Assignment 0

#$&*

course Phy 201

*********************************************

Question: This question, related to the use of the TIMER program in an experimental situation, is posed in terms of a familiar first-semester system.

Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a straight wooden incline about 50 cm long. If the computer timer indicates that on five trials the times of an object down an incline are 2.42sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then:

Are the discrepancies in timing on the order of 0.1 second, 0.01 second, or 0.001 second?

****

The discrepencies in timing are on the order of .001 seconds.

#$&*

To what extent do you think the discrepancies in the time intervals could be explained by each of the following:

• The lack of precision of the TIMER program. Base your answer on the precision of the TIMER program as you have experienced it. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

The timer program is extremely precise. I think the discrepancies in timing are do to precision or lack there of approximately 5% of the time because the timer is so precise.

#$&*

• The uncertainty associated with human triggering (uncertainty associated with an actual human finger on a computer mouse). What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****I think that discrepencies due to human error happen 15% of the time because humans are not very precise due to neural and motor control

#$&*

• Actual differences in the time required for the object to travel the same distance. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

I think that 15% of differences are due to this factor because the perception of start and finish can vary along with delayed reaction on timer starting and finishing.

#$&*

• Differences in positioning the object prior to release. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

I think that differences in positioning only contribute 5% to discrepencies because this can easily be made precise and there is not a large margin for error.

#$&*

• Human uncertainty in observing exactly when the object reached the end of the incline. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

20% of the time I think human uncertainty contributes to uncertainty because there is a wide range of perception.

****

#$&*

*********************************************

Question: If you had carefully timed the ball and obtained the results given above, how confident would you be that the mean of those five intervals was within 0.1 seconds of the actual mean? (Note that the mean of the given intervals is 2.43 seconds, as rounded to three significant figures)? Briefly explain your thinking.

****

I would be very confident that the mean would be within.1 of the actual mean because I believe that the timer is precise to the .1 second.

#$&*

How confident would you be that the 2.43 second mean is within .01 second? Briefly explain your thinking.

****

I would be less confident that the mean is with .01 of that actual mean because there is more room for error as precision increase.

#$&*

How confident would you be that the 2.43 second mean is within .03 second?

****

I would be slightly more confident that the mean is within .03 of a second

#$&*

At what level do you think you can be confident of the various degrees of uncertainty?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.1 second of the actual mean?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.01 second of the actual mean?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.03 second of the actual mean?

Give your three answers and briefly explain your thinking:

****

I would only be 90% confident that the mean is within .1 seconds. I would be less than 90% confident that it is within .01 or .03.

#$&*

*********************************************

Question: What, if anything, could you do about the uncertainty due to each of the following? Address each specifically.

• The lack of precision of the TIMER program.

****

You cannot do anything to adjust the precision of the timer program.

#$&*

• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

****

To adjust the precision of the human triggering you can increase the sensitivity of the mouse.

#$&*

• Actual differences in the time required for the object to travel the same distance.

****

You cannot adjust actual time required.

#$&*

• Differences in positioning the object prior to release.

****

To adjust differences in positioning you can create a mark on the starting point and face the object the same way each time.

#$&*

• Human uncertainty in observing exactly when the object reached the end of the incline.

****

To adjust for human uncertainty you can use slow motion or a pause feature.

#$&*

*********************************************

Question: If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

You would divide the distance traveled by the time required resulting in the speed in inces/sec for example

confidence rating #$&*:3

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

#$&*

*********************************************

Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: It’s average velocity is 8 cm/sec. This is relevant to traveling long distances and wanting to know your average velocity over the trip

confidence rating #$&*:3

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

#$&*

*********************************************

Question: If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: The average velocity for the first half is 6.6 cm/sec and 1cm/sec for the second half.

confidence rating #$&*:3

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

#$&*

*********************************************

Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:I think that doubling the length will result in less than half the frequency.

confidence rating #$&*:2

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

#$&*

*********************************************

Question: `qNote that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Because (0,0) is equilibrium this is where the point is equally from extremes in all directions.

confidence rating #$&*:3

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

#$&*

*********************************************

Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: This isn’t possible because for the frequency to cross the vertical axis it would have to be negative at some point and you can’t have negative frequency.

