ = ma ⎜ n + 1 ⎟
ma ⎢⎣
2!
⎝
⎠
⎦
2
2
j
ωj =
i.e.
jπ
T
n + 1 ma
which can be written as:
ωj =
jπ
l
T
ρ
where, ρ = m a and l = (n + 1)a
4.22
From the first equation, we have:
LI&&r −1 =
q& r −1 − q& r
C
By substitution of q& r = I r −1 − I r and q& r −1 = I r −2 − I r −1 into the above equation, we
have:
I r −2 − 2 I r −1 + I r
(4.22.1)
C
If, in the normal mode, the currents oscillate at a frequency ω , we may write the
displacements as:
LI&&r −1 =
I r −2 = Ar −2eiωt , I r −1 = Ar −1eiωt and I r = Ar eiωt
Using these values of I in equation (4.22.1) gives:
A − 2 Ar −1 + Ar iωt
− ω 2 LAr −1eiωt = r −2
e
C
or
− Ar −2 + (2 − LCω 2 ) Ar −1 − Ar = 0
(4.22.2)
By comparison of equation (4.22.2) with equation (4.14) in text book, we may find
the expression of I r is the same as that of yr in the case of mass-loaded string, i.e.
rjπ iωt
e
n +1
Where D is constant, and the frequency ω is given by:
I r = Ar eiωt = D sin
© 2008 John Wiley & Sons, Ltd
ω 2j =
1 ⎛
jπ ⎞
⎜1 − cos
⎟
LC ⎝
n +1⎠
where, j = 1,2,3...n
4.23
By substitution of y into
∂2 y
, we have:
∂t 2
∂ 2 y ∂ 2 iωt ikx
= 2 (e e ) = −ω 2ei (ωt + kx )
2
∂t
∂t
By substitution of y into
∂2 y
, we have:
∂x 2
∂ 2 y ∂ 2 iωt ikx
=
(e e ) = − k 2ei (ωt +kx )
∂x 2 ∂x 2
If ω = ck , we have:
∂2 y 2 ∂2 y
−c
= (−ω 2 + c 2 k 2 )ei (ωt +kx ) = 0
2
2
∂t
∂x
i.e.
2
∂2 y
2 ∂ y
=
c
∂t 2
∂x 2
© 2008 John Wiley & Sons, Ltd
SOLUTIONS TO CHAPTER 5
5.1
Write u = ct + x , and try
∂2 y
with y = f 2 (ct + x) , we have:
∂x 2
∂y ∂f 2 (u )
∂ 2 y ∂ 2 f 2 (u )
=
, and
=
∂x 2
∂u 2
∂u
∂x
Try
1 ∂2 y
with y = f 2 (ct + x) , we have:
c 2 ∂t 2
2
∂y
∂f (u )
∂2 y
2 ∂ f 2 (u )
c
=c 2
, and
=
∂t 2
∂u 2
∂t
∂u
so:
1 ∂ 2 y 1 2 ∂ 2 f 2 (u ) ∂ 2 f 2 (u )
= c
=
c 2 ∂t 2 c 2
∂u 2
∂u 2
Therefore:
∂2 y 1 ∂2 y
=
∂x 2 c 2 ∂t 2
5.2
If y = f1 (ct − x) , the expression for y at a time t + Δt and a position x + Δx , where
Δt = Δx c , is given by:
yt +Δt , x+ Δx = f1[c(t + Δt ) − ( x + Δx)]
= f1[c(t + Δx c) − ( x + Δx)]
= f1[ct + Δx − x − Δx]
= f1[ct − x] = yt , x
i.e. the wave profile remains unchanged.
If y = f 2 (ct + x) , the expression for y at a time t + Δt and a position x + Δx , where
Δt = − Δx c , is given by:
© 2008 John Wiley & Sons, Ltd
yt + Δt , x + Δx = f1[c(t + Δt ) + ( x + Δx)]
= f1[c(t − Δx c) + ( x + Δx)]
= f1[ct − Δx + x + Δx]
= f1[ct + x] = yt , x
i.e. the wave profile also remains unchanged.
5.3
y
x
∂y
∂y
=c
∂t
∂x
x
5.4
The pulse shape before reflection is given by the graph below:
l
The pulse shapes after of a length of Δl of the pulse being reflected are shown below:
(a) Δl = l 4
1
l
2
© 2008 John Wiley & Sons, Ltd
3
l
4
Z1
Z2 = ∞
(b) Δl = l 2
Z1
Z2 = ∞
Z1
Z2 = ∞
Z1
Z2 = ∞
(c) Δl = 3l 4
3
l
4
1
l
2
(d) Δl = l
l
5.5
The boundary condition yi + yr = yt gives:
A1ei (ωt −kx ) + B1ei (ωt +kx ) = A2ei (ωt −kx )
At x = 0 , this equation gives:
A1 + B1 = A2
© 2008 John Wiley & Sons, Ltd
(5.5.1)
The boundary condition Ma = T
∂
∂
yt − T ( yi + yr ) gives:
∂x
∂x
Ma = −ikTA2 e i (ωt −kx ) + ikTA1e i (ωt −kx ) − ikTB1e i (ωt + kx )
At x = 0 , a = &y&t = &y&i + &y&r , so the above equation becomes:
− ω 2 MA2 = −iω
i
i.e.
T
T
T
A2 + iω A1 − iω B1
c
c
c
T
T
T⎞
⎛
A1 − i B1 = ⎜ − ωM + i ⎟ A2
c
c
c⎠
⎝
Noting that T c = ρc , the above equation becomes:
iρcA1 − iρcB1 = (− ωM + iρc )A2
(5.5.2)
By substitution of (5.5.1) into (5.5.2), we have:
iρcA1 − iρcB1 = (− ωM + iρc )( A1 + B1 )
i.e.
− iq
B1
=
A1 1 + iq
where q = ωM 2 ρc
By substitution of the above equation into (5.5.1), we have:
A1 −
iq
A1 = A2
1 + iq
i.e.
A2
1
=
A1 1 + iq
5.6
Writing q = tan θ , we have:
A2
1
1
cosθ
=
=
=
= cosθe −iθ
A1 1 + iq 1 + i tan θ cosθ + i sin θ
and
− iq
− i tan θ
− i sin θ
B1
=
=
=
= sin θe −i (θ +π
A1 1 + iq 1 + i tan θ cosθ + i sin θ
which show that A2 lags A1 by
© 2008 John Wiley & Sons, Ltd
2)
θ and that B1 lags A1 by (π 2 + θ ) for 0 < θ < π 2
The reflected energy coefficients are given by:
2
B1
= sin θe −i (θ + π
A1
2) 2
= sin 2 θ
and the transmitted energy coefficients are given by:
2
2
A2
= cos θe −iθ = cos 2 θ
A1
5.7
Suppose T is the tension of the string, the average rate of working by the force over one period
of oscillation on one-wavelength-long string is given by:
ω
2π
W =
2π ω 1 k
∫ ∫
0
0
−T
∂y ∂y
dxdt
∂x ∂t
By substitution of y = a sin(ωt − kx) into the above equation, we have:
ω 2π ω 1 k
− T [−ka sin(ωt − kx)][ωa sin(ωt − kx)]dxdt
2π ∫0 ∫0
ω 2 k 2 a 2T 2π ω 1 k 2
=
∫0 ∫0 sin (ωt − kx)dxdt
2π
ω 2 k 2 a 2T 2π ω 1 k 1 − cos(2ωt − 2kx)
dxdt
=
∫0 ∫0
2π
2
ω 2 k 2 a 2T 1 2π 1
=
⋅ ⋅
⋅
2π
2 ω k
ωka 2T
W=
=
2
Noting that k = ω c and T = ρc , the above equation becomes
2
W=
ω 2 a 2 ρc 2
2c
=
ω 2 a 2 ρc
2
which equals the rate of energy transfer along the string.
5.8
Suppose the wave equation is given by: y = sin(ωt − kx) . The maximum value of transverse
harmonic force Fmax is given by:
TAω
⎛ ∂y ⎞
⎡∂
⎤
Fmax = T ⎜ ⎟ = T ⎢ A sin(ωt − kx)⎥ = TAk =
c
⎝ ∂x ⎠ max
⎦ max
⎣ ∂x
i.e.
© 2008 John Wiley & Sons, Ltd
0.3
0.3
T Fmax
=
=
=
c
Aω 0.1× 2π × 5 π
Noting that
ρc = T c , the rate of energy transfer along the string is given by:
P=
ρcω 2 A2
2
=
1 T 2 2 1 0.3
3π
× (2π × 5) 2 × 0.12 =
[W ]
ω A = ×
2c
2 π
20
so the velocity of the wave c is given by:
c=
2P
2 × 3π 20
30
=
= [ms −1 ]
2 2
2
2
0.01× (2π × 5) × 0.1
ρω A
π
5.9
This problem is not viable in its present form and it will be revised in the next printing. The first
part in the zero reflected amplitude may be solved by replacing Z3 by Z1, which then equates r
with R′ because each is a reflection at a Z1Z 2 boundary. We then have the total reflected
amplitude as:
R + tTR′(1 + R′2 + R′4 + L) = R +
tTR′
1 − R′ 2
Stokes’ relations show that the incident amplitude may be reconstructed by reversing the paths of
the transmitted and reflected amplitudes.
T is transmitted back along the incident direction as tT in Z1 and is reflected as TR′ in
Z2 .
R is reflected in Z1 as ( R) R = R 2 back along the incident direction and is refracted as TR
in the TR′ direction in Z 2 .
We therefore have tT + R = 1 in Z1 , i.e. tT = 1 − R
2
2
and T ( R + R′) = 0 in Z 2 giving
R = − R′ , ∴ tT = 1 − R 2 = 1 − R′2 giving the total reflected amplitude in Z1 as R + R′ = 0
with R = − R′ .
1
θ1 θ1
θ2
R
R2
tT
TR ′
T
θ1 θ1
θ2 θ2
TR
R
T
Z1
Z2
Z1
Fig Q.5.9(a)
© 2008 John Wiley & Sons, Ltd
Fig Q.5.9(b)
Note that for zero total reflection in medium Z1, the first reflection R is cancelled by the sum of all
subsequent reflections.
5.10
The impedance of the anti-reflection coating Z coat should have a relation to the impedance of air
Z air and the impedance of the lens Z lens given by:
Z coat = Z air Z lens =
1
nair nlens
So the reflective index of the coating is given by:
ncoat =
1
Z coat
= nair nlens = 1.5 = 1.22
and the thickness of the coating d should be a quarter of light wavelength in the coating, i.e.
d=
λ
4ncoat
=
5.5 × 10 −7
= 1.12 × 10 −7 [m]
4 × 1.22
5.11
By substitution of equation (5.10) into
∂y
, we have:
∂x
∂y ωn
ωt
=
( An cos ωnt + Bn sin ωnt ) cos n
c
c
∂x
so:
∂2 y
ωn2
ωnt
ωn2
=
−
(
A
cos
t
+
B
sin
t
)
sin
=
−
y
ω
ω
n
n
n
n
c2
c
c2
∂x 2
Noting that k =
ωn
c
, we have:
∂2 y
ωn2
ωn2
2
+
k
y
=
−
y
+
y=0
c2
c2
∂x 2
5.12
2
By substitution of the expression of ( yn ) max into the integral, we have:
© 2008 John Wiley & Sons, Ltd
l
l
ωx
1
1
ρωn2 ∫ ( yn2 ) max dx = ρωn2 ( An2 + Bn2 ) ∫ sin 2 n dx
0
0
2
2
c
l 1 − cos( 2ω x c )
1
n
dx
= ρωn2 ( An2 + Bn2 ) ∫
0
2
2
⎛
1
2ω l ⎞
c
sin n ⎟⎟
= ρωn2 ( An2 + Bn2 )⎜⎜ l −
4
c ⎠
⎝ 2ωn
Noting that
ωn =
nπc
2ωnl
, i.e. sin
= sin 2nπ = 0 , the above equation becomes:
l
c
l
1
1
ρωn2 ∫ ( yn2 ) max dx = ρlωn2 ( An2 + Bn2 )
0
2
4
which gives the expected result.
5.13
Expand the expression of y ( x, t ) , we have:
y ( x, t ) = A cos(ωt − kx) + rA cos(ωt + kx)
= A cos ωt cos kx + A sin ωt sin kx + rA cos ωt cos kx − rA sin ωt sin kx
= A(1 + r ) cos ωt cos kx + A(1 − r ) sin ωt sin kx
which is the superposition of standing waves.
5.14
The wave group has a modulation envelope of:
Δk ⎞
⎛ Δω
A = A0 cos⎜
t−
x⎟
2 ⎠
⎝ 2
where Δω = ω1 − ω2 is the frequency difference and Δk = k1 − k2 is the wave number
difference. At a certain time t , the distance between two successive zeros of the modulation
envelope Δx satisfies:
Δk
Δx = π
2
Noting that k = 2π
λ , for a small value of Δλ λ , we have: Δk ≈ (2π λ2 )Δλ , so the above
equation becomes:
2πΔλ
Δx ≈ π
2λ2
i.e.
Δx ≈
which shows that the number of wavelengths
modulating envelop is ≈ λ Δλ
© 2008 John Wiley & Sons, Ltd
λ
λ
Δλ
λ contained between two successive zeros of the
5.15
The expression for group velocity is given by:
vg =
dω d
dv
=
(kv) = v + k
dk dk
dk
By substitution of the expression of v into the above equation, we have:
⎡ sin( ka 2) ⎤
⎢c ka 2 ⎥
⎣
⎦
2
sin( ka 2)
(ka 4) cos(ka 2) − (a 2) sin( ka 2)
=c
+ ck
ka 2
(ka 2) 2
vg = c
d
sin( ka 2)
+k
ka 2
dk
ka
sin( ka 2)
sin( ka 2)
+ c cos − c
ka 2
ka 2
2
ka
= c cos
2
At long wavelengths, i.e. k → 0 , the limiting value of group velocity is the phase velocity c .
=c
5.16
Noting that the group velocity of light in gas is given on page 131 as:
⎛
λ ∂ε r
Vg = v⎜⎜1 +
⎝ 2ε r ∂λ
⎞
⎟⎟
⎠
we have:
⎛
λ ∂ε r ⎞
λ ∂ε r ⎞
⎟⎟ε r = v⎛⎜ ε r +
Vg ε r = v⎜⎜1 +
⎟
2 ∂λ ⎠
⎝
⎝ 2ε r ∂λ ⎠
⎡⎛
B
B
⎞ λ ∂ ⎛
2 ⎞⎤
= v ⎢⎜ A + 2 − Dλ2 ⎟ +
⎜ A + 2 − Dλ ⎟⎥
λ
λ
⎠⎦
⎠ 2 ∂λ ⎝
⎣⎝
⎡⎛
B
⎞ λ ⎛ 2B
⎞⎤
= v ⎢⎜ A + 2 − Dλ2 ⎟ + ⎜ − 3 − 2 Dλ ⎟⎥
λ
⎠ 2⎝ λ
⎠⎦
⎣⎝
⎡⎛
B
⎞ ⎛ B
⎞⎤
= v ⎢⎜ A + 2 − Dλ2 ⎟ + ⎜ − 2 − Dλ2 ⎟⎥
λ
⎠ ⎝ λ
⎠⎦
⎣⎝
= v( A − 2 Dλ2 )
5.17
c2
⎛ω ⎞
The relation ε r = 2 = 1 − ⎜ e ⎟ gives:
v
⎝ω⎠
2
ω 2c 2
v
2
= ω 2 − ωe2
By substitution of v = ω k , the above equation becomes:
© 2008 John Wiley & Sons, Ltd
ω 2 = ωe2 + c 2 k 2
As
(5.17.1)
ω → ωe , we have:
c2
⎛ω ⎞
=1− ⎜ e ⎟ <1
2
v
⎝ω ⎠
2
i.e. v > c , which means the phase velocity exceeds that of light c .
From equation (5.17.1), we have:
d (ω 2 ) = d (ωe2 + c 2 k 2 )
2ωdω = 2kc 2 dk
i.e.
which shows the group velocity v g is given by:
vg =
dω
k c2 c
= c2 =
= c<c
dk
ω v v
i.e. the group velocity is always less than c .
5.18
From equation (5.17.1), we know that only electromagnetic waves of
ω > ωe can propagate
through the electron plasma media.
20
For an electron number density ne ~ 10 , the electron plasma frequency is given by:
ne
10 20
−19
= 1.6 × 10 ×
= 5.65 × 1011[rad ⋅ s −1 ]
ωe = e
−31
−12
meε 0
9.1× 10 × 8.8 × 10
Now consider the wavelength of the wave in the media given by:
λ=
v 2πv 2πv 2πc 2π × 3 × 108
=
<
<
=
= 3 × 10 −3[m]
ω
ωe
ωe
f
5.65 × 1011
−3
which shows the wavelength has an upper limit of 3 × 10 m .
5.19
The dispersion relation
ω 2 c 2 = k 2 + m 2c 2 h 2 gives
d (ω 2 c 2 ) = d (k 2 + m 2c 2 h 2 )
i.e.
© 2008 John Wiley & Sons, Ltd
2ω
dω = 2kdk
c2
ω dω
i.e.
= c2
k dk
Noting that the group velocity is dω dk and the particle (phase) velocity is
ω k , the above
2
equation shows their product is c .
5.20
The series in the problem is that at the bottom of page 132. The frequency components can be
expressed as:
R = na
sin(Δω ⋅ t 2)
cos ω t
Δω ⋅ t 2
which is a symmetric function to the average frequency
ω0 . It shows that at Δt =
∴ Δt ⋅ Δω = 2π
In k space, we may write the series as:
2π
, R = 0,
Δω
y (k ) = a cos k1 x + a cos(k1 + δk ) x + L + a cos[k1 + (n − 1)δk ]x
As an analogy to the above analysis, we may replace ω by k and t by x , and R is zero at
Δx =
2π
, i.e. ΔkΔx = 2π
Δk
5.21
The frequency of infrared absorption of NaCl is given by:
ω=
2T ⎛ 1
1 ⎞
1
1
⎞
⎜⎜
⎟⎟ = 2 × 15 × ⎛⎜
+
+
= 3.608 × 1013[rad ⋅ s −1 ]
− 27
− 27 ⎟
a ⎝ mNa mCl ⎠
35 × 1.66 × 10 ⎠
⎝ 23 × 1.66 × 10
The corresponding wavelength is given by:
2πc
2π × 3 × 108
λ=
=
≈ 52[ μm]
ω
3.608 × 1013
which is close to the experimental value: 61μm
The frequency of infrared absorption of KCl is given by:
ω=
2T ⎛ 1
1 ⎞
1
1
⎞
⎜⎜
⎟⎟ = 2 × 15 × ⎛⎜
+
+
= 3.13 × 1013[rad ⋅ s −1 ]
−27
−27 ⎟
a ⎝ mK mCl ⎠
35 × 1.66 × 10 ⎠
⎝ 39 × 1.66 × 10
The corresponding wavelength is given by:
2πc
2π × 3 × 108
λ=
=
≈ 60[ μm]
ω
3.13 × 1013
which is close to the experimental value: 71μm
© 2008 John Wiley & Sons, Ltd
5.22
Before the source passes by the observer, the source has a velocity of u , the frequency noted by
the observer is given by:
ν1 =
c
ν
c−u
After the source passes by the observer, the source has a velocity of − u , the frequency noted by
the observer is given by:
ν2 =
c
ν
c+u
So the change of frequency noted by the observer is given by:
c ⎞
2νcu
⎛ c
Δν = ν 2 −ν 1 = ⎜
−
⎟ν = 2
(c − u 2 )
⎝ c−u c +u ⎠
5.23
By superimposing a velocity of − v on the system, the observer becomes stationary and the
source has a velocity of u − v and the wave has a velocity of c − v . So the frequency registered
by the observer is given by:
ν ′′′ =
5.24
The relation between wavelength
So the Doppler Effect
ν′ =
c−v
c−v
=
ν
c − v − (u − v) c − u
λ and frequency ν of light is given by:
c
ν=
λ
νc
c −u
can be written in the format of wavelength as:
c
c2
=
λ ′ λ (c − u )
λ′ =
i.e.
c −u
λ
c
λ ′ > λ , so we have:
u
Δλ = λ ′ − λ = − λ
c
Noting that wavelength shift is towards red, i.e.
i.e.
u=−
cΔλ
λ
=−
3 × 108 × 10 −11
= −5[ Kms −1 ]
−7
6 × 10
−1
which shows the earth and the star are separating at a velocity of 5 Kms .
5.25
© 2008 John Wiley & Sons, Ltd
Suppose the aircraft is flying at a speed of u , and the signal is being transmitted from the aircraft
at a frequency of ν and registered at the distant point at a frequency of
Effect gives:
ν ′ =ν
ν ′ . Then, the Doppler
c
c −u
Now, let the distant point be the source, reflecting a frequency of ν ′ and the flying aircraft be the
receiver, registering a frequency of ν ′′ . By superimposing a velocity of − u on the flying
aircraft, the distant point and signal waves, we bring the aircraft to rest; the distant point now has a
velocity of − u and signal waves a velocity of − c − u . Then, the Doppler Effect gives:
ν ′′ = ν ′
−c −u
c+u
c+u
=ν
=ν ′
− c − u − ( −u )
c
c −u
which gives:
u=
Δν
15 × 103
ν ′′ −ν
c=
c=
× 3 × 108 = 750[ms −1 ]
2ν + Δν
2 × 3 × 109
ν ′′ +ν
i.e. the aircraft is flying at a speed of 750 m s
5.26
Problem 5.24 shows the Doppler Effect in the format of wavelength is given by:
λ′ =
c−u
λ
c
where u is the velocity of gas atom. So we have:
Δλ = λ ′ − λ =
u
λ
c
i.e.
Δλ
2 × 10 −12
u = λ′ − λ =
c=
× 3 × 108 = 1× 103[ms −1 ]
−7
6 × 10
λ
The thermal energy of sodium gas is given by:
1
3
mNa u 2 = kT
2
2
where k = 1.38 × 10
−23
[ JK −1 ] is Boltzmann’s constant, so the gas temperature is given by:
T=
mNau 2 23 × 1.66 × 10−27 × 10002
=
≈ 900[ K ]
3k
3 × 1.38 × 10 −23
5.27
A point source radiates spherical waves equally in all directions.
⎛ vc ⎞
v′ = ⎜
⎟ : Observer is at rest with a moving source.
⎝ c − u′ ⎠
© 2008 John Wiley & Sons, Ltd
u
θ
u′ = u cos θ
s′
o′
⎛ c − v′ ⎞
v′′ = ⎜
⎟ : Source at rest with a moving observer.
⎝ c ⎠
v
θ
v′ = v cosθ
s′
o′
⎛ c − v′ ⎞
v′′′ = ⎜
⎟ : Source and observer both moving.
⎝ c − u′ ⎠
u
v
s′
θ
u′ = u cosθ
α
v′ = v cosθ
o′
5.28
By substitution of equation (2) into (3) and eliminating x′ , we can find the expression of t ′
given by:
t′ =
1⎡x
⎤
− k ( x − vt )⎥
⎢
v ⎣ k′
⎦
Now we can eliminate x′ and t ′ by substituting the above equation and the equation (2) into
equation (1), i.e.
x 2 − c 2t 2 = k 2 ( x − vt ) 2 −
c2 ⎡ x
⎤
− k ( x − vt )⎥
2 ⎢
v ⎣ k′
⎦
2
i.e.
2
⎡
⎡ 2 2 ⎛ c2 ⎞ 2 ⎤ 2
⎡
c2 ⎛ 1
c2 ⎛ 1
⎞ ⎤ 2
⎞⎤
2
⎢1 − k + 2 ⎜ − k ⎟ ⎥ x + 2kv ⎢k + 2 ⎜ − k ⎟⎥ xt + ⎢k v ⎜⎜ 2 − 1⎟⎟ − c ⎥t = 0
v ⎝ k′
v ⎝ k′
⎠ ⎦⎥
⎠⎦
⎣
⎝v
⎠
⎣
⎦
⎣⎢
which is true for all x and t if and only if the coefficients of all terms are zeros, so we have:
1⎞
c2 ⎛
1− k = 2 ⎜ k − ⎟
v ⎝
k′ ⎠
2
⎛ c2 ⎞
c2
⎜⎜ 2 − 1⎟⎟kk ′ = 2
v
⎝v
⎠
© 2008 John Wiley & Sons, Ltd
2
k 2 (c 2 − v 2 ) = c 2
The solution to the above equations gives:
k = k′ =
where,
1
1− β 2
β =v c
5.29
Source at rest at x1 in O frame gives signals at intervals measured by O as Δt = t 2 − t1
where t 2 is later than t1 . O′ moving with velocity v with respect to O measures these
intervals as:
v
Δx) with Δx = 0
c2
∴ Δt ′ = kΔt
t2′ − t1′ = Δt ′ = k (Δt −
l = ( x2 − x1 ) as seen by O , O′ sees it as ( x′2 − x1′ ) = k[( x2 − x1 ) − v(t 2 − t1 )] .
