# Physics and Astronomy Labs/Beads on string drop

## The introduction is under construction

Student tasks yet to be done:

• Astronomy students will be tasked with describing the procedure used to take data, and attempting to explain how a connection to Galileo justifies this as an "astronomy experiment".
• Somebody needs to link the data taken from the cell phone video to Wikiversity's Vernier scale where this data is analyzed.
• The audiovideo file shows the string falling in real time and in slow motion. We need to verify that the time-scale on the screenshot is accurate because preliminary estimates indicate that there is a 15% error in either the paperclip positions or the time measured between strikes. This might be related to a well-known slinky-drop video. See Hooke's law and Young's modulus for efforts to measure the stiffness of the string, which in turn can be used to calculate the propagation time of a compressional wave that propagates downward when the string is released. This would cause a delay in the time for the early paperclips to drop, giving the illusion of stronger gravity than is the case.
• Mechanical engineering students will be tasked with writing a calculation that explains why the distance of the n-th paperclip from the bottom of the string should vary as n-squared. They will also show how to use how a plot of n versus time can be used to calculate g.

## Measuring earth's gravitational field, g

Use the cell phone audio data to find g using,

${\displaystyle g={\frac {2x_{1}}{T^{2}}}={\frac {2x_{n}}{n^{2}T^{2}}}}$ ,

where T is the time between two consecutive hits by the paperclips, x1 is the initial distance of the first paperclip above the coffee can, and we hope to obtain a value for the acceleration of gravity at Earth's surface approximately equal to g ≈ 9.8 m/s2.

Exercise: Prove the two equations shown above. (click for hint)
1. Use ${\displaystyle x=x_{0}+v_{0}t+{\tfrac {1}{2}}at^{2}}$
2. Begin with the first paperclip at x1 and explain why vi equals zero.
3. Explain why the time for the n-th paperclip to fall is nT.
4. Write down the formula that relates xn to x1 and n.
5. The rest is algebra

## Future project: Make a model Galileo's ramp

Astronomy students learn how Galileo, Kepler and Newton (and others) developed a theory for how planets revolve around the Sun. Of these people, Galileo did the kinds of experiments best suited for labs in a college science course. Unfortunately this image is not available on commons. But at least two Wikipedia pages that could use such an image:

I asked commons about the legality of making and then photographing a 3D model of the device, either out cardboard or using Solidworks software. They seemed to think it was OK to post an such an image on commons.

click to view question posed to commons and their response

Using copyrighted images to make 3D models

Two questions:

1. Suppose I take a copyrighted image such as this, take measurements from the image and construct a 3D model using Solidworks, rotate the image so that it bears little resemblance to the photo, and put it on commons. Is this a violation? My instincts tell me it is, but I am hoping I am wrong.
2. Same image, same question. But this time its a hand-drawn sketch made by a student, also rotated so as to not look like the photo. How bad does it have to be before the sketch is "original" and therefore permissible on commons?--Guy vandegrift (discuss) 01:34, 31 January 2015 (UTC)
If your 3D model is independent of the creative inspiration of the photo, you don't have to worry about the copyright of the photo. Same thing with a sketch. The problem is, with only one photo, you're showing in detail what the photo shows in detail and you have no detail on what the photo doesn't show. In this case, maybe it's simple enough.--Prosfilaes (discuss) 16:35, 31 January 2015 (UTC)
The information about spatial positions and shapes of 3D objects in this particular photo can not be copyrighted itself. The copyrightable elements in the photo are only angle at which it was taken, a particular illumination and other elements which can be changed by the photographer. So, it would be ok to create a 3D model as you described above. The same logic applies to the sketch. Ruslik (discuss) 17:32, 31 January 2015 (UTC)
Thanks.--Guy vandegrift (discuss) 21:39, 31 January 2015 (UTC)

## Recent data to be processed

### Engineers measured intervals on 2/5/15

They used the peak, not the leading edge. The class took median values and by consensus concluded the average spaceing was consistent with ${\displaystyle T_{1}=.10143\pm .00917\ seconds}$    and   ${\displaystyle g=15.6\pm 2.5\ m/s^{2}}$

## Data on string and paperclips

• Each paperclip except one weighed 1.065 g. One weighed 1.466 g. The string was 5.04 m long and had a mass of 2.578 g.

## Where did we situate those paperclips?

Gaussian parametric versus a nonparametric methods: An introduction to hypothesis testing

There was some confusion about the distance to the first paperclip. A preliminary investigation suggested that 3 inches was a suitable gap, but we constructed the device using meter sticks. Three inches is about 7.6 cm. Did we round the first gap to 8 cm, or did we use 7.6 cm? The drop seems to have moved the paperclips when we remeasured the lengths on the next day. Since the answer was either 7.6 or 8, it was time for some hypothesis testing.

• Hypothesis A was that the first gap was constructed at 7.6cm and hypothesis B was that it was 8cm. All other paperclip positions were calculated (in cm) as either 7.6n2 or 8n2, where n is an integer. But for a variety of reasons, the paperclips were not at these exact locations the next day. We remeasured all the lengths. Now we pose the question: Is the data more consistent with hypothesis A or hypothesis B?

This led to an interesting discussion about outliers and the sum of squares, often abbreviated as SS. Depending on how the analysis is performed, we got different results. Three methods suggested the correct scenario B and one suggests scenario A as being more likely.[1]

#### Explanation of columns in the spreadsheet

The spreadsheet that follows shows the measured paperclip locations, as measured one day after the string was dropped and handled in such a way that it is understandable that the paperclips had moved. The spreadsheet also displays the calculated paperclip positions under our two alternative hypotheses (A and B).

