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course Phy 121
The following data set represents the position of an object (first-semester students can think of an object rolling down an incline, while second-semester students can think of the water surface in a leaky uniform cylinder). The data represents position in centimeters vs. clock time in seconds.
Recall that by convention this means that the clock time is the first number, position the second.
10 , 81.659
20 , 64.649
30 , 49.178
40 , 36.972
50 , 25.821
60 , 15.453
70 , 9.39
80 , 4.96
90 , .487
100 , .067
A graph of this position vs. clock time data is depicted below. It should be clear that the position coordinate is decreasing at clock time increases, and also that the position is decreasing more and more slowly. That is, position is decreasing in time at a decreasing rate.
It is possible to fit a good curve to the graph. In this case a quadratic curve (a second-degree polynomial) yields an excellent fit to the data. The closeness of the quadratic fit is a strong indication that velocity is a linear function of clock time, changing at a constant rate, and that acceleration is therefore constant, unchanging from one clock time to the next.
In reality, this should be very nearly the case either for a ball rolling down a ramp or for a leaky cylinder.
The graph below shows the data points with the 'best-fit' quadratic curve.
It should be clear that the points do not all lie exactly on the curve.
There are two points that most clearly lie above the curve. Give the coordinates of those two points below, one point on each line with the two coordinates in comma-delimited format:
40 , 36.9
80 , 4.9
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There are two points that most clearly lie below the curve. Give the coordinates of those two points below, one point on each line with the two coordinates in comma-delimited format:
30 , 49.2
70 , 9.3
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It should be clear that the t = 30 point lies a bit below the curve, while the t = 40 point lies a bit above. Imagine a straight line connecting these two points. Would you say that the straight line is more or less steep than the curve, and why would you say this?
Straight line is less steep than the curve
T point 40 lies above t point 30 below- equal out graph
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The graph below looks very similar to the preceding graph, except that in this graph the points are connected by straight lines. Rather lying close to a smooth curve, as in the preceding graph, the data points are joined by what is might be called a 'slightly broken' or 'slightly jagged' line, which makes it change in slope at every data point.
Let's assume that the curve does in fact represent the actual behavior of the system being observed, and that the slight discrepancies between the curve and the data points is due to uncertainties in observation.
It should therefore be clear that experimental uncertainty causes the steepness of most or all of the line segments to differ at least slightly from the steepness exhibited by the curve over the same time interval.
Refer to either or both of the preceding graphs to answer the following:
• Between t = 70 and t = 80, do you think the line segment is steeper than the curve or less steep, and why?
• Between t = 80 and t = 90, do you think the line segment is steeper than the curve or less steep, and why?
• Do you think there is a connection between these two answers? That is, is the reason for your first answer related to the reason for your second?
Answer these questions below:
segment between t=70 to t=80 is less steep, point at t=80 above the line
segment between t=80 to t=90 is more steep (t 90 below line)
I do not know that there is a distinct connection between the two.
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Open the Data Analysis Program and copy into the data window the comma-delimited data given at the beginning, then proceed as follows:.
• Click on Difference Quotient and when prompted verify that your data is in the correct format.
• The program will give you three tables, the third being a table of difference quotient vs. midpoint values.
Copy the table of difference quotient vs. midpoint values into the box below:
15,-1.701
25,-1.547
35,-1.221
45,-1.115
55,-1.037
65,-.6063
75,-.443
85,-.4473
95,-.042
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A difference quotient is calculated for every time interval represented by the graph.
The difference quotient for a given interval is the average rate of change of the quantity in the second column with respect to the quantity in the first column. By the definition of rate of change, this means that the difference quotient for an interval is
• difference quotient = change in second-column quantity / change in first-column quantity.
In this case the second-column quantity is position and the first-column quantity is clock time, so the difference quotient represents change in position / change in clock time. This is the average rate of change of position with respect to clock time:
• average rate of change of position with respect to clock time = change in position / change in clock time.
Since the position is represented by the vertical coordinate on this graph, the change in position for the interval between two points is represented by the change between the vertical coordinates of the two graph points. This is the 'rise' between the graph points.
Since clock time is represented by the horizontal coordinate on this graph, the change in clock time for an interval is represented by the change between the horizontal coordinates of the two corresponding graph points. This is the 'run' between the graph points.
So for an interval the average rate of change of position with respect to clock time is represented by the rise / run between the corresponding graph points. As you know, the rise / run between two graph points is the slope of the line segment connecting those points.
Average rate of change of position with respect to clock time is called average velocity, so for a given interval we have
• average velocity = change in position / change in clock time = rise / run = graph slope.