Measuring l ′ puts t 2′ = t1′ or Δt ′ = 0
v
v
⎡
⎤
∴ Δt ′ = k ⎢Δt − 2 ( x2 − x1 )⎥ = 0 i.e. Δt = 2 ( x2 − x1 ) = t 2 − t1
c
c
⎣
⎦
⎡
⎤ x −x
v2
′
′
′
∴ l = x2 − x1 = k[( x2 − x1 ) − v(Δt )] = k ⎢( x2 − x1 ) − 2 ( x2 − x1 )⎥ = 2 1
c
k
⎣
⎦
∴ l′ = l k
5.30
Two events are simultaneous (t1 = t 2 ) at x1 and x2 in O frame. They are not simultaneous
in O′ frame because:
v ⎞
v ⎞
⎛
⎛
t1′ = k ⎜ t1 − 2 x1 ⎟ ≠ t2′ = k ⎜ t 2 − 2 x2 ⎟ i.e. x1 ≠ x2
c
c
⎝
⎠
⎝
⎠
5.31
The order of cause followed by effect can never be reversed.
2 events x1 , t1 and x2 , t 2 in O frame with t 2 > t1 i.e. t 2 − t1 > 0 ( t 2 is later).
© 2008 John Wiley & Sons, Ltd
v
v
⎡
⎤
⎡
⎤
t2′ − t1′ = k ⎢(t2 − t1 ) − 2 ( x2 − x1 )⎥ i.e. Δt ′ = k ⎢Δt − 2 Δx ⎥ in O′ frame.
c
c
⎣
⎦
⎣
⎦
v ⎛ Δx ⎞
v
Δt ′ real requires k real that is v < c , Δt ′ is + ve if Δt > ⎜ ⎟ where
is + ve
c⎝ c ⎠
c
but < 1 and
Δx
is shortest possible time for signal to traverse Δx .
c
SOLUTIONS TO CHAPTER 6
6.1
Elementary kinetic theory shows that, for particles of mass m in a gas at temperature T , the
energy of each particle is given by:
1 2 3
mv = kT
2
2
where v is the root mean square velocity and k is Boltzmann’s constant.
Page 154 of the text shows that the velocity of sound c is a gas at pressure P is given by:
c2 =
γP γPV γRT γNkT
=
=
=
ρ
M
M
M
where V is the molar volume, M is the molar mass and N is Avogadro’s number, so:
Mc 2 = γNkT = αkT ≈
5
kT
3
6.2
The intensity of sound wave can be written as:
I = P 2 ρ 0c
where P is acoustic pressure,
ρ 0 is air density, and c is sound velocity, so we have:
P = Iρ 0c = 10 × 1.29 × 330 ≈ 65[ Pa]
which is 6.5 × 10 −4 of the pressure of an atmosphere.
6.3
The intensity of sound wave can be written as:
I=
1
ρ 0cω 2η 2
2
where η is the displacement amplitude of an air molecule, so we have:
© 2008 John Wiley & Sons, Ltd
η=
1
2πν
2I
1
2 ×10
=
×
= 6.9 ×10−5 [m]
ρ 0c 2π × 500 1.29 × 330
6.4
The expression of displacement amplitude is given by Problem 6.3, i.e.:
η=
1
2πν
2 × 10 −10 I 0
1
2 × 10 × 10 −10 × 10−2
=
×
≈ 10 −10 [m]
2π × 500
1.29 × 330
ρ0c
6.5
The audio output is the product of sound intensity and the cross section area of the room, i.e.:
P = IA = 100 I 0 A = 100 × 10 −2 × 3 × 3 ≈ 10[W ]
6.6
The expression of acoustic pressure amplitude is given by Problem 6.2, so the ratio of the pressure
amplitude in water and in air, at the same sound intensity, are given by:
pwater
=
pair
I ( ρ 0 c) water
I ( ρ 0 c) air
=
( ρ 0 c) water
1.45 × 10 6
=
≈ 60
( ρ 0 c) air
400
And at the same pressure amplitudes, we have:
I water
( ρ 0 c) air
400
=
=
≈ 3 × 10 −4
6
I air
( ρ 0 c) water 1.45 × 10
6.7
If η is the displacement of a section of a stretched spring by a disturbance, which travels along it
in the x direction, the force at that section is given by: F = Y
∂η
, where Y is young’s
∂x
modulus.
The relation between Y and s , the stiffness of the spring, is found by considering the force
required to increase the length L of the spring slowly by a small amount l << L , the force F
being the same at all points of the spring in equilibrium. Thus
∂η l
⎛Y ⎞
and F = ⎜ ⎟l
=
∂x L
⎝L⎠
If l = x in the stretched spring, we have:
⎛Y ⎞
F = sx = ⎜ ⎟ x and Y = sL .
⎝L⎠
If the spring has mass m per unit length, the equation of motion of a section of length dx is
given by:
© 2008 John Wiley & Sons, Ltd
m
∂ 2η
∂F
∂ 2η
dx
dx
Y
dx
=
=
∂t 2
∂x
∂x 2
∂ 2η Y ∂ 2η sL ∂ 2η
=
=
∂t 2 m ∂x 2 m ∂x 2
or
sL
m
a wave equation with a phase velocity
6.8
At x = 0 ,
η = B sin kx sin ωt
At x = L ,
M
∂ 2η
∂η
= − sL
2
∂t
∂x
− Mω 2 sin kL = − sLk cos kL
i.e.
(which for k = ω v ,
ρ = m L and v = sL ρ from problem 7 when l << L )
becomes:
ωL
v
tan
ωL
v
For M >> m , v >> ωL and writing
=
ρL m
sL2
=
=
2
Mv
M M
(6.8.1)
ωL v = θ where θ is small, we have:
tan θ = θ + θ 3 3 + ...
and the left hand side of equation 6.8.1 becomes
θ 2 [1 + θ 2 3 + ...] = (ωL v) 2 [1 + (ωL v) 2 3 + ...]
Now v = ( sL
ρ )1 2 = ( sL2 m)1 2 = L( s m)1 2 and ωL v = ω m s
So eq. 6.8.1 becomes:
ω 2 m s (1 + ω 2 m 3s + ...) = m M
or
ω 2 (1 + ω 2 m 3s) = s M
Using
(6.8.2)
ω 2 = s M as a second approximation in the bracket of eq. 6.8.2, we have:
© 2008 John Wiley & Sons, Ltd
⎛
⎝
ω 2 ⎜1 +
ω2 =
i.e.
1 m⎞ s
⎟=
3M⎠ M
s
1
M+ m
3
6.9
The Poissons ratio
σ = 0.25 gives:
λ
2(λ + μ )
= 0.25
λ=μ
i.e.
So the ratio of the longitudinal wave velocity to the transverse wave velocity is given by:
λ + 2μ
vl
=
=
vt
μ
μ + 2μ
= 3
μ
In the text, the longitudinal wave velocity of the earth is 8kms −1 and the transverse wave
velocity is 4.45kms −1 , so we have:
λ + 2μ
8
=
4.45
μ
λ = 1.23μ
i.e.
so the Poissons ratio for the earth is given by:
σ=
λ
1.23μ
=
≈ 0.276
2(λ + μ ) 2 × (1.23μ + μ )
6.10
At a plane steel water interface, the energy ratio of reflected wave is given by:
I r ⎛ Z steel − Z water
=⎜
I i ⎜⎝ Z steel + Z water
2
2
⎞ ⎛ 3.9 × 107 − 1.43 × 106 ⎞
⎟⎟ = ⎜⎜
⎟ ≈ 86%
7
6 ⎟
⎠ ⎝ 3.9 × 10 + 1.43 × 10 ⎠
At a plane steel water interface, the energy ratio of transmitted wave is given by:
4 Z ice Z water
4 × 3.49 × 106 × 1.43 × 106
It
=
=
≈ 82.3%
I i ( Z ice + Z water ) 2 (3.49 × 106 + 1.43 × 106 ) 2
6.11
Solution follow directly from the coefficients at top of page 165.
© 2008 John Wiley & Sons, Ltd
Closed end is zero displacement with
Open end:
nr
= 1 (antinode, η is a max)
ni
Pressure: closed end:
Open end :
nr
= −1 (node).
ni
pr
= 1 . Pressure doubles at antinode
pi
pr
= −1 (out of phase – cancels to give zero pressure, i.e. node)
pi
6.12
(a) The boundary condition
∂η
= 0 at x = 0 gives:
∂x
(− Ak sin kx + Bk cos kx) sin ωt x=0 = 0
i.e. B = 0 , so we have: η = A cos kx sin ωt
The boundary condition
∂η
= 0 at x = L gives:
∂x
− kA sin kx sin ωt x=l = 0
kA sin kL sin ωt = 0
2l
2π
which is true for all t if kl = nπ , i.e.
l = nπ or λ =
n
λ
i.e.
The first three harmonics are shown below:
© 2008 John Wiley & Sons, Ltd
η
n = 1:
0
l 2
l
x
l
x
l
x
η
n = 2:
0
l 4
3l 4
η
n = 3:
0
l 6
(b) The boundary condition
l 2
5l 6
∂η
= 0 at x = 0 gives:
∂x
(− Ak sin kx + Bk cos kx) sin ωt x=0 = 0
i.e. B = 0 , so we have: η = A cos kx sin ωt
The boundary condition η = 0 at x = L gives:
A sin kx sin ωt x=l = 0
A cos kl sin ωt = 0
i.e.
⎛
⎝
which is true for all t if kl = ⎜ n +
1⎞
4l
2π
1⎞
⎛
l = ⎜ n + ⎟π or λ =
⎟π , i.e.
2⎠
λ
2⎠
2n + 1
⎝
The first three harmonics are shown below:
© 2008 John Wiley & Sons, Ltd
η
n = 0:
0
l
x
l
x
l
x
η
n = 1:
0
l 3
η
n = 2:
0
l 5
3l 5
6.13
The boundary condition for pressure continuity at x = 0 gives:
[ A1e i (ωt −k1x ) + B1e i (ωt −k1x ) ] x=0 = [ A2 ei (ωt −k2 x ) + B2 e i (ωt −k2 x ) ]x=0
A1 + B1 = A2 + B2
i.e.
(6.13.1)
In acoustic wave, the pressure is given by: p = Zη& , so the continuity of particle velocity η& at
x = 0 gives:
A1ei (ωt −k1x ) + B1ei (ωt −k1x )
Z1
i.e.
=
x =0
A2ei (ωt −k2 x ) + B2ei (ωt −k2 x )
Z2
Z 2 ( A1 − B1 ) = Z1 ( A2 − B2 )
x =0
(6.13.2)
At x = l , the continuity of pressure gives:
[ A2ei (ωt −k2 x ) + B2ei (ωt −k2 x ) ]x=l = A3ei (ωt −k3x )
i.e.
A2 e − ik2l + B2 eik2l = A3
The continuity of particle velocity gives:
© 2008 John Wiley & Sons, Ltd
x =l
(6.13.3)
A2ei (ωt −k2 x ) + B2ei (ωt −k2 x )
Z2
=
x =l
A3ei (ωt −k3 x )
Z3
x =l
Z 3 ( A2 e − ik2l − B2 e ik2l ) = Z 2 A3
i.e.
(6.13.4)
By comparison of the boundary conditions derived above with the derivation in page 121-124, we
can easily find:
Z1 A32
4r31
=
2
2
2
Z 3 A1 (r31 + 1) cos k 2l + (r21 + r32 ) 2 sin 2 k 2l
where r31 =
Z3
Z
Z
, r21 = 2 , and r32 = 3 .
Z1
Z1
Z2
If we choose l = λ2 4 , cos k 2l = 0 and sin k2l = 1 , we have:
Z1 A32
4r31
=
=1
2
Z 3 A1 (r21 + r32 ) 2
when r21 = r32 , i.e.
Z2 Z3
or Z 22 = Z1Z 3 .
=
Z1 Z 2
6.14
The differentiation of the adiabatic condition:
P ⎡ V0 ⎤
=⎢
⎥
P0 ⎣V0 (1 + δ ) ⎦
γ
gives:
∂P ∂p
∂ 2η
=
= −γP0 (1 + δ ) −(γ +1) 2
∂x ∂x
∂x
since
δ = ∂η ∂x .
Since (1 + δ )(1 + s ) = 1 , we may write:
∂ 2η
∂p
= −γP0 (1 + s )γ +1 2
∂x
∂x
and from Newton’s second law we have:
∂p
∂ 2η
= − ρ0 2
∂x
∂t
so that
2
γP
∂ 2η
2
2
γ +1 ∂ η
=
c
(
1
+
s
)
, where c0 = 0
0
2
2
∂t
∂x
ρ0
© 2008 John Wiley & Sons, Ltd
which shows the sound velocity of high amplitude wave is given by c0 (1 + s )
( γ +1) 2
6.15
The differentiation of equation ω 2 = ωe2 + 3aTk 2 gives:
2ω
ω dω
i.e.
k dk
where a represents Boltzmann constant,
dω
= 6aTk
dk
= 3aT
ω k is the phase velocity, dω dk is the group
velocity.
6.16
The fluid is incompressible so that during the wave motion there is no change in the volume of the
fluid element of height h , horizontal length Δx and unit width. The distortion η in the
element Δx is therefore directly translated to a change in its height h and its constant volume
requires that:
hΔx = (h + α )( Δx + Δη ) = hΔx + hΔη + αΔx + αΔη
Because
α << h and Δη << Δx , the second order term αΔη is ignorable, we then have
α = −h Δη Δx = −h ∂η ∂x , and from now on we replace Δx by dx .
We see that for
α + ve (or increase in height), we have ∂η ∂x − ve , that is, a compression.
On page 153 of the text, the horizontal motion of the element is shown to be due to the difference
in forces acting on the opposing faces of the element hΔx , that is:
−
∂F
∂ 2η
Δx = ρh 2 Δx
∂x
∂t
where the force difference, dF is − ve when measured in the + ve
compression.
Thus:
−
∂F
∂ 2η
∂P
dx = ρh 2 dx = − h av dx
∂x
∂t
∂x
x direction for a
(6.16.1)
where the pressure must be averaged over the height of the element because it varies with the
liquid depth. This average value is found from the pressure difference (to unit depth) at the liquid
surface to between the two values of α on the opposing faces of the element. This gives:
dPav = ρgdα
© 2008 John Wiley & Sons, Ltd
dPav
dα
dx = ρg
dx
dx
dx
so
and we have from eq. 6.16.1:
−
dα
∂F
∂P
∂ 2η
∂ 2η
dx = −h av dx = − hρg
dx = h 2 ρg 2 dx = ρh 2 dx
dx
∂x
∂x
∂x
∂t
The last two terms equate to give the wave equation. For horizontal motion as:
∂ 2η
∂ 2η
=
gh
∂t 2
∂x 2
with phase velocity v =
gh .
The horizontal motion translates directly to the vertical displacement
α to give an equation of
wave motion:
∂ 2α
∂ 2α
=
gh
∂t 2
∂x 2
with a similar phase velocity v =
gh
6.17
(a) Since h >> λ , i.e. kh >> 1 , we have: tanh kh ≈ 1 , therefore:
⎡ g Tk ⎤
g Tk
g Tk
gT
v 2 = ⎢ + ⎥ tanh kh ≈ +
≥2
⋅
=2
k ρ
k ρ
ρ
⎣k ρ ⎦
i.e. the velocity has a minimum value given by:
v4 =
when
4 gT
ρ
g Tk
gρ
T
=
, i.e. k 2 =
or λc = 2π
k
ρ
T
ρg
(b) If T is negligible, we have:
v2 ≈
and when
g
tanh kh
k
λ >> λc , k → 0 , and for a shallow liquid, h → 0 . Noting that when hk → 0 ,
tanh kh → kh , we have:
v=
g
tanh kh ≈
k
g
kh = gh
k
(c) For a deep liquid, h → +∞ i.e. tanh kh → 1 , the phase velocity is given by:
v 2p =
© 2008 John Wiley & Sons, Ltd
g
g
i.e. v p =
tanh kh ≈
k
k
g
k
and the group velocity is given by:
vg = v p +
kdv p
1
g
= vp − k 3 =
dk
2 k
g 1 g 1 g
−
=
k 2 k 2 k
(d) For the case of short ripples dominated by surface tension in a deep liquid, i.e.
and h → +∞ , we have:
v 2p = lim
h→+∞
Tk
ρ
tanh kh =
Tk
ρ
i.e. v p =
Tk
ρ
and the group velocity is given by:
vg = v p +
© 2008 John Wiley & Sons, Ltd
kdv p
Tk k
=
+
ρ 2
dk
3 Tk 3
T
=
= v
ρk 2 ρ 2 p
g
Tk
<<
k
ρ
SOLUTIONS TO CHAPTER 7
7.1
The equation
I r −1 − I r =
at the limit of dx → 0 becomes:
d
d
qr = C0 dx Vr
dt
dt
dI
dV
= C0
dx
dt
(7.1.1)
The equation
L0 dx
at the limit of dx → 0 becomes:
d
I r = Vr − Vr +1
dt
∂V
∂I
= L0
∂x
∂t
(7.1.2)
The derivative of equation (7.1.1) on t gives:
∂2I
∂ 2V
= C0 2
∂x∂t
∂t
(7.1.3)
The derivative of equation (7.1.2) on x gives:
∂ 2V
∂2I
=
L
0
∂x 2
∂x∂t
(7.1.4)
Equation (7.1.3) and (7.1.4) give:
∂ 2V
∂ 2V
=
L
C
0 0
∂x 2
∂t 2
The derivative of equation (7.1.1) on x gives:
∂2I
∂ 2V
=
C
0
∂x 2
∂x∂t
(7.1.5)
The derivative of equation (7.1.2) on t gives:
∂ 2V
∂2I
= L0 2
∂x∂t
∂t
(7.1.6)
Equation (7.1.5) and (7.1.6) give:
© 2008 John Wiley & Sons, Ltd
∂2I
∂2I
=
L
C
0 0
∂x 2
∂t 2
7.2
l
2a
I
2d
Fig Q.7.2.1
A pair of parallel wires of circular cross section and radius a are separated at a distance 2d
between their centres.
To find the inductance per unit length we close the circuit by joining the sides of a section of
length l .
The self inductance of this circuit is the magnetic flux through the circuit when a current of 1 amp
flows around it.
If the current is 1 amp the field outside the wire at a distance r from the centre is
where
μ0 I 2πr ,
μ0 is the permeability of free space. For a clockwise current in the circuit (Fig Q.7.2.1)
both wires contribute to the magnetic flux B which points downwards into the page and the total
flux through the circuit is given by:
2l ∫
2 d −2 a
a
μ 0 Idr μ 0lI ⎛ 2d ⎞
ln⎜
=
⎟ for d >> a
π
2πr
⎝ a ⎠
Hence the self inductance per unit length is:
L=
μ 0 ⎛ 2d ⎞
ln⎜ ⎟
π ⎝ a ⎠
To find the capacitance per unit length of such a pair of wires we first find the electrostatic
potential at a distance r from a single wire and proceed to find the potential from a pair of wires
via the principle of electrostatic images.
If the radius of the wire is a and it carries a charge of
λ per unit length then the electrostatic
flux E per unit length of the cylindrical surface is: 2πrE ( r ) = λ
ε 0 , where ε 0 is the
permittivity of free space. Thus E ( r ) = λ 2πε 0 r for r > a and we have the potential:
© 2008 John Wiley & Sons, Ltd
φ (r ) = −
λ
λ
⎛b⎞
ln(r ) + constant =
ln⎜ ⎟ for φ = 0 at r = b .
2πε 0
2πε 0 ⎝ r ⎠
y
p
r
a
r
+λ
−λ
o
r
θ
x
p
p
d
d
Fig Q.7.2.2
The conducting wires are now represented in the image system of Fig Q.7.2.2. The equipotential
surfaces will be seen to be cylindrical but not coaxial with the wires. Neither the electric field nor
the charge density is uniform on the conducting surface.
The surface charge is collapsed onto two line carrying charges ± λ per unit length. The y axis
represents an equipotential plane.
The conducting wires, of radius a , are centred a distance d from the origin ο ( x = y = 0) .
The distances ± p can be chosen of the line charges so that the conducting surfaces lie on the
equipotentials of the image charge. Choosing the potential to be zero at ∞ on the y axis, the
potential at point p in the xy plane is given by:
φp =
⎛1⎞
⎛1⎞
⎛ r2 ⎞
λ
λ
λ
ln⎜⎜ ⎟⎟ −
ln⎜⎜ ⎟⎟ =
ln⎜⎜ 22 ⎟⎟
2πε 0 ⎝ r1 ⎠ 2πε 0 ⎝ r2 ⎠ 4πε 0 ⎝ r1 ⎠
In Fig Q.7.2.2:
r22 = 2αγ + δ
r12 = 2βγ + δ
where α = ( d + p ) ; β = ( d − p ) ; γ = ( d + r cos θ ) and
δ = (r 2 + p 2 − d 2 ) .
If the position of the image charge is such that p = ( d − a ) , then, at r = a :
2
φ (a) =
2
2
⎛d + p⎞
λ
⎟ , independent of θ , and the right-hand conductor is an equipotential of
ln⎜⎜
4πε 0 ⎝ d − p ⎟⎠
the image charge. Symmetry requires that the potential at the surface of the other conductor is
© 2008 John Wiley & Sons, Ltd
− φ (a ) and the potential difference between the conductors is:
V=
⎛d + p⎞
λ
⎟
ln⎜⎜
2πε 0 ⎝ d − p ⎟⎠
Gauss’s theorem applied to one of the equipotentials surrounding each conductor proves that the
surface charge on each conductor is equal to the image charge.
