qa_02

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course PHY 231

002. Velocity*********************************************

Question: `q001. Note that there are 17 questions in this assignment.

If an object moves 12 meters in 4 seconds, then what is its average velocity? Explain how you obtained your result in terms of commonsense ideas.

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Your solution:

12 meters / 4 seconds = 3 meters per second

confidence rating #$&*:

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Given Solution:

Moving 12 meters in 4 seconds, we move an average of 3 meters every second.

We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will span 3 meters, corresponding to the distance moved in 1 second, on the average.

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q002. How is the preceding problem related to the concept of a rate?

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Your solution:

According to the dictionary, rate is a certain quantity or amount of one thing considered in relation to a unit of another thing and used as a standard or measure:

In our case, the amount of speed traveled per certain amount of time and meters per second is a standard measurement.

confidence rating #$&*:

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Given Solution:

A rate is obtained by dividing the change in a quantity by the change in another quantity on which is dependent. In this case we divided the change in position by the time during which that change occurred.

More specifically

The rate of change of A with respect to B is defined to be the quantity (change in A) / (change in B).

An object which moves 12 meters in 3 seconds changes its position by 12 meter during a change in clock time of 3 seconds. So the question implies

Change in position = 12 meters

Change in clock time = 3 seconds

When we divide the 12 meters by the 3 seconds we are therefore dividing (change in position) by (change in clock time). In terms of the definition of rate of change:

the change in position is the change in A, so position is the A quantity.

the change in clock time is the change in B, so clock time is the B quantity.

So

(12 meters) / (3 seconds) is

(change in position) / (change in clock time) which is the same as

average rate of change of position with respect to clock time.

Thus

average velocity is average rate of change of position with respect to clock time.

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Self-critique (if necessary):

I think we are describing the same idea, but you have a better talent at describing terms than me.

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Self-critique rating: OK

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Question: `q003. We are still referring to the situation of the preceding questions:

Is object position dependent on time or is time dependent on object position?

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Your solution:

I will say that object position is dependent on time since we can control the distance, as to how fast or in which direction, but we can not control time - we can not slow down or speed up time.

confidence rating #$&*:

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Given Solution:

Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else (this might not be so at the most fundamental level, but for the moment, unless you have good reason to do otherwise, this should be your convention).

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q004. We are still referring to the situation of the preceding questions, which concern average velocity:

So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations.

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Your solution:

I think I understand this concept.

confidence rating #$&*:

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Given Solution:

Be sure you have reviewed all the definitions and concepts associated with velocity. If there’s anything you don’t understand, be sure to address it in your self-critique.

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object? What is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas.

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Your solution:

-6 meters / 3 seconds = -2 meters per second

Since speed = velocity (I do not see a difference between speed and velocity, similarly, I do not see a difference between the word ""big"" and ""large""), then average speed = average velocity. Hence, average velocity is -2 meters per second

confidence rating #$&*:

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Given Solution:

Speed is the average rate at which distance changes with respect to clock time. Distance cannot be negative and the clock runs forward. Therefore speed cannot be negative.

Velocity is the average rate at which position changes with respect to clock time, and since position changes can be positive or negative, so can velocity.

In general distance has no direction, while velocity does have direction.

Putting it loosely, speed is just how fast something is moving; velocity is how fast and in what direction.

In this case, the average velocity is

vAve = `ds / `dt = -6 m / (3 s) = -2 m/s.

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Self-critique (if necessary):

Ah, ok I got it, distance has no direction and can not be negative.

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Self-critique rating: 3/3

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Question: `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which its position changes, then what expression stands for the average velocity vAve of the object during this time interval?

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Your solution:

vAve = 'ds / 'dt

confidence rating #$&*:

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Given Solution:

Average velocity is rate of change of position with respect to clock time.

Change in position is `ds and change in clock time is `dt, so average velocity is expressed in symbols as

vAve = `ds / `dt.

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q007. How do you write the expressions `ds and `dt on your paper?

