Frequency of an open air column Given only the length of an organ pipe to be $2.14 m$, is it possible to find what frequency it vibrates at? If I use the equation $f=\frac{v}{\lambda}$, does the $v$ apply to the speed of sound in the organ pipe or in air?
 A: The speed of sound should apply to $v$ because the sound waves are travelling through the air after it leaves the organ pipe. 
The speed of sound is approximated by the following formula:
$$
v = 331.3 + 0.606T
$$
Where $T$ is the temperature in degrees Celsius, and $v$ is the velocity in meters per second. In your case, suppose you're at room temperature (~25 degrees Celsius), then the speed of sound would be:
$$
\begin{align}
v &= 331.3 + 0.606(25)
\\&=346.45m/s
\end{align}
$$
Now, to solve for the frequency:
$$
\begin{align}
f &= \frac{v}{\lambda}\\\\
&=\frac{345.45ms^{-1}}{4.24m}\\\\
&=81.71s^{-1}\\\\
&=8.17\times10^1 \ Hertz
\end{align}
$$
A: $v$ applies to the speed of sound in the equation $f=\frac{v}{\lambda}$. Assuming air to be an ideal gas we can use the following equation to calculate the speed of sound in air:
\begin{equation}
v=331.3\sqrt{1+\frac{T}{273.15}}
\end{equation}
where $T$ is the air temperature in degrees Celsius.
The wavelength should be twice the length of the organ pipe, thus, the frequency is:
\begin{equation}
f=\frac{v}{\lambda}=\frac{331.3\sqrt{1+\frac{T}{273.15}}}{2L} 
\end{equation}
Assuming that the air temperature is about standard room temperature (~20°C), the frequency is equal to
\begin{equation}
f=\frac{331.3\sqrt{1+\frac{20}{273.15}}}{2*2.14m}=80.190332Hz \approx 80Hz
\end{equation}
