1

On Free Mechanical Vibrations• As derived in section 4.1( following Newton’s 2nd

law of motion and the Hooke’s law), the D.E. for the mass-spring oscillator is given by:

m.equilibriu fromnt displaceme theis system. the toforces externalother all :

(friction)t coefficien damping theis the: )(stiffnessconstant sHooke' theis :

spring the toattached mass theis : where,)('"

y :Fbkm

tFkybymy

e

e

2

In the simplest case, when b = 0, and Fe = 0, i.e. Undamped, free

vibration, we can rewrite the D.E:• As

frequency.angular theasknown is

. 2

frequency and , 2 period

hmotion wit harmonic Simple called is This

. tan and

with .), sin()( :as written be alsocan Which . sin cos)( :clearly is

solution general a , where,0 "

2

122

21

21

2

ccccA

tAtytctcty

mkyy

3

When b 0, but Fe = 0, we have damping on free vibrations.

• The D. E. in this case is:

:cases possible threehave We. 4nt discrimina on the depends

clearlysolution The . 421

2

asseen easily are roots the, 0

is eq.auxiliary theand , 0'"

2

2

2

mkb

mkbmm

bkbrmr

kybymy

4

Case I: Underdamped Motion (b2 < 4mk)

.factor damping a with wavesinusoidal a is This

). sin()(or ) sin cos()(

issolution generalA . i are roots the

hence ,421 and

2 :let We

21

2

t

tt

Ae

tAetytctcety

bmkmm

b

5

Case II: Overdamped Motion (b2 > 4mk)

• In this case, we have two distinct real roots, r1 & r2. Clearly both are negative, hence a general solution:

oscillate.not doessolution the thatfollowsit zero, onemost at has 0)(' Since . as 0)( 21

21

tytececty trtr

No local max or min

One local maxOne local min

6

Case III: Critically Damped Motion (b2 = 4mk)

• We have repeated root -b/2m. Thus the a general solution is:

motions.Overdamped of that similar to behave solutionsits solution, onemost at has 0)('

and , tas 0)(limthat see Again we . )( 2/

22/

1

tyty

tececty mbtmbt

7

Example• The motion of a mass-spring system with damping

is governed by

• This is exercise problem 4, p239.• Find the equation of motion and sketch its graph

for b = 10, 16, and 20.

. 0)0(' and , 1)0(; 0)(64)(')("

yytytbyty

8

• 1. b = 10: we have m = 1, k = 64, and • b2 - 4mk = 100 - 4(64) = - 156, implies = (39)1/2 .

Thus the solution to the I.V.P. is

Solution.

. 539 tan

where, ) 39sin(398

) 39sin395 39(cos)(

2

1

5

5

cc

te

ttety

t

t

9

When b = 16, b2 - 4mk = 0, we have repeated root -8,

• thus the solution to the I.V.P is

tetty 8)81()(

1

t

y

10

• r1 = - 4 and r2 = -16, the solution to the I.V.P. is:

When b = 20, b2 - 4mk = 64, thus two distinct real roots are

:like looksgraph The .31

34)( 164 tt eety

1

1t

y

11

• with the following D. E.

Next we consider forced vibrations

(*). ofequation ushomogeneno associated theofsolution genernal a is

& (*), osolution t particular afor stands

where, form in the written becan (*) oSolution t

d).underdampe (i.e. 4 0

and 0 , 0 that assume We. cos'" (*)

2

0

0

h

p

ph

y

y

yyy

mkb

FtFkybymy

12

We know a solution to the above equation has the form

• where:

• In fact, we have

ph yyy

, sin cos)( thatknow we

ts,coefficien edundetermin of method by the and

, 2

4sin)(

21

2)2/(

tBtBty

tm

bmkAety

p

tmbh

. ) (

, ) (

) (2222

022222

20

1

bmkbFB

bmkmkFB

13

Thus in the case 0 < b 2 < 4mk (underdamped), a general

solution has the form:

). sin( ) (

)(

and , 2

4sin)(

where, )()()(

22220

2)2/(

tbmk

Fty

tm

bmkAety

tytyty

p

tmbh

ph

14

Remark on Transient and Steady-State solutions.

