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course Mth 279
5/14
Query 21 Differential Equations*********************************************
Question: A 10 kg mass stretches a spring 9.8 cm beyond its original rest position.
A driving force F(t) = 20 N * cos((8 s^-1) * t) begins at t = 0, where the downward direction is regarded as positive.
Write down and solve the appropriate differential equation, obtaining the position function for the motion of the mass.
Plot your solution, and find the maximum distance of the mass from its equilibrium position.
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Your solution:
my'' + (gamma)y' + ky = F(t)
Assuming that (gamma)y' = 0
and Hook's law:
mg = k(9.81cm)
(10kg)(9.81m/s^2) = k (9.81cm)(1m/100cm)
k = 1000
10y'' + 1000y = 20cos(8t)
y'' + 100y = 2cos(8t)
y(t) = y_c(t) + y_p(t)
y'' + 100y = 0
y_c(t) = Asin(10t) + Bcos(10t)
y_p(t) = Asin(8t) + Bcos(8t)
y'' + 100y = 2cos(8t)
(Asin(8t) + Bcos(8t))'' + 100(Asin(8t) + Bcos(8t)) = 2cos(8t)
(8Acos(8t) - 8Bsin(8t))' + 100(Asin(8t) + Bcos(8t)) = 2cos(8t)
-64Acos(8t) - 64Bsin(8t) + 100Asin(8t) + 100Bcos(8t) = 2cos(8t)
36Asin(8t) + 36Bcos(8t) = 2cos(8t)
36A = 0
A = 0
36B = 2
B = 2/36 = 1/18
y_p(t) = (1/18)cos(8t)
y(t) = Asin(10t) + Bcos(10t) + (1/18)cos(8t)
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Your solution is correct for driving force 2 N cos( 8 t).
For the given force 20 N cos(8 t) your particular solution would be 10 times as great, so your value of B would be 18 * 1/18 = 5/9.
The solution would then be
y(t) = Asin(10t) + Bcos(10t) + (5/9)cos(8t)
= Asin(10t) + Bcos(10t) + (5/9)cos(8t)
Solving for initial conditions the result is
y = -5/9 cos(10 t) - 40/9 sin(10 t) + 5/9 cos(8 t).
The term -40/9 sin(10 t) is a sine function with period pi / 5 (approximately .63) and amplitude 40/9.
-5/9 cos(10 t) + 5/9 cos(8 t) will come in and out of phase with angular frequency 10 rad / sec - 8 rad / sec = 2 rad / sec, and will therefore do so with period pi.
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Question: The motion of a mass is governed by the equation
m y '' + 2 gamma y ' + omega_0^2 y = F(t),
with m = 2 kg, gamma = 8 kg / s and k = 80 N / m and F(t) = 20 N * e^(- t s^-1).
Solve the equation for the function y(t).
What is the long-term behavior of this system?
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Your solution:
2y'' + 2(8)y' + Omega_0^2 * y = 20e^(-t)
Omega_0 = sqrt(k/m)
2y'' + 2(8)y' + 80/2 * y = 20e^(-t)
y'' + 8y' + 20y = 20e^(-t)
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80/2 = 40, not 20
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y(t) = y_c(t) + y_p(t)
y'' + 8y' + 20y = 0
Using the quadratic equation we get -4 +- 2i
y_c(t) = e^(-4t)(Asin(2t) + Bcos(2t))
y_p(t) = Ae^(-t)
y'' + 8y' + 20y = 20e^(-t)
(Ae^(-t))'' + 8(Ae^(-t))' + 20(Ae^(-t)) = 20e^(-t)
(-Ae^(-t))' - 8(Ae^(-t) + 20Ae^(-t) = 20e^(-t)
Ae^(-t) - 8(Ae^(-t) + 20Ae^(-t) = 20e^(-t)
13Ae^(-t) = 20e^(-t)
13A = 20
A = 20/13
y_p(t) = (20/13)e^(-t)
y(t) = e^(-4t)(Asin(2t) + Bcos(2t)) + (20/13)e^(-t)
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I didn't check all your details but you're doing the right stuff.
