INVESTIGACIN DE OPERACIONES
EMPLEO
DE
MTODOS
CUANTITATIVOS
EN
LA
DEFINICIN DE ALTERNATIVAS DE SOLUCIN. ESTUDIO
DE CASO POLIGOM
Resumen / Abstract
El presente trabajo muestra un caso de estudio sobre l
PLANEACIN DE EVALUACIN: QUMICA
UNIDAD 4- ANLISIS DE RIESGOS Y PUNTOS CRTICOS DE CONTROL (ARYPCC)
ACTIVIDADES
Individualmente realizar
Investigacin de los
orgenes y aplicaciones del
HACCP e identifica
=
12
m(
km
)A2 sin2(t + )
=
12
kA2 sin2(t + ) (854)
The total energy is thus:
E=
12
kA2 sin2(t + ) + 1 2
kA2 cos2(t + )
=
12
kA2 (855)
and is constant in time! Kinda spooky how that works out.
Note th
also have: d2xI dt2 + 20 dxI dt + 2 0xI = F0 m cos(t) (900) (the
inhomogeneous ODE for a damped harmonically driven SHO) with
solution xI(t) yet to be determined. Suppose also that we have found at
le
(or general complex) position!. So we take the real part of either of
these solutions: x+(t) = |A+|e+i(t+) = |A+|(cos(t + +) +
isin(t + +) = |A+|cos(t + +) (814) and x(t) = |A|
ei(t+) = |A|(cos(t + )i
x+(t) = A+e+it (805) x(t) = Aeit (806) that follow, one for each
possible value of . Note that we will always get n independent
solutions for an nth order linear ODE, because we will always have
solve
of sharp bumps could leave your shocks still compressed and unable to
absorb the impact of the last one and keep your tires still on the
ground.
Damped oscillation is ubiquitous. Pendulums, once start
experimental results in an understandable if approximate way, but
hardly exact.
The exact (relevant) laws of nature are those of electromagnetism and
quantum mechanics, where the many electron problem
and the moment of inertia (found using the parallel axis theorem) to
obtain the frequency of the oscillator183. 183I know, I know, you had
hoped that you could nally forget all of that stressful stu w
In this course we will study simple harmonic oscillators, both with and
without damping (and harmonic driving) forces. The principle
examples we will study will be masses on springs and various pendul
This equation is maximum when = 0, at resonance. At that point
the peak average power can be written: Pavg(0) = Pmax = F2 0
4m0 = F2 0 2b (912)
d) This shape of Pavg() is very important to semi-quanti
cosine or sine solution instead of an exponential solution at the
beginning of the previous section.
Note Well! An unfortunately commonly made mistake for SHO
problems is for students to take Fx = ma
= x(0) = 0.1 m and v0 = 1 m/sec. Then A = rx2 0 + v2 0 2 =
p(0.01 + 0.01) = 0.12 m. Also = tan11 1= tan1(1) = 4
(calculator result) or 5 4 or 3 4 . In this case either: x(t) =
+0.12cos(10t + 4 ) (830)
xed standard pace, would consistently cover twenty miles in ve
summer hours, or by increasing the length of their pace slightly,
twenty four miles in the same number of hours. Note well that the mile
where:
A=
F0/m p(2 2 0)2 + (20)2
(904)
and the phase angle is:
= tan1 20 2 2 0 (905) b) The average power delivered to
the mass by the driving force is a quantity of great interest. This is the
rate
410 Week 9: Oscillations
m
l
Figure 123: A simple pendulum is just a point like mass suspended on
a long string and displaced sideways by a small angle. We will assume
no damping forces and that there
point is that one can always make cos() and sin() change sign by
adding or subtracting , so that = + (for example), and as long
as you change the sign of A as well, you will get exactly the same
physi
Since we are going to quite a bit with harmonic oscillators from now
on, we should take a few moments to plot x(t), v(t), and a(t). We
remarked above that omega had to have units of t1. The following
-0.5
0
0.5
1
1.5
t
x
Figure 126: Two identical oscillators, one undamped (with 0 = 2, or
if you prefer with an undamped period T0 = 1) and one weakly
damped (b/m = 0.3).
Several of these properties ar
time almost precisely matches the human heartbeat. Unfortunately,
the normal human heartbeat varies by a factor of around three as
one moves from resting to heavy exercise, a range that is further
inc
You should now be able to see that in fact, the total energy of the
oscillator is constant in the small angle approximation.
Of course, it is actually constant even for large oscillations, but
proving
build up much amplitude. If you apply the torques with the wrong
phase you will not manage to get the same amplitude that youd get
pumping in phase with the motion.
Thats really it, qualitatively. Res
constants of integration, just as one can for ordinary 1D kinematic
solutions).
Lets see how to accomplish this algebraically. Suppose we are given
x(0) = x0 and v(0) = v0 the initial position and vel
In principle, if you have satised the prerequisites for this course you
have almost certainly studied imaginary numbers and complex
numbers176 in a high school algebra class and perhaps again in
colle
times that you can immediately recognize them when they are given in
a problem and just write down the answer. They are the four
possibilities:
Week 9: Oscillations 407
Example 9.1.5: Given: x0 > 0, v
9.5.1: Simple Models for Molecular Bonds
Consider, then a very crude microscopic model for a solid. We know
that this solid is made up of many elementary particles, and that those
elementary particles
U(r)
Figure 130: Two generic classical potential energy functions
associated with atomic bonds on a common scale Umin = 1, rb = 1.0,
and a = 6.2, the latter a value that makes the two potential have
r
As before, we exclude the trivial solution x(t) = 0 as being too boring
to solve for (requiring that A 6= 0, that is) and are left with the
characteristic equation for : 2 + b m + k m = 0 (885) This
q