Assessment:2

#$&*

*********************************************

Question: `qOn a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: This is also not possible for the graph to intersect the horizontal axis because that means the length would be negative and the length cannot be negative.

confidence rating #$&*:2

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

#$&*

*********************************************

Question: `qIf a ball rolls between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: The points are 30 cm apart. 6cm/sec*5seconds =30cm.

confidence rating #$&*:3

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

#$&*

.............................................

Given Solution:

`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm.

The formal calculation goes like this:

• We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval.

• It follows by algebraic rearrangement that `ds = vAve * `dt.

• We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that

• `ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.

The details of the algebraic rearrangement are as follows:

• vAve = `ds / `dt. We multiply both sides of the equation by `dt:

• vAve * `dt = `ds / `dt * `dt. We simplify to obtain

• vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt

Be sure to address anything you do not fully understand in your self-critique.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

#$&*

*********************************************

Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: I understand significant figures, but I am not sure under what circumstances you use them and when you do not.

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

@&

You should always be aware of significant figures.

In labs especially you should try to answer accordingly.

*@

Assignment 0

#$&*

course Phy 201

*********************************************

Question: This question, related to the use of the TIMER program in an experimental situation, is posed in terms of a familiar first-semester system.

Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a straight wooden incline about 50 cm long. If the computer timer indicates that on five trials the times of an object down an incline are 2.42sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then:

Are the discrepancies in timing on the order of 0.1 second, 0.01 second, or 0.001 second?

****

The discrepencies in timing are on the order of .001 seconds.

#$&*

To what extent do you think the discrepancies in the time intervals could be explained by each of the following:

• The lack of precision of the TIMER program. Base your answer on the precision of the TIMER program as you have experienced it. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

The timer program is extremely precise. I think the discrepancies in timing are do to precision or lack there of approximately 5% of the time because the timer is so precise.

#$&*

• The uncertainty associated with human triggering (uncertainty associated with an actual human finger on a computer mouse). What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****I think that discrepencies due to human error happen 15% of the time because humans are not very precise due to neural and motor control

#$&*

• Actual differences in the time required for the object to travel the same distance. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

I think that 15% of differences are due to this factor because the perception of start and finish can vary along with delayed reaction on timer starting and finishing.

#$&*

• Differences in positioning the object prior to release. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

I think that differences in positioning only contribute 5% to discrepencies because this can easily be made precise and there is not a large margin for error.

#$&*

• Human uncertainty in observing exactly when the object reached the end of the incline. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

20% of the time I think human uncertainty contributes to uncertainty because there is a wide range of perception.

****

#$&*

*********************************************

Question: If you had carefully timed the ball and obtained the results given above, how confident would you be that the mean of those five intervals was within 0.1 seconds of the actual mean? (Note that the mean of the given intervals is 2.43 seconds, as rounded to three significant figures)? Briefly explain your thinking.

****

I would be very confident that the mean would be within.1 of the actual mean because I believe that the timer is precise to the .1 second.

#$&*

How confident would you be that the 2.43 second mean is within .01 second? Briefly explain your thinking.

****

I would be less confident that the mean is with .01 of that actual mean because there is more room for error as precision increase.

#$&*

How confident would you be that the 2.43 second mean is within .03 second?

****

I would be slightly more confident that the mean is within .03 of a second

#$&*

At what level do you think you can be confident of the various degrees of uncertainty?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.1 second of the actual mean?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.01 second of the actual mean?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.03 second of the actual mean?

Give your three answers and briefly explain your thinking:

****

I would only be 90% confident that the mean is within .1 seconds. I would be less than 90% confident that it is within .01 or .03.

#$&*

*********************************************

Question: What, if anything, could you do about the uncertainty due to each of the following? Address each specifically.

• The lack of precision of the TIMER program.

****

You cannot do anything to adjust the precision of the timer program.

