physics-of-vibration-and-waves-solutions-pain

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

SOLUTIONS TO CHAPTER 1
1.1
In Figure 1.1(a), the restoring force is given by:
F = −mg sinθ
By substitution of relation sinθ = 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 = sinθ = x l , or x = lθ , we have the restoring force
given by:
F = −mgθ
Now, the equation of motion using angular displacement θ can by derived from Newton’s
second law:
F = m&x&
i.e. −mgθ = mlθ&&
&& g
i.e. θ + θ = 0
l
which shows the frequency is given by:
ω 2 = 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&&
&& C
i.e. θ + θ = 0
I
ω2 = C I

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
i.e. &x&+ 2Tx lm = 0
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 ρ
Ag
Axg
= = =
x
x
ρ
2 2
2
Newton’s second law gives:
−G = m&x&
i.e. − 2ρAxg = ρAl&x&
&x& g
i.e. + 2 x = 0
l
which show the frequency is given by:
ω 2 = 2g l
In Figure 1.1(f), by taking logarithms of equation pVγ = constant , we have:
ln p +γ lnV = constant
dV
dp γ
so we have: + = 0
V
p
i.e.
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&

2
−γp A x = ρ &&
i.e. Alx
V
x γ
pA
i.e. + x = 0
l ρ
V
&&
which show the frequency is given by:
pA
ρ
γ
l V
ω 2 =
In Figure 1.1 (g), the volume of liquid displaced is Ax , so the restoring force is − ρgAx . Then,
F = −ρgAx = m&x&
x g A ρ
i.e. + x = 0
m
&&
ω 2 = gρA m
1.2
Write solution x = a cos(ωt +φ ) in form: x = a cosφ cosωt − asinφ sinωt and
compare with equation (1.2) we find: A = a cosφ and B = −asinφ . We can also
find, with the same analysis, that the values of A and B for solution
x = asin(ωt −φ ) are given by: A = −asinφ and B = acosφ , and for solution
x = acos(ωt −φ ) are given by: A = a cosφ and B = asinφ .
Try solution x = a cos(ωt +φ ) in expression &x&+ω 2x , we have:
&x&+ω 2x = −aω 2 cos(ωt +φ ) +ω2a cos(ωt +φ ) = 0
Try solution x = asin(ωt −φ ) in expression &x&+ω 2x , we have:
&x&+ω 2x = −aω2 sin(ωt −φ ) +ω 2asin(ωt −φ ) = 0
Try solution x = a cos(ωt −φ ) in expression &x&+ω 2x , we have:
&x&+ω 2x = −aω 2 cos(ωt −φ ) +ω2a cos(ωt −φ ) = 0

1.3
(a) If the solution x = asin(ωt +φ ) satisfies x = a at t = 0 , then, x = asinφ = 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
asin(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 = acos(ωt +φ ) satisfies x = a at t = 0 , then, x = a cosφ = 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
acosωt = + 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 = asin(ωt −φ ) satisfies x = a at t = 0 , then,
x = asin(−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 asin(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
solution of equation acosωt = + 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 = asin(ωt +φ ) satisfies x = −a at t = 0 , then,
x = asinφ = −a i.e. φ = −π 2 . When the pendulum swings to the position

solution of equation asin(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 = acos(ωt +φ ) satisfies x = −a at t = 0 , then,
x = a cosφ = −a i.e. φ =π . When the pendulum swings to the position
solution of equation acos(ωt +π ) = + a 2 , i.e. ωt = 3π 4 . Similarly, we can
If the solution x = asin(ωt −φ ) satisfies x = −a at t = 0 , then,
x = asin(−φ ) = −a i.e. φ =π 2 . When the pendulum swings to the position
solution of equation asin(ωt −π 2) = + a 2 , i.e. ωt = 3π 4 . Similarly, we can
If the solution x = a cos(ωt −φ ) satisfies x = −a at t = 0 , then,
x = acos(−φ ) = −a i.e. φ =π . When the pendulum swings to the position
solution of equation acos(ωt −π ) = + a 2 , i.e. ωt = 3π 4 . Similarly, we can
1.4
The frequency of such a simple harmonic motion is given by:
s
e
πε π
e e = = = rad s
4.5 10 [ ]
19 2
−
(1.6 ×
10 )
4 8.85 10 (0.05 10 ) 9.1 10
4
16 1
12 9 3 31
3
0
2
0
−
− − −
≈ × ⋅
× × × × × × ×
r m
m
ω
Its radiation generates an electromagnetic wave with a wavelength λ given by:

= = − π
ω
c ≈ × m = nm
2 2 3 10 8
4.2 10 [ ] 42[ ]
× × ×
16
4.5 10
8
0
×
π
λ
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 s m s m a ω2 = ′ = 2 .
(b) The frequency of the system is given by: s m b ω2 =
(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 = −2sx and the stiffness of
the system is given by: s′ = −F x = 2s . Hence the frequency is given by
s m s m c ω2 = ′ =2 .
Therefore, we have the relation: 2 : 2 : 2 = s 2m: s m: 2s m = 1: 2 : 4 a b c ω ω ω
1.6
At time t = 0 , 0 x = x gives:
0 asinφ = x (1.6.1)
0 x& = v gives:
0 aω cosφ = v (1.6.2)
From (1.6.1) and (1.6.2), we have
0 0 tanφ =ωx v and 2 2 1 2
a = (x 2
+ v ω )
0 0
1.7
The equation of this simple harmonic motion can be written as: x = asin(ωt +φ ) .
The time spent in moving from x to x + dx is given by: t dt = dx v , where t v is
the velocity of the particle at time t and is given by: v = x = aω cos(ωt +φ ) t & .

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 = 2dt where the period is given by:
T
dt
2 2
ω
2π T = , so we have:
ω
dx
dx
dx
dx
2 cos( ) cos( ) a 1 sin2 ( t
) a 2 x
2
a t
a t
T
−
=
− +
=
+
=
+
= =
π ω ω φ π ω φ π ω φ π
η
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
i.e.
ω 2 sin(ω +φ )
=
t
A g
Therefore, the minimum amplitude, which makes the mass lose contact with the
platform, is given by:
A g g ≈
min 2 4 2 2 2 2 m
0.01[ ]
9.8
= = =
ω π π
4 5
f
× ×
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

of this element given by:
2 2
v mv
l
KE KE dy 1
m
= = ⎛ l l l
1
spring dy y dy mv
2
dy y
l
= ′ ′ 2
= ⎛ v
1
2
1
2
⎞
⎟⎠
⎛
⎞
⎜⎝
⎟⎠
⎜⎝
l
KE m v m dy
The total kinetic energy of the spring is given by:
⎞
⎛
⎞
∫ ∫ ∫ = = ⎟⎠
⎜⎝
⎟⎠
⎜⎝
l
dy y
l
0 0
2
0
2
3
6
2 2
The total kinetic energy of the system is the sum of kinetic energies of the spring and
the suspended mass, and is given by:
KE = 1 mv + 1
Mv = 1
M +
m v tot 2 2 ( 3) 2
2
2
6
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:
3
2
s
+
M m
ω =
1.11
In Figure 1.1(a), the restoring force of the simple pendulum is −mg sinθ , then, the
stiffness is given by: s = mg sinθ x = mg l . So the energy is given by:
E = mv + sx = mx& + mg
1 1
1
1
x
2 2 2 2
2
2
2
2
l
The equation of motion is by setting dE dt = 0 , i.e.:
0
mx mg
d &
⎟⎠
1 2 1
2 = ⎛ + x
2
2
⎞
⎜⎝
l
dt
&x& g
i.e. + 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:
E = 1 Iθ& + 1
Cθ
2 2
2
2
0
d &
⎟⎠
1 2 1
2 = ⎛ Iθ + Cθ
dt
2
2
⎞
⎜⎝

&& C
i.e. θ + θ = 0
E = 1 mv + sx = ρAlx& + ρgAx = ρAlx& + ρgAx
I
In Figure 1.1(c), the energy is directly given by:
E = 1 mv + 1
sx
2 2
2
2
0
d &
⎟⎠
1 2 1
2 = ⎛ mx + sx
dt
2
2
⎞
⎜⎝
&x& s
i.e. + x = 0
m
In Figure 1.1(c), the restoring force is given by: − 2Tx l , then the stiffness is given
by: s = 2T l . So the energy is given by:
x mx T
l
1 x
2 1
2
1
1
2 2 2 2 2 2
2
1
2
2
2
l
E = mv + sx = mx& + T = & +
0
mx T
d &
1 2 2 = ⎟⎠
⎛ + x
2
⎞
⎜⎝
l
dt
&x& T
i.e. + 2 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 = 2ρgAl l = 2ρgA. So the energy is given by:
2 1
2
1
1
2 2 2 2 2 2
2
1
2
2
2
0
d ρ & ρ
1 2 2 = ⎟⎠
⎛ Alx + gAx
dt
2
⎞
⎜⎝
&x& g
i.e. + 2 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:
2 2
E mv sx Alx pA x
V
= 1 2 + 1
2 γ
= 1
ρ & 2
+
1
2
2
2
2
0
1 2 2
d γ
Alx 1
pA x
2
2
⎞
2 = ⎟ ⎟⎠
⎛
⎜ ⎜⎝
+
V
dt
ρ &
x γ
pA
i.e. + x = 0
l ρ
V
&&
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 mv + 1
sx = 1
mx& + 1
ρgAx
2 2 2 2
2
2
2
2
0
d & ρ
⎟⎠
1 2 1
2 = ⎛ mx + gAx
dt
2
2
⎞
⎜⎝
x A g ρ
i.e. + x = 0
m
&&
1.12
The displacement of the simple harmonic oscillator is given by:
x = asinωt (1.12.1)
so the velocity is given by:
x& = aω cosωt (1.12.2)
From (1.12.1) and (1.12.2), we can eliminate t and get:
2
+ = t + t =
x ω ω
& (1.12.3)
sin2 cos2 1
2
2 2
2
a
x
a
ω
which is an ellipse equation of points (x, x&) .
The energy of the simple harmonic oscillator is given by:

⎛
y
− + ⎟ ⎟⎠
x
y
sin sin cos cos
2
φ φ φ φ
2
2
⎞
xy
x
y
xy
y
x
sin sin 2 sin sin cos cos 2 cos cos
⎛
x
= + − + + −
φ φ φ φ φ φ φ φ
2
xy
y
(sin cos ) (sin cos ) 2 (sin sin cos cos )
x
= + + + − +
2
2
2
2
φ φ φ φ φ φ φ φ
xy
y
x
= + − −
E = 1 mx& 2 + 1
sx 2
(1.12.4)
2
2
Write (1.12.3) in form x&2 =ω2 (a2 − x2 ) and substitute into (1.12.4), then we have:
E = 1 mx& + 1
sx = 1
mω ( a − x ) + 1
sx
2 2 2 2 2 2
2
2
2
2
Noting that the frequency ω is given by: ω 2 = s m , we have:
E = 1 s a − x + sx = 1
sa
( ) 1
2
2 2 2 2
2
2
which is a constant value.
1.13
The equations of the two simple harmonic oscillations can be written as:
sin( ) 1 y = a ωt +φ and sin( ) 2 y = a ωt +φ +δ
The resulting superposition amplitude is given by:
[sin( ) sin( )] 2 sin( 2)cos( 2) 1 2 R = y + y = a ωt +φ + ωt +φ +δ = a ωt +φ +δ δ
and the intensity is given by:
I = R2 = 4a2 cos2 (δ 2)sin2 (ωt +φ +δ 2)
i.e. I ∝ 4a2 cos2 (δ 2)
Noting that sin2 (ωt +φ +δ 2) varies between 0 and 1, we have:
0 ≤ I ≤ 4a2 cos2 (δ 2)
1.14
2 cos( )
1 2
1 2
2
2
2
1
1 2 1 2
1 2
1
2
1
2
2
2
2
2
2
2
2
1
1 2
1 2
2
2
2
1
1
2
2
2
1 2
1 2
1
2
2
2
2
2
2
1
2
2
1
1
2
2
1
2
2
1
φ φ
⎟ ⎟⎠
⎜ ⎜⎝
⎞
⎜ ⎜⎝
−
a a
a
a
a a
a
a
a a
a
a
a a
a
a
a
a
a
a
On the other hand, by substitution of :
x = +
1 1
1
sinωt cosφ cosωt sinφ
a

