course Phy 201
6/5 8:30
Most 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 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.
The first experiment with the pendulum was fairly straightforward. The main part with uncertainty with the data was attempting to determine the number of cycles of the pendulum or the time per cycle based only on the graph. Because the graph was not perfect or completely accurate, there was room for a lot of error. The second experiment dealt with the timer program. There was also a lot of uncertainty because the program did not give completely accurate measures of the clock time. I am still not completely sure how accurate the program was, or why the inaccuracy happens. The final lab was part of the first seed question, with the videos. Because of the reduced resolution of the picture and the use of the timer program, it was hard to tell how accurate the time and location of the objects was. This, in turn, made any calculations done with the data even more uncertain.
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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
All of the intervals are near to the same length of time, so there is a chance the TIMER program was inaccurate and gave incorrect values for the interval.
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• 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:
Because the intervals are short, the response time to get the exact time would have to be very fast. There is some hesitation in the human reflexes, so it could be a factor in causing the intervals to be shown as different lengths of time.
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• 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:
The numbers in the intervals do not show the decreasing interval that is expected when a ball is rolling down an incline, because of the acceleration due to gravity. There should be some differences in the numbers, but the pattern of the data does not indicate the normal progression.
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• Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
Your answer:
If the object were placed farther up the incline for one test then it would have farther to roll and more time to pick up speed. Also, the lengths of the intervals would be different causing some differences in the numbers. It is possible that this caused some of the variation in the measured time.
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• 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:
It is possible that the differences in the numbers were caused by the uncertainty of the end of each interval. If the ball was moving so fast, it would be very difficult to tell the end of one interval even if it were marked clearly. The ball passes so fast that even a slight hesitation will give a large variation in the interval time.
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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:
The timer program is pretty accurate for intervals that are longer. The intervals in this experiment are longer than 2 seconds so there is some precision, though anything after the tenths place should not really be accepted. The uncertainty should be taken into account for calculations, but it should not cause big problems.
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• 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:
The hesitation in human reflexes should also be considered; it may not cause too many problems because the reaction time is not that slow. The times would probably be within .02 of the actual times for the interval.
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• 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:
As the ball gains speed as it rolls there is the potential for the intervals to take different lengths of time. It is also expected in the results, since the test is normally to look at the acceleration of the object. The differences in time definitely will cause some uncertainty, but the acceleration should be regular enough that it should not be a problem.
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• Differences in positioning the object prior to release.
To what extent to you think this factor would contribute to the uncertainty?
Your answer:
The position of the object would cause many variations in the length of the time intervals for the test. Even a slight difference in position will change the length of the intervals and will make the data invalid because the length of the run will be different and the amount of acceleration will change.
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• 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:
The uncertainty of human observation will cause the differences in the intervals because the eye cannot be absolutely when the ball crosses the end of an interval and enters a new one. This combined with the reaction time will affect how close to the real time the interval is. It is best to use an average of the times in this case because the uncertainty is so great from the variation in the length of the intervals (length, not time).
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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
There is not much you could do to make the program more accurate. You could use a different program whose times are more certain and close to the real thing. Another solution is to get a longer track and make the length of the intervals longer, to give the program longer time intervals and not have to deal with very short and uncertain intervals.
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• 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:
There is not much you can do to speed up human reaction time. If you wanted a more precise triggering system you could set up another system so that the clock is stopped when the interval ends.
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• 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:
There is nothing you can do to correct the acceleration that the ball undergoes as it rolls. Also, you really shouldn’t change it because the change in the rate of the ball is natural. However, you should know how much of the uncertainty is due to acceleration so that you can study it later. If you remove the differences due to acceleration then the data you get will be less accurate than they were before.
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• Differences in positioning the object prior to release.
What do you think you could do about the uncertainty due to this factor?
Your answer:
You need to make a specific place from which to release the ball on every test so that the results, intervals and acceleration are consistent every time.
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• 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
You should mark the exact end of the incline and stop the clock when the ball touches that place. There is not much you can do other than that, just try to be consistent when the clock is stopped (like a certain location for the ball).
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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:
To get the average speed on the incline, you divide the total distance covered by the object and divide it by the total time it took for the object to cover that distance. The units for the rate will be in distance per time, so, like, cm/s.
confidence rating #$&* 3
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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.
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Your solution:
The average speed is distance / time = 40 cm / 5 sec = 8 cm/s. We have done this calculation many times involving the object going down the incline. You measure the length of the ramp, from the starting point of the object to the place where the cock is stopped. Then you divide that value by the time it took the object to cover the distance.
confidence rating #$&* 3
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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:
Half of the distance is 20 cm; 3 seconds for the first half and 2 seconds for the other half.
