Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the power input needed to produce a \(90.0\;{\rm{dB}}\) sound intensity level for a \({\rm{12}}{\rm{.0}}\;{\rm{cm}}\)diameter speaker that has an efficiency of \(1.00\% \). (This value is the sound intensity level right at the speaker.)

The input power is\(1.767 \times {10^{ - 3}}\;{\rm{W}}\).

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Intensity of Sound

The intensity of sound depends on the amplitude, and the loudness is the effect of intensity on human ears.

Given Data

The sound intensity is\(90.0\;{\rm{dB}}\).

The diameter of the speaker is\(12.0\;{\rm{cm}}\).

The efficiency is\(1.00\% \).

Calculation of the power of the sound

The intensity of the unaided hearing is,

\({{\rm{d}}_{\rm{1}}}{\rm{ = 10log}}\frac{{{{\rm{I}}_{\rm{0}}}}}{{{\rm{1}}{{\rm{0}}^{{\rm{12}}}}}}\)and\({I_0} = {10^{12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

Now,

\(\begin{array}{c}90 = 10\log \frac{{{I_0}}}{{{{10}^{ - 12}}}}\\9 = \log \frac{{{I_0}}}{{{{10}^{ - 12}}}}\\{I_0} = {10^9} \times {10^{ - 12}}\\{I_0} = {10^{ - 3}}\end{array}\)

The power output is,

\(\begin{array}{c}P = {I_0}A\\P = {10^{ - 3}} \times \pi \times {\left[ {\frac{{0.15}}{2}} \right]^2}\\P = 1.767 \times {10^{ - 5}}\;{\rm{W}}\end{array}\)

The input power is,

\(\begin{array}{c}P' = \frac{P}{{efficiency}}\\ = \frac{{1.767 \times {{10}^{ - 5}}}}{{0.01}}\\ = 1.767 \times {10^{ - 3}}\;{\rm{W}}\end{array}\)

The input power is\(1.767 \times {10^{ - 3}}\;{\rm{W}}\).

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Intensity of Sound

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Calculation of the power of the sound

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College Physics (Urone)

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Intensity of Sound

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Calculation of the power of the sound

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