on free mechanical vibrations
DESCRIPTION
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 PresentationTRANSCRIPT
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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:
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
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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
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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
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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
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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
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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
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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
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• 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
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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
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• 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
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• 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
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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
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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
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Remark on Transient and Steady-State solutions.
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• 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
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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.
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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
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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
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• 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
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• 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
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• 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
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• 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:
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(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 (**).
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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
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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.
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• 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
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• 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
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• Which implies y = (D - D2)[x] -1 - t2.
To find y, note that (3) - (4) yields : (D2 - D)[x] + y = -1 - t2.
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• This is simply a mapping of functions to functions
• This is an integral operator.
Chapter 7: Laplace Transforms
f FLLf F
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• 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
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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
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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
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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
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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
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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, ).
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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”.
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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
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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
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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 "
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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
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• 1. e -2t sin 2t + e 3t t2.
• 2. t n.
• 3. t sin (bt).
Some Examples.