The capacitance per unit length is now given by:
C=
λ
V
=
2πε 0
2πε 0
for d >> a
≈
⎛ 2d ⎞
⎛d + p⎞
⎟⎟ ln⎜ ⎟
ln⎜⎜
⎝ a ⎠
⎝d − p⎠
and
12
⎛ μ 0 ⎛ 2d ⎞ ⎞
⎜ ln⎜ ⎟ ⎟
L ⎜ π ⎝ a ⎠⎟
Z 0 for the parallel wires =
=
C ⎜ 2πε 0 ⎟
⎜ ln(2d a ) ⎟
⎝
⎠
12
⎛ μ ⎞
⎛ 2d ⎞
≈ ⎜⎜ 2 0 ⎟⎟ ln⎜ ⎟
⎝ a ⎠
⎝ π ε0 ⎠
7.3
The integral of magnetic energy over the last quarter wavelength is given by:
2
0
0
⎞
λL0V02+
1
1 ⎛ 2V0+
V02+ 1 + cos 4πx λ
2
⎟
⎜
=
=
−
cos
2
L
I
dx
L
kx
dx
L
dx
=
∫−λ 4 2 0
∫−λ 4 2 0 ⎜⎝ Z 0
∫−λ 4 0 Z 02
⎟
2
4 Z 02
⎠
0
The integral of electric energy over the last quarter wavelength is given by:
0
0
λC0V02+
1
1
2
2
2 1 − cos 4πx λ
(
)
C
V
dx
C
V
kx
dx
C
V
dx
=
=
=
−
2
sin
2
∫−λ 4 2 0
∫−λ 4 2 0 0+
∫−λ 4 0 0+
2
4
0
Noting that Z 0 =
L0 C0 , we have:
0
λL0V02+ λC0V02+
1
1
2
L
I
dx
=
=
=∫
C0V 2 dx
2
∫−λ 4 2 0
−λ 4 2
4Z 0
4
0
7.4
The maximum of the magnetic energy is given by:
( Em ) max
2
⎡ 1 ⎛ 2V
⎞ ⎤
2L V 2
⎛1
2⎞
0+
= ⎜ L0 I ⎟ = ⎢ L0 ⎜⎜
cos kx ⎟⎟ ⎥ = 0 2 0 + = 2C0V02+
Z0
⎝2
⎠ max ⎢⎣ 2 ⎝ Z 0
⎠ ⎥⎦ max
The maximum of the electric energy is given by:
⎛1
⎞
⎡1
2⎤
( Ee ) max = ⎜ C0V 2 ⎟ = ⎢ C0 (2V0 + sin kx ) ⎥ = 2C0V02+
⎝2
⎠ max ⎣ 2
⎦ max
The instantaneous value of the two energies over the last quarter wavelength is given by:
© 2008 John Wiley & Sons, Ltd
2
⎞ 1
1 ⎛ 2V
( Em + Ee )i = L0 ⎜⎜ 0 + cos kx ⎟⎟ + C0 (2V02+ sin kx) 2
2 ⎝ Z0
⎠ 2
= 2C0V02+ cos 2 kx + 2C0V02+ sin 2 kx
= 2C0V02+
So we have:
( Em ) max = ( Ee ) max = ( Em + Ee )i = 2C0V02+
7.5
For a real transmission line with a propagation constant γ , the forward current wave I x + at
position x is given by:
I x + = I 0+ e −γx = Ae −γx
where I 0+ = A is the forward current wave at position x = 0 . So the forward voltage wave at
position x is given by:
Vx+ = Z 0 I x + = Z 0 Ae −γx
The backward current wave I x − at position x is given by:
I x − = I 0 − e + γx = Be + γx
where I 0 − = B is the backward current wave at position x = 0 . So the backward voltage wave
at position x is given by:
Vx − = − Z 0 I x − = − Z 0 Be + γx
Therefore the impedance seen from position x is given by:
Zx =
Vx+ + Vx− Z 0 Ae −γx − Z 0 Be + γx
Ae −γx − Be + γx
=
=
Z
0
I x+ + I x−
Ae −γx + Be +γx
Ae −γx + Be +γx
If the line has a length l and is terminated by a load Z L , the value of Z L is given by:
ZL =
VL Vl + + Vl −
Ae −γl − Be + γl
=
= Z0
I L Il+ + Il−
Ae −γl + Be +γl
7.6
The impedance of the line at x = 0 is given by:
⎛
A−γx − Be +γx ⎞
A− B
⎟ = Z0
Z i = ⎜⎜ Z 0 −γx
+γx ⎟
A+ B
⎝ Ae + Be ⎠ x=0
© 2008 John Wiley & Sons, Ltd
Noting that:
Z L = Z0
Ae −γl − Beγl
Ae −γl + Beγl
we have:
( Z 0 − Z L ) Ae −γl = ( Z 0 + Z L ) Beγl
A ( Z 0 + Z L ) 2γl
=
e
B (Z0 − Z L )
i.e.
so we have:
Zi = Z0
A B −1
Z (eγl − e −γl ) + Z L (eγl + e −γl )
Z sinh γl + Z L cosh γl
= Z 0 0 γl
= Z0 0
−γl
γl
−γl
A B +1
Z 0 ( e + e ) + Z L (e − e )
Z 0 cosh γl + Z L sinh γl
7.7
If the transmission line of Problem 7.6 is short-circuited, i.e. Z L = 0 , The expression of input
impedance in Problem 7.6 gives:
Z sc = Z 0
Z 0 sinh γl
= Z 0 tanh γl
Z 0 cosh γl
If the transmission line of Problem 7.6 is open-circuited, i.e. Z L = ∞ , The expression of input
impedance in Problem 7.6 gives:
Z oc = Z 0
Z L cosh γl
= Z 0 coth γl
Z L sinh γl
By taking the product of these two impedances we have:
Z sc Z oc = Z 02 , i.e. Z 0 = Z sc Z oc
which shows the characteristic impedance of the line can be obtained by measuring the
impedances of short-circuited line and open-circuited line separately and then taking the square
root of the product of the two values.
7.8
The forward and reflected voltage waves at the end of the line are given by:
Vl + = −Vl − = V0+ e − ikl
where V0+ is the forward voltage at the beginning of the line. So the reflected voltage wave at
the beginning of the line is given by:
V0− = Vl − e − ikl = −V0+ e − i 2 kl
© 2008 John Wiley & Sons, Ltd
The forward and reflected current waves at the end of the line are given by:
I l + = Vl + Z 0 = V0+ e − ikl Z 0 = I 0+ e −ikl
I l − = − Vl − Z 0 = Vl + Z 0 = I 0+ e −ikl
where I 0+ is the forward current at the beginning of the line. So the reflected current wave at the
beginning of the line is given by:
I 0− = I l − e − ikl = I 0+ e − i 2 kl
Therefore the input impedance of the line is given by:
Zi =
V0+ + V0− V0+ (1 − e −i 2 kl ) V0+ (eikl − e − ikl )
sin kl
L
2πl
=
=
= iZ 0
= i 0 tan
ikl
−ikl
−i 2 kl
λ
I 0+ + I 0− I 0+ (1 + e ) I 0+ (e + e )
cos kl
C0
The variation of the ratio Z i
L0 C0 with l is shown in the figure below:
Zi
L0 C0
…
…
−λ 2
− 3λ 4
λ 4
−λ 4
0
3λ 4
λ 2
l
7.9
The boundary condition at Z 0 Z m junction gives:
V0+ + V0− = Vm 0+ + Vm 0−
I 0+ + I 0− = I m 0+ + I m 0−
where V0 + , V0 − are the voltages of forward and backward waves on Z 0 side of Z 0 Z m
junction; I 0+ , I 0− are the currents of forward and backward waves on Z 0 side of Z 0 Z m
junction; Vm 0+ , Vm 0 − are the voltages of forward and backward waves on Z m side of Z 0 Z m
© 2008 John Wiley & Sons, Ltd
junction; I m 0 + , I m 0 − are the currents of forward and backward waves on Z m side of Z 0 Z m
junction;
The boundary condition at Z m Z L junction gives:
VmL+ + VmL− = VL
I mL+ + I mL− = I L
where VmL + , VmL − are the voltages of forward and backward waves on Z m side of Z m Z L
junction; I mL+ , I mL− are the currents of forward and backward waves on Z m side of Z m Z L
junction; VL , I L are the voltage and current across the load.
If the length of the matching line is l , we have:
Vm 0+ = VmL+ e ikl
I m 0+ = I mL + e ikl
Vm 0− = VmL − e − ikl
I m 0− = I mL− e − ikl
In addition, we have the relations:
VL
= ZL
IL
V0
= Z0
I0
Vm 0 +
V
V
V
= − m 0 − = mL + = − mL − = Z m
I m0+
I m 0 − I mL +
I mL −
The above conditions yield:
VmL + = Vm 0+ e − ikl
I mL+ = I m 0+ e − ikl
VmL− =
© 2008 John Wiley & Sons, Ltd
Z L − Zm
VmL+
ZL + Zm
I mL− =
Zm − ZL
I mL+
Zm + ZL
Vm 0− = VmL− e −ikl =
ZL − Zm
Z − Z m −i 2 kl
VmL+ e −ikl = Vm 0+ L
e
Z L + Zm
ZL + Zm
I m 0− = I mL− e −ikl =
Zm − ZL
Z − Z L −i 2 kl
I mL+ e −ikl = I m 0+ m
e
ZL + Zm
ZL + Zm
Impedance mating requires V0− = 0 and I 0− = 0 , i.e.:
V0+ = Vm 0+ + Vm 0−
I 0+ = I m 0+ + I m 0−
i.e.
⎛ Z − Z m −i 2 kl ⎞
⎟⎟
V0+ = Vm 0+ ⎜⎜1 + L
e
+
Z
Z
L
m
⎝
⎠
⎛ Z − Z L −i 2 kl ⎞
⎟⎟
I 0+ = I m 0+ ⎜⎜1 + m
e
⎝ Z L + Zm
⎠
By dividing the above equations we have:
Z0 = Zm
( Z L + Z m )eikl + ( Z L − Z m )e − ikl
Z cos kl + iZ m sin kl
= Zm L
−ikl
ikl
( Z L + Z m )e + ( Z m − Z L )e
Z L sin kl + iZ m cos kl
which is true if kl = π 2 , or l = λ 4 and yields:
Z m2 = Z 0 Z L
7.10
Analysis in Problem 7.8 shows the impedance of a short-circuited loss-free line has an impedance
given by:
Z i = iZ 0 tan
2πl
λ
so, if the length of the line is a quarter of one wavelength, we have:
Z i = iZ 0 tan
2π λ
π
= iZ 0 tan = ∞
λ 4
2
If this line is bridged across another transmission line, due to the infinite impedance, the
transmission of fundamental wavelength
harmonic wavelength
λ will not be affected. However for the second
λ 2 , the impedance of the bridge line is given by:
© 2008 John Wiley & Sons, Ltd
Z i = iZ 0 tan
2π λ
= iZ 0 tan π = 0
λ 24
which shows the bridge line short circuits the second harmonic waves.
7.11
For Z 0 to act as a high pass filter with zero attenuation, the frequency
ω2 >
1
, where
2 LC
Z0 = L C .
The exact physical length of Z 0 is determined by
ω . Choosing the frequency ω1 determines
k1 = 2π λ1 .
For a high frequency load Z L and a loss- free line, we have, for the input impedance:
⎛ Z cos kl + iZ 0 sin kl ⎞
⎟⎟
Z in = Z 0 ⎜⎜ L
Z
kl
iZ
kl
+
cos
sin
L
⎝ 0
⎠
For n even, we have:
cos k1l = cos
2π nλ1
= cos nπ = 1
λ1 2
For n odd, we have:
cos k1l = cos
2π nλ1
= cos nπ = −1
λ1 2
The sine terms are zero.
So Z in = Z L for n odd or even, and the high frequency circuits, input and load, are uniquely
matched at
ω1 when the circuits are tuned to ω1 .
7.12
The phase shift per section β should satisfy:
cos β = 1 +
i.e.
i.e.
© 2008 John Wiley & Sons, Ltd
Z1
iω L
ω 2 LC
= 1+
= 1−
2Z 2
2 iωC
2
1 − cos β =
2 sin 2
β
2
=
ω 2 LC
2
ω 2 LC
2
For a small β , β ≈ sin β , so the above equation becomes:
ω 2 LC
⎛β ⎞
2⎜ ⎟ =
2
⎝2⎠
2
β = ω LC = ω v = k
i.e.
where the phase velocity is given by v = 1
LC and is independent of the frequency.
7.13
The propagation constant γ can be expanded as:
γ = ( R0 + iωL0 )(G0 + iωC0 )
⎞
⎞⎛ G
⎛ R
= ω L0C0 ⎜⎜ 0 + i ⎟⎟⎜⎜ 0 + i ⎟⎟
⎝ ωL0 ⎠⎝ ωC0 ⎠
⎛ R
G ⎞
R G0
= ω L0C0 i 2 + ⎜⎜ 0 + 0 ⎟⎟i + 0
⎝ ωL0 ωC0 ⎠ ωL0 ωC0
2
⎛
R
G0 ⎞ ⎛ R02
G02
R G0 ⎞
⎜⎜ i + 0 +
⎟⎟ − ⎜⎜ 2 2 +
⎟⎟
+ 0
2
⎝ 2ωL0 2ωC0 ⎠ ⎝ 4ω L0 4ωC0 ωL0 ωC0 ⎠
= ω L0C0
Since R0
ωL0 and G0 ωC0 are both small quantities, the above equation becomes:
⎛
γ = ω L0C0 ⎜⎜ i +
⎝
where
α=
R0
2
C0 G0
+
L0
2
R0
G ⎞
+ 0 ⎟⎟ = α + ik
2ωL0 2ωC0 ⎠
L0
, and k = ω L0C0 = ω v
C0
If G = 0 , we have:
ω L0C0
ω L0C0
k
ωL
=
=
= 0
2α R0 C0 L0 + G0 L0 C0 R0 C0 L0
R0
which is the Q value of this transmission line.
7.14
Suppose
R0 G0
=
= K , where K is constant, the characteristic impedance of a lossless line is
L0 C0
given by:
Z0 =
© 2008 John Wiley & Sons, Ltd
R0 + iωL0
=
G0 + iωC0
KL0 + iωL0
L0
=
KC0 + iωC0
C0
which is a real value.
7.15
Try solution ψ = ψ m eγx in wave equation:
∂ 2ψ 8π 2 m
+ 2 ( E − V )ψ = 0
∂x 2
h
we have:
8π 2 m
γ = 2 (V − E )
h
2
For E > V (inside the potential well), the value of γ is given by:
γ in = ±i
2π
h
2 m( E − V )
So the ψ has a standing wave expression given by:
ψ = Ae
i
2π
m (V − E ) x
h
+ Be
−i
2π
m (V − E ) x
h
where A , B are constants.
For E < V (outside the potential well), the value of γ is given by:
γ out = ±
So the expression of ψ is given by:
2π
ψ = Ae h
2π
h
2m(V − E )
m (V − E ) x
+ Be
−
2π
m (V − E ) x
h
where A , B are constants. i.e. the x dependence of ψ is e ± γx , where
γ =
2π
h
2m(V − E )
7.16
Form the diffusion equation:
∂H
1 ∂2H
=
∂t
μσ ∂x 2
we know the diffusivity is given by: d = 1
μσ . The time of decay of the field is approximately
given by Einstein’s diffusivity relation:
t=
where L is the extent of the medium.
© 2008 John Wiley & Sons, Ltd
L2
L2
=
= L2 μσ
d 1 μσ
For a copper sphere of radius 1m , the time of decay of the field is approximately given by:
t = L2 μσ = 12 × 1.26 × 10 −6 × 5.8 × 107 ≈ 73[ s ] < 100[ s ]
7.17
Try solution f (α , t ) =
r
π
e −( rα ) in
2
∂f (α , t )
, we have:
∂t
∂f (α , t ) ∂ ⎡ r −( rα )2 ⎤
= ⎢
e
⎥
∂t
∂t ⎣ π
⎦
=
=
r
e −( rα ) (−2rα )α
2
π
1 − 2r 2α 2
e −( rα )
π
2r α 2 − 1 −( rα )
=
e
4t πdt
2
Try solution f (α , t ) =
r
π
e −( rα ) in
2
2
1 dr −( rα )2
dr
+
e
dt
π dt
dr
dt
2
(A.7.17.1)
∂f (α , t )
, we have:
∂x
2r 3α −( rα )2
∂f (α , t )
=−
e
∂x
π
so:
∂ 2 f (α , t )
2r 3 −( rα )2 2r 3α −( rα )2
=−
−
(−2rα )r
e
e
∂x 2
π
π
=−
2r 3
(1 − 2r 2α 2 )e −( rα )
π
2r α 2 − 1 −( rα )
=
e
4td πdt
2
2
2
(A.7.17.2)
By comparing the above derivatives, A.7.17.1 and 2, we can find the solution
f (α , t ) =
r
π
e −( rα )
satisfies the equation:
∂f
∂2 f
=d 2
∂t
∂x
© 2008 John Wiley & Sons, Ltd
2
SOLUTIONS TO CHAPTER 8
8.1
Write the expressions of E x and H y as:
E x = E0 sin
2π
(v t − z )
λE E
2π
H y = H 0 sin
(v t − z )
λH H
where λE and λH are the wavelengths of electric and magnetic waves respectively,
and vE and vH are the velocities of electric and magnetic waves respectively.
By substitution of the these expressions into equation (8.1a), we have:
−μ
2π
λH
vH H 0 cos
μ
i.e.
vH H 0
λH
2π
λH
cos
(v H t − z ) = −
2π
λH
2π
(v H t − z ) =
λE
E0 cos
E0
λE
cos
2π
λE
2π
λE
(v E t − z )
(v E t − z )
which is true for all t and z , provided:
vH = vE =
E0
μH 0
λ H = λE
and
so, at any t and z , we have:
φE = φH =
E0
t−z
μH 0
Therefore E x and H y have the same wavelength and phase.
8.2
Energy Force ⋅ Distance Force
=
=
= pressure
Volume
L3
Area
© 2008 John Wiley & Sons, Ltd
W
C
W
A
C
Currents in W into page. Field lines at A cancel. Those at C force wires together.
Reverse current in one wire. Field lines at A in same direction, force wires apart.
Fig Q.8.2.a
W
C
A
Motion
Field lines at C in same direction as those from current in wire –
in opposite direction at A . Motion to the right
Fig Q.8.2.b
8.3
The volume of a thin shell of thickness dr is given by: 4πr 2 dr , so the electrostatic
energy over the spherical volume from radius
+∞ 1
by: ∫
ε 0 E 2 (4πr 2 )dr , which equals mc 2 , i.e.:
a 2
+∞ 1
2
2
2
∫a 2 ε 0 E (4πr )dr = mc
a
to infinity is given
By substitution of E = e 4πε 0 r 2 into the above equation, we have:
∫
+∞
a
1
e2
ε0
(4πr 2 )dr = mc 2
2 (4πε 0 r 2 ) 2
e2
i.e.
8πε 0
∫
+∞
a
1
dr = mc 2
2
r
e2
i.e.
8πε 0 a
= mc 2
Then, the value of radius a is given by:
a=
e2
(1.6 × 10 −19 ) 2
=
≈ 1.41× 10 −15 [m]
8πε 0 mc 2 8π × 8.8 × 10 −12 × 9.1× 10−31 × (3 × 108 ) 2
© 2008 John Wiley & Sons, Ltd
Another approach to the problem yields the value:
a = 2.82 × 10−15 [m]
8.4
The magnitude of Poynting vector on the surface of the wire can be calculated by
deriving the electric and magnetic fields respectively.
The vector of magnetic field on the surface of the cylindrical wire points towards the
azimuthal direction, and its magnitude is given by Ampere’s Law:
I
H = Hθ eθ =
eθ
2πr
where r is the radius of the wire’s cross circular section, and I is the current in the
wire.
Ohm’s Law, J = σE , shows the vector of electric field on the surface of the
cylindrical wire points towards the current’s direction, and its magnitude equals the
voltage drop per unit length, i.e.:
V
IR
E = Eze z = e z = e z
l
l
where, l is the length of the wire, and the V is the voltage drop along the whole
length of the wire and is given by Ohm’s Law: V = IR , where R is the resistance of
the wire.
Hence, the Poynting vector on the surface of the wire points towards the axis of the
wire is given by:
S = E × H = E z e z × H θ eθ = − E z Hθ e r
which shows the Poynting vector on the surface of the wire points towards the axis of
the wire, which corresponds to the flow of energy into the wire from surrounding
space. The product of its magnitude and the surface area of the wire is given by:
IR I
S × 2πrl = E z Hθ × 2πrl =
2πrl = I 2 R
l 2πr
which is the rate of generation of heat in the wire.
8.5
By relating Poynting vector to magnetic energy, we first need to derive the magnitude
of Poynting vector in terms of magnetic field.
The electric field on the inner surface of the solenoid can be derived from the integral
format of Faraday’s Law:
∂H
∫ Edl = −μ ∫ ∂t dS
where S is the area of the solenoid’s cross section. E is the electric field on the
inner surface of the solenoid, and H is the magnetic field inside the solenoid.
For a long uniformly wound solenoid the electric field uniformly points towards
azimuthal direction, i.e. E = Eθ eθ , and the magnetic field inside the solenoid
© 2008 John Wiley & Sons, Ltd
uniformly points along the axis direction, i.e. H = H z e z . So the above equation
becomes
∂H z 2
πr
∂t
μr ∂H z
i.e.
Eθ = −
2 ∂t
where r is the radius of the cross section of the solenoid.
Hence, the Poynting vector on the inner surface of the solenoid is given by:
μr
∂H z
E × H = Eθ eθ × H z e z = − H z
er
2
∂t
which points towards the axis of the solenoid and corresponds to the inward energy
flow. The product of its magnitude and the surface area of the solenoid is given by:
μr
∂H z
∂H z
× 2πrl = πr 2lH z
S × 2πrl =
Hz
2
∂t
∂t
where l is the length of the solenoid.
On the other hand, the time rate of change of magnetic energy stored in the solenoid
of a length l is given by:
Eθ × 2πr = − μ
d ⎛1
∂H z
2
2 ⎞
2
⎜ μH × πr l ⎟ = πr lH z
dt ⎝ 2
∂t
⎠
which equals S × 2πrl
8.6
For plane polarized electromagnetic wave ( E x , H y ) in free space, we have the
relation:
Ex
=
Hy
μ0
ε0
Its Poynting vector is given by:
S = Ex H y = Ex
where
c =1
Ex
μ0 ε 0
=
1
ε0 2
Ex =
ε 0 E x2 = cε 0 E x2
μ0
μ0ε 0
μ0ε 0 is the velocity of light.
Noting that:
2
⎞
1
1 ⎛ μ
1
ε 0 E x2 = ε 0 ⎜⎜ 0 H y ⎟⎟ = μ0 H y2
2
2 ⎝ ε0
2
⎠
we have:
1
⎛1
⎞
S = E x H y = c⎜ ε 0 E x2 + μ0 H y2 ⎟ = cε 0 E x2
2
⎝2
⎠
© 2008 John Wiley & Sons, Ltd
Since the intensity in such a wave is given by:
I = S av = cε 0 E 2 =
1
2
cε 0 Emax
2
we have:
S =
2 12
2
S =
S 1 2 ≈ 27.45S 1 2 [Vm −1 ]
8
−12
cε 0
3 × 10 × 8.8 × 10
Emax =
H max =
1
2
2
× 3 × 108 × 8.8 × 10−12 × Emax
≈ 1.327 × 10 −3 Emax
2
2 12
2
ε0
Emax =
S =
S 1 2 ≈ 7.3 × 10 −2 S 1 2 [ Am −1 ]
8
-7
cμ0
3 × 10 × 4π × 10
μ0
8.7
The average intensity of the beam and is given by:
I=
Power
Energy
0.3
=
=
= 1.53 × 108 [Wm − 2 ]
−3 2
−4
area
area × pulse duration π × (2.5 × 10 ) × 10
Using the result in Problem 8.6, the root mean square value of the electric field in
the wave is given by:
E2 =
I
cε 0
=
1.53 × 108
≈ 2.4 × 105 [Vm −1 ]
8
−12
3 × 10 × 8.8 × 10
8.8
Using the result of Problem 8.6, the amplitude of the electric field at the earth’s
surface is given by:
E0 = 27.45S 1 2 = 27.45 × 1350 ≈ 1010[Vm −1 ]
and the amplitude of the associated magnetic field in the wave is given by:
H 0 = 7.3 × 10 −2 × 1350 ≈ 2.7[ Am −1 ]
The radiation pressure of the sunlight upon the earth equals the sum of the electric
field energy density and the magnetic field energy density, i.e.
1
1
prad = ε 0 E02 + μ0 H 02 = ε 0 E02 = 8.8 × 10 −12 × 10102 = 8.98 × 10 −6 [ Pa]
2
2
8.9
The total radiant energy loss per second of the sun is given by:
Eloss = S × 4πr 2 = 1350 × 4π × (15 × 1010 ) 2 = 3.82 × 1026 [ J ]
which is associated with a mass of:
m = Eloss c 2 =
© 2008 John Wiley & Sons, Ltd
3.82 × 10 26
= 4.2 × 109 [kg ]
(3 × 108 ) 2
8.10
At a point 10km from the station, the Poynting vector is given by:
S=
P
105
=
= 1.6 × 10 −4 [W m 2 ]
2πr 2 2 × π × (10 × 103 ) 2
Using the result in Problem 8.6, the amplitude of electric field is given by:
E0 = 27.45 × S 1 2 = 27.45 × 1.6 × 10 −4 = 0.346[V m]
The amplitude of magnetic field is given by:
H 0 = 7.3 × 10 −2 S 1 2 = 7.3 × 10 −2 × 1.6 × 10−4 = 9.2 × 10−4 [ A m]
8.11
The surface current in the strip is given by:
I = Qv
where Q is surface charge per unit area on the strip and is given by: Q = εE x , and
v is the velocity of surface charges along the transmission line.
Since the surface charges change along the transmission line at the same speed as the
electromagnetic wave travels, i.e. v = c =
1
με
, the surface current becomes:
I = Qv = εE x
1
=
με
ε
E
μ x
Analysis in page 207 shows, for plane electromagnetic wave,
μ H y = ε E x , so the
surface current is now given by:
ε μ
H = Hy
μ ε y
I=
On the other hand, the voltage across the two strips is given by:
V = Ex L = Ex
where L = 1 is the distance between the two strips.
