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course Mth 279
3/31 20
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 + k y = F(t)
Assume gamma y = 0
Hooks law: m g = k (9.81 cm)
(10 kg) (9.81 m / s^2) =k (9.81 cm) (1 m / 100 cm)
k = 1000
10 y + 1000 y = 20 cos(8 t)
y + 100 y = 2 cos(8 t)
y + 100 y = 0
Characteristic equation:
r^2 + 100 = 0 r = 10 i
y_c(t) = A cos(10 t) + B sin(10 t)
y_p(t) = A cos(8 t) + B sin(8 t)
y + 100 y = 2 cos(8 t)
[A cos(8 t) + b sin(8 t)] + 100[A cos(8 t) + B sin(8 t)] = 2 cos(8 t)
[- 8 A sin(8t) + 8 B cos(8 t)] + 100Acos(8t) + 100 B sin(8 t) = 2 cos(8t)
-64 A cos(8 t) - 64 B sin(8 t) + 100 A cos(8 t) + 100 B sin(8 t) = 2 cos(8 t)
36 A cos( 8 t) + 36 B sin(8 t) = 2 cos(8 t)
B = 0
A = 1 / 18
y(t) = A sin(10 t) + B sin(10 t) + 1 / 18 cos(8 t)
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Given Solution:
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If you apply the initial conditions you can solve for A and B to get a specific solution.
The function with angular frequency 10 will come in and out of phase with the function whose angular frequency is 8, resulting in 'beats' with period 2 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:
m y + 2 gamma y + omega_0 y = F (t)
omega_0 = sqrt( k / m) = sqrt ( 80 / 2) = sqrt(40)
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The last term is omega^2 y, not omega y.
omega^2 = k / m = 80 N/m / (2 kg) = 40 s^-2.
So your equation should be
2 y + 16 y + 40 y = 20 e^-t
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2 y + 16 y + sqrt(40) y = 20 e^-t
y + 8 y + sqrt(10) y = 10 e^-t
Characteristic equation:
r^2 + 8 r + sqrt(10) = 0
I used wolfram alpha to solve this for r
r= - 4 - sqrt( 16 - sqrt(10) ) = -4 - 3.58 i
r = sqrt(16 - sqrt(10) - 4 = - 4 + 3.58 i
y_c(t) = e^(-4 t) A sin(3.58 t) + B cos(3.58 t)
y_p(t) = A e^(-t)
y + 8 y + sqrt(10) y = 10 e^-t
[A e^-t] + 8[A e^-t] + sqrt(10) A e^-t = 10 e^-t
[-A e^-t] +8[-A e^-t] + sqrt(10) A e^-t = 10 e^-t
A e^-t -8 A e^-t + sqrt(10) A e^-t = 10 e^-t
A - 8 A + sqrt(10) A = 10
A = -2.6
y_p(t) = -2.6 e^-t
y(t) = e^(-4 t) A sin(3.58 t) + B cos(3.58 t) - 2.6 e^-t
???? This problem included very messy values from the quadratic formula. Im thinking that I did something wrong when assuming that omega_0 = sqrt (k / m).
If this is so was the overall though process and direction correct?
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The equation is a lot less messy if you use omega^2 y instead of omega y.
Your method is OK, but correcting that value I believe the solution will be
y(t0 = A e^(-4t) cos(2t) + B e^(-5 t) sin(2t) + 10/13 e^-t.
The sine and cosine terms, being multiplied by e^(-4 t) and e^(-5 t) repectively, will approach zero much faster than the e^-t term, which will quickly come to dominate the solution.
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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:
y + 2 delta y + omega_0 ^ 2 y = F cos(omega_1 t)
Characteristic equation:
r^2 + 2 delta r + omega_0 ^ 2 = 0
??? Im confused about how to continue without any values for delta or omega.
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You need to continue the solution. You will get a symbolic solution, which can then be analyzed under the assumption that omega_1 approaches omega_0.
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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:
L Q + R Q + Q / C = 10 e^-t
Assume R = 0
4 micro = 4 * 10^ -6
L Q + Q / C = 10 e^-t
Q + (4 * 10^6)Q = 10 e^-t
r^2 + (4 * 10^6) r = 0
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You don't get an r term here. You get just r^2 + 4 * 10^6 = , with solutions r = +- 2 * 10^3.
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This leads to fundamental set {cos(2 * 10^3 t), sin(2 * 10^3 t)}.
The particular solution can be found by the usual methods.
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r = 0
r = - 1 / (4 * 10^6)
y_c(t) = A cos ( r t) + B sin( r t)
y_p(t) = A t e^-t + B e^-t
[A t e^-t + B e^-t] +
???
Not sure if I began this problem correctly.
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You're doing the right things, but you have a couple of easily-corrected errors on a couple of problems.
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Check my notes and let me know if you have questions.
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