#$&*
course mth 279
Class Notes and q_a_ for first class.________________________________________
This document is supplemented by Chapter 1 of the text.
The Class Notes document is followed by a set of q_a_ questions.
Grading of homework:
These submissions, and assigned text problems, will count as 20% of your test grade. If they are late, you get half credit. Your work will be due between 5 and 7 days following posting of the associated class notes, as specified with the posting.
The bulk of your grade on q_a_ questions will be evaluated in terms of a good attempt to answer the questions. A good attempt can constitute any combination of a reasonable attempt at a solution, or a good question about the problem, relating it to what you have read in the text and/or in the Class Notes. A good question will show that you have thought about and attempted to assimilate the information in the text and the Class Notes.
A large portion of the subsequent class will address these questions.
A typical submission will be accompanied by a request for revision, which will typically be due 5 - 7 days after the new class notes have been posted.
You will receive 1, 2, 3 or 4 points on each assignment, as follows:
4 points: Good first attempt at all reasonable questions, plus substantially correct revisions as requested, both on time.
3 points: Reasonable attempt at all reasonable questions, plus reasonably correct revisions if requested, both on time.
2 points: Reasonable attempt on at least half the questions, reasonable revisions of at least half, both on time.
1 point: Any submission revealing some significant effort.
0 points: No significant effort.
Note the term 'reasonable questions'. The instructor, and the author of your text, include some questions of a very challenging nature. You shouldn't let yourself get bogged down for an excessive length of time on any question before you make some attempt at understanding the question, assembling relevant information and posing a good question of your own. The tougher the problem, the more lenient your instructor will be in defining the word 'good'.
Extra points may be awarded for work which exceeds expectations on 'unreasonable' questions.
Course grading:
The course will be graded based on tests and a final exam, which will be combined into a single 'test aveage'. The exam will count either as one test or as 35% of your 'test average', whichever is to your advantage. There will likely be 3 tests and one exam.
Your 'test average' will count as 80% of your grade. Your homework will count as 20%.
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The questions in this first q_a_ assume familiarity with first-semester physics, first-year calculus, and precalculus. Most students will begin a differential equations course needing review of some topics which are usually covered in prerequisite courses, and this will be considered during the first couple of weeks of the course. Your instructor will note apparent these areas and make suggestions for review. By the third or fourth week of the course (or in the case of a summer course by the end of the second or beginning of the third week) you will be responsible for having reviewed these topics.
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Class Notes 110110
The equation
F_net = m a
expresses Newton's Second Law, which relates the acceleration of a mass m to the net force acting on that mass.
Since the acceleration function a(t) is equal to the second derivative x '' (t) of the position function x(t), Newton's Second Law can be written in terms of derivatives as
F_net = m x '' ,
This is always so.
So the equation
m x '' = F_net
is very important in applications.
We will use this equation to understand some basic things about differential equations.
SHM (simple harmonic motion) occurs then F_net = - k x. In this case our equation becomes
m x '' = - k x.
We rearrange this to the form
x '' = - k / m * x.
To clarify some terminology:
This equation relates a function x(t) to at least on of its derivatives, so we call it a differential equation.
The highest derivative in this equation is the second derivative. So we call this a second order differential equation.
An example of a first-order differential equation might be
x ' = t + x.
Note that this equation contains the variable t as well as the function x and its derivative x '.
We could write this equation as
x ' = f(t, x),
where f(t, x) = t + x.
This equation looks very simple, and it isn't all that complicated, but it's harder than it looks. You are unlikely to be able to find a solution to this equation until we have developed some more terminology to classify equations, and the the machinery to solve this particular type of equation.
An equation we can easily solve is
x ' = t.
Remember that x ' is the derivative of the x(t) function with respect to t. Thus x(t) is an antiderivative of x ' (t), and if we integrate x ' (t) with respect to t we get x(t). There is also a constant involved in the integration, but we will take care of that when we integrate the equation.
Integrating both sides of the equation, we therefore get
x (t) = 1/2 t^2 + c,
where c is a constant. Both sides of the equation would have integration constants, so c represents the integration constant from the right-hand side, minus the constant from the left-hand side.
