Query 21

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course MTH 279

4/2 7pm

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'' + ky = 20cos8t

y = -5/9 cos20t + 5/9 cos8t

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For this situation k = 98 N / (9.8 cm) = 98 N / (.098 m) = 1000 N / meter, and the mass is 10 kg, so

omega = sqrt(k/m) = sqrt(1000 N / meter / (10 kg) ) = sqrt(100 s^-2) = 10 rad / sec, not 20 rad / sec.

I agree with your particular solution, but I do believe the resulting solution for the given initial conditions is

y = A cos(10 t) + B sin(10 t) + 5/9 cos(8 t).

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I feel okay about the equation I made, but I couldn't think of a good way to find the maximum displacement.

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Self-critique rating: 3

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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:

y = Ae^(-4t)cos2sqrt(6)t + Be^(-4t)sin2sqrt(6)t + 10/13e^-t

Over time, the system would diminish in the amplitude of its motion until it ceased to move.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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 = F(omega_0^2 - omega_1^2)/ [4delta*omega_1^2 - omega_1^2(omega_0^2 -omega_1^2) + omega_0^2(omega_0^2 - omega_1^2)]cos(omega_1t) + F2deltaomega_1/ [4delta*omega_1^2 - omega_1^2(omega_0^2 -omega_1^2) + omega_0^2(omega_0^2 - omega_1^2) sin(omega_1t) + e^(-deltat)[Acos(sqrt(delta^2 - omega_0^2)t + Bsin(sqrt(delta^2 - omega_0^2)]

To outline my work;

I first found the complementary solution to the system by using the quadratic formula to find :

delta+/- sqrt(delta^2 - omega_0^2) * i (assuming omega_0^2 > delta^2)

Which led to complementary solution :

y_c = e^(-deltat)[Acos(sqrt(delta^2 - omega_0^2)t + Bsin(sqrt(delta^2 - omega_0^2)]

The particular solution I found using the method of undetermined coefficients. In general: Ccos(omega_1t) + Dsin(omega_1t)

Of course, solving for C and D was very messy work, but to give you an idea of what I did, I first took the first and second derivs of y_p, factored out common terms, set everything that had a cos(omega_1t) in it equal to F, and after a bunch of messy algebra, I eventually got A and B. I tried not to simplyify too much after that for fear of messing it up.

The limiting function as omega_1 approaches omega_0 would cause the system to eventually stop rising in amplitude (due to the damping)

The limiting function as delta approaches 0 would cause the system to display a beat pattern (since damping would no longer be present) since omega_1 and omega_0 aren't the same.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

While I understood the process of the problem, I do feel like there was much room for error in the calculations.

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Self-critique rating:

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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:

I = 9cos.5t -18sin.5t -10e^-t +e^-t

It appears to me that after time, the current would simply display a steady oscillatory motion since the e^-t terms would drop out.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I feel like my approach to the equation was correct, but I don;t know about my interpretation. I haven't studied circuits yet.

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You've done good work here and your solutions are substantially correct. There are some discrepancies but I can't identify their source without more detail. However I think you're OK on these topics.

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