Therefore the characteristic impedance of the transmission line is given by:
Z=
8.12
Write equation (8.6) in form:
© 2008 John Wiley & Sons, Ltd
V
E
= x =
I Hy
μ
ε
∂2
∂⎛∂
∂
⎞
E x = μ ⎜ εE x ⎟ + μ (σE x )
2
∂z
∂t ⎝ ∂t
∂t
⎠
which can be dimensionally expressed as:
voltage length inductance displacement current inductance current area
=
×
+
×
length × length
length
time × area
length
time
Multiplied by a dimension term, length, the above equation has the dimension:
voltage
current
= inductance ×
area
time × area
which is the dimensional form (per unit area) of the equation:
dI
V =L
dt
where V is a voltage, L is a inductance and I is a current.
8.13
Analysis in page 210 and 211 shows, in a conducting medium, the wave number of
electromagnetic wave is given by:
k=
ωμσ
2
where ω is angular frequency of the electromagnetic wave, μ and σ are the
permeability and conductivity of the conducting medium.
Differentiation of the above equation gives:
dk =
1
2 μσ
1 μσ
dω =
dω
2 ωμσ 2
2 2ω
ω
2ω
2
dω
ω=2
=2
=2
μσ
ωμσ
dk
k
i.e.
which shows, when a group of electromagnetic waves of nearly equal frequencies
propagates in a conducting medium, where the group velocity and the phase velocity
can be treated as fixed values, the group velocity, vg = dω dk , is twice the wave
velocity, v p = ω k .
8.14
(a)
σ
σ
0.1
=
=
× 36π × 109 = 720 > 100
ωε 2πνε r ε 0 2π × 50 ×103 × 50
which shows, at a frequency of 50kHz , the medium is a conductor
© 2008 John Wiley & Sons, Ltd
(b)
σ
σ
0.1
=
=
× 36π × 109 = 3.6 × 10 −3 < 10−2
4
ωε 2πνε r ε 0 2π ×10 × 106 × 50
which shows, at a frequency of 104 MHz , the medium is a dielectric.
8.15
The Atlantic Ocean is a conductor when:
σ
σ
=
> 100
ωε 2πνε r ε 0
i.e.
ν<
σ
2π × 100 × ε r ε 0
=
4.3
× 36π × 109 ≈ 10[ MHz ]
2π × 100 × 81
Therefore the longest wavelength that could propagate under water is given by:
ν=
λmax =
i.e.
v
λmax
=
c
ε r λmax
= 10× 106
c
3 × 108
=
≈ 3[m]
ε r × 10 × 106
81 × 10 × 106
8.16
When a plane electromagnetic wave travelling in air with an impedance of Z air is
reflected normally from a plane conducting surface with an impedance of Z c , the
transmission coefficient of magnetic field is given by:
TH =
Ht
Hi
Using the relations: Et = Z c H t , Ei = Z air H i , and
Et
2Z c
=
, the above equation
Ei Z c + Z air
becomes:
TH =
H t Et Z air Z air 2 Z c
2 Z air
=
=
=
Hi
Ei Z c
Z c Z c + Z air Z c + Z air
The impedance of a good conductor tends to zero, i.e. Z c → 0 , so we have:
TH ≈
2Z air
= 2 or H t ≈ 2 H i
Z air
After reflection from the air-conductor interface, standing waves are formed in the air
© 2008 John Wiley & Sons, Ltd
with a magnitude of H i + H r in magnetic field and a magnitude of Ei + Er in
electric field.
Using the relations: H i + H r = H t , Ei + Er = Et , the standing wave ratio of magnetic
field to electric field in air is given by:
Hi + Hr Ht
1
=
=
Ei + Er
Et Z c
which is a large quantity due to Z c → 0 .
As an analogy, for a short-circuited transmission line, the relation between forward
(incident) and backward (reflected) voltages is given by: V+ + V− = 0 or Vi + Vr = 0 ,
the forward (incident) current is given by: I i = I + = V+ Z 0 , and the backward
(reflected) current is given by: I r = I − = −V− Z 0 , so the transmitted current is given
by:
It = Ii + I r =
V+ V− V+ − V+
V
−
=
−
= 2 + = 2Ii
Z0 Z0 Z0
Z0
Z0
When a plane electromagnetic wave travelling in a conductor with an impedance of
Z c is reflected normally from a plane conductor-air interface, the transmission
coefficient of electric field is given by:
TE =
Et
2 Z air
=
Ei Z c + Z air
The impedance of a good conductor tends to zero, i.e. Z c → 0 , so we have:
TE ≈
2Z air
= 2 or Et ≈ 2 Ei
Z air
As an analogy, for a open-circuited transmission line, the forward (incident) voltage
equals the backward (reflected) voltages, i.e. Vi = V+ = Vr = V− , the transmitted
voltage is given by:
Vt = Vi + Vr = V+ + V+ = 2V+
8.17
Analysis in page 215 and 216 shows, in a conductor, magnetic field H y lags electric
© 2008 John Wiley & Sons, Ltd
field E x by a phase angle of φ = 45o , so we can write the electric field and magnetic
field in a conductor as:
E x = E0 cos ωt and H y = H 0 cos(ωt − φ )
so the average value of the Poynting vector is the integral of the Poynting vector
E x H y over one time period T divided by the time period, i.e.:
S av =
1 T
Ex H y
T ∫0
1 T
E0 cos ωt H 0 cos(ωt − φ )dt
T ∫0
EH T1
= 0 0 ∫ [cos(2ωt − φ ) + cos φ ]dt
T 0 2
1 E0 H 0
1
=
T cos φ = E0 H 0 cos 45o [Wm 2 ]
2 T
2
=
Noting that the real part of impedance of the conductor is given by:
(real part of Z c ) =
E0 =
i.e.
E0
E
cos φ = 0 cos 45o
H0
H0
H0
× (real part of Z c )
cos 45o
so we have:
1
E0 H 0 cos 45o
2
1 H 02
=
× real part of Z c × cos 45o
2 cos 45o
1
= H 02 × (real part of Z c )[Wm 2 ]
2
S av =
We know from analysis in page 216 that, at a frequency ν = 3000 MHz , the value of
ωε σ for copper is 2.9 ×10−9 , hence, at of frequency of 1000 MHz , the value of
ωε σ for copper is given by 2.9 ×10−9 3 = 9.7 ×10 −10 , and μ r ≈ ε r ≈ 1 . So, the real
part of impedance of the large copper sheet is given by:
(real part of Z copper ) =
=
μr
2
× 376.6 ×
εr
2
2
Z copper
2
ωε
2
= 376.6 ×
× 9.7 × 10 −10 = 8.2 × 10 −3[Ω]
σ
2
Noting that, at an air-conductor interface, the transmitted magnetic field in copper
H copper doubles the incident magnetic field H 0 , i.e. H copper = 2H 0 , the average
© 2008 John Wiley & Sons, Ltd
power absorbed by the copper per square metre is the average value of transmitted
Poynting vector, which is given by:
S copper =
1 2
H copper × (real part of Z copper )
2
1
(2 H copper ) 2 × (real part of Z copper )
2
2
= 2 H copper
× (real part of Z copper )
=
2
⎛ E ⎞
= 2 × ⎜ 0 ⎟ × (real part of Z copper )
⎝ 376.6 ⎠
2
⎛ 1 ⎞
−3
−7
= 2×⎜
⎟ × 8.2 × 10 = 1.16 × 10 [W ]
376
.
6
⎝
⎠
8.18
Analysis in page 222 and 223 shows that when an electromagnetic wave is reflected
normally from a conducting surface its reflection coefficient I r is given by:
I r = 1− 2
2ωε 0
σ
Noting that ε r = 1 , the fractional loss of energy is given by:
⎛
2ωε 0
1 − I r = 1 − ⎜⎜1 − 2
σ
⎝
⎞
8ωε 0
8ω ε ε r
8ωε
⎟=
=
=
⎟
σ
σ
σ
⎠
8.19
Following the discussion of solution to problem 8.17, we can also find the average
value of Poynting vector in air.
The electric and magnetic field of plane wave in air have the same phase, so the
Poynting vector in air is given by:
S air = E x H y = E0 cos ωt × H 0 cos ωt = E0 H 0 cos 2 ωt
and its average value is given by:
S air =
1 T
Ex H y
T ∫0
1 T
E0 cos ωt H 0 cos ωtdt
T ∫0
E H T1
= 0 0 ∫ [1 + cos(2ωt )]dt
T 0 2
1 E0 H 0
1
1
E02
1
=
T = E0 H 0 = ×
=
= 1.33 × 10 −3[Wm −1 ]
2 T
2
2 376.6 2 × 376.6
=
© 2008 John Wiley & Sons, Ltd
So, the ratio of transmitted Poynting vector in copper to the incident Poynting vector
in air is given by:
S copper
S air
=
1.16 × 10 −7
= 8.81× 10 −5
1.33 × 10 −3
which equals the fractional loss of energy I r given by:
Ir =
8ωε
σ
= 8 × 9.7 × 10 −10 = 8.81× 10 −5
8.20
E x and H y are in complex expression, we have:
12
⎛ σ ⎞ −kz −i (ωt −kz ) iπ
1
1
E x H y* = Ae −kz ei (ωt −kz ) A⎜⎜
e
⎟⎟ e e
2
2
⎝ ωμ ⎠
4
12
1 ⎛ σ ⎞ −2 kz iπ
= A2 ⎜⎜
⎟ e e
2 ⎝ ωμ ⎟⎠
4
So, the average value of the Poynting vector in the conductor is given by:
12
⎛1
⎞ 1 ⎛ σ ⎞ −2 kz
S av = real part of ⎜ E x H y* ⎟ = A2 ⎜⎜
⎟⎟ e [Wm −2 ]
⎝2
⎠ 2 ⎝ 2ωμ ⎠
The mean value of the electric field vector, E x , is a constant value, which contributes
to the same electric energy density at the same amount of time, i.e.:
2
1
1 T1
(average electric energy density) = ε E x = ∫ εE x2 dt
T 0 2
2
i.e.
2
2 − 2 kz
Ex = A e
T 1 + cos 2ωt
A2e −2 kz
1 T
2
2 − 2 kz 1
dt =
cos ωtdt = A e
T ∫0
T ∫0
2
2
Ex =
or:
2 −kz
Ae
2
Noting that:
12
∂S av
1 ⎛ σ ⎞ −2 kz
= −2k × A2 ⎜⎜
⎟ e
∂z
2 ⎝ 2ωμ ⎟⎠
12
⎛ ωμσ ⎞ ⎛ σ ⎞ −2 kz
= −A ⎜
⎟⎟ e
⎟ ⎜⎜
⎝ 2 ⎠ ⎝ 2ωμ ⎠
2
A2σ −2 kz
=−
e
= −σ E x
2
12
2
© 2008 John Wiley & Sons, Ltd
we find the value of ∂S av ∂z is the product of the conductivity σ and the square of
the mean value of the electric field vector E x . The negative sign in the above
equation shows the energy is decreasing with distance.
8.21
Noting that the relation between refractive index n of a dielectric and its impedance
Z d is given by: n =
Z0
, where Z 0 is the impedance in free space, so, when light
Zd
travelling in free space is normally incident on the surface of a dielectric, the reflected
intensity is given by:
⎛E
I r = ⎜⎜ r
⎝ Ei
2
2
2
⎞ ⎛ Zd − Z0 ⎞ ⎛ 1 − Z0 Zd ⎞ ⎛ 1 − n ⎞
⎟⎟ = ⎜⎜
⎟⎟ = ⎜⎜
⎟⎟ = ⎜
⎟
⎠ ⎝ Zd + Z0 ⎠ ⎝ 1 + Z0 Zd ⎠ ⎝ 1 + n ⎠
2
and the transmitted intensity is given by:
2
2
⎞
Z 0 Et2 Z 0 ⎛ 2Z d ⎞
Z ⎛
2
2 ⎞
4n
⎜⎜
⎟⎟ = 0 ⎜⎜
⎟⎟ = n⎛⎜
=
It =
⎟ =
2
Z d Ei
Zd ⎝ Zd + Z0 ⎠
Z d ⎝ 1 + Z0 Z d ⎠
(1 + n) 2
⎝1+ n ⎠
2
8.22
If the dielectric is a glass (nglass = 1.5) , we have:
2
I r _ glass
⎛ 1 − nglass ⎞ ⎛ 1 − 1.5 ⎞ 2
⎟ =
= 4%
=⎜
⎜ 1 + n ⎟ ⎜⎝ 1 + 1.5 ⎟⎠
glass
⎠
⎝
I t _ glass =
4nglass
(1 + nglass )
2
=
4 × 1.5
= 96%
(1 + 1.5) 2
Problem 8.15 shows water is a conductor up to a frequency of 10MHz, i.e. water is a
dielectric at a frequency of 100MHz and has a refractive index of:
nwater = ε r = 81 = 9
So, the reflectivity is given by:
2
I r _ water
⎛ 1 − nwater ⎞ ⎛ 1 − 9 ⎞
⎟⎟ = ⎜
= ⎜⎜
⎟ = 64%
⎝ 1 + nwater ⎠ ⎝ 1 + 9 ⎠
2
and transmittivity given by:
I t _ water =
8.23
© 2008 John Wiley & Sons, Ltd
4nwater
4×9
=
= 36%
2
(1 + nwater )
(1 + 9) 2
The loss of intensity is given by:
I loss = 1 − I t1I t 2
where I t1 is the transmittivity from air to glass and I t 2 is the transmittivity from
glass to air. Following the discussion in problem 8.21, we have:
Z E 2 Z ⎛ 2Z 0
I t 2 = i t2 = d ⎜⎜
Z t Ei
Z0 ⎝ Z0 + Zd
2
2
2
⎞
⎞
2
1⎛ 2 ⎞
4n
Z ⎛
⎟⎟ = d ⎜⎜
⎟⎟ = ⎜⎜
⎟⎟ =
= I t1
(1 + n) 2
Z0 ⎝ 1 + Zd Z0 ⎠
n ⎝1+1 n ⎠
⎠
So we have:
I loss = 1 − I t21 = 1 − 0.96 2 = 7.84%
8.24
Noting that c = 1
μ0ε 0 = λω 2π , the radiating power can be written as:
P=
dE q 2ω 4 x02
=
dt 12πε 0c 3
=
ω 2 x02
ω 2q 2
2
12πε 0c ⋅ c
=
4π 2ω 2 x02
μ0ε 0 I 02
2 2
12πε 0λ ω
1 2π
= ×
2 3
i.e.
2π
R=
3
μ0 ⎛ x0 ⎞ 2
⎜ ⎟ I
ε0 ⎝ λ ⎠ 0
2
μ0 ⎛ x0 ⎞
⎛x ⎞
⎜ ⎟ = 787⎜ 0 ⎟ [Ω]
ε0 ⎝ λ ⎠
⎝λ⎠
2
2
By substitution of given parameters, the wavelength is given by:
λ=
ν
c
=
3 × 108
= 600[m] >> x0 = 30[m]
5 × 105
So the radiation resistance and the radiated power are given by:
2
2
⎛ 30 ⎞
⎛x ⎞
R = 787 × ⎜ 0 ⎟ = 787 × ⎜
⎟ = 1.97[Ω]
⎝ 600 ⎠
⎝λ⎠
P=
© 2008 John Wiley & Sons, Ltd
1 2 1
RI 0 = × 1.97 × 20 2 ≈ 400[W ]
2
2
SOLUTIONS TO CHAPTER 9
9.1
Substituting the expression of z into
∂2 z ∂2 z
+
, we have:
∂x 2 ∂y 2
∂2z ∂2z
+ 2 = −(k12 + k22 ) Aei[ωt −( k1x+k2 y )] = −(k12 + k22 ) z
2
∂x ∂y
Noting that k = ω
2
2
c 2 = k12 + k 22 , we have:
∂2 z ∂2 z
ω2
+
=
−
z
∂x 2 ∂y 2
c2
Substituting the expression of z into
1 ∂2 z
, we have:
c 2 ∂t 2
1 ∂2 z
ω 2 [ωt −( k1x+k2 y )]
ω2
= − 2 Ae
=− 2 z
c 2 ∂t 2
c
c
So we have:
∂2 z ∂2 z 1 ∂2 z
+
=
∂x 2 ∂y 2 c 2 ∂t 2
9.2
Boundary condition z = 0
at y = 0 gives:
A1ei (ωt −k1x ) + A2ei (ωt −k1x ) = 0 i.e. A1 = − A2
so the expression of z can be written as:
z = A1{ei[ωt −( k1x+k2 y )] − ei[ωt −( k1x−k2 y )]} = A1[ei (ωt −k1x ) (e −ik2 y − eik2 y )] = −2iA1 sin( k 2 y )ei (ωt −k1x )
Therefore, the real part of z is given by:
zreal = +2 A1 sin k 2 y sin(ωt − k1 x)
Using the above expression, boundary condition z = 0
at y = b gives:
z = −2iA1 sin k2bei (ωt −k1x ) = 0
which is true for any t and x , provided: sin k 2b = 0 , i.e. k 2 =
© 2008 John Wiley & Sons, Ltd
nπ
.
b
9.3
As an analogy to discussion in text page 242, electric field E z between these two planes is the
superposition of the incident and reflected waves, which can be written as:
E z = E1e
i[( k x x + k y y ) −ωt ]
+ E2 e
i [( − k x x + k y y ) −ωt ]
where k x = k cos θ and k y = k sin θ
Boundary condition E z = 0 at x = 0 gives:
( E1 + E2 )e
i ( k y y −ωt )
=0
which is true for any t and y if E1 = − E2 = E0 , so we have:
Ez = E0e
i[( k x x + k y y )−ωt ]
− E0e
i[( − k x x + k y y )−ωt ]
= E0 (eikx x − e −ikx x )e
i ( k y y −ωt )
Using the above equation, boundary condition E z = 0 at x = a gives:
Ez = E0 (eikxa − e − ikxa )e
sin k x ae
i.e.
i ( k y y −ωt )
i ( k y y −ωt )
=0
=0
which is true for any t and y if sin k x a = 0 , i.e. k x = nπ a .
By substitution of the expressions for
λc and λg into
1
λ
2
c
+
1
λ2g
, we have:
2
2
k x2 + k y2
1
k2
⎛ ω ⎞
⎛ kx ⎞ ⎛ k y ⎞
⎟
⎜
=
=
=⎜
+ 2 =⎜
⎟ = 2
⎟ +⎜
2
2
2
⎟
(2π )
(2π )
λc λg ⎝ 2π ⎠ ⎝ 2π ⎠
λ0
⎝ 2πc ⎠
1
2
1
9.4
Electric field components in x, y , z directions in problem 9.3 are given by:
E x = E y = 0 and Ez = E0 (eikx x − e − ikx x )e
i ( k y y −ωt )
By substitution of these values into equation 8.1, we have:
∂
∂
i ( k y −ωt )
Hx =
E z = ik y E0 (eikx x − e −ikx x )e y
∂t
∂y
∂
∂
i ( k y −ωt )
− μ H y = − E z = −ik x E0 (eik x x + e −ikx x )e y
∂t
∂x
−μ
which yields:
© 2008 John Wiley & Sons, Ltd
Hx =
k y E0
μω
Hy = −
(eik x x − e −ik x x )e
k x E0
μω
i ( k y y −ωt )
(eik x x + e −ik x x )e
+C
i ( k y y −ωt )
+D
where C and D are constants, which shows the magnetic fields in both x and y directions
have non-zero values.
9.5
line integral
∫ B ⋅ dl
Current into
paper
a
B
Current out of
paper
b
∫ B ⋅ dl = μI
∴ B = μI b
Closed circuit formed by connecting ends of line length l threaded by flux:
B=
μIla
b
a
b
b
capacitance C per unit length = ε
a
∴ inductance L per unit length = μ
∴ Z0 =
L a
=
C b
μ
Ω
ε
9.6
Text in page 208 shows the time averaged value of Poynting vector for an electromagnetic wave in
a media with permeability of μ and permittivity of ε is given by:
I=
1
cεE02
2
Noting that, in the waveguide of Problem 9.5, the area of cross section is given by: A = ab , and
the velocity of the electromagnetic wave is given by: c = 1
με , the power transmitted by a
single positive travelling wave is given by:
P = IA =
© 2008 John Wiley & Sons, Ltd
ε
1
1
1
1
εE02 = abE02
cεE02 ab = ab
2
2
2
μ
με
9.7
The wave equation of such an electromagnetic wave is given by:
∇ 2E =
1 ∂ 2 E( y , z )
c2
∂t 2
∂ 2E ∂ 2E ∂ 2E 1 ∂ 2E
+
+
=
∂x 2 ∂y 2 ∂z 2 c 2 ∂t 2
i.e.
By substitution of the solution E = E ( y, z )n cos(ωt − k x x) into the above equation, we have:
⎡ 2
ω 2 ∂2
∂ 2 E ( y, z ) ∂ 2 E ( y, z ) ⎤
−
k
E
(
y
,
z
)
+
+
cos(
ω
t
−
k
x
)
=
−
E ( y, z ) cos(ωt − k x x)
x
⎢ x
∂y 2
∂z 2 ⎥⎦
c 2 ∂t 2
⎣
which is true for any t and x if:
∂ 2 E ( y, z ) ∂ 2 E ( y, z ) ⎛ 2 ω 2 ⎞
+
= ⎜⎜ k x − 2 ⎟⎟ E ( y, z )
∂y 2
∂z 2
c ⎠
⎝
∂ 2 E ( y, z ) ∂ 2 E ( y, z )
+
= −k 2 E ( y, z )
2
2
∂y
∂z
or:
where k =
2
ω2
c
2
− k x2
9.8
Using the result of Problem 9.7, the electric field in x direction can be written as:
E x = F ( y, z ) cos(ωt − k x z )
and equation:
∂ 2 F ( y, z ) ∂ 2 F ( y, z )
+
= − k 2 F ( y, z )
2
2
∂y
∂z
is satisfied.
Write F ( y, z ) in form: F ( y, z ) = G ( y ) H ( z ) and substitute to the above equation, we have:
1 ∂ 2G ( y )
1 ∂ 2 H ( z)
+
= −k 2
2
2
G ( y ) ∂y
H ( z ) ∂z
The solution to the above equation is given by:
G ( y ) = C1e
ik y y
+ C2 e
− ik y y
and H ( z ) = D1e
2
2
2
where C1 , C2 , D1 , D2 are constants and k y + k z = k
© 2008 John Wiley & Sons, Ltd
ik z z
+ D2e −ik z z
So the electric field in x direction is given by:
E x = F ( y, z ) cos(ωt − k x x) = G ( y ) H ( z ) cos(ωt − k x x)
= (C1e
ik y y
+ C2 e
−ik y y
)( D1eik z z + D2e −ik z z ) cos(ωt − k x x)
(9.8.1)
Using boundary condition E x = 0 at y = 0 in equation (9.8.1) gives: C1 = −C2 .
Using boundary condition E x = 0 at z = 0 in equation (9.8.1) gives: D1 = − D2 .
So equation (9.8.1) becomes:
E x = C1 D1 (e
or
ik y y
−e
− ik y y
)(eikz z − e −ik z z ) cos(ωt − k x x)
E x = A sin k y y sin k z z cos(ωt − k x x)
(9.8.2)
where A is constant.
Using boundary condition E x = 0 at y = a in equation (9.8.2) gives: sin k y a = 0 , i.e.
k y = mπ a , where m = 1,2,3,L .
Using boundary condition E x = 0 at z = b in equation (9.8.2) gives: sin k z b = 0 , i.e.
k z = nπ b , where n = 1,2,3,L .
Finally, we have:
E x = A sin
mπy
nπz
sin
cos(ωt − k x x)
a
b
where
⎛ m2 n2 ⎞
k 2 = k y2 + k z2 = π 2 ⎜⎜ 2 + 2 ⎟⎟
b ⎠
⎝a
9.9
From problem 9.7 and 9.8, we know:
⎛ m2 n2 ⎞
k x2 = ω 2 c 2 − k 2 = ω 2 c 2 − π 2 ⎜⎜ 2 + 2 ⎟⎟
b ⎠
⎝a
For k x to be real, we have:
⎛ m2 n2 ⎞
k x2 = ω 2 c 2 − π 2 ⎜⎜ 2 + 2 ⎟⎟ > 0
b ⎠
⎝a
i.e.