So x(t) = 1/2 t^2 + c is the general solution to our equation x ' = t.
The integration constant allows us to impose one condition on our solution. For example if we want to impose an initial condition that x(0) = 12 (we call this an initial condition because it applies at t = 0), we can accomodate that:
x(t) = 1/2 t^2 + c so
x(0) = 1/2 * 0^2 + c = c.
Thus, if x(0) = 12, we have
12 = c.
Our specific solution (called a particular solution) to the equation is therefore
x(t) = 1/2 t^2 + 12.
If x(t) represents the position function for a particle with constant acceleration a, then the fact that acceleration is the second derivative of the position leads us to the equation
x '' (t) = a = constant.
An antiderivative of x '' (t) is x ' (t), and an antiderivative of the constant quantity a is a t. So a single integration of our equation yields
x ' (t) = a t + c_1,
where c_1 is a constant number.
Integrating once more we get
x(t) = a * t^2 / 2 + c_1 * t + c_2,
where c_1 is the integration constant from our previous integration and c_2 is the constant from our present integration.
The two constants allow us to impose two conditions on our solution. For example we could impose the conditions x ' (0) = 5 and x(0) = 7. In terms of motion, x ' represents velocity and x represents position, so our conditions specify an initial velocity and an initial position on our motion.
The first of our integrals is x ' (t) = a t + c_1. You can verify that the condition x ' (0) = 5 gives us c_1 = 5.
Our position function is therefore x(t) = a t^2 / 2 + 5 t + c_2. You can verify that our condition x(0) = 7 gives us c_2 = 7 so that our position function becomes
x(t) = 1/2 a t^2 + 5 t + 7.
In general if x(0) = x_0 and v(0) = x ' (0) = v_0, our general solution x(t) = a * t^2 / 2 + c_1 * t + c_2 becomes
x(t) = 1/2 a t^2 + v_0 t + x_0.
You should recognize this as a standard equation of uniformly accelerated motion in introductory physics.
In general a first-order differential equation will yield one arbitrary constant, allowing us to impose one condition on the solution, and a second-order equation will yield two arbitrary constants, allowing us to impose two conditions.
If an equation is of order n, then a general solution will include n arbitrary constants, allowing us to impose n conditions on our solution.
Returning to the equation for SHM:
x '' = - k / m * x,
we don't yet have a method for solving this equation. There is a simple method (let x = A e^(r t) and see what this tells us about the two arbitrary constants A and r). You might have seen this method in your first-year calculus course, which often includes a brief introduction to differential equations. The method is not difficult, and we will see it in a later chapter.
Here we simply assert that the general solution to the equation can be expressed as
x(t) = B cos( sqrt(k/m) t ) + C sin ( sqrt(k/m) t ).
If you plug this function into the equation x '' = -k/m x, you can easily verify that x '' (t) is in fact equal to -k /m * x(t).
As expected, this solution includes two arbitrary constants B and C.
Basic trigonometric identities allow us to rearrange the expression B cos( sqrt(k/m) t ) + C sin ( sqrt(k/m) t ) into the expression A cos( sqrt(k/m) t + phi ), where A and phi are constants that can be expressed in terms of B and C in the original form. So another way of writing the solution is
x(t) = A cos( sqrt(k/m) t + phi).
You should verify that this is also a solution of the equation.
Both solutions can be written in terms of omega = sqrt(k /m):
x(t) = B cos( omega * t) + C sin( omega * t)
x(t) = A cos(omega * t + phi) ).
The second solution is easily interpreted as the x component of a position vector r ( t ) whose terminal point moves with angular velocity omega around a circle of radius A, with initial angular position phi.
x(t) is therefore the general equation of motion for a simple harmonic oscillator with amplitude A and angular frequency omega = sqrt(k / m).
You should verify that x(t) = A cos(omega * t + phi), with omega = sqrt(k/m), is a solution of the equation x '' = -k / m * x.
An object moving against a frictional force F_frict and an air resistance - k v which is proportional to its speed has
F_net = -F_frict - c v, so that
m x '' = -F_frict - c v and
x '' = -F_frict / m - c / m * v.
