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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?
They are in the reverse order- 0.001, 0.01. 0.1
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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?
I think the precision of the timer program is consistent and therefore, 0%. However, the precision in which the individual controlling the timer program may be different with each trial.
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No measuring instrument is perfect.
As it turns out the TIMER program is designed to measure only in increments of .001 second (Java version) or about .015 second (.exe version).
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· 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 90% of uncertainty is associated with human triggering. Our brains may fire signals at different speeds to initiate the ball rolling. We also may not process that signal as quickly with each trial.
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· 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 0% is associated with travel time. Since gravity pulls with a constant force, the marble’s mass is constant, and the distance is constant, the ball will travel at the same speeds given these constant variables.
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· 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?
0% of discrepancy is due to this because the object is a sphere and maintains the same shape throughout. The ball will always have to overcome the same amount of inertia to begin rolling and positional changes will not affect this.
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You can't position the sphere at exactly the same initial position every time.
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· 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?
Because this experiment requires real time, snap judgements, 10% of the error is associated with observation. We have a good judgement of when we exactly let go of the ball, but we may be off by a significant margin when judging when the marble reaches the end of the book.
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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.
Very confident since our brains frequently fire and make snap decisions in less than .1seconds. Therefore, our brains may be trained to make these snap judgments often.
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How confident would you be that the 2.43 second mean is within .01 second? Briefly explain your thinking.
Not very confident. .01 is a very little amount of time and our human error is most likely larger than that.
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How confident would you be that the 2.43 second mean is within .03 second?
Somewhat confident.
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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 think I could be 90% confident the mean is within .1seconds. Our brain always make snap judgements like this.
I would not 90% confident with .01seconds since that is an extremely short amount of time.
I could not be 90% confident with .03seconds, but I would be more sure than .01seconds.
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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.
Ensure your finger triggers the program in a consistent manner.
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· The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
Perform and practice lots of trials to train your brain to trigger at consistent times.
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· Actual differences in the time required for the object to travel the same distance.
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Begin and end at designated starting and ending points and be consistent with those points.
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· Differences in positioning the object prior to release.
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Find a mark on the marble and ensure the mark faces the same direction each time. Ensure your hands are also in the same position with each trial.
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· Human uncertainty in observing exactly when the object reached the end of the incline.
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Train your eye to with a lot of practice.
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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:
Take the distance travelled divided by the time. Add the three most precise measurements and divide by 3 to find the average speed.
confidence rating #$&*:3
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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.
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Your solution:
8cm/sec. We had to find the speeds associated with each of trials and the average of our trails by using this arithmetic.
confidence rating #$&*:
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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:
1st- 6.67 cm/sec
2nd- 10cm/sec
confidence rating #$&*:3
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Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
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Your solution:
In my experience, I found more than half occurred.
Confidence Assessmen
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Question: `qNote that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.
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Your solution:
Because the two variables have no relationship.
confidence rating #$&*:2
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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:
At a certain short length, the pendulum will not swing to create a cycle. Length and frequency may not be related a certain point.
confidence rating #$&*:2
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Question: `qOn a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?
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Your solution:
At a certain length, the string is too long to complete a cycle. Length and frequency are only related until a certain threshold of length. If the pendulum becomes too long, cycles cannot be completed.
confidence rating #$&*:2
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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?
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Your solution:
30cm
confidence rating #$&*:3
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Given Solution:
`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm.
The formal calculation goes like this:
We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval.
It follows by algebraic rearrangement that `ds = vAve * `dt.
We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that
`ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.
The details of the algebraic rearrangement are as follows:
vAve = `ds / `dt. We multiply both sides of the equation by `dt:
vAve * `dt = `ds / `dt * `dt. We simplify to obtain
vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt
Be sure to address anything you do not fully understand in your self-critique.
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Your solution:
confidence rating #$&*:
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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 understand the average rates (change A/ change B) and the associated notes, but I do not fully understand how the change in the position of an object over time is equal to avg. velocity x time?
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&#This looks good. See my notes. Let me know if you have any questions. &#