Math 2250Computer Lab 3 This project deals with using Maple to plot solutions to certain differential equations with constant coefficients in order to visualize damped unforced motion and undamped and damped forced motion. Once again, you should create a document which consists of a Maple worksheet with textual inserts that explain your steps and answer any questions posed in the assignment. We will look at the graphs of some solutions to differential equations that might arise when we model mechanical systems or electrical circuits. The first half of this worksheet has some examples. Place the cursor at the end of the lines in "Math" mode (these are italicized) to execute the lines.In class we discussed free damped motion and saw that there were three cases to consider: overdamped, critically damped, and underdamped. For our first example, we will consider the overdamped case.Assume we have a mass attached to both a spring and a dashpot (shock absorber). Let m =1, b = 5, and k = 4. Here is an example of how to enter the DE. Don't forget to type the DE in "Math" mode. You can use "F5" to toggle between "Text" and "Math" modes.
The notation ":=" names the equation, which than can be used later. We can use Maple to solve the following initial value problem. The command "dsolve" will do the trick. You need to specify the equation, the initial conditions, and what function you want to solve for. The "D" is operator notation and D(x)(0)=1 means x'(0)=1, so the object in this case starts at x =1 (x(0)=1) with an initial velocity of 1. This can be done in a few ways. We can enter the initial conditions and then use the numeric labels on the right to call up those conditions (using "Ctrl L"), or we can type them directly. Let's type them directly.
LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2I1EhRic=print(); # input placeholderWe know from the solution that we are in the overdamped case (why?). The position function, x(t), is the sum of two exponential functions so there will be no oscillatory behavior. Now plot the solution curve. (This is the graph of the position function x(t).). The code "rhs" means the right hand side of the equation. You can also right click to isolate the rhs and then right click to plot. Set the values for t so that you get a good graph.LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2I1EhRic=%;
The equilibrium position is x = 0, and we can see that x(t) \342\206\222 0 as t \342\206\222 \342\210\236 without any oscillations - the graph never crosses the t axis. For overdamped, it may or may not pass through equilibrium (cross the axis), but it will never cross more than once.The critically damped case is similar. Try the following example with m = 1, b = 8, and k = 16.
Notice the factor of "t" in the solution, hence the clue that it is critically damped (repeated root). Critically damped, like overdamped, may or may not pass through the equilibrium, but will never do so more than once. What influence does 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 have on the system, and thus the solution?Here is an example of the underdamped case. Assume m = 1, b = 2, and k = 100.print(); # input placeholder
The products of exponential functions and sines and cosines tells us we have damped oscillations. The oscillations are quite evident from the graph of the position function.
%;Now we will consider forced motion. If there is no damping and the forcing function is of the form Acos\317\211t, then we expect to see oscillations. The phenomenon of resonance occurs when the frequency of the forcing function is the same as the natural frequency of the system. To visualize the phenomenon of resonance we will solve and graph the solution, x(t).
With the solution in this form we can see that x(t) will be bounded by -t and t (because sin(t) is bounded by -1 and 1). Since this is the case, we will plot the three functions together (even though the t and -t are not part of the solution, we are plotting them to see the linearly increasing amplitude). %;
"Real life" situations will always involve some damping. However, if the damping forces are relatively small, then we would expect some oscillations of very large amplitude. What happens if the damping force is not too "small" ? We will look at the following example to answer this question.We will solve the DE, plot the solution, and analyze the results as we have done in the previous examples.
Remembering that the solution LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkjbWlHRiQ2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JKG1mZW5jZWRHRiQ2JC1GIzYkLUYsNiVRInRGJ0YvRjIvRjNRJ25vcm1hbEYnRj1GPQ== can be expressed as the sum of the solution of the associated homogeneous DE (call it xh) and a particular solution(call it xp), can you see in the above output which piece is xp? Use Maple to find the associated homogeneous solution. Notice that xh \342\206\222 0 as t \342\206\222 \342\210\236 because of the negative exponential (so xh, the associated homogeneous equations solution, is called the transient solution). This means that as t\342\206\222 \342\210\236, x(t) approaches the particular solution, also known as the steady state solution. To visualize this phenomenon graph the solution, x(t),(in green) and the steady state solution, xp,(in blue) on the same set of axes.%;
Please complete the following exercises on a different Maple worksheet. Please put your name on it!
Exercise 1) Using Maple, solve the following free (no driving force) spring-mass problems. For each situation determine whether the motion is overdamped, critically damped, or underdamped. Justify your conclusions to receive full credit. a) m = 2, b = 4, k = 10 with initial displacement of 1 and initial velocity of 3. b) m = 4, b = 12, k = 5 with initial displacement of 1 and initial velocity of -3. c) m = 4, b = 4, k = 1 with initial displacement of 4 and initial velocity of -1.Exercise 2) Using Maple, and taking a case of overdamped, then a case of critically damped and then a case of underdamped, experiment with the values for initial displacement and initial velocity to determine in each part how many times the mass can pass through the equilibrium point (where x = 0). What I would like to see is a conclusion and justification for each case: overdamped, critically damped and underdamped. Hint: You can use the values of Exercise 1 for your m, b, and k.Exercise 3) Suppose that m =5 kg and that k = 500 N/m for a spring-mass system with no damping. Assume that the external force is given by 50 cos 10t and the initial position and velocity are both zero. Using Maple, solve the differential equation that models this situation. Plot the solution and analyze the solution in terms of its graph.Choose a range for t that gives a clear picture of the situation.Exercise 4) Suppose that in an RC circuit we have R = 10 , C = 0.02 , Q(0)=0, and V(t) = 50cost . a) Use Maple to solve for Q(t) and I(t).b)Graph I(t) (the current) and the steady state current on the same set of axes with 0 < t < 10. Label your graph (by hand is fine) to indicate which curve is the graph of I(t) and which curve is the graph of the steady state current. c) Approximate the amplitude of the steady state current from your graph. Then calculate the amplitude of the steady state current and compare with your approximation..Exercise 5) A mass m is attached to a spring (with given spring constant k) and a dashpot (with given damping constant c). The mass is set in motion with the given initial position and initial velocity. Find the position function x(t) and determine whether the motion is overdamped, critically damped, or underdamped. Plot x(t) over an appropriate range that illustrates the behavior of the system. a) m = LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1JI21uR0YkNiRRIjFGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRictRiM2JC1GLzYkUSIyRidGMkYyLyUubGluZXRoaWNrbmVzc0dRIjFGJy8lK2Rlbm9tYWxpZ25HUSdjZW50ZXJGJy8lKW51bWFsaWduR0Y/LyUpYmV2ZWxsZWRHUSZmYWxzZUYnRjI=, c= 3, k = 4; x(0) = 2, x'(0) = 0 b) m = 1, c = 8, k = 63; x(0) = 5, x'(0) = -10 c) m = 4, c = 20, k = 169; x(0) = 4, x'(0) = 16Exercise 6) The following equations model a driven undamped mass-spring system. The mass is set in motion with the given initial position and initial velocity. Find and graph the position function x(t) over an appropriate domain. What (if any) characteristics do the solution curves exhibit? a) 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 ; 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 b) 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 ; 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