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On Free Mechanical Vibrations

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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 ... – PowerPoint PPT presentation

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Title: On Free Mechanical Vibrations


1
On Free Mechanical Vibrations
  • As derived in section 4.1( following Newtons 2nd
    law of motion and the Hookes law), the D.E. for
    the mass-spring oscillator is given by

2
In the simplest case, when b 0, and Fe 0,
i.e. Undamped, free vibration, we can rewrite the
D.E
  • As

3
When b ? 0, but Fe 0, we have damping on free
vibrations.
  • The D. E. in this case is

4
Case I Underdamped Motion (b2 lt 4mk)
5
Case II Overdamped Motion (b2 gt 4mk)
  • In this case, we have two distinct real roots, r1
    r2. Clearly both are negative, hence a general
    solution

One local max
One local min
No local max or min
6
Case III Critically Damped Motion (b2 4mk)
  • We have repeated root -b/2m. Thus the a general
    solution is

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.

8
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

9
When b 16, b2 - 4mk 0, we have repeated
root -8,
  • thus the solution to the I.V.P is

y
1
t
10
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

y
1
t
1
11
Next we consider forced vibrations
  • with the following D. E.

12
We know a solution to the above equation has the
form
  • where
  • In fact, we have

13
Thus in the case 0 lt b 2 lt 4mk (underdamped), a
general solution has the form
14
Remark on Transient and Steady-State solutions.
15
Introduction
  • Consider the following interconnected fluid tanks

B
A
8 L/min
Y(t)
X(t)
24 L
6 L/min
24 L
6 L/min
X(0) a
Y(0) b
2 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 gt
0.
17
Set up the differential equations
  • For tank A, we have
  • and for tank B, we have

18
This gives us a system of First Order Equations
19
On the other hand, suppose
  • We have the following 2nd order Initial Value
    Problem
  • Let us make substitutions
  • Then the equation becomes

20
A system of first order equations
  • Thus a 2nd order equation is equivalent to a
    system of 1st order equations in two unknowns.

21
General Method of Solving System of equations is
the Elimination Method.
  • Let us consider an example solve the system

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

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

26
Example
  • Rewrite the system in operator form
  • (D2 - 1)x (D 1)y -1, ...(3)
  • (D - 1)x Dy 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.

27
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.

28
To find y, note that (3) - (4) yields (D2 -
D)x y -1 - t2.
  • Which implies y (D - D2)x -1 - t2.

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

F
L
f
f
F
30
More precisely
  • 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
    Lf.

31
Example
  • 1. Consider f(t) 1, for all t gt 0. We have

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).

33
Example 4, P.375
  • A function is piecewise continuous on 0, ?), if
    it is piecewise continuous on 0,N for any N gt
    0.

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.

35
Existence Theorem of Laplace Transform.
  • Theorem If f(t) is piecewise continuous on
    0, ?) and of exponential order ?, then Lf(s)
    exists for all s gt ? .
  • Proof. We shall show that the improper integral
    converges for s gt ? . 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 gt ? , then we have
    Lc1 f1 c2 f2 c1
    Lf1 c2 Lf2 for any constants c1 and c2 .
  • 2. Laplace Transform converts differentiation
    into multiplication by s.

37
Properties of Laplace Transform
  • Recall
  • Proof.

38
How about the derivative of f(t)?
39
Generalization to Higher order derivatives.
40
Derivatives of the Laplace Transform
41
Some Examples.
  • 1. e -2t sin 2t e 3t t2.
  • 2. t n.
  • 3. t sin (bt).
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