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80GP

Expert-verifiedFound in: Page 473

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**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?**

** **

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

**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 data:

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

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

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

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

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 \(4\;{\rm{Hz}}\).

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