⎛
x
⎛
y
− + ⎟ ⎟⎠
x
y
sin sin cos cos
φ φ φ φ
⎞
⎛
t t
x
sin (sin cos sin cos ) cos (cos sin cos sin )
= − + −
ω φ φ φ φ ω φ φ φ φ
(sin cos )sin ( )
= + −
ω ω φ φ
= −
1
E = mx + sx + my +
sy
1
& &
ω ω ω ω ω ω ω ω
= + + +
1
ω ω
= +
1
1
1
2 2 2 2
cos 1
2 2 2 2 2 2 2 2 2 2 2 2
ma t ma t mb t mb t
1
2 2 2 2
ma mb
ω
y = +
2 2
2
sinωt cosφ cosωt sinφ
a
into expression
2
2
x
1
⎞
y
+ − 1
2
⎟⎠
⎟ 2
1
y
2
2
1
⎞
⎛
⎜ ⎜⎝
cos cos sin sin ⎟ ⎟⎠
⎜ ⎜⎝
φ − φ φ φ
a
a
a
a
, we have:
sin ( )
2 1
2
2 1
2 2 2
2
1 2 2 1
2 2
2 1 1 2
2
2
2
1
1
2
2
1
2
2
1
φ φ
⎟ ⎟⎠
⎜ ⎜⎝
⎞
⎜ ⎜⎝
−
t t
a
a
a
a
From the above derivation, we have:
2
+ − xy
φ −φ = φ −φ
2 cos( ) sin ( )
2 1
2
1 2
1 2
2
y
2
2
x
2
1
a a
a
a
1.15
By elimination of t from equation x = asinωt and y = bcosωt , we have:
2
2
+ =
1 2
2
y
b
x
a
which shows the particle follows an elliptical path. The energy at any position of x ,
y on the ellipse is given by:
( )
2
2
2
cos
sin 1
2
sin 1
2
2
2
1
2
2
2
2
2 2 2
m a b
= +
The value of the energy shows it is a constant and equal to the sum of the separate
energies of the simple harmonic vibrations in x direction given by 2 2
1 mω a and in
2
1 mω b .
y direction given by 2 2
2
At any position of x , y on the ellipse, the expression of m(xy& − yx&) can be
written as:

m(xy& − yx&) = m(−abω sin2ωt − abω cos2ωt) = −abmω(sin2ωt + cos2ωt) = −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 1 2a wide and
2 2a high, where 1 max a = x and 2 max a = y , see Figure 1.8.
When x = 0 in equation (1.3) the positive value of sin( ) 2 2 1 y = a φ −φ . The value of
max 2 y = a . So sin( ) 0 max 2 1 = φ −φ = y y x which defines 2 1 φ −φ .
1.17
In the range 0 ≤φ ≤π , the values of i cosφ are −1 ≤ cos ≤ +1 i φ
. For n random
values of i φ , statistically there will be n 2 values −1 ≤ cos ≤ 0 i φ
and n 2 values
0 ≤ cos ≤ 1 i φ
. The positive and negative values will tend to cancel each other and the
n
Σ →
sum of the n values: cos 0
i i
1
j ≠ =
i φ
→ Σ=
, similarly cos 0
1
n
j
j φ
. i.e.
Σ Σ →
cos cos 0
1 =
1
≠ =
n
j
j
n
j i i
i φ φ
1.18
The exponential form of the expression:
a sinωt + a sin(ωt +δ ) + a sin(ωt + 2δ ) +L+ a sin[ωt + (n −1)δ ]
is given by:
aeiωt + aei(ωt+δ ) + aei(ωt+2δ ) +L+ aei[ωt+(n−1)δ ]
From the analysis in page 28, the above expression can be rearranged as:
⎡
⎞
⎛ −
+
ae ω δ nδ
2 sin 2
sin 2
1
δ
i t n ⎥⎦
⎤
⎢⎣
⎟⎠
⎜⎝
with the imaginary part:
δ ω n n t a ⎥⎦
sin δ
2
sin 2
⎡
−
sin +
⎛ 1
2
δ
⎤
⎢⎣
⎞
⎟⎠
⎜⎝
which is the value of the original expression in sine term.

1.19
From the analysis in page 28, the expression of z can be rearranged as:
z aeiωt eiδ ei δ ei n δ aeiωt nδ = + + + − = L
zz aeiωt nδ ae iωt n a n = ⋅ − =
(1 2 ( 1) ) sin 2
sin δ
2
The conjugate of z is given by:
z ae−iωt nδ =
* sin 2
sin δ
2
so we have:
2
sin δ
2
sin 2
sin δ
2
sin 2
sin 2
sin 2
2
* 2
δ
δ
δ

2.1
The system is released from rest, so we know its initial velocity is zero, i.e.
dx
0
0
=
dx
t= dt
(2.1.1)
Now, rearrange the expression for the displacement in the form:
x F +
G e( − p + q ) t +
F − G e ( − p −
q )
t 2 2
=
(2.1.2)
Then, substitute (2.1.2) into (2.1.1), we have
⎤
⎡ −
( ) ( p q ) t ( ) ( p q )
t
0
2 2 0
0
= ⎥⎦
⎢⎣
+ − −
+
= − +
=
− + − −
= t
t
p q F G e p q F G e
dt
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:
r
(r2 4ms)1 2
G
F
−
=
2.2
The first and second derivatives of x are given by:
x ⎡ r ⎤
& = B − +
2
(A Bt) e rt m
m
2
−
⎥⎦
⎢⎣
⎤
⎡
x rB 2
(A Bt) e rt m
r
&& = − + +
m
m
2
2
4
− ⎥⎦
⎢⎣
We can verify the solution by substitution of x , x& and &x& into equation:
m&x&+ rx& + sx = 0
then we have equation:
⎛
− A Bt
( ) 0
s r
4
2
⎞
= + ⎟ ⎟⎠
⎜ ⎜⎝
m
which is true for all t, provided the first bracketed term of the above equation is zero, i.e.

⎟⎠
x r i C e r m i t r
r m i t = − ′ & = ⎛− + ′ − + ′ − − ′ at t = 0
+ ⎛− − ′ ⎟⎠
0
s r
4
2
− =
m
i.e. r2 4m2 = s m
2.3
The initial displacement of the system is given by:
( ω ω ) cosφ 1 2
x = e−rt 2m C ei ′t +C e−i ′t = A at t = 0
So:
cosφ 1 2 C +C = A
(2.3.1)
Now let the initial velocity of the system to be:
ω ( ω ) ω ( ω ) ω sinφ
2 2
2
2
2
1 i C e A
m
m
⎞
⎜⎝
⎞
⎜⎝
− r + ′ − = − ′
i.e. cosφ ω ( ) ω sinφ
2 1 2 A i C C A
m
If r m is very small orφ ≈π 2 , the first term of the above equation approximately equals zero,
so we have:
sinφ 1 2 C −C = iA
(2.3.2)
From (2.3.1) and (2.3.2), 1 C and 2 C are given by:
( )
φ
( ) φ
C A cos φ i sin
φ
A e
2 2
cos sin
φ φ
i
i
1
C A i A e
= −
−
=
=
+
=
2 2
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 = C e−t RC 1

where 1 C is arbitrary in value. Use initial condition, we get 1 0 C = q ,
i.e. q = q e−t RC 0
which shows the relaxation time of the process is RC s.
2.5
(a) 2 2 2
2 2 -6
0 ω -ω′ =10 ω = r 4m => 500 0 ω m r =
π
1 2 4 3
E = mx&2 + sx2 = sx = × × − = × − J
E E E −
0
ω′ ≈ , so:
The condition also shows 0 ω
500 0 Q =ω′m r ≈ω m r =
Use
ω
τ
′
′ = 2
, we have:
π π
r
π
r
δ τ = =
′
2 ω
500
= ′ =
m Q
m
(b) The stiffness of the system is given by:
2 1012 10 10 100[ 1]
0
s =ω m = × − = Nm−
and the resistive constant is given by:
6 10
10 10 7 1
r m ω
0 − −
2 10 [ ]
500
−
= × ⋅
×
= = N sm
Q
(c) At t = 0 and maximum displacement, x& = 0 , energy is given by:
100 10 5 10 [ ]
1
2
1
2
1
2
2
max
Time for energy to decay to e−1of initial value is given by:
0.5[ ]
t m 10
=
= = −
2 10
7
10
ms
r
×
−
(d) Use definition of Q factor:
E
Q E
− Δ
= 2π
where, E is energy stored in system, and − ΔE is energy lost per cycle, so energy loss in the
first cycle, 1 − ΔE , is given by:
2 2 5 10 5
2 10 [ ]
500
3
1 J
Q
−
= ×
×
− Δ = −Δ = π = π × π
2.6

The frequency of a damped simple harmonic oscillation is given by:
ω′ =ω − r => 2
m Q 0 ω
= we find fractional change in the resonant frequency is given by:
2
2
2
0
2
4m
2
ω′ 2
−ω 2
= r 0
=> 4m
2
( ) 0
2
0 4
-
ω ω
ω ω ω
′ +
Δ = ′ =
m
r
Use ω′ ≈ω and
r
r
ω ω
ω
Δ Q
( 2 ) 1
− ≈ =
2
0
2
2
0
0
0
8
8
′ −
=
m
ω ω
ω
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:
0 E = nex /ε ，the restoring force per unit area is given by:
2 2
ε
F = qE = − xn e l
0
restoring force per unit area = electronsmass per unit area × electrons acceleration
i.e.
2 2
ε
F xn e l nlm x = − = e × &&
0
2
+ x =
m
i.e. 0
0
x ne
ε
&&
From the above equation, we can see the displacement distance of electrons, x , oscillates with
angular frequency:
2
0
2
= ne
ε
ω
e m
e
2.9
As the string is shortened work is done against: (a) gravity (mg cosθ ) and (b) the centrifugal
force (mv2 r = mlθ&2 ) along the time of shortening. Assume that during shortening there are

many swings of constant amplitude so work done is:
A = −(mgcosθ + mlθ&2 )Δl
where the bar denotes the average value. For small θ , cosθ =1−θ 2 2 so:
A = −mgΔl + (mgθ 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 = (mgθ 2 2 −mlθ&2 )Δl
Now the pendulum motion has energy:
E = ml2θ&2 + mgl (1 − cos θ )
,
2
that is, kinetic energy plus the potential energy related to the rest position, for small θ this
becomes:
&
ml2θ 2 mglθ 2 E = +
2 2
which is that of a simple harmonic oscillation with linear amplitude 0 θ l .
Taking the solution θ θ cosωt 0 = which gives 2 2
θ&2 =ω 2θ with
θ 2 =θ and 2 2
0
0
ω = g l we may write:
ml2ω 2θ mglθ E = =
2 2
2
0
2
0
and
⎞
⎛
ω 2θ ω θ ω θ
l ml l ml ml E Δ ⋅ − = Δ ⎟ ⎟⎠
⎜ ⎜⎝
Δ = −
4 2 4
2
0
2 2
0
2 2
0
so:
E Δ
= −
l
l
Δ
E
1
2
Now ω = 2πν = g l so the frequency ν varies with l−1 2 and
E
E
l
l Δ
=
Δ
= −
Δ
1
2
ν
ν
so:
E
= constant
ν

3.1
The solution of the vector form of the equation of motion for the forced oscillator:
−
t F
iF
iF e
i t
= = − − + −
m r s F eiωt
0 &x& + x& + x =
is given by:
cos( ) sin( )
ω φ
( )
0 ω φ
ω
ω φ
ω ω
t
Z
Z
Z
m m m
x
Since F eiωt
0 represents its imaginary part: F sinωt 0 , the value of x is given by the
imaginary part of the solution, i.e.:
x F
= − t −
cos(ω φ )
Z
ω
m
The velocity is given by:
v x F
= = sin(ωt −φ )
Z
m
&
3.2
The transient term of a forced oscillator decay with e−rt 2m 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)
For small damping, we have
s m 0 ω ≈ω = (3.2.2)
We also have steady state displacement given by:
sin( ) 0 x = x ωt −φ
where the maximum displacement is:
x F
=
0
ω ω
0 r ( m 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:

0
F
x =
0 0
2kmω
t
3.3
Write the equation of an undamped simple harmonic oscillator driven by a force of
frequency ω in the vector form, and use F eiωt
0 to represent its imaginary part
F sinωt 0 , we have:
m s F eiωt
0 &x& + x = (3.3.1)
We try the steady state solution x = Aeiωt and the velocity is given by:
x& = iωAeiωt = iωx
so that:
&x& = i2ω2x = −ω2x
and equation (3.3.1) becomes:
ω m s eiωt F eiωt 0
(−A 2 + A ) =
which is true for all t when
− Aω 2m+ As = F
0
i.e.
F
0ω
2
s −
m
A =
F ω
ω 2
i.e. ei t
0
−
s m
x =
The value of x is the imaginary part of vector x , given by:
x F ω
t
sin 2
0
−
s ω
m
=
i.e.
x F t where
sin
ω
2 −
2
0
0
ω ω
( )
=
m
2 = s
m
0 ω
Hence, the amplitude of x is given by:
A F
( 2 2 )
0
0
ω −ω
=
m

and its behaviour as a function of frequency is shown in the following graph:
By solving the equation:
⎡
A
F
0 ω m
2
0
0 ω
x F t
ω
0 ω
= A t B t A
0 = = ⎥⎦
cos
⎡
dx
ω ω
x F 0
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 t D t 0 0 = cosω + sinω
where, = s m 0 ω
, 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 F t 2 2 0 0
sin ω ω
ω ω
C t D t
= (3.3.2)
m
0 cos sin
( )
0
ω
+ +
−
3.4
In equation (3.3.2), x = 0 at t = 0 gives:
cos sin 0
sin
( )
0
2 2 0 0
0
0
⎤
⎢⎣
+ +
−
=
=
m
t
t ω ω
ω ω
(3.4.1)
In equation (3.3.2), x& = 0 at t = 0 gives:
0
A t B t F
0
ω
( )
sin cos
( )
2 2 0
0
0
2 2 0 0 0 0
0
0
0
+ =
−
⎤
= ⎥⎦
⎢⎣
− +
−
=
= =
B
m
m
F t
dt
t t
ω
ω ω
ω ω ω ω
ω ω
i.e.
B F (3.4.2)
ω
−
0
( 2 2 )
0 0
ω ω ω
= −
m
By substitution of (3.4.1) and (3.4.2) into (3.3.2), we have:
⎞
⎟ ⎟⎠
⎛
1 ω
⎜ ⎜⎝
ω
= 0 sin t −
sin
t
m
( −
)
0
2 2
0
ω
ω
ω ω
(3.4.3)
By substitution of ω =ω + Δω 0 into (3.4.3), we have:

x F ω
0
= − t t t t t
x F ω
0
= − t t t t
x
F
π
m
0
2 ω
2
0
0
F
π
m
0ω
2
0
⎞
⎟ ⎟⎠
⎛
⎜ ⎜⎝
Δ + Δ −
+ Δ
m
0
0 0
0
0 sin cos sin cos sin
( )
ω
ω
ω ω ω ω
ω ω ω
(3.4.4)
Since 1 0 Δω ω << and Δωt << 1, we have:
ω ≈ , sin Δωt ≈ Δωt , and cosΔωt ≈ 1
0 ω
Then, equation (3.4.4) becomes:
⎞
⎟ ⎟⎠
⎛
⎜ ⎜⎝
0 sin + Δ cos −
sin
Δ
m
0
0 0
0
2
ω
ω
ω ω ω
ω ω
⎞
⎛
x F ω
0 0
i.e. ⎟ ⎟⎠
⎜ ⎜⎝
= − 0 sin t + Δ
t cos
t
Δ
−
Δ
m
0 0
2
ω ω ω
ω
ω ω
⎞
⎛
x F 0
= t − t t
sin ω
0 cos
2
i.e. ⎟ ⎟⎠
⎜ ⎜⎝
m
0
0
0
ω
ω
ω
x F ω ω ω
0 t t t
m
i.e. = (sin −
cos )
2 ω
2 0 0 0
0
The behaviour of displacement x as a function of t 0 ω
is shown in the following
F t
0
2mω
t 0 ω
0
− F t
0
2mω
0
F
7
ω
π
m
0
2
F 2
5
ω
π
m
0
2
F 2
3
ω
π
m
2
0
F
0
2
0 2
ω
π
m
2
0
0
F
0 3
ω
π
m
2
0
0
graph:
3.5
The general expression of displacement of a simple damped mechanical oscillator
driven by a force F cosωt 0 is the sum of transient term and steady state term, given

by:
ωm s ω
⎞
⎛ −
φ tan 1 , so the general expression of velocity is given by:
C r i v&
C r ms i v& (3.5.1)
v ω
F r sm
= = ω ω
i t i t
ω (ω φ )
− + − x = −
m
rt
Ce iF e 2 m
i
0
Z
ω
where, C is constant, Z r2 (ωm s ω)2 m = + − , s m r2 4m2 t ω = − and
⎟⎠
⎜⎝
= −
r
rt
i e F
C r i v x&
= = ⎛− + i t
2 0 ( )
2
⎞
ω ω φ ω − + − + ⎟⎠
⎜⎝
t e
m
i t
m
Z
m
and the general expression of acceleration is given by:
⎞
rt
e i F
i r
2 2 0 ( )
2
2
4
ω ω ω ω φ
ω − + − + ⎟ ⎟⎠
⎛
⎜ ⎜⎝
= − − i t
t e
m
i t
m
t
Z
m
m
⎞
⎛
rt
e i F
i r
i.e. 2 0 ( )
2
2
2
2 ω − + ω ω ω −φ + ⎟ ⎟⎠
⎜ ⎜⎝
−
−
= i t
t e
m
i t
m
Z
m
m
From (3.5.1), we find the amplitude of acceleration at steady state is given by:
F
v F
ω ω
& = =
0 0
2 2
r m s
(ω ω)
Z
m + −
dv&
At the frequency of maximum acceleration: = 0
dω
⎤
⎡
F
ω
d
i.e. 0 =
0
ω r m s
2 ( )2
⎥ ⎥
⎦
⎢ ⎢
⎣
+ ω − ω
d
2
r ms s
i.e. 2 − 2 + 2 =
0 2
ω
2
2
s
−
i.e. 2
2
2
sm r
ω =
Hence, we find the expression of the frequency of maximum acceleration given by:
2
2
2
s
−
2
sm r
ω =
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:
F
m
s mF
sm sm sm
0
r m s
0 0
2 2
( ) ( )
=
+ −
=
+ −
&

The limit of the acceleration amplitude at high frequencies is given by:
v ω
F 0
lim &
lim =
F
m
F
2
0
r m s
0 lim
r m s
2
+ ⎛ −
2 2
2 2
( )
⎞
⎟⎠
⎜⎝
=
+ −
=
→∞ →∞ →∞
ω ω
ω ω
ω ω ω
So we have:
v lim
v r sm
= →∞ =
ω
& &
3.6
The displacement amplitude of a driven mechanical oscillator is given by:
x F
0
2 2
ω r + (ωm −
s ω)
=
i.e.
x F
0
2 2 2 2
r + ( m −
s)
=
ω ω
(3.6.1)
The displacement resonance frequency is given by:
ω = s − (3.6.2)
2
r
2
2m
m
By substitution of (3.6.2) into (3.6.1), we have:
2 2
2
x F
2
r 2
s
0
⎞
⎛
+ ⎟ ⎟⎠
r
r
2 2 ⎟ ⎟⎠
⎜ ⎜⎝
⎞
⎛
⎜ ⎜⎝
−
=
m
m
m
i.e.
2
2
x F
0
r
4m
r s
m
−
=
which proves the exact amplitude at the displacement resonance of a driven
mechanical oscillator may be written as:
= 0
r
x F
ω ′
where,
2
2
ω′ 2
= s −
r
4m
m
3.7
(a) The displacement amplitude is given by:
x F
0
2 2
ω r + (ωm −
s ω)
=

At low frequencies, we have:
x F 0
lim lim =
v F
v ω
F 0
lim &
lim =
F
s
F
0
r m s
0
r m s
2 2 2 2
lim
2 2 0
0 0 ( )
+ ( −
)
=
+ −
=
ω→ ω→ ω ω ω ω→ ω ω
(b) The velocity amplitude is given by:
v F
0
2 2
r + (ωm −
s ω)
=
At velocity resonance: ω = s m , so we have:
F
r
F
0
r sm sm
0
r m s
s m
r
0
2 2
2 2
( ) ( )
=
+ −
=
+ −
=
ω=
ω ω
(c) From problem 3.5, we have the acceleration amplitude given by:
v F
0
ω
2 2
r m s
(ω ω)
+ −
& =
At high frequency, we have:
F
m
F
0
r m s
0
r m s
2 2 2 2
2 2
( )
lim
( )
+ −
=
+ −
=
→∞ →∞ ω ω →∞ ω ω
ω ω ω
From (a), (b) and (c), we find x
lim
ω→
0
, r v and v&
lim are all constants, i.e. they are all
ω→∞
frequency independent.
3.8
The expression of curve (a) in Figure 3.9 is given by:
( )
F m
−
= − = (3.8.1)
a ω ω ω
2 2 2 2 2
0
2
2 2
0 0
x F X
m
2
0
m ( )
r
Z
m
ω ω
ω − +
where = s m 0 ω
dxa
a x has either maximum or minimum value when = 0
dω
( )
⎤
⎡
−
F m
d
ω ω
i.e. 0
2 2 2 2 2
0
m r
( )
2
2 2
0 0 = ⎥⎦
⎢⎣
− +
d
ω ω ω
ω
i.e. m2 ( ω −ω ) −ω 2 r 2 =
0
0
2 2 2
0
Then, we have two solutions of ω given by:
r 2 0
m
1 0
ω
ω = ω − (3.8.2)

and
= − = ⎛ω − ω
= + = ⎛ω + ω
ω = :
2 0r
m
2 0
ω
ω = ω + (3.8.3)
Since r is very small, rearrange the expressions of 1 ω
and 2 ω
, we have:
r
m
2 2
r
m
r
m
r
m
⎞
2 4 2 0 2
0
2 0
1 0 − ≈ − ⎟⎠
⎜⎝
ω
ω ω
r
m
2 2
r
m
r
m
r
m
⎞
2 4 2 0 2
0
2 0
2 0 + ≈ − ⎟⎠
⎜⎝
ω
ω ω
1 ω
The maximum and the minimum values of x can found by substitution of (3.8.2)
a and (3.8.3) into (3.8.1), so we have:
when ω = :
r
F
x F a
0
r r m
0
2ω ω 2ω
0
2
0 0
≈
−
=
which is the maximum value of a x , and
when 2 ω
r
F
x F a
0
r r m
0
2ω ω 2ω
0
2
0 0
≈ −
+
= −
which is the minimum value of a x .
3.9
The undamped oscillatory equation for a bound electron is given by:
&&+ x ω 2
x = ( − eE m)cosωt (3.9.1)
0 0
Try solution x = Acosωt in equation (3.9.1), we have:
− 2 + = −
2
0
( ω ω )Acosωt ( eE m)cosωt 0
which is true for all t provided:
(−ω 2 +ω ) = −
A eE m 0
2
0
i.e.
A eE
( 2 2 )
0
0
ω −ω
= −
m
So, we find a solution to equation (3.9.1) given by:
x eE ω
t
= − (3.9.2)
m
ω ω
cos
( 2 2 )
0
0−
For an electron number density n , the induced polarizability per unit volume of a

medium is given by:
x F X
= − = (3.10.3)
nex
χ = − (3.9.3)
E
e
0 ε
By substitution of (3.9.2) and E E cosωt 0 = into (3.9.3), we have
( 2 2 )
0 0
2
nex
0 ε ε ω ω
χ
−
= − =
m
ne
E
e
3.10
The forced mechanical oscillator equation is given by:
mx rx sx F cosωt 0 &&+ & + =
which can be written as:
2
0 &&+ & + = (3.10.1)
mx rx mω x F cosωt 0
where, = s m 0 ω
. Its solution can be written as:
t
t F X
= 0 − (3.10.2)
Z
x F r
Z
m
m
m
ω
ω
ω
ω
sin 0
cos 2
2
ωm s ω
⎞
⎛ −
where, X ωm s ω m = − , Z r2 (ωm s ω)2 m − + = , ⎟⎠
⎜⎝
= −
r
φ tan 1
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:
t
t F m
ω −
ω
cos ( ) 2 2 2 2 2
m ω
m r
Z
m
ω ω ω
ω
ω
cos
( )
0
2
2 2
0 0
2
0
− +
The damped oscillatory electron equation can be written as:
2
0 &&+ & + = − (3.10.4)
mx rx mω x eE cosωt 0
Comparing (3.10.1) with (3.10.4), we can immediately find the displacement x for a
damped oscillatory electron by substituting 0 0 F = −eE into (3.10.3), i.e.:
x eE m ω
t
( ω −
ω
)
= − (3.10.5)
2 2 2 2 2
0
m ω ω ω
r
cos
( )
2
2 2
0 0
− +
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:

2
− +
ω ω
1 1 ( ) 2 2 2 2 2
ε χ cos
r ω
ne m
( )
2 2 2 2 2
0
2
ω −
ω
[ ( ) ]
0
2 2
0
2
nex
E
ε m ω ω ω
r
0 ε
χ
− +
= − =
So we have:
t
ne m
m r
[ ( ) ]
ε ω ω ω
0
2
0
2 −
2
0
= + = +
3.11
The instantaneous power dissipated is equal to the product of frictional force and the
instantaneous velocity, i.e.:
2
P rx x r F
( ) cos2 ( )
= = 0 ωt −φ
2
Z
m
& &
The period for a given frequency ω is given by:
2π T =
ω
Therefore, the energy dissipated per cycle is given by:
2
0
E Pdt r F
∫ ∫
= = −
2
0
rF
π ω
2
0 2
= − −
rF
2
0
π
ω
rF
2
2
0
2
0
2
0
2
2
2
2
ω φ
[1 cos2( )]
2
cos ( )
m
m
m
T
m
π
Z
Z
t dt
Z
t dt
Z
ω
ω φ
π ω
=
=
∫
(3.11.1)
The displacement is given by:
x F
= t −
0 sin(ω φ )
ω
Z
m
so we have:
x F
max = (3.11.2)
m Z
ω
0
2
max E =πrωx
3.12
The low frequency limit of the bandwidth of the resonance absorption curve 1 ω
satisfies the equation:

m − s = −r 1 1 ω ω
which defines the phase angle given by:
ω m −
s ω
tan 1 1 1
=
= −
1 r
φ
The high frequency limit of the bandwidth of the resonance absorption curve 2 ω
satisfies the equation:
m− s = r 2 2 ω ω
which defines the phase angle given by:
ω m −
s ω
tan 2 2 1
=
=
2 r
φ
3.13
For a LCR series circuit, the current through the circuit is given by
I I eiωt
0 =
The voltage across the inductance is given by:
L dI = L d
iωt = ω iωt = ω 0 0
I e i LI e i LI
dt
dt
i.e. the amplitude of voltage across the inductance is:
0 V LI L =ω (3.13.1)
The voltage across the condenser is given by:
q i t i t
I e iI
C
1 1 1
= ∫ = ∫ ω = ω = − 0
i C
e dt
C
Idt
C C
ω ω
i.e. the amplitude of the voltage across the condenser is:
V I C ω
= 0 (3.13.2)
C
When an alternating voltage, amplitude 0 V is applied across LCR series circuit,
current amplitude 0 I is given by: e I V Z 0 0 = , where the impedance e Z is given
by:
2
2 1
⎞
⎟⎠
= + ⎛ −
Z R L m ω
⎜⎝
C
ω
At current resonance, 0 I has the maximum value:
I V0
0 = (3.13.3)
R
and the resonant frequency 0 ω
is given by:

V V C
1 0
0
ω or
0 − =
C
L
ω
1
0 ω = (3.13.4)
LC
0 V
R
0
V ω
=
L L
0
V
L
V
C
= = = = =
0 0 0
0
0
L
LC
0 V
R
R
LC
R
V L
RC
RC
ω
ω
Noting that the quality factor of an LCR series circuit is given by:
L Q 0 ω
R
=
so we have:
0 V V QV L C = =
3.14
In a resonant LCR series circuit, the potential across the condenser is given by:
V I C ω
= (3.14.1)
C
where, I is the current through the whole LCR series circuit, and is given by:
I I eiωt
0 = (3.14.2)
The current amplitude 0 I is given by:
I V0
0 = (3.14.3)
e Z
where, 0 V is the voltage amplitude applied across the whole LCR series circuit and is
a constant. e Z is the impedance of the whole circuit, given by:
2
2 1
⎞
ω (3.14.4)
⎟⎠
= + ⎛ −
Z R L e ω
⎜⎝
C
From (3.14.1), (3.14.2), (3.14.3), and (3.14.4) we have:
V V ω i ω
t
C
i t
C e V e
C
+ ⎛ −
C R L
ω
ω ω
2 0
2
0
1
=
⎞
⎟⎠
⎜⎝
=
dVC , i.e.:
which has the maximum value when 0 = 0
dω

