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Unformatted text preview: n,
, of the total population at the end of the previous day
365
(neglect leapyear effects). Every day 275 immigrants enter Freedonia.
(a) Write a difference equation for the population at the beginning of the nth day after
January 1, 2000 with the immigration rate as the excitation of the system. (b) By finding the zeroexctiation and zerostate responses of the system determine the
population of Freedonia be at the beginning of the year 2050. Solutions 343 M. J. Roberts  8/16/04 (b) The beginning of the year 2050 is the 18250th day. 61. A car rolling on a hill can be modeled as shown in Figure E61. The excitation is the
force, f ( t) , for which a positive value represents accelerating the car forward with the
motor and a negative value represents slowing the car by braking action. As it rolls, the
car experiences drag due to various frictional phenomena which can be approximately
modeled by a coefficient, k f , which multiplies the car’s velocity to produce a force which
tends to slow the car when it moves in either direction. The mass of the car is m and
gravity acts on it at all times tending to make it roll down the hill in the absence of other
N⋅s
forces. Let the mass, m, of the car be 1000 kg, let the friction coefficient, k f , be 5
m
π
and let the angle, θ , be .
12
(a) Write a differential equation for this system with the force, f ( t) , as the excitation and
the position of the car, y( t) , as the response.
(b) If the nose of the car is initially at position, y(0) = 0 , with an initial velocity,
[y′(t)]t = 0 = 10 m , and no applied acceleration or braking force, graph the velocity of the car,
s
y′ ( t) , for positive time.
(c) If a constant force, f ( t) , of 200 N is applied to the car what is its terminal velocity ? f(t)
y(t) (θ) θ sin mg Figure E61 Car on an inclined plane
(a) Summing forces, (b) The zeroexcitation response can be found by setting the force, f ( t) , to zero. f ( t) − mg sin(θ ) − k f y′ ( t) = m y′′ ( t) − kf The homogeneous solution is y h ( t) = K h1 + K h 2e m . The particular solution must be in the
form of a linear function of t, to satisfy the differential equation.
t Solutions 344 M. J. Roberts  8/16/04
t
− 200
y( t) = 1.0346 × 10 1 − e − 507.28 t 5 t
t
− −t 1.0346 × 10 5 − 200 507.28 = 517.28e 200 − 507.28 = 517.28 e 200 − 1 + 10
y′ ( t) =
e −
200 y’(t)
1000 t 550 (c) The differential equation is
m y′′ ( t) + k f y′ ( t) + mg sin(θ ) = f ( t) We can rewrite the equation as
m y′′ ( t) + k f y′ ( t) = f ( t) − mg sin(θ )
treating the force due to gravity as part of the excitation. Then the impulse response is the
solution of
m h′′ ( t) + k f h′ ( t) = δ ( t)
which is of the form,
kf − t
h( t) = K h1 + K h 2e m u( t) . The impulse response is
− 1− e
h( t) =
kf kf
m t u( t) . Now, if we say that the force, f ( t) , is a step of size, 200 N, the excitation of the system is
x( t) = 200 u( t) − mg sin(θ ) .
But this is going to cause a problem. The problem is that the term, − mg sin(θ ) , is a constant,
therefore presumed to have acted on the system for all time before time, t = 0. The
implication from that is that the position at time, t = 0, is at infinity. Since we are only
interested in th...
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 Spring '09
 ATOUSA

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