By what fraction will the frequencies produced by a wind instrument change when air temperature goes from \({\rm{10}}{\rm{.0^\circ C}}\) to \({\rm{30}}{\rm{.0^\circ C}}\)? That is, find the ratio of the frequencies at those temperatures.

The initial temperature is \({{\rm{t}}_{\rm{1}}}{\rm{ = 10}}{\rm{.0^\circ C = }}\left( {{\rm{10 + 273}}} \right)\;{\rm{K = 283}}\;{\rm{K}}\)

The final temperature is \({{\rm{t}}_{\rm{2}}}{\rm{ = 30}}{\rm{.0^\circ C = }}\left( {{\rm{30 + 273}}} \right)\;{\rm{K = 303}}\;{\rm{K}}\)

For a closed tube, the resonance frequencies are,

\({{\rm{f}}_{\rm{n}}}{\rm{ = n}}\frac{{\rm{v}}}{{{\rm{2l}}}}{\rm{,}}\;{\rm{n = 1,3,}}\;{\rm{5,}}\;...\)

For an open tube, the resonance frequencies are,

\({{\rm{f}}_{\rm{n}}}{\rm{ = n}}\frac{{\rm{v}}}{{{\rm{2l}}}}{\rm{,}}\;{\rm{n = 1,2,}}\;{\rm{3,}}\;....\)

The speed of sound at t₁temperature is

\({\rm{v = 331}}\sqrt {\frac{{{{\rm{t}}_{\rm{1}}}}}{{{\rm{273}}}}} \)

The speed of sound at t₂ temperature is

\({\rm{v ' = 331}}\sqrt {\frac{{{{\rm{t}}_{\rm{2}}}}}{{{\rm{273}}}}} \)

The frequencies are,

\(\begin{aligned}\frac{{\rm{f}}}{{{\rm{f '}}}}{\rm{ = }}\frac{{\rm{v}}}{{{\rm{v'}}}}\\{\rm{ = }}\sqrt {\frac{{{{\rm{t}}_{\rm{1}}}}}{{{{\rm{t}}_{\rm{2}}}}}} \\{\rm{ = }}\sqrt {\frac{{{\rm{303}}}}{{{\rm{283}}}}} \\{\rm{ = 1}}{\rm{.03}}{\rm{.}}\end{aligned}\)

Hence, the ratio of the frequencies is 1.03.