2 1
0 ω = ,
Q 0L
0
0
1 2
2
V
0 =
⎞
⎟⎠
+ ⎛ −
⎜⎝
C
C R L
d
d
ω
ω ω
ω
i.e. 1 1 0
2 2
2 2
2
⎞
+ ⎛ −
2 = − + ⎟⎠
⎜⎝
C
L
C
R L
ω
ω
ω
ω
R ω L L
i.e. 2 + 2 2 2 − 2 = 0
C
2
1 1
R
1 Q
i.e. ω = − =ω −
2
2 0 0
2
2
L
LC
where
LC
ω
=
R
3.15
In a resonant LCR series circuit, the potential across the inductance is given by:
V LI L =ω (3.15.1)
where, I is the current through the whole LCR series circuit, and is given by:
I I eiωt
0 = (3.15.2)
The current amplitude 0 I is given by:
I V0
0 = (3.15.3)
e Z
where, 0 V is the voltage amplitude applied across the whole LCR series circuit and is
a constant. e Z is the impedance of the whole circuit, given by:
2
2 1
⎞
ω (3.15.4)
⎟⎠
= + ⎛ −
Z R L e ω
⎜⎝
C
From (3.15.1), (3.15.2), (3.15.3), and (3.15.4) we have:
V LV ω i ω
t
L
i t
ω
L e V e
C
+ ⎛ −
R L
ω
ω
2 0
2
0
1
=
⎞
⎟⎠
⎜⎝
=
dVL , i.e.:
which has the maximum value when 0 = 0
dω

LC R C LC R C −
2 1
0 ω = ,
Q 0L
0
0
1 2
2
LV
0 =
⎞
⎟⎠
ω
+ ⎛ −
⎜⎝
C
R L
d
d
ω
ω
ω
i.e. 1 1 0
2 2
2 2
2
⎞
+ ⎛ −
2 = + − ⎟⎠
⎜⎝
C
L
C
R L
ω
ω
ω
ω
L
i.e. 2 + 2 − 2 =
0
2 2
C
C
R
ω
i.e.
2
0
0
2
0
1 1
2 2 2 0 2
R
2
1 1
2
2
1
1
2
1
2
1
Q
L
L
=
−
=
−
=
−
=
ω
ω
ω ω
where
LC
ω
=
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:
δω = r (3.16.1)
m
On the other hand, the relation between frequency and wavelength of light is given
by:
f = c (3.16.2)
λ
where, c is speed of light in vacuum. From (3.16.2) we find at frequency resonance:
f = − c
δ 20
δλ
λ
where 0 λ
is the wavelength at frequency resonance. Then, the relation between
angular frequency bandwidth δω and the width of spectral line δλ is given by:
π
= 2 f = 2 c δλ
(3.16.3)
δω π δ 20
λ
From (3.16.1) and (3.16.3) we have:
r λ
r
0
δλ = = 0
=
m Q
λ
cm
0
20
2
λ
ω
π
So the width of the spectral line from such an atom is given by:
−
= ×
0.6 10 −
14
1.2 10 [ ]
×
5 10
7
6
λ
0 m
Q
×
= =
δλ

3.17
According to problem 3.6, the displacement resonance frequency r ω
⎛
− −
ω s ω
and the
corresponding displacement amplitude max x are given by:
2
2
r
2
0 2m
ωr = ω −
= =
0 0
r m ω ω ω ω ′
r
F
x F
Z
=
max
2
ω′ = ω − r ,
where, Z r2 (ωm s ω)2 m = + − , 2
2
0 4m
= s 0 ω
m
Now, at half maximum displacement:
0 max 0
m ω ω′
r
x F
F
Z
= =
2 2
i.e. ω r2 + (ωm− s ω)2 = 2ω′r
⎛
⎤
r m s s ⎟ ⎟⎠
r
⎡
⎞
+ ⎛ −
i.e. 2
4 r
2
2 2
2 2
4
m
m
⎞
⎜ ⎜⎝
= −
⎥ ⎥⎦
⎢ ⎢⎣
⎟⎠
⎜⎝
ω
ω ω
4
2
r
sr
s
ω r sm ω
i.e. ( 2 ) 4 0
4
3
2
2
2
2
2
−
4 +
+ − + =
m
m
m
m
2 2
⎞
⎛
− − ⎟ ⎟⎠ ⎞
⎛
⎞
r
s
r
r
s
r
i.e. 0
4
3
2 2
2
2
2
2
2
2
2
2
2
2
4 = ⎟ ⎟⎠
⎜ ⎜⎝
⎜ ⎜⎝
− + ⎟ ⎟⎠
⎜ ⎜⎝
m
m
m
m
m
m
m
2
⎞
⎛
r
i.e. ( ) 3 0
2 2 2 2 = ⎟ ⎟⎠
⎜ ⎜⎝
ω −ω − ω′
m
r
r
i.e. ω 2 −ω 2 = ± 3 ω′2
r (3.17.1)
m
If 1 ω
and 2 ω
are the two solutions of equation (3.17.1), and 2 1 ω >ω , then:
r
ω −ω = 3 ω′
2 2 2
2
r (3.17.2)
m
r
ω −ω = − 3 ω′
2 2 2
1
r (3.17.3)
m

Since the Q-value is high, we have:
x F r
⎛
P neE t eE r
m
= 0 >> 1
r
Q
ω
2
ω >> r
2
0 m
i.e. 2
ω ≈ω′ ≈ r
i.e. 0 ω
Then, from (3.17.2) and (3.17.3) we have:
− + ≈ r
(ω ω )(ω ω ) 2 3 ω
2
2 1 2 1 0
m
and 1 2 0 ω +ω ≈ 2ω
Therefore, the width of displacement resonance curve is given by:
3r
m
2 1 ω −ω ≈
3.18
In Figure 3.9, curve (b) corresponds to absorption, and is given by:
t
t F ω
r
sin 2 2 2 2 2
m r
Z
m
ω
ω ω ω
ω
ω
sin
( )
0
2
0
2
0
− +
= =
and the velocity component corresponding to absorption is given by:
v x F ω
r ω
t
m ω ω ω
r
cos
( 2 2 )2 2 2
0
2
2
0
− +
= & =
For Problem 3.10, the velocity component corresponding to absorption can be given
by substituting 0 0 F = −eE into the above equation, i.e.:
v x eE r ω
t
ω
= & = − (3.18.1)
m ω ω ω
r
cos
( 2 2 )2 2 2
0
2
2
0
− +
For an electron density of n , the instantaneous power supplied equal to the product
of the instantaneous driving force neE cosωt 0 − and the instantaneous velocity, i.e.:
t
ne E ω
r
m r
t
ω
m r
ω
ω ω ω
ω
ω ω ω
ω
2
2 2 2 2 2
0
2
2 2
0
2
2 2 2 2 2
0
2
2
0
0
cos
( )
cos
( )
( cos )
− +
=
⎞
⎟ ⎟⎠
⎜ ⎜⎝
− +
= − × −

The average power supplied per unit volume is then given by:
ne E r
ω
π ω
∫
P Pdt av
2 2
0
π
ω
ne E ω
r
ω
2 2 2 2 2
0
2
2
2
0
2
2 2 2 2 2
0
2
2 2
0
2
2
0
2 ( )
cos
2
2 ( )
m r
t
m r
ω ω ω
ω
ω ω ω
π
π ω
− +
=
=
− +
=
∫
which is also the mean rate of energy absorption per unit volume.

4.1
The kinetic energy of the system is the sum of the separate kinetic energy of the two
masses, i.e.:
E 1 mx my m x y x y mX 1
mY k
( ) 1
2
1
1
1
⎤
= + = ⎡ + + −
2 2 2 2 2 2
1
= + − + + −
= + + −
1
= ⎡ + + −
1
mg
mg
x s y x mg
l
X mg
l
= + ⎛ +
⎡
⎤
⎡
m Y c X a E & & & & & & + = ⎥⎦
k q q q q Y X Y
⎡
⎡
E bX dY mg ⎟⎠
4
4
( ) 1
2
2
2
2
& & & & & & & & + = ⎥⎦
⎢⎣
The potential energy of the system is the sum of the separate potential energy of the
two masses, i.e.:
1
( ) 1
2
2 2 2 2
2 2 2
1
( ) 1
2
⎤
2 2 2
1
⎞
2 2
2
4
( ) ( )
2
1
2
( ) ( )
2
( )
2
2
1
2
s Y
l
mg
x y x y s x y
l
x y s x y
l
y s x y
l
E mg p
⎟⎠
⎜⎝
− + ⎥⎦
⎢⎣
Comparing the expression of k E and p E with the definition of X E and Y E given
by (4.3a) and (4.3b), we have:
= 1 ,
a m
2
b mg
d = mg +
= 1 , and s
= , c m
l
4
4
l
2
Noting that:
1 2 1 2
X m x y m X q
2
( )
2
⎞
⎟⎠
⎞
⎜⎝⎛ = + ⎟⎠
= ⎛
⎜⎝
1 2 1 2
( )
2
Y m x y m q
⎞
− = ⎛ ⎟⎠
= ⎛
and Y
c
⎟⎠
⎜⎝
⎞
⎜⎝
1 2
i.e. q X m X
⎜⎝⎛ = and q Y m Y
2
−
⎞
⎟⎠
1 2
2
−
⎞
⎟⎠
= ⎛
⎜⎝
we have the kinetic energy of the system given by:
1
2 2
2 2
2 2
2
2 1
2
2 1
4
1
4
m
X m
m
⎤
⎢⎣
+ ⎥⎦
⎢⎣
= + =
and
s
X g
l
= + ⎛ + ⎥⎦
2 2
2 2
X mg
2 2 2 1
2
2
+ ⎛ + 2
⎥⎦
2
⎞
Y
p 4 q q q l
m
q Y g
m
s
l
l m
⎞
⎜⎝
⎤
⎢⎣
⎟⎠
⎜⎝
⎤
⎢⎣
= + =

which are the expressions given by (4.4a) and (4.4b)
4.2
The total energy of Problem 4.1 can be written as:
E E E mx my mg k p = + = & + & + + + −
E E mx mg x pot kin ⎟⎠
( ) 1 s x
E E my mg y pot kin ⎟⎠
( ) 1 s y
+ = & + ⎛ + , 2 2
1 1
1
x y s x y
2 2 ( 2 2 ) ( )2
2
2
2
l
The above equation can be rearranged as the format:
kin pot x kin pot y pot xy E = (E + E ) + (E + E ) + (E )
where, 2 2
l
2 2
⎞
⎜⎝
+ = & + ⎛ + ,
l
2 2
⎞
⎜⎝
and E sxy pot = −2
4.3
x = −2a , y = 0 :
x = 0 , y = −2a :
-X + Y
-X - Y
4.4
For mass 1 m , Newton’s second law gives:
m x = sx 1&&1
For mass 2 m , Newton’s second law gives:
m x = −sx 2&&2
≡
x=0 -2a -a -a
+
-a a
y=0
≡
-2a -a -a
+
-a a

Provided x is the extension of the spring and l is the natural length of the spring,
we have:
2 2 13 2 27 2
s m m
(2 ) 4 × (1.14 × 10 ) × (23 × 35) × (1.67 ×
10 ) 1
Na Cl πν π
2 −
= = Nm
m m
x − x = l + x 2 1
By elimination of 1 x and 2 x , we have:
s − − = &&
x s
x x
m
m
2 1
x& m +
m
&i.e. 0
+ sx
m m
1 2 =
1 2
which shows the system oscillate at a frequency:
ω2 = s
μ
where,
m m
+
1 2
m m
1 2
μ =
For a sodium chloride molecule the interatomic force constant s is given by:
120[ ]
(23 35) 1.67 10
27
−
−
≈
+ × ×
=
+
Na Cl
ω μ
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:
mx s ( y x )
sx
= − −
&&
my s x y
( )
= −
&&
i.e.
mx s x y sx
&&
( ) 0
− − =
+ − − =
my 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:
i t
ω
i t
x =
Ae
y Be
ω
=
where A and B are the displacement amplitude of x and y at frequency ω .

Using these solutions, the equations of motion become:
x 2 a cos ( )
t t 2 1 1 2
a t t
2 sin ( )
2
m A s A B sA e
[ − ω
+ ( − ) + ] =
0
2
i t
[ − − ( − )] =
0
i t
m B s A B e
ω
ω
ω
We may, by dividing through by meiωt , rewrite the above equations in matrix form
as:
0
2
2
2
A
⎤
= ⎥⎦
⎡
⎢⎣
⎤
⎥⎦
⎡
⎢⎣
s m − −
s m
− −
B
s m s m
ω
ω
(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−ω2 ) − s2 m2 = 0
i.e. ω 4 − (3s m)ω 2 + s2 m2 = 0
i.e.
s
2
m
ω 2 = (3± 5)
In the slower mode, ω 2 = (3 − 5) s 2m . By substitution of the value of frequency
into equation (4.5.1), we have:
5 1
2
A ω
2
2
2
−
=
s −
m
=
−
=
s
s
s m
B
ω
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:
5 1
2
A ω
2
2
2
+
= −
s −
m
=
−
=
s
s
s m
B
ω
4.6
The motions of the two pendulums in Figure 4.3 are given by:
m a
cos ( )
y a t t a ω t ω
t
m a
sin ( )
ω ω ω ω
ω ω
ω ω ω ω
2 sin sin
2
2
2 cos cos
2
2
2 1 1 2
=
− +
=
=
− +
=
a ω
where, the amplitude of the two masses, 2 a cosω t and 2 a sinω t , are constants
m m over one cycle at the frequency .
Supposing the spring is very weak, the stiffness of the spring is ignorable, i.e. s ≈ 0.

