xuebaunion@vip.163.com

3551 Trousdale Rkwy, University Park, Los Angeles, CA

留学生论文指导和课程辅导

无忧GPA：https://www.essaygpa.com

工作时间：全年无休-早上8点到凌晨3点

微信客服：xiaoxionga100

微信客服：ITCS521

程序代写案例-PHYSICS 3 CLASS

时间：2021-04-07

Tuesday, 10 December 2019

2 Questions: 09:30 – 10:50

3 Questions: 09:30 – 11:30

4 Questions: 09:30 – 12:10

PHYSICS 3 CLASS EXAM

Physics 3 – Theoretical Physics 3 – Chemical Physics 3

Physics with Astrophysics 3 – Combined Physics 3

[ PHYS4011, PHYS4031, PHYS4003, PHYS4017 ]

Mathematical methods I — Waves &

diffraction — Circuits & systems —

Numerical methods

Answer the question for each lecture course you have enrolled in

this semester.

Answer each question in a separate booklet

Candidates are reminded that devices able to store or display text or images

may not be used in examinations without prior arrangement.

Approximate marks are indicated in brackets as a guide for candidates.

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-986

F

u

n

d

a

m

e

n

ta

l

co

n

st

a

n

ts

n

a

m

e

sy

m

b

o

l

v

a

lu

e

sp

ee

d

of

li

gh

t

c

2

.9

98

×

10

8

m

s−

1

p

er

m

ea

b

il

it

y

of

fr

ee

sp

ac

e

µ

0

4

pi

×

10

−

7

H

m

−

1

p

er

m

it

ti

v

it

y

of

fr

ee

sp

ac

e

0

8

.8

54

×

10

−

1

2

F

m

−

1

el

ec

tr

on

ic

ch

ar

ge

e

1

.6

02

×

10

−

1

9

C

A

vo

ga

d

ro

’s

n

u

m

b

er

N

0

6

.0

22

×

10

2

3

m

ol

−

1

el

ec

tr

on

re

st

m

as

s

m

e

9

.1

10

×

10

−

3

1

k

g

p

ro

to

n

re

st

m

as

s

m

p

1

.6

73

×

10

−

2

7

k

g

n

eu

tr

on

re

st

m

as

s

m

n

1

.6

75

×

10

−

2

7

k

g

F

ar

ad

ay

’s

co

n

st

an

t

F

9

.6

49

×

10

−

4

C

m

ol

−

1

P

la

n

ck

’s

co

n

st

an

t

h

6

.6

26

×

10

−

3

4

J

s

fi

n

e

st

ru

ct

u

re

co

n

st

an

t

α

7

.2

97

×

10

−

3

el

ec

tr

on

ch

ar

ge

to

m

as

s

ra

ti

o

e/

m

e

1

.7

59

×

10

1

1

C

k

g

−

1

q

u

an

tu

m

/c

h

ar

ge

ra

ti

o

h

/e

4

.1

36

×

10

−

1

5

J

s

C

−

1

el

ec

tr

on

C

om

p

to

n

w

av

el

en

gt

h

λ

e

2

.4

26

×

10

−

1

2

m

p

ro

to

n

C

om

p

to

n

w

av

el

en

gt

h

λ

p

1

.3

21

×

10

−

1

5

m

R

y

d

b

er

g

co

n

st

an

t

R

1

.0

97

×

10

7

m

−

1

B

oh

r

ra

d

iu

s

a

0

5

.2

92

×

10

−

1

1

m

B

oh

r

m

ag

n

et

on

µ

B

9

.2

74

×

10

−

2

4

J

T

−

1

n

u

cl

ea

r

m

ag

n

et

on

µ

N

5

.0

51

×

10

−

2

7

J

T

−

1

p

ro

to

n

m

ag

n

et

ic

m

om

en

t

µ

p

1

.4

11

×

10

−

2

6

J

T

−

1

u

n

iv

er

sa

l

ga

s

co

n

st

an

t

R

8

.3

14

J

K

−

1

m

o

l−

1

n

or

m

al

vo

lu

m

e

of