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Your solution:

ds / dt

confidence rating #$&*:

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Given Solution:

You use the Greek capital Delta when writing on paper or when communicating outside the context of this course; this is the symbol that looks like a triangle. See Introductory Problem Set 1.

`d is used for typewritten communication because the symbol for Delta is not interpreted correctly by some Internet forms and text editors. You should get in the habit of thinking and writing the Delta symbol when you see `d.

You may use either `d or Delta when submitting work and answering questions.

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Self-critique (if necessary):

I don't know, I see that most of my professors white ""ds"" or ""dt"" rather than writing out delta s or delta t, so I see both ways as correct.

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Self-critique rating: 2/3

@&

The correct distinction is the following:

"ds" and "dt" refer to infinitesimal changes in s and t, which are the results of a limiting process.

" `ds " and " `dt " are finite changes in s and t.

*@

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Question: `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move?

How is this problem related to the concept of a rate?

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Your solution:

5 meters/second * 10 seconds = 50 meters

It is related to rate, in that we are using a rate of m/s and seconds to determine the meters traveled.

confidence rating #$&*:

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Given Solution:

In this problem you are given the rate at which position changes with respect to time, and you are given the time interval during which to calculate the change in position.

The definition of rate of change states that the rate of change of A with respect to B is (change in A) / (change in B), which we abbreviate as `dA / `dB. `dA stands for the change in the A quantity and `dB for the change in the B quantity.

For the present problem we are given the rate at which position changes with respect to clock time. The definition of rate of change is stated in terms of the rate of change of A with respect to B.

So we identify the position as the A quantity, clock time as the B quantity.

The basic relationship

ave rate = `dA / `dB

can be algebraically rearranged in either of two ways:

`dA = ave rate * `dB or

`dB = `dA / (ave rate)

Using position for A and clock time for B the above relationships are

ave rate of change of position with respect to clock time = change in position / change in clock time

change in position = ave rate * change in clock time

change in clock time = change in position / ave rate.

In the present situation we are given the average rate of change of position with respect to clock time, which is 5 meters / second, and the change in clock time, which is 10 seconds.

Thus we find

change in position = ave rate * change in clock time = 5 cm/sec * 10 sec = 50 cm.

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q009. If vAve stands for the rate at which the position of the object changes with respect to clock time (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position?

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Your solution:

VAve = 'ds / 'dt

VAve * 'dt = 'ds

confidence rating #$&*:

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Given Solution:

To find the change in a quantity we multiply the rate by the time interval during which the change occurs.

The velocity is the rate, so we obtain the change in position by multiplying the velocity by the time interval:

`ds = vAve * `dt.

The units of this calculation pretty much tell us what to do:

We know what it means to multiply pay rate by time interval (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour).

When we multiply vAve, for example in in units of cm / sec or meters / sec, by `dt in seconds, we get displacement in cm or meters. Similar reasoning applies if we use different measures of distance.

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem.

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Your solution:

As stated previously, the definition for rate is a certain quantity or amount of one thing considered in relation to a unit of another thing and used as a standard or measure:

In our case, the ""one thing"" is 'ds and the ""another thing"" is 'dt. As by definition, when we have 'ds per 'dt, we automatically have vAve

confidence rating #$&*:

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Given Solution:

vAve is the average rate at which position changes. The change in position is the displacement `ds, the change in clock time is `dt, so vAve = `ds / `dt.

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result?

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Your solution:

We multiply both sides by 'dt o solve this equation for `ds. Our result is: 'ds = vAve * 'dt

confidence rating #$&*:

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Given Solution:

To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps:

vAve = `ds / `dt. Multiply both sides by `dt:

vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1

vAve * `dt = `ds . Switching sides we have

`ds = vAve * `dt.

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q012. How is the preceding result related to our intuition about the meanings of the terms average velocity, displacement and clock time?

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Your solution:

When we divide the TOTAL displacement over TOTAL time, we obtain AVERAGE velocity. So total/total = average.

confidence rating #$&*:

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Given Solution:

For most of us our most direct intuition about velocity probably comes from watching an automobile speedometer.

We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea.

Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval.

