Title: On Free Mechanical Vibrations
1On 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
2In the simplest case, when b 0, and Fe 0,
i.e. Undamped, free vibration, we can rewrite the
D.E
3When b ? 0, but Fe 0, we have damping on free
vibrations.
- The D. E. in this case is
4Case I Underdamped Motion (b2 lt 4mk)
5Case 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
6Case 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.
8Solution.
- 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
9When b 16, b2 - 4mk 0, we have repeated
root -8,
- thus the solution to the I.V.P is
y
1
t
10When 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
11Next we consider forced vibrations
12We know a solution to the above equation has the
form
13Thus in the case 0 lt b 2 lt 4mk (underdamped), a
general solution has the form
14Remark on Transient and Steady-State solutions.
15Introduction
- 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
16Suppose 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.
17Set up the differential equations
- For tank A, we have
- and for tank B, we have
18This gives us a system of First Order Equations
19On the other hand, suppose
- We have the following 2nd order Initial Value
Problem - Let us make substitutions
- Then the equation becomes
20A system of first order equations
- Thus a 2nd order equation is equivalent to a
system of 1st order equations in two unknowns.
21General 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 ().
24To find x(t), we solve the 2nd eq.Y?(t) 4x(t)
- 7y(t) 10t for x(t),
25Generalization
- 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
26Example
- 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.
27The 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.
28To find y, note that (3) - (4) yields (D2 -
D)x y -1 - t2.
- Which implies y (D - D2)x -1 - t2.
29Chapter 7 Laplace Transforms
- This is simply a mapping of functions to
functions - This is an integral operator.
F
L
f
f
F
30More 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.
31Example
- 1. Consider f(t) 1, for all t gt 0. We have
32Other 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).
33Example 4, P.375
- A function is piecewise continuous on 0, ?), if
it is piecewise continuous on 0,N for any N gt
0.
34Function 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.
35Existence 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, ?).
36A 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.
37Properties of Laplace Transform
38How about the derivative of f(t)?
39Generalization to Higher order derivatives.
40Derivatives of the Laplace Transform
41Some Examples.
- 1. e -2t sin 2t e 3t t2.
- 2. t n.
- 3. t sin (bt).