Pain Solution Manual

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 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 >

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 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 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 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 > 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 ,

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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 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 > 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