course Phy231
2/2/10 1pm
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 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 I would think that most of the error does not come from the timer program itself. It seems fairly accurate. It is especially better with larger amounts of time than fractions of a second.
· 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 think this would explain most the discrepancies because human error probably is the cause of all the difference. It is definitely improbable that every time that a person could perfectly time release the ball and hit the timer and then seeing the ball reach the end and precisely hit the timer again.
· 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 This could play a factor dependent on whether or not the person actually just released the ball or slightly pushed it. I would imagine in a perfect situation it would take the ball the same amount of time, but of course this was not a perfect situation.
· Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv If the ball was released at an angle then it would take the ball longer to travel down the book because it would travelling a longer distance. The possibility of this happening would be fairly high due to human error.
· 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 I think this definitely could add to the discrepancies because this would require a person to precisely recognize when the ball reaches the end of the book and hit the timer all at once.
*********************************************
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 I'm assuming maybe you would want a percentage now since I explained previously what I thought about each scenario. I would say about 5%.
· 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 My guess would be approximately 10%.
· 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 15%
· Differences in positioning the object prior to release.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv 35%
· 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 35%
*********************************************
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 Not much, the program is what it is. If that's what you've decided to use then there isn't much that you can do about it.
· 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 Possibly practice so more. Maybe with some practice you could cut back on mistakes with tapping the mouse.
· 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 Try to release it exactly the same way everytime. Don't push but just release. This could help fix this problem.
· 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 Pick a specific release position and try to use it everytime. You could also possibly use a guideline somehow to make sure that the ball is following a straight line.
· 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 I don't think there is anything that you can do about this factor. This is just plain human observation and that will always have error.
*********************************************
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: I will take the average of all the times and then take the distance and divide by the average time. For instance, with the info you gave me the avg. speed would be 20.49cm/sec.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: 40cm/5sec= 8cm/sec this is the average velocity. This is how I determined the average velocity of my experiment. I just averaged my times and divided the distance by my average time.
I believe my spool traveled faster than this though. I think it travelled about 14cm/sec.
confidence rating #$&*: 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:
20cm/3sec= 6.67cm/sec so this is the velocity of the first half
20cm/2sec= 10cm/sec is the velocity of the second half
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: According to my measurements more than half of the shorter pendulum.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
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: A point on the x-axis will have a coordinate of whatever position it is on the x-axis and zero. The reason this is because the x-axis is horizontal and the y-axis is vertical. So a point on the x-axis is on the horizontal plane and doesn't have any vertical height. So therefore, a point on the y-axis has no horizontal length and has a coordinate of zero and its position on the y-axis.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
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: It seems to me that the frequency would have reached zero, so that would mean that the pendulum length had gotten so long that the pendulum couldn't swing. I'm not sure if that's possible or not.
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution: 6cm/sec * 5sec= 30cm
confidence rating #$&*: 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:
I believe I understand all of this.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I think so far I fully understand. I do better with the examples after the reading sections than I do with the explanations with the letters substituting in for the numbers, but other than that I think I've got it.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
*********************************************
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
They all seem to make sense. I had trouble at first understanding the problem with the three contestants, but after I examined it for a few minutes I understood the problem. The diagram helped alot.
SOME COMMON QUESTIONS:
*********************************************
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.
*********************************************
QUESTION: I understood the main points of changing the different units, but I’m not sure when in the problem I should change the number to 10 raised to a certain power. In example 1-8 I did not understand why they changed 70 beats/min to 2 x 10^9 s.
2 * 10^9 is about the number of seconds in 70 years.
70 beats / min were not changed to 2 * 10^9 seconds; in changing the beats / minute to beats in a lifetime, there was a step where it was necessary to multiply by 2 * 10^9 seconds.
The example actually used 80 beats / min as a basis for the solution. This was converted to beats / second by the calculation
80 beats / min * 1 minute / (60 seconds), which would yield about 1.33 beats / second.
This was then multiplied by 2 * 10^9 seconds to get the number of beats in a lifetime:
2 * 10^9 seconds * 1.33 beats / second = 3 * 10^9 beats.
In the given solution 80 beats / min * 1 minute / (60 seconds) was not actually calculated; instead 80 beats / min * 1 minute / (60 seconds) was multiplied by 2 * 10^9 seconds in one step
80 beats / min * 1 minute / (60 seconds) * 2 * 10^9 seconds = 3 * 10^9 beats.
In your instructor's opinion the unit 'beats' should have been left in the result; the text expressed the result simply as 3 * 10^9, apparently ignoring the fact that the unit 'beats' was included in the quantities on the left-hand side.
Also the text identified this number as 3 trillion. In the British terminology this would be correct; in American terminology this number would be 3 billion, not 3 trillion.
COMMENT:
I thought that these problems were pretty basic and felt that I understood them well. However, when I got to questions 14 (determine your own mass in kg) and 15 (determining how many meters away the Sun is from the Earth), I did not understand how to complete these. I know my weight in pounds, but how can that be converted to mass in kilograms? I can look up how to convert miles to meters, but is this something I should already know?
INSTRUCTOR RESPONSE:
Both of these questions could be answered knowing that an object with a mass of 1 kg has a weight of 2.2 lb, and that an inch is 2.54 centimeters. This assumes that you know how many feet in a mile, and that the Sun is 93 million miles away. All these things should be common knowledge, but it doesn't appear to be so.
For my own weight I would reason as follows:
I weigh 170 lb and every kg of my mass weighs 2.2 lb. I'll have fewer kg of mass than I will pounds of weight, so it's reasonable to conclude that my mass is 170 / 2.2 kg, or about 78 kg.
More formally 170 lb * (1 kg / (2.2 lb) ) = 170 / 2.2 kg = 78 kg, approx.. (technical point: this isn't really right because pounds and kilograms don't measure the same thing--pounds measure force and kg measure mass--but we'll worry about that later in the course).
Converting 93 million miles to kilometers:
93 million miles * (5280 feet / mile) * (12 inches / foot) * (2.54 cm / inch) * (1 meter / (100 cm) ) = 160 billion meters (approx.) or 160 million kilometers.
Please feel free to include additional comments or questions:"
&#Your work looks very good. Let me know if you have any questions. &#