Since v = x ' this equation becomes
x '' = -F_frict / m - c / m * x '.
We won't at this point discuss how to classify or solve this equation. However note the following:
This is a second-order equation, with x '' expressed in terms of t, x, and x '.
In fact x '' is expressed only in terms of x ', with no specified dependence on t or x.
The equation is certainly of the form
x '' = f(t, x, x '),
though in this case t and x aren't actually part of the definition of the function f.
If the object in the preceding is also subject to a linear restoring force F = - k x (e.g., consider a mass oscillating on a smooth tabletop while attached to a spring; friction and air resistance are still present but now we have the restoring force of the spring to consider), the equation would be
x '' = -F_frict / m - c / m * x ' - k x.
This is still of the form
x '' = f(t, x, x').
In this case the function f(t, x, x') includes the expressions x and x ', but there is no t dependence.
Now suppose the table is gradually tilted. The gravitational force on the mass will then have a component parallel to the object's motion, which will increase in magnitude as the table gets steeper. The equation of motion could then be of the form
x '' = -F_frict / m - c / m * x ' - k x + m g cos ( b t ),
where g is the acceleration of gravity and b is a constant which determines how quickly the angle of the incline changes.
This equation is also of the form
x '' = f(t, x, x ' ),
where now all three of the quantities t, x and x ' are included in the expression for t.
[Physics students may note that the frictional force will no longer be constant when as table is tilted through different angles, so F_frict also becomes a function of t. This was not mentioned before for fear of overly complicating the equation, but the equation would be x '' = - mu g sin( b t) - c / m * x ' - k x + m g cos(b t). This is still of the form x '' = f(t, x, x').]
[Instructor note: Off the top of my head it doesn't appear that the two equations given here have a closed-form solution. Most differential equations don't. However we can always generate approximate solutions, and we can often solve a similar equation in closed form and then using reasonable approximations consider how solutions to our actual equation will differ.]
Let's return to the equation x ' = x + t. This equation can be solved, and we'll soon learn how to solve it. But unless you've studied differential equations to some extent, you probably can't solve it right now. So we can pretend that it's just not solvable.
However we can develop an approximate solution.
First note that the equation is of first order, so we can impose one condition on the equation. Let's say that our condition is that x = .37 when t = .30.
We can reason as follows:
If x = .37 and t = .30, then x ' = x + t = .37 + .30 = .67.
x ' tells us how quickly x changes with respect to t, so that as long as x ' doesn't change too significantly,
`dx = x ' * `dt
is a good approximation to how much x will change as a result of a change in t.
Without a lot of further analysis, let's just see what happens if we assume that a change `dt = .1 doesn't significantly affect the value of x '. This might or might not be the case, depending on your interpretation of the word 'significant' for the specific situation.
Now, if x ' = .67, then `dt = .1 gives us a slope of .67 and a 'run' of .1, which will result in a 'rise' equal to slope * run = .67 * .1 = .067.
Remember that we started with x = .37. Adding a 'run' of .067 we end up with x = .37 + .067 = .437.
An interval `dt = .1 increases the value of t from .30 to .40.
The approximate value of x, expected at t = .40, is therefore .437.
We can summarize the same series of calculations more formally as follows:
Our equation is x ' = f(t, x) = x + t.
At (t, x) = (.30, .37) we have x ' = f( .30, .37) = .67.
Using `dt = .1, our approximation for the change in x is
`dx = x ' * `dt = .67 * .1 = .067.
Our approximation is therefore
(t_new, x_new) = (30 + .1, .37 + .067) = (.40, .437).
There is no need to stop at this point, except to note that between our original point and our new point, x ' will indeed change. Remember that we did assume that x ' didn't change significantly. When x = .437 and t = .40, our value of x ' will be x ' = x + t = .837. Comparing this with .67, we see that there is about a 20% difference, and our estimate of `dx is probably off by about 10%. Depending on what we're trying to accomplish with our solution, this might or might not be significant. Taking into account other uncertainties in our situation we might decide that we need to do better. It would be easy to do so by reducing our `dt. For example, reducing `dt to .01 would result in the approximate point (.3767, .31), at which x ' would be .6867. We would expect our approximation to be off by only about 1%. The disadvantage is that we would not yet have a value of x corresponding to t = .40. Another option, based on `dt = .1, would be to average our two values .67 and .837 of x ', and use this average to recalculate our new x value.