Therefore, when m = n = 1 ,
© 2008 John Wiley & Sons, Ltd
m2 n2
ω ≥ πc 2 + 2
a
b
ω has the lowest possible value (the cut-off frequency) given by:
ωmin = πc
1
1
+ 2
2
a b
9.10
The dispersion relation of the waves of Problem 9.7 – 9.9 is given by:
⎛ m2 n2 ⎞
k x2 = ω 2 c 2 − π 2 ⎜⎜ 2 + 2 ⎟⎟
b ⎠
⎝a
The differentiation of this equation gives:
2k x dk x =
ω dω
i.e.
k x dk x
2
ωdω
c2
= c 2 or v p vg = c 2
9.11
Using boundary condition z = 0 at x = 0 in the displacement equation gives:
( A1 + A4 )ei (ωt −k2 y ) + ( A2 + A3 )ei (ωt +k2 y ) = 0
which is true for any t and y if:
A1 = − A4 and A2 = − A3
so we have:
z = A1{ei[ωt −( k1x+k2 y )] − ei[ωt −( − k1x+k2 y )]} + A2{ei[ωt −( k1x−k2 y )] − ei[ωt −( − k1x−k2 y )]
= −2 A1i sin k1 xei (ωt −k2 y ) + 2 A2i sin k1 xei (ωt +k2 y )
= −2i sin k1 x[ A1e
i ( ωt − k 2 y )
− A2e
i ( ωt + k 2 y )
(9.11.1)
]
Using boundary condition z = 0 at y = 0 in equation (9.11.1) gives:
− 2i sin k1 x( A1 − A2 )eiωt = 0
which is true for any t and x if:
A1 = A2
Therefore, equation (9.11.1) becomes:
z = −4 A1 sin k1 x sin k 2 yeiωt
and the real part of z is given by:
zreal = −4 A1 sin k1 x sin k 2 y cos ωt
Using boundary condition z = 0 at x = a in equation (9.11.2) gives:
sin k1a = 0 , i.e. k1 =
n1π
, where n1 = 1,2,3,L .
a
© 2008 John Wiley & Sons, Ltd
(9.11.2)
Using boundary condition z = 0 at y = b in equation (9.11.2) gives:
sin k 2b = 0 , i.e. k 2 =
n2π
, where n2 = 1,2,3,L .
b
9.12
Multiplying the equation of geometric progression series by e
e − hν
kT
N = e − hν
kT
∑N
n
= N 0 [ e − hν
+ e −2 hν
kT
− hν kT
+ e −3hν
kT
kT
on both sides gives:
+ L + e − ( n +1) hν
kT
]
n
so we have:
N − e − hν
kT
N = N 0 [1 − lim e − ( n+1) hν
kT
n→∞
] = N0
i.e.
N=
N0
1 − e −hν
kT
The total energy over all the n energy states is given by:
E = ∑ En = ∑ N n nhν = hν ∑ N n n
n
n
= hνN 0 (e
−hν kT
n
+ 2e
Multiplying the above equation by e
Ee − hν
−2 hν kT
− hν kT
+ 3e −3hν
+ L + ne −nhν
kT
)
on both sides gives:
= hνN 0 [e −2 hν
kT
+ 2e −3hν
= hνN 0 lim[e − hν
kT
+ e −2 hν
kT
kT
kT
+ 3e −4 hν
kT
+ L + ne − ( n+1) hν
kT
]
so we have:
E − Ee − hν
kT
= hνN 0e −hν
= hνN 0e −hν
= hνN 0e −hν
n→∞
kT
lim[1 + e −hν
n→∞
kT
+ e − 2 hν
kT
⎛ 1 − e −nhν kT
n
− nhν kT
lim⎜⎜
−
h
ν
kT
n→∞ 1 − e
e
⎝
1
kT
1 − e −hν kT
kT
E = N0
i.e.
kT
+ e −3hν
kT
+ L + e − nhν
+ L + e −( n−1) hν
kT
− ne −nhν
kT
− ne − ( n+1) hν
kT
]
⎞
⎟⎟
⎠
hνe − hν kT
(1 − e −hν kT ) 2
Hence, the average energy per oscillator is given by:
ε =
E
=
N
N0
hνe − hν kT
e −hν kT
hν
(1 − e −hν kT ) 2
= hν
= hν kT
−hν kT
N0
1− e
e
−1
−hν kT
1− e
By expanding the denominator of the above equation for hν << kT , we have:
© 2008 John Wiley & Sons, Ltd
kT
]
ε =
hν
[1 + hν kT + (hν kT ) 2 + L] − 1
2
≈
hν
= kT
hν kT
which is the classical expression of Rayleigh-Jeans for an oscillator with two degrees of freedom.
Alternative derivation for E and
ε:
∞
E = Nε = N nhν where nhν =
∑ (nhνe
n =0
∞
∑e
−nhν kT
)
− nhν kT
n =0
∞
∂
log
e −nhν
∑
−1
∂ (kT )
n =0
∂
1
=−
log
−1
∂ (kT )
1 − e −hν kT
∴ nhν = −
=
kT
hνe −hν kT
1 − e −hν kT
N 0 hνe − hν kT
∴ E = N nhν =
(1 − e −hν kT ) 2
and
ε = nhν =
hν
e
−hν kT
−1
9.13
One solution of this Schrodinger’s time-independent equation can be written as:
ψ = X ( x)Y ( y ) Z ( z )
Substituting this expression into the Schrodinger’s equation and dividing
ψ on both sides of the
equation, we have:
1 ∂ 2 X ( x)
1 ∂ 2Y ( y )
1 ∂ 2Z ( z)
8π 2 m
+
+
=
−
E
X ( x) ∂x 2
Y ( y ) ∂y 2
Z ( z ) ∂z 2
h2
which yields:
∂ 2 X ( x)
+ E x X ( x) = 0
∂x 2
∂ 2Y ( y )
+ E yY ( y ) = 0
∂y 2
∂ 2Z ( z)
+ Ez Z ( z) = 0
∂z 2
where E x , E y , E z are constants and satisfy: E x + E y + E z =
By solving the above three equations, we have:
© 2008 John Wiley & Sons, Ltd
8π 2 m
E
h2
i Ex x
+ Dx e
− i Ex x
i Ey y
+ Dy e
−i E y y
X ( x) = C x e
Y ( y) = C y e
Z ( z) = Cz e
i Ez z
+ Dz e
−i Ez z
and
ψ = (C x ei
Ex x
+ Dx e
− i Ex x
)(C y e
i Ey y
+ Dy e
−i Ey y
)(C z e
i Ez z
+ Dz e
− i Ez z
)
where C x , Dx , C y , D y , C z , Dz are constants.
Boundary condition
ψ = 0 at x = 0 gives C x = − Dx , boundary condition ψ = 0 at
y = 0 gives C y = − Dy , boundary condition ψ = 0 at z = 0 gives C z = − Dz , so we
have:
ψ = C x C y C z (e i
Ex x
−e
− i Ex x
)(e
i Ey y
−e
−i Ey y
)(e
i Ez z
−e
− i Ez z
)
= A sin E x x sin E y y sin E z z
2
⎛ lπ ⎞
⎟⎟ ,
Using the above expression ψ , boundary condition ψ = 0 at x = Lx gives: E x = ⎜⎜
L
⎝ x⎠
boundary condition
ψ = 0 at x = Ly
⎛ nπ
x = Lz gives: E z = ⎜⎜
⎝ Lz
⎛ rπ
gives: E y = ⎜
⎜L
⎝ y
2
⎞
⎟ , boundary condition ψ = 0 at
⎟
⎠
2
⎞
⎟⎟ , where l , r , n = 0,1,2,L , so we have:
⎠
2
⎛ lπ ⎞ ⎛ rπ ⎞ ⎛ nπ ⎞
8π 2 m
⎜⎜ ⎟⎟ + ⎜ ⎟ + ⎜⎜
⎟⎟ =
E
⎜ ⎟
h2
⎝ Lx ⎠ ⎝ Ly ⎠ ⎝ Lz ⎠
2
i.e.
E=
When Lx = Ly = Lz = L , E =
2
h 2 ⎛⎜ l 2 r 2 n 2 ⎞⎟
+
+
8m ⎜⎝ L2x L2y L2z ⎟⎠
h2 2 2
(l + r + n 2 ). If E = E0 for l = 1 , r = n = 0 , the
8mL2
next energy levels are given by:
E = 3E0 for l = r = n = 1 .
E = 6E0 for l = r = 1, n = 2 ; l = n = 1, r = 2 and n = r = 1, l = 2 , which is a three-fold
degenerate state.
E = 9E0 for l = r = 2, n = 1 ; l = n = 2, r = 1 and n = r = 2, l = 1 which is a three-fold
© 2008 John Wiley & Sons, Ltd
degenerate state.
E = 11E0 for l = r = 1, n = 3 ; l = n = 1, r = 3 and n = r = 1, l = 3 which is a three-fold
degenerate state.
E = 12E0 for l = r = n = 2 .
E = 14E0
for
l = 1, r = 2, n = 3
l = 1, n = 3, r = 2
;
;
l = 2, n = 1, r = 3
;
l = 2, n = 3, r = 1 ; l = 3, n = 1, r = 2 and l = 3, n = 2, r = 1 which is a six-fold degenerate
state.
9.14
Planck’s Radiation Law is given by:
Eν dν =
8πν 2 hν
dν
c 3 e hν kT − 1
At low energy levels hν << kT , by expansion of e
hν kT
in series, the above equation
becomes:
Eν dν =
8πν 2
c3
hν
hν 1 ⎛ hν ⎞
1+
+ ⎜
⎟ + L −1
kT 2 ⎝ kT ⎠
8πν 2 hν
8πν 2 kT
dν
dν =
≈ 3
c3
c hν kT
2
dν
which is Rayleigh-Jeans expression
9.15
Using the variable x = ch
λkT , energy per unit range of wavelength can be written as:
Eλ =
8πch(kTx)5
8π (kTx)5
=
(ch)5 (e x − 1) (ch) 4 (e x − 1)
Substitute the expression of Eλ into integral
© 2008 John Wiley & Sons, Ltd
∫
∞
0
Eλ dλ and, we have:
∫
∞
0
Eλ dλ = ∫
∞
0
=∫
∞
0
8π (kTx)5
ch
d
x
4
(ch) (e − 1) xkT
8π (kT ) 4 x 3
dx
(ch)3 (e x − 1)
8π (kT ) 4 π 4
=
×
(ch)3
15
8π 5 k 4 4
=
T
15c 3h 3
i.e.
∫
∞
0
where a =
Eλ dλ = aT 4
8π 5 k 4
15c 3h3
9.16
Using the expression of Eλ in Problem 9.15, the wavelength
λm at which Eλ is maximum
should satisfy the equation:
d
Eλ = 0
dx
which yields:
d
x5
=0
dx (e x − 1)
5 x 4 (e x − 1) − e x x 5
=0
(e x − 1) 2
i.e.
⎛ x⎞ x
⎜1 − ⎟e = 1
⎝ 5⎠
i.e.
where x =
ch
λkT
9.17
The most sensitive wavelength to the human eye can be given by substituting the sun’s
temperature T = 6000[ K ] into equation ch
λm = 5kT , i.e.:
ch
3 × 108 × 6.63 × 10 −34
λm =
≈ 4.7 × 10 −7 [m]
=
−23
5kT 5 × 1.38 × 10 × 6000
which is in the green region of the visible spectrum.
© 2008 John Wiley & Sons, Ltd
9.18
Substituting the tungsten’s temperature T = 2000[ K ] into equation ch
λm =
λm = 5kT , i.e.:
ch
3 × 108 × 6.63 × 10 −34
≈ 14 × 10 −7 [m]
=
5kT 5 × 1.38 × 10 −23 × 2000
which is well into infrared.
9.19
As an analogy to the derivation of number of points in
ν state shown in text page 250, 251, the
number of points in k space between k and k + dk is given by:
1 (volume of spherical shell)
8
volume of cell
4πk 2 dk ⎛ L ⎞
⋅⎜ ⎟
8
⎝π ⎠
=
3
Noting that for each value of k there are two allowed states, the total number of states in k
space between k and k + dk is given by:
4πk 2 dk ⎛ L ⎞
⋅⎜ ⎟
P (k ) = 2 ⋅
8
⎝π ⎠
From E = (h
2
3
2m* )k 2 , we have k = (2m* h 2 ) E . By substitution into the above equation,
we get the number of states S ( E )dE in the energy interval dE given by:
S ( E )dE = 2 ⋅
=
4π (2m* h 2 )3 2 Ed E ⎛ L ⎞
⎜ ⎟
8
⎝π ⎠
L3 (2m* h 2 )3 2 E
L3 (2m* h 2 )3 2 E
dE
=
dE
2π 2
2π 2 E
Provided m ≈ m and A = L , we have:
*
3
S (E) =
A ⎛ 2m ⎞
⎜
⎟
2π 2 ⎝ h 2 ⎠
32
E
Since Fermi energy level satisfies the equation:
∫
Ef
0
S ( E )dE = N
By substitution of the expression of S ( E ) , we have:
∫
Ef
0
i.e.
© 2008 John Wiley & Sons, Ltd
3
A ⎛ 2m ⎞
⎜
⎟
2π 2 ⎝ h 2 ⎠
32
E dE = N
A ⎛ 2m ⎞
⎟
⎜
2π 2 ⎝ h 2 ⎠
32
E 3f 2
32
=N
which gives:
h 2 ⎛ 3π 2 N ⎞
⎜
⎟
Ef =
2m* ⎜⎝ A ⎟⎠
provided m ≈ m
*
© 2008 John Wiley & Sons, Ltd
23
SOLUTIONS TO CHAPTER 10
10.1
The wave form in the upper figure has an average value of zero and is an odd function
of time, so its Fourier series has a constant of zero and only sine terms. Since the
wave form is constant over its half period, the Fourier coefficient bn will be zero if
n is even, i.e. there are only odd harmonics and the harmonics range from 1,3,5 to
infinity.
The wave form in the lower figure has a positive average value and is a even function
of time, so its Fourier series has a constant of positive value and only cosine terms.
Since τ T ≠ 1 2 , there are both odd and even harmonics. The harmonics range from
1,2,3 to infinity.
10.2
Such a periodic waveform should satisfy: f ( x) = − f ( x − T 2) , where T is the
period of the waveform. Its Fourier coefficient of cosine terms can be written as:
2 T
2πnx
f ( x) cos
dx
∫
T 0
T
T
2⎡ T2
2πnx
2πnx ⎤
dx + ∫ f ( x) cos
dx ⎥
= ⎢ ∫ f ( x) cos
T
0
2
T⎣
T
T
⎦
T
2⎡ T2
2πnx
2πnx
⎤
dx + ∫ − f ( x − T 2) cos
d ( x − T 2) ⎥
= ⎢ ∫ f ( x) cos
T
0
2
T⎣
T
T
⎦
an =
If n is even, we have
cos
2πn( x − T 2)
2πnx
⎛ 2πnx
⎞
= cos⎜
− nπ ⎟ = cos
T
T
⎝ T
⎠
Hence, by substituting into an and using u = x − T 2 , we have:
an =
T 2
2⎡ T2
2πnx
2πnu ⎤
(
)
cos
−
du ⎥ = 0
dx
f (u ) cos
f
x
∫
∫
⎢
0
0
T
T
T⎣
⎦
Similarly, the coefficient of sine terms is given by:
2 T
2πnx
f ( x) sin
dx
∫
0
T
T
T
2⎡ T2
2πnx
2πnx ⎤
dx + ∫ f ( x) sin
dx ⎥
= ⎢ ∫ f ( x) sin
T
0
2
T⎣
T
T
⎦
T
2⎡ T2
2πnx
2πnx
⎤
dx + ∫ − f ( x − T 2) sin
d ( x − T 2) ⎥
= ⎢ ∫ f ( x) sin
T
0
2
T⎣
T
T
⎦
bn =
© 2008 John Wiley & Sons, Ltd
If n is even, using u = x − T 2 , we have:
bn =
T 2
2⎡ T2
2πnx
2πnu ⎤
du ⎥ = 0
dx − ∫ f (u ) sin
f ( x) sin
∫
⎢
0
T
T
T⎣0
⎦
Therefore, if n is even, the Fourier coefficients of both cosine and sine terms are
zero, i.e. there are no even order frequency components.
10.3
The constant term of the Fourier series is given by:
h
1
1 2π
1 π
a0 =
ydx =
h sin xdx =
∫
∫
π
2
2π 0
2π 0
The Fourier coefficient of cosine term is given by:
1 2π
h π
an = ∫ y cos nxdx = ∫ sin x cos nxdx
π
when n = 1 , we have:
=
∫
π
when n > 1 , we have:
an =
π
h
a1 =
0
π
h
π∫
h
2π
π
0
h
2π
sin x cos xdx =
π
∫
0
sin 2 xdx = 0
sin x cos nxdx
0
∫
π
0
sin(1 + n) x + sin(1 − n) xdx
0
π
=−
h ⎡ 1
1
⎤
cos(1 + n) x +
cos(1 − n) x ⎥
⎢
2π ⎣1 + n
1− n
⎦0
which gives:
h 2
h 2
h 2
, a3 = 0 , a4 = −
, a5 = 0 , a6 = −
,…
π 1⋅ 3
π 3⋅5
π 5⋅7
The Fourier coefficient of sine term is given by:
1 2π
h π
bn = ∫ y sin nxdx = ∫ sin x sin nxdx
a2 = −
when n = 1 , we have:
π
0
h
π
b1 =
π∫
bn =
h
when n > 1 , we have:
=
h
2π
0
π
∫
π
0
π
∫
0
π
sin x sin xdx =
h
0
π
π∫
0
sin 2 xdx =
h
2
sin x sin nxdx
cos(1 − n) x − cos(1 + n) xdx
π
h ⎡ 1
1
⎤
sin(1 − n) x +
sin(1 + n) x ⎥
=
⎢
2π ⎣1 − n
1+ n
⎦0
=0
© 2008 John Wiley & Sons, Ltd
Overall, the Fourier series is given by:
∞
∞
1
y = a0 + ∑ an cos nx + ∑ bn sin nx
2
1
1
=
h⎛
π
2
2
2
⎞
cos 6 x L⎟
cos 4 x −
cos 2 x −
sin x −
⎜1 +
π ⎝ 1⋅ 2
5⋅7
3⋅5
1⋅ 3
⎠
10.4
Such a wave form is a even function with a period of π . Hence, there are only
constant term and cosine terms.
The constant term is given by:
1
1 π
2h
a0 = ∫ h sin xdx =
2
π 0
π
which doubles the constant shown in Problem 10.3
The coefficient of cosine term is given by:
an =
=
4h
=
2h
π
π
4h
π
∫
π 2
∫
0
π 2
∫
0
π 2
0
sin x cos
2πnx
π
dx
sin x cos 2nxdx
[sin(1 + 2n) x + sin(1 − 2n) x]dx
π 2
1
2h ⎡ 1
⎤
=− ⎢
cos(1 − 2n) x ⎥
cos(1 + 2n) x +
π ⎣1 + 2n
1 − 2n
⎦0
which gives:
h 2
h 2
h 2
, a2 = −
, a3 = −
,…
π 1⋅ 3
π 3⋅5
π 5⋅7
Therefore the Fourier series is given by:
a1 = −
y=
=
∞
2πnx
1
a0 + ∑ an cos
π
2
1
h⎛
2
2
2
⎞
cos 6 x L⎟
cos 4 x −
cos 2 x −
⎜1 −
π ⎝ 1⋅ 3
5⋅7
3⋅5
⎠
Compared with Problem 10.3, the modulating ripple of the first harmonic
h
sin x
2
disappears.
10.5
f (x) is even function in the interval ± π , so its Fourier series has a constant term
given by:
1
1
a0 =
2π
2
© 2008 John Wiley & Sons, Ltd
π
∫π
−
f ( x)dx =
1
2π
π
∫π
−
x 2 dx =
π2
3
The coefficient of cosine term is given by:
2πnx
dx
2π
2 π
2 π 2
= ∫ x 2 cos nxdx =
x d sin nx
0
π
nπ ∫0
π
π
2 ⎡ 2
4 π
=
x sin nx − ∫ sin nxdx 2 ⎤ = 2 ∫ xd cos nx
0
⎢
⎥
0
⎦ nπ 0
nπ ⎣
π
4
4
4
π
= 2 ⎡ x cos nx 0 − ∫ cos nxdx ⎤ = 2 cos nπ = (−1) n 2
⎢
⎥
0
⎦ n
nπ⎣
n
an =
2
π
π∫
0
f ( x) cos
Therefore the Fourier series is given by:
f ( x) =
∞
4
2πnx 1 2 ∞
1
= π + ∑ (−1) n 2 cos nx
a0 + ∑ an cos
2π
3
2
n
1
1
10.6
The square wave function of unit height f (x) has a constant value of 1 over its first
half period [0, π ] , so we have:
f (π 2) = 1
By substitution into its Fourier series, we have:
f (π 2) ≈
5π ⎞
4⎛ π 1
3π 1
+ sin
⎜ sin + sin
⎟ =1
π⎝ 2 3
2 ⎠
2 5
i.e.
1 1 1 π
1− + − =
3 5 7 4
10.7
It is obvious that the pulse train satisfies f (t ) = f (−t ) , i.e. it is an even function. The
cosine coefficients of its Fourier series are given by:
τ
2πnx
2
2π
4 τ
2πnx
4 T
an = ∫ cos
dx = ⋅
=
nτ
sin
sin
T 0
T
T 2πn
T 0 nπ
T
10.8
2π
2π
nτ →
nτ , so we have:
T
T
2
2π
2 2π
4τ
an =
nτ ≈
nτ =
sin
⋅
nπ
T
nπ T
T
As τ becomes very small, sin
We can see as τ → 0 , an → 0 , which shows as the energy representation in time
© 2008 John Wiley & Sons, Ltd
domain → 0 , the energy representation in frequency domain → 0 as well.
10.9
The constant term of the Fourier series is given by:
1
1 T2 1
1 τ 1
1
a0 = ∫
dt = ∫
dt =
2
T −T 2 2τ
T −τ 2τ
T
The coefficient of cosine term is given by:
τ
2πnt
1
2πnτ
2πnt 4 1 T
4 τ 1
an = ∫
= ⋅ ⋅
=
sin
sin
cos
0
T 2τ
T
T 2τ 2πn
T 0 nπτ
T
As τ → 0 , we have:
2πnτ
1 2πnτ 2
≈
⋅
=
nπτ
T
nπτ T
T
Now we have the Fourier series given by:
an =
f (t ) =
1
sin
∞
2πnt 1 2 ∞
2πnt
1
= + ∑ cos
a0 + ∑ an cos
T
T T n=1
T
2
n =1
10.10
Following the derivation in the problem, we have:
1 +∞
F (ω )eiωt dω
∫
−
∞
2π
+∞ 1
iωT
iωt
∫−∞ iω (1 − e )e dω
+∞ 1
iωT
iωt
∫−∞ iω (1 − e )e dω
+∞⎡ 1
1 iω ( t −T ) ⎤
iωt
∫−∞ ⎢⎣ iω e − iω e ⎥⎦ dω
f (t ) =
1
2π
1
=
2π
1
=
2π
=
Using the fact that for T very large:
+∞ 1
+∞ 1
−iωT
iω ( t −T )
∫−∞ iω e dω = ∫−∞ iω e dω = −π
we have:
1 1 +∞ 1 iωt
f (t ) = +
e dω
2 2π ∫−∞ iω
10.11
Following the derivation in the problem, we have:
© 2008 John Wiley & Sons, Ltd
+∞
F (ν ) = ∫ f (t ′)e −i 2πνt′ dt ′
−∞
=∫
+τ 2
−τ 2
=
=
f 0e −i 2π (ν −ν 0 )t′ dt ′
+τ 2 −i 2π (ν −ν ) t ′
f0
0
e
d [−i 2π (ν −ν 0 )t ′]
− i 2π (ν −ν 0 ) ∫−τ 2
f0
(e −i 2π (ν −ν 0 )τ 2 − ei 2π (ν −ν 0 )τ 2 )
− i 2π (ν −ν 0 )
= f 0τ
sin[π (ν −ν 0 )τ ]
π (ν −ν 0 )τ
which shows the relative energy distribution in the spectrum given by:
F (ν ) = ( f 0τ ) 2
2
sin 2 [π (ν −ν 0 )τ ]
[π (ν −ν 0 )τ ]2
follows the intensity distribution curve in a single slit diffraction pattern given by:
I = I0
sin 2 (πd sin θ λ )
(πd sin θ λ ) 2
10.12
The energy spectrum has a maximum when:
sin 2 [π (ν max −ν 0 )τ ]
=1
[π (ν max −ν 0 )τ ]2
i.e.
