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course PHY 232
Question: Suppose you measure the length of a pencil. You use both a triply-reduced ruler and the original ruler itself, and you make your measurements accurate to the smallest mark on each. You then multiply the reading on the triply-reduced ruler by the appropriate scale factor.
• Which result is likely to be closer to the actual length of the pencil?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
First of all: I’m going to assume that the ruler is metric and includes millimeter markings.
Second of all: The original ruler will be the most accurate, because the triply reduced ruler will make reading millimeters nearly impossible, whereas the normal original ruler will make reading millimeters very easy.
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• What factors do you have to consider in order to answer this question and how do they weigh into your final answer?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The ability of the human eye to accurately read to 1/3 millimeters. This is very influential upon my final answer “because the triply reduced ruler will make reading millimeters nearly impossible, whereas the normal original ruler will make reading millimeters very easy.”
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Question: Answer the same questions as before, except assume that the triply-reduced ruler has no optical distortion, and that you also know the scale factor accurate to 4 significant figures.
• Which result is likely to be closer to the actual length of the pencil?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I’m going to interpret the question as meaning that the scaled-down ruler will be accurately readable by the human eye down to its smallest unit.
Under this understanding, the triply-reduced and the original rulers will be equally close to the actual length of the pencil.
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• What factors do you have to consider in order to answer this question and how do they weigh into your final answer?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I considered the ability of the human eye to read the scaled-down ruler, which I think the question was saying would not be a factor. Therefore, if both rulers can be read equally well and there is no distortion on the scaled-down ruler, then they will read the same length.
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Question: Suppose you are to measure the length of a rubber band whose original length is around 10 cm, measuring once while the rubber band supports the weight of a small apple and again when it supports the weight of two small apples. You are asked to report as accurately as possible the difference in the two lengths, which is somewhere between 1 cm and 2 cm. You have available the singly-reduced copy and the triply-reduced copy, and your data from the optical distortion experiment.
• Which ruler will be likely to give you the more accurate difference in the lengths?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The original ruler
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• Explain what factors you considered and how they influence your final answer.
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The original ruler’s markings will be larger, so assigning the final ‘0’ or ‘5’ based on visual estimation will be easier to assign.
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Question: Later in the course you will observe how the depth of water in a uniform cylinder changes as a function of time, when water flows from a hole near the bottom of the cylinder. Suppose these measurements are made by taping a triply-reduced ruler to the side of a transparent cylinder, and observing the depth of the water at regular 3-second intervals.
The resulting data would consist of a table of water depth vs. clock times, with clock times 0, 3, 6, 9, 12, ... seconds. As depth decreases the water flows from the hole more and more slowly, so the depth changes less and less quickly with respect to clock time.
Experimental uncertainties would occur due to the optical distortion of the copied rulers, due to the spacing between marks on the rulers, due to limitations on your ability to read the ruler (your eyes are only so good), due to timing errors, and due to other possible factors.
Suppose that depth changes vary from 5 cm to 2 cm over the first six 3-second intervals.
Assume also that the timing was very precise, so that there were no significant uncertainties due to timing.
• Based on what you have learned in experiments done through Assignment 1, without doing extensive mathematical analysis, estimate how much uncertainty would be expected in the observed depths, and briefly explain the basis for your estimates. Speculate also on how much uncertainty would result in first-difference calculations done with the depth vs. clock time data, and how much in second-difference calculations.
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I’m going to assume that we are reading the ruler to decimeters (the reader will make use of the ruler’s millimeter markings, and determine whether the actual depth is closer to the nearest millimeter marking or the halfway point between the two adjacent markings). There would be 0.5 millimeters of uncertainty in the observed depths because the depth could’ve been incorrectly, but understandably, judged to be closer to the halfway mark when it was really closer the millimeter mark or vise versa.
If two of these depths, taken to the decimeter, were added, the uncertainty would be plus or minus 1 millimeter; the maximum amount of justifiable error would occur if both depth was incorrectly estimated to be 0.5 mm larger or smaller than they really were, causing the resulting sum to be 1 mm off.
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• How would these uncertainties affect a graph of first difference vs. midpoint clock time, and how would they affect a graph of second difference vs. midpoint clock time?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Overall, a maximum difference of 0.5 millimeters suggests that it is unlikely that there will not be a noticeable difference in a first-difference graph from an accurate first-difference graph, because the difference graph will average 5 mm per data point (5 cm - 2 cm = 3 cm / 6 = ˝ cm = 5 mm). In a worst-case scenario, however, it is possible that this uncertainty could result in mis-identification of a type of function (i.e. the curve of a quadratic difference vs. clock point is flattened out and mistaken for a linear graph). The sign of the graph’s derivative will not be affected, however, by a difference of 0.5 mm.
In the second difference graph, the uncertainty is plus or minus 1 mm. A second difference graph makes estimating normal x values difficult, so it is hard to say whether or not 1 mm will be influential. However, if the data points fall in the same range as the first-difference graph, then the second difference graph will be more affected by uncertainty than the first difference graph.
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• How reliably do you think the first-difference graph would predict the actual behavior of the first difference?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
More reliably than the second-difference graph.
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• Answer the same for the second-difference graph.
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Less reliably than the first-difference graph.
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• What do you think the first difference tells you about the system? What about the second difference?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The first difference shows that the system is not ideal. The second difference shows that system is not accurate enough for effective data analysis.
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Question: Suppose the actual second-difference behavior of the depth vs. clock time is in fact linear. How nearly do you think you could estimate the slope of that graph from data taken as indicated above (e.g., within 1% of the correct slope, within 10%, within 30%, or would no slope be apparent in the second-difference graph)?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
If the graph is linear, and depth goes from 5 cm to 2 cm over six equal intervals of three seconds each, as was stated earlier, than the actual slope would be (5cm - 2cm) / (0 sec - 18 sec) = -1/6 (cm/sec) = -0.167 (cm/sec).
If the depth uncertainty is plus or minus .5 mm, as I suggested earlier, than the maximum uncertainty in slope would be 0.1 cm (0.05 cm + 0.05 cm). So the worst-case scenario slope may be about - (1.1)/6 (cm/sec) = -0.183 (cm/sec). This figure is 0.016 (cm/sec) off from the correct slope, therefore, the slope taken from the graph would be within about 10% of the actual slope.
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Again no extensive analysis is expected, but give a brief synopsis of how you considered various effects in arriving at your estimate.
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
See above.
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&#Very good responses. Let me know if you have questions. &#