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Physics Principles with Applications
Found in: Page 473
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: 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.

Step 2:Given information                                 

Given data:

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

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

Step 3: 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:

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

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

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