This average velocity is also equal to the slope between the corresponding graph points.
below:
• In terms of the table, explain how the first difference quotient was calculated.
• Explain how this calculation is related to the graph.
• Explain what the first difference quotient tells us about the object being observed.
difference quotient= taking the difference of the intervals in columns and dividing them by each other, average velocity (Change in x and y values) is like slope (rise over run)
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Explain below why the difference quotients for this data are negative, what this has to do with the graph and what this tells us about the behavior of the object being observed:
slope of the graph is negative, decreasing object
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Are the difference quotients increasing or decreasing, and overall do they appear to be doing so at an increasing, decreasing or constant rate?
Seem to be decreasing at a decreasing rate
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The last table shown in the results of your data analysis program depicts difference quotient vs. midpoint clock time.
As we have seen, if we interpret the original data as position vs. clock time for some object, the difference quotients represent the approximate average velocities, over the various time intervals, of the object being observed.
As we have seen, small errors in observation affect our difference quotients and cause them to diverge at least slightly from the actual behavior of the system. In this case, the difference quotients give us a pretty good approximation to the average velocities, but they do fluctuate above or below the actual average velocities.
The average clock time for an interval will be the clock time at the midpoint of the interval.
So the last table lists average velocities vs. midpoint clock times.
Explain why we might therefore say that this table therefore gives us the most accurate velocity vs. clock time approximation we can obtain directly from our data points.
This table is most accurate because it is records the midpoint clock times.
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Clear everything from the data window but the table of difference quotient vs. midpoint clock time values. Your window should then contain 9 lines of comma-delimited data, and nothing else.
These data represent approximate velocity vs. clock time for your system, but there are fluctuations due to uncertainties in the original observations.
The graph below represents the data in your window. Check to see that the coordinates of the points below seem to correspond to the data in the data window.
Note that the velocities are negative, for reasons that should be clear, while the speeds are decreasing (remember that speed is just how fast and doesn't care about positive or negative; more formally speed is the absolute value, or magnitude, of velocity). The result is negative velocities that approach zero. The velocities are increasing through negative numbers toward zero, so the graph has an upward slope.
Explain below how the graph depicts the trend of the velocity of the system.
Explain also how it depicts the fluctuations that result from the original uncertainties in observations.
The graph shows decreasing speeds approaching 0. Some points are higher and some are lower. Is not a straight line.
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The 'broken-line' graph below depicts straight lines between points. How do you think these straight lines differ from the actual behavior of the system?
Could be faster at one end or slower at the other
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The graph below 'smooths out' the random fluctuations due to experimental error, and gives us a good idea of the actual velocity vs. clock time behavior of the object being observed.
Explain how the best-fit straight line is more likely to represent the actual velocity of the object than the broken-line graph shown previously.
This graph is straight and smooths out the data. It makes it easier to read.
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Do you think the best-fit straight line is completely accurate in its representation of the actual velocity vs. clock time behavior of the object being observed, or do you think that the random uncertainties leave us with some uncertainty about the actual velocities?
I don’t think any one of the graphs can be without any uncertainty, however I do believe this graph comes pretty close
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Click on the Difference Quotient button again, verify that your data is in the appropriate format, and copy the new difference quotient vs. midpoint clock time table into the box below:
20, -.0154
30, -.0326
40, -.0106
50, -0.007800
60, -.04307
70, -.01633
80, 0.0004299
90, -.04053
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One thing you will notice is that while our original data showed clock times ranging from 10 to 100 seconds, the results here show clock times only from 20 to 90 seconds. What happened to the extra 10 seconds on each end?
Gets rid of the two end points when getting an average
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Look at the difference quotients obtained in this last step. Do you see any clear trend to this data?
I’m not sure that there is a trend to this data, If so I couldn’t find it.
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In terms of the velocity vs. clock time table from which these new difference quotients were calculated, explain how the first of these new difference quotients was calculated, and what this calculation means.
Explain how this calculation is related to the velocity vs. clock time graph.
Was it more related to the broken-line graph or the graph of the best-fit line?
Explain what the first of the difference quotients tells us about the object being observed.
The first difference quotient was found when I took the difference in the first two changes in position and divided by the changes in time.
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Once again recall that
• difference quotient = change in second-column quantity / change in first-column quantity.
In this case the second-column quantities are velocities, while again the first-column quantities are clock times.
So each of these calculations will represent change in velocity / change in clock time, or acceleration.
Again the calculations will also represent graph slopes.