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Self-critique (if necessary):

I didn't completely understand the question, as to specifically what you are asking, but the answer seems to fall within guidelines of your answer.

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Self-critique rating: 3/3

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Question: `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?

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Your solution:

We multiply both sides by 'dt and we obtain vAve * 'dt = 'ds. Then we divide both sides by vAve and we obtain 'dt = 'ds / vAve.

confidence rating #$&*:

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Given Solution:

To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps:

vAve = `ds / `dt. Multiply both sides by `dt:

vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1

vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve.

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Self-critique (if necessary):

OK

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Self-critique rating: OK

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Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?

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Your solution:

I guess, that out of the three (velocity, displacement and clock time), if we know two of them, we can automatically using algebra obtain the third one.

confidence rating #$&*:

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Given Solution:

If we want to know how long it will take to make a trip at a certain speed, we know to divide the distance in miles by the speed in mph.

If we divide the number of miles we need to travel by the number of miles we travel in hour, we get the number of hours required. This is equivalent to the calculation `dt = `ds / vAve.

We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval.

When dealing with displacement, velocity and time interval, we can always check our thinking by making the analogy with a simple example involving miles, hours and miles/hour.

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Self-critique (if necessary):

I think my answer falls along the lines of your answer, but I don't really see the difference between this question and question #12

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Self-critique rating:

3/3

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: `q015. A ball falls 20 meters from rest in 2 seconds. What is the average velocity of its fall?

Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change.

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Your solution:

As shown in problem #13, vAve = `ds / `dt, so...

'ds / 'dt = 20 meters / 2 seconds = 10 meters per second on average = vAve

confidence rating #$&*:

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Question: `q016. A car moves at an average speed of 20 m/s for 6 seconds. How far does it move?

Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change.

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Your solution:

Average velocity is meters / seconds

Meters / seconds * seconds = meters

20 m/s * 6 seconds = 120 meters

confidence rating #$&*:

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Question: `q017. An object's position changes by amount `ds during a time interval `dt. What is the expression for its average velocity during this interval?

Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change.

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Your solution:

Velocity is change in distance divided by change in time which equals to 'ds / 'dt

confidence rating #$&*:

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Self-critique rating: 3/3"

3/3

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: `q015. A ball falls 20 meters from rest in 2 seconds. What is the average velocity of its fall?

Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change.

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Your solution:

As shown in problem #13, vAve = `ds / `dt, so...

'ds / 'dt = 20 meters / 2 seconds = 10 meters per second on average = vAve

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Question: `q016. A car moves at an average speed of 20 m/s for 6 seconds. How far does it move?

Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change.

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Your solution:

Average velocity is meters / seconds

Meters / seconds * seconds = meters

20 m/s * 6 seconds = 120 meters

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Question: `q017. An object's position changes by amount `ds during a time interval `dt. What is the expression for its average velocity during this interval?

Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change.

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Your solution:

Velocity is change in distance divided by change in time which equals to 'ds / 'dt

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Self-critique rating: 3/3"

#*&!

Self-critique (if necessary):

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Self-critique rating:

&#Your work looks good. See my notes. Let me know if you have any questions. &#

course Mth 279

Class Notes 110110The 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 one 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 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)

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Your solution:

• x = cos(t), x’ = -sin(t), x’’ = -cos(t)

o m(-cos(t)) = -k(cos(t))

o m = k SATISFIED

• x = sin( sqrt( k / m) * t),

o x’ = k/m cos( sqrt( k / m) * t), x’’ = -(k/m)^2 sin( sqrt( k / m) * t)

o m[-(k/m)^2 sin( sqrt( k / m) * t)] = -k[sin( sqrt( k / m) * t)]

o m(k/m)^2 = k k = m SATISFIED

• x = 3 cos( sqrt( k / m) * t ) + 5 sin (sqrt(k / m) t)

o x’ = -3 sqrt(k/m) sin( sqrt( k / m) * t ) + 5 sqrt(k/m) cos( sqrt( k / m) * t )

o x’’ = -3(k/m)cos( sqrt( k / m) * t ) -5(k/m)sin( sqrt( k / m) * t )

o 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)]

o m (k/m) [-3cos( sqrt( k / m) * t ) -5sin( sqrt( k / m) * t )] = k[-3 cos( sqrt( k / m) * t ) - 5 sin (sqrt(k / m) t)]

o ((mk)/m) = k

o k = k

o ???????

