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: my "
 by '
 ky

F e
( t ) , where m : is the mass attached to the spring k : is the Hooke' s constant (stiffness ) b : the is the damping coefficien t (friction)
F e
: all other external forces to the system.
y : is the displaceme nt from equilibriu m.
1

In the simplest case, when b = 0, and F e
= 0, i.e. Undamped, free vibration, we can rewrite the D.E:
• As y "
 
2 y

0 , where
  k
, a general solution m is clearly : y ( t )
 c
1 cos
 t
 c
2 sin
 t . Which can also be written as : y ( t )

A sin(
 t
 
.), with
A
 c
1
2  c
2
2 and tan
  c c
2
1 .
This is called Simple harmonic motion wit period

2


 is known
, and frequency as the angular


2

. frequency.
h
2

When b

0, but F e
= 0, we have damping on free vibrations.
• The D. E. in this case is: my "
 by '
 ky

0 , and the auxiliary eq.
is mr
2  br
 k

0 , the roots are easily seen as
 b
2 m

1
2 m b
2 
4 mk . The solution clearly depends on the discrimina nt b
2 
4 mk .
We have three possible cases :
3

Case I: Underdamped Motion (b 2 < 4mk)
We let the
:
 
 b
2 roots are m
 and
 i

 
2
1 m
. A general
4 mk
 b
2 solution is
, hence y ( t )
 e
 t
( c
1 cos
 t
This is a sinusoidal
 c
2 sin
 t ) or y ( t )

Ae
 t sin(
 t
wave with a damping factor Ae
 t
.
 
).
4

Case II: Overdamped Motion (b 2 > 4mk)
• In this case, we have two distinct real roots, r
1
& r
2
Clearly both are negative, hence a general solution:
. y ( t ) y ' ( t )

 c
1 e r
1 t
0
 has c
2 e r
2 t  at most
0 as t
 
.
Since one zero, it follows that the solution does not oscillate.
One local max
No local max or min
One local min
5

Case III: Critically Damped Motion (b 2 = 4mk)
• We have repeated root -b/2m. Thus the a general solution is: y ( t ) that
 c
1 e
 bt / 2 m lim y ( t )


0 c
2 te
 bt as t
/ 2 m
.
Again we
 
, and see y ' ( t )

0 has at most one solution, its solutions behave similar to that of Overdamped motions.
6

Example
• The motion of a mass-spring system with damping is governed by y " ( t )
 by ' ( t )

64 y ( t ) y ( 0 )

1 , and y ' ( 0 )

0

0 .
;
• This is exercise problem 4, p239.
• Find the equation of motion and sketch its graph for b = 10, 16, and 20.
7

Solution.
• 1. b = 10: we have m = 1, k = 64, and
• b 2 - 4mk = 100 - 4(64) = - 156, implies

= (39) 1/2 .
Thus the solution to the I.V.P. is y ( t )
 e

5 t
(cos 39 t

5
39 sin 39 t )



8
39

 e

5 t sin( 39 t
 
) , where tan
  c
1 c
2

39
.
5
8

When b = 16 , b 2 - 4mk = 0 , we have repeated root -8 ,
• thus the solution to the I.V.P is y ( t )

( 1

8 t ) e

8 t y
1 t
9

When b = 20, b 2 - 4mk = 64, thus two distinct real roots are
• r
1
= - 4 and r
2
= -16, the solution to the I.V.P. is: y ( t )

4
3 e

4 t 
1
3

 e

16 t
.
The graph looks like : y
1
1 t
10

Next we consider forced vibrations
• with the following D. E.
(*) my "
We
 assume by '
 ky
that F
0


F
0
0 cos

,
  t .
0 and
0
 b
2 
4 mk (i.e.
underdampe d).
Solution t y
 y h o (*) can be written in the form
 y p
, where y p stands for a particular solution t o (*), & y h is a genernal solution of the associated homogeneno us equation of (*).
11

We know a solution to the above equation has the form y
 y h
 y p
• where: y h
( t )

Ae

( b / 2 m ) t sin




4 mk
 b
2
2 m t
  



, and by the method of undetermin ed coefficien ts, we know that y p
( t )

B
1 cos
 t

B
2 sin
 t ,
• In fact, we have
B
1

( k
F
0

( k m


2 m
)
2 

2
) b
2

2
, B
2

( k
 m

F
0 b

2
)
2  b
2

2
.
12

Thus in the case 0 < b 2 < 4mk
(underdamped), a general solution has the form: y ( t )
 y h
( t )
 y p
( t ) , where y h
( t )

Ae

( b / 2 m ) t sin


4 mk
 b
2
2 m t
  



, and y p
( t )

( k
 m

F
0
2
)
2  b
2

2 sin(
 t
 
).
13

Remark on Transient and Steady-
State solutions.
14

Introduction
• Consider the following interconnected fluid tanks
6 L/min
A
X(t)
24 L
8 L/min
B
Y(t)
24 L
6 L/min
X(0)= a Y(0)= b
2 L/min
15

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.
16

Set up the differential equations
• For tank A, we have: dx dt

2
24 y ( t )

8
24 x ( t )
• and for tank B, we have dy dt

8
24 x ( t )

2
24 y ( t )

6
24 y ( t ).
17

This gives us a system of First
Order Equations x '
 
1
3 x

1
12 y , and y '

1
3 x

1
3 y .
This is equivalent to a
2nd order equation : since x

3 y '
 y implies x '

3 y "
 y ' , put them into first equation, we get
3 y "

2 y '

1
4 y

0 .
18

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: x t
1
 y t and x t
2
( )

' ( ),
• Then the equation becomes:
19

A system of first order equations x
1
' ( )
 x t
2
( ) x
2
' ( )
 
3
2
( )