#$&*

• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

****

To adjust the precision of the human triggering you can increase the sensitivity of the mouse.

#$&*

• Actual differences in the time required for the object to travel the same distance.

****

You cannot adjust actual time required.

#$&*

• Differences in positioning the object prior to release.

****

To adjust differences in positioning you can create a mark on the starting point and face the object the same way each time.

#$&*

• Human uncertainty in observing exactly when the object reached the end of the incline.

****

To adjust for human uncertainty you can use slow motion or a pause feature.

#$&*

*********************************************

Question: If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

You would divide the distance traveled by the time required resulting in the speed in inces/sec for example

confidence rating #$&*:3

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

#$&*

*********************************************

Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: It’s average velocity is 8 cm/sec. This is relevant to traveling long distances and wanting to know your average velocity over the trip

confidence rating #$&*:3

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

#$&*

*********************************************

Question: If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: The average velocity for the first half is 6.6 cm/sec and 1cm/sec for the second half.

confidence rating #$&*:3

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

#$&*

*********************************************

Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:I think that doubling the length will result in less than half the frequency.

confidence rating #$&*:2

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

#$&*

*********************************************

Question: `qNote that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Because (0,0) is equilibrium this is where the point is equally from extremes in all directions.

confidence rating #$&*:3

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

#$&*

*********************************************

Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: This isn’t possible because for the frequency to cross the vertical axis it would have to be negative at some point and you can’t have negative frequency.

Assessment:2

#$&*

*********************************************

Question: `qOn a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: This is also not possible for the graph to intersect the horizontal axis because that means the length would be negative and the length cannot be negative.

confidence rating #$&*:2

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

#$&*

*********************************************

Question: `qIf a ball rolls between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: The points are 30 cm apart. 6cm/sec*5seconds =30cm.

confidence rating #$&*:3

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

#$&*

.............................................

Given Solution:

`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm.

The formal calculation goes like this:

• We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval.

• It follows by algebraic rearrangement that `ds = vAve * `dt.

• We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that

• `ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.

The details of the algebraic rearrangement are as follows:

• vAve = `ds / `dt. We multiply both sides of the equation by `dt:

• vAve * `dt = `ds / `dt * `dt. We simplify to obtain

• vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt

Be sure to address anything you do not fully understand in your self-critique.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

#$&*

*********************************************

Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: I understand significant figures, but I am not sure under what circumstances you use them and when you do not.

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

@&

You should always be aware of significant figures.

In labs especially you should try to answer accordingly.

*@

#*&!

Assignment 0

#$&*

course Phy 201

*********************************************

Question: This question, related to the use of the TIMER program in an experimental situation, is posed in terms of a familiar first-semester system.

Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a straight wooden incline about 50 cm long. If the computer timer indicates that on five trials the times of an object down an incline are 2.42sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then:

Are the discrepancies in timing on the order of 0.1 second, 0.01 second, or 0.001 second?

****

The discrepencies in timing are on the order of .001 seconds.

#$&*

To what extent do you think the discrepancies in the time intervals could be explained by each of the following:

• The lack of precision of the TIMER program. Base your answer on the precision of the TIMER program as you have experienced it. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

The timer program is extremely precise. I think the discrepancies in timing are do to precision or lack there of approximately 5% of the time because the timer is so precise.

#$&*

• The uncertainty associated with human triggering (uncertainty associated with an actual human finger on a computer mouse). What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****I think that discrepencies due to human error happen 15% of the time because humans are not very precise due to neural and motor control

#$&*

• Actual differences in the time required for the object to travel the same distance. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

I think that 15% of differences are due to this factor because the perception of start and finish can vary along with delayed reaction on timer starting and finishing.

#$&*

• Differences in positioning the object prior to release. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

****

I think that differences in positioning only contribute 5% to discrepencies because this can easily be made precise and there is not a large margin for error.

#$&*

• Human uncertainty in observing exactly when the object reached the end of the incline. What percent of the discrepancies in timing do you think are due to this factor, and why do you think so?