Noting that 2 = g l
1 ω
E s a 1
mg
2 2 2 2 2
= = =
ω ω ω
1
x x x m a m
E = ma ω ω − ω t = E + ω −
ω
t
x a
= − = − −
and 2 ( 2 )
2 ω = g l + s m , we have:
g ω
2
2
⎛ +
ω ω
2 1 2
2
2
⎞
= ≈ ≈
ω ω = ⎟⎠
⎜⎝
l 1 2 a
Hence, the energies of the masses are given by:
( )
( a t) ma t
l
E s a mg
a t ma t
l
1
2 2 2 2 2
2 sin 2 sin
ω ω ω
1
y y y m a m
2
2
2 cos 2 cos
2
2
= = =
The total energy is given by:
2 2 2 (cos2 sin2 ) 2 2 2 x y a m m a E = E + E = ma ω ω t + ω t = ma ω
Noting that ( ) 2 2 1 ω = ω −ω m , we have:
[1 cos( ) ]
2
2 sin ( )
[1 cos( ) ]
2
2 cos ( )
ω ω ω ω ω
2 1 2 1
2 2 2
2 1 2 1
2 2 2
E ma t E t
y a
which show that the constant energy E is completed exchanged between the two
pendulums at the beat frequency ( ) 2 1 ω −ω .
4.7
By adding up the two equations of motion, we have:
( )( ) 1 2 1 2 m &x&+m &y& = − m x +m y g l
By multiplying the equation by 1 ( ) 1 2 m +m on both sides, we have:
m x +
m y
1 2
l
1 2
m x m y
&&+ &&
1 2
1 2
m m
g
m m
+
= −
+
m x +
m y
l
m &x& +
m &y&
i.e. 1 2 1 2
=
0
1 2
1 2
+
+
+
m m
g
m m
2 0
1 X&& +ω X = (4.7.1)
where,
X m x +
m y
= and 2 = g l
1 2
m +
m
1 2
1 ω

On the other hand, the equations of motion can be written as:
( )
x s
l
= − − −
1
x y
m
( )
y s
l
2
&&
y g
x y
m
x g
= − + −
&&
By subtracting the above equations, we have:
⎛
x y g x y s
s
l
⎟⎠
− ⎟ ( ) ( )
1 2
x y
m
m
⎞
⎜ ⎜⎝
&&− && = − − − +
⎤
⎡
⎞
⎛
&x& &y& g
i.e. 1 1 ( ) 0
1 2
= − ⎥⎦
⎢⎣
⎟ ⎟⎠
⎜ ⎜⎝
− + + + x y
m m
s
l
2 0
2 Y&&+ω Y = (4.7.2)
where,
⎞
⎛
ω g
y x Y − = and ⎟ ⎟⎠
⎜ ⎜⎝
= + +
1 2
2
2
1 1
m m
s
l
2 1 ω
ω
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 and 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 t
cos
ω
ω
0 1
cos
=
Y Y t
0 2
=
i.e.
X t
m x +
m y
1 2
m m
cos
ω
0 1
x y Y t
0 2
1 2
cos
ω
− =
=
+
By substitution of initial conditions: t = 0 , x = A and y = 0 into the above
equations, we have:
X =
m M A
=
Y A
0
0 1 ( )

where, 1 2 M = m +m
so the equations of motion in original coordinates x , y are given by:
[ cos( ω ω ) cos( ω ω
) ]
= − + +
a m a m
1 2
[ (cos ω cos ω sin ω sin ω ) (cos ω cos ω sin ω sin ω
)]
= + + −
m a m a m a m a
1 2
m m t t m m t t
x A
A
A
[( )cos ω cos ω ( )sin ω sin ω
]
= + + −
m a m a
1 2 1 2
A t t A
cos cos ( )sin sin
ω ω ω ω
= + −
E 1
m x s x m x mg
x m x m ω
x
l
= & + = & + = &
+
x x a
E 1
m y s y m y mg
= + = + = +
m
A t
M
m x +
m y
1 2
m m
x y A t
2
1
1
1 2
cos
cos
ω
ω
− =
=
+
The solutions to the above equations are given by:
m t m t
( cos ω cos ω
)
= +
(cos cos )
1 2
x A
y A m
1
1 1 2 2
t t
M
M
ω ω
= −
Noting that a m ω =ω −ω 1 and a m ω =ω +ω 2 , where ( ) 2 2 1 ω = ω −ω m and
( ) 2 1 2 ω = ω +ω a , the above equations can be rearranged as:
m m t t
M
M
m t t t t m t t t t
M
m t m t
M
m a 1 2
m a
and
[cos( ) cos( ) ]
= − − +
ω ω ω ω
a m a m
t t
y A m
A m
2 sin sin
M
t t
M
ω ω
m a
1
1
=
4.9
From the analysis in Problem 4.6, we know, at weak coupling conditions, t m cosω
and t m sinω are constants over one cycle, and the relation: g l a ω ≈ , so the energy
of the mass 1 m , x E , and the energy of the mass 2 m , y E , are the sums of their
separate kinetic and potential energies, i.e.:
2 2
ω
1
2
& & &
1
1
1
2 2
2
1
1
2 2
21
2 2
1
2
1
2 2
1
2 2
1
1
2
1
2
2
1
2
2
2
1
2
1
2
2
1
2
2
2
y m y m y
l
y y a

By substitution of the expressions of x and y in terms of t acosω and t a ω
E 1
m A t t A
= ⎡− + −
ω ω ω ω ω ω
x a m a a m a
m A t t A
⎡ + −
ω ω ω ω ω
a m a m a
m A
2
= + + −
ω ω ω
a m m
m A
2
= + + −
ω ω ω
a m m
m A
1
1
1
1
E
ω ω
= + + −
E m A m
2 sin cos 1
1
+ ⎡ ⎥⎦
ω ω ω ω ω ω
= ⎡
y a m a a m a
m m A
2
2 ω sin ω (cos ω sin ω
)
= +
a m a a
m m A
2
2 sin
ω ω
a m
m A m m
1
⎞
⎛
2 2 1 2
⎞
ω ω
= ⎛
E m m
1 2
= ⎛
m(x y) mg ( ) ( ) 2 ( ) cosω 0 &&− && + − + & − & + − =
sin
given by Problem 4.8 into the above equations, we have:
[ ]
[ ]
[ 2 cos2
]
2
[m m m m t]
M
m m m m t
M
m m m m t t
M
m m t m m t
M
m m t t
M
m m t t
M
a m
2 cos( )
2
2 (cos sin )
2
( ) cos ( ) sin
2
cos cos ( )sin sin
2
cos sin ( )sin cos
2
1 2 2 1
2
2
2
2 1
1 2
2
2
2
2 1
2
1
2 2
1 2
2
2
2
2 1
2
1
2 2
1 2
2 2
2 1 2
2
1
2
1 2
2
1
2
1 1 2
ω ω
= + +
⎤
⎥⎦
⎢⎣
⎤
+ ⎥⎦
⎢⎣
and
t t m A m
[ ]
2 1 cos2
[ ]
2
⎞
⎟⎠
[ t]
M
t
M
t
M
t t t
M
t t
M
M
a m
2 1 cos( )
2
2 sin sin
2
2
2 2 1
1
2
2
2
2
2
1
2 2 2
2
2
2
2
1
2
2 1
1
2
1
2
ω ω
− − ⎟⎠
⎜⎝
− ⎟⎠
⎜⎝
⎜⎝
=
⎤
⎥⎦
⎢⎣
⎤
⎢⎣
where,
E 1 m A a ω
2 2
2 1
=
4.10
Add up the two equations and we have:
m(&&+ x && y) + mg ( x + y ) + r ( x & + y & ) =
F cosω t
0 l
mX rX mg cosω 0 && + & + = (4.10.1)
i.e. X F t
l
Subtract the two equations and we have:
x y r x y s x y F t
l

mY rY mg 2 cosω 0 = ⎟⎠
&&+ & + ⎛ + (4.10.2)
i.e. s Y F t
l
⎞
⎜⎝
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 F cosωt 0 .
By neglecting the effect of r , equation of (4.10.1) and (4.10.2) become:
mX mg
X F cos
t
l
⎞
s Y F t
+ ≈
&&
mY mg
+ ⎛ +
l
ω
ω
2 cos
0
0
≈ ⎟⎠
⎜⎝
&&
Suppose the above equations have solutions: X X cosωt 0 = and Y Y cosωt 0 = , by
substitution of the solutions to the above equations, we have:
cos cos
⎞
⎛− +
ω ω ω
⎞
s Y t F t
m mg
m mg
⎛− + +
l
X t F t
l
2 cos cos
ω ω ω
0 0
2
0 0
2
≈ ⎟⎠
⎜⎝
≈ ⎟⎠
⎜⎝
These equations satisfy any t if
0 0
m mg
⎛− +
ω
m mg
⎞
⎛− 2
+ +
0 0
2
⎞
2s Y F
l
X F
l
≈ ⎟⎠
⎜⎝
≈ ⎟⎠
⎜⎝
ω
i.e.
X F
m g l
( )
2
Y F
0
( 2 )
0
2
0
0
ω
ω
+ −
≈
−
≈
m g l s m
so the expressions of X and Y are given by:
t
X x y F
2
0
( )
Y x y F
cos
0
m g l s m
t
m g l
ω
ω
2
ω
ω
cos
( + 2 −
)
= − ≈
−
= + ≈
By solving the above equations, the expressions of x and y are given by:

⎡
⎡
F
F
y
2
−
2
ω ω
⎤
⎤
⎥⎦
⎡
⎡
⎢⎣
−
−
−
x F
y ≈
F
⎥⎦
⎢⎣
−
+
−
≈
2 2
2
2 2
1
0
2 2
2
2 2
1
0
cos 1 1
2
cos 1 1
2
ω ω ω ω
ω
ω ω ω ω
ω
t
m
t
m
where,
2 = g
l
1 ω
and
s
2 g 2
2 ω = +
m
l
The ratio of y x is given by:
ω −
ω
2 2
1
2
2
2
1
2
2
1 1
ω ω ω ω
2 2
2
2 2
1
2 2
2
2 2
1
2 2
2
2 2
1
0
2 2
2
2 2
1
0
1 1 2
cos 1 1
2
cos 1 1
2
ω ω ω
ω ω ω ω
ω ω ω ω
ω
ω ω ω ω
ω
+ −
=
−
−
−
−
+
= −
⎤
⎤
⎥⎦
⎢⎣
−
−
−
⎥⎦
⎢⎣
−
+
−
≈
t
m
t
m
x
y
2
2 ω +ω
0 ω
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 1 ω
ω < or 2 ω
ω > , i.e. outside frequency
range 2 1 ω −ω 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:
Mx + kx = F cosωt +T 0 && (4.11.1)
m&y& = −T (4.11.2)
s( y − x) = T (4.11.3)
2
1
2
2
2
1
ω ω
+
x
1 ω
1
-1
2
2
1
2 ω

Eliminating T , we have:
cos ( ) 0 M&x&+ kx = F ωt + s y − x
so for x = 0 at all times, we have
cos 0 0 F ωt + sy =
that is
y = − F0 cosω
t
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 F cosωt 0 = −
4.12
Noting the relation: V = q C , the voltage equations can be written as:
L dI
a
L dI
dt
q
− =
1 2
q
C
q
q
C
dt
C
C
b
− =
2 3
so we have:
& & &&
− =
q q LCI
& & &&
a
b
− =
1 2
q q LCI
2 3
i.e.
& & &&
− =
q q LCI
& & &&
a
b
− =
1 2
q q LCI
2 3
By substitution of a q = −I 1 & , a b q = I − I 2 & and b q = I 3 & into the above equations, we
have:
I I I LCI
− − + =
a a b a
I I I LCI
&&
a b b b
&&
− − =
i.e.
&&
LCI I I
+ 2 − =
0
− + =
a a b
&&
LCI I I
2 0
b a b

By adding up and subtracting the above equations, we have:
&& &&
LC I I I I
( + ) + + =
0
a b a b
&& &&
LC I I I I
( − ) + 3( − ) =
0
a b a b
Supposing the solutions to the above normal modes equations are given by:
I I A t a b + = cosω
I I B t a b − = cosω
so we have:
2
A LC A t
ω ω
( ) cos 0
− ω + ω
=
B 2
LC B t
( − + 3 ) cos =
0
which are true for all t when
ω 2 = 1 and B = 0
LC
or
ω 2 = 3 and A = 0
LC
which show that the normal modes of oscillation are given by:
a b I = I at
2 1
1 ω =
LC
and
a b I = −I at
2 3
2 ω =
LC
4.13
From the given equations, we have the relation between 1 I and 2 I given by:
1
I i M
ω
+
2 I
Z i L
2
s ω
=
so:
1
⎛
E i L I i MI i L M
p p 2
⎟⎠
⎟ 2
1 2 I
Z i L
s
⎞
⎜ ⎜⎝
+
= − = +
ω
ω
ω ω ω
i.e.
p Z i L
s
i L M
E
I
ω
ω
ω
+
= +
2
2 2
1
which shows that 1 E I , the impedance of the whole system seen by the generator, is
the sum of the primary impedance, p iωL , and a ‘reflected impedance’ from the

secondary circuit of ω2M2 Zs , where s s Z = Z + iωL 2 .
4.14
Problem 4.13 shows the impedance seen by the generator Z is given by:
i ω L Z − ω L L +
ω
M
Z i L
2 1 1 1 1 1
p Z i L
s
Z i L M
ω
ω
ω
+
= +
2
2 2
Noting that p s M = L L and 2 2
p s p s L L = n n , the impedance can be written as:
s
i L Z
p
i ω L Z − ω M +
ω
M
s
p
s
p p s
Z i L
Z i L
Z i L
Z
ω
ω
ω
ω
+
=
+
=
+
=
2
2
2
2 2 2 2
2
2
2 2 2
2
so we have:
2
s == + = +
p p
2 2
Z i L n
L
2
ω
p 2
p p
Z
n
i L Z i L L
Z
s
s
+
=
ω ω ω
which shows the impedance Z is equivalent to the primary impedance iωL
p connected in parallel with an impedance (n n )2Z .
p s 2
4.15
Suppose a generator with the internal impedance of 1 Z is connected with a load with
an impedance of 2 Z via an ideal transformer with a primary inductance of p L and
the ratio of the number of primary and secondary transformer coil turns given by
p s n n , 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
L p p
Z
n
2 2
Z i L n
s
= +
ω
At the maximum output power: 1 Z Z L = , i.e.:
1 1 1 1
1
= + =
ω
2
L p p
2 2
Z
Z i L n
Z
n
s
which is the relation used for matching a load to a generator.

