Question

Question: Two identical tubes, each closed one end, have a fundamental frequency of 349 Hz at \({\bf{25}}.{\bf{0^\circ C}}\). The air temperature is increased to \({\bf{31}}.{\bf{0^\circ C}}\) in one tube. If the two pipes are now sounded together, what beat frequency results?

Short answer

The difference in frequency is \(4\;{\rm{Hz}}\).

Step-by-step solution

Determination of difference in frequency

The frequency of a sound wave is the ratio of its speed and wavelength (which is four times the length of pipe). The ratio is used to find the difference in frequency for different temperatures.

Given information

Given data:

The fundamental frequency is at \(25^\circ {\rm{C}}\) is \({f_{25}} = 349\;Hz\).

The air temperature is increased by \(\Delta T = 31^\circ {\rm{C}}\).

Find the ratio of frequencies at different temperature

The fundamental frequency of a tube is \({f_1} = \frac{v}{{4l}}\). The ratio of frequencies at different temperature is:

\(\begin{array}{c}\frac{{{f_{31}}}}{{{f_{25}}}} = \frac{{\left( {\frac{{{v_{31}}}}{{4l}}} \right)}}{{\left( {\frac{{{v_{25}}}}{{4l}}} \right)}}\\ = \frac{{{v_{31}}}}{{{v_{25}}}}\end{array}\)

Find the difference of frequency

The difference in frequency can be calculated as:

\(\begin{array}{c}\Delta f = {f_{31}} - {f_{25}}\\ = {f_{25}}\left( {\frac{{{f_{31}}}}{{{f_{25}}}} - 1} \right)\\ = \left( {349\;{\rm{Hz}}} \right)\left( {\frac{{{v_{31}}}}{{{v_{25}}}} - 1} \right)\\ = \left( {349\;{\rm{Hz}}} \right)\left( {\frac{{331 + 0.6\left( {31} \right)}}{{331 + 0.6\left( {25} \right)}} - 1} \right)\\\; \approx 4\;{\rm{Hz}}\end{array}\)

Thus, the difference in frequency is \(4\;{\rm{Hz}}\).