clc207 se 1112s2 student

Feb/Mar 2012 Time Allowed: 2 hrs __________________________________________________________________ INSTRUCTIONS TO CANDIDATES 1 This examination paper consists of FIFTEEN (15) pages including this

page. 2 This examination paper consists of 2 sections. Section A: 5 Questions Section B: 5 Questions 3 Answer ALL questions in Section A. Answer ANY THREE (3) questions in Section B. 4 All working and presentation must be shown neatly and clearly.

Do NOT use pencils. 5 Begin each question on a new page of the answer booklet. 6 State the questions attempted on the cover page of the answer booklet. 7 Mathematical formula lists are provided on 4 to 15.

2011/2012 SEMESTER 2 – SEMESTRAL EXAMINATION

Course: Diploma in Electronics, Computer & Communications Engineering Diploma in Telematics & Media Technology Diploma in Mechatronics Engineering Diploma in Manufacturing Engineering Diploma in Digital & Precision Engineering Diploma in Aeronautical & Aerospace Technology Diploma in Nanotechnology & Materials Science Module: EG2001/EG2691/EGB201/ EGC201/EGD201/EGF201 – Engineering Mathematics 2A EGJ201 – Mathematics 2A

Semestral Exam / EG2001 / EG2691 / EGB201 / EGC201 / EGD201 / EGF201 / EGJ201 /

SECTION A – ANSWER ALL QUESTIONS ( 40 Marks )

Question 1 Evaluate 2 cos3xe xdx . ( 8 marks )

Solution: Diff Diff

2 xf x e cos 3g x x

2' 2 xf x e ' 3sin 3g x x

2'' 4 xf x e '' 9 cos 3g x x

4n 9m

2 cos 3xe x dx 2 23sin 3 ( 2 cos3 )

9 4

x xe x e xC

2

3sin 3 2 cos 313

xex x C

Question 2

Solve the differential equation 2x ydye

dx by separation of variables, given the

initial condition 0 2y . ( 8 marks )

Solution:

2

2

y x

xy

e dy e dx

e dy e dx

2xye e c Given 2 0y x when ,

2 2

22

e e c

c e

Therefore 222 xye e e


Question 3

Given the differential equation 2 3xy x y , 0x .

(a) Show that the integrating factor is 3

1

x. ( 4 marks )

(b) Hence, solve the differential equation. ( 4 marks ) Solution: (a)

3

2

3

3ln

1ln

3

3

Integrating factor

dxx

x

x

xy y x

y y xx

I x e

e

e

3

1

x

(b)

3 4 2

3 2

3 2

3 2

1 3 1

1

1 1

y yx x x

d y

dx x x

ydx c

xx x

y cx x


Question 4

Find the Laplace transform of

(a) cos 3ty t e t ( 4 marks )

(b) 4 3 2 ty t t e ( 4 marks )

Solution:

(a)

1

2 2 2 21

cos3 cos3

1

3 1 3

t

s s

s s

e t t

ss

s s

L L

(b)

4 3 4 35

3

5

4!2 2 2

48

3

t t

s s

t e t es

s

L L

Question 5

Find the inverse Laplace transform of 1

.1 2 3

Y ss s s

( 8 marks )

Solution:

1

1 2 3 1 2 3

A B CY s

s s s s s s

Using Cover Up method:

4

1

2313

1

3

1

3212

1

12

1

3121

1

C

B

A

11 1312 4

1 2 3Y s

s s s

2 31 1 1

12 3 4t t ty t e e e

END OF SECTION A


SECTION B – ANSWER ANY 3 QUESTIONS ( 60 Marks ) Question 6 (a) A mathematical model for the rate at which the population of a certain

community changes is given by cosdP

Pk tdt

, where k is a positive constant

and P(t) is the population (in ten thousands) at time t (in years). Find P t

given that 0 5P . ( 6 marks )

(b) From Kirchhoff’s second law, a series circuit containing a resistor R and an inductor L is related by the differential equation

di

L Ri v tdt

where i t is the current and v t is the voltage at t seconds. A voltage

117 0 25

0 25

tv t

t

is applied to the circuit in which the inductance is 15 henries and the resistance is 1.8 ohms.