@&

You have arrived at an identity which is equivalent to the original equations, confirming that your original equation is true for all values of the variables.

In other words, x = 3 cos( sqrt( k / m) * t ) + 5 sin (sqrt(k / m) t) is a solution to the equation.

*@

• x = B sin(sqrt(k / m) * t) + C cos( sqrt( k / m) * t + 3)

o x’ = B sqrt(k/m) cos(sqrt(k / m) * t) - C sqrt(k/m) sin(sqrt(k / m) * t + 3)

o x’’ = -B (k/m) sin(sqrt(k / m) * t) - C (k/m) cos(sqrt(k / m) * t + 3)

o m[B sin(sqrt(k / m) * t) + C cos( sqrt( k / m) * t + 3)] = -k[-B (k/m) sin(sqrt(k / m) * t) - C (k/m) cos(sqrt(k / m) * t + 3)]

@&

You appear to have substituted x where you should have substituted x '', and vice versa.

*@

o m^2 = k^2

confidence rating #$&*:

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Given Solution:

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.

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 do not understand why I got the third equation to be false. I did it twice and got k = k both times.

@&

k = k is a true statement, equivalent to your original equation.

*@

I guess the fourth does not satisfy because m could be positive while k is negative of vice-versa???

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Self-critique rating:

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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:

If this were the case, we would be taking the integral with respect to x and then t for the separate terms. However, we must take the full integral with respect to ONE variable. In this case, we will take it with respect to t.

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):

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Self-critique rating: OK

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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:

1. x = A cos(omega * t + theta_0), x’ = -A*omega sin(omega * t + theta_0), x’’ = -A*omega^2 cos(omega * t + theta_0)

2. m[-A*omega^2 cos(omega * t + theta_0)] = -k[A cos(omega * t + theta_0)]

-m*omega^2 = -k

omega = sqrt(k/m)

3. Setting a value for theta_0 does not change the value of omega.

4. theta_0 does not determine the graph.

5. theta_0 does not determine the graph.

@&

The value of theta_0 has an important effect on the graph. The value of theta_0 affects the horizontal shift of the graph.

You will need to review basic graphing, as I've indicated in notes on previous assignments.

*@

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):

A little unsure on 4 and 5.

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Self-critique rating:

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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:

1. The oscillator reaches a maximum displacement of .3 at clock time t = 0.

2. The oscillator reaches a maximum displacement of .3 , and at clock time t = 0 its position is x = .15.

3. The oscillator has a maximum velocity of 2, and is at its maximum displacement of .3 at clock time t = 0.

4. 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) ).

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Your solution:

5 x'' = - 2000 x mx’’ = -kx

x = A cos(omega * t + theta_0)

omega = sqrt(k/m) omega = sqrt(2000/5) = 20rad/s

x = A cos(20 * t + theta_0)

Max values are where x’(t) = 0. Where the slope = 0.

x’(t) = -20A cos(20 * t + theta_0) sin(x) = 0 when x = 0 or n*pi

confidence rating #$&*:

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Given Solution:

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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Self-critique (if necessary):

I need clarification on this line: “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.” ????

I am really having trouble on this concept????

@&

It isn't clear from your note whether you are having trouble with the second-derivative test (this is a first-semester calculus procedure and numerous excellent resources are available online for review of the concept) or its application to this specific problem.

As with the graphing question, you have a little time before functions of this nature become important in the course, but you need to identify the things that need review and start that process.

If you want to submit a copy of this problem, including your solution and the given solution, along with a more specific question or questions, I'll be glad to respond.

*@

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Self-critique rating:

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Question:

`q005. Describe the motion of the oscillator in each of the situations of the preceding problem. SI units for position and velocity are respectively meters and meters / second.

