91Ó°ÊÓ

The string's frequency is proportional to the square root of the ratio of tensile force and mass of the string. Using this relation, the relative mass of the remaining string can be determined with respect to the first string.

The string of a violin is tuned to a frequency \(1\frac{1}{2}\) times

Let the frequencies of four strings in wire are \({f_A}\), \({f_B}\), \({f_C}\) and \({f_D}\)and among them \({f_A}\) is the lowest. The length of all the strings is same and have the same tension.

For the string \(B\),

\(\begin{array}{c}{f_B} = 1.5{f_A}\\\frac{1}{{2l}}\sqrt {\frac{{{F_T}}}{{{\mu _B}}}} = \left( {1.5} \right)\frac{1}{{2l}}\sqrt {\frac{{{F_T}}}{{{\mu _A}}}} \\{\mu _B} = \frac{{{\mu _A}}}{{{{\left( {1.5} \right)}^2}}}\\ = 0.44{\mu _A}\end{array}\)

For the string C,

\(\begin{array}{c}{f_C} = 1.5{f_B}\\{f_C} = {\left( {1.5} \right)^2}{f_B}\\\frac{1}{{2l}}\sqrt {\frac{{{F_T}}}{{{\mu _C}}}} = {\left( {1.5} \right)^2}\frac{1}{{2l}}\sqrt {\frac{{{F_T}}}{{{\mu _C}}}} \\{\mu _C} = \frac{{{\mu _A}}}{{{{\left( {1.5} \right)}^4}}}\\{\mu _C} = 0.20{\mu _A}\end{array}\)

For the string D,

\(\begin{array}{c}{f_D} = 1.5{f_C}\\{f_D} = {\left( {1.5} \right)^3}{F_A}\\\frac{1}{{2l}}\sqrt {\frac{{{F_T}}}{{{\mu _D}}}} = {\left( {1.5} \right)^3}\frac{1}{{2l}}\sqrt {\frac{{{F_T}}}{{{\mu _A}}}} \\{\mu _D} = \frac{{{\mu _A}}}{{{{\left( {1.5} \right)}^6}}}\\{\mu _D} = 0.088{\mu _A}\end{array}\)

Thus, the masses of the string relative to string A are \({\mu _B} = 0.44{\mu _A}\), \({\mu _C} = 0.20{\mu _A}\), and \({\mu _D} = 0.088{\mu _A}\).