1st average velocity: 20cm / 3 s = 20/3 cm/s or 6.67 cm/s
2nd average velocity: 20cm / 2 s = 10 cm/s
confidence rating #$&* 3
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Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
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Your solution:
If you double the length of the pendulum, you will probably get more than half of the number of cycles. This is because the data of the length vs. cycles graph follow a negative exponential curve, so every time you double the length it does not decrease by exactly half.
confidence rating #$&* 2
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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:
The two points described above would be (x,0) and (0,y), respectively. This happens because, at those points, the graph crosses the y and x axes. Whenever the coordinate has a 0 in the pair, it means that the graph has an axis intercept there.
confidence rating #$&* 3
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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?
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Your solution:
If the curve crosses the y-axis, it would mean that the length of the pendulum is zero, because to cross the y axis, the x coordinate must be 0. More than likely, the y coordinate would also be 0 because the pendulum cannot really swing if it does not have any length. With the intersection you can probably assume that the pendulum is stationary.
confidence rating #$&* 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:
If the graph crosses the x axis it would mean that the frequency of the pendulum was 0 cycles. This would mean that the pendulum was not moving at all. (??? Or it could possible mean that the pendulum did not do 1 cycle in the time allotted for the interval, however this is unlikely.) Crossing the x axis means that the frequency was 0.
confidence rating #$&*: 3
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Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?
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Your solution:
6 cm/s = x cm / 5 sec x cm = 6cm/s * 5s = 30 cm apart
confidence rating #$&* 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.
Your solution: OK
Confidence Assessment: OK
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 was looking at the components of vectors. I think that they are numbers that can be used to describe a vector and can be used to do other operations with vectors. I also understand that you can use the components to find the magnitude and direction as well as using them to multiply by a scalar. Basically, I know what the components can do, in theory, but what are they for??????How do you use them in real-world problems?
You'll see numerous examples later in the course.
For instance if I pull on an object with a 50 lb force in one direction, and you exert a 60 lb force in a direction that makes an angle of 130 deg with mine, then we have to use vector components to determine the magnitude and direction of the net force.
My other question is: are we going to get more practice with the various word problems from the text book, like other examples?????? I was looking at the ones from the end of each section and I could follow the reasoning for a little bit but then I got confused. Am I going to get other practice problems so that I learn how to solve all different types of problems??????
Absolutely.
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STUDENT QUESTION
I understand that we cannot measure to exact precision, but when we are dealing with estimated uncertainty, do we always
just increment our lowest unit by one and that is our uncertainty? Is there a standard that is used to figure out this?
INSTRUCTOR RESPONSE
The standard answer is that we assume an uncertainty of +- 1 of our smallest unit of precision. However, depending on how well we can 'see' that smallest unit, we can get pretty close to +- 1/2 of a unit.
STUDENT QUESTION
I fully understand how to calculate uncertainty, but what if the uncertainty isn’t given? For example,
problem 6 asks us for the uncertainty of 1.67. Do we just use .01 as the uncertainty?
INSTRUCTOR RESPONSE
Depending on the nature of the instrument and the observation, +- .01 might be necessary, but we could go to +-.005 if can regard 1.67 as an accurate roundoff.
Without very good reason, though, +-.01 would be the safer assumption.
QUESTION RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)
I don’t fully understand the dot product rule
INSTRUCTOR RESPONSE
The dot product of vectors A = a_1 i + a_2 j + a_3 k and B = b_1 i + b_2 j + b_3 k is a_1 * b_1 + a_2 * b_2 + a_3 * b_3. The dot product is simply a number.
The magnitude of A is | A | = sqrt( a_1 ^ 2 + a^2 ^ 2 + a_3 ^ 2); the magnitude of B is found in a similar manner.
The dot product is equal to | A | * | B | * cos(theta), where theta is the angle between the two vectors.
If you have the coefficients of the i, j and k vectors, it is easy to calculate the dot product, and it's easy to calculate the magnitudes of the two vectors. Setting the two expressions for the dot product equal to one another, we can easily solve for cos(theta), which we can then use to find theta.
More importantly for physics, we can find the projection of one vector on another. The projection of A on B is just the component of A in the direction of B, equal to | A | cos(theta). The projection of one vector on another is important in a number of situations (e.g., the projection of the force vector on the displacement, multiplied by the displacement, is the work done by the force on the interval corresponding to the displacement).
Dot products are a standard precalculus concept. Check the documents at the links below for an introduction to vectors and dot products. You are welcome to complete these documents, in whole or in part, and submit your work. If you aren't familiar with dot products, it is recommended you do so.
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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:
In my problems (I am working from the University Physics text- exercise 1.14) they are asking for the ratio of length to width of a rectangle based on the fact that both of the measurements have uncertainty. ?????Is there anything special you have to do when adding or multiplying numbers with uncertainty?????? I know that there are rules with significant figures, but I don’t understand if the same is true for uncertain measurements.
