The following data corresponds to the measurements from the bridge of a guitar to each of the frets:
Fret # |
Length to the fret |
Fret # |
Length to the fret |
0 | 64.5 | 11 | 34.2 |
1 | 60.7 | 12 | 32.3 |
2 | 57.4 | 13 | 30.4 |
3 | 54.1 | 14 | 28.7 |
4 | 51.1 | 15 | 27.1 |
5 | 48.3 | 16 | 25.6 |
6 | 45.5 | 17 | 24.2 |
7 | 43 | 18 | 22.8 |
8 | 40.6 | 19 | 21.6 |
9 | 38.4 | 20 | 20.4 |
10 | 36.3 |
Students might conjecture that the graph corresponds to a branch of a parabola or that corresponds to a exponential function. Depending on the conjecture the teacher might suggest the students to compose a table of second differences for the parabola or ratios for the exponential.
Computing the ratios of the values corresponding to 2 successive frets we get the following sequence:
Fret# | Length to Fret | Ratio | Fret# | Length to Fret | Ratio |
0 | 64.5 | 0.941085271 | 11 | 34.2 | 0.944444444 |
1 | 60.7 | 0.945634267 | 12 | 32.3 | 0.941176471 |
2 | 57.4 | 0.942508711 | 13 | 30.4 | 0.944078947 |
3 | 54.1 | 0.944547135 | 14 | 28.7 | 0.944250871 |
4 | 51.1 | 0.945205479 | 15 | 27.1 | 0.944649446 |
5 | 48.3 | 0.942028986 | 16 | 25.6 | 0.9453125 |
6 | 45.5 | 0.945054945 | 17 | 24.2 | 0.94214876 |
7 | 43 | 0.944186047 | 18 | 22.8 | 0.947368421 |
8 | 40.6 | 0.945812808 | 19 | 21.6 | 0.944444444 |
9 | 38.4 | 0.9453125 | 20 | 20.4 | |
10 | 36.3 | 0.94214876 |
The ratio seems to be around the value 0.944. In fact, if we average the values in the third column we get the value 0.94406996. The function seems to be exponential with first value 64.5. If we produce the values of this function in excel we have:
x | x | ||
0 |
64.5 |
11 |
34.217766 |
1 |
60.888 |
12 |
32.3015711 |
2 |
57.478272 |
13 |
30.4926831 |
3 |
54.2594888 |
14 |
28.7850929 |
4 |
51.2209574 |
15 |
27.1731277 |
5 |
48.3525838 |
16 |
25.6514325 |
6 |
45.6448391 |
17 |
24.2149523 |
7 |
43.0887281 |
18 |
22.858915 |
8 |
40.6757593 |
19 |
21.5788157 |
9 |
38.3979168 |
20 | |
10 |
36.2476335 |
And if we graph simultaneously the original data and the values given by this new table we can see that the values of the function:
approximates quite well the string data:
What if we change the length of the open string to 50 or 10 or 1?
x |
|||
0 |
50 |
1 |
10 |
1 |
47.2 |
0.944 |
9.44 |
2 |
44.5568 |
0.891136 |
8.91136 |
3 |
42.0616192 |
0.84123238 |
8.41232384 |
4 |
39.7061685 |
0.79412337 |
7.9412337 |
5 |
37.4826231 |
0.74965246 |
7.49652462 |
6 |
35.3835962 |
0.70767192 |
7.07671924 |
7 |
33.4021148 |
0.6680423 |
6.68042296 |
8 |
31.5315964 |
0.63063193 |
6.30631928 |
9 |
29.765827 |
0.59531654 |
5.9531654 |
10 |
28.0989407 |
0.56197881 |
5.61978813 |
11 |
26.5254 |
0.530508 |
5.30508 |
12 |
25.0399776 |
0.50079955 |
5.00799552 |
13 |
23.6377388 |
0.47275478 |
4.72754777 |
14 |
22.3140255 |
0.44628051 |
4.46280509 |
15 |
21.06444 |
0.4212888 |
4.21288801 |
16 |
19.8848314 |
0.39769663 |
3.97696628 |
17 |
18.7712808 |
0.37542562 |
3.75425617 |
18 |
17.7200891 |
0.35440178 |
3.54401782 |
19 |
16.7277641 |
0.33455528 |
3.34555283 |
Then the graphs are:
The students might want to look at better graphs of the functions and graph them in another application like the the Graphical Calculator where the functions can be seen from different zooms.
If the students decide to explore the possibility of the data fitting a parabola we can ask them to make second differences, discuss what do the second differences look like if the curve is a parabola.
The table is:
fret |
data |
1st Differences |
2nd Differences |
0 |
64.5 |
3.8 |
0.5 |
1 |
60.7 |
3.3 |
7.10543E-15 |
2 |
57.4 |
3.3 |
0.3 |
3 |
54.1 |
3 |
0.2 |
4 |
51.1 |
2.8 |
7.10543E-15 |
5 |
48.3 |
2.8 |
0.3 |
6 |
45.5 |
2.5 |
0.1 |
7 |
43 |
2.4 |
0.2 |
8 |
40.6 |
2.2 |
0.1 |
9 |
38.4 |
2.1 |
7.10543E-15 |
10 |
36.3 |
2.1 |
0.2 |
11 |
34.2 |
1.9 |
7.10543E-15 |
12 |
32.3 |
1.9 |
0.2 |
13 |
30.4 |
1.7 |
0.1 |
14 |
28.7 |
1.6 |
0.1 |
15 |
27.1 |
1.5 |
0.1 |
16 |
25.6 |
1.4 |
3.55271E-15 |
17 |
24.2 |
1.4 |
0.2 |
18 |
22.8 |
1.2 |
-3.55271E-15 |
19 |
21.6 |
1.2 |
1.2 |
20 |
20.4 |
The teacher can also plot the points one by one in the graphical calculator and ask students to construct the parabola that best fits the data. Then, to draw the function:
and to discuss again the fitness of the graphs.
The parabola that we found to best fit the data was:
The following is a graph of both functions: