The frequency of the note F4 is 349 Hz. (a) If an organ pipe is open at one end and closed at the other, what length must it have for its fundamental mode to produce this note at 20.0°C? (b) At what air temperature will the frequency be 370 Hz, corresponding to a rise in pitch from F to F-sharp? (Ignore the change in length of the pipe due to the temperature change.)

The length of the pipe is given as $\mathit{L}\mathbf{=}\mathit{v}\mathbf{4}{\mathit{f}}_{\mathbf{1}}$were. ${\mathit{f}}_{\mathbf{1}}$is the fundamental frequency of an organ pipe, V is the speed of the sound, L is length of the pipe.

The new speed of the sound for the original temperature is ${\mathit{v}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{4}\mathit{f}\mathit{L}$were, v' is the new speed of the sound for the original temperature, F is the frequency.

The fundamental frequency of an organ pipe is 349 Hz and the speed of the sound is 344 Substitute 349 Hz for $f_1$ and 344${\mathrm{ms}}^{-1}$for v to find the L

$\mathrm{L}=344\times 4\times 349=0.246\mathrm{m}$

Thus, the length of the pipe for the given fundamental note is 0.246 m

The frequency of fundamental note is 370 Hz, The new speed of the sound for the original temperature is $v^{\text{'}}=4fL$Substitute 370 Hz for f and 0.246 m for L to find the v'

${\mathrm{v}}^{\text{'}}=4\left(370\right)(0.246)=364.08{\mathrm{ms}}^{-1}$

Formula to calculate air temperature corresponding to the pitch is

Substitute 364.08 $\mathrm{T}={\mathrm{v}}^{\text{'}}-\mathrm{v}0.6{\mathrm{ms}}^{-1}$ for v' and 344 ${\mathrm{ms}}^{-1}$for v to find T

role="math" localid="1664346666978" $$\begin{gathered} \mathrm{T}=364.08-344 \\ =35.5^{0}C \end{gathered}$$

Therefore, the air temperature corresponding the pitch is$35.5^{0}C$