In any case, continuing our original approximation, we can use our new point (.437, .40) to predict a new point:
Our equation is x ' = f(t, x) = x + t.
At (t, x) = (.40, .437) we have x ' = f( .40 , .437) = .837.
Using `dt = .1, our approximation is for the change in x is
`dx = x ' * `dt = .837 * .1 = .084,
where we have rounded off our `dx to two significant figures (more significant figures would be meaningless because of the error inherent in the approximation process).
Our new approximation is therefore
(t_new, x_new) = (.40 + .1, .437 + .084) = (.50, .53),
where again we have rounded to 2 significant figures.
We now have three points: (.37, .30), (.437, .40) and (.50, .53).
If you plot these points on a graph of x vs. t, you will see that they indicate a curve which is concave upward. This is what we would expect based on our calculations, since according to our calculations the values of x ' are positive and increasing (positive x ' means a positive slope, increasing x ' means increasing slope).
This curve is an approximate solution curve for our function.
Our curve is only approximate, since it is based on the assumption that x ' doesn't change between points.
Approximation errors accumulate with every step, multiplying in such a way that our curve varies more and more from the actual curve.
We could get a better approximation to the curve by using a smaller increment `dt (e.g., `dt = .01). It would take us 20 steps instead of 2 steps to get to the t = .50 point, but we would end up with a much more accurate solution curve.
Or after our first approximation of the new point, we could calculate the slope at the new point then re-predict that point based on the average of our two slopes. Using this method with `dt = .1 would give us a far more accurate solution curve in fewer steps than would be required with `dt = .01. There would be about twice as much calculation per step, but we would still be ahead.
Another way of analyzing solution curves is to use direction fields.
For any point (t, x) of the x vs. t plane, the equation x ' = x + t can be evaluated to obtain a value of the slope x '. Through any such point we can sketch a short line segment of the corresponding slope, and this segment will be the slope of a solution curve through that point.
If we do this for a grid of points, we get a direction field.
For the equation x ' = x + t, it is easy to evaluate x ' for the points on the grid
(0, 0), (0, 1/4), (0, 1/2), (0, 3/4), (0, 1)
(1/4, 0), (1/4, 1/4), (1/4, 1/2), (1/4, 3/4), (1/4, 1)
(1/2, 0), (1/2, 1/4), (1/2, 1/2), (1/2, 3/4), (1/2, 1)
(3/4, 0), (3/4, 1/4), (3/4, 1/2), (3/4, 3/4), (3/4, 1)
(1, 0), (1, 1/4), (1, 1/2), (1, 3/4), (1, 1).
The respective slopes range from 0 to 2. The respective slopes are
0, 1/4, 1/2, 3/4, 1
1/4, 1/2, 3/4, 1, 5/4
1/2, 3/4, 1, 5/4, 3/2
3/4, 1, 5/4, 3/2, 7/4
1, 5/4, 3/2, 7/4, 2
Plotted on a graph of x vs. t, we get something like the picture below:
All we have done is evaluate our function f(x, t) at a number of points within our region of interest.
It's easy to make a similar plot for any function f(x, t). All we need to do is plug in x and t values.
Having plotted our direction field, we can then sketch a solution curve starting from any point in the field.
The curve through (.30, .37) is approximated based on the hand-sketched direction field.
The t = .5 value of x predicted by the hand-sketched curve on the hand-sketched direction field is .63. Our calculated prediction was x = .53, and our approximations were seen to underestimate the actual changes in x, so a prediction of .63 might not be bad. However our calculated predication of the total change in x (which is .53 - .37 = .16) is probably not off by much more than 20%, and the sketch predicts a change of .63 - .37 = .26, so we can't claim a lot of accuracy for the hand sketch. The main value of the hand sketch is to provide a geometric picture of the behavior of this differential equation.
