Transcript Superposition Principle & the Method of Undetermined

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 y"by' ky  Fe (t ), where
m : is themass attachedto thespring
k : is theHooke's constant(stiffness)
b : theis thedampingcoefficient (friction)
Fe : all otherexternalforcesto thesystem.
y : is thedisplacement fromequilibrium.
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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
k
y" y  0, where 
, a generalsolut ion
m
is clearly: y (t )  c1 cos t  c2 sin  t . Whichcan
2
also be writt enas : y (t )  A sin( t   .), with
A  c1  c2
2
2
and t an 
c1
.
c2
T hisis called Simple harmonicmot ion with
2

period
, and frequency
.

2
 is known as theangular frequency.
2
When b  0, but Fe = 0, we have
damping on free vibrations.
• The D. E. in this case is:
m y"by' ky  0 , and theauxiliaryeq. is
m r2  br  k  0 , therootsare easily seen as
b
1
2


b  4m k . T hesolutionclearly
2m 2m
depends on thediscriminant b 2  4m k .
We have threepossible cases :
3
Case I: Underdamped Motion (b2 < 4mk)
b
1
We let :  
and  
4m k  b 2 , hence
2m
2m
the rootsare   i . A generalsolutionis
y (t )  e t (c1 cos  t  c2 sin  t ) or y (t )  Ae t sin( t   ).
T hisis a sinusoidal wave with a dampingfactor Ae t .
4
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:
y (t )  c1e r1t  c2 e r2t  0 as t   . Since
y ' (t )  0 has at most one zero,it follows t hat
t hesolut iondoes not oscillat e.
One local max
No local
max or min
One 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:
y (t )  c1e
 bt / 2 m
 c2te
 bt / 2 m
. Again we see
that lim y (t )  0 as t  , and
y ' (t )  0 has at most onesolution,its
solutionsbehavesimilar tothatof Overdamped
motions.
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Example
• The motion of a mass-spring system with damping
is governed by
y" (t )  by' (t )  64 y (t )  0 ;
y (0)  1 , and y ' (0)  0 .
• This is exercise problem 4, p239.
• Find the equation of motion and sketch its graph
for b = 10, 16, and 20.
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Solution.
• 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
y (t )  e



5t
(cos 39 t 
5
sin
39
39 t )
8  5t
e sin( 39 t   ) , where
39 
c1
t an 

c2
39
.
5
8
When b = 16, b2 - 4mk = 0, we
have repeated root -8,
• thus the solution to the I.V.P is
y(t )  (1  8t )e
8t
y
1
t
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When b = 20, b2 - 4mk = 64, thus
two distinct real roots are
• r1 = - 4 and r2 = -16, the solution to the I.V.P. is:
 4  4t  1  16t
y(t )   e   e . T hegraph lookslike :
 3
 3
y
1
t
1
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Next we consider forced vibrations
• with the following D. E.
m y"by' ky  F0 cos t .
(*)
We assume that F0  0 ,   0 and
0  b  4m k (i.e.underdamped).
Solution to (*) can be writtenin theform
y  yh  y p , where
2
y p standsfor a particularsolution ot (*),&
yh is a genernalsolutionof theassociated
homogenenous equationof (*).
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We know a solution to the above
equation has the form
y  yh  y p
• where:
2


4
m
k

b
( b / 2 m ) t
y h (t )  Ae
sin 
t   ,


2
m


and by themethodof undetermined coefficients,
we know that
y p (t )  B1 cos  t  B2 sin  t ,
• In fact, we have
2
F0 (k  m  )
F0b
B1 
, B2 
.
2 2
2 2
2 2
2 2 12
(k  m  )  b 
(k  m  )  b 
2
Thus in the case 0 < b < 4mk
(underdamped), a general
solution has the form:
y (t )  yh (t )  y p (t ) , where
2


4
m
k

b
(b / 2 m )t
yh (t )  Ae
sin 
t    , and


2
m


F0
y p (t ) 
sin( t   ).
2 2
2
2
(k  m  )  b 
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Remark on Transient and SteadyState solutions.
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Introduction
• Consider the following interconnected fluid
tanks
B
A
8 L/min
6 L/min
X(t)
Y(t)
24 L
24 L
6 L/min
X(0)= a
Y(0)= b
2 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:
dx 2
8