2
1
( ), with the initial conditions become: x
1
0

1 , and x
2
0

3
• Thus a 2nd order equation is equivalent to a system of 1st order equations in two unknowns.
20

General Method of Solving System of equations: is the Elimination Method .
• Let us consider an example: solve the system x ' ( t )

3 x ( t )

4 y ( t )

1 , y ' ( t )

4 x ( t )

7 y ( t )

10 t .
21

• 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:
22

(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
(D 2 + 4D - 5)[y] =14 - 30t. This is just a 2nd order nonhomogeneous equation.
• The corresponding auxiliary equation is
• r 2 + 4r - 5 = 0, which has two solution r = -5, and r = 1, thus y h
= c
1 e -5t + c
2 e
• general solution is y = c
1 e -5t + c
2 t . And the e t + 6t + 2.
• To find x(t), we can use (**).
23

To find x(t), we solve the 2nd eq.
Y

(t) = 4x(t) - 7y(t)+ 10t for x(t),
• We obtain: x ( t )

1
4
{ y ' ( t )

7 y ( t )

10 t }

1
4
{[(

5 ) c
1 e

5 t  c
2 e t 
6 ]

7 [ c
1 e

5 t  c
2 e t 
6 t

2 ]

10 t }

1
2 c
1 e

5 t 
2 c
2 e t 
8 t

5 ,
24

Generalization
• Let
L
1
, L
2
, L
3
, and L
4 denote linear differential operators with constant coefficients, they are polynomials in D. We consider the 2x2 general system of equations:
L x
1

L
2
[ ]
 f
1
, .........(1)
L x
3

L
4
[ ]
 f
2
Since L L i j

L L j i
.........(2)
, we can solve the above equations by first eliminating the varible y. And solve it for x. Finally solve for y.
25

Example: x t
 y t
 x t
 y t x t
 y t
 x t
 t
2
.
 
1 ,
• Rewrite the system in operator form:
• (D 2 - 1)[x] + (D + 1)[y] = -1, .……..(3)
• (D - 1)[x] + D[y] = t 2 ……………...(4)
• To eliminate y, we use D(3) - (D + 1)(4) ;
• which yields:
• {D(D 2 - 1) - (D + 1)(D - 1)}[x] = -2t - t 2 . Or
• {(D(D 2 - 1) - (D 2 - 1)}[x] = -2t - t 2 . Or
• {(D - 1)(D 2 - 1)}[x] = -2t - t 2 .
26

The auxiliary equation for the corresponding homogeneous eq. is
(r - 1)(r 2 - 1) = 0
• Which implies r = 1, 1, -1. Hence the general solution to the homogeneous equation is
• x h
= c
1 e t + c
2 te t + c
3 e -t .
• Since g(t) = -2t - t 2 , we shall try a particular solution of the form :
• x p
= At 2 + Bt + C, we find A = -1, B = -4, C = -6,
• The general solution is x = x h
+ x p
.
27

To find y, note that (3) - (4) yields : (D 2 - D)[x] + y = -1 - t 2 .
• Which implies y = (D - D 2 )[x] -1 - t 2 .
28

Chapter 7: Laplace Transforms
• This is simply a mapping of functions to functions f f
L
• This is an integral operator .
F
F
29

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 ) :
 
0
  e st f ( t ) dt .
• The domain of F(s) is all values of s for which the integral (*) exists.
F is also denoted by L {f}.
30

Example
• 1. Consider f(t) = 1, for all t > 0. We have
F ( s )
 
0
 e
 st 
( 1 ) dt

N lim
 

0
N e
 st dt

N lim
 
 e s
 st this holds for all s
0
N

N lim
  


1 s
 e

N t s 



1 s

0 .
(Note on the domain of F(s).) The Laplace Transform is an improper integral.
31

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). f ( t ) t , 0

2 , 1
 t
 t

1 ,

2,
(t 2)
2
, 2
 t

3 .
32

Example 4, P.375
• A function is piecewise continuous on [0,

), if it is piecewise continuous on [0,N] for any N > 0.
Determine the Laplace tranform of f ( t )


2 , 0


0 , 5 e
4 t
, 10


 t t t .


5 ,
10 ,
33

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.
34

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,

).
35

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 f
1 and f
2 both exist for s >

, then we have
L {c
1 f
1
+ c
2 f
2
} = c
1
L {f any constants c
1 and c
2
1
} + c
2
.
L {f
2
} for
• 2.
Laplace Transform converts differentiation into multiplication by “s”.
36

Properties of Laplace Transform
• Recall :
L { f
Theorem
}( s )
: If L { f
 
0
  e st
}( s )
 f ( t ) dt .
F ( s ) exists for s
 
, then for all s
   a , we have
(1) L { e at f ( t )}( s )

F ( s
 a ) .
This means that L transform multiplica tion by e at
to translat ion (a shift) by a .
• Proof.
37

How about the derivative of f(t)?
Theorem : Let f ( t ) be continuous on [0,

) and f ' (t) be piecewise continuous exponentia on [0,

) , with both f and f l order

. Then for s
 
, we have
(2) L { f ' }( s )
 sL { f }( s )
 f ( 0 ) .
' of
Proof.
38

Generalization to Higher order derivatives.
Since f "

(f ')' , we see easily tha
L { f " }( s )

 sL { s
 f sL {
' }( f s )
}( s

)
 f ' f
( 0 )
( 0 )

 t 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 continuous of exponentia l order

.
Let F(s)
 on [0,

)
L{f}(s) , then for s
 
, we have L{t n f(t)}(s)

(

1 ) n d n
F
(s)

( ds n d n

1 ) n ds n
( L { f } (s) ).
Proof.
40

Some Examples.
• 1. e -2t sin 2t + e 3t t 2 .
• 2. t n .
• 3. t sin (bt).
41