Do these new difference quotients represent the slopes of the broken-line graph or the straight-line approximation of the best-fit v vs. t line?
difference quotients could be the slopes of the broken lines in the graph
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Here is the graph of the 'new' difference quotient vs. clock time data that should be in your window. Check to be sure the coordinates of these points match the information in your window.
If you were to draw a straight line to fit the acceleration vs. clock time information of this graph, would it be clear where to draw it, whether it should be increasing or decreasing and if so by how much?
Based on the graph of acceleration vs. clock time estimates, is there a clear trend in the graph of our acceleration estimates, or is it difficult to tell whether the acceleration is increasing, decreasing or tending to remain the same?
I cannot tell…I don’t know if it is in any of these categories. The points seem to be all over the place.
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The 'broken-line' graph of this data is also shown. Note that this broken line is much more jagged than that of the 'broken-line' graph obtained for the velocity.
Look again at the graph of the original position vs. clock time data, in which the data points were connected by straight lines. This is actually a broken-line graph, but the lines aren't nearly 'as broken' as in the velocity vs. midpoint clock time graph.
The results we obtained in our second application of the difference quotient routine are based on the 'more badly broken' of the two graphs.
The minor uncertainties in our original position vs. clock time data caused more significant uncertainties in the estimates of velocity obtained by our first application of difference quotients, though our velocity trend was still recognizable.
These uncertainties were magnified when we calculated the difference quotients of these estimated velocities, making any trend observed in the resulting acceleration estimates very uncertain.
Summarize your understanding of how repeated application of the difference quotient can magnify uncertainties below.
Using difference quotients showed the uncertainty in the position vs. time on a bigger scale.
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The actual 'best-fit' line for this data is shown below; however it would have been difficult to predict this line from the picture of the data points.
Can you see how the slight jaggedness of the broken-line graph for position vs. t is magnified into the significant jaggedness of the approximate velocity vs. t graph and is then magnified again to give us the extreme jaggedness of our acceleration vs. t graph?
Describe this progression in your own words.
velocity vs. time graph, uncertainties more noticeable in the broken line graph, kept being shown on a large scale
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Speculate on your answer to the following question:
• To what extent do you think that a lack of a clear trend in our acceleration graph indicates that there is no actual clear trend in the acceleration of the object, to what extent would it be the result of our inability to observe the system accurately, and to what extent would it be the result of our reliance on the difference quotient rather than some other method of analysis?
I think it has played a role in the process
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The data below represent an observation of the same system as before, but as they might be with experimental uncertainties only about 1/3 as great.
10 , 80.718
20 , 63.848
30 , 49.15
40 , 36.13
50 , 25.042
60 , 15.81
70 , 9.005
80 , 4.286
90 , .955
100 , .279
The graphs shown below depict the results of a similar analysis based on these observations.
Original data with 'ideal' curve (note that data points lie closer to the actual curve than in the first data set).
The curve in this graph, and in the original position vs. clock time graph, is quadratic in nature and the closeness of the fit in both cases strongly indicates a constant acceleration.
Line-to-line graph, slightly 'broken':
Graph of difference quotient vs. midpoint clock time, representing approximate velocity vs. clock time. Note that this graph is less jagged than the corresponding graph for the original data set.
This graph also strongly indicates that the velocity is a very nearly linear function of clock time, which is a strong indication that acceleration is constant or nearly constant.
Difference quotients vs. midpoint clock time based on velocity vs. clock time estimates. These represent the second iteration of the 'difference quotient' routine, starting with the original data.
While still quite jagged, the data points cluster much more closely than before around the 'ideal' value of -.02, and would provide a good indication of the average acceleration. However they still do not provide a strong indication whether or not the acceleration is in fact constant.
In iterated calculations based on the difference quotient, the constant nature of the acceleration still gets 'lost in the noise'.
In both this model and our original model, the position vs. clock time information is in fact very consistent with a constant acceleration close to -.02 cm/s^2. It should be clear whether this acceleration graph or the first gives ua a better estimate of the acceleration of the system.
Describe in your own words the difference in the quality of the information we can obtain from the velocity and acceleration graphs as a result of reducing experimental uncertainty to 1/3 its original value.
When you reduce this you are getting a more accurate estimate of acceleration and velocity (average)
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It should also be clear that successive applications of the difference quotient are very susceptible to uncertainties in observation, and that mathematical methods of 'smoothing' our data can be very effective.
Comment on the potential usefulness of mathematical smoothing vs. repeated application of difference quotients.
Difference quotients gives more of an uncertainty to the graphs than using the mathematical soothing.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
• Approximately how long did it take you to complete this experiment?
2 hours
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&#Your work on this lab exercise is good. Let me know if you have questions. &#