π (ν min −ν 0 )τ = 0 or ν min = ν 0
The frequencies for the minima of the energy spectrum are given by:
F (ν min ) = ( f 0τ ) 2
i.e.
sin 2 [π (ν min −ν 0 )τ ]
=0
[π (ν min −ν 0 )τ ]2
n
n
π (ν min
−ν 0 )τ = nπ or ν min
−ν 0 =
n
τ
where n = −∞,L,−3,−2,−1,1,2,3,L,+∞
Hence, the total width of the first maximum of the energy spectrum is given by:
2
1
+1
−1
or Δν =
−ν min
=
2Δν = ν min
τ
Using the differentiation of the relation ν =
Δν =
we have
© 2008 John Wiley & Sons, Ltd
c
λ
c
λ2
τ
:
Δλ
c
λ2
Δλ =
1
τ
or cτ =
λ2
Δλ
which is the coherence length l of Problem 10.11.
10.13
Use the relation Δλ =
λ20
c
Δν , we have:
(6.936 × 10 −7 ) 2
Δλ =
× 10 4 ≈ 1.6 × 10−17 [m]
8
3 × 10
Then, using the result in Problem 10.12, the coherence length is given by:
l=
λ20 (6.936 ×10 −7 ) 2
=
= 3 × 10 4 [m]
Δλ
1.6 × 10 −17
10.14
Referring to pages 46 and 47 of the text, and in particular to the example of the
radiating atom, we see that the energy of the damped simple harmonic motion:
E = E0e −ω 0 t Q = E0 e −1 where Q ω0 = t , the period for which the atom radiates
before cut off at e −1 .
The length of the wave train radiated by the atom is l = ct where c is the velocity
of light and l is the coherence length which contains Q radians.
Since the coherence length is finite the radiation cannot be represented by a single
angular frequency ω0 but by a bandwidth Δω centred about ω0 .
Now Q = ω0 Δω so Q ω0 = t = 1 Δω . Writing t = Δt we have ΔωΔt = 1 or
ΔνΔt = 1 2π .
The bandwidth effect on the spectral line is increased in a gas of radiating atoms at
temperature T . Collisions between the atoms shorten the coherence length and the
Doppler effect from atomic thermal velocities adds to Δν .
10.15
The Fourier transform of f (t ) gives:
+∞
+∞
+∞
−∞
−∞
−∞
F (ν ) = ∫ f (t )e −i 2πνt dt = ∫ f 0ei 2πν 0t e −t τ e −i 2πνt dt = ∫ f 0e[ i 2π (ν 0 −ν )−1 τ ]t dt
Noting that t ≥ 0 for f (t ) , we have:
© 2008 John Wiley & Sons, Ltd
F (ν ) = ∫
+∞
0
=
f 0e[ i 2π (ν 0 −ν )−1 τ ]t dt
{
+∞
f0
e[ i 2π (ν 0 −ν )−1 τ ]t
0
i 2π (ν 0 −ν ) − 1 τ
=−
}
f0
f0
=
i 2π (ν 0 −ν ) − 1 τ 1 τ + i 2π (ν −ν 0 )
Hence, the energy distribution of frequencies in the region ν −ν 0 is given by:
2
f0
f2
f 02
f 02
F (ν ) =
= 02 =
=
1 τ + i 2π (ν −ν 0 )
r
(1 τ ) 2 + [2π (ν −ν 0 )]2 (1 τ ) 2 + (ω − ω0 ) 2
2
10.16
In the text of Chapter 3, the resonance power curve is given by the expression:
F02
F02 r
F02 r
Pav =
cos φ =
=
2Z m
2Z m2 2[r 2 + (ωm − s ω ) 2 ]
In the vicinity of ω0 = s m , we have ω ≈ ω0 , so the above equation becomes:
Pav ≈
F02 r
F02 r
f 02
2
=
=
= F (ν )
2
2
2
2
2
2
2
2[r + (ωm − s ω ) ] 2[r + m (ω − ω0 ) ] (1 τ ) + (ω − ω0 )
where f 02 =
F02 r
m
and τ =
2
2m
r
The frequency at half the maximum value of F (ν ) is given by:
2
F (ν ) =
2
i.e.
f 02
( f oτ ) 2
=
(1 τ ) 2 + (ω − ω0 ) 2
2
ω − ω0 = ± 1 τ
so the frequency width at half maximum is given by:
Δω = 2 τ or Δν = 1 πτ
In Problem 10.12 the spectrum width is given by:
Δν ′ = 1 τ
so we have the relation between the two respectively defined frequency spectrum
widths given by:
Δν ′
Δν =
π
© 2008 John Wiley & Sons, Ltd
Noting that Δλ =
λ20
c
Δν and Δλ ′ =
λ20
c
Δν ′ , we have:
Δλ =
Δλ ′
π
where Δλ and Δλ ′ are the wavelength spectrum widths defined here and in
Problem 10.12, respectively.
If the spectrum line has a value Δλ = 3 ×10−9 m in Problem 10.12, the coherence
length is given by:
l=
(5.46 × 10−7 ) 2
λ20
λ2
= 0 =
≈ 32 × 10 −6 [m]
−9
3 × 10 × π
Δλ ′ Δλπ
10.17
The double slit function (upper figure) and its self convolution (lower figure) are
shown below:
τ
τ
d
2τ
d −τ
2τ
d −τ
2τ
Fig. A.10.17
10.18
The convolution of the two functions is shown in Fig. A.10.18.1.
=
⊗
d
d
Fig. A.10.18.1
© 2008 John Wiley & Sons, Ltd
d
d
The respective Fourier transforms of the two functions are shown in Fig. A.10.18.2.
F(
2d
=
)
d
1d
2d
F(
)
=
d
1d
Fig. A.10.18.2
Hence, the Fourier transform of the convolution of the two functions is the product of
the Fourier transform of the individual function, which is shown in Fig. A.10.18.3
© 2008 John Wiley & Sons, Ltd
⊗
F(
d
=
)
d
) × F(
F(
)
d
d
2d
=
2d
×
1d
1d
2d
=
1d
Fig. A.10.18.3
10.19
The area of the overlap is given by:
1
⎛1
⎞
A = 2⎜ r ⋅ 2θ − ⋅ 2r sin θ ⋅ r cosθ ⎟
2
⎝2
⎠
2
= r (2θ − 2 sin θ cos θ )
where cos θ =
R
R2
and sin θ = 1 − 2
4r
2r
Hence the convolution is given by:
1
⎡
⎤
2
2 R
⎛
⎞
R
R
2⎢
−1
⎥
O( R) = r 2 cos
− 2⎜⎜1 − 2 ⎟⎟
⎢
⎥
2r
4
r
2
r
⎝
⎠
⎥⎦
⎣⎢
© 2008 John Wiley & Sons, Ltd
The convolution O(R ) in the region [0,2r ] is sketched below:
O(R)
πr
2
0
2r
Fig. A.10.19
© 2008 John Wiley & Sons, Ltd
R
SOLUTIONS TO CHAPTER 11
11.1
r2
2R1
A′
r2
2R2 A
r
R2
z
B′
O
B
f
d
R1
Fig A.11.1
In a bi-convex lens, as shown in Fig A.11.1, the time taken by the wavefront to travel
through path AB is the same as through path A′B′ , so we have:
r2 ⎞
r2
r 2 ⎞ ⎛ 1 ⎞⎛
nd r 2 1 ⎛ n ⎞⎛
⎟⎟ + ⎜ ⎟⎜⎜ z +
⎟
+
⋅ = ⎜ ⎟⎜⎜ d +
−
2 R2 2 R1 ⎠ ⎝ c ⎠⎝
2 R1 ⎟⎠
c 2 R2 c ⎝ c ⎠⎝
which yields:
⎛1
1 ⎞ r2
z = (n − 1)⎜⎜ − ⎟⎟
⎝ R1 R2 ⎠ 2
P=
i.e.
⎛1
1
1 ⎞
= (n − 1)⎜⎜ − ⎟⎟
f
⎝ R1 R2 ⎠
11.2
A′
z B′
r
A
d
B
f
O
Fig A.11.2
As shown in Fig A.11.2, the time taken by the wavefront to travel through path AB
© 2008 John Wiley & Sons, Ltd
is the same as through path A′B′ , so we have:
n0 d (n0 − αr 2 )d z
=
+
c
c
c
which yields:
z = αr 2 d
f =
i.e.
1
2αd
11.3
Choosing the distance PF = λ 2 then, for the path difference BF − BF ′ , the phase
difference is π radians.
Similarly for the path difference AF ′ − AF the phase difference is π radians.
Thus for the path difference AF ′ − BF ′ the phase difference is 2π radians and the
resulting amplitude of the secondary waves is zero.
x
λ
Writing F ′F = x 2 , we then have in the triangle F ′FP : sin θ = , so the width of
2
2
the focal spot is x =
λ
sin θ
.
11.4
If a man’s near point is 40cm from his eye, his eye has a range of accommodation of:
1
= 2.5[dioptres]
0.4
Noting that a healthy eye has a range of accommodation of 4 dioptres, he needs
spectacles of power:
P = 4 − 2.5 = 1.5[dioptres]
If anther man is unable to focus at distance greater than 2m, his eye’s minimum
accommodation is:
1
= 0.5[dioptres]
2
Therefore, he needs diverging spectacles with a power of -0.5 dioptres for clear image
of infinite distance.
11.5
Noting that
l′
y′ l ′
= , we have the transverse magnification given by: M T = . The
y l
l
angular magnification is given by: M α =
© 2008 John Wiley & Sons, Ltd
β
y l
d
=
= 0 . Using the thin lens power
γ y d0
l
equation: P =
1
1
1
1
− , we have = P + , i.e. M α = d 0 ( P + 1 l ′) = Pd 0 + 1
l
l′
− l′ − l
11.6
The power of the whole two-lens telescope system is zero, so we have:
P = P1 + P2 − LP1P2 = 0
where L is the separation of the two lenses. Noting that P1 =
1
1
and P2 = , we
f0
fe
have:
1 1
1 1
+ −L
=0
f0 fe
fo fe
which gives: L = f 0 + f e
Suppose the image height at point I is h , we have:
α = d 2 L = h f o and α ′ = D 2 L = h f e
which yields:
Mα =
α′
f
D
= o =
α
fe
d
11.7
As shown from Figure 11.20, the magnification of objective lens is given by:
Mo = −
x′
1
. Suppose the objective lens is a thin lens, we have: Po =
i.e.
f o′
f o′
M o = − Po x′ .
Similarly, the magnification of eye lens is given by: M e =
is a thin lens, we have: Pe =
1
i.e. M o = Pe d o .
f e′
So we have the total magnification given by:
M = M o M e = − Po Pe d o x′
© 2008 John Wiley & Sons, Ltd
do
. Suppose the eye lens
f e′
11.8
P α
Air
C
O
I
n
Glass
Fig.A.11.8(a)
As shown in Fig.A.11.8(a), Snell’s law gives:
n sin ∠OPC = sin α = sin ∠IPC
In triangle OCP , we have:
OC
PC
=
sin ∠OPC sin ∠POC
R
R
=
n sin ∠OPC sin ∠POC
R
R
=
sin ∠IPC sin ∠POC
∠IPC = ∠POC
Δ ’s OPC and PIC are similar
i.e.
i.e.
i.e.
i.e.
OC
PC
=
PC
IC
i.e.
2
IC =
i.e.
PC
R2
=
= nR
OC
Rn
Alternative proof using Fermat’s Principle:
P α
Air
I
O
C
Glass
© 2008 John Wiley & Sons, Ltd
B
n
Fig.A.11.8(b)
Equate optical paths IP = n 2 OP . Let IC = k , CB = l and PB = d , then:
2
2
⎡⎛ R ⎞ 2
⎤
IP = (k + c) + d = n (OP) = n ⎢⎜ + l ⎟ + d 2 ⎥
⎠
⎣⎢⎝ n
⎦⎥
2
i.e.
i.e.
2
2
2
2
2
k 2 + 2kl + l 2 + d 2 = R 2 + 2nRl + n 2 (l 2 + d 2 )
k 2 + 2kl + R 2 = R 2 + 2nRl + n 2 R 2
k = IC = nR
that is
11.9
(a) The powers of the two spherical surfaces are given by:
n′ − n 1.5 − 1
n′ − n 1 − 1.5
P1 =
=
= −0.5 and P2 =
=
=0
R
−1
R
∞
Suppose a parallel incident ray (α1 = 0) strikes the front surface of the system at a
height of y1 . By using matrix method, we can find the ray angle α 2′ and height y2′
at the back surface of the system given by:
⎡α 2′ ⎤
⎡α1 ⎤ ⎡1 P2 ⎤ ⎡ 1 0⎤ ⎡1 P1 ⎤ ⎡α1 ⎤
⎢ y′ ⎥ = R2T12 R1 ⎢ y ⎥ = ⎢0 1 ⎥ ⎢− d ′ 1⎥ ⎢0 1 ⎥ ⎢ y ⎥
⎦⎣ 1
⎦⎣ 1 ⎦
⎣ 2⎦
⎣ 1⎦ ⎣
⎦⎣
0⎤ ⎡1 − 0.5⎤ ⎡ 0 ⎤ ⎡− 0.5 y1 ⎤
⎡1 0 ⎤ ⎡ 1
=⎢
=
⎥
⎢
⎥⎢
1 ⎥⎦ ⎢⎣ y1 ⎥⎦ ⎢⎣ 1.15 y1 ⎥⎦
⎣0 1⎦ ⎣− 0.3 1⎦ ⎣0
So the focal length is given by:
f =
y1
y
= 1 = 2[m]
α 2′ 0.5 y1
The principal plane is located at a distance d to the left side of the right-end surface
of the system, which is given by:
d=
y′2 − y1 1.15 y1 − y1
=
= 0.3[m]
0.5 y1
α 2′
(b) The powers of the four spherical surfaces are given by:
n′ − n 1.5 − 1
P1 =
=
=0
R
∞
n − n′ 1 − 1.5
P2 =
=
=1
R
− 0.5
n′ − n 1.5 − 1
P3 =
=
= −0.5
R
−1
© 2008 John Wiley & Sons, Ltd
P4 =
n − n′ 1 − 1.5
=
=0
R
∞
Suppose a parallel incident ray (α1 = 0) strikes the front surface of the system at a
height of y1 . By using matrix method, we can find the ray angle α 4′ and height y4′
at the back surface of the system given by:
⎡α 4′ ⎤
⎡α1 ⎤
=
R
T
R
T
R
T
R
4
34
3
23
2
12
1
⎢ y′ ⎥
⎢y ⎥
⎣ 4⎦
⎣ 1⎦
0⎤ ⎡1 P3 ⎤ ⎡ 1
0⎤ ⎡1 P2 ⎤ ⎡ 1 0⎤ ⎡1 P1 ⎤ ⎡α1 ⎤
⎡1 P4 ⎤ ⎡ 1
=⎢
⎥
⎢
⎢
⎥⎢
⎥⎢
⎢
⎥
⎥⎢
⎥⎢ ⎥
⎥
⎣0 1 ⎦ ⎣− d 3′ 1⎦ ⎣0 1 ⎦ ⎣− d 2′ 1⎦ ⎣0 1 ⎦ ⎣− d1′ 1⎦ ⎣0 1 ⎦ ⎣ y1 ⎦
0 ⎤ ⎡ 1 − 0 .5 ⎤ ⎡ 1
0⎤ ⎡1 1⎤ ⎡ 1
0 ⎤ ⎡1 0 ⎤ ⎡ 0 ⎤
⎡1 0 ⎤ ⎡ 1
=⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
⎢
1 ⎦ ⎣− 0.2 1⎦ ⎣0 1⎦ ⎣− 0.15 1⎥⎦ ⎢⎣0 1⎥⎦ ⎢⎣ y1 ⎥⎦
⎣0 1⎦ ⎣− 0.15 1⎦ ⎣0
⎡ 0.6 y1 ⎤
=⎢
⎥
⎣0.71 y1 ⎦
So the focal length is given by:
f =
y1
y
= 1 = 1.67[m]
α 4′ 0.6 y1
The principal plane is located at a distance d to the left side of the right-end surface
of the system, which is given by:
d=
y4′ − y1 0.71y1 − y1
=
= 0.48[m]
0.6 y1
α 4′
(c) The powers of the four spherical surfaces are given by:
n′ − n 1.5 − 1
P1 =
=
=0
R
∞
n − n′ 1 − 1.5
P2 =
=
=1
R
− 0.5
n′ − n 1.5 − 1
P3 =
=
=1
R
0.5
n − n′ 1 − 1.5
P4 =
=
=0
R
∞
Suppose a parallel incident ray (α1 = 0) strikes the front surface of the system at a
height of y1 . By using matrix method, we can find the ray angle α 4′ and height y4′
at the back surface of the system given by:
© 2008 John Wiley & Sons, Ltd
⎡α 4′ ⎤
⎡α1 ⎤
⎢ y′ ⎥ = R4T34 R3T23 R2T12 R1 ⎢ y ⎥
⎣ 4⎦
⎣ 1⎦
0⎤ ⎡1 P3 ⎤ ⎡ 1
0 ⎤ ⎡1
⎡1 P4 ⎤ ⎡ 1
=⎢
⎥
⎢
⎢
⎥⎢
⎢
⎥
⎥
⎣0 1 ⎦ ⎣− d 3′ 1⎦ ⎣0 1 ⎦ ⎣− d 2′ 1⎦ ⎣0
0⎤ ⎡1 1⎤ ⎡ 1
0 ⎤ ⎡1
⎡1 0 ⎤ ⎡ 1
=⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥⎢
⎣0 1⎦ ⎣− 0.15 1⎦ ⎣0 1⎦ ⎣− 0.6 1⎦ ⎣0
⎡ 1.4 y1 ⎤
=⎢
⎥
⎣0.19 y1 ⎦
So the focal length is given by:
f =
P2 ⎤ ⎡ 1
1 ⎥⎦ ⎢⎣− d1′
1⎤ ⎡ 1
1⎥⎦ ⎢⎣− 0.15
0⎤ ⎡1 P1 ⎤ ⎡α1 ⎤
1⎥⎦ ⎢⎣0 1 ⎥⎦ ⎢⎣ y1 ⎥⎦
0⎤ ⎡1 0⎤ ⎡ 0 ⎤
1⎥⎦ ⎢⎣0 1⎥⎦ ⎢⎣ y1 ⎥⎦
y1
y
= 1 = 0.71[m]
α 4′ 1.4 y1
The principal plane is located at a distance d to the left side of the right-end surface
of the system, which is given by:
d=
© 2008 John Wiley & Sons, Ltd
y4′ − y1
0.19 y1 − y1
=
= 0.58[m]
1.4 y1
α 4′
SOLUTIONS TO CHAPTER 12
12.1
R2
R1
t2
t
t1
r
Fig.A.12.1
As shown in Fig.A.12.1, the air gap thickness t is given by:
t = t2 − t1 =
r2
r2
−
2 R2 2 R1
Noting that there is a π rad of phase shift upon the reflection at the lower surface of
the air gap, the thickness of air gap at dark rings should satisfy:
2t = nλ
r2 r2
− = nλ
R2 R1
i.e.
which yields the radius rn of the nth dark ring given by:
rn2 =
R1 R2 nλ
R1 − R2
12.2
The matrix relating reflection coefficient r and transmission coefficient t for the
λ 4 film is given by:
⎡ cos δ
M =⎢
⎣in2 sin δ
i sin δ n2 ⎤ ⎡ 0
=
cos δ ⎥⎦ ⎢⎣in2
i n2 ⎤
0 ⎥⎦
where the phase change δ = π 2 for the λ 4 film.
Following the analysis in text page 352, we can find the coefficient A and B are
given by:
A = n1 ( M 11 + M 12 n3 ) = in1n3 n2
© 2008 John Wiley & Sons, Ltd
B = ( M 21 + M 22 n3 ) = in2
A perfect anti-reflector requires:
R=
A − B in1n3 n2 − in2
=
=0
A + B in1n3 n2 + in2
which gives:
n22 = n1n3
12.3
As shown in page 357 of the text, the intensity distribution of the interference pattern
is given by:
I = 4a 2 cos 2
δ
2
where δ is the phase difference between the two waves transmitted from the two
radio masts to a point P and is given by:
δ = kf sin θ =
2π
λ
f sin θ =
so we have:
I = 4a 2 cos 2
2π
× 400 × sin θ = 4π sin θ
3 × 10 1500 × 103
8
4π sin θ
= 2 I 0 [1 + cos(4π sin θ )]
2
where I 0 = a 2 represents the radiated intensity of each mast.
The intensity distribution is shown in the polar diagram below:
θ = 90o
I max = 4I 0
I
θ = 150o
θ = 30o
θ
θ = 180
θ = 0o
o
θ = 330o
θ = 210o
θ = 270o
Fig.A.12.3
© 2008 John Wiley & Sons, Ltd
12.4
(a)
Analysis is the same as Problem 12.3 except:
2π λ
δ = δ 0 + kf sin θ = π +
⋅ sin θ = π + π sin θ
λ 2
Hence, the intensity distribution is given by:
⎛δ ⎞
⎛ π + π sin θ ⎞
2 ⎛ π sin θ ⎞
I = 4a 2 cos 2 ⎜ ⎟ = 4a 2 cos 2 ⎜
⎟ = 4 I s sin ⎜
⎟
2
⎝2⎠
⎝
⎠
⎝ 2 ⎠
where I s = a 2 is the intensity of each source.
The polar diagram for I versus θ is shown below:
θ = 90o
I max = 4 I s
I
θ
θ = 180
θ = 0o
o
θ = 270o
Fig.A.12.4(a)
(b)
In this case, the phase difference is given by:
π 2π λ
π + π sin θ
δ = δ 0 + kf sin θ = +
⋅ sin θ =
2 λ 4
2
Hence, the intensity distribution is given by:
I = 4a 2 cos 2
δ
2
= 4a 2 cos 2
π + π sin θ
4
π
⎡
⎤
= 4 I s ⎢cos 2 (1 + sin θ )⎥
4
⎣
⎦
where I s = a 2 is the intensity of each source.
The polar diagram for I versus θ is shown below:
© 2008 John Wiley & Sons, Ltd
θ = 90o
I = 2I s
I max = 4 I s
I = 2I s
θ
θ = 0o
θ = 180o
I
θ = 270o
Fig.A.12.4(b)
12.5
(a)
θ
θ
θ
d
d
Fig.A.12.5(a)
Fig.A.12.5(a) shows elements of a vertical column and a horizontal row of radiators in
a rectangular lattice with unit square cells of side d . Rays leave each lattice point at
an angle θ to reach a distant point P . If P is simultaneously the location of the
mth spectral order of interference from the column radiation and the nth spectral
order of interference from the row radiation, we have from pages 364/5 the relations:
d sin θ = mλ and d cosθ = nλ
Thus
sin θ
m
= tan θ =
cosθ
n
where m and n are integers.
(b)
© 2008 John Wiley & Sons, Ltd
θ
θ
A
d
C
d sin θ
B
Fig.A.12.5(b)
Waves scattered elastically (without change of λ ) by successive planes separated by
a distance d in a crystal reinforce to give maxima on reflection when the path
difference 2d sin θ = nλ . In Fig.A.12.5(b), the path difference ABC between the
incident and the reflected rays = 2d sin θ .
12.6
Using the Principal Maximum condition:
f sin θ = nλ
at θ = ±
π
2
, we have: f = nλ , which shows the minimum separation of equal
sources is given by: f = λ .
When N = 4 , the intensity distribution as a function of θ is given by:
sin 2 (4π sin θ )
I = Is
sin 2 (π sin θ )
The N − 1 = 3 points of zero intensity occur when:
λ λ 3λ
f sin θ = , ,
4 2 4
1 1 3
i.e.
sin θ = , ,
4 2 4
The position of the N − 2 = 2 points of secondary intensity maxima should occur
between the zero intensity points and should satisfy:
dI
=0
dθ
i.e.
i.e.
© 2008 John Wiley & Sons, Ltd
dI
d
=
dθ d θ
⎡ sin 2 (4π sin θ ) ⎤
⎢ I s sin 2 (π sin θ ) ⎥ = 0
⎣
⎦
6 cos 2 (π sin θ ) − 1 = 0
which yields: sin θ =
⎛ 1 ⎞
arccos⎜ ±
⎟
π
6⎠
⎝
1
i.e. secondary intensity maxima occur when θ = 21.5o and θ = 39.3o .
