Question

A certain organ pipe, open at both ends, produces a fundamental frequency of 271 Hz in air.
If the pipe is filled with helium at the same temperature, what fundamental frequency f He will it produce? Take the molar mass of air to be 28.8 g/mol and the molar mass of helium to be 4.00g/mol .

Answer 1

Concepts and reason

The concepts required to solve this problem are the velocity of waves in a medium and the relation between the velocity and frequency.

Use the expression for velocity of the wave in a medium and the velocity and frequency relation to calculate the fundamental frequency in the Helium medium.

Fundamentals

Write the expression for the velocity of the wave in a medium.

$c = \sqrt {\frac{{\gamma RT}}{M}}$

Here, $R$ is the gas constant, $\gamma$ is the adiabatic index, $T$ is the absolute temperature and $M$ is the Molar mass.

Write the expression for the frequency of the wave:

$f = \frac{c}{\lambda }$

Here, $\lambda$ is the wavelength and $c$ is the speed of the wave.

The velocity of the wave in a medium is given by the relation as follows:

$c = \sqrt {\frac{{\gamma RT}}{M}}$

Here, $R$ is the gas constant, $\gamma$ is the adiabatic index, $T$ is the absolute temperature and $M$ is the Molar mass.

Write the expression for the frequency of the wave:

$f = \frac{c}{\lambda }$

Here, $\lambda$ is the wavelength and $c$ is the speed of the wave.

Therefore, by the expression of frequency and velocity it can be deduced that the frequency of the wave is given by the relation as follows:

$f = \frac{1}{\lambda }\sqrt {\frac{{\gamma RT}}{M}}$

The obtained relation of frequency of the wave is given by the relation as follows:

$f = \frac{1}{\lambda }\sqrt {\frac{{\gamma RT}}{M}}$

Here, $\lambda$ is the wavelength, $R$ is the gas constant, $\gamma$ is the adiabatic index, $T$ is the absolute temperature and $M$ is the molar mass.

For air, the equation for the frequency of the wave can be written as follows:

${f_{{\rm{air}}}} = \frac{1}{\lambda }\sqrt {\frac{{{\gamma _{{\rm{air}}}}RT}}{{{M_{{\rm{air}}}}}}}$

Here, ${\gamma _{{\rm{air}}}}$ is the adiabatic index of the air and ${M_{{\rm{air}}}}$ is the molar mass of the air.

For helium, the expression for the frequency in the helium medium can be written as follows:

${f_{{\rm{He}}}} = \frac{1}{\lambda }\sqrt {\frac{{{\gamma _{{\rm{He}}}}RT}}{{{M_{{\rm{He}}}}}}}$

Here, ${\gamma _{{\rm{He}}}}$ is the adiabatic index of Helium and ${M_{{\rm{He}}}}$ is the molar mass of Helium.

Take the ratio of the frequency in helium to the frequency in air.

$\frac{{{f_{{\rm{He}}}}}}{{{f_{{\rm{air}}}}}} = \frac{{\frac{1}{\lambda }\sqrt {\frac{{{\gamma _{{\rm{He}}}}RT}}{{{M_{{\rm{He}}}}}}} }}{{\frac{1}{\lambda }\sqrt {\frac{{{\gamma _{{\rm{air}}}}RT}}{{{M_{{\rm{air}}}}}}} }}$

Rearrange for ${f_{{\rm{He}}}}$ .

${f_{He}} = {f_{{\rm{air}}}}\sqrt {\frac{{{\gamma _{{\rm{He}}}}{M_{{\rm{air}}}}}}{{{\gamma _{{\rm{air}}}}{M_{{\rm{He}}}}}}}$

Substitute $271{\rm{ Hz}}$ for ${f_{{\rm{air}}}}$ , $1.4$ for ${\gamma _{{\rm{air}}}}$ , $28.8{\rm{ g/mol}}$ for ${M_{{\rm{air}}}}$ , $1.67$ for ${\gamma _{{\rm{He}}}}$ and $4{\rm{ g/mol}}$ for ${M_{{\rm{He}}}}$ .

$\begin{array}{c}\\{f_{{\rm{He}}}} = \left( {271{\rm{ Hz}}} \right)\sqrt {\frac{{\left( {1.67} \right)\left( {28.8{\rm{ g/mol}}} \right)}}{{\left( {1.4} \right)\left( {4{\rm{ g/mol}}} \right)}}} \\\\ = \left( {271{\rm{ Hz}}} \right)\sqrt {8.59} \\\\ = \left( {271{\rm{ Hz}}} \right)\left( {2.93} \right)\\\\ = 794.2{\rm{ Hz}}\\\end{array}$

Ans:

The fundamental frequency produced by the organ pipe in Helium medium is ${\rm{794}}{\rm{.2 Hz}}$ .