What is the length of a tube that has a fundamental frequency of176 Hzand a first overtone of352 Hzif the speed of sound is 343 m/s?

The fundamental frequency is \({{\rm{f}}_{\rm{1}}}{\rm{ = 176}}\;{\rm{Hz}}\).

The first overtone is \({{\rm{f}}_{\rm{2}}}{\rm{ = 352}}\;{\rm{Hz}}\)

The speed of the sound is \({\rm{v = 343}}\;{\rm{m/s}}\).

The resonance frequencies of a tube open at both the ends are,

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

For a closed tube, the resonance frequencies are,

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

The ratio of the first overtone to the fundamental is

\(\frac{{{\rm{352}}}}{{{\rm{176}}}}{\rm{ = 2}}{\rm{.}}\)

The ratio is even number, so the tube is open at both the ends.

We can write,

\(\begin{aligned}{{\rm{f}}_{\rm{1}}}{\rm{ = }}\frac{{\rm{v}}}{{{\rm{2l}}}}{\rm{ = 176}}\\\frac{{{\rm{343}}}}{{{\rm{2l}}}}{\rm{ = 176}}\\{\rm{l = 0}}{\rm{.9744}}\;{\rm{m}}\end{aligned}\)

Hence, the length of the tube is 0.974 m.