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Your solution:

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Question:

Part II: Solutions of equations requiring only direct integration.

`q006. Find the general solution of the equation x ' = 2 t + 4, and find the particular solution of this equation if we know that x ( 0 ) = 3.

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Your solution:

Find the integral of both sides:

x(t) = t^2 + 4t + C

x(0) = 0^2 + 4(0) + C = 3 C = 3

x(t) = t^2 + 4t + 3

confidence rating #$&*:

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Given Solution:

Integrating both sides we obtain

x(t) = t^2 + 4 t + c,

where c is an arbitrary constant.

The condition x(0) = 3 becomes

x(0) = 0^2 + 4 * 0 + c = 3,

so that c = 3 and our particular solution is

x(t) = t^2 + 4 t + 3.

We check our solution.

Substituting x(t) = t^2 + 4 t + 3 back into the original equation:

(t^2 + 4 t + 3) ' = 2 t + 4 yields

2 t + 4 = 2 t + 4,

verifying the general solution.

The particular solution satisfies x(0) = 3:

x(0) = 0^2 + 4 * 0 + 3 = 3.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question:

`q007. Find the general solution of the equation x ' ' = 2 t - .5, and find the particular solution of this equation if we know that x ( 0 ) = 1, while x ' ( 0 ) = 7.

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Your solution:

x’(t) = t^2 - ½ t + C1

x(t) = 1/3 t^3 - Ό t^2 + C1*t + C2

x’(0) = 0 - 0 + C1 = 7 C1 = 7

x’(t) = t^2 - ½ t + 7

x(0) = 0 - 0 + 0 + C2 = 1 C2 = 1

x(t) = 1/3 t^3 - Ό t^2 + 7*t + 1

confidence rating #$&*:

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Given Solution:

Integrating both sides we obtain

x ' = t^2 - .5 t + c_1,

where c_1 is an arbitrary constant.

Integrating this equation we obtain

x = t^3 / 3 - .25 t^2 + c_1 * t + c_2,

where c_2 is an arbitrary constant.

Our general solution is thus

x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2.

The condition x(0) = 1 becomes

x(0) = 0^3 / 3 - .25 * 0^2 + c_1 * 0 + c_2 = 1

so that c_2 = 1.

x ' (t) = t^2 - .5 t + c_1, so our second condition x ' (0) = 7 becomes

x ' (0) = 0^2 - .5 * 0 + c_1 = 7

so that c_1 = 7.

For these values of c_1 and c_2, our general solution x(t) = t^3 / 3 - .25 t^2 + c_1 * t + c_2 becomes the particular solution

x(t) = t^3 / 3 - .25 t^2 + 7 t + 1.

You should check to be sure this solution satisfies both the given equation and the initial conditions.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question:

`q008. Use the particular solution from the preceding problem to find x and x ' when t = 3. Interpret your results if x(t) represents the position of an object at clock time t, assuming SI units.

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Your solution:

x(t) = 1/3 t^3 - Ό t^2 + 7*t + 1

x(3) = 1/3 * 27 - Ό * 9 + 21 + 1 = 28.75

x’(3) = 3^2 - ½ *3 + 7 = 14.5

confidence rating #$&*:

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Given Solution:

Our solution was

x(t) = t^3 / 3 - .25 t^2 + 7 t + 1.

Thus

x ' (t) = t^2 - .5 t + 7.

When t = 3 we obtain

x(3) = 3^3 / 3 - .25 * 3^2 + 7 * 3 + 1 = 28.75

and

x ' (3) = 3^2 - .5 * 3 + 7 = 14.5.

A graph of x vs. t would therefore contain the point (3, 28.75), and the slope of the tangent line at that point would be 14.5.

x(t) would represent the position of an object. x(3) = 28.75 represents an object whose position with respect to the origin is 28.75 meters when the clock reads 3 seconds.

x ' (t) would represent the velocity of the object. x ' (3) = 14.5 indicates that the object is moving at 14.5 meters / second when the clock reads 3 seconds.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question:

`q009. The equation x '' = -F_frict / m - c / m * x ', where the derivative is understood to be with respect to t, is of at least one of the forms listed below. Which form(s) are appropriate to the equation?

• x '' = f(x, x')

• x '' = f(t)

• x '' = f(x, t)

• x '' = f(x', t)

• x '' = f(x, x ' t)

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Your solution:

x’ MUST be on the right side of the equation. So these forms work: x '' = f(x, x'), x '' = f(x', t), x '' = f(x, x ' t).

confidence rating #$&*:

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Given Solution:

The right-hand side of the equation includes the function x ' but does not include the variable t or the function x.

So the right-hand side can be represented by any function which includes among its variables x '. That function may also include x and/or t as a variable.

The forms f(t) and f(x, t) fail to include x ', so cannot be used to represent this equation.

All the other forms do include x ' as a variable, and may therefore be used to represent the equation.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question:

`q010. If F_frict is zero, then the function x in the equation

x '' = -F_frict / m - c / m * x '

represents the position of an object of mass m, on which the net force is - c * x '.

Explain why the expression for the net force is -c * x '.

Explain what happens to the net force as the object speeds up.

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Your solution:

ma = Fnet = mx’’ x’’ = Fnet/m

With friction: x '' = -F_frict / m - c / m * x '

In this expression, the Fnet is -F_frict / m - c / m. If Ffrict = 0, then Fnet = -c/m.

As the object speeds up, the net force increses.

confidence rating #$&*:

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Given Solution:

Newton's Second Law gives us the general equation

m x '' = F_net

so that

x '' = F_net / m.

It follows that

x '' = -F_frict / m - c / m * x '

represents an object on which the net force is -F_frict - c x '.

If F_frict = 0, then it follows that the net force is -c x '.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question:

`q011. We continue the preceding problem.

• If w(t) = x '(t), then what is w ' (t)?

• If x '' = - b / m * x ', then if we let w = x ', what is our equation in terms of the function w?

• Is it possible to integrate both sides of the resulting equation?

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Your solution:

• w’(t) = x’’(t)

• w’(t) = - b / m * w(t)

• No because we do not know enough about the function w(t).

confidence rating #$&*:

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Given Solution:

If w(t) = x ' (t) then w ' (t) = (x ' (t) ) ' = x '' ( t ).

If x '' = - b / m * x ', then if w = x ' it follows that x '' = w ', so our equation becomes

w ' (t) = - b / m * w (t)

The derivative is with respect to t, so if we wish to integrate both sides we will get

w(t) = integral ( - b / m * w(t) dt),

The variable of integration is t, and we don't know enough about the function w(t) to perform the integration on the right-hand side.

[ Optional Preview:

There is a way around this, which provides a preview of a technique we will study soon. It isn't too hard to understand so here's a preview:

w ' (t) means dw / dt, where w is understood to be a function of t.

So our equation is dw/dt = -b / m * w.

It turns out that in this context we can sort of treat dw and dt as algebraic quantities, so we can rearrange this equation to read

dw / w = -b / m * dt.

Integrating both sides we get

integral (dw / w) = -b / m integral( dt )

so that

ln | w | = -b / m * t + c.

In exponential form this is

w = e^(-b / m * t + c).

There's more, but this is enough for now ... ].

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Self-critique (if necessary):

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Self-critique rating:OK

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Question:

Part III: Direction fields and approximate solutions

`q012. Consider the equation x ' = (2 x - .5) * (t + 1). Suppose that x = .3 when t = .2.

If a solution curve passes through (t, x) = (.2, .3), then what is its slope at that point?

What is the equation of its tangent line at this point?

If we move along the tangent line from this point to the t = .4 point on the line, what will be the x coordinate of our new point?

If a solution curve passes through this new point, then what will be the slope at the this point, and what will be the equation of the new tangent line?

If we move along the new tangent line from this point to the t = .6 point, what will be the x coordinate of our new point?

Is is possible that both points lie on the same solution curve? If not, does each tangent line lie above or below the solution curve, and how much error do you estimate in the t = .4 and t = .5 values you found?

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Your solution:

x ' = (2 x - .5) * (t + 1)

The derivative = rate of change = slope. At (.2, .3):

x ' = (2 (.3) - .5) * (.2 + 1) = .12

x- x1 = m(t - t1)

x - .3 = .12(t - .2)

Solve for x:

x = .12t - .024 + .3

x = .12t + .28

confidence rating #$&*:

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Given Solution:

At the point (.2, .3) in the (t, x) plane, our value of x ' is

x ' = (2 * .3 - .5) * (.2 + 1) = .12, approximately.

This therefore is the slope of any solution curve which passes through the point (.2, .3).

The equation of the tangent plane is therefore

x - .3 = .12 * (t - .2)

so that

x = .12 t - .24.

If we move from the t = .2 point to the t = .4 point our t coordinate changes by `dt = .2, so that our x coordinate changes by `dx = (slope * `dt) = .12 * .2 = .024. Our new x coordinate will therefore be .3 + .024 = .324.

This gives us the new point (.4, .324).

At this point we have

x ' = (2 * .324 - .5) * (.4 + 1) = .148 * 1.4 = .207.

If we move to the t = .6 point our change in t is `dt = .2. At slope .207 this would imply a change in x of `dx = slope * `dt = .207 * .2 = .041. Our new x coordinate will therefore be .324 + .041 = .365.

Our t = .6 point is therefore (.6, .365).

From our two calculated slopes, the second of which is significantly greater than the first, it appears that in this region of the x-t plane, as we move to the right the slope of our solution curve in fact increases. Our estimates were based on the assumption that the slope remains constant over each t interval. We conclude that our estimates of the changes in x are probably a somewhat low, so that our calculated points lie a little below the solution curve.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question:

`q013. Consider once more the equation x ' = (2 x - .5) * (t + 1).

Note on notation:

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)

can be specified succinctly in set notation as

{ (t, x) | t = 0, 1/4, ..., 1, x = 0, 1/4, ..., 1}.

( A more standard notation would be { (i / 4, j / 4) | 0 <= i <= 4, 0 <= j <= 4 } )

Find the value of x ' at every point of this grid and sketch the corresponding direction field. To get you started the values corresponding to the first, second and last rows of the grid are

-.5, -.625, -.75, -.875, -1

0, 0, 0, 0, 0

...

...

1.5, 1.875, 2.25, 2.625, 3

So you will only need to calculate the values for the third and fourth rows of the grid.

• List your values of x ' at the five points (0, 0), (1/4, 1/4), (1/2, 1/2), (3/4, 3/4) and (1, 1).

• Sketch the curve which passes through the point (t, x) = (.2, .3).

• Describe your curve. Is it increasing or decreasing, and is it doing so at an increasing or decreasing rate?

• According to your curve, what will be the value of x when t = 1?

• Sketch the curve which passes through the point (t, x) = (.5, .7). According to your curve, what will be the value of x when t = 1?

• Describe your curve and compare it with the curve you sketched through the point (.2, .3).

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Your solution:

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

@&

You need to make your best attempt to answer this question.

The assigned introductory videos are relevant to this problem.

You should submit a copy of the problem and your best attempt at a solution, and/or questions.

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Question:

`q014. We're not yet done with the equation x ' = (2 x - .5) * (t + 1).

x ' is the derivative of the x(t) function with respect to t, so this equation can be written as

dx / dt = (2 x - .5) * (t + 1).

Now, dx and dt are not algebraic quantities, so we can't multiply or divide both sides by dt or by dx. However let's pretend that they are algebraic quantities, and that we can. Note that dx is a single quantity, as is dt, and we can't divide the d's.

• Rearrange the equation so that expressions involving x are all on the left-hand side and expressions involving t all on the right-hand side.

• Put an integral sign in front of both sides.

• Do the integrals. Remember that an integration constant is involved.

• Solve the resulting equation for x to obtain your general solution.

• Evaluate the integration constant assuming that x(.2) = .3.

