Question

A bass clarinet can be modeled as a 120-cm-long open-closed tube. A bass clarinet player starts playing in a 20°C room, but soon the air inside the clarinet warms to where the speed of sound is 352 m/s. Does the fundamental frequency increase or decrease? By how much?

Short answer

The warmer air temperature increases the fundamental frequency by$1.9Hz$.

Step-by-step solution

Given information

The length of the open-closed tube is $L=120cm=1.20m$

The speed of sound is$v=352m/s$.

The fundamental frequency at v=343 m/s

Assume that the strings of the violin have the same tension and same length and therefore the same fundamental wavelength.

The fundamental frequency at the cool room temperature where the speed of sound is $343m/s$.

role="math" localid="1650031107837" $$\begin{gathered} f_{cool}=\frac{v}{4L} \\ =\frac{343}{4\times 1.2} \\ =71.46Hz \end{gathered}$$

Increase in the fundamental frequency

Take the ratio $f_{hot}$and $f_{cool}$.

$$\begin{gathered} \frac{f_{hot}}{f_{cool}}=\frac{{\displaystyle \raisebox{1ex}{$v_{hot}$}\!\left/ \!\raisebox{-1ex}{$4L$}\right.}}{{\displaystyle \raisebox{1ex}{$v_{cool}$}\!\left/ \!\raisebox{-1ex}{$4L$}\right.}} \\ =\frac{352}{343} \\ {f}_{hot}=\frac{352}{343}\times 71.46 \\ =73.3Hz \end{gathered}$$

Thus, the warmer air temperature increases the fundamental frequency.

role="math" localid="1650031592581" $$\begin{gathered} f_{hot}-f_{cool}=73.33-71.46 \\ \approx 1.9Hz \end{gathered}$$

Therefore, the warmer air temperature increases the fundamental frequency by$1.9Hz$.