4.16
From the second equation, we have:
I E
I E ≤ =
M R R
2 i M R R
2 2
⎡
− = ⎥⎦⎤
2 1 2
π
= ⎡ −
2
1 (2 2)
ω ω − = ⎥⎦
⎡
+ = ⎥⎦
π
2 1 cos 3 2 ω 1 2
ω
= ⎡ −
2
1 (2 2)
ω ω + = ⎥⎦
2
I Z
2
I
1 Z
M
= −
By substitution into the first equation, we have:
I Z I E
Z Z
− 1 2
+ = 2 2
Z
M
M
2 I
i.e. 1
Z Z Z
M ⎟ ⎟⎠
1 2
Z
I E
M
⎞
⎛
⎜ ⎜⎝
−
=
Noting that Z i M M = ω and 2 I has the maximum value when 0 1 2 X = X = , i.e.
1 1 Z = R and 2 2 Z = R , we have:
1
1 2
1
M R R
1 2
1
1 2
1
1 2
I
R R
M
I E
M
I E
i M
+
=
−
=
ω
ω ω
ω
ω
ω
E , when
which shows 2 I has the maximum value of 1
1 2 2
I
R R
M R R
ω = 1 2 , i.e.
M
ω
1 2 ωM = R R
4.17
By substitution of j = 1 and n = 3 into equation (4.15), we have:
2
0
2
0
2
0
2 1 cos ω ω
2
4
⎤
⎢⎣
⎢⎣
2 1 cos 2 ω
2
0
π
= ⎡ −
2
2
2
1 0
ω ω = ⎥⎦
4
⎤
⎢⎣
2
0
2
0
2
0
2
4
⎤
⎢⎣
⎤
⎢⎣

In equation (4.14), we have 0 4 0 A = A = when n = 3 , and noting that 2 = T ma
2 1 0
⎛
2
1 ω = (2 − 2)ω ,` 2
2
1 ω = (2 − 2)ω into equation (4.17.1), we have:
2
1 ω = (2 − 2)ω into equation (4.17.3), we have:
2
1 ω = (2 − 2)ω , the relative displacements are given by:
2
2 ω = 2ω into equation (4.17.1), we have:
0 ω
,
equation (4.14) gives:
⎛
when r 1: ω
2
= 2 0 2 1 2
− = 0 0
⎟⎠
⎟ ⎞
⎜ ⎜⎝
− A + − A A
ω
⎛
2
⎞
− A A
ω (4.17.1)
i.e. 2 0 2 1 2
0
= − ⎟ ⎟⎠
⎜ ⎜⎝
ω
2
⎛
ω (4.17.2)
1 = − ⎟ ⎟⎠
when r = 2 : 2 0 2 2 3
0
⎞
⎜ ⎜⎝
− A + − A A
ω
2
⎛
ω
2 = − ⎟ ⎟⎠
when r = 3: 2 0 2 3 4
0
⎞
⎜ ⎜⎝
− A + − A A
ω
2
⎛
ω (4.17.3)
2 = ⎟ ⎟⎠
i.e. 2 0 2 3
0
⎞
⎜ ⎜⎝
− A + − A
ω
Write the above equations in matrix format, we have:
0
− −
1 2 1
− − −
0 1 2
A
1
A
2
3
2
0
2
2
0
2
2
0
2
=
⎞
⎟ ⎟ ⎟
⎠
⎞
⎛
⎟ ⎟ ⎟
⎜ ⎜ ⎜
⎝
⎠
⎜ ⎜ ⎜
⎝
− −
A
ω ω
ω ω
ω ω
which has non zero solutions provided the determinant of the matrix is zero, i.e.:
(2 −ω 2 ω ) − 2(2 −ω ω 2 ) =
0
0
2 3 2
0
The solutions to the above equations are given by:
2
0
ω 2
= 2ω , and 2
2 0
ω 2
= (2 + 2)ω
1 0
4.18
By substitution of 2
0
2 0 1 2 A − A = i.e. : 1: 2 1 2 A A =
0
2 0 2 3 − A + A = i.e. : 2 :1 2 3 A A =
Hence, when 2
0
: : 1: 2 :1 1 2 3 A A A =
0

2
2 ω = 2ω into equation (4.17.2), we have:
2
2 ω = 2ω , the relative displacements are given by:
2
2 ω = (2 + 2)ω into equation (4.17.1), we have:
2
1 ω = (2 + 2)ω into equation (4.17.3), we have:
2
1 ω = (2 + 2)ω , the relative displacements are given by:
ω 2 = (2 − 2)ω 2
2
0
A2 = 0
0
0 1 3 − A + A = i.e. : 1: 1 1 3 A A = −
Hence, when 2
0
: : 1: 0 : 1 1 2 3 A A A = −
0
2 0 1 2 − A − A = i.e. : 1: 2 1 2 A A = −
0
2 0 2 3 − A − A = i.e. : 2 :1 2 3 A A = −
Hence, when 2
0
: : 1: 2 :1 1 2 3 A A A = −
The relative displacements of the three masses at different normal frequencies are
shown below:
ω 2 = 2ω 2
0
ω 2 = (2 + 2)ω
0
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:

− = −
− = − − + −
mx s y x
( )
= −
&&
My s y x s z y
( ) ( )
= − − + −
&&
mz s z y
( )
= − −
&&
If the system has a normal frequency of ω , and the displacements of the three masses
can by written as:
i t
ω
i t
x η
e
1
=
y e
ω
i t
η
2
=
z e
ω
η
3
=
By substitution of the expressions of displacements into the above equations of
motion, we have:
i t i t
ω ω
m e s e
( )
ω η η η
i t i t i t
ω ω ω
M e s e s e
( ) ( )
ω η η η η η
i t i t
ω ω
m e s e
( )
ω η η η
3 3 2
2
2 2 1 3 2
2
1 2 1
2
− = − −
i.e.
s m s e
[( − ω ) η − η
] =
0
s s M s e
[ − η + (2 − ω ) η − η
] =
0
i t
[ ( ) ] 0
3
2
2
2 3
2
1
1 2
2
− + − =
i t
i t
s s m e
ω
ω
ω
η ω η
which is true for all t if
s m s
( − ω ) η − η
=
0
s s M s
(2 ) 0
− η + − ω η − η
=
s s m
( ) 0
3
2
2
2 3
2
1
1 2
2
− η + − ω η
=
The matrix format of these equations is given by:
0
s − m −
s
s s M s
− − −
0
2
0
η
1
η
2
3
2
2
2
=
⎞
⎟ ⎟ ⎟
⎠
⎞
⎛
⎟ ⎟ ⎟
⎜ ⎜ ⎜
⎝
⎠
⎛
⎜ ⎜ ⎜
⎝
− −
η
ω
ω
ω
s s m
which has non zero solutions if and only if the determinant of the matrix is zero, i.e.:
(s −mω2 )2 (2s −Mω 2 ) − 2s2 (s −mω2 ) = 0
i.e. (s −mω2 )[(s −mω 2 )(2s −Mω 2 ) − 2s2 ] = 0
i.e. (s −mω2 )[mMω 4 − s(M + 2m)ω 2 ] = 0
i.e. ω2 (s −mω2 )[mMω 2 − s(M + 2m)] = 0
The solutions to the above equation, i.e. the frequencies of the normal modes, are
given by:
2 0
1 ω = ,
2 = s
m
2 ω
and
2 s(M +
2m)
3
mM
ω =

At the normal mode of ω2 = 0 , all the atoms are stationary, 1 2 3 η =η =η , i.e. all the
masses has the same displacement;
At the normal mode of
A C rj r
ω2 = s , 0 2 η = and 1 3 η = −η , 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;
At the normal mode of
2 s(M + 2m)
ω = , : : M : 2m:M 1 2 3 η η η = − , 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 j ω
the amplitude of the rth mass is
1
sin
+
=
n
π
where C is a constant. For j = n 2 adjacent masses have a π 2
phase difference, so the ratios: : : 1: 0 :1 1 1 = − r− r r+ A A A , with the rth masses
stationary and the amplitude r−1 A anti-phase with respect to r+1 A , so that:
2j ω
As n 2→n , A begins to move, the coupling between masses tightens and when
r 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 2 = 4 T ma is reached. The end points are fixed and this restricts the
j motion of the masses near the end points at all frequencies except the lowest.
4.21
By expansion of the expression of , we have:
→0 r A
j →n
r+1 A
r−1 A
r A
j = n 2

2 1 cos 2 4 6
2 ⎛
j n j n j n
( 1)
( 1)
T
2 ( 1)
j
T
= − L
⎤
⎥⎦
⎡
⎢⎣
−
+
+
+
−
+
⎞
= ⎟⎠
⎜⎝
+
6!
4!
2!
1
ma
n
ma
j
π π π π
ω
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:
2 2
2
⎡ +
j n T
T
2 ( 1)
⎛
⎤
j
2! 1
⎞
⎟⎠
⎜⎝
+
= ⎥⎦
⎢⎣
=
n
ma
ma
j
π π
ω
i.e.
T
ma
j
j +1
n
=
π
ω
ω T
ρ
j
π
l
j =
where, ρ = m a and l = (n +1)a
4.22
From the first equation, we have:
LI && qr & − qr
&
C
r
= −
−
1
1
By substitution of r r r q = I − I & −1 and −1 −2 −1 = − r r r q& I I into the above equation, we
have:
LI Ir Ir Ir
&& 2 (4.22.1)
C
r
− +
= − −
−
2 1
1
If, in the normal mode, the currents oscillate at a frequency ω , we may write the
displacements as:
= i t
, i t
−2 −2 r r I A e ω
−1 −1 = and i t
r r I A e ω
r rI = A e ω
Using these values of I in equation (4.22.1) gives:
ω 2 LA e i ω t A − 2
A +
A r r r i ω
t
r e
− = − −
C
−
2 1
1
or
2
− A + (2 − LCω ) A − A = 0 (4.22.2)
r− 2 r− 1
r By comparison of equation (4.22.2) with equation (4.14) in text book, we may find
the expression of r I is the same as that of r y in the case of mass-loaded string, i.e.
I A e i ω t D rjπ i ω
t
r r e
n
1
sin
+
= =
Where D is constant, and the frequency ω is given by:

⎞
⎟⎠
⎛
⎜⎝
j
+
= −
1
2 1 1 cos
n
j LC
π
ω
where, j = 1,2,3...n
4.23
By substitution of y into 2
2
t
y
∂
∂
, we have:
y = − +
∂ ω ω ω
2 ( )
2
2
2
2
(ei teikx ) ei t kx
∂
=
t ∂
t
∂
2
x
y
∂
∂
By substitution of y into 2
, we have:
y = − +
∂ ω ω
2 ( )
2
2
2
2
(ei teikx ) k ei t kx
∂
=
x ∂
x
∂
If ω = ck , we have:
c y
∂ c k ei t+kx
( 2 2 2 ) ( ) 0
2
2
2
2
2
= − + =
∂
∂
−
∂
x
t
y ω ω
i.e.
c y
2
2
2
2
2
x
t
y
∂
∂
=
∂
∂

5.1
Write u = ct + x , and try 2
2
1
∂
2
x
y
∂
∂
with ( ) 2 y = f ct + x , we have:
f u
2 f ( u
)
∂ ( ) 2 , and 2
u
y
x
∂
∂
=
∂
2
2
2
u
x
y
∂
∂
=
∂
∂
Try 2
2
t
y
c ∂
with ( ) 2 y = f ct + x , we have:
c f u
2 ( )
c f u
∂ ( ) 2 , and 2
u
y
t
∂
∂
=
∂
2
2
2
2
u
t
y
∂
∂
=
∂
∂
so:
f u
c f u
y
1 1 ( ) ( )
2
2
2
2
2
2
2
2 2
2
2
∂
c ∂
u
u
t c
∂
=
∂
∂
=
∂
Therefore:
2
2
2 y
1
2 2
t
y
x c
∂
∂
=
∂
∂
5.2
If ( ) 1 y = f ct − x , the expression for y at a time t + Δt and a position x + Δx , where
Δt = Δx c , is given by:
y f c t t x x
[ ( ) ( )]
= + Δ − + Δ +Δ +Δ
, 1
f c t x c x x
[ ( ) ( )]
= + Δ − + Δ
1
f ct x x x
[ ]
= + Δ − − Δ
t x
t t x x
1
f [ ct x ]
y
= − =
1 ,
i.e. the wave profile remains unchanged.
If ( ) 2 y = f ct + x , the expression for y at a time t + Δt and a position x + Δx , where
Δt = −Δx c , is given by:

y
c y
x
y
∂
=
∂
t
∂
∂
y f c t t x x
[ ( ) ( )]
= + Δ + + Δ +Δ +Δ
, 1
f c t x c x x
[ ( ) ( )]
= − Δ + + Δ
1
f ct x x x
[ ]
= − Δ + + Δ
t x
t t x x
1
f [ ct x ]
y
= + =
1 ,
i.e. the wave profile also remains unchanged.
5.3
5.4
The pulse shape before reflection is given by the graph below:
The pulse shapes after of a length of Δl of the pulse being reflected are shown below:
(a) Δl = l 4
x
x
l
l
2
3
l
4
1
1 Z = ∞ 2 Z