Direction fields are easy enough to sketch using a grid (though of course the process could be tedious). The process becomes much easier if we use isoclines:
f(x, t) = x + t has constant value c when x + t = c.
x + t = c when x = -t + c.
x = -t + c describes a straight line in the x vs. t plane, with slope -1 and y-intercept c.
It's easy to sketch these lines for a set of c values.
For c = 0, 1/2, 1, 3/2, 2 our lines would look something like the ones in the picture below:
The slope corresponding to each value of c is equal to c (this since x + t = c, and the slope is x ' = x + t). A number of segments of more or less appropriate slope have been sketched along each line. All the slopes along each line are the same, which makes it fairly easy to draw a decent slope field.
The figure isn't very well drawn; for example the c = 1/2 and c = 3/2 lines are badly misplaced, and the c = 1 line clearly missed the point (1, 0) on the t axis. The artist blames an unbuttoned sleeve for blocking his vision. Use of a straightedge and more care would have also been helpful.
The figure below depicts three solution curves. A solution curve can be sketched starting at any point, and moving either to the right or to the left. Once the slope field is drawn, the solution curves are fairly easy to sketch.
Had the equation been x ' = x + t^2 instead of x + t, the constant-slope condition would have been x + t^2 = c. This would give us x = -t^2 + c, which for each value of c is a parabola rather than a straight line.
In general, the equation x ' = f(x, t) yields constant-slope curves f(x, t) = c.
f(x, t) = x + t gives us a series of straight lines, along each of which the slope is constant.
f(x, t) = x + t^2 gives us a series of parabolas, along each of which the slope is constant.
In general a function f(x, t) gives us a series of curves f(x, t) = c, along each of which the slope is constant.
These curves are called isoclines ('iso' for 'one', 'cline' for inclination).
Spreadsheets and computer programs have obvious applications to the calculation schemes we have introduced here.
Let's consider for a moment the case of a second-order equation of the form
x '' = f(x, x', t).
We can still imagine a geometric picture depicting the behavior of the function f(x, x' t). However our picture graph would be in 3 dimensions, with an axis for each quantity x, x ' and t. Values of x '' would dictate changes in x ', not it x; and changes in x ' would then dictate changes in x. Our solution curves would become surfaces in a 3-dimensional space, our 3-dimensional space would no longer represent the 3-dimensional space we live in, and x vs. t solution curves would represent the intersections of these surfaces with the planes x ' = constant.
It's even more fun to try to imagine what happens when we go into higher dimensions.
However the scheme for numerically approximating the solution curves of a second-order equation isn't that much more complicated than the scheme we saw here.
Q_A_Questions
There are three parts to this set of questions.
• The questions in each part tend to be progressive in difficulty.
• If you get bogged down on one question, move on to another.
• If you get bogged down on one part, move the the next.
Part I, Part II and Part III follow below, in order. The links below might or might not be useful in helping you navigate.
• Part I: The Equation m x '' = - k x
• Part II: Solution of Equations requiring only Direct Integration
• Part III: Direction Fields and Approximate Solutions
Part I: The equation m x '' = - k x
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Question: `q001. Show whether each of the following functions all satisfy the equation m x '' = -k x:
• x = cos(t)
• x = sin( sqrt( k / m) * t)
• x = 3 cos( sqrt( k / m) * t ) + 5 sin (sqrt(k / m) t)
• x = B sin(sqrt(k / m) * t) + C cos( sqrt( k / m) * t + 3)
If x = cos(t) then x ' = - sin(t) and x '' = - cos(t). Substituting the expressions for x and x '' into the equation we obtain
m * (-cos(t)) = - k * cos(t).
Dividing both sides by cos(t) we obtain m = k. If m = k, then the equation is satisfied. If m inot equal to k, it is not.
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Your solution:
Show whether each of the following functions all satisfy the equation m x '' = -k x:
• x = cos(t)
If x = cos(t) then x ' = - sin(t) and x '' = - cos(t). Substituting the expressions for x and x '' into the equation we obtain
m * (-cos(t)) = - k * cos(t).
