course Phy 121
I'm not 100% sure if this is how I submit these
open query
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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Within that range of times, I do not think the TIMER program itself would show great error. More often than not, the discrepancies within TIMER were shown past the thousandths place.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv I REALLY believe this is why there are discrepancies within these times. I would venture to say it is completely impossible for a human to start a timer EXACTALY when they release the ball and equally impossible to stop it when the ball reaches the end exactly.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv I see no reason why an item on a smooth surface would have time variations in rolling a fixed distance if no environmental factors came into play.
• Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv This may have had a minor impact, definitely one large enough to justify the small time variations above. Without extreme positioning measures, it would be impossible to place the ball in the same place.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv This may have played a fairly large roll, especially if the end was defined as when the ball hit the ground as opposed to passing a fixed point.
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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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv If one was rounding to 4 significant figures or less, I doubt the lack of precision in TIMER would be an issue. Beyond 4 figures, or past the thousandths place, TIMERS lack of precision would have a large baring on results.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv One could definitely not trust results very much if human error can be a large factor.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv This should not ever been an issue if using a smooth ramp and having no environmental factors.
• Differences in positioning the object prior to release.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv This would definitely make someone looking at the data have to question extreme precision, especially if the ball was placed by hand.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Probably equally as much as the initial placement of the ball would contribute to uncertainty would discrepancies in where the ball should end contribute.
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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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Only use 3 maybe 4 significant figures past the decimal point.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Use an electronic timer that began when the ball passed point A and ended when the ball passed point B.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Ensure there are no environmental contributors and that the ramp was smooth and level.
• Differences in positioning the object prior to release.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Have a sort of holding system in place such as maybe a fixed ruler that can pivot up or something along those lines.
• 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: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv The same solution as starting the TIMER, using an electronic means of stopping the timer at the end point.
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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 we took the distance (ie: 10m), and took the time (ie: 5sec), we could find an average velocity down the incline by dividing 10/5 and getting 2m/s. To increase accuracy, we could measure multiple times.
Confidence Assessment: I’m pretty positive this is a valid means of coming across the results you want.
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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: 8cm/sec. This is the exact way I found my results in my incline experiment.
Confidence Assessment: Pretty sure.
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: The first half was 6.67cm/sec and the second half was 10cm/sec.
Confidence Assessment: Pretty sure.
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: Assuming nothing changes in the amplitude or speed its being swung, I would assume that double length would equate to half the frequency.
Confidence Assessment: Not 100%, but close.
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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: If something has an Y coord. Of 0, then it has to be an X intercept meaning it has no relation to the Y axis. If something has an X coord. Of 0, then it must be a Y intercept and would have no relation to the X axis.
Confidence Assessment: I may have over simplified the problem.
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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: It would mean that the pendulum has no length thus is not a pendulum at all. I do not think we would ever see a true 0 mark on a pendulum graph.
Confidence Assessment: This problem was not very clear at first, so I’m not overly confident in my answer.
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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: It would mean at that specific length, the pendulum would not swing.
Confidence Assessment: Very similar confidence as the previous question.
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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: 30cm, 6cm/sec x 5 sec = 30cm.
Confidence Assessment: Pretty sure.
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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:
Confidence Assessment: I understood these concepts.
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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: #12) Write the following as full (decimal) numbers with standard units: (a) 286.6mm, (b) 85uV, (c) 760mg, (d) 60.0ps, (e) 22.5fm, (f) 2.50 gigavolts. More than anything on this problem, I have no idea what the question is asking so I don’t even know where to begin my approach. Does it want scientific notation, unit conversions, etc.?
The standard unit of length is the meter (m).
The standard unit of electrostatic potential is the volt (V).
The standard unit of mass is the kilogram (kg).
The standard unit of time is the second (s).
(a) 'mm' stands for 'millimeter'; a millimeter is 10^-3 meterr so 286.6mm = 286.6 * 10^-3 meter = 0.2866 meter.
(b) the Greek 'mu' stands for 10^-6, so 85 `mu V is 85 * 10^-6 V = 8.5 * 10^-5 V.
(c) 760mg = 760 * 10^-3 g = 0.760 g. The gram is the standard unit in the cgs system.
In SI units, the unit of mass is the kilogram, and 1 g = 10^-3 kg, so 0.760 g = 0.760 * 10^-3 kg = 7.60 * 10^-4 kg.
(d) 'ps' stands for 'picosecond'; 'pico' is a prefix meaning 10^-12. Thus 60.0ps = 60.0 * 10^-12 s = 6.00 * 10^-11s.
(e) 'fm' means 'femtometer'; 'femto' is a prefix meaning 10^-15 so 22.5fm = 22.5 * 10^-15 m = 2.25 * 10^-14 m.
(f) 'giga' is a prefix meaning 10^9 (a billion) so 2.50 gigavolts. = 2.50 * 10^9 volts.
Confidence Assessment: None.
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Question: `qTell your instructor about something in the problems you understand up to a point but don't fully understand. Explain what you did understand, and ask the best question you can about what you didn't understand.
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Your solution: The problem above was the only one I was unable to answer.
SOME COMMON QUESTIONS:
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QUESTION: I didn’t 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.
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&#Good responses. See my notes and let me know if you have questions. &#