15

• Consider the following interconnected fluid tanksIntroduction

A B8 L/min

X(t) Y(t)

24 L 24 L

X(0)= a Y(0)= b6 L/min

2 L/min

6 L/min

16

Suppose both tanks, each holding 24 liters of a brine solution, are interconnected by pipes as shown . Fresh water flows into tank A at a rate of 6 L/min, and fluid is drained out of tank B at the same rate; also 8 L/min of fluid are pumped from tank A to tank B, and 2 L/min from tank B to tank A. The liquids inside each tank are kept well stirred, so that each mixture is homogeneous. If initially tank A contains a kg of salt and tank B contains b kg of salt, determine the mass of salt in each tanks at any time t > 0.

17

Set up the differential equations• For tank A, we have:

• and for tank B, we have

)(248)(

242 txty

dtdx

).(246)(

242)(

248 tytytx

dtdy

18

This gives us a system of First Order Equations

.041'2"3

get weequation,first into put them ,'"3'implies '3 since :equationorder 2nd

a toequivalent is This . 31

31'

and , 121

31'

yyy

yyxyyx

yxy

yxx

19

• We have the following 2nd order Initial Value Problem:

• Let us make substitutions:

• Then the equation becomes:

On the other hand, suppose

3)0(' ,1)0( ;0)(2)('3)("

yytytyty

x t y t x t y t1 2( ) ( ) ( ) ' ( ), and

20

• Thus a 2nd order equation is equivalent to a system of 1st order equations in two unknowns.

A system of first order equations

x t x tx t x t x t

x x

1 2

2 2 1

1 2

3 2

0 1 0 3

'( ) ( )' ( ) ( ) ( ),

( ) , ( )

with the initial conditions become:

and

21

• Let us consider an example: solve the system

General Method of Solving System of equations: is the Elimination Method.

.10)(7)(4)(',1)(4)(3)('

ttytxtytytxtx

22

• We want to solve these two equations simultaneously, i.e.

• find two functions x(t) and y(t) which will satisfy the given equations simultaneously

• There are many ways to solve such a system.• One method is the following: let D = d/dt,• then the system can be rewritten as:

23

(D - 3)[x] + 4y = 1, …..(*)-4x + (D + 7)[y]= 10t .…(**)

• The expression 4(*) + (D - 3)(**) yields:• {16 + (D - 3)(D + 7)}[y] = 4+(D - 3)(10t), or

(D2 + 4D - 5)[y] =14 - 30t. This is just a 2nd order nonhomogeneous equation.

• The corresponding auxiliary equation is • r2 + 4r - 5 = 0, which has two solution r = -5, and

r = 1, thus yh = c1e -5t + c2 e t. And the

• general solution is y = c1e -5t + c2 e t + 6t + 2.• To find x(t), we can use (**).

24

To find x(t), we solve the 2nd eq.Y(t) = 4x(t) - 7y(t)+ 10t for x(t),

• We obtain:

, 58221

}10]26[7]6)5{[(41

}10)(7)('{41)(

25

1

25

125

1

tecec

ttecececec

ttytytx

tt

tttt

25

Generalization• Let L1, L2, L3, and L4 denote linear differential

operators with constant coefficients, they are polynomials in D. We consider the 2x2 general system of equations:

L x L y fL x L y f

L L L Li j j i

1 2 1

3 4 2

[ ] [ ][ ] [ ]

, .........(1) .........(2)

Since , we can solve the

above equations by first eliminating the varible y. And solve it for x. Finallysolve for y.

26

• Rewrite the system in operator form:• (D2 - 1)[x] + (D + 1)[y] = -1, .……..(3)• (D - 1)[x] + D[y] = t2 ……………...(4)

• To eliminate y, we use D(3) - (D + 1)(4) ;• which yields:• {D(D2 - 1) - (D + 1)(D - 1)}[x] = -2t - t2. Or • {(D(D2 - 1) - (D2 - 1)}[x] = -2t - t2. Or• {(D - 1)(D2 - 1)}[x] = -2t - t2.

Example:x t y t x t y t

x t y t x t t

"( ) ' ( ) ( ) ( ) ,

' ( ) ' ( ) ( ) .