Dividing both sides by cos(t) we obtain m = k. If m = k, then the equation is satisfied. If m is not equal to k, it is not.
• x = sin( sqrt( k / m) * t)
x’ = sqrt( k / m) * cos( sqrt( k / m) * t), x’’ = -( k / m) * sin( sqrt( k / m) * t). Substituting the expressions for x and x '' into the equation we obtain
m * -( k / m) * sin( sqrt( k / m) * t) = - k * sin( sqrt( k / m) * t)
Dividing both sides by sin( sqrt( k / m) * t) and after rearranging the m’s and k’s, we obtain -(m/m) = -k/k, which yields 1 = 1, which works.
• x = 3 cos( sqrt( k / m) * t ) + 5 sin (sqrt(k / m) t)
x’ = -3 sqrt(k/m) sin( sqrt( k / m) * t ) + 5 sqrt(k/m) cos (sqrt(k / m) t)
x’’ = -3 (k/m cos( sqrt( k / m) * t ) - 5 k/m sin (sqrt(k / m) t)
Plugging into original equation:
m * (-3 (k/m cos( sqrt( k / m) * t ) - 5 k/m sin (sqrt(k / m) t)) = k * 3 cos( sqrt( k / m) * t ) + 5 sin (sqrt(k / m) t),
dividing both sides by 3 cos( sqrt( k / m) * t ) + 5 sin (sqrt(k / m) t) after factoring out the -km from the left hand side yields -m*k /m = -k, and after factoring we have all of the m’s and k’s cancel each other out as in the last part. Thus 1 = 1
• x = B sin(sqrt(k / m) * t) + C cos( sqrt( k / m) * t + 3)
x’ = sqrt(k / m) B cos(sqrt(k / m) * t) - sqrt(k / m) C sin( sqrt( k / m) * t + 3)
x’’ = k / m B sin(sqrt(k / m) * t) - k / m C cos( sqrt( k / m) * t + 3)
Plugging this back into the original equation we obtain,
m * (- k / m B sin(sqrt(k / m) * t) - k / m C cos( sqrt( k / m) * t + 3)) =
-k * B sin(sqrt(k / m) * t) + C cos( sqrt( k / m) * t + 3)
Then we divide both sides by B sin(sqrt(k / m) * t) + C cos( sqrt( k / m) * t + 3) after factoring out the -k/m on the left hand side, this gives us -m* k/m = k, thus again 1 = 1
confidence rating #$&*:
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Given Solution:
If x = sin(sqrt(k/m) * t) then x ' = sqrt(k / m) cos(sqrt(k/m) * t) and x '' = -k / m sin(sqrt(k/m) * t). Substituting this into the equation we have
m * (-k/m sin(sqrt(k/m) * t) ) = -k sin(sqrt(k/m) * t
Simplifying both sides we see that the equation is true.
The same procedure can and should be used to show that the third equation is true, while the fourth is not.
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Self-critique (if necessary): I didn’t get the fourth one right. I must have made an algebra error…
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Self-critique rating: 2
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Question:
`q002. An incorrect integration of the equation x ' = 2 x + t yields x = x^2 + t^2 / 2. After all the integral of x is x^2 / 2 and the integral of t is t^2 / 2.
Show that substituting x^2 + t^2 / 2 (or, if you prefer to include an integration constant, x^2 + t^2 / 2 + c) for x in the equation x ' = 2 x + t does not lead to equality.
Explain what is wrong with the reasoning given above.
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Your solution:
You have two variables in the equation. You can only take a derivative with respect to one variable.
To show this, we plug in the respective equations for x’ and x, yielding
2x + t = 2 * (x^2 + t^2/2), leading to 2x + t = 2x^2 + t^2, however, these two sides are not equal. QED.
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
The given function is a solution to the equation, provided its derivative x ' satisfies x ' = 2 x + t.
It would be tempting to say that the derivative of x^2 is 2 x, and the derivative of t^2 / 2 is t.
The problem with this is that the derivative of x^2 was taken with respect to x and the derivative of t^2 / 2 with respect to t.
We have to take both derivatives with respect to the same variable.
Similarly we can't integrate the expression 2 x + t by integrating the first term with respect to x and the second with respect to t.
Since in this context x ' represent the derivative of our solution function x with respect to t, the variable of integration therefore must be t.
We will soon see a method for solving this equation, but at this point we simply cannot integrate our as-yet-unknown x(t) function with respect to t.
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Self-critique (if necessary): Hm… my explanation was a bit weak.
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Self-critique rating: 3
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Question:
`q003. The general solution to the equation
m x '' = - k x
is of the form x(t) = A cos(omega * t + theta_0), where A, omega and theta_0 are constants. (There are reasons for using the symbols omega and theta_0, but for right now just treat these symbols as you would any other constant like b or c).
Find the general solution to the equation 5 x'' = - 2000 x:
• Substitute A cos(omega * t + theta_0) for x in the given equation.
• The value of one of the three constants A, omega and theta_0 is dictated by the numbers in the equation. Which is it and what is its value?
• One of the unspecified constants is theta_0. Suppose for example that theta_0 = 0. What is the remaining unspecified constant?
• Still assuming that theta_0 = 0, describe the graph of the solution function x(t).
• Repeat, this time assuming that theta_0 = 3 pi / 2.
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Your solution:
• Substitute A cos(omega * t + theta_0) for x in the given equation.
5 x'' = - 2000 A cos(omega * t + theta_0)
• The value of one of the three constants A, omega and theta_0 is dictated by the numbers in the equation. Which is it and what is its value?
It’s omega, it’s value is 20
x’ = -A omega sin(omega * t + theta_0)
x’’ = - A omega^2 cos(omega * t + theta_0)
Thus,
- A omega^2 cos(omega * t + theta_0) = - 400 A cos(omega * t + theta_0), dividing both sides by A cos(omega * t + theta_0) yields - omega^2 = -400, dividing both sides by -1 and taking the square root of 400 we obtain solutions +-20, but I will assume omega must be positive,
so omega = 20
• One of the unspecified constants is theta_0. Suppose for example that theta_0 = 0. What is the remaining unspecified constant?
A
• Still assuming that theta_0 = 0, describe the graph of the solution function x(t).
I can’t figure out how to calculate the value of A, but the graph of x(t) would be a cosine function with amplitude of A, having a period of 2 pi, and not having any vertical or phase shift
• Repeat, this time assuming that theta_0 = 3 pi / 2.
This would not affect amplitude, still A, the period would still be 2 pi, but now it would have a phase shift 3 pi / 2 to the left
confidence rating #$&*:
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Given Solution:
If x = A cos(omega * t + theta_0) then x ' = - omega A sin(omega * t + theta_0) and x '' = -omega^2 A cos(omega * t + theta_0).
Our equation therefore becomes
m * (-omega^2 A cos(omega * t + theta_0) ) = - k A cos(omega * t + theta_0).
Rearranging we obtain
-m omega^2 A cos(omega * t + theta_0) = -k A cos(omega * t + theta_0)
so that
-m omega^2 = - k
and
omega = sqrt(k/m).
Thus the constant omega is determined by the equation.
The constants A and theta_0 are not determined by the equation and can therefore take any values.
No matter what values we choose for A and theta_0, the equation will be satisfied as long as omega = sqrt(k / m).
Our second-order equation
m x '' = - k x
therefore has a general solution containing two arbitrary constants.
In the present equation m = 5 and k = 2000, so that omega = sqrt(k / m) = sqrt(2000 / 5) = sqrt(400) = 20.
Our solution x(t) = A cos(omega * t + theta_0) therefore becomes
x(t) = A cos(20 t + theta_0).
If theta_0 = 0 the function becomes x(t) = A cos( 20 t ). The graph of this function will be a 'cosine wave' with a 'peak' at the origin, and a period of pi / 10.
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Self-critique (if necessary): I forgot to include the period change as a result of omega.
