course Phy 231
I was not sure about the frequency vs length graph questions!
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.
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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 to what extent do you think the discrepancies could be explained by each of the following:
The lack of precision of the TIMER program.
To what extent to you think the discrepancies are explained by this factor?
Your answer:
The software does the best that it can based upon human entry
The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
To what extent to you think the discrepancies are explained by this factor?
Your answer:
This shows in the work and also the timer lab when clicking it 20 times in a row there are different values because sometimes the finger is faster than the next.
Actual differences in the time required for the object to travel the same distance.
To what extent to you think the discrepancies are explained by this factor?
Your answer:
I dont think this is true, if the ball is placed in the same exact spot each time then it should take the same time or roughly very very close to each trial
Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
Your answer:
I think the ball is put in the same spot most of the time or at least within millimeters.
Human uncertainty in observing exactly when the object reached the end of the incline.
To what extent to you think the discrepancies are explained by this factor?
Your answer:
This can easily be a factor because when the human eye thinks its the end and clicks the timer button is not going to be exactly the same at every trial
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Question: How much uncertainty do you think each of the following would actually contribute to the uncertainty in timing a number of trials for the ball-down-an-incline lab?
The lack of precision of the TIMER program.
To what extent to you think this factor would contribute to the uncertainty?
Your answer:
Not very much
The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
To what extent to you think this factor would contribute to the uncertainty?
Your answer:
This is the most likely to happen
Actual differences in the time required for the object to travel the same distance.
To what extent to you think this factor would contribute to the uncertainty?
Your answer:
Not likely
Differences in positioning the object prior to release.
To what extent to you think this factor would contribute to the uncertainty?
Your answer:
Not likely
Human uncertainty in observing exactly when the object reached the end of the incline.
To what extent to you think this factor would contribute to the uncertainty?
Your answer:
This also is probably a big factor because the eye is the only judge and is not going to be exactly the same
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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.
What do you think you could do about the uncertainty due to this factor?
Your answer:
Nothing really to do because it is software
The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
What do you think you could do about the uncertainty due to this factor?
Your answer:
Try to make each time the button is clicked your finger is the same distance away from the mouse and you move at the same speed with each click.
Actual differences in the time required for the object to travel the same distance.
What do you think you could do about the uncertainty due to this factor?
Your answer:
Not much you can do with this, just make sure the track is clean and that there is nothing in the way of the ball
Differences in positioning the object prior to release.
What do you think you could do about the uncertainty due to this factor?
Your answer:
Clearly mark where the ball is to be placed for each trial so that the distance is as close to exact as possible
Human uncertainty in observing exactly when the object reached the end of the incline.
What do you think you could do about the uncertainty due to this factor?
Your answer:
Not much you can do with this one unless a video is shot and the frame is frozen when the ball hits a certain spot at each different time. Other than that the human eye does the best job it can but there is still room for small error.
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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.
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Your solution:
If you know the distance the ball went and the time that was required then you would divide the distance by the time and that would give you an answer in distance over time or speed
Confidence Assessment:
OK
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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.
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Your solution:
The average velocity would be 40cm over 5 seconds, which equals 8cm per second. This answer is connected to the experience because we rolled a ball down a ramp and recorded the distance and how long it took to reach the end.
Confidence Assessment:
OK
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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?
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Your solution:
Because it was a 40 cm distance being split up into two it would be 20 and 20cm. Because the first half took 3 seconds it would be 20cm over 3 seconds which would equal 6.67cm/s for the first half. The second half would be the remaining 5 seconds out of the 8 so it would be 20cm over 5 seconds which would equal 4cm per second.
Confidence Assessment:
OK
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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?
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Your solution:
I would not say that it cuts the frequency directly in half because according to my data most of them were a little more than half, but regardless increasing the length does reduce the frequency.
Confidence Assessment:
OK
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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.
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Your solution:
This is so because it lies directly on the y or x axis and it does not lie in the x-y plane. Because it does not fall in the x-y plane the other coordinate would have to be zero because it is lying on the axis.
Confidence Assessment:
OK
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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?
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Your solution:
If the graph went through the vertical axis then it would have to be a length of zero which would mean the pendulum is just lying in equilibrium or that it is not swinging. If the line went through the vertical axis then the length would have to be a negative length which I do not think is possible.
Confidence Assessment:
Some What, I am not sure if I answered this right
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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?
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Your solution:
If it intersected the horizontal axis then it would mean that the length is so large that it cannot swing all the way back in the given time length causing it to have a frequency less than 0. Although, I do not think that this is possible because once the pendulum is dropped at some length the frequency has to be greater than 0 at that point unless for some reason it was physically pushed the opposite way causing it to swing up
Confidence Assessment:
Some What, not sure if this is what you are looking for
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Question: `qIf a ball rolls down 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?
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Your solution:
Because it takes 5 seconds than 6 times 5 equals 30 cm apart using the average velocity.
Confidence Assessment:
Good
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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.
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Your solution:
OK
Confidence Assessment:
Good
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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.
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Your solution:
I have not completely read the whole chapter and work every problem yet, but so far so good. Vectors are pretty fresh in my head and that material has been ok so far. I hope by tom I will have finished chapter one and be ready to move to this weeks assignments.
Confidence Assessment:
OK
SOME COMMON QUESTIONS:
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QUESTION: I didnt understand how to calculate uncertainty for a number such as 1.34. When given examples we had problems such as 1.34 ±0.5 and with that we had a formula (0.5/1.34)*100. So I do not understand how to compute uncertainty when no estimated uncertainty is given.
INSTRUCTOR RESPONSE:
The +- number is the uncertainty in the measurement.
The percent uncertainty is the uncertainty, expressed as a percent of the number being observed.
So the question in this case is simply, 'what percent of 1.34 is 0.5?'.
0.5 / 1.34 = .037, approximately. So 0.5 is .037 of 1.34.
.037 is the same as 3.7%.
I recommend understanding the principles of ratio, proportion and percent as opposed to using a formula. These principles are part of the standard school curriculum, though it does not appear that these concepts have been well mastered by the majority of students who have completed the curriculum. However most students who have the prerequisites for this course do fine with these ideas, after a little review. It will in the long run save you time to do so.
There are numerous Web resources available for understanding these concepts. You should check out these resources and let me know if you have questions.
Please feel free to include additional comments or questions:
Your answers were good.
Pendulum frequency approaches zero as pendulum length increases; but to actually get 0 frequency would require infinite length, which is not possible. So the graph does not intersect the horizontal axis.