Question

A particular guitar wire is \(30.0 cm\) long and vibrates at a frequency of \(196 Hz\) when no finger is placed on it. The next higher notes on the scale are \(220 Hz\),\(247 Hz\), \(262 Hz\) and \(294 Hz\). How far (in 𝑐𝑚) from the end of the string must the finger be placed to play these notes? Report the smallest of all the calculated lengths.

Answer 1

Given, length of the guitar wire,
\(l=30~cm=0.3~cm\)
As the fundamental frequency in stretched
string is given by,
\(f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}\)
\(\Rightarrow fl=\text{constant}\)
\((i)\) \(f_{1}l_{1}=f_{2}l_{2}\Rightarrow 196\times30=220l_{2}\)
\(\Rightarrow l_{2}=26.7~cm\)
\((ii)\) \(196\times30=247l_{2}\Rightarrow l_{2}=23.8cm\)
\((iii)\) \(196\times30=262l_{3}\Rightarrow 22.4~cm\)
\((iv)\) \(196\times30=294l_{4}\Rightarrow l_{4}=20~cm\)
Therefore, fingers must be placed \(20 cm\) far from the end of the string to play these notes.
Final answer: 20