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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:. In the simplest case, when b = 0, and F e = 0, i.e. Undamped, free vibration, we can rewrite the D.E:. As . - PowerPoint PPT Presentation
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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.