id

ea

l

ga

s

–

2

.2

41

×

10

−

2

m

3

m

ol

−

1

B

ol

tz

m

an

n

co

n

st

an

t

k

B

1

.3

81

×

10

−

2

3

J

K

−

1

F

ir

st

ra

d

ia

ti

on

co

n

st

an

t

2pi

h

c2

c 1

3

.7

42

×

10

−

1

6

W

m

2

S

ec

on

d

R

ad

ia

ti

on

co

n

st

an

t

h

c/

k

B

c 2

1

.4

39

×

10

−

2

m

K

W

ie

n

d

is

p

la

ce

m

en

t

co

n

st

an

t

b

2

.8

98

×

10

−

3

m

K

S

te

fa

n

-B

ol

tz

m

an

n

co

n

st

an

t

σ

5

.6

70

×

10

−

8

W

m

−

2

K

−

4

gr

av

it

at

io

n

al

co

n

st

an

t

G

6

.6

73

×

10

−

1

1

m

3

k

g

−

1

s−

2

im

p

ed

an

ce

of

fr

ee

sp

ac

e

Z

0

3

.7

67

×

10

2

W

D

e

ri

v

e

d

u

n

it

s

q

u

a

n

ti

ty

d

im

e

n

si

o

n

s∗

d

e

ri

v

e

d

u

n

it

en

er

gy

M

L

2

T

−

2

J

fo

rc

e

M

L

T

−

2

N

fr

eq

u

en

cy

T

−

1

H

z

gr

av

it

at

io

n

a

l

fi

el

d

st

re

n

g

th

L

T

−

2

N

k

g

−

1

gr

av

it

at

io

n

a

l

p

o

te

n

ti

al

L

2

T

−

2

J

k

g

−

1

p

ow

er

M

L

2

T

−

3

W

en

tr

op

y

M

L

2

T

−

2

J

K

−

1

h

ea

t

M

L

2

T

−

2

J

ca

p

a

ci

ta

n

ce

M

−

1

L

−

2

T

4

I

2

F

ch

ar

ge

I

T

C

cu

rr

en

t

I

A

el

ec

tr

ic

d

ip

ol

e

m

o

m

en

t

L

T

I

C

m

el

ec

tr

ic

d

is

p

la

ce

m

en

t

L

−

2

T

I

C

m

−

2

el

ec

tr

ic

p

ol

a

ri

sa

ti

on

L

−

2

T

I

C

m

−

2

el

ec

tr

ic

fi

el

d

st

re

n

gt

h

M

L

T

−

3

I

−

1

V

m

−

1

el

ec

tr

ic

(d

is

p

la

ce

m

en

t)

fl

u

x

T

I

C

el

ec

tr

ic

p

ot

en

ti

al

M

L

2

T

−

3

I

−

1

V

in

d

u

ct

an

ce

M

L

2

T

−

2

I

−

2

H

m

ag

n

et

ic

d

ip

ol

e

m

om

en

t

L

2

I

A

m

2

m

ag

n

et

ic

fi

el

d

st

re

n

gt

h

L

−

1

I

A

m

−

1

m

ag

n

et

ic

fl

u

x

M

L

2

T

−

2

I

−

1

W

b

m

ag

n

et

ic

in

d

u

ct

io

n

M

T

−

2

I

−

1

T

m

ag

n

et

is

at

io

n

L

−

1

I

A

m

−

1

p

er

m

ea

b

il

it

y

M

L

T

−

2

I

−

2

H

m

−

1

p

er

m

it

ti

v

it

y

M

−

1

L

−

3

T

4

I

2

F

m

−

1

re

si

st

an

ce

M

L

2

T

−

3

I

−

2

W

re

si

st

iv

it

y

M

L

3

T

−

3

I

−

2

W

m

∗

M

=

m

as

s,

L

=

le

n

g

th

,

T

=

ti

m

e,

I

=

cu

rr

en

t

SECTION I – Mathematical methods I

1 A damped, driven, harmonic oscillator is described by the differential equation

d2y(t)

dt2

+

dy(t)

dt

+ y(t) = F (t) ,

where F (t) is the time-dependent driving force. For a harmonic driving force

F (t) = A sin(ωt+ φ) ,

the steady-state solution is given by

y(t) =

A

Z(ω)

sin [ωt+ φ+ ∆φ(ω)] ,

where Z(ω) and ∆φ(ω) are functions of ω .

