#$&*
course Mth 279
7/22
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:
1) find k?
mg - ky = 0
(10*9.8) - k (0.098) = 0
k = 1000
2) model equation for spring mass systems:
my’’ + ky = F(t)
plug in what we know
10y’’ + 1000y = 20cos(8t)
y(0) = 0 and y’(0) = 0
3)
my’’ + gamma*y’ + ky = F(t)
y_c?
10r^2 + 1000 = 0
r^2 = - 100
r = +- 10i
y_c = C_1 cos(10t) + C_2 sin(10t)
4)y_p?
g(t) = 20cos(8t)
y_p = Acos(8t) + Bsin(8t)
y’ = -8Asin(8t) + 8Bcos(8t)
y’’ = -64cos(8t) - 64Bsin(8t)
plugging in
10(-64cos(8t) - 64Bsin(8t)) + 1000(Acos(8t) + Bsin(8t)) = 20cos(8t)
360Acos(8t) + 360Bsin(8t) = 20cos(8t)
360A = 20
360B = 0
A = 1/18
B = 0
y(t) = C_1 cos(10t) + C_2 sin(10t) + 1/18cos(8t)
Now find C_1 and C_2 using initial conditions
y(0) = C_1cos(0) + C_2sin(0) + 1/18cos(0) = 0
0 = C_1 + 1/18
C_1 = -1/18
y’(0) = -10C_1sin(10t) _ 10C_2cos(10t) + 8/18sin(8t) = 0
0 = 10C_2
C_2 = 0
y(t) = -1/18cos(10t) + 1/18cos(8t)
Unsure how to really plot this to find max distance? Do you just plug in values for t? But the graph oscillates so how would you go about graphing something like this on a test without a graphing calculator???
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The -1/18 cos(10 t) graph has period pi/5 while the 1/18 cos(8t) graph has period pi/4.
You would start by sketching several periods of each of the graphs.
You will see how they come in and out of phase, and they have the same amplitude (which is 1/18) .
The two graphs have equal and opposite values at t = 0 so y(0) will be zero.
They will also have equal and opposite phases at t values which are common multiples of pi/5 and pi/4. The smallest common multiple of 1/5 and 1/4 is 1, so the smallest common multiple of pi/5 and pi/4 is pi. Thus at t = -2 pi, -pi, 0, pi, 2pi, etc., the values of the two trigonometric functions will be equal and opposite and therefore add up to zero.
Halfway between these t values the graphs come into phase, with peaks meeting peaks The result is a big peak in the graph of the y(t) function, which occurs when t = -3/2 pi, -1/2 pi, 1/2 pi, 3/2 pi, etc..
From t = 0 to t = pi the graph of -1/18 cos(10 t) will have five peaks and five valleys. The peaks at t = 0 and t = pi, as we have seen, are countered by the simultaneous peaks of 1/18 cos(4 t). As we move from t = 0 to t = pi/2, the peaks of the second function progressively become aligned with those of the first. Then between t = pi/2 and t = pi the peaks become less and less aligned, until they are again opposite at t = pi.
The same thing happens repeatedly with every subsequent interval of pi.
I suggest you sketch the graphs of the two functions to see how they come into and out of phase, and without plugging in to get actual values construct the graph of the sum of these functions. You won't be completely accurate, but you should clearly see when the two functions 'cancel' one another and when they reinforce.
Having done the graphs by hand, you can then plot them on your calculator and look at the result with meaning and understanding (something that students miss because of the widespread misuse of graphing calculators to substitute for, rather than complement, the sort of understanding required to comprehend the physical world).
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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’ + (sqrt(40))^2y = 20e^-t
y’’ + 8y’ + 20y = 10e^-t
y_c?
r^2 + 8r + 20 = 0
r = -8 +-sqrt(8^2 - 4(20))/2 = -4+- 2i
y_c = C_1cos(2t) + C_2sin(2t)
@&
e^( (-4 + 2 i) t) = e^(-4 t) (cos(2 t) + i sin(2 t) ), and
e^( (-4 + 2 i) t) = e^(-4 t) (cos(2 t) - i sin(2 t) ).
Linear combinations of these functions lead to solution set
{e^(-4 t) cos(2 t), e^(-4 t) sin(2 t) }.
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y_p?
g(t) = 10e^-t
y_p = Ae^-t
y’ = -Ae^-t
y’’ = Ae^-t
plugging in
Ae^-t + 8(-Ae^-t) + 20(Ae^-t) = 10e^-t
A -8A + 20A = 10
13A = 10
A = 10/13
y_p = 10/13 e^-t
y(t) = C_1cos(2t) + C_2sin(2t) + 10/13e^-t
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Your particular solution is good.
