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17.3-26PE

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College Physics (Urone)
Found in: Page 629

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Short Answer

An 8-hour exposure to a sound intensity level of \(90.0\;{\rm{dB}}\) may cause hearing damage. What energy in joules falls on a \(0.800\;{\rm{cm}}\) diameter eardrum so exposed?

The energy received by the ear is \(1.45 \times {10^{ - 3}}\;{\rm{J}}\).

See the step by step solution

Step by Step Solution

Given Data

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

The diameter of the ear is \(0.800\;{\rm{cm}}\).

The exposure was for \(8\;{\rm{hours}}\).

Definition of sound intensity

The power carried by sound waves per unit area perpendicularly to that region is defined as sound intensity, also known as acoustic intensity. The watt per square meter is the SI intensity measure, including sound intensity.

Calculation of the energy

The intensity of the sound is,

\(d = 10\log \frac{I}{{{I_0}}}\)

Now, for\(90\;{\rm{dB}}\),

\(\begin{array}{c}90 = 10\log \frac{{{I_0}}}{{{{10}^{ - 12}}}}\\9 = \log \frac{{{I_0}}}{{{{10}^{ - 12}}}}\\I = {10^9} \times {10^{ - 12}}\\I = {10^{ - 3}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{array}\)

The energy input by the ear in every second is,

\(\begin{array}{c}Q = {I_0}A\\Q = {10^{ - 3}} \times \pi \times {\left( {\frac{{0.8}}{2} \times {{10}^{ - 2}}} \right)^2}\\Q = 5.02 \times {10^{ - 8}}\;{\rm{W}}\end{array}\)

The total energy in that time is,

\(\begin{array}{c}Qt = 5.02 \times {10^{ - 8}} \times 8 \times 3600\\ = 1.45 \times {10^{ - 3}}\;{\rm{J}}\end{array}\)

Thus, the energy is \(1.45 \times {10^{ - 3}}\;{\rm{J}}\).

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