ph1 query 0

#$&*

course Phy 201

ph1 query 0Most queries in this course will ask you questions about class notes, readings, text problems and experiments. Since the first two assignments have been lab-related, the first two queries are related to the those exercises. While the remaining queries in this course are in question-answer format, the first two will be in the form of open-ended questions. Interpret these questions and answer them as best you can.

Different first-semester courses address the issues of experimental precision, experimental error, reporting of results and analysis in different ways and at different levels. One purpose of these initial lab exercises is to familiarize your instructor with your work and you with the instructor 's expectations.

Comment on your experience with the three lab exercises you encountered in this assignment or in recent assignments.

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

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?

****

.1 second

#$&*

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 was not completely accurate when I used it in lab although part of which could be due to human error. The timer program is computer organized so I believe the discrepancies are not due to this (2%).

#$&*

• 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 most of the discrepancies (>50%) could be due to human error and the speed at which the person pressed the timer.

#$&*

• 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?

****

The timing of the speed of the ball should not have been affected when the ramp was the same every time.

#$&*

• 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?

****

The person could have dropped the ball differently so that the speed or distance could change. A slight difference in the way the ball dropped could change the seed by a second.

#$&*

• 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?

****

This goes along with human error on the triggering of the timer system. Human error is the biggest percent of change and people see differently.

#$&*

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

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.

****

#$&*

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

****

I don’t think the 2.43 second mean is within the .01 second of the real mean. Human error has to account for more that.

#$&*

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

****

I wouldn’t be confident enough to say that the real mean is within .03 of 2.43.

#$&*

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:

****

For .1 I can be 90% confident that 2.43 is within .1 of the actual mean because all of the numbers in the set were within this. I cannot be 90% confident for either .03 and .01

#$&*

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

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.

****

For this you could go into the program and try to fix the precision or count a .001 difference.

#$&*

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

****

For this you could have multiple people clicking and see the average in hope that it will be closer.

#$&*

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

****

For this you could make sure no variable come into play like wind and temperature.

#$&*

• Differences in positioning the object prior to release.

****

For this you could make an extending area that drops the ball automatically to exclude human error.

#$&*

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

****

For this you could put a wire across the finish line so that when it touches the wire the timer is pressed.

#$&*

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

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: Speed=distance/ time, so you can multiply the distance the ball traveled and the time to get speed.

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: The average velocity is 8cm/sec on the incline.

confidence rating #$&*: 2

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

#$&*

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

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 of the first half is 6.67cm/sec and the second half of the incline is 10cm/sec.

confidence rating #$&*: 2

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

#$&*

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

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: Doubling the length will result in a little more 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: in a graph equation y=mx+b it shows that there is a x coordinate for y=0 and a y coordinate for x=0. This is because a line is continuous and always has to land on zero for both sides.

confidence rating #$&*: 1

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

#$&*

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

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:

confidence rating #$&*:

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

#$&*

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

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 would mean that the length hit zero compared to the negative frequency.

confidence rating #$&*: 1

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

#$&*

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

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: 30cm apart

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.

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

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 was not certain on how to do the cos section.

#$&*

end document

Self-critique (if necessary):

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

Self-critique rating:

ph1 query 0

#$&*

course Phy 201

Comment on your experience with the three lab exercises you encountered in this assignment or in recent assignments.

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

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.

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

****

.1 second

#$&*

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

****

#$&*

****

I think most of the discrepancies (>50%) could be due to human error and the speed at which the person pressed the timer.

#$&*

• 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?

****

The timing of the speed of the ball should not have been affected when the ramp was the same every time.

#$&*

• 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?

****

The person could have dropped the ball differently so that the speed or distance could change. A slight difference in the way the ball dropped could change the seed by a second.

#$&*

• 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?

****

This goes along with human error on the triggering of the timer system. Human error is the biggest percent of change and people see differently.

#$&*

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

****

#$&*

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

****

I don’t think the 2.43 second mean is within the .01 second of the real mean. Human error has to account for more that.

#$&*

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

****

I wouldn’t be confident enough to say that the real mean is within .03 of 2.43.

#$&*

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:

****

For .1 I can be 90% confident that 2.43 is within .1 of the actual mean because all of the numbers in the set were within this. I cannot be 90% confident for either .03 and .01

#$&*

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

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.

****

For this you could go into the program and try to fix the precision or count a .001 difference.

#$&*

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

****

For this you could have multiple people clicking and see the average in hope that it will be closer.

#$&*

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

****

For this you could make sure no variable come into play like wind and temperature.

#$&*

• Differences in positioning the object prior to release.

****

For this you could make an extending area that drops the ball automatically to exclude human error.

#$&*

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

****

For this you could put a wire across the finish line so that when it touches the wire the timer is pressed.

#$&*

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Speed=distance/ time, so you can multiply the distance the ball traveled and the time to get speed.

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: The average velocity is 8cm/sec on the incline.

confidence rating #$&*: 2

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

#$&*

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

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 of the first half is 6.67cm/sec and the second half of the incline is 10cm/sec.

confidence rating #$&*: 2

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

#$&*

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Doubling the length will result in a little more 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: in a graph equation y=mx+b it shows that there is a x coordinate for y=0 and a y coordinate for x=0. This is because a line is continuous and always has to land on zero for both sides.

@&
There is no assumption regarding a line connecting the points. However I believe you understand this aspect of the graph.
*@

confidence rating #$&*: 1

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

#$&*

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

#$&*

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: This would mean that the length hit zero compared to the negative frequency.

confidence rating #$&*: 1

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

#$&*

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

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: 30cm apart

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:

I was not certain on how to do the cos section.

@&
It isn't clear to what you are referring here.
*@

#$&*

ph1 query 0

ph1 query 0

@& There is no assumption regarding a line connecting the points. However I believe you understand this aspect of the graph. *@

@& It isn't clear to what you are referring here. *@

@& Your work on this assignment appears to be OK. I've just sent you an email about the question you asked at the beginning of the document. *@

@&
There is no assumption regarding a line connecting the points. However I believe you understand this aspect of the graph.
*@

@&
It isn't clear to what you are referring here.
*@

@&
Your work on this assignment appears to be OK.

I've just sent you an email about the question you asked at the beginning of the document.
*@