y (t )  x(t )
dt 24
24
• and for tank B, we have
dy 8
2
6
 x(t )  y (t )  y (t ).
dt 24
24
24
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This gives us a system of First
Order Equations
1
1
x'   x  y , and
3
12
1
1
y '  x  y . T hisis equivalent to a
3
3
2nd order equation: since x  3 y ' y implies
x'  3 y" y ' , put themintofirst equation,we get
1
3 y"2 y ' y  0.
4
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On the other hand, suppose
• We have the following 2nd order Initial Value
Problem:
y" (t )  3 y' (t )  2 y(t )  0;
y(0)  1, y' (0)  3
• Let us make substitutions:
x1 (t )  y(t ) and x2 (t )  y'(t ),
• Then the equation becomes:
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A system of first order equations
x1 '(t )  x2 (t )
x2 '(t )  3x2 (t )  2 x1 (t ), with the
initial conditions become:
x1 (0)  1, and x2 (0)  3
• Thus a 2nd order equation is equivalent to a
system of 1st order equations in two unknowns.
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General Method of Solving System of
equations: is the Elimination Method.
• Let us consider an example: solve the system
x' (t )  3x(t )  4 y (t )  1,
y ' (t )  4 x(t )  7 y (t )  10t.
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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.
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• To find x(t), we can use (**).
To find x(t), we solve the 2nd eq.
Y(t) = 4x(t) - 7y(t)+ 10t for x(t),
• We obtain:
1
x(t )  { y ' (t )  7 y (t )  10t}
4
1
 {[(5)c1e 5t  c2 et  6]  7[c1e 5t  c2et  6t  2]  10t}
4
1 5t
 c1e  2c2et  8t  5 ,
2
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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:
L1[ x ]  L2 [ y ]  f 1 ,
L3 [ x ]  L4 [ y ]  f 2
.........(1)
.........(2)
Since Li L j  L j Li , we can solve the
above equations by first eliminating the
varible y. And solve it for x. Finally
solve for y.
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Example:
•
•
•
•
•
•
•
•
x"(t )  y' (t )  x(t )  y(t )  1,
x ' (t )  y ' (t )  x (t )  t 2 .
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.
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The auxiliary equation for the
corresponding homogeneous eq. is
(r - 1)(r2 - 1) = 0
• 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.
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To find y, note that (3) - (4)
yields : (D2 - D)[x] + y = -1 - t2.
• Which implies y = (D - D2)[x] -1 - t2.
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Chapter 7: Laplace Transforms
• This is simply a mapping of functions to
functions
ff
L
F
F
• This is an integral operator.
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More precisely
• Definition: Let f(t) be a function on [0, ).
The Laplace transform of f is the function
F defined by the integral
(*)

F (s) :  e f (t )dt.
 st
0
• The domain of F(s) is all values of s for
which the integral (*) exists. F is also
denoted by L{f}.
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Example
• 1. Consider f(t) = 1, for all t > 0. We have

F ( s)   e
0
 st
N
 (1)dt  lim  e dt
 st
N  0
 st N
N t

e
1 e  1
 
 lim
 lim  
N 
s 0 N  s
s  s
this holds for all s  0. (Noteon thedomain
of F(s).) T heLaplaceT ransformis an improper
integral.
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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).
0  t  1,
1  t  2,
t,
f (t )  2,
2
(t - 2) ,
2  t  3.
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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.
Det erminet heLaplacet ranformof
2,
0  t  5,

f (t )  0, 5  t  10,
e 4t , 10  t.

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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
(*)
f (t )  Me t , for all t  T .
• That is the function f(t) grows no faster than a
function of the form Me t .
• For example: f(t) = e 3t cos 2t, is of order  = 3.
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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”.
36
Properties of Laplace Transform
• Recall :

L{ f }( s )   e
0
 st
f (t )dt .
T heorem: If L{ f }( s )  F ( s ) existsfor
s   , thenfor all s    a , we have
(1)
L{e at f (t )}(s )  F ( s  a) .
T hismeans that L transformmultiplication
by e at to translation (a shift) by a .
• Proof.
37
How about the derivative of f(t)?
T heorem: Let f (t ) be continuouson [0,) and f ' (t) be
piecewise continuouson [0,) , with both f and f ' of
exponentia
l order  . T henfor s   , we have
(2)
L{ f '}( s)  sL{ f }( s)  f (0) .
Proof.
38
Generalization to Higher order
derivatives.
Since f "  (f ')', we see easily that
L{ f "}( s)  sL{ f '}( s)  f ' (0)
 ssL{ f }( s)  f (0)  f ' (0)
 s 2 L{ f }( s)  sf (0)  f ' (0) .
In general we have, by induction on n ,
L{ f ( n ) }( s)  s n L{ f }( s)  s ( n 1) f (0)  s ( n  2) f ' (0)    f ( n 1) (0).
39
Derivatives of the
Laplace Transform
Theorem: Suppose f (t ) is piecewise continuouson [0,  )
of exponential order  . Let F(s)  L{f}(s), thenfor s   ,
we have
n
n
d
F
d
L{t n f(t)}(s)  (  1 )n n (s)  (  1 )n n ( L{ f }(s)).
ds
ds
Proof.
40
Some Examples.
• 1. e -2t sin 2t + e 3t t2.
• 2. t n.
• 3. t sin (bt).
41