Question

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

Short answer

The ratio of the frequencies is 1.03.

Step-by-step solution

Given Data

The initial temperature is ${\mathrm{t}}_1=10{.0}^{\circ }\mathrm{C}=\left(10+273\right)\mathrm{K}=283\mathrm{K}$

The final temperature is ${\mathrm{t}}_2=30{.0}^{\circ }\mathrm{C}=\left(30+273\right)\mathrm{K}=303\mathrm{K}$

Calculation of the ratio of the frequencies  

For a closed tube, the resonance frequencies are,

${\mathrm{f}}_{\mathrm{n}}=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{l}},\mathrm{n}=1,3,5,...$

For an open tube, the resonance frequencies are,

${\mathrm{f}}_{\mathrm{n}}=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{l}},\mathrm{n}=1,2,3,....$

The speed of sound at t₁temperature is

$\mathrm{v}=331\sqrt{\frac{{\mathrm{t}}_1}{273}}$

The speed of sound at t₂ temperature is

${\mathrm{v}}^{\prime }=331\sqrt{\frac{{\mathrm{t}}_2}{273}}$

The frequencies are,

$\begin{array}{r}\frac{\mathrm{f}}{{\mathrm{f}}^{\prime }}=\frac{\mathrm{v}}{{\mathrm{v}}^{\prime }}\\ =\sqrt{\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}}\\ =\sqrt{\frac{303}{283}}\\ =1.03.\end{array}$

Hence, the ratio of the frequencies is 1.03.