(b) Δl = l 2
(c) Δl = 3l 4
(d) Δl = l
1 Z = ∞ 2 Z
3
l
4
1
l
2
1 Z = ∞ 2 Z
l
1 Z = ∞ 2 Z
5.5
The boundary condition i r t y + y = y gives:
Aei ωt−kx + B ei ωt+kx = A ei ωt−kx
( )
2
( )
1
( )
1
At x = 0 , this equation gives:
1 1 2 A + B = A (5.5.1)

∂
∂
The boundary condition ( ) t i r y y
Ma = −ikTA ei ωt−kx + ikTA ei ωt−kx − ikTB ei ωt+kx
A 2
= −
sin
tan
i
iq
−
= e i
θ θ
=
− + −
=
B
1 sin
Ma T +
= gives:
x
y T
x
∂
−
∂
( )
1
( )
1
( )
2
At x = 0 , t i r a = &y& = &y& + &y& , so the above equation becomes:
−ω MA = −iω T + ω − ω
A i T
c
2 B
2 2 1 1
c
A i T
c
i T − B ⎟⎠
= ⎛−ω M +
i T
c
i.e. 1 1 2 A
c
A i T
c
⎞
⎜⎝
Noting that T c = ρc , the above equation becomes:
( ) 1 1 2 iρcA − iρcB = −ωM + iρc A (5.5.2)
( )( ) 1 1 1 1 iρcA − iρcB = −ωM + iρc A + B
i.e.
iq
iq
B
−
=
1 1
A
+
1
where q =ωM 2ρc
By substitution of the above equation into (5.5.1), we have:
A iq =
A A
iq
1 1 +
1 2
−
i.e.
A
2
A +
iq
=
1
1
1
5.6
Writing q = tanθ , we have:
θ θ
cos
θ
1
= cos
θ θ
θ
e i
A iq i +
i
=
+
=
+
cos sin
1 tan
1
1
1
and
( 2)
1
cos sin
1 tan
1
θ θ π
θ θ
θ
+
−
=
+
+
i
i
i
iq
A
which show that 2 A lags 1 A by θ and that 1 B lags 1 A by (π 2 +θ ) for 0 <θ <π 2

The reflected energy coefficients are given by:
2
0
ω
∫ ∫
W = − T − ka t − kx a t −
kx dxdt
k a T π ω
k
t kx dxdt
∫ ∫
= −
k a T t kx dxdt
2 2 2
= ⋅ ⋅ ⋅
F T y ω
max sin( )
θ (θ π 2) 2 2θ
2
B
1 = sin e−i + = sin
A
1
and the transmitted energy coefficients are given by:
θ θ 2 2θ
2
A
2 = cos e−i = cos
A
1
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:
∂
y
x
∂
∂
ω 2
= − π ω
∫ ∫ ∂
π
0
1
2 0
dxdt
t
W T y k
By substitution of y = asin(ωt − kx) into the above equation, we have:
2
1
1 cos(2 2 )
2 1
2
2
2
2
sin ( )
2
[ sin( )][ sin( )]
2
2
2
0
1
0
2 2 2
2
0
1
0
2
2 2 2
1
0
ka T
k
k a T
k
k
ω
π
ω
π
ω
ω
π
ω
ω
π
ω
ω ω ω
π
π ω
π ω
=
− −
=
∫ ∫
Noting that k =ω c and T = ρc2 , the above equation becomes
2 2 2 2a2 c
c
W ω a ρ c ω ρ
= =
2 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 max F is given by:
A t kx TAk TA
x
c
T
⎛
∂
x
⎤
⎡ −
∂
ω = = ⎥⎦
⎢⎣
∂
⎞
= ⎟⎠
⎜⎝
∂
=
max max
i.e.

P ρ ω
π
= c A = T = × × × × =
c P
1 R
1 θ
1 θ
2 θ
0.3
0.3 max =
0.1 × 2 ×
5
= =
A
ω π π
F
T
c
Noting that ρc = T c , the rate of energy transfer along the string is given by:
0.3 (2 5) 0.1 3
[ ]
20
1
2
1
2
2
2 2 2 2
2 2
A W
c
π
π
ω
so the velocity of the wave c is given by:
2 2 3 20 1
2 2 2 2
30[ ]
×
π
= = ms
0.01 (2 5) 0.1
= −
× × ×
A
π π
ρω
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 1 2 Z Z boundary. We then have the total reflected
amplitude as:
2
R tTR R 2 R 4
R tTR
1
(1 )
− ′
R
′
+ ′ + ′ + ′ +L = +
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 1 Z and is reflected as TR′ in
2 Z .
R is reflected in 1 Z as (R)R = R2 back along the incident direction and is refracted as TR
in the TR′ direction in 2 Z .
We therefore have tT + R2 = 1 in 1 Z , i.e. tT = 1− R2 and T(R + R′) = 0 in 2 Z giving
R = −R′ , ∴tT = 1− R2 = 1− R′2 giving the total reflected amplitude in 1 Z as R + R′ = 0
with R = −R′ .
T
R2
tT R
1 θ
1 θ
TR′ θ
θ
2 2 T TR
Fig Q.5.9(a) Fig Q.5.9(b)
1 Z
2 Z
1 Z

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 coat Z should have a relation to the impedance of air
air Z and the impedance of the lens lens Z given by:
2
2
y n n
∂
k n ω
= , we have:
Z = Z Z = 1
coat air lens n n
air lens
So the reflective index of the coating is given by:
= 1 = = 1.5 = 1.22 air lens
coat n n
coat
Z
n
and the thickness of the coating d should be a quarter of light wavelength in the coating, i.e.
1.12 10 [ ]
5.5 ×
10
4 1.22
4
7
7
m
λ
n
d
coat
−
−
= ×
×
= =
5.11
By substitution of equation (5.10) into
y
∂
∂
x
, we have:
y n
A t B t t
ω
= ( cos + sin )cos
n ω
c
∂
x c
ω ω
n n n n
∂
so:
y
A t B t t
c c
x c
n n n n
n
2
2
2
2
( cos sin )sin
ω ω
ω ω
ω
= − + = −
∂
Noting that
c
2
2
+ = − + =
∂
∂ y
0 2
2
2
2
2
c
y
c
k y
x
y n n ω ω
5.12
By substitution of the expression of max
( 2 ) n y into the integral, we have:

y dx A B x
( ) 1
∫ ∫
1
2 2
n c
l n , the above equation becomes:
ω = , i.e. sin 2 = sin 2 π = 0
y x t A t kx rA t kx
( , ) = cos( ω − ) + cos( ω
+
)
A t kx A t kx rA t kx rA t kx
cos cos sin sin cos cos sin sin
= + + −
ω ω ω ω
= + + −
2 2 2 2
A B x c dx
( ) 1 cos(2 )
⎞
⎟ ⎟⎠
∫
⎛
A B l c l
⎜ ⎜⎝
−
= +
1
= +
1
= + −
c
dx
c
n
n
n n n
l n
n n n
l n
n n n
l
n n
ω
ω
ρω
ω
ρω
ω
ρω ρω
sin 2
2
( )
4
2
2
( ) sin
2
2
2 2 2
0
2 2 2
0
0 max
Noting that
l
n
π
ω n
c
1 2 2 2
( )
( ) 1
4
2
2 2
0 max
n n n
l
n n ρω ∫ y dx = ρlω A + B
which gives the expected result.
5.13
Expand the expression of y(x,t) , we have:
A r t kx A r t kx
(1 )cos ω cos (1 )sin ω
sin
which is the superposition of standing waves.
5.14
The wave group has a modulation envelope of:
⎞
⎟⎠
A A ⎛ t Δ
= k x
⎜⎝
−
Δ
2 2
cos 0
ω
where 1 2 Δω =ω −ω is the frequency difference and 1 2 Δk = k − k is the wave number
difference. At a certain time t , the distance between two successive zeros of the modulation
envelope Δx satisfies:
Δ k x
2
Δ =π
Noting that k = 2π λ , for a small value of Δλ λ , we have: Δk ≈ (2π λ2 )Δλ , so the above
equation becomes:
π
π λ
Δ x 2 λ
2
Δ ≈
2
i.e.
λ
Δ
λ
λ
Δx ≈
which shows that the number of wavelengths λ contained between two successive zeros of the
modulating envelop is ≈ λ Δλ

5.15
The expression for group velocity is given by:
v c ka g
c ka
c ka
⎡
= + −
⎛
∂
⎞
⎛
∂
V v v r
2
= = − ⎛
ε e
r v
v d g = = ( ) = +
kv v k dv
dk
d
dk
ω
dk
By substitution of the expression of v into the above equation, we have:
sin( 2)
sin( 2)
sin( 2)
2
cos
c ka
sin( 2)
k d
⎤
ck ka ka a ka
( 4)cos( 2) ( 2)sin( 2)
c ka c ka
sin( 2)
2
2
cos
2
( 2)
2
2
2
2
2
c ka
ka
ka
ka
ka
ka
dk
ka
=
−
= +
⎥⎦
⎢⎣
= +
At long wavelengths, i.e. k →0, the limiting value of group velocity is the phase velocity c .
5.16
Noting that the group velocity of light in gas is given on page 131 as:
⎞
⎟ ⎟⎠
⎛
⎜ ⎜⎝
∂
λ r
r
ε
∂
= +
λ
ε
g V v
2
1
we have:
λ ε
ε
λ
2 2
⎞
v ⎡
A B D ⎞
∂
⎛ A + B −
D
= ⎛ + −
v ⎡
A B D B D
= ⎛ + −
v ⎡
A B D B D
= ⎛ + −
( 2 )
2 2
2
2
1
2
2
+ ⎛− − 2
⎟⎠
2
2
3
2
2
2
2
2
2
λ
λ
λ
λ
λ
⎞
λ
λ
λ
λ
λ
λ
λ λ
λ
λ
λ
λ
ε ε
λ
ε
ε
v A D
r r
r
r
g r
= −
⎤
⎥⎦
⎢⎣
⎞
⎟⎠
⎜⎝
⎞
⎜⎝
⎤
⎥⎦
⎢⎣
⎟⎠
⎜⎝+ ⎛− − ⎟⎠
⎞
⎜⎝
⎤
⎥⎦
⎢⎣
⎞
⎟⎠
⎜⎝
∂
+ ⎟⎠
⎜⎝
⎟⎠
⎜⎝
∂
+ = ⎟ ⎟⎠
⎜ ⎜⎝
∂
= +
5.17
The relation
2
2
⎞
ω
1 ⎟⎠
⎜⎝
ω
c
gives:
2 2
v e
c ω ω
2 2
2
ω
= −
By substitution of v =ω k , the above equation becomes:

e n
= = × × rad s
ω
v
2 2 c2k 2 e ω =ω + (5.17.1)
As e ω
ω → , we have:
2
⎞
= − ⎛
ωe
1 1
2
2
< ⎟⎠
⎜⎝
ω
c
v
i.e. v > c , which means the phase velocity exceeds that of light c .
From equation (5.17.1), we have:
d( 2 ) d( 2 c2k 2 ) e ω = ω +
i.e. 2ωdω = 2kc2dk
which shows the group velocity g v is given by:
v = d = = = c
<
g c c
v
c k c
v
dk
2
2
ω
ω
i.e. the group velocity is always less than c .
5.18
From equation (5.17.1), we know that only electromagnetic waves of e ωω > can propagate
through the electron plasma media.
For an electron number density ~ 1020 e n , the electron plasma frequency is given by:
20
1.6 10 − 10 = × 11 ⋅
1
5.65 10 [ ]
31 12
9.1 10 8.8 10
19
0
−
− −
× × ×
m
e
e
e ε
Now consider the wavelength of the wave in the media given by:
2 2 2 2 3 10 3
3 10 [ ]
11
× ×
5.65 10
8
v v c m
f
π
π
e e
= × −
×
π
= = < < =
π
ω
ω
ω
λ
which shows the wavelength has an upper limit of 3×10−3m.
5.19
The dispersion relation ω 2 c2 = k 2 + m2c2 h2 gives
d(ω 2 c2 ) = d(k 2 + m2c2 h2 )
2 ω
ω =
2
2 i.e. d kdk
c

d
k
i.e. c2
⎞
⎛
T
2 1 1 2 15 1 13 1
⎞
⎛
T
2 1 1 2 15 1 13 1
dk
=
ω ω
Noting that the group velocity is dω dk and the particle (phase) velocity is ω k , the above
equation shows their product is c2 .
5.20
The series in the problem is that at the bottom of page 132. The frequency components can be
expressed as:
R na t ω
t
sin( ω
2)
Δ ⋅
t
ω
cos
2
Δ ⋅
=
which is a symmetric function to the average frequency 0 ω
. It shows that at
Δt = 2 , R = 0 ,
π
Δ
ω
∴Δt ⋅Δω = 2π
In k space, we may write the series as:
y(k) a cos k x a cos(k k)x a cos[k (n 1) k]x 1 1 1 = + +δ +L+ + − δ
As an analogy to the above analysis, we may replace ω by k and t by x , and R is zero at
k
x
Δ
Δ =
2π
, i.e. ΔkΔx = 2π
5.21
The frequency of infrared absorption of NaCl is given by:
3.608 10 [ ]
1
35 1.66 10
23 1.66 10
27 27
−
⎞
− − ⋅ × = ⎟⎠
⎛
⎜⎝
× ×
+
× ×
× × = ⎟ ⎟⎠
⎜ ⎜⎝
= + rad s
a m m
Na Cl
ω
The corresponding wavelength is given by:
52[ ]
2 2 × 3 ×
10
λ ≈
13
3.608 10
8
c π
μm
π
ω
×
= =
which is close to the experimental value: 61μm
The frequency of infrared absorption of KCl is given by:
3.13 10 [ ]
1
35 1.66 10
39 1.66 10
27 27
−
⎞
− − ⋅ × = ⎟⎠
⎛
⎜⎝
× ×
+
× ×
× × = ⎟ ⎟⎠
⎜ ⎜⎝
= + rad s
a m m
K Cl
ω
The corresponding wavelength is given by:
60[ ]
2 2 × 3 ×
10
λ ≈
13
3.13 10
8
c π
μm
π
ω
×
= =
which is close to the experimental value: 71μm

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:
′ = ν
ν can be written in the format of wavelength as:
c
−
ν ν
c u
= 1
After the source passes by the observer, the source has a velocity of − u , the frequency noted by
the observer is given by:
c
+
ν ν
c u
= 2
So the change of frequency noted by the observer is given by:
2
cu
c
c
⎞
⎛
2 1 c2 u2
( )
c u
c u
−
= ⎟⎠
⎜⎝
+
−
−
Δ = − =
ν
ν ν ν ν
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:
c v
c v
−
−
ν ν
c u
c v u v
−
=
− − −
′′′ =
( )
5.24
The relation between wavelength λ and frequency ν of light is given by:
ν = c
λ
So the Doppler Effect
c
−
c u
2
c u
c c
( −
)
=
λ′ λ
c − u
i.e. λ λ
c
′ =
Noting that wavelength shift is towards red, i.e. λ′ > λ , so we have:
Δ = ′ − = − u
λ λ λ λ
c
8 11
u = − c Kms
3 × 10 ×
10 1
λ
i.e. 5[ ]
6 10
7
−
−
−
= −
×
= −
Δ
λ
which shows the earth and the star are separating at a velocity of 5Kms−1 .
5.25

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 ν ′ . Then, the Doppler
Effect gives:
′′ −
ν
ν ν
u = c c ms
−
T mNau ≈
v vc ⎞
: Observer is at rest with a moving source.
⎛
− ′
c
−
c u
ν ′ =ν
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:
c +
u
c u
c u
′ − − ν ′′ =ν ν ν
c
c u
c u u
−
=
′ + =
( )
− − − −
which gives:
3
× × = −
3 10 750[ ]
15 ×
10
2 3 10
2
8 1
9
× ×
=
Δ
+ Δ
=
′′ +
ν ν
ν ν
i.e. the aircraft is flying at a speed of 750m s
5.26
Problem 5.24 shows the Doppler Effect in the format of wavelength is given by:
c − u
λ λ
c
′ =
where u is the velocity of gas atom. So we have:
u
λ λ λ λ
c
Δ = ′ − =
i.e.
12
−
× × = ×
2 10 8 3 1
3 10 1 10 [ ]
×
6 10
7
−
−
×
=
Δ
λ
u = ′ − = c ms
λ
λ λ
The thermal energy of sodium gas is given by:
3
1 2 =
m u kT Na 2
2
where k = 1.38×10−23[JK−1] is Boltzmann’s constant, so the gas temperature is given by:
900[ ]
2 27 2
23 × 1.66 × 10 ×
1000
= = −
3 3 1.38 10
23
K
k
× ×
5.27
A point source radiates spherical waves equally in all directions.
⎟⎠
⎜⎝
′ =
c u

s′
θ
u
u′ = u cosθ
v c v ⎞
: Source at rest with a moving observer.
⎟⎠
⎛ − ′
⎜⎝
′′ =
c
v
s v′ = v cosθ ′
v c v ⎞
: Source and observer both moving.
⎟⎠
− ′
⎛
− ′
⎜⎝
′′′ =
c u
θ o′
v
o′
5.28
By substitution of equation (2) into (3) and eliminating x′ , we can find the expression of t′
given by:
2
⎡
⎞
⎛
⎤
⎡
k ⎞
xt k v c
⎛ −
′
⎤
k ⎞
x kv k c
⎛ −
′
k c
2 = ⎥⎦
⎡
− + c t
⎤
⎥⎦
x
′ = 1 k(x vt)
⎡ − −
′
⎢⎣
k
v
t
Now we can eliminate x′ and t′ by substituting the above equation and the equation (2) into
equation (1), i.e.
2
2
x
⎡ − −
′
x c t k x vt c
2 − 2 2 = 2 ( − )2 − k ( x vt
2
)⎥⎦
⎤
⎢⎣
k
v
i.e.
2
2
1 1 2 1 1 2 2 0
2
2 2
2
2
2
2
⎤
⎢⎣
− ⎟ ⎟⎠
⎜ ⎜⎝
− + ⎥⎦
⎢⎣
⎟⎠
⎜⎝
+ +
⎥ ⎥⎦
⎢ ⎢⎣
⎟⎠
⎜⎝
v
v k
v k
which is true for all x and t if and only if the coefficients of all terms are zeros, so we have:
2
2
k c
1 2 ⎛
1 ⎞
2
⎟⎠
⎜⎝
′
− = −
k
k
v
2
2
2
1
⎟⎠
c 2
⎟ ′ = kk c
v
v
⎞
⎛
⎜ ⎜⎝
−
α o′
s′
u
θ
u′ = u cosθ v′ = v cosθ

t k t v i.e. 2 2 1 2 1 (x x ) t t
∴Δ ′ = ⎡Δ − x − x
2
⎤
⎡
l x x k x x v t k x x v 2 1
∴ ′ = ′ − ′ = − − Δ = − − −
2 1 2 1 2 1 [( ) ( )] ( ) ( )
t k t v i.e. 1 2 x ≠ x
x t k t v
c
⎜⎝⎛ − = ′ 1 1 2 1 2 2 2 2 x
k 2 (c2 − v2 ) = c2
The solution to the above equations gives:
1
−β
1 2
k = k′ =
where, β = v c
5.29
Source at rest at 1 x in O frame gives signals at intervals measured by O as 2 1 Δt = t − t
where 2 t is later than 1 t . O′ moving with velocity v with respect to O measures these
intervals as:
t′ − t′ = Δt′ = k Δt − v Δ with Δx = 0
( ) 2 1 2 x
c
∴Δt′ = kΔt
( ) 2 1 l = x − x as seen by O, O′ sees it as ( ) [( ) ( )] 2 1 2 1 2 1 x′ − x′ = k x − x − v t − t .
Measuring l′ puts 2 1 t′ = t′ or Δt′ = 0
⎤
( ) 0 2 2 1 = ⎥⎦
⎢⎣
c
Δt = v − = −
c
x x x x
k
c
2 2 1
−
= ⎥⎦
⎢⎣
∴l′ = l k
5.30
Two events are simultaneous ( ) 1 2 t = t at 1 x and 2 x in O frame. They are not simultaneous
in O′ frame because:
⎞
⎟⎠
≠ ′ = ⎛ − ⎟⎠
⎜⎝
⎞
c
5.31
The order of cause followed by effect can never be reversed.
2 events 1 1 x ,t and 2 2 x ,t in O frame with 2 1 t > t i.e. 0 2 1 t − t > ( 2 t is later).

t k t v2 in O′ frame.
⎥⎦
t ′ − t ′ = k ⎡( t − t ) − v ( x − x
) 2 1 2 1 2 2 1 i.e. Δx
⎤
⎥⎦
⎢⎣
c
⎤
Δ ′ = ⎡Δ − Δx
⎢⎣
c
t v ⎛ Δ
x
⎞
where
t′ Δ real requires k real that is c v < , t′ Δ is ve + if ⎟⎠
⎜⎝
Δ >
c
c
v
c
is + ve
but < 1 and
c
is shortest possible time for signal to traverse Δx .
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:
3
1 2 =
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:
c P PV γ γ γ
NkT
M
RT
2 = = = =
M
M
γ
ρ
where V is the molar volume, M is the molar mass and N is Avogadro’s number, so:
2 = γ =α ≈ 5
Mc NkT kT kT
3
6.2
The intensity of sound wave can be written as:
= 2 ρ
I P c 0
where P is acoustic pressure, 0 ρ
is air density, and c is sound velocity, so we have:
10 1.29 330 65[ ] 0 P = Iρ c = × × ≈ 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 cω η
2 2
2 0
where η is the displacement amplitude of an air molecule, so we have:

− − −
2 10 1
water
p
I = × −
6.9 10 [ ]
2 ×
10
1.29 330
2 1
2 500
2
1 5
0
m
c
×
×
×
= =
πν ρ π
η
6.4
The expression of displacement amplitude is given by Problem 6.3, i.e.:
10 [ ]
10 2
2 × 10 × 10 ×
10
1.29 330
2 500
2
1 10
0
0
10
m
c
I −
≈
×
×
×
=
×
=
πν ρ π
η
6.5
The audio output is the product of sound intensity and the cross section area of the room, i.e.:
100 100 10 2 3 3 10[ ]
0 P = IA = I A = × − × × ≈ 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:
60
I c
( ) 6
1.45 10
water
c
ρ
ρ
0 ≈
400
( )
water
( )
( )
0
0
0
×
= = =
air
air
air
c
I c
p
ρ
ρ
And at the same pressure amplitudes, we have:
4
c
( ) ≈ × −
6
ρ
0
400
0 3 10
1.45 10
( )
×
= air
=
water
water
air
c
I
I
ρ
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:
x
F Y
∂
∂
=
η
, where Y is young’s
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
l
L
∂η
x
=
∂
F Y ⎟⎠
= ⎛
and l
L
⎞
⎜⎝
If l = x in the stretched spring, we have:
F sx Y ⎟⎠
= = ⎛ ⎞
and Y = sL .
x
L
⎜⎝
If the spring has mass m per unit length, the equation of motion of a section of length dx is
given by:

∂ η ∂
η
∂
=
m 2
dx
x
dx F
t
dx Y
x
2
2
2
∂
=
∂
∂
2
sL
Y
∂ η ∂
η η
or 2
2
2
2
2
t ∂
m x
m x
∂
=
∂
=
∂
a wave equation with a phase velocity
sL
m
6.8
At x = 0,
η = Bsin kx sinωt
At x = L ,
∂ η η
x
sL
t
M
∂
∂
= −
∂
2
2
i.e. −Mω 2 sin kL = −sLk cos kL
(which for k =ω v , ρ = m L and v = sL ρ from problem 7 when l << L )
becomes:
m
tan (6.8.1)
M
2
L = = L
=
ω ω ρ
M
sL
Mv
L
v
v
2
For M >> m, v >>ωL and writing ωL v =θ where θ is small, we have:
tanθ =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 2m s (1+ω 2m 3s +...) = m M
or
ω 2 (1+ω 2m 3s) = s M (6.8.2)
Using ω2 = s M as a second approximation in the bracket of eq. 6.8.2, we have:

⎛
r
I
t
I
s
M
m = ⎟⎠
M
⎞
ω 2 ⎛ 1 +
1
⎜⎝
3
i.e.
s
1
M m
3
2
+
ω =
6.9
The Poissons ratio σ = 0.25 gives:
0.25
λ
2( )
=
λ + μ
i.e. λ = μ
So the ratio of the longitudinal wave velocity to the transverse wave velocity is given by:
2 +
=
2 = 3
λ μ
+
=
μ μ
μ
μ
v
l
v
t
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:
2 = 8
4.45
λ μ
+
μ
i.e. λ = 1.23μ
so the Poissons ratio for the earth is given by:
0.276
1.23
μ
2 (1.23 )
λ
2( )
≈
× +
=
+
=
μ μ
λ μ
σ
6.10
At a plane steel water interface, the energy ratio of reflected wave is given by:
86%
2 7 6
3.9 × 10 − 1.43 ×
10 2
7 6
3.9 10 1.43 10
⎞
≈ ⎟ ⎟⎠
⎛
⎜ ⎜⎝
× + ×
⎞
= ⎟ ⎟⎠
⎜ ⎜⎝
Z −
Z
steel water
+
=
steel water
i
Z Z
I
At a plane steel water interface, the energy ratio of transmitted wave is given by:
82.3%
6 6
4 × 3.49 × 10 × 1.43 ×
10
2 ≈
(3.49 10 1.43 10 )
4
Z Z
ice water
( )
6 6 2
× + ×
=
+
=
ice water
i
Z Z
I
6.11
Solution follow directly from the coefficients at top of page 165.

n (node).
Closed end is zero displacement with = −1
n (antinode, η is a max)
r
n
p . Pressure doubles at antinode
r
p
p (out of phase – cancels to give zero pressure, i.e. node)
r
p
r
n
i
Open end: = 1
i
Pressure: closed end: = 1
i
Open end : = −1
i
6.12
(a) The boundary condition = 0
∂
x
η
∂
at x = 0 gives:
( sin cos )sin 0 0 − + = x= Ak kx Bk kx ωt
i.e. B = 0, so we have: η = Acos kx sinωt
∂
x
η
The boundary condition = 0
∂
at x = L gives:
− sin sin = 0 x=l kA kx ωt
i.e. kAsin kLsinωt = 0
which is true for all t if kl = nπ , i.e. π
π 2 l = n or
λ
λ = 2l
n
The first three harmonics are shown below:

0 l 2 l x n =1:
η
0 l
l 4 3l 4 x
0 x
∂
x
η
η
η
l 6 l 2 5l 6 l
n = 2 :
n = 3:
(b) The boundary condition = 0
∂
at x = 0 gives:
( sin cos )sin 0 0 − + = x= Ak kx Bk kx ωt
i.e. B = 0, so we have: η = Acos kx sinωt
The boundary condition η = 0 at x = L gives:
sin sin = 0 x=l A kx ωt
i.e. Acoskl sinωt = 0
kl = ⎛ n +
1 ⎞
, i.e. π
which is true for all t if π⎟⎠
⎜⎝
2
2 π
l n 1 or
λ
⎞
⎟⎠
= ⎛ +
⎜⎝
2
λ l
4
+
2 1
=
n
The first three harmonics are shown below:

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