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Physics Principles with Applications
Found in: Page 328
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

What is the resultant sound level when an 81 dB sound and an 87 dB sound are heard simultaneously?

The resultant sound level when both sounds heard simultaneously is \(87.97\;{\rm{dB}}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Determination of sound intensity by sound level

The intensity of the sound is determined by sound level by using its relation with threshold intensity in logarithmic form.

Given information

Given data:

The sound level for sound 1 is \({\beta _1} = 81\;{\rm{dB}}\).

The sound level for sound 2 is \({\beta _2} = 87\;{\rm{dB}}\).

Evaluation of the resultant sound level

The intensity of sound 1 is calculated below:

\(\begin{aligned}{c}{\beta _1} &= 10\log \left( {\frac{{{I_1}}}{{{I_0}}}} \right)\\{I_1} &= {I_0}{10^{\left( {\frac{{{\beta _1}}}{{10}}} \right)}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{I_1} &= {I_0}{10^{\left( {\frac{{81\;{\rm{dB}}}}{{10}}} \right)}}\\ &= 1.259 \times {10^8}{I_0}\end{aligned}\)

The intensity of sound 2 is calculated below:

\(\begin{aligned}{c}{\beta _2} &= 10\log \left( {\frac{{{I_2}}}{{{I_0}}}} \right)\\{I_2} &= {I_0}{10^{\left( {\frac{{{\beta _2}}}{{10}}} \right)}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{I_2} &= {I_0}{10^{\left( {\frac{{87\;{\rm{dB}}}}{{10}}} \right)}}\\ &= 5.012 \times {10^8}{I_0}\end{aligned}\)

The total intensity of both sounds is calculated below:

\(I = {I_1} + {I_2}\)

Substitute the values in the above equation.

\(\begin{aligned}{l}I &= \left( {1.259 \times {{10}^8}{I_0}} \right) + \left( {5.012 \times {{10}^8}{I_0}} \right)\\I &= 6.271 \times {10^8}{I_0}\end{aligned}\)

The resultant sound level when both sounds heard simultaneously is calculated below:

\(\beta = 10\log \left( {\frac{I}{{{I_0}}}} \right)\)

Substitute the values in the above equation.

\(\begin{aligned}{l} &= 10\log \left( {\frac{{6.271 \times {{10}^8}{I_0}}}{{{I_0}}}} \right)\\ &= 87.97\;{\rm{dB}}\end{aligned}\)

Hence, the resultant sound level when both sounds heard simultaneously is \(87.97\;{\rm{dB}}\).

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