Don’t Fret!
MCR 3U
Name:__________________
Instruments such as guitars, violins, and pianos all produce sounds by vibrating strings. The
frequency of the notes produced depends on several factors, including the length of the
string.
Instruments such as pianos and harps have many strings of varying yet fixed lengths. On a
guitar, the player can change the length of each string by pressing on frets.
In this activity, we will investigate how the frequency of the notes produced changes with
the frets. We will use a phone app called ‘Frequency’. Several different apps are
available.
Part 1: Collect Data
Pick any string to work with.
Record the frequency when played ‘open’.
Record the frequency under Fret 0.
Next hold down Fret 1, and record this frequency.
Repeat this until Fret 12.
Part 2: Graphical Model
Create a scatter-plot from your table.
Draw a smooth line through the points.
700.0
Fret
Ratio
650.0
Frequency
(Hz)
0
329.6
------
1
349.2
1.0595
2
370.0
1.0595
3
392.0
1.0595
400.0
4
415.3
1.0595
350.0
5
440.0
1.0595
300.0
6
466.1
1.0595
250.0
7
493.8
1.0595
200.0
8
523.2
1.0595
150.0
9
554.3
1.0595
100.0
10
587.3
1.0595
50.0
11
622.2
1.0595
12
659.2
1.0595
600.0
550.0
Frequency
Frequency (Hz)
500.0
450.0
0
0.0
0
1
2
3
4
5
6
7
Fret
Fret
8
9
10
11
12
13
Part 3: Algebraic Model
In the right-hand column of the table, calculate the ratio between successive
frequencies.
See table
What can you conclude based on these values?
Exponential
Determine an equation relating Frequency (F) and fret number (n).
F=
329.6(1.0595)n
Part 4: Analysis
Compare your equation to that of another group. How are they the same / different?
Same base value, different ‘a’ value
Compare the frequencies for Fret 0 and Fret 12. How do they relate? How do the
corresponding notes relate? (You may need to consult a musician for this one)
Fret 12 is double the frequency of Fret 0. They are the same note, one octave apart.
How might your equation relate to the answer above?
The base of the power is the 12th root of 2. ( √2 = 1.0595)
12
Suppose a different string has an open (Fret 0) frequency of 147 Hz.
What would you expect the frequency of Fret 5 to be?
𝐹 = 147(1.0595)5 = 196.2 𝐻𝑧
On the same string, a note has a frequency of 247 Hz. Which fret produces this note?
247 = 147(1.0595)𝑛
𝑛=9
(guess and test)