(a) Using a simple argument involving φ, or otherwise, verify that, for a driv-

ing force

F (t) = A cos(ωt+ φ) ,

the steady-state solution is given by

y(t) =

A

Z(ω)

cos [ωt+ φ+ ∆φ(ω)] .

[1]

(b) Show that the differential equation is linear by showing that, if y1(t) and

y2(t) are solutions for the driving forces F1(t) and F2(t) , respectively, then

ay1(t) + by2(t) is a solution for the driving force aF1(t) + bF2(t) . [4]

(c) What is the steady-state solution for the driving force

F (t) = 4 sin(2t) + 3 cos(t) ?

Do not attempt to find and/or evaluate Z(...) and ∆φ(...) . [1]

For the remainder of this question, the driving force is given by a sawtooth

function:

F (t) =

{

2t for − 1 < t ≤ 1 ,

periodically repeating elsewhere .

(d) Sketch a plot of F (t). What is its period? [2]

(e) Define even and odd functions. Does F (t) fall into either of these symme-

try categories? [2]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9863/8 Q 1 continued over. . .

Q 1 continued

(f) Outline how one can obtain the displacement of the oscillator, y(t), in

response to the sawtooth driving force by using the solution given in part (a)

in response to a sinusoidal driving force. [2]

(g) The current density J associated with a magnetic field B is given by the

equation

∇×B = µ0J ,

where µ0 is the magnetic permeability.

For the magnetic field (in Cartesian coordinates)

B = B0z

−1

y−x

z2

,

where B0 is a constant, show that the associated current density is

J = −B0

µ0

z−2

xy

2z

.

Determine the component of the current density J in the direction (1,1,0) at

the position x = y = z = 0.5 m for a magnetic field constant B0 = 1.2 mT. [5]

(h) Show that the magnetic field above can be written as

B = B0

(

−ρ

z

eˆφ + zeˆz

)

,

where ρ, φ and z are cylindrical coordinates. [3]

[Total: 20]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9864/8 Paper continued over. . .

SECTION II – Waves & diffraction

2 (a) Show that the wave-equation for a transverse wave oscillating in the y-

direction and travelling in the x-direction along a stretched string is given

by

∂2y

∂t2

=

T

ρ

∂2y

∂x2

,

where y is the wave amplitude, T is the tension in the string and ρ is the mass

per unit length. Make sure that you justify the steps of your proof, explaining

clearly any assumptions made. [6]

(b) Demonstrate that y(x, t) = A ei(ωt−kx) is a possible travelling wave solu-

tion of the wave equation of part a) and show that the velocity of waves on

the string is given by c =

√

T/ρ . [4]

(c) A stretched string lying along the negative x-axis is terminated at x = 0

by a “damper” with impedance Zd. The damper provides a force proportional

in magnitude to the transverse velocity of the end point of the string but acting

in the opposite direction. The damper thus absorbs some of the wave energy

incident on it.

(Note that you could choose to model the “damper” by a suitably-chosen

second string. In that case the second string would convey away some of the

energy from the first string rather than dissipate it. The effect on the first

string would be the same.)

A single frequency transverse wave propagates along the string from the

negative x-direction. Apply an appropriate boundary condition at x = 0 to

show that the amplitude reflection coefficient of the wave is given by

r =

Zs − Zd

Zs + Zd

,

where Zs and Zd are the impedances of the string and the damper respectively. [7]

(d) Is the stretched string a dispersive or non-dispersive medium? Justify

your answer. Suggest an alteration to the stretched string case that would

change its dispersive behaviour. [3]

[Total: 20]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9865/8 Paper continued over. . .

SECTION III – Circuits & systems

3 (a) Show that the s-domain transfer function for the circuit shown below is

given by

G(s) =

1

1 + sτ

,

where τ = RC .

[3]

(b) Sketch a pole-zero diagram for the transfer function of part (a). [2]

In an effort to design an oscillator, an engineer proposed the following

circuit in which three copies of the building block of part (a) are cascaded and

a feedback loop created using an inverting operational amplifier circuit.

The engineer expected that the gain of this amplifier could be adjusted (by

choice of opamp feedback resistor) so that the circuit oscillated.

(c) What condition would have to be achieved for the circuit to oscillate? By

considering the phase response of a single block, find the expected frequency

of oscillation in terms of the time constant τ . [5]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9866/8 Q 3 continued over. . .

Q 3 continued

(d) Defining K = Rf/R1, write down the s-domain form of the open-loop

transfer function. [1]

(e) Set down the Characteristic Equation for the feedback system. Explain

briefly the significance for feedback loop stability of the positions on the com-

plex plane of the roots of the Characteristic Equation.

[3]

(f) By using the technique of equating real and imaginary parts, find the value

of K for which the circuit just starts to oscillate. Find also the frequency of

oscillation. Does this result agree with your prediction from part (c)? [6]

[Total: 20]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9867/8 Paper continued over. . .

SECTION IV – Numerical Methods

4 (a) State briefly what Gaussian elimination is used for. [1]

Describe how the Newton-Raphson method works and how it can be used

to find the root of a continuous function f . State any assumption that is

necessary. [5]

When this method is used, one finds

xn+1 = xn − f(xn)

f ′(xn)

.

Show that the error in the Newton-Raphson method converges quadratically. [4]

(b) The diffusion equation is given by:

∂u

∂t

= D

∂2u

∂x2

.

What class of second order partial differential equation is the diffusion equa-

tion?

[1]

The finite difference formulas for space and time are

∂2u

∂x2

=

unj+1 − 2unj + unj−1

(∆x)2

,

∂u

∂t

=

un+1j − unj

∆t

,

where x is denoted by index j, t is denoted by n, ∆x is the step size in x, and

∆t is the step size in t.

Show that

un+1j = R(u

n

j+1 + u

n

j−1) + (1− 2R)unj ,

where

R = D

∆t

(∆x)2

.

[3]

For the finite difference formula presented above, what is the condition for

the maximum time step at which numerical stability is assured? Evaluate this

maximum time step for a case in which ∆x = 0.06 m and D = 10−6 m2 s−1 [3]

The method used above is an explicit method. Name another type of

finite difference method that can be used to solve the diffusion equation that

is stable for all values of ∆t. Why is this method always stable? [3]

[Total: 20]

End of Paper

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9868/8 END

学霸联盟

2 Questions: 09:30 – 10:50

3 Questions: 09:30 – 11:30

4 Questions: 09:30 – 12:10

PHYSICS 3 CLASS EXAM

Physics 3 – Theoretical Physics 3 – Chemical Physics 3

Physics with Astrophysics 3 – Combined Physics 3

[ PHYS4011, PHYS4031, PHYS4003, PHYS4017 ]

Mathematical methods I — Waves &

diffraction — Circuits & systems —

Numerical methods

Answer the question for each lecture course you have enrolled in

this semester.

Answer each question in a separate booklet

Candidates are reminded that devices able to store or display text or images

may not be used in examinations without prior arrangement.

Approximate marks are indicated in brackets as a guide for candidates.