The angular distribution of the intensity is shown below:
θ = 90o
I max = 16 I s
I
θ = 150o
θ = 30o
θ
θ = 180
θ = 0o
o
θ = 330o
θ = 210o
θ = 270o
12.7
The angular width of the central maximum δθ is the angular difference between +1
and -1 order zero intensity position and should satisfy:
sin δθ =
2λ 2 × 0.21
=
= 1.875 × 10− 3 or δθ = 6′
Nf
32 × 7
The angular separation between successive principal maxima Δθ is given by:
sin Δθ =
© 2008 John Wiley & Sons, Ltd
λ
f
=
0.21
= 0.03 or Δθ = 1o 42′
7
12.8
120
o
90 o
60
o
120
30 o
150 o
o
90 o
60 o
30 o
150 o
0 o 180 o
180 o
d λ =1
210 o
240 o
120
o
270
o
90 o
330 o
240 o
o
120
30 o
150 o
d λ=2
210 o
300 o
60
0o
o
270
o
90 o
300 o
60 o
30 o
150 o
0 o 180 o
180 o
d λ =3
210 o
240 o
270
o
330 o
300 o
330 o
0o
d λ =4
210 o
240 o
270
o
330 o
300 o
Fig.A.12.8
The above polar diagrams show the traces of the tip of the intensity of diffracted light
I for monochromatic light normally incident on a single slit when the ratio of slit
width to the wavelength d λ changes from 1 to 4. It is evidently shown that the
polar diagram becomes concentrated along the direction θ = 0 as d λ becomes
larger.
12.9
It is evident that α = 0 satisfies the condition: α = tan α .
By substitution of α = 3π 2 − δ into the condition: α = tan α we have:
3π 2 − δ = tan(3π 2 − δ )
© 2008 John Wiley & Sons, Ltd
i.e.
3π 2 − δ = cot δ
i.e.
(3π 2 − δ ) sin δ = cos δ
when δ is small, we have:
(3π 2 − δ )δ = 1 −
δ2
2
The solution to the above equation is given by: δ = 0.7π .
Using the similar analysis for α = 5π 2 − δ and α = 7π 2 − δ , we can find
δ = 0.041π and δ = 0.029π respectively. Therefore the real solutions for α are
α = 0,±1.43π ,±2.459π ,3.471π , etc .
12.10
If only interference effects are considered the intensity of this grating is given by:
sin 2 3β
I = I0
sin 2 β
The intensity of the principal maximum is given by: I max = 9I 0 when β = 0 .
The β for the secondary maximum should satisfy:
d ⎛ sin 2 3β
⎜
dβ ⎜⎝ sin 2 β
⎞
⎟⎟ = 0
⎠
sin 2 β = 1
i.e.
β sec_ max = (2n + 1)
i.e.
π
2
, where n is integer
Hence, at the secondary maximum:
I sec_ max = I 0
sin 2 3β sec_ max
sin β sec_ max
2
= I0 =
1
I max
9
12.11
Suppose a monochromatic light incident on a grating, the phase change dβ required
to move the diffracted light from the principal maximum to the first minimum is given
by:
πf λ
π
⎛ πf sin θ ⎞ πf
⋅
=
dβ = d ⎜
d (sin θ ) =
⎟=
λ Nf N
⎝ λ ⎠ λ
© 2008 John Wiley & Sons, Ltd
Since N is a very large number, we have:
dβ =
π
≈0
N
Then, suppose a non-monochromatic light, i.e. λ is not constant, incident on the
same grating, the phase change dβ required to move the diffracted light from the
principal maximum to the first minimum should be the same value as given above, so
we have:
⎛ πf sin θ ⎞ πf
⎛1⎞
dβ = d ⎜
d (sin θ ) + πf sin θd ⎜ ⎟
⎟=
⎝ λ ⎠ λ
⎝λ⎠
πf
πf sin θ
π
=
cos θdθ −
dλ = ≈ 0
2
λ
λ
N
which gives:
dθ = (nN cot θ ) −1
12.12
(a)
The derivative of the equation:
f sin θ = nλ
gives:
f cosθdθ = ndλ
when θ is a small angle we have:
dθ n
=
dλ f
When the diffracted light from the grating is projected by a lens of focal length F
on the screen, the relation between linear spacing on the screen l and the diffraction
angle θ is given by:
l = Fθ
Its derivative over λ gives:
dl
dθ nF
=F
=
dλ
dλ
f
(b)
Using the result given above, the change in linear separation per unit increase in
spectral order is given by:
dl Fdλ 2 × (5.2 × 10 −7 − 5 × 10 −7 )
= 2 × 10 −2 [m]
=
=
n
f
2 × 10 −6
© 2008 John Wiley & Sons, Ltd
12.13
(a)
Using the resolving power equation:
λ
= nN
dλ
we have:
λ
(5.89 × 10 −7 + 5.896 × 10 −7 ) 2
N=
≈ 328
=
ndλ 3 × (5.896 × 10 − 7 − 5.89 × 10 − 7 )
(b)
Using the resolving power equation:
λ
= nN
dλ
we have:
dλ =
λ
nN
=
6.5 × 10−7
= 2.4 × 10 −12 [m]
3 × 9 × 10 4
12.14
When the objects O and O′ are just resolved at I and I ′ the principal
maximum of O and the first minimum of O′ are located at I . Rayleigh’s criterion
thus defines the path difference:
O′BI − O′AI = O′B − O′A = 1.22λ ( BI = AI )
Also OB = OA giving
(O′B − OB) + (OA − O′A) = 1.22λ
Fig.Q.12.14 shows OA parallel to O′A and OB parallel to O′B , so:
OA − O′A = O′C = OO′ sin i = s sin i
and
O′B − OB = O′C ′ = OO′ sin i = s sin i
We therefore write:
1.22λ
1.22λ
or s =
if i = 45o
s≈
2 sin i
2 sin i
A
C
O′
C′
ii
i
O
B
© 2008 John Wiley & Sons, Ltd
Fig.A.12.14
SOLUTIONS TO CHAPTER 13
13.1
For such an electron, the uncertainty of momentum Δp roughly equals the magnitude of
momentum p , and the uncertainty of radius Δr roughly equals the magnitude of radius r . So
we have:
p ≈ Δp =
h
h
≈
Δr r
By substitution of the above equation into the expression of electron energy, we have:
h2 r 2
p2
e2
e2
−
=
−
E=
2m 4πε 0 r
2m
4πε 0 r
(13.1.1)
The minimum energy occurs when dE dr = 0 , i.e.:
d ⎛ h2 r 2
e2 ⎞
⎟=0
⎜⎜
−
4πε 0 r ⎟⎠
dr ⎝ 2m
−
i.e.
e2
h2
+
=0
mr 3 4πε 0 r 2
which yields the minimum Bohr radius given by:
r=
4πε 0h 2 ε 0 h 2
=
me 2
πme 2
By substitution into equation 13.1.1, we find the electron’s ground state energy given by:
2
h 2 ⎛ πme 2 ⎞
e 2 πme 2 − me 4
⎟
⎜⎜
−
=
E0 =
2m ⎝ ε 0 h 2 ⎟⎠ 4πε 0 ε 0 h 2 8ε 02 h 2
13.2
Use the uncertainty relation ΔpΔx ≈ h we have:
Δx ≈
h
h
≥
Δp p
Photons’ energy converted from mass m is given by:
E = pc = mc 2
© 2008 John Wiley & Sons, Ltd
So, the momentum of these photons is given by:
p = mc
Therefore, these photons’ spatial uncertainty should satisfy:
h
mc
Δx ≥
which shows the short wavelength limit on length measurement, i.e. the Compton wavelength, is
given by:
λ=
h
mc
By substitution of electron mass: me = 9.1× 10
−31
[kg ] into the above equation, we have the
Compton wavelength for an electron given by:
λ=
h
6.63 × 10 −34
=
≈ 2.42 × 10 −12 [m]
me c 9.1× 10 −31 × 3 × 108
13.3
The energy of a simple harmonic oscillation at frequency
E=
The relation: ( Δx )(Δp ) ≈
2
2
ω should satisfy:
p2 1
Δp 2 1
+ mω 2 x 2 ≥
+ mω 2 Δx 2
2m 2
2m 2
h2
h2
2
gives: Δp ≈
, by substitution into the above equation,
4
4Δx 2
we have:
Δp 2 1
h2
h2 ⎛ 1
1
2
2
2
2⎞
+ mω 2 Δx 2 =
+
m
ω
Δ
x
≥
2
⎜ mω Δx ⎟
2
2
2m 2
2
⎠
8mΔx
8mΔx ⎝ 2
1
1
= hω = hν
2
2
1
i.e. the simple harmonic oscillation has a minimum energy of
hν .
2
E≥
13.4
When an electron passes through a slit of width Δx , the intensity distribution of diffraction
pattern is given by:
I = I0
sin 2 α
α
2
, where
α=
The first minimum of the intensity pattern occurs when
α =π ,
α=
i.e.
Noting that
π
Δx sin θ
λ
λ = h p , we have:
© 2008 John Wiley & Sons, Ltd
π
Δx sin θ = π
λ
π
α=
h
Δxp sin θ = π
ΔxΔp = h
i.e.
where Δp = p sin θ is the change of the electron’s momentum in the direction parallel to the
plane of the slit. This relation is in accordance with Heisenberg’s uncertainty principle.
13.5
The angular spread due to diffraction can be seen as the half angular width of the principal
maximum Δθ of the diffraction pattern. Use the same analysis as Problem 13.4, we have:
α=
π
π
d sin θ ≈ dΔθ = π
λ
λ
Δθ =
i.e.
λ
d
=
10 −5
= 0.1 ≈ 5o 44′
10 −4
13.6
The energy of the electron after acceleration across a potential difference V is given by:
E = eV , so its momentum is given by: p = 2me E = 2me eV , therefore its de Broglie
wavelength is given by:
h
λ= =
p
h
=
2me eV
6.63 × 10 −34
2 × 9.1× 10
−31
−19
× 1.6 × 10 V
= 1.23 × 10 −9V −1 2 [m]
13.7
From problem 13.6 we have:
V
12
=
1.23 × 10 −9
λ
1.23 × 10 −9
=
= 4.1
3 × 10 −10
∴V = 16.81[V]
13.8
The energy per unit volume of electromagnetic wave is given by: E =
1
ε 0 E02 , where E0 is the
2
electric field amplitude. For photons of zero rest mass, the energy is given by: E = mc = pc ,
2
where p is the average momentum per unit volume associated with this electromagnetic wave.
So we have:
1
ε 0 E02 = pc
2
© 2008 John Wiley & Sons, Ltd
1
p = ε 0 E02 c
2
i.e.
The dimension the above equation is given by:
F ⋅ V 2 ⋅ m -2 C ⋅ V ⋅ m -2 C ⋅ W ⋅ m -2 A ⋅ s ⋅ kg ⋅ m 2 ⋅ m -2
=
=
=
= kg ⋅ m -1 ⋅ s -1
m ⋅ s -1
m ⋅ s -1
m ⋅ s -1 ⋅ A
m ⋅ s -1 ⋅ s 3 ⋅ A
which is the dimension of momentum.
13.9
When the wave is normally incident on a perfect absorber, all the photons’ velocity changes from
c to 0, the radiation pressure should equal the energy density of the incident wave, i.e.:
1
P = cp − 0 = cp = ε 0 E02
2
When the wave is normally incident on a perfect reflector, all the photons’ velocity changes to the
opposite directing but keeps the same value, hence, the radiation pressure is given by:
P = cp − (−cp) = 2cp = ε 0 E02
13.10
Using the result of Problem 13.9, we have the radiation pressure from the sun incident upon the
perfectly absorbing surface of the earth given by:
1 1
1 I 1 1.4 × 103
P = × ε 0 E02 = × = ×
= 1.5 × 10− 6 [Pa ] ≈ 10−11[atm]
8
3 2
3 c 3 3 × 10
13.11
Using the result of Problem 13.3, we have the minimum energy, i.e. the zero point energy, of such
an oscillation given by:
1
1
hν = × 6.63 × 10 −34 × 6.43 × 1011 = 2.13 × 10 −22 [J ] = 1.33 × 10 −3[eV]
2
2
13.12
The probability of finding the mass in the box is given by the integral:
2
1 ⎛ π 2 x2 ⎞
x
dx
(
)
=
ψ
∫−a
∫−a a ⎜⎜⎝1 − 8a 2 ⎟⎟⎠ dx
a
2
a
1 ⎛ π 2 x2 π 4 x4 ⎞
⎜1 −
⎟dx
+
−a a ⎜
4a 2 64a 4 ⎟⎠
⎝
=∫
a
= 2−
2π 2 2π 4
+
≈ 0.96
12 320
The general expression of the wave function is given by:
ψ = Ceikx + De −ikx , where A, B are
constants. Using boundary condition at x = a and x = − a , we have:
© 2008 John Wiley & Sons, Ltd
ψ (a) = Ceika + De −ika = 0
ψ (−a) = Ce − ika + Deika = 0
which gives: C = D , so we have:
ψ = Ceikx + Ce −ikx = A cos kx
where A = 2C .
Boundary condition:
ψ = 0 at
x = a gives: cos ka = 0 , i.e.:
1⎞π
⎛
k = ⎜ n + ⎟ , where n = 0,1,2,3,L
2⎠ a
⎝
Hence, the ground state equation is given by letting n = 0 , i.e.:
⎛ πx ⎞
⎟
⎝ 2a ⎠
ψ = A cos⎜
By normalization of the wave function, we have:
∫
+∞
−∞
ψ ( x) dx = 1
2
⎛ πx ⎞
A2 cos 2 ⎜ ⎟dx = 1
−∞
⎝ 2a ⎠
∫
+∞
i.e.
∫
+a
i.e.
−a
A2
1 + cos(πx a)
dx = 1
2
A =1
i.e.
a
Therefore the normalized ground state wave function is:
ψ ( x) = (1 a ) cos(πx 2a )
which can be expanded as:
1
ψ ( x) =
a
⎡ 1 ⎛ πx ⎞ 2
⎤
1 ⎛ π 2 x2 ⎞
⎜1 −
⎟
⎢1 − ⎜ ⎟ + L⎥ ≈
8a 2 ⎟⎠
a ⎜⎝
⎢⎣ 2 ⎝ 2a ⎠
⎥⎦
13.13
At ground state, i.e. at the bottom of the deep potential well, n1 = n2 = n3 = 1 .
By normalization of the wave function at ground state, we have:
© 2008 John Wiley & Sons, Ltd
∫∫∫ ψ ( xyz)
2
dV = ∫
c b
∫∫
a
0 0 0
A sin
πx
a
sin
πy
b
sin
πz
2
dxdydz
c
a⎛
a⎛
a⎛
πx ⎞
πy ⎞
πz ⎞
= A2 ∫ ⎜ sin ⎟ dx ⋅∫ ⎜ sin ⎟ dy ⋅ ∫ ⎜ sin ⎟ dz
0
0
0
c ⎠
b ⎠
a⎠
⎝
⎝
⎝
a⎛ 1
b⎛ 1
c⎛ 1
2πz ⎞
2πy ⎞
1
2πx ⎞
1
1
= A2 ∫ ⎜ − cos
dy ⋅ ∫ ⎜ − cos
dx ⋅∫ ⎜ − cos
⎟
⎟
⎟dz
0 2
0 2
0 2
2
c ⎠
2
b ⎠
2
a ⎠
⎝
⎝
⎝
a b c
= A2 ⋅ ⋅ ⋅ = 1
2 2 2
2
2
2
A = 8 abc
i.e.
13.14
Text in page 426 shows number of electrons per unit volume in energy interval dE is given by:
2 × 4π ( 2 m 3 )1 2 E 1 2
dn =
dE
h3
and the total number of electrons given by:
16π (2me3 )1 2 EF3 2
N=
3h3
so we have the total energy of these electrons given by:
dn
dE
0
dE
3 12 32
E f 2 × 4π ( 2m )
E
=∫
dE
3
0
h
16π (2m3 )1 2 EF5 2 3
=
= NEF
5h3
5
Ef
U = ∫ Edn = ∫ E
13.15
Noting that Copper has one conduction electron per atom and one atom has a mass of
m0 = 1.66 × 10 −27 kg , the number of free electrons per unit volume in Copper is given by:
ρ
9 × 103
n0 =
≈ 8 × 10 28 [m -3 ]
=
− 27
64m0 64 × 1.66 × 10
Using the expression of number of electrons per unit volume in text of page 426, we have the
Fermi energy level of Copper given by:
⎛ n ⋅ 3h 3
EF = ⎜ 0
⎜ 16π 2m3
e
⎝
© 2008 John Wiley & Sons, Ltd
⎞
⎟
⎟
⎠
23
⎡ 8 × 10 28 × 3 × (6.63 × 10 −34 )3 ⎤
=⎢
⎥
⎢⎣ 16π × 2 × (9.1× 10 −31 )3 ⎥⎦
23
≈ 1.08 × 10 −18 [J] = 7[eV]
13.16
values of x , V − E , and m into the expression e
By substitution of
−2αx
, we have:
For an electron:
e −2αx = e
−2
[
] = e −2 ⎡⎢⎣
[
2 m p (V − E ) h x
2 me (V − E ) h x
(
)
(
)
2×9.1×10−31×1.6×10−19 6.63×10−34 2π ⎤×2×10−10
⎥⎦
= e −2.05 ≈ 0.1
For a proton:
e −2αx = e
−2
]
=e
−2 ⎡ 2×1.67×10−27 ×1.6×10−19 6.63×10−34 2π ⎤×2×10−10
⎥⎦
⎢⎣
= e −87.4 ≈ 10 −38
13.17
Text in page 432-434 shows the amplitude reflection and transmission coefficients for such a
particle are given by:
r=
C
2k1
B k1 − k 2
=
and t =
=
A k1 + k 2
A k1 + k 2
where,
k1 =
2mE
and k2 =
h
2m( E − V )
h
If V is a very large negative value at x > 0 , we have the amplitude reflection coefficient given
by:
r = lim
V →−∞
2mE − 2m( E − V )
= −1
2mE + 2m( E − V )
and the amplitude transmission coefficient given by:
t = lim
V →−∞
2 2mE
=0
2mE + 2m( E − V )
i.e. the amplitude of reflected wave tends to unity and that of transmitted wave to zero.
13.18
The potential energy of one dimensional simple harmonic oscillator of frequency
V =
ω is given by:
1
mω 2 x 2
2
By substitution into Schrödinger’s equation, we have:
d 2ψ 2m ⎡
1
⎤
+ 2 ⎢ E − mω 2 x 2 ⎥ψ = 0
2
dx
h ⎣
2
⎦
(13.18.1)
Try
ψ ( x) = a
π e −a x
© 2008 John Wiley & Sons, Ltd
2 2
2
in dψ dx :
dψ
= −a 2 x
dx
a
π
e
− a2 x2
2
so:
d 2ψ
= −a 2
dx 2
a
π
e
− a2 x2
2
+a x
4
a
2
e
π
− a 2 x2
2
= (a x − a )
4
2
2
a
π
e
− a2 x2
2
In order to satisfy the Schrödinger’s equation (13.18.1), we should have:
m 2ω 2
2mE
= a4
a = 2 and
2
h
h
2
which yields:
E0 =
ψ ( x) = a 2 π 2axe −a x
2 2
Try
a
dψ
= 2a
e
dx
2 π
− a2 x2
2
2
h 2a 2 1
= hω
2m 2
in dψ dx :
− 2a x
a
3 2
2 π
e
− a2 x2
2
= ( 2a − 2a x )
3 2
a
2 π
e
− a2 x2
2
so:
a
d 2ψ
= −4 a 3 x
e
2
dx
2 π
− a2 x2
2
a
+ ( 2a x − 2a x )
5 3
3
2 π
e
− a2 x2
2
= ( 2a x − 6a x )
5 3
In order to satisfy the Schrödinger’s equation (13.18.1), we should have:
3a 2 =
m 2ω 2
2mE
and
= a4
h2
h2
which yields:
3h 2 a 2 3
= hω
E1 =
2m
2
13.19
When n = 0 :
N 0 = (a π 1 2 200!)1 2 = a
π
H 0 (ax) = (−1) 0 e a x e − a x = 1
2 2
2 2
Hence:
ψ 0 = N 0 H 0 (ax)e −a x
2 2
2
= a
π e −a x
When n = 1 :
N1 = (a π 1 2 211!)1 2 = a 2 π
© 2008 John Wiley & Sons, Ltd
2 2
2
3
a
2 π
e
− a2 x2
2
H1 (ax) = (−1)1 e a x
2 2
2 2
2 2
2 2
d
e −a x = −e a x ⋅ e −a x ⋅ (−2ax) = 2ax
d (ax)
Hence:
ψ 1 = N1H1 (ax)e −a x
2 2
2
= a 2 π 2axe −a x
2 2
2
When n = 2 :
N 2 = (a π 1 2 22 2!)1 2 = a 8 π =
H 2 (ax) = (−1) 2 e a x
2 2
a 2 π
2
2 2
d2
e −a x = −2 + 4a 2 x 2
2
d (ax)
Hence:
ψ 2 = N 2 H 2 (ax)e −a x
2 2
2
= a 2 π (2a 2 x 2 − 1)e −a x
2 2
2
When n = 3 :
N 3 = (a π 1 2 233!)1 2 = a 48 π =
H 3 (ax) = (−1)3 e a x
2 2
a 3 π
4
2 2
d3
e −a x = 8a 3 x 3 − 12ax
3
d (ax)
Hence:
ψ 3 = N 3 H 3 (ax)e −a x
2 2
2
= a 3 π (2a 3 x 3 − 3ax)e −a x
2 2
2
13.20
The reflection angle
θ r and reflection wavelength λd should satisfy Bragg condition:
2a sin θ r = λr
where a is separation of the atomic plane of the nickel crystal. Hence the reflected electron
momentum pr should satisfy:
pr =
h
λr
=
Hence, the reflected electron energy is given by:
© 2008 John Wiley & Sons, Ltd
h
2a sin θ r
Er =
=
pr2
h2
=
2me 8me a 2 sin 2 θ r
(6.63 × 10−34 ) 2
8 × 9.1× 10 −31 × (0.91× 10−10 ) 2 × sin 2 65o
= 8.88 × 10 −19 [J] = 55.5[eV]
The difference between the incident and scattered kinetic energies is given by:
Er − Ei 55.5 − 54
=
× 100% = 2.8% < 3.9%
54
Ei
13.21
For
ψ = sin ka :
Since
ψ ,ψ * ,V are all periodic functions with a period of a , we have:
ΔE =
*
∫ψ Vψdx
∫ψ ψdx
*
a
∞
= −∑
m=1
∫ sin
0
2
2πmx
dx
a
a
2
∫ sin kxdx
kxVm cos
0
1 − cos (2πnx a )
2πmx
dx
cos
0
a
2
= −∑Vm
a 1 − cos ( 2πnx a )
m =1
dx
∫0
2
1 a
2πnx
2πmx
dx
− ∫ cos
⋅ cos
∞
0
a
a
2
= −∑ Vm
a2
m =1
∞
∫
a
1 a⎡
2π (m + n) x
2π (m − n) x ⎤
cos
+ cos
∞
∫
⎥⎦dx
⎢
0
a
a
2 ⎣
= ∑Vm
a
m=1
The above equation has non-zero term only when m = n , so we have:
a
ΔE = Vn
For
ψ = cos ka :
© 2008 John Wiley & Sons, Ltd
⎛
∫ ⎜⎝ cos
0
4πnx ⎞
+ 1⎟dx
a
⎠ = V a = 1V
n
n
2a
2a 2
∫ψ Vψdx
*
ΔE =
*
∫ψ ψdx
a
∞
= −∑
m =1
∫ cos
0
2
2πmx
dx
a
a
2
∫ cos kxdx
kxVm cos
0
1 + cos (2πnx a)
2πmx
dx
cos
a
2
= −∑Vm
a 1 + cos ( 2πnx a )
m =1
dx
∫0
2
1 a
2πnx
2πmx
dx
cos
⋅ cos
∞
∫
0
a
a
= −∑Vm 2
a2
m =1
∞
∫
a
0
1 a⎡
2π (m + n) x
2π (m − n) x ⎤
cos
+ cos
∫
⎥⎦dx
⎢
a
a
2 0⎣
= −∑Vm
a
m =1
The above equation has non-zero term only when m = n , so we have:
∞
a
ΔE = −Vn
© 2008 John Wiley & Sons, Ltd
⎛
∫ ⎜⎝ cos
0
4πnx ⎞
+ 1⎟dx
a
⎠ = −V a = − 1 V
n
n
2a
2a
2
SOLUTIONS TO CHAPTER 14
14.1
For θ 0 < 30o , we have:
⎛ 1
30o ⎞
⎟ = 1.017T0
T < T0 ⎜⎜1 + sin 2
2 ⎟⎠
⎝ 4
T − T0
= 1.7% < 2%
T0
i.e.
For θ 0 = 90o , we have:
⎛ 1
90o ⎞
⎟ = 1.125T0
T = T0 ⎜⎜1 + sin 2
2 ⎟⎠
⎝ 4
T − T0
= 12.5%
T0
i.e.
14.2
Multiplying the equation of motion by 2 dx dt and integrating with respect to t
gives:
2
x
⎛ dx ⎞
m⎜ ⎟ = A − 2 ∫ f ( x)dx
0
⎝ dt ⎠
where A is the constant of integration. The velocity
dx
is zero at the maximum
dt
x0
displacement x = x0 , giving A = 2 ∫ f ( x)dx .
0
2
i.e.
x0
x
⎛ dx ⎞
m⎜ ⎟ = 2∫ f ( x)dx − 2∫ f ( x)dx = 2 F ( x0 ) − 2 F ( x)
0
0
⎝ dt ⎠
i.e.
dx
=
dt
2
[ F ( x0 ) − F ( x)]
m
Upon integration of the above equation, we have:
© 2008 John Wiley & Sons, Ltd
t=∫
m
2
dx
F ( x0 ) − F ( x)
If x = 0 at time t = 0 and τ 0 is the period of oscillation, then x = x0 at t = τ 0 4 ,
so we have:
τ0 = 4
m x0
dx
∫
0
2
F ( x0 ) − F ( x)
14.3
By substitution of the solution into &x& :
∞
⎡
n2
n
n2
n ⎤
&x& = ∑ ⎢− an
cos φ − bn
sin φ ⎥
9
3
9
3 ⎦
n =1 ⎣
Since s3 << s1 , we have s ( x) ≈ s1 x , so:
∞ ⎡
⎛
⎛
n2 ⎞
n
n2 ⎞
n ⎤
&x& + s ( x) = ∑ ⎢an ⎜⎜ s1 − ⎟⎟ cos φ + bn ⎜⎜ s1 − ⎟⎟ sin φ ⎥ = F0 cos ωt
9 ⎠
3
9 ⎠
3 ⎦
n =1 ⎣
⎝
⎝
⎡ ⎛
⎛
n2 ⎞
n
n2 ⎞
n ⎤
⎟⎟ sin φ ⎥ = F0 cos φ
⎜
⎟
⎜
a
s
cos
φ
b
s
+
−
−
⎢ n⎜ 1
∑
n⎜ 1
⎟
9⎠
3
9 ⎠
3 ⎦
n =1 ⎣
⎝
⎝
∞
i.e.
i.e.
The above equation is true only if bn = 0 and the even numbered cosine terms are
zero. By neglecting the zero terms, we have:
a3 ( s1 − 1) cos φ + a9 ( s1 − 9) cos 3φ + L = F0 cos φ
i.e.
i.e.
a3 ( s1 − 1) cos φ + a9 ( s1 − 9)(4 cos3 φ − 3 cos φ ) + L = F0 cos φ
[a3 ( s1 − 1) − 3a9 ( s1 − 9)] cos φ + 4a9 ( s1 − 9) cos3 φ + L = F0 cos φ
As we can see, only a3 and a9 are the main coefficients in the solution, i.e. the
fundamental frequency term and its third harmonic term are the significant terms in
the solution.
14.4
Since V = V0 at r = r0 , by expanding V at r0 , we have:
⎛ d 2V
⎛ dV ⎞
V = V0 + ⎜
⎟ (r − r0 ) + ⎜⎜ 2
⎝ dr ⎠ r0
⎝ dr
© 2008 John Wiley & Sons, Ltd
⎞
⎟⎟ (r − r0 ) 2 + L
⎠ r0
Noting that:
⎛ r 6 r 12 ⎞
⎛ dV ⎞
⎜
⎟ = 12V0 ⎜⎜ 07 − 013 ⎟⎟ = 0
⎝ dr ⎠ r0
⎝ r0 r0 ⎠
⎛ d 2V
⎜⎜ 2
⎝ dr
⎛ 13r 12 7r 6 ⎞
⎞
V
⎟⎟ = 12V0 ⎜⎜ 140 − 80 ⎟⎟ = 72 02
r0 ⎠
r0
⎠ r0
⎝ r0
We have:
V = V0 +
72V0
(r − r0 ) 2 + L
r02
The expression of potential energy for harmonic oscillation is given by:
1
72V
V = V0 + sx 2 , hence s = 2 0 , and the oscillation frequency is given by:
r0
2
ω2 =
s 72V0
=
m mr02
14.5
The restoring force of this oscillator is given by:
dV ( x)
F ( x) = −
= − kx + ax 2
dx
Hence, the equation of motion is given by:
m&x& = F ( x)
&x& +
i.e.
k
a
x − x2 = 0
m
m
At ω02 = ω 2 = k m , using α = a m , the equation of motion becomes:
&x& + ω02 x − αx 2 = 0
try the solution x = A cos ω0t + B sin 2ω0t + x1 in the above equation with x1 =
A2
2ω02
αA 2
αx
and B = − 2 = − 1 , we have:
6ω0
3
x& = x&1 − ω0 A sin ω0t + 2ω0 B cos 2ω0t
4
&x& = &x&1 − ω02 A cos ω0t − 4ω02 B sin 2ω0t = &x&1 − ω02 A cos ω0t − αω02 x1 sin 2ω0t
3
(a)
x 2 = x12 + A2 cos 2 ω0t + B 2 sin 2 2ω0t + 2 AB cos ω0t sin 2ω0t + 2( A cos ω0t + B sin 2ω0t ) x1
© 2008 John Wiley & Sons, Ltd
− αx 2 = −αx12 − αA2 cos 2 ω0t − αB 2 sin 2 2ω0t − 2αAB cos ω0t sin 2ω0t
− 2α ( A cos ω0t + B sin 2ω0t ) x1
where
− αA2 cos 2 ω0t = −2αω02 x1 cos 2 ω0t
− αB 2 sin 2 2ω0t = −
α2
9
x12 sin 2 2ω0t
⎛2
⎞
− 2αAB cos ω0t sin 2ω0t = ⎜ α 2 A cos ω0t sin 2ω0t ⎟ x1
⎝3
⎠
− 2αAx1 cos ω0t − 2αBx1 sin 2ω0t = −2αAx1 cos ω0t
2
+ α 2 x12 sin 2ω0t
3
(b)
(e)
(c)
(d)
(f)
Using (a)(b)(c)(d) the coefficients of x1 are:
4
3
ω02 − αω02 sin 2ω0t − 2αω02 cos 2 ω0t − 2α 2 A cos ω0t sin 2ω0t − 2αA cos ω0t
which with α << ω02 leaves ω02 x1 as the only significant term.
Similarly using (e) and (f) the coefficients of x12 are:
−α −
α2
2
sin 2ω0t + α 2 sin 2ω0t
9
3
with − αx12 the dominant term. (Note x12 ∝
1
ω04
).
We therefore have &x&1 + ω02 x1 − αx12 = 0 as a good approximation to the original
equation.
14.6
Extending the chain rule at the bottom of page 472 and noting that the fixed point x0
is the origin of the cycle: x1* → x2* → x1* → x2* which are fixed points for f 2 when
λ>
3
and also noting that x1* = f ( x2* ) and x2* = f ( x1* ) , we have:
4
f 2 ( x2* ) = f ′( x1* ) f ′( x2* ) and f 2 ( x1* ) = f ′( x2* ) f ′( x1* )
'
So the slopes of f 2 at x1* and x2* are equal.
© 2008 John Wiley & Sons, Ltd
'
14.7
The fractal dimension of the Koch Snowflake is:
d=
log 4
= 1.262
log 3
The Hausdorff–Besicovitch definition uses a scaling process for both integral and
fractal dimensions to produce the relation:
c = ad
where c is the number of copies(including the original) produced when a shape of
dimensions d has its side length increased by a factor a .
Thus, for a = 2
a line d = 1 has c = 2
(i)
a square d = 2 has c = 4
(ii)
(iii) a cube d = 3 has c = 8
(iv)
an equilateral triangle with a horizontal base produces 3 copies to give
d=
log 3
= 1.5849 (a fractal)
log 2
© 2008 John Wiley & Sons, Ltd
SOLUTIONS TO CHAPTER 15
15.1
In the energy conservation equation the internal energy:
e = cV T =
1 p
γ −1 ρ
so the two terms:
e+
p
ρ
=
γ
p
γ −1 ρ
In the reservoir there is no flow energy, so its total energy is internal =
γ
p0
γ − 1 ρ0
=
1 2
c ,
γ −1 0
where c0 is the velocity of sound in the reservoir.
When the diaphragm where is flow along the tube of velocity u and energy 1 2 u
2
so the total
energy of flow along the tube is:
1 2
γ p* 1 *2
1 *2 ⎛ 1
1 ⎞ *2
= c +
u +
c = ⎜⎜ +
⎟⎟c
*
2
γ −1 ρ
2
γ −1
⎝ 2 γ −1⎠
γp *
where u = c , u = c = * .
ρ
*
2
*2
Hence,
1 2 2 + (γ − 1) *2
γ + 1 *2
c0 =
c =
c
γ −1
2(γ − 1)
2(γ − 1)
If the wavefront flows at a velocity u1 with a local velocity of sound c1 , the energy
conservation condition gives:
1 2
1 2
γ + 1 *2
u1 +
c1 =
c
2
γ −1
2(γ − 1)
2
i.e.
i.e.
© 2008 John Wiley & Sons, Ltd
1
1 ⎛ c1 ⎞
γ + 1 ⎛ c* ⎞
⎜ ⎟
⎜⎜ ⎟⎟ =
+
2 γ − 1 ⎝ u1 ⎠
2(γ − 1) ⎜⎝ u1 ⎟⎠
2
γ +1
1
1
+
=
2
2 (γ − 1) M s 2(γ − 1) M *2
2
M* =
i.e.
(γ + 1) M s2
(γ − 1) M s2 + 2
15.2
Energy conservation gives
1 2 1 2
1 2 1 2
γ + 1 *2
c1 + u1 =
c2 + u2 =
c
γ −1
2
γ −1
2
2(γ − 1)
from Problem 15.1
So
c12 +
γ −1
2
u12 = c22 +
γ −1
2
u22 =
γ +1
2
c* 2
(A)
Momentum conservation gives
c12 + γu12 =
u
ρ2 2
(c2 + γu22 ) = 1 (c22 + γu22 )
ρ1
u2
(B)
2
2
Combine equations A and B to eliminate c1 and c2 and rearrange terms to give
γ +1
2
c*2 −
γ −1
2
u12 + γu12 =
γ +1
i.e.
i.e.
u1 ⎡ γ + 1 *2 γ − 1 2
⎤
c −
u2 + γu22 ⎥
⎢
2
u2 ⎣ 2
⎦
2
(c*2 + u12 ) =
u1 ⎛ γ + 1 ⎞ *2
2
⎜
⎟ (c + u 2 )
u2 ⎝ 2 ⎠
(u1 − u2 )c*2 = (u2u12 − u1u22 ) = u1u2 (u1 − u2 )
c*2 = u1u2
i.e.
15.3
The three conservation equations are given by:
ρ1u1 = ρ 2u2
(15.3.1)
p1 + ρ1u12 = p2 + ρ 2u22
(15.3.2)
1 2
1
p
p
u1 + e1 + 1 = u22 + e2 + 2
2
ρ1 2
ρ2
(15.3.3)
Using equation 15.3.1 and 15.3.2 to eliminate u1 gives:
© 2008 John Wiley & Sons, Ltd
ρ 22 − ρ1ρ 2 2
u2 = p2 − p1
ρ1
(15.3.4)
Using equation 15.3.1 and 15.3.3 and the relation
e = cV T =
1 p
γ −1 ρ
to eliminate u1 gives:
ρ 22 − ρ12 2
γ ⎛ p2 p1 ⎞
⎜ − ⎟
u2 =
2
2 ρ1
γ − 1 ⎜⎝ ρ 2 ρ1 ⎟⎠
(15.3.5)
Then, using equation 15.3.4 and 15.3.5 to eliminate u2 , we have:
ρ 2 + ρ1
2
=
γ
γ −1
(ρ1 p2 − ρ 2 p1 )
ρ 2 ρ1 + 1
i.e.
2
=
1
p2 − p1
γ
p2 p1 − ρ 2 ρ1
γ − 1 p2 p1 − 1
[(1 + ρ 2 ρ1 )(γ − 1) − 2γ ] p2 p1 = (1 + ρ 2 ρ1 )(γ − 1) − 2γβ
i.e.
[ β (γ − 1) − (γ + 1)] p2 p1 = (γ − 1) − β (γ + 1)
i.e.
which yields:
p2 β − α
=
p1 1 − βα
where
α = (γ − 1) (γ + 1) and β = ρ 2 ρ1 .
15.4
Using the result of Problems 15.1 and 15.2, we have:
u2
u
(γ + 1) M s2
⎛u ⎞
= ⎜ 1* ⎟ = 1 = 1 =
u1u2 u2 (γ − 1) M s2 + 2
⎝c ⎠
2
M
*2
(15.4.1)
Using equation 15.3.1 and 15.3.2 to eliminate
ρ1 gives:
p1 − p2
u
= 1− 1
2
u2
ρ 2u 2
© 2008 John Wiley & Sons, Ltd
(15.4.2)
Using equation 15.3.1 and 15.3.3 to eliminate
ρ1 gives:
⎞
γ ⎛ u1
1
1 ⎛ u12 ⎞
⎜⎜1 − 2 ⎟⎟ =
⎜⎜ p1 − p2 ⎟⎟
2
2 ⎝ u 2 ⎠ ρ 2u 2 γ − 1 ⎝ u 2
⎠
(15.4.3)
Using equation 15.4.2 and 15.4.3 to eliminate
ρ 2 gives:
y
⎛1
⎞u
⎜ + y ⎟ 1 = +1
⎝α
⎠ u2 α
(15.4.4)
where y = p2 p1 and
α = (γ − 1) (γ + 1) .
Then, using equation 15.4.1 and 15.4.4 to eliminate u1 u2 we have:
Ms =
u1
=
c1
y +α
1+ α
(15.4.5)
Frome equation 15.4.4 and 15.4.5 we have:
u2
1 + αy
=
c1
1+ α y + α
(15.4.6)
Hence, from equations 15.4.5 and 15.4.6 we have the flow velocity behind the shock give by:
u = u1 − u2 =
c1 (1 − α )( y − 1)
1(1 + α )( y + α )
15.5
In the case of reflected shock wave, as shown in Fig.(b), the shock strength is p3 p2 and the
velocity of sound ahead of the shock front is c2 . Hence, using the result of Problem 15.4, we
have the flow velocity ur behind the reflected wave given by:
(1 − α )( p3 p2 − 1)
ur
=
c2
(1 + α )( p3 p2 + α )
(15.5.1)
In Fig.(a) the flow velocity u behind of the incident shock front is given by:
© 2008 John Wiley & Sons, Ltd
u
(1 − α )( y − 1)
=
c1
(1 + α )( y + α )
(15.5.2)
Using equation 15.5.1 and 15.5.2 together with the relation u + ur = 0 , c2 c1 = (T1 T2 )
12
and
⎛ 1 + αy ⎞
T2
⎟⎟ , where y = p2 p1 , we have:
= y⎜⎜
T1
⎝α + y ⎠
( y − 1) 2 ( p3 p2 + α )
= y (1 + αy )
( p3 p2 − 1) 2
which yields:
p3 (2α + 1) y − α
=
p2
αy + 1
15.6
p3 − p1 p3 p2 − p1 p2 p3 p2 − 1 y y p3 p2 − 1
=
=
=
p2 − p1
y −1
1 − p1 p2
1 −1 y
By substitution of y = p2 p1 and
p3 (2α + 1) y − α
=
into the above equation, we have:
p2
αy + 1
(2α + 1) y 2 − αy
−1
p3 − p1
(2α + 1) y 2 − 2αy − 1
αy + 1
=
=
p2 − p1
y −1
( y − 1)(αy + 1)
In the limit of very strong shock, i.e. y >> 1 , we have:
p3 − p1 (2α + 1) y 2
1
≈
= 2+
2
p2 − p1
αy
α
15.7
ut = −(u + tut ) f ′
ut = −
i.e.
From equation 15.9 in the text, we have:
ux =
© 2008 John Wiley & Sons, Ltd
f′
1 + tf ′
uf ′
1 + tf ′
so we have:
ut + uu x = −
uf ′
uf ′
+
=0
1 + tf ′ 1 + tf ′
15.8
Using
u = −2ν ψ x ψ , we have:
ut = −2ν
ψ xtψ −ψ tψ x
ψ2
u x = −2ν
ψ xxψ −ψ x2
ψ2
⎡ψ
3ψ ψ
2ψ 3 ⎤
u xx = −2ν ⎢ xxx − x 2 xx + 3x ⎥
ψ
ψ ⎦
⎣ψ
By substitution of the above expression into Burger’s equation, we have:
− 2ν
3
3ψ xψ xx 2ψ x3 ⎞
ψ xtψ −ψ tψ x
2 ⎛ ψ xxx
2 ψψ xψ xx −ψ x
⎜
+
+
−
+ 3 ⎟⎟ = 0
ν
ν
4
2
⎜ ψ
ψ2
ψ3
ψ2
ψ ⎠
⎝
ψψ xt −ψ tψ x
ψψ xxx −ψ xψ xx
=ν
2
ψ2
ψ
i.e.
⎛ψ t
⎜⎜
⎝ψ
i.e.
⎞ ⎛ ψ xx ⎞
⎟⎟ = ⎜⎜ν
⎟⎟
⎠x ⎝ ψ ⎠x
which yields:
ψ t = νψ xx
15.9
Using the relation tanh′φ ≡ sech2φ and sech′φ = sechφtanhφ, where φ = α(x – ct), we have the
derivatives:
ut = 4α3 c sech2φ tanhφ = 2αuc tanhφ
ux = –4α3 sech2φ tanhφ = –2αu tanhφ
uxx = 8α4 sech2φ tanh2φ – 4α4 sech4φ = 4α2u tanh2φ – u2
uxx = –16α5 sech2φ tanhφ + 48α5 sech4φ tanhφ = –8α3u tanhφ + 12αu2 tanhφ
Then, using the relation c = 4α
© 2008 John Wiley & Sons, Ltd
2
we have:
ut + 6uu x + u xxx
= 2αuc tanh φ − 12αu 2 tanh φ − 8α 3u tanh φ + 12αu 2 tanh φ
=0
15.10
At the peak of Figure 15.5(a),
φ = 2α ( x − ct ) = 0 , and near the base of Figure 15.5(a),
φ = 2α ( x − ct ) >> 0 , i.e. e −φ ≈ 0 . Hence, the solution of the KdV equation can be written as:
u ( x, t ) = 2
∂2
∂ 2 −2α ( x−ct )
−2α ( x −ct )
log[
1
+
e
]
≈
2
e
= 8α 2e −2α ( x−ct )
2
2
∂x
∂x
Then, we have the derivatives,
ut = 16α 3ce −2α ( x−ct )
u xxx = −64α 5e −2α ( x−ct )
Therefore, if c = 4α ,
2
ut + u xxx = 0
15.11
Using the substitution z = x − ct and the relation tanh′φ ≡ sech2φ and sech′φ ≡ –sechφtanhφ,
where
φ = α ( z − z0 ) , we have the derivatives:
ut = –4α3c sech2φ tanhφ = 2αuc tanhφ
ux = –4α3 sech2φ tanhφ = –2αu tanhφ
uxx = 4α4 sech4φ + 4α2u tanh2φ = 4α2u tanh2φ + u2
uxx = –16α5 sech2φ tanhφ + 48α5 sech4φ tanhφ = –8α3u tanhφ – 12αu2 tanhφ
Then, using the relation c = 4α , we have:
2
ut − 6uu x + u xxx
= 2αuc tanh φ + 12αu 2 tanh φ + 8α 3u tanh φ − 12αu 2 tanh φ
= −8α 3u tanh φ + 8α 3u tanh φ
=0
15.12
If v + v x = u , then:
2
© 2008 John Wiley & Sons, Ltd
u x = 2vvx + vxx
u xx = 2(vx2 + vvxx ) + vxxx
u xxx = 2(2v x vxx + vvxxx + v x vxx ) + v xxxx = 6vx vxx + 2vvxxx + v xxxx
ut = 2vvt + vxt
The left side terms of equation mark in equation 15.13 give:
⎛∂
⎞
2
⎜ + 2v ⎟(vt − 6v v x + vxxx )
x
∂
⎝
⎠
= vxt − 6(2vvx2 + v 2vxx ) + vxxxx + 2v(vt − 6v 2vx + v xxx )
= 2vvt − 12v 3vx + vxt − 12vvx2 − 6v 2vxx + 2vvxxx + v xxxx
The right side terms of equation mark in equation 15.13 give:
ut − 6uu x + u xxx
= 2vvt + vxt − 6(v 2 + vx )(2vvx + vxx ) + 6vx vxx + 2vvxxx + vxxxx
= 2vvt − 12v 3v x + vxt − 12vvx2 − 6v 2v xx + 2vvxxx + vxxxx
So we have:
⎛∂
⎞
2
⎜ + 2v ⎟(vt − 6v vx + vxxx ) = ut − 6uu x + u xxx
∂
x
⎝
⎠
15.13
By substitution of v = ψ x
ψ into u ( x) = vx + v 2 , we have:
ψ xxψ − ψ x2 ψ x2 ψ xx
u ( x) = vx + v =
+ 2 =
ψ2
ψ
ψ
2
hence,
ψ xx − u ( x)ψ = ψ xx −ψ
ψ xx
=0
ψ
15.14
ψx = –αA sechα (x – x0) tanhα (x – x0)
ψxx = α2A sechα(x – x0) – 2α2A sech3α (x – x0)
Using the soliton solution
u = –2α2 sech2α (x – x0)
we have:
© 2008 John Wiley & Sons, Ltd
ψxx + (λ – u(x))ψ
= α2A sechα(x – x0) – 2α2A sech3α (x – x0) + [–α2 sech2α (x – x0)]A sechα (x – x0)
=0
15.15
Using transformation u → u − λ and x → x + 6λt , where u
*
*
*
and x
*
are the variables
before transformation, we have: u = u + λ and x = x − 6λt , hence:
*
*
ut = (u * − λ )t = ut*
u x = (u * − λ ) x* xx∗ = u *x*
u xx = u *x*x* x*x = u *x*x*
u xxx = u x∗*x*x* x*x = u ∗x*x*x*
Using the above relations, equation ut + 6uu x + u xxx = 0 is transformed to its original form:
ut* + 6u *u *x * + u *x * x * x * = 0
15.16
A
y = a sin πx
D
B
Fig.A.15.16
In Fig.A.15.16, A is the peak and B is the base point of the leading edge of the right going
wave: y = a sin πx . The phase velocity of any point on the leading edge is given by:
∂y ⎞
⎛
v = c0 ⎜1 + εa ⎟
∂x ⎠
⎝
12
© 2008 John Wiley & Sons, Ltd
⎛ 1 ∂y ⎞
⎛ 1
⎞
= c0 ⎜1 + εa ⎟ = c0 ⎜1 + εaπ cos πx ⎟
⎝ 2 ∂x ⎠
⎝ 2
⎠
π
⎛
⎞
⎜ cos πx = cos = 0 ⎟
2
⎝
⎠
⎛ 1
⎞
and phase velocity at B : vB = c0 ⎜1 − εaπ ⎟ (cosπx = cosπ = -1)
⎝ 2
⎠
∴ phase velocity at A : v A = c0
∴ phase velocity of A related to B :
⎛ 1
⎞ 1
v A − vB = c0 − c0 ⎜1 − εaπ ⎟ = c0εaπ = du
⎝ 2
⎠ 2
The phase distance dx between A and B = π 2
∴ Time for A to reach the position vertically above B is :
1
1
dx π
=
=
[secs]
du 2 1 2 c0εaπ c0εa
© 2008 John Wiley & Sons, Ltd