By touching a string lightly at its center while bowing, a violinist can produce a note exactly one octave above the note to which the string is tuned—that is, a note with exactly twice the frequency. Why is this possible?

The frequency of a string is directly proportional to the square root of its tension.

The frequency of the string is given by

$\mathrm{f}=\frac{1}{2\mathrm{L}}\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}$

Here, L is the length of the string, T is the tension, $\mu$is the linear mass density.

The frequency of the string vibration is inversely proportional to the length of the string.

If the length of the string is short then the frequency is high. When the violinist touches center of the string, a node is produced at that point and the length of the vibrating element is half of the actual element.

The formula to calculate the frequency is,

$\mathrm{f}=\frac{1}{2\mathrm{L}}\sqrt{\frac{\mathrm{F}}{\mathrm{\mu }}}$ …... (1)

For center position, the length becomes half, Substitutefor in equation (1)

$\mathrm{f}\text{'}=\frac{1}{2\left({\displaystyle \frac{\mathrm{L}}{2}}\right)}\sqrt{\frac{\mathrm{F}}{\mathrm{\mu }}}$ ….. (2)

Here. f' is the frequency of vibration when plucked at the center of the string.

Divide equations (2) and (1)

$$\begin{gathered} \frac{\mathrm{f}\text{'}}{\mathrm{f}}=\frac{{\displaystyle \frac{1}{\mathrm{L}}}\sqrt{{\displaystyle \frac{\mathrm{F}}{\mathrm{\mu }}}}}{{\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{\mathrm{F}}{\mathrm{\mu }}}}} \\ \mathrm{f}\text{'}=2\mathrm{f} \end{gathered}$$

Therefore, the frequency of the vibration gets double when the finger touched at the middle vibrating string of guitar.