• Write out the resulting particular solution.

• Sketch the graph of this function for 0 <= t <= 1. Describe your graph.

• How does the value of your x(t) function at t = 1 compare to the value your predicted based on your previous sketch?

• How do your values of x(t) at t = .4 and t = .6 compare with the values you estimated previously?

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Your solution:

confidence rating #$&*:

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Given Solution:

The equation is easily rearranged into the form

dx / (2 x - .5) = (t + 1) dt.

Integrating the left-hand side we obtain 1/2 ln | 2 x - .5 |

Integrating the right-hand side we obtain t^2 / 2 + 4 t + c, where the integration constant c is regarded as a combination of the integration constants from the two sides.

Thus our equation becomes

1/2 ln | 2 x - 5 | = t^2 / 2 + t + c.

Multiplying both sides by 2, then taking the exponential function of both sides we get

exp( ln | 2 x - 5 | ) = exp( t^2 + 2 t + c ),

where as before c is an arbitrary constant.

Since the exponential and natural log are inverse functions the left-hand side becomes | 2 x + .5 |.

The right-hand side can be written e^c * e^(t^2 + 8 t), where c is still an arbitrary constant. e^c can therefore be any positive number, and we replace e^c with A, understanding that A is a positive constant.

Our equation becomes

| 2 x - .5 | = A e^(t^2 + 2 t).

For x > -.25, as is the case for our given value x = .3 when t = .2, we have

2 x - .5 = A e^(t^2 + 2 t)

so that

x = A e^(t^2 + 2 t) + .25.

Using x = .3 and t = .2 we find the value of A:

.05 = A e^(.2^2 + 2 * .2)

so that

A = .05 / e^(.44) = .03220, approx..

Our solution function is therefore

x(t) = .05 / e^(.44) * e^(t^2 + 2 t) + .25, or approximately

x(t) = .03220 e^(t^2 + 2 t) + .25

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Self-critique (if necessary):

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Question:

`q015. OK, this time we are really going to be done with this equation. Again, x ' = (2 x - .5) * (t + 1)

• Along what line or curve is x ' = 1?

• Along what line or curve is x ' = 0?

• Along what line or curve is x ' = 2?

• Along what line or curve is x ' = -1?

• Sketch these three lines and/or curves for 0 <= t <= 1.

• Along each of these lines x ' is constant. Along each sketch 'slope segments' with slopes equal to the corresponding value of x '.

• How consistent is your sketch with your previous sketch of the direction field?

• Sketch a solution curve through the point (.2, .25), and estimate the coordinates of the t = 1 point on this curve.

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Your solution:

confidence rating #$&*:

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Given Solution:

x ' = 1 when

(2 x - .5) * (t + 1) = 1.

Solving for x we obtain

x = 1/2 ( 1 / (t + 1) + .5) = 1 / (2(t + 1)) + .25.

The resulting curve is just the familiar curve x = 1 / t, vertically compressed by factor 2 then shifted -1 unit in the horizontal and .25 unit in the vertical direction, so its asymptotes are the lines t = -1 and x = .25. The t = 0 and t = 1 points are (0, .75) and (1, .5).

Similarly we find the curves corresponding to the other values of x ':

For x ' = 0 we get the horizontal line x = .25. Note that this line is the horizontal asymptote to the curve obtained in the preceding step.

For x ' = 2 we get the curve 1 / (t + 1) + .25, a curve with asymptotes at t = -1 and x = .25, including points (0, 1.25) and (1, .75).

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Self-critique (if necessary):

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Self-critique rating:

These questions can be challenging if you are rusty on first-year calculus and trigonometric functions.

Fortunately we won't be using these functions a lot near the beginning of the course, but they become very important later, and if you're rusty, or if you never really mastered these functions, you want to get a good start now.

If you are rusty, 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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f`q001

@&

You're doing well with much of this.

However you didn't answer that last sequence of questions, and you do need a refresher on graphing. The graphs of trigonometric functions aren't too important in Module 1, but you will need solid knowledge of the topic in Module 2.

See my notes.

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