(i) Using the integrating factor method, find the current i t in the

circuit if 0 0i . ( 12 marks )

(ii) Describe the behaviour of the current. ( 2 marks ) Solution:

(a) cosdP

Pk tdt

1cosdP k tdt

P

1ln sinP k t c

sink tP ce Since 0 5P , 5c ,

sin5 k tP e

(b)(i) 15 1.8 0.12

15

v tdi dii v t i

dt dt

Since 0.12P t , integrating factor, 0.12 0.12dt tI t e e


For 0 25t :

0.12 0.12 117

15t te i e dt

0.12117

15 0.12te c

0.12 0.1265t ti t e e c

0.1265 ti t ce

Since 0 0i , 65c ,

0.1265 1 ti t e for 0 25t

For 25t :

0.12 0te i dt k

0.12ti t ke

When t = 25, 0.12 25 325 65 1 65 1i e e

0.12 253 365 1 65 1e ke k e

Hence

0.12

3 0.12

65 1 0 25

65 1 25

t

t

e ti t

e e t

(b)(ii) The current increases on 0 25t and decreases on 25t .


Question 7

(a) The acceleration of a falling object is given by 2

2

d xg

dt

m/s2, where g is a

constant. Find the position x (in metres, m) of the object at time t (in seconds, s) if its initial velocity is 0m/s and the object falls from an initial position of 2m. ( 8 marks )

(b) A variable force F (in Newton, N) is applied to the object in part (a) and the corresponding acceleration now becomes:

2

2

d x Fg

dt m

where m is the mass (in kilogram, kg) of the object. Find the position of the object at time t if 2m kg, 10F t N, the initial velocity of the object is 12m/s and its initial position is 2x m. What can we say about the motion of the object here as compared to the answers in part (a)? ( 12 marks ) Solution:

(a) Integrating 2

2

d xg

dt with respect to t , we have:

dx

g dt gt Cdt

where C is a constant.

Since initial velocity is 0dx

dt m/s, we have:

0 0 0g C C

Hence, dx

gtdt

.

Integrating dx

gtdt

with respect to t , we have:

2

2

gtx gt dt K ( K is a constant)

Since the initial position of the object is 2m, we have:

0

2 22

gK K

Hence the position of the object at time t is: 212

2x t gt


(b) Given 2m kg and 10F t N, we have:

2

2

105

2

d x F tg g g t

dt m

Integrating with respect to t , we have:

255

2

dxg t dt gt t C

dt where C is a constant.

Since the initial velocity of the object is 12m/s, we have:

50 0 12 12

2g C C

Hence, 2512

2

dxgt t

dt .

Integrating with respect to t again, we have:

2 2 35 512 12

2 2 6

gx gt t dt t t t K

( K is a constant)

If the initial position is 2m, then we have:

50 0 12 0 2 2

2 6

gK K

Hence the position of the object at time t is:

2 3512 2

2 6

gx t t t t

Observe that for t sufficiently large, x t is negative (notice that 3t will become the

dominating factor in the expression). As such, the object travels in the opposite direction as compared to the motion of the object described in part (a).


Question 8 The equation of motion of a mass attached to a spring and suspended in a viscous liquid is given by

2

2 4 20 0d x dx

dt dtx

where x is the displacement of the mass at time t.

(a) If the mass is initially released from rest at a point 0.3m above the

equilibrium position, find x t . ( 16 marks )

(b) Sketch tx and describe the motion of the mass. ( 4 marks )

Solution:

(a) The initial conditions are :

0 0.3

0 0

[1 m]

[1 m]

x

x

Auxiliary equation:

2 4 20 0m m

24 4 4 1 202 4

2 1m j

General solution:

21 2cos 4 sin 4tx t e c t c t

Applying initial conditions: 0 0.3, 0 0 x x

0

1 20 cos 0 sin 0 0.3x e c c

1 0.3c

Differentiating: 2 21 2 1 22 cos4 sin 4 4 sin 4 4 cos4t tx t e c t c t e c t c t

00 x

0 01 2 1 20 2 cos 0 sin 0 4 sin 0 4 cos 0 0x e c c e c c

042 21 cc

2

2

2 0.3 4 0

0.60.15

4

c

c

Equation of Motion:

2 0.3cos 4 0.15sin 4tx t e t t

20.15 2 cos 4 sin 4te t t


(b)

The object/mass goes through an underdamped motion or oscillatory motion with decreasing amplitude.


Question 9

A system consists of a single mass lying on a smooth surface and connected to two fixed points by three springs. When a sinusoidal force is applied to the system, the displacement of the mass from its equilibrium position satisfies the equation:

2

24 8 cos 2

d x dxx t

dt dt

Given that the system is initially at rest in the equilibrium position, use Laplace transform to solve the equation for .x t

( 20 marks ) Solution:

Let x t X s

2

24 8 cos 2

d x dxx t

dt dt

Taking Laplace transform

22

0 0 4 0 84

ss X s sx x sX s x X s

s

Initial conditions: 0 0, 0 0x x

22

4 84

ss X s sX s X s

s

2 24 4 8

sX s

s s s

2 24 4 8

As B Cs DX s

s s s

1 1 1 220 5 20 5

2 24 4 8

s sX s

s s s

2 22 2

21 3

20 4 5 4 20[ 2 4] 10[ 2 4]

ssX s

s s s s

2 21 1 1 3cos 2 sin 2 cos 2 sin 2

20 10 20 20t tx t t t e t e t


Question 10

(a) Find the Laplace transform of the function g t where

223 costg t t e u t t u t ( 6 marks )

(b) Use Laplace transform to solve the following initial-value problem:

4 2 , 0 0, 0 1y y u t y y ( 14 marks )

Solution:

(a)

2

2

2

2

3 33

3 33

3 1

3 2

2

2

3 cos

2cos

2 1sin

12 1 1

1 1

t

ss t

ss

ss

g t t e u t t u t

e e e ts

e e e tss

e ess s

L L L LL L

L

(b) 4 2 , 0 0, 0 1y y u t y y

Taking Laplace transform:

4 2y y u t L L L

2 210 0 4 ss Y s sy y Y s e

s

2 211 4 ss Y s Y s e

s

2 214 1 ss Y s e

s

22

1 11

4sY s e

s s

1 2 1

2 2

1 1

4 4sy t e

s s s

L L

22

2

1 14 4

1

44

1 4

, , 0

A Bs C

s ss s

A s Bs C s

A B C


1 11 2 1 24 4

2 2

14

14

1

4 4

2 2 cos 2 2

1 cos 2 2

s sse e

ss s s

u t u t t

t u t

L L

1 11 1

2 22 2 2

1 2sin 2

4 2 t

s s

L L

1 14 21 cos 2 2 sin 2 y t t u t t

END OF SECTION B


Formula List

Rules on Differentiation Product Rule:

where u, v are functions of x Quotient Rule:

Chain Rule:

Integration By Parts:

Standard Derivatives

c = constant

Standard Integrals (The constant of integration is omitted)

( )

( )

or =

or =


END OF PAPER

Trigonometry Basic Identities:

Compound Angle Relations:

Sine Rule:

Cosine Rule:

Double Angle Relations:

Factor Formulae:

Algebraic Identities

Indices and Surds

Logarithms

1f t F sL s f tF = L

1 or U t1

, 0ss

, 0, 1, 2, 3nt n 1

!n

n

s

ate1

s a

sinkt 2 2

k

s k

coskt 2 2

s

s k

atte 2

1

( )s a

sint kt 2 2 2

2

( )

ks

s k

cost kt2 2

2 2 2( )

s k

s k

( )nt f t ( 1) ( )n

nn

dF s

ds

ate f t s s a

f t

L

sinate kt 2 2

k

s a k

cosate kt 2 2

s a

s a k

U t a 1, 0ase s

s

U t a f t a ase f t L

U t f t f tL

y t s y tY = L

'dy

y tdt

0sY s y

2

2''

d yy t

dt 2 0 ' 0s Y s s y y

( ) ( )ny t( )1

10( )

0

( ) ( ) |kn

n n ktk

k

ds F s s f t

dt

clc207 se 1112s2 student

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