For example:
If there is a 5% uncertainty in length and no significant uncertainty in width, then area will be uncertain by 5%.
If there is a 5% uncertainty in length and a 3% uncertainty in width, then it is possible for the area result to be as much as 1.05 * 1.03 = 1.08 times the actual area, or as little as .95 * .97 = .92 times the actual area. Thus the area is uncertain by about 8%.
This generalizes. The percent uncertainty in the product or quotient of two quantities is equal to the sum of the percent uncertainties in the individual quantities (assuming the uncertainties are small compared to the quantities themselves).
The argument is a little abstract for this level, but the proof that it must be so, and the degree to which it actually is so, can be understood in terms of the product rule (fg) ' = f ' g + g ' f. However we won't go into those details at this point.
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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.
QUESTIONS AND RESPONSES
1)In the text question five asks for the percent uncertainty of a measurement given 1.57 m^2
I think that we figure this by an uncertainty of .01/1.57m^2 = .6369 or approximately one. ??????Am I correct in how I
calculate this??????? Can I assume that if the number given was 1.579 then we would calculate it by .001/1.57 = .1 % approximately or am I incorrect?????
You're on the right track.
There are two ways to look at this.
1.57 m^2 represents a quantity which rounds off to 1.57, so presumably lies between 1.565 and 1.575.
This means that the quantity is within .005 of 1.57.
.005 / 1.57 = .003, approx., so the uncertainty is .003 of 1.57, which is the same as 0.3%, of 1.57.
Another way to look at it:
1.57 could be interpreted to mean a number between 1.56 and 1.58. The uncertainty would then be .01, which is .01 / 1.57 = .006, or 6%, of 1.57.
2)In the text question number 11 the book asks what is the percent uncertainty in the volume of a sphere whose radius is
r=2.86 plus or minus .09.
I know that the Volume of a sphere is 4/3 pi r^3, so I calculated the volume to be 4/3 pi (2.86)^3 = 97.99 and to get the
percent uncertainty I tried to divide 0.09/97.99 * 100 =.091846, but the book answer is 9% ??????I am not sure what i am doing wrong here?????????????????
Again there are two ways to approach this.
I believe the book tells you that the uncertainty in the square of a number is double the uncertainty in the number, and the uncertainty in the cube of the number is trip the uncertainty in the number.
An uncertainty of .09 in a measurement of 2.86 is .09 / 2.86 = .03, approx., or about 3%. As you state, you cube the radius to find the volume. When 2.86 is cubed, the resulting number has three times the uncertainty, or about 9%.
Another approach:
Calculate the volume for r = 2.86.
Then calculate the volume for r = 2.86 - .09 = 2.77.
You will find that the resulting volumes differ by about 9%.
You could just as well have calculated the volume for r = 2.86 + .09 = 2.95. Again you would find that the volume differs from the r = 2.86 volume by about 9%.
QUESTION RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)
I understand everything but the part on measuring the individual i j k vectors by using cosine.
INSTRUCTOR RESPONSE
It's not completely clear what you are asking, but I suspect it has to do with direction cosines.
The vector A = a_1 i + a_2 j + a_3 k makes angles with the directions of the x axis, the y axis and the z axis.
Let's consider first the x axis.
The direction of the x axis is the same as the direction of the unit vector i.
The projection of A on the x direction is just a_1. This is obvious, but it can also be found by projecting the A vector on the i vector.
This projection is just | A | cos(alpha), where alpha is the angle between A and the x direction.
Now A dot i = A = (a_1 i + a_2 j + a_3 k) dot i = A = a_1 i dot i + a_2 j dot i + a_3 k dot i = a_1 * 1 + a_2 * 0 + a_3 * 0 = a_1.
It's also the case that A dot i = | A | | i | cos(alpha). Since | i | = 1, it follows that A dot i = | A | cos(alpha), so that
• cos(alpha) = A dot i / | A | = a_1 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).
Making the convention that alpha is the angle made by the vector with the x direction, we say that cos(alpha) is the direction cosine of the vector with the x axis.
If beta and gamma are, respectively, the angles with the y and z axes, reasoning similar to the above tells us that
• cos(beta) = a_2 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ) and
• cos(gamma) = a_3 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).
cos(alpha), cos(beta) and cos(gamma) are called the 'direction cosines of the vector A' with respect to the three coordinate axes.
Recall that alpha, beta and gamma are the angles made the the vector with the three respective coordinate axes.
If we know the direction cosines and the magnitude of the vector, we can among other things find its projection on any of the coordinate axes.
Please feel free to include additional comments or questions:
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&#Your work looks good. See my notes. Let me know if you have any questions. &#
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