Compare with the following solution:
With gamma / m in units of s^-1, omega_0^2 / m in units of s^-2, and F(t) / m in units of m / s^2 we have
y '' + 4 y ' + 20 y = 10 e^(-t).
The homogeneous equation has solution set
{e^(-2 t) cos(t), e^(-2 t) sin(t)
so the complementary solution is
y_C = A e^(-2 t) cos(t) + B e^(-2 t) sin(t)
We expect particular solution of the form
y_P = A e^(- t)
with derivative
y_P ' = - A e^(-t)
and second derivative
y_P '' = A e^(- t).
Substituting this into our equation y '' + 4 y ' + 20 y = 10 e^(-t) we have
A e^(- t) - 4 A e^(-t) + 20 A e^(-t) = 10 e^(-t)
so
17 A = 10
and
A = 10/17.
Our particular solution is thus
y_P = 10/17 e^(-t)
and our general solution is
y(t) = A e^(-2 t) cos(t) + B e^(-5 t) sin(t) + 10/17 e^(-t).
We can write this as
e^(-2 t) ( A cos(t) + B sin(t) ) + 10/17 e^(- t).
As t gets large, e^(-2 t) approaches zero much more quickly than e^(-t).
Since A cos(t) + B sin(t) is bounded (for example its magnitude can certainly not exceed | A | + | B | ), whatever the initial conditions might be the expression e^(-2 t) ( A cos(t) + B sin(t) ) will at some point be far exceeded in magnitude by e^(-t). So the long-term behavior of the solution will be that of 10/17 e^(-t).
In fact by the time the cosine and sine functions have completed one cycle, which requires time 2 pi, e^(-2 t) will be e^(-4 pi ) = e^(-12.6 ) = 3 * 10^-6 = .000003, while e^(-t) will be .0018, over 500 times as great as e^(-2t).
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Question:
Solve the equation
y '' + 2 delta y ' + omega_0^2 y = F cos( omega_1 * t), y(0) = 0, y ' (0) = 0.
Give an outline of your work. A very similar problem was set up and partially solved in class on 110309, and your text gives the solution but not the steps.
Find the limiting function as omega_1 approaches omega_0, and discuss what this means in terms of a real system.
Find the limiting function as delta approaches 0, and discuss what this means in terms of a real system.
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Your solution:
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Question: An LC circuit with L = 1 Henry and C = 4 microFarads is driven by voltage V_S(t) = 10 t e^(-t).
Write and solve the differential equation for the system.
Interpret your result.
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Your solution:
LQ'' + RQ' + Q/C = V
Assuming R = 0
Q'' + Q/4 = 10te^(-t)
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C is 4 microCoulombs so you would need to use 4 * 10^-6 as your capacitance.
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y(t) = y_c(t) + y_p(t)
From Q'' + (1/4)Q = 0
the Quadratic equation yeilds roots: +- i/2
y_c(t) = Asin(t/2) + Bcos(t/2)
y_p(t) = Ate^(-t) + Be^(-t)
Q'' + Q/4 = 10te^(-t)
(Ate^(-t) + Be^(-t))'' + (1/4)(Ate^(-t) + Be^(-t)) = 10te^(-t)
(-Ate^(-t) + Ae^(-t) - Be^(-t))' + (1/4)(Ate^(-t) + Be^(-t)) = 10te^(-t)
(Ate^(-t) - 2Ae^(-t) + Be^(-t) + (1/4)(Ate^(-t) + Be^(-t)) = 10te^(-t)
(5/4)Ate^(-t) - 2Ae^(-t) + (5/4)Be^(-t) = 10te^(-t)
(5/4)A = 10
A = 8
-2A + (5/4)B = 0
-2(8) + (5/4)B = 0
B = 12.8
y_p(t) = 8te^(-t) + 12.8e^(-t)
y(t) = Asin(t/2) + Bcos(t/2) + 8te^(-t) + 12.8e^(-t)
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Given Solution:
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Good overall, but check my notes.
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