# A One-String Musical Toy

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## Introduction: A One-String Musical Toy

This toy would not only amuse children but also teach them that there’s a relationship between notes and numbers. More specifically, the instrument demonstrates that the frequency of a note depends on the length of the string which produces the note. (Pythagoras was one of the first to have observed this phenomenon).

### Supplies:

That’s what you’ll need to make this device:

Materials

- wooden bar

- a piece of plywood

- a piece of plastic

- fishing line

- hardwood pins

- M6 bolt with nut

Tools

- saw

- drill with drilling bits

- exacto knife

- wire cutters

- long nose pliers

- soldering gun with solder

- file

### Teacher Notes

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## Step 1: Calculating the Frequency

The fundamental frequency of a vibrating string is calculated as:

f1 = v / 2*L, where

v - speed of wave in the spring

L - length of the string

The value of v is calculated as:

v = sqrt ( T / (m/L)), where

T - tension of the spring

m/L - mass of the string per unit length

The mass per unit length is calculated as:

m/L = A*ro*UL, where

A - the section of the string

ro - the density of the material of which the string is made

UL - unit of length

The string used in this project is made of nylon (ro = 1150 kg/m^3); its diameter is 0.25 mm (2.5*10E-4 m); the unit of length is taken 1 m.

The section of the string is calculated as:

A = (pi*d^2) / 4

A = 4.91*10E-8 m^2

The mass per unit length is:

m/L = 4.91*10E-8 m^2 x 1150 kg/m^3 x 1 m = 5.64*10E-5 kg/m

The tension was applied to the string by means of a weight with the known mass - 2.5 lb (it was a disk of a hand weight). Expressed in kilograms this mass would be, rounded to 2 signs, 1.13 kg.

Then, the speed can be calculated as:

v = sqrt (1.13 kg x 9.81 kg/m*s^2 / 5.64*10E-5) = 443 m/s

Now, f1 can be found, taking into account that the length of the string is 0.8 m:

f1 = 443 m/s / 2 x 0.8 m = 277 1/s, or 277 Hz

This frequency corresponds to the note C#4; in this case, the string is fixed only on its ends. When the string is also fixed to the movable bridge put at half-length of the string, f1 will be equal to 554 Hz (note C#5).

These are the notes situated between the above mentioned ones and their respective frequencies rounded to integers:

D4 294, D#4 311, E4 330, F4 349, F#4 370, G4 392, G#4 415, A4 440, A#4 466, B4 494, C5 523, C#5 554

You can observe that the ratio between two neighbouring frequencies is constant and equal to 1.06 (rounded to 2 signs). For example:

554/523 = 1.06; 349/330 = 1.06

Therefore, the length between the middle of the string and its left end could be divided in 12 intervals in such manner that the ratio of two neighbouring intervals is 1.06. Let’s designate the middle as X0 = 400 mm measured from the right support, then:

X1 / 400 = 1.06 ——> X1 = 424 mm

The values (in mm) of the other Xs are as follows:

X2 = 449, X3 = 476, X4 = 505, X6 = 567, X7 = 601, X8 = 637, X9 = 675, X10 = 716, X11 = 759, X12 = 805 (in fact, it means the open string, it’s not pressed to the movable bridge.

Thus, for example, the length of X3 should correspond to A#4. Indeed:

f (X3) = 443 / 2 x 0.476 = 465 Hz

The string should be plucked in such manner that the vibration passes in vertical plane; the sound in this case is more pleasant than when the string moves from side to side (this movement produces a ‘trembling’ sound). When the right edge of the movable bridge is put on the mark corresponding to a note, the top of the bridge defines the length of the string corresponding to this note.

## Step 2: Design of the Resonator

I used an available octagon shaped tin can to make a Helmholtz resonator for my instrument.

The resonant frequency of this device is calculated as:

f = (v / 2*pi) x sqrt ( A/ V*L), where

v - speed of sound in air (considered to be 343 m/s)

A - section of the resonator’s neck

V - volume of the resonator’s chamber

L - length of the neck

The volume of an octagon is calculated as:

V = 2*( 1 + sqrt2 )*a^2*h, where

a - side of the octagon (35 mm in this case)

h - height of the octagon (85 mm)

Thus, the volume is 502760 mm^3 and it’s a fixed value. On the contrary, the ratio A/L can be changed to change the resonant frequency; the formula for the frequency can be modified to separate this ratio:

A/L = 4*pi^2*f^2*V / v

for the frequency f = 277 Hz this ratio will be 12.94 (rounded to integers 13)

I made the central hole with the diameter of 10 mm in the tin’s cover; the section of this hole is 78.5 mm^2. Therefore, the height of the neck should be 6 mm. I made a sleeve with the diameter of 10 mm and height of 6 mm and soldered it to the cover.

## Step 3: Base and the String Supports

The base is made of a wooden bar with dimensions 800 x 40 x 20 mm that was available at my workshop. The left and right string supports are made of 19 mm thick plywood and fixed to the base’s ends with wood screws.

The left and right resonator supports are also made of 19 mm thick plywood and glued to the bar.

A piece of fishing line passes with moderate tension through the holes in the left string support, left resonator support and the movable bridge; thus, the bridge can be moved along the base.

## Step 4: Installing the Resonator

The resonator sits on its supports through two spacers (I made them as 10 x 10 mm squares of 2 mm thick plastic). If the bottom of the resonator contacts the entire surface of the supports, its flexibility decreases; that’s why I thought that some spacers were desirable.

The distance between the centre of the resonator and the right string support is 100 mm. The amplitude of the vibration of the string should be 1/4 of its maximal amplitude which is at the centre of the string; I found it sufficient to make vibrate the bottom of the resonator. Those who like to experiment could make a movable support for the resonator and observe how its position influences the sound.

The link between the string and the resonator is made of hardwood pin and fixed with a screw to the centre of the resonator’s bottom; the string passes through the 1 mm hole in the pin. Once the string is tensioned, it should be fixed in the link’s hole with a thin ‘wedge’ (I used an end of a toothpick).

## Step 5: Fixing the String

The right end of the string passes through the 1 mm hole in the corresponding hardwood pin and is fixed with a knot. The pin is fixed in the string support with a metal pin (2 mm diameter, 20 mm length) passing through the 2 mm hole in the pin.

After this, the string passes through the 1 mm hole in the link between the string and the resonator.

The left end of the string passes through the 2 mm hole in the bolt; a thin plastic sleeve (for example, a piece of cable insulation) must be put into the hole, after which the string passes through the sleeve and is fixed with a knot. The nut is situated next to the bolt’s head at this stage of the assembly; the bolt should move freely in the corresponding hole.

A weight with the known value (I used a 2.5 lb (1.13 kg) disk of hand weight) is attached to the head of the bolt, and the base is put vertically in such manner that the weight is suspended on the string. The latter stretches until the weight is balanced by the tension; the bolt goes out of the hole; the nut should be screwed until it touches the string support thus keeping the string under tension. Put some glue on the thread to prevent the nut from loosening. 