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-986

F

u

n

d

a

m

e

n

ta

l

co

n

st

a

n

ts

n

a

m

e

sy

m

b

o

l

v

a

lu

e

sp

ee

d

of

li

gh

t

c

2

.9

98

×

10

8

m

s−

1

p

er

m

ea

b

il

it

y

of

fr

ee

sp

ac

e

µ

0

4

pi

×

10

−

7

H

m

−

1

p

er

m

it

ti

v

it

y

of

fr

ee

sp

ac

e

0

8

.8

54

×

10

−

1

2

F

m

−

1

el

ec

tr

on

ic

ch

ar

ge

e

1

.6

02

×

10

−

1

9

C

A

vo

ga

d

ro

’s

n

u

m

b

er

N

0

6

.0

22

×

10

2

3

m

ol

−

1

el

ec

tr

on

re

st

m

as

s

m

e

9

.1

10

×

10

−

3

1

k

g

p

ro

to

n

re

st

m

as

s

m

p

1

.6

73

×

10

−

2

7

k

g

n

eu

tr

on

re

st

m

as

s

m

n

1

.6

75

×

10

−

2

7

k

g

F

ar

ad

ay

’s

co

n

st

an

t

F

9

.6

49

×

10

−

4

C

m

ol

−

1

P

la

n

ck

’s

co

n

st

an

t

h

6

.6

26

×

10

−

3

4

J

s

fi

n

e

st

ru

ct

u

re

co

n

st

an

t

α

7

.2

97

×

10

−

3

el

ec

tr

on

ch

ar

ge

to

m

as

s

ra

ti

o

e/

m

e

1

.7

59

×

10

1

1

C

k

g

−

1

q

u

an

tu

m

/c

h

ar

ge

ra

ti

o

h

/e

4

.1

36

×

10

−

1

5

J

s

C

−

1

el

ec

tr

on

C

om

p

to

n

w

av

el

en

gt

h

λ

e

2

.4

26

×

10

−

1

2

m

p

ro

to

n

C

om

p

to

n

w

av

el

en

gt

h

λ

p

1

.3

21

×

10

−

1

5

m

R

y

d

b

er

g

co

n

st

an

t

R

1

.0

97

×

10

7

m

−

1

B

oh

r

ra

d

iu

s

a

0

5

.2

92

×

10

−

1

1

m

B

oh

r

m

ag

n

et

on

µ

B

9

.2

74

×

10

−

2

4

J

T

−

1

n

u

cl

ea

r

m

ag

n

et

on

µ

N

5

.0

51

×

10

−

2

7

J

T

−

1

p

ro

to

n

m

ag

n

et

ic

m

om

en

t

µ

p

1

.4

11

×

10

−

2

6

J

T

−

1

u

n

iv

er

sa

l

ga

s

co

n

st

an

t

R

8

.3

14

J

K

−

1

m

o

l−

1

n

or

m

al

vo

lu

m

e

of

id

ea

l

ga

s

–

2

.2

41

×

10

−

2

m

3

m

ol

−

1

B

ol

tz

m

an

n

co

n

st

an

t

k

B

1

.3

81

×

10

−

2

3

J

K

−

1

F

ir

st

ra

d

ia

ti

on

co

n

st

an

t

2pi

h

c2

c 1

3

.7

42

×

10

−

1

6

W

m

2

S

ec

on

d

R

ad

ia

ti

on

co

n

st

an

t

h

c/

k

B

c 2

1

.4

39

×

10

−

2

m

K

W

ie

n

d

is

p

la

ce

m

en

t

co

n

st

an

t

b

2

.8

98

×

10

−

3

m

K

S

te

fa

n

-B

ol

tz

m

an

n

co

n

st

an

t

σ

5

.6

70

×

10

−

8

W

m

−

2

K

−

4

gr

av

it

at

io

n

al

co

n

st

an

t

G

6

.6

73

×

10

−

1

1

m

3

k

g

−

1

s−

2

im

p

ed

an

ce

of

fr

ee

sp

ac

e

Z

0

3

.7

67

×

10

2

W

D

e

ri

v

e

d

u

n

it

s

q

u

a

n

ti

ty

d

im

e

n

si

o

n

s∗

d

e

ri

v

e

d

u

n

it

en

er

gy

M

L

2

T

−

2

J

fo

rc

e

M

L

T

−

2

N

fr

eq

u

en

cy

T

−

1

H

z

gr

av

it

at

io

n

a

l

fi

el

d

st

re

n

g

th

L

T

−

2

N

k

g

−

1

gr

av

it

at

io

n

a

l

p

o

te

n

ti

al

L

2

T

−

2

J

k

g

−

1

p

ow

er

M

L

2

T

−

3

W

en

tr

op

y

M

L

2

T

−

2

J

K

−

1

h

ea

t

M

L

2

T

−

2

J

ca

p

a

ci

ta

n

ce

M

−

1

L

−

2

T

4

I

2

F

ch

ar

ge

I

T

C

cu

rr

en

t

I

A

el

ec

tr

ic

d

ip

ol

e

m

o

m

en

t

L

T

I

C

m

el

ec

tr

ic

d

is

p

la

ce

m

en

t

L

−

2

T

I

C

m

−

2

el

ec

tr

ic

p

ol

a

ri

sa

ti

on

L

−

2

T

I

C

m

−

2

el

ec

tr

ic

fi

el

d

st

re

n

gt

h

M

L

T

−

3

I

−

1

V

m

−

1

el

ec

tr

ic

(d

is

p

la

ce

m

en

t)

fl

u

x

T

I

C

el

ec

tr

ic

p

ot

en

ti

al

M

L

2

T

−

3

I

−

1

V

in

d

u

ct

an

ce

M

L

2

T

−

2

I

−

2

H

m

ag

n

et

ic

d

ip

ol

e

m

om

en

t

L

2

I

A

m

2

m

ag

n

et

ic

fi

el

d

st

re

n

gt

h

L

−

1

I

A

m

−

1

m

ag

n

et

ic

fl

u

x

M

L

2

T

−

2

I

−

1

W

b

m

ag

n

et

ic

in

d

u

ct

io

n

M

T

−

2

I

−

1

T

m

ag

n

et

is

at

io

n

L

−

1

I

A

m

−

1

p

er

m

ea

b

il

it

y

M

L

T

−

2

I

−

2

H

m

−

1

p

er

m

it

ti

v

it

y

M

−

1

L

−

3

T

4

I

2

F

m

−

1

re

si

st

an

ce

M

L

2

T

−

3

I

−

2

W

re

si

st

iv

it

y

M

L

3

T

−

3

I

−

2

W

m

∗

M

=

m

as

s,

L

=

le

n

g

th

,

T

=

ti

m

e,

I

=

cu

rr

en

t

SECTION I – Mathematical methods I

1 A damped, driven, harmonic oscillator is described by the differential equation

d2y(t)

dt2

+

dy(t)

dt

+ y(t) = F (t) ,

where F (t) is the time-dependent driving force. For a harmonic driving force

F (t) = A sin(ωt+ φ) ,

the steady-state solution is given by

y(t) =

A

Z(ω)

sin [ωt+ φ+ ∆φ(ω)] ,

where Z(ω) and ∆φ(ω) are functions of ω .

(a) Using a simple argument involving φ, or otherwise, verify that, for a driv-

ing force

F (t) = A cos(ωt+ φ) ,

the steady-state solution is given by

y(t) =

A

Z(ω)

cos [ωt+ φ+ ∆φ(ω)] .

[1]

(b) Show that the differential equation is linear by showing that, if y1(t) and

y2(t) are solutions for the driving forces F1(t) and F2(t) , respectively, then

ay1(t) + by2(t) is a solution for the driving force aF1(t) + bF2(t) . [4]

(c) What is the steady-state solution for the driving force

F (t) = 4 sin(2t) + 3 cos(t) ?

Do not attempt to find and/or evaluate Z(...) and ∆φ(...) . [1]

For the remainder of this question, the driving force is given by a sawtooth

function:

F (t) =

{

2t for − 1 < t ≤ 1 ,

periodically repeating elsewhere .

(d) Sketch a plot of F (t). What is its period? [2]

(e) Define even and odd functions. Does F (t) fall into either of these symme-

try categories? [2]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9863/8 Q 1 continued over. . .

Q 1 continued

(f) Outline how one can obtain the displacement of the oscillator, y(t), in

response to the sawtooth driving force by using the solution given in part (a)

in response to a sinusoidal driving force. [2]

(g) The current density J associated with a magnetic field B is given by the

equation

∇×B = µ0J ,

where µ0 is the magnetic permeability.

For the magnetic field (in Cartesian coordinates)

B = B0z

−1

y−x

z2

,

where B0 is a constant, show that the associated current density is

J = −B0

µ0

z−2

xy

2z

.

Determine the component of the current density J in the direction (1,1,0) at

the position x = y = z = 0.5 m for a magnetic field constant B0 = 1.2 mT. [5]

(h) Show that the magnetic field above can be written as

B = B0

(

−ρ

z

eˆφ + zeˆz

)

,

where ρ, φ and z are cylindrical coordinates. [3]

[Total: 20]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9864/8 Paper continued over. . .