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What is the long term behavior of this system???
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As t gets large, e^(-4 t) approaches zero much more quickly than e^(-t).
Since A cos(2t) + B sin(2t) is bounded (for example its magnitude can certainly not exceed | A | + | B | ), whatever the initial conditions might be the expression e^(-4 t) ( A cos(2t) + B sin(2t) ) will at some point be far exceeded in magnitude by e^(-t). So the long-term behavior of the solution will be that of 10/13 e^(-t).
In fact by the time the cosine and sine functions have completed one cycle, which requires time 2 pi, e^(-4 t) will be e^(-8 pi ) = e^(-25 ) = 10^-11 = .00000000001, very roughly, while e^(-t) will be .0018, overwhelmingly greater than e^(-4t).
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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:
r^2 + 2dr + w^2 = 0
r = -2d +- sqrt((2d)^2 - 4w^2)/2
r = -d +- sqrt(d^2 -w^2)
y_c = C_1sin(sqrt(d^2 -w^2)t) ???
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You aren't including the delta in your function.
You get e^(-delta t) (cos(sqrt(delta^2 - omega-0^2) t) + sin(sqrt(delta^2 - omega-0^2) t ), etc..
You have to subscript your angular frequency omega, since the equation includes two angular frequencies (one for the natural oscillation of the system, the other for the driving force).
You end up with complementary solution
c_1 e^(-delta t) * cos( sqrt( omega_0^2 - delta^2) t) + c_2 e^(-delta t) * sin( sqrt( omega_0^2 - delta^2) t).
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y_p?
g(t) = Fcos(wt)
y_p = Asin(wt) + Bcos(wt)
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Your particular solution will be of the form
y_P = A cos(omega_1 * t) + B sin(omega_1 * t)
You have to subscript your angular frequency, as indicated in my previous note.
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y’ = wAcos(wt) - wBsin(wt)
y’’ = -w^2Asin(wt) - w^2Bsin(wt)
plugging in
-w^2Asin(wt) - w^2Bsin(wt) + 2d(wAcos(wt) - wBsin(wt)) + w_0^2(Asin(wt) + Bcos(wt)) = Fcos(wt)
I am starting to get lost in my work and where to go next. Not even sure that what I did was correct???
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You do want to correct what you've done so far. You are on the right track but you've missed some important details. Shouldn't be difficult, though, to get to this point.
The next step is to write down your complete solution and analyze its behavior, which does require some heavy trigonometry. That might entail some review.
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Given 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_s(t)
L = 1 R = 0
resulting in
Q” + 1/4*10^-6 * Q = 10te^-6
Q” + 250000Q =0
r^2 + 250000 = 0
r = +- 500i
y_c = C_1sin(500t) + C_2cos(500t)
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This is of the form c_1 sin(omega t) + c_2 cos(omega t), with omega = 500.
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y_p?
g(t) = 10te^-t
y_p = Ate^-t + Be^-t
y’ = -Ate^-t + Ae^-t + (-Be^-t)
y” = Ate^-t -Ae^-t -Ae^-t + Be^-t = Ate^-t - 2Ae^-t + Be^-t
plugging in
Ate^-t - 2Ae^-t + Be^-t + 250000( Ate^-t + Be^-t) = 10te^-t
250001A = 10
-2A + 250001B = 0
A = 10/250001
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Not sure where you did it, but somewhere you used some nonexact values. Using exact values this is 10 / 250000, not 10 / 250001, and can safely be written as the exact decimal number .00004.
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-2(10/250001) + 250001B = 0
B = 20/(250001)^2
y_p = 10/250001 te^-t + 20/(250001^2) e^-t
y(t) = C_1sin(500t) + C_2cos(500t) + 10/250001 te^-t + 20/(250001^2) e^-t
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This can be written, exactly, as
y_P = 4 * 10^-5 e^(-t) + 3.2 * 10^-10 e^(-t).
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Not sure how to interpret the result seeing we have unknowns C_1 and C_2???
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c_1 and c_2 depend on the state of the system when the driving voltage is applied.
There is no resistance in this circuit. The driving voltage starts at zero, reaches a maximum then decreases asymptotically to zero. Maximum charge, current and voltage occur at t = 1, after nearly 100 cycles of the circuit's natural frequency 500 / (2 pi) seconds.
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:"
&#Good responses. See my notes and let me know if you have questions. &#