Question

A violin string is 30 cm long. It sounds the musical note A (440 Hz) when played without fingering. How far from the end of the string should you place your finger to play the note C (523 Hz)?

Short answer

The violinist must place his finger$4.8cm$from the end.

Step-by-step solution

Given information

The length of the violin string is $L=30cm=0.30m$.

The frequency of musical note A is $f_A=440Hz$.

Distance from the end of the string

A string fixed at both ends forms standing waves.

A simple string sounds the fundamental frequency of $f_1=\frac{v}{2L}$.

When the length of the string is $L_A=30cm$, the note has a frequency role="math" localid="1650034394357" $f_{1A}=\frac{v}{2L_A}\cdots \left(1\right)$.

For a different length role="math" localid="1650034659883" $L_C$, the note has a frequency role="math" localid="1650034674350" $f_{1C}=\frac{v}{2L_C}\cdots \left(2\right)$.

From (1) and (2),

role="math" localid="1650034707562" $$\begin{gathered} \frac{f_{1A}}{f_{1C}}=\frac{{\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2L_A$}\right.}}{{\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2L_C$}\right.}} \\ =\frac{L_A}{L_C} \\ {L}_C=\frac{f_{1A}}{f_{1C}}{L}_A \end{gathered}$$

The second frequency will be role="math" localid="1650034718418" $f_{1C}=523Hz$.

role="math" localid="1650034728404" $$\begin{gathered} L_C=\frac{440}{523}\times 30 \\ =25.2cm \end{gathered}$$

The full string is $30cm$long, so the violinist must place his finger $30-25.2=4.8cm$ from the end.