12

27

• Which implies r = 1, 1, -1. Hence the general solution to the homogeneous equation is

• xh = c1e t + c2te t + c3e -t.• Since g(t) = -2t - t2, we shall try a particular

solution of the form : • xp = At2 + Bt + C, we find A = -1, B = -4, C = -6,

• The general solution is x = xh + xp.

The auxiliary equation for the corresponding homogeneous eq. is

(r - 1)(r2 - 1) = 0

28

• Which implies y = (D - D2)[x] -1 - t2.

To find y, note that (3) - (4) yields : (D2 - D)[x] + y = -1 - t2.

29

• This is simply a mapping of functions to functions

• This is an integral operator.

Chapter 7: Laplace Transforms

f FLLf F

30

• Definition: Let f(t) be a function on [0, ). The Laplace transform of f is the function F defined by the integral

• The domain of F(s) is all values of s for which the integral (*) exists. F is also denoted by L{f}.

More precisely

.)(:)( (*)0

dttfesF st

31

Example• 1. Consider f(t) = 1, for all t > 0. We have

integral.improperan is Transform Laplace The F(s).) of

domain on the (Note .0 allfor holds

11limlim

lim)1()(

0

00

sthis

sse

sse

dtedtesF

tN

N

Nst

N

N st

N

st

32

Other examples,• 2. Exponential function f(t) = e t .• 3. Sine and Cosine functions say: f(t) = sin ßt,• 4. Piecewise continuous (these are functions

with finite number of jump discontinuities).

. 3t2 , 2)-(t

2,t1 ,2 )(,10 ,

2

tftt

33

Example 4, P.375

• A function is piecewise continuous on [0, ), if it is piecewise continuous on [0,N] for any N > 0.

.10 ,

,105 ,0,50 ,2

)(

of tranformLaplace theDetermine

4 te

tt

tft

34

Function of Exponential Order • Definition. A function f(t) is said to be of

exponential order if there exist positive constants M and T such that

• That is the function f(t) grows no faster than a function of the form

• For example: f(t) = e 3t cos 2t, is of order = 3.

. allfor , )( (*) TtMetf t

. tMe

35

Existence Theorem of Laplace Transform.

• Theorem: If f(t) is piecewise continuous on [0, ) and of exponential order , then L{f}(s) exists for all s > .

• Proof. We shall show that the improper integral converges for s > . This can be seen easily, because [0, ) = [0, T] [ T, ). We only need to show that integral exists on [ T, ).

36

A table of Laplace Transforms can be found on P. 380

• Remarks: • 1. Laplace Transform is a linear operator. i.e.

If the Laplace transforms of f1 and f2 both exist for s > , then we have L{c1 f1 + c2 f2} = c1 L{f1 } + c2 L{f2 } for any constants c1 and c2 .

• 2. Laplace Transform converts differentiation into multiplication by “s”.

37

Properties of Laplace Transform

• Recall :

• Proof.

. by shift) (aion translat toby

tionmultiplica transform that means This. )())}(({ (1)

have we, allfor then , for exists )()}({ If :Theorem

. )()}({ 0

ae

LasFstfeL

asssFsfL

dttfesfL

at

at

st

38

How about the derivative of f(t)?

Proof. . )0()}({)}('{ (2)

have we, sfor Then . order lexponentiaof ' and both with , )[0,on continuous piecewise be ' and )[0,on continuous be )(Let :Theorem

fsfsLsfL

ff(t)ftf

39

Generalization to Higher order derivatives.

).0()0(')0()}({)}({

,n on induction by , have wegeneralIn . )0(')0()}({

)0(')0()}({ )0(')}('{)}("{

teasily tha see we, Since

)1()2()1()(

2

nnnnn ffsfssfLssfL

fsfsfLs

ffsfsLsfsfsLsfL

(f ')'f "

40

Derivatives of the Laplace Transform

Proof.

).}{(1 1 have we

, sfor then ,Let . order lexponentia of) [0,on continuous piecewise is )( Suppose :

(s)fLdsd)((s)

dsFd)(f(t)}(s) L{t

L{f}(s)F(s)tfTheorem

n

nn

n

nnn

41

• 1. e -2t sin 2t + e 3t t2.

• 2. t n.

• 3. t sin (bt).

Some Examples.