20% of the time I think human uncertainty contributes to uncertainty because there is a wide range of perception.

****

#$&*

*********************************************

Question: If you had carefully timed the ball and obtained the results given above, how confident would you be that the mean of those five intervals was within 0.1 seconds of the actual mean? (Note that the mean of the given intervals is 2.43 seconds, as rounded to three significant figures)? Briefly explain your thinking.

****

I would be very confident that the mean would be within.1 of the actual mean because I believe that the timer is precise to the .1 second.

#$&*

How confident would you be that the 2.43 second mean is within .01 second? Briefly explain your thinking.

****

I would be less confident that the mean is with .01 of that actual mean because there is more room for error as precision increase.

#$&*

How confident would you be that the 2.43 second mean is within .03 second?

****

I would be slightly more confident that the mean is within .03 of a second

#$&*

At what level do you think you can be confident of the various degrees of uncertainty?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.1 second of the actual mean?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.01 second of the actual mean?

• Do you think you could be 90% confident that the 2.43 second mean is within 0.03 second of the actual mean?

Give your three answers and briefly explain your thinking:

****

I would only be 90% confident that the mean is within .1 seconds. I would be less than 90% confident that it is within .01 or .03.

#$&*

*********************************************

Question: What, if anything, could you do about the uncertainty due to each of the following? Address each specifically.

• The lack of precision of the TIMER program.

****

You cannot do anything to adjust the precision of the timer program.

#$&*

• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

****

To adjust the precision of the human triggering you can increase the sensitivity of the mouse.

#$&*

• Actual differences in the time required for the object to travel the same distance.

****

You cannot adjust actual time required.

#$&*

• Differences in positioning the object prior to release.

****

To adjust differences in positioning you can create a mark on the starting point and face the object the same way each time.

#$&*

• Human uncertainty in observing exactly when the object reached the end of the incline.

****

To adjust for human uncertainty you can use slow motion or a pause feature.

#$&*

*********************************************

Question: If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

You would divide the distance traveled by the time required resulting in the speed in inces/sec for example

confidence rating #$&*:3

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

#$&*

*********************************************

Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: It’s average velocity is 8 cm/sec. This is relevant to traveling long distances and wanting to know your average velocity over the trip

confidence rating #$&*:3

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

#$&*

*********************************************

Question: If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: The average velocity for the first half is 6.6 cm/sec and 1cm/sec for the second half.

confidence rating #$&*:3

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

#$&*

*********************************************

Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:I think that doubling the length will result in less than half the frequency.

confidence rating #$&*:2

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

#$&*

*********************************************

Question: `qNote that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Because (0,0) is equilibrium this is where the point is equally from extremes in all directions.

confidence rating #$&*:3

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

#$&*

*********************************************

Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: This isn’t possible because for the frequency to cross the vertical axis it would have to be negative at some point and you can’t have negative frequency.

Assessment:2

#$&*

*********************************************

Question: `qOn a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: This is also not possible for the graph to intersect the horizontal axis because that means the length would be negative and the length cannot be negative.

confidence rating #$&*:2

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

#$&*

*********************************************

Question: `qIf a ball rolls between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: The points are 30 cm apart. 6cm/sec*5seconds =30cm.

confidence rating #$&*:3

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

#$&*

.............................................

Given Solution:

`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm.

The formal calculation goes like this:

• We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval.

• It follows by algebraic rearrangement that `ds = vAve * `dt.

• We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that

• `ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.

The details of the algebraic rearrangement are as follows:

• vAve = `ds / `dt. We multiply both sides of the equation by `dt:

• vAve * `dt = `ds / `dt * `dt. We simplify to obtain

• vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt

Be sure to address anything you do not fully understand in your self-critique.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

#$&*

*********************************************

Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: I understand significant figures, but I am not sure under what circumstances you use them and when you do not.

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

@&

You should always be aware of significant figures.

In labs especially you should try to answer accordingly.

*@

#*&!#*&!

&#Good work. See my notes and let me know if you have questions. &#