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Self-critique rating: 3
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Question:
`q004. In the preceding equation we found the general solution to the equation 5 x'' = - 2000 x. Assuming SI units, this solution applies to a simple harmonic oscillator of mass 5 kg, which when displaced to position x relative to equilibrium is subject to a net force F = - 2000 N / m * x. With these units, sqrt(k / m) has units of sqrt( (N / m) / kg), which reduce to radians / second. Our function x(t) describes the position of our oscillator relative to its equilibrium position.
Evaluate the constants A and theta_0 for each of the following situations:
• The oscillator reaches a maximum displacement of .3 at clock time t = 0.
• The oscillator reaches a maximum displacement of .3 , and at clock time t = 0 its position is x = .15.
• The oscillator has a maximum velocity of 2, and is at its maximum displacement of .3 at clock time t = 0.
• The oscillator has a maximum velocity of 2, which occurs at clock time t = 0. (Hint: The velocity of the oscillator is given by the function x ' (t) ).
As seen in the preceding problem, a general solution to the equation is
x = A cos(omega * t + theta_0),
where omega = sqrt(k / m). For the current equation 5 x '' = -2000 x, this gives us omega = 20. In the current context omega = 20 radians / second.
So
x(t) = A cos( 20 rad / sec * t + theta_0 ).
Maximum displacement occurs at critical values of t, values at which x ' (t) = 0.
Taking the derivative of x(t) we obtain
x ' (t) = - 20 rad / sec * A sin( 20 rad/sec * t + theta_0).
The sine function is zero when its argument is an integer multiple of pi, i.e., when
20 rad/sec * t + theta_0 = n * pi, where n = 0, +-1, +-2, ... .
A second-derivative test shows that whenever n is an even number, our x(t) function has a negative second derivative and therefore a maximum value.
We can therefore pick any even number n and we will get a solution.
If maximum displacement occurs at t = 0 then we have
20 rad / sec * 0 + theta_0 = n * pi
so that
theta_0 = n * pi, where n can be any positive or negative even number.
We are free to choose any such value of n, so we make the simplest choice, n = 0. This results in theta_0 = 0.
Now if x = .3 when t = 0 we have
A cos(omega * 0 + theta_0) = .3
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Your solution:
• The oscillator reaches a maximum displacement of .3 at clock time t = 0.
A = 0.3.. I think I’m heading into this wrong, and I really don’t understand what this question is asking for… I’m a highly visual learner, so I find it really hard to understand the concepts in this format… I think the dvd’s might help once I get those in
• The oscillator reaches a maximum displacement of .3 , and at clock time t = 0 its position is x = .15.
Oi, same story here.
• The oscillator has a maximum velocity of 2, and is at its maximum displacement of .3 at clock time t = 0.
ditto
• The oscillator has a maximum velocity of 2, which occurs at clock time t = 0. (Hint: The velocity of the oscillator is given by the function x ' (t) ).
Oh man…
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These questions can be challenging if you are rusty on first-year calculus and trigonometric functions.
To help get the rust out, try to answer the following without the use of a calculator:
Quick refresher on trigonometric functions:
What are the maximum and minimum possible values of cos(theta)?
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For what values of theta does cos(theta)
take its maximum value, and for what values of theta is cos(theta) equal to zero?
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For what values of t does y = cos(5 t) take its maximum value? For what t values does this function take its minimum value?
#$&*
What is the maximum possible value of y = 4 cos(83 t)?
#$&*
Find at least one value of t for which y = cos(2 t + pi) takes the value zero.
#$&*
Find at least one value of t for which y = 3 cos(2 t + pi/4) takes its maximum value.
#$&*
Find at least one value of t for which the derivative of y = cos(t) is zero.
#$&*
Find at least one value of t for which the derivative of y = cos(t) is zero, and the second derivative is negative.
#$&*
Find the critical points of y = 4 sin(2 pi t + pi/2), then using a second-derivative test find the values of t at which this function takes its maximum and minimum values.
Feel free to submit a copy of these questions with your answers.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): ….
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Self-critique rating: 0
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As is the case for most students, you might be a little rusty, or need a bit of a primer, with the behavior of trigonometric functions.
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