#$&* course Mth 279 7/30 7:03 pmI did these last week but forgot to submit them Query 21 Differential Equations
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: The motion of a mass is governed by the equation m y '' + 2 gamma y ' + omega_0^2 y = F(t), with m = 2 kg, gamma = 8 kg / s and k = 80 N / m and F(t) = 20 N * e^(- t s^-1). Solve the equation for the function y(t). What is the long-term behavior of this system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: my'' + 2 gamma y' + omega_0^2 y = F(t) 2y'' + 16 y' + 4(80)(2) y = 20 e^(-t) y'' + 8y' + 320 y = 10 e^(-t) [ -8 +- sqrt(64 - 4(320))]/2 y1 = -4 +- sqrt(304) i yc = c1 e^(-4t) cos(sqrt(304) t) + c2 e^(-4t) sin(sqrt(304) t) yp = Ae^(-4t) yp' = -4Ae^(-4t) yp'' = 16A e^(-4t) y'' + 8y' + 320y = 10 e^(-4t) 16A e^(-4t) + (8) -4Ae^(-4t) + (320) Ae^(-4t) = 10e^(-4t) 16A - 32A + 320A = 10 A = 10/304 yp = 10/304 e^(-4t) y = c1 e^(-4t) cos(sqrt(304) t) + c2 e^(-4t) sin(sqrt(304) t) + 10/304 e^(-4t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: step 1: I would find yc by using the quadratic formula. I believe y1 will be a complex solution. yc will be in the form of c1e^(alpha *t)cos(beta*t) + c2 e^(alpha *t)sin(beta*t) step 2: I would find yp. Because g(t) is in the form Acos(t), I can use the method of undetermined coefficients. It will be in the form of Asin(omega *t) + Bcos(omega*t). Then I would solve for A and B step 3: I would add y = yc + yp As omega_0 approaches omega_1, the oscillations of the mass are getting linearly larger. It is as if a mass is oscillating on a spring while the other end of the spring is being oscillated at the same frequency. Because the whole system and the mass are both in phase, this increases the amplitude. As delta approaches 0, the system is no longer dampened and will continue to oscillate forever. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I'' + R/L I' + 1/(LC) I = 1/L Vs' I'' + 1/.004 I = 1/.004 Vs' I'' + 250 I = -2500 e^(-t) I^2 + 250 = 0 y1 = 5sqrt(10) i yc = c1 cos(sqrt(10) 5t) + c2 sin(sqrt(10) 5t) yp = Ae^(-t) yp' = -Ae^(-t) yp'' = Ae^(-t) Ae^(-t) + 250 * Ae^(-t) = -2500 e^(-t) 251A = -2500 A = -2500/251 yp = -2500/251 e^(-t) I = c1 cos(sqrt(10) 5t) + c2 sin(sqrt(10) 5t) -2500/251 e^(-t) This means that the current is oscillating but there are ""beats"" because of yp