If you need more help interpreting the table that follows click here
• The first column represents n, and labels the 8 that were attached to the string at various distances from the bottom end.
• The second column is n2, used by Excel to calculate other lengths.
• The column "A" represents the distanced under scenario A; with the first paperclip at 7.6 cm (3 in).
• The column "B" represents the distanced under scenario B; with the first paperclip at 8 cm.
• The column "x" represents the measured distances of each paperclip, measured the next day.
• εA is properly called a deviation, but is often casually referred to as an error. For example, the deviation was 0 for the first paperclip since both the measurement and hypothesis A correspond to x1 = 8 cm.
• εB depicts the error (deviation) under scenario B. For example, the deviation for the first paperclip −0.4 cm for the first paperclip because 7.6−8=0.4
• εA2 is the square of the error (deviation) under scenario A. Since the square of a negative number is positive, the sum of these values are meaningful measure of the overall fit to the data.[2] That sum is shown at the bottom, where the sum was made in two ways:
1. First the sum of squares is made for all eight values of n (i.e. the distances to all eight paperclips). This sum of squares is often written as SS and is the sum of the squared deviations.
2. Then the sum is calculated for the first 7 paperclips. This was done because the error for n=8 is much large. Such data points are known as outliers, and our data set has only one outlier.

Subsequent columns show the square of the error for each paperclip under both scenarios. The sum of these squares is often called SS in statistics books. Since the paperclip at n=8 is an outlier[3], we calculate SS for all 8 paperclips, as well as SS for the first 7 paperclips.

#### Using Excel to conduct an informal hypothesis test

This table from Excel[4] shows the observed values of x, as well as the values associated with two possible scenarios[5]. This was an unusual hypothesis test because we are certain the attempt was made to situate the paperclips at either xA or xB. Here ε represents the deviation (so called "error") and SS is the sum of the squares of the deviations. We have one outlier at n=8, so the test was performed with and without that outlier. The last two rows represent what I believe[6] is a simple nonparametric analysis: The column entry equals 1 if that the hypothesis is favored.

 n n2 Acm Bcm xAcm xBcm xcm εAcm εBcm εA2cm2 εB2cm2 * A likely B likely 1 1 8 7.6 8 7.6 8 0 −0.4 0 0.16 * 1 0 2 4 8 7.6 32 30.4 31.25 0.75 −0.85 0.5625 0.7225 * 1 0 3 9 8 7.6 72 68.4 71 1 −2.6 1 6.76 * 1 0 4 16 8 7.6 128 121.6 127 1 −5.4 1 29.16 * 1 0 5 25 8 7.6 200 190 198 2 −8 4 64 * 1 0 6 36 8 7.6 288 273.6 286 2 −12.4 4 153.76 * 1 0 7 49 8 7.6 392 372.4 386 6 −13.6 36 184.96 * 1 0 8 64 8 7.6 512 486.4 487 25 −0.6 625 0.36 * 0 1 * * * * * * * * * * * * * * SS 1-8 671.56 439.88 * 7 1 SS 1-7 10.56 254.56 *

### Questions for the students

click for questions worth asking students regarding this hypothesis test
1. Let's set aside the paperclip for n = 8 and look at 1-7. Note that all the errors in scenario A are positive and all those in scenario B are negative. Which best describes what this fact alone seems to suggest?
1. That the spacing might be less than 7.6 cm
2. That the spacing might be between 7.6 and 8 cm
3. That the spacing might be greater than 8 cm
4. Asolutely no clue is available from the fact at all the errors in one column are positive while all the errors in the other column are negative.
1. Based on n from 1 through 7, which is the most likely scenario?
2. Now consider all data points (n=1 to 8). Most computer programs would base their conclusion on this sum of squares (SS). From this number alone, which would be considered the most likely scenario?
3. Now look at the last two columns on the right. They compare the errors for A and B for each individual paperclip, line by line. The value of 1 or 0 is determined by which scenario has the smaller error for that particular paperclip. What does it mean if a given paper clip has a 1 in the A column and a 0 in the B column?
4. Looking at these two columns alone, without looking at the residual (SS), which would you estimate to be the more likely scenario if only cases 1 through 7 are considered?
5. Without looking at the residual (SS), what do the last two columns suggest about the two scenarios if all 8 paperclips are considered?

The last two columns represent the starting point of something I think is called nonparametric[7] analysis. Place an A or B in each of the cells in the table shown below, depending on which scenario is favored by that analysis:

 nonparametric[8] Sum of squares n= 1 to 7 n = 1 to 8

### Plotting activities

1. Make a plot of x versus n that shows both scenarios (A and B) as well as the actual measurements.
2. Since x is proportional to n, it is better to plot the square root of x versus n. Make such a plot and show the measured values as well as both scenarios.

## Footnotes

1. Several students recalled that scenario B (8cm) had occurred. A student asked if I did the calculation for purely pedagogical purposes. I thought it was 8 cm, but wasn't sure, especially when I measured the largest three distances the total error to be more consistent with 7.6.
2. The student who constructed this spreadsheet had instead used the absolute value of the errors to form this measure. But the absolute value function, f(x) =|x| has a sharp corner at x=0. Applied mathematicians can be prissy about functions that are not smooth.
3. An outlier is a data point that seems more in error than the rest.
4. The Excel spreadsheet was converted to wiktext using http://excel2wiki.net/ It was necessary to change the table's header to suit Wikiversity because the website's converter uses a table template unavailable on Wikiversity.
5. Both hypothesis are that xn=n2x1 where x1 is either 7.6 or 8.
6. I am working to confirm this belief.
7. I think I am using the term nonparametric correctly.
8. I think I am using the term nonparametric correctly.