1-Mar-2017: Free Fall Lab

Lab 2: Determination of g and some statistics for analyzing data
1-Mar-2017

Part 1

Idea: I have to demonstrate the idea that a falling body will accelerate at 9.81 m/s^2 when there are no other external forces except gravity.
Summary: First, with regards to materials, in this lab we used a special paper tape with markings already recorded by a dropped spark generator (shown below), as well as a meter stick to record the distance between each mark, and lastly Excel to plot our data.

1. Then, we used our meter stick to record the position of each dot from the 0 cm mark.
2. We opened up a data set in Excel and set column A to record our time. We set A1 to be Time, A2 = 0, and A3 = A2+1/60.
3. Column B recorded our distance, as B2 was 0 (for the 0 mark) and we went all the way down to B16, filling in our measured distances in each cell starting with the position of the second dot.
4. Column C was simply our change in distance, so from C2 downwards we set C2 = (B3-B2)
5. Column D would give the time for the middle of each 1/60th second interval, so from D2 we entered =A2+1/120
6. Lastly, column E gave the nid-interval speed, so from E2 down we entered =C2/(1/60)
7. Using this data table, we selected columns D and E, and used a linear fit, and obtained the equation of this line.
8. Subsequently, we graphed columns A and B, abut instead used a polynomial fit of order 2, and got the equation for that as well.

Measured Data

This is the data based off the recorded distances in cm for each mark on the paper tape, and subsequently steps 2-6 in the previous section in which we found the mid-interval time & speed.

Analysis: 

1) In the first graph, we plot the mid interval speed vs time because doing so will give us the slope, which is also the acceleration of the free falling object. If our measurements were accurate, then there is reason to believe that the slope should read a gravitational acceleration close to 9.8 m/s^2.
2) The second graph, distance over time, will show simply the slope, also the velocity, of the falling object.

Graph/Calculations:

This graph depicts Columns D and E with a linear fit

As I could see, the slope of the linear graph depicts a downward acceleration of the object at 958.15 cm/s^2. This was obviously off from the expected gravitational acceleration of around 980 cm/s^2.

This graph depicts Columns A and B, with a power fit of order 2

The slope of this graph depicts the velocity of the falling object.

Questions/Analysis

1)Velocity in middle of time interval vs average velocity for time interval

2) From my velocity/time graph, I can also get my acceleration by

compared to the accepted value of 980 cm/s^2, the value is also off by around 2.2 cm/s^2.
3) From a position.time graph, I can again get my acceleration by

compared to the accepted value of 980 cm/s^2, the value is also off by around 2.5 cm/s^2.

Conclusions for Part 1

1) There were clear patterns to the data. First, the increase in X in column C of my data table remained noticeably consistent all the way through, and of course even without a line of best fit, the graph points that columns A and B, D and E gave showed definitive patterns. Linear and power, respectively.

2) As the lines of best fit showed, the patterns I expected to find were all there.

Experimental uncertainty:

I found the relative difference of my resulting acceleration to that of the accepted value, how much they differ in percentage:

Part 2

Idea: I have to analyze the class data for gravity and find the Standard Deviation of the Mean of the class data.
Summary:

1. I opened up a new file in Excel, and entered the class values for g from cells A2 to A11. In cell A12, I average these values with =average(A2:A11)
2. Column B is for my deviation from the mean, so from B2 to B11, I enter and fill =A2-$A$12

Data Table

Based on the class data, the chart gives me the average, average deviation squared, and lastly, the standard deviation

Analysis: 

Thus the standard deviation for g was 6.11, meaning that within the entire probability range of the entire class, from (961.5-6.11)=955.2 to (967.6+6.11)=973.7, the range of values from 961.5 to 967.6 would be the most likely.

Conclusion/Questions

1) We observed that for the values of g, the higher its deviation, the lower the value of g.

2) Our average, 961 cm/s^2, was less than the accepted value of 980 cm/s^2 by approximately 19 cm/s^2.

3)  In all of our values of g, none of us ever got to or exceeded the accepted value. All of our data for g pointed to values less than 9.8 m/s^2.

4) For the discrepancy between my measurements and those of the class, the fact that there is likely to be a  ±1mm of uncertainty in the measurements of the marks (due to simple human error in judging the distance) could certainly be a factor. This would directly affect my entire data table, and subsequently the velocity/time and position/time graph.This would be a random error.

The second would be in possibly the marked paper itself. As the object falls, and creates sparks that leave marks on the paper, it makes slight contact with the paper as it falls, introducing an external force apart from gravity, friction, to the object. This would lead to inaccuracies in the marks themselves, that we would not be aware of, so it is subsequently a systematic error.

5) In this part of the lab, we were to analyze multiple results of the same experiment in order to firstly be able to analyze the standard deviation, and in addition, distinguish possibilities in conditions that may have led to discrepancies in each of our values of g. We accomplished this by finding the average of our values, the average deviations, which allowed us to visualize the probability range of our values within 68%. Subsequently, we were able to notice a trend in all of our data being lesser than the accepted value of g, which allowed us to conclude that there must have been a systematic error in the experiment that affected everyone on top of individual random errors.