SECTION II – Waves & diffraction

2 (a) Show that the wave-equation for a transverse wave oscillating in the y-

direction and travelling in the x-direction along a stretched string is given

by

∂2y

∂t2

=

T

ρ

∂2y

∂x2

,

where y is the wave amplitude, T is the tension in the string and ρ is the mass

per unit length. Make sure that you justify the steps of your proof, explaining

clearly any assumptions made. [6]

(b) Demonstrate that y(x, t) = A ei(ωt−kx) is a possible travelling wave solu-

tion of the wave equation of part a) and show that the velocity of waves on

the string is given by c =

√

T/ρ . [4]

(c) A stretched string lying along the negative x-axis is terminated at x = 0

by a “damper” with impedance Zd. The damper provides a force proportional

in magnitude to the transverse velocity of the end point of the string but acting

in the opposite direction. The damper thus absorbs some of the wave energy

incident on it.

(Note that you could choose to model the “damper” by a suitably-chosen

second string. In that case the second string would convey away some of the

energy from the first string rather than dissipate it. The effect on the first

string would be the same.)

A single frequency transverse wave propagates along the string from the

negative x-direction. Apply an appropriate boundary condition at x = 0 to

show that the amplitude reflection coefficient of the wave is given by

r =

Zs − Zd

Zs + Zd

,

where Zs and Zd are the impedances of the string and the damper respectively. [7]

(d) Is the stretched string a dispersive or non-dispersive medium? Justify

your answer. Suggest an alteration to the stretched string case that would

change its dispersive behaviour. [3]

[Total: 20]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9865/8 Paper continued over. . .

SECTION III – Circuits & systems

3 (a) Show that the s-domain transfer function for the circuit shown below is

given by

G(s) =

1

1 + sτ

,

where τ = RC .

[3]

(b) Sketch a pole-zero diagram for the transfer function of part (a). [2]

In an effort to design an oscillator, an engineer proposed the following

circuit in which three copies of the building block of part (a) are cascaded and

a feedback loop created using an inverting operational amplifier circuit.

The engineer expected that the gain of this amplifier could be adjusted (by

choice of opamp feedback resistor) so that the circuit oscillated.

(c) What condition would have to be achieved for the circuit to oscillate? By

considering the phase response of a single block, find the expected frequency

of oscillation in terms of the time constant τ . [5]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9866/8 Q 3 continued over. . .

Q 3 continued

(d) Defining K = Rf/R1, write down the s-domain form of the open-loop

transfer function. [1]

(e) Set down the Characteristic Equation for the feedback system. Explain

briefly the significance for feedback loop stability of the positions on the com-

plex plane of the roots of the Characteristic Equation.

[3]

(f) By using the technique of equating real and imaginary parts, find the value

of K for which the circuit just starts to oscillate. Find also the frequency of

oscillation. Does this result agree with your prediction from part (c)? [6]

[Total: 20]

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9867/8 Paper continued over. . .

SECTION IV – Numerical Methods

4 (a) State briefly what Gaussian elimination is used for. [1]

Describe how the Newton-Raphson method works and how it can be used

to find the root of a continuous function f . State any assumption that is

necessary. [5]

When this method is used, one finds

xn+1 = xn − f(xn)

f ′(xn)

.

Show that the error in the Newton-Raphson method converges quadratically. [4]

(b) The diffusion equation is given by:

∂u

∂t

= D

∂2u

∂x2

.

What class of second order partial differential equation is the diffusion equa-

tion?

[1]

The finite difference formulas for space and time are

∂2u

∂x2

=

unj+1 − 2unj + unj−1

(∆x)2

,

∂u

∂t

=

un+1j − unj

∆t

,

where x is denoted by index j, t is denoted by n, ∆x is the step size in x, and

∆t is the step size in t.

Show that

un+1j = R(u

n

j+1 + u

n

j−1) + (1− 2R)unj ,

where

R = D

∆t

(∆x)2

.

[3]

For the finite difference formula presented above, what is the condition for

the maximum time step at which numerical stability is assured? Evaluate this

maximum time step for a case in which ∆x = 0.06 m and D = 10−6 m2 s−1 [3]

The method used above is an explicit method. Name another type of

finite difference method that can be used to solve the diffusion equation that

is stable for all values of ∆t. Why is this method always stable? [3]

[Total: 20]

End of Paper

Mathematical methods I — Waves & diffraction — Circuits & systems — Numerical methods/312-9868/8 END

学霸联盟