Download The Damped Nonlinear Pendulum

PERTEMUAN XVII-XVIII
The Damped Nonlinear Pendulum
Growth and decay models
Simple exponential growth models have the differential equation
p'( t ) = kp( t ), p( 0 ) = p0.
These models are application to certain biological models ( e.g. bacteria growth
) over a short period of time and for money compounded continuously. These
are separable differential equations with a solution of p( t ) = p 0ekt. The usual
case is to be given the initial population, p0 , and the population at some other
time, t. This enables you to solve for the constant, k, and then solve the
problem. The typical solution looks like:
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For example, if the initial population, p( 0 ), of a bacteria culture is 2,000 and
p( 3 ) = 4,000, find the population at time t = 6 days. Mathematica finds the
solution to be 8,000. Please note that the time to double is a constant 3 days.
This is characteristic of exponential growth problems.
Radioactive decay models
Radioactive decay models, on the other hand, are very accurate over long
periods of time. They are the primary method for determining age of prehistoric
fossils and ancient artifacts. The basic differential equation is:
q'( t ) = -kq( t ), q( 0 ) = q0, where k > 0.
This is also separable with a solution
q( t ) = q0 e-kt.
The typical solution has a graph:
For example, if the quantity of radioactive material at t = 0 is 100 g and
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q( 1000 ) = 80 g, find its half-life.
Here q0 = 100 and solving for k with a Solve[{q[ t ] == 80, t == 1000}, k ]
gives k = Log[5/4]/1000. Solving for t gives a half-life of 3106.28 years.
Newton's Law of Cooling
When an object has a temperature greater than the ambient temperature, it cools
accroding to Newton's Law of cooling which states that the rate of cooling is
proportional to the difference in the temperatures, that is:
y'( t ) = k( y( t ) - a ), y( 0 ) = y0,
where y( t ) is the temperature of the object at any time t and a is the ambient
temperature. The solution to this separable differential equation is
y( t ) = a + ( y0 - a )e-kt, k > 0.
A typical solution has this graph:
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The horizontal asymptote is the ambient temperature. For example, a turkey is
taken from the oven at 300o F and placed at room temperature of 70o F. In t = 2
minutes, the turkey's temperature is 200o F. Find how long it takes the turkey to
cool to 150o F. The solution is 3.7 minutes. By the way Mathematica needs
some help in solving some of these log functions. You may have to solve some
of it yourself and then have Mathematica solve the rest.
Chemical mixture problems
Another application of linear differential equations is a chemical mixture
problem where a fixed concentration of a solid dissolved in a liquid is poured
into mixture with a fixed amount being drawn off at a constant rate. The
differential equation is: a'( t ) = r1 - r2, where a( t ) is the amount of solid in the
mixture, a'( t ) is the rate at which the amount is changing, r 1 is the rate at
which the solid is entering the mixture and r2 is the rate leaving.
For example, a 100 gal. vat contains 100 gal. of water with 20 lbs. of salt
dissolved in it. If a saline solution of 50 lbs.of salt per gallon of water is
entering the vat at 25 gal/sec. and the resulting mixture is being drawn from the
vat at 25 gal/sec., how many lbs. of salt are in the vat at any time t? What is the
concentration of salt as t increases without bound?
The rate at which the salt is entering is
50 lbs/gal*25 gal/sec = 1250 lbs./sec.
The rate at which the salt is leaving is
a( t )/100*25.
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The differential equation is
a'( t ) = 1250 - a( t )/4.
This is linear and is solved with an integrating factor or DSolve. In either case
the solution is
a( t ) = 5000 - 4980e-t/4.
The graph is as shown.
Series Circuits
Yet another example is from series circuits.With only a resistor and inductor,
the linear differential equation
Li'( t ) + ri( t ) = e( t ),
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Where




L is the inductance in henries,
i'( t ) is the rate of change of the current in amps,
r is the resistance in ohms,
i( t ) is the current and e( t ) is the voltage.
This is called an LR series circuit.
With a capacitor and resistor the differential equation is
rq'( t ) + ( 1/c )q( t ) = e( t ),
where
 q is the charge.
This is called an RC series circuit.
A typical problem for an LR series circuit would be given a circuit with
inductance of 2 henries, a resistor of 20 ohms and a voltage of 20 volts, find the
current i( t ) at any time time if the initial current is 0.
The solution is 3/5 - 3/( 5e10t ) with a graph:
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The steady state current is 0.6 amps.
Things to know and do

Set up, solve and graph an exponential growth problem

Set up, solve and graph a radioactive decay problem

Set up, solve and graph a chemical mixture problem

Set up, solve and graph a Newton's Law of cooling problem

Set up, solve and graph an LR and an RC circuit problem
Do the following:
1. If the initial population of bacteria is 10000 and it increases to 12000 after 1
hour, find how long it takes to triple. Graph p( t ).
2. If the initial amount of radioactive material is 10 lbs. and its half-life is 100
yrs., how long does it take to decay to 2 lbs.? Graph q( t ).
3. If there is 100 lbs. of sugar dissolved in a 1000 gal. tank of water and sugar
is entering at the rate of 10 lbs/gal of water at 20 gal/sec, find the amount of
sugar in the tank at any time t if the resulting mixture is being pumped out at 20
gal/sec. Graph a( t ).
4. Molten lava at 1000o F is brought to the surface of the Earth with a
temperature of 100o F. At t = 1 hour, the temperature of the lava is 800o F. How
long does it take to drop to 120o F? Graph the temperature at any time t.
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5. Given an RC circuit with r = 10 ohms and c = 2 farads.If the voltage is 120
v, find the charge, q( t ) at any time t. Graph q( t ).
Logistics Function
The exponential growth function given in the previous section is clearly not
realistic for biological organisms as the growth continues to infinity. At some
point in time the food supply will run out or overcrowded conditions will have
an adverse effect on the population growth. A more realistic description of the
population growth is the nonlinear differrential equation, p'( t ) = kp( t )( a - bp(
t ) ), where a and b are constants determined by the conditions of the problem.
The solution to this differential equation is called a logistics function. If p( 0 ) =
p0, then Mathematica gives the solution as:
.
Please note that p( 0 ) = p0 and the limit as t goes to infinity is a/b. This ratio of
a/b is the carrying capacity of the environment.
There are other applications of the logistics function beyond population growth.
For example, if 1% of a population has a communicable disease within a
confined population, then a model for the spread of the disease could be the
logistics function with p0 = 0.01, a = 1 and b = 1. This means that the rate of
growth is proportional not only to the people with the disease, but to the
product of those with it and those who do not yet have the disease. The solution
is p( t ) = 0.01/(0.01 + 0.99e-t ). The graph:
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The number of people infected approaches 100% asymptotically. This model
also applies to the spread of a rumor throughout a population. What happens if
the constants a and b are changed? Let's try a = 4. An increase of a ought to
lead to an increase in the rate of growth given our original equation. In
addition, the limit should go to 4/1 or 4. Here is the graph:
Reversing the ratio of a and b so that a = 1 and b = 4 gives much slower
growth:
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The point of inflection gives information about the future growth of the
population, especially when a new species is introduced into an environment.
This gives the time of maximum growth. Mathematica finds this to be: t =
.
http://www.math.ubc.ca/~feldman/demos/pendulum.html
TERIMA KASIH
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