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Q. 19

Expert-verifiedFound in: Page 483

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

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?

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

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

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

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" ${f}_{cool}=\frac{v}{4L}\phantom{\rule{0ex}{0ex}}=\frac{343}{4\times 1.2}\phantom{\rule{0ex}{0ex}}=71.46Hz$

Take the ratio ${f}_{hot}$ and ${f}_{cool}$.

$\frac{{f}_{hot}}{{f}_{cool}}=\frac{\raisebox{1ex}{${v}_{hot}$}\!\left/ \!\raisebox{-1ex}{$4L$}\right.}{\raisebox{1ex}{${v}_{cool}$}\!\left/ \!\raisebox{-1ex}{$4L$}\right.}\phantom{\rule{0ex}{0ex}}=\frac{352}{343}\phantom{\rule{0ex}{0ex}}{f}_{hot}=\frac{352}{343}\times 71.46\phantom{\rule{0ex}{0ex}}=73.3Hz$

Thus, the warmer air temperature increases the fundamental frequency.

role="math" localid="1650031592581" ${f}_{hot}-{f}_{cool}=73.33-71.46\phantom{\rule{0ex}{0ex}}\approx 1.9Hz$

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

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