Question

A person standing a certain distance from an airplane with four equally noisy jet engines is experiencing a sound level of 140 dB. What sound level would this person experience if the captain shut down all but one engine? [Hint: Add intensities, not dBs.]

Short answer

The sound level produced due to one engine is \(133.98\;{\rm{dB}}\).

Step-by-step solution

Understanding the intensity of combined sound

The resultant intensity of combined sound can be evaluated directly by adding intensities of individual sound intensity.

Given data

The combined sound level produced by all four jet engines is \(\beta = 140\;{\rm{dB}}\).

Evaluating the intensity of sound produced by four engines

Let \({I_2}\) be the intensity of sound produced by each engine, then the total intensity of sound produced by all four jet engines is \({I_1} = 4{I_2}\).

The standard value for the threshold intensity is \({I_0} = 1 \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}\).

The intensity of sound due to four engines is calculated below:

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

Substitute the values in the above equation.

\(\begin{aligned}{c}\log \left( {\frac{{{I_1}}}{{1 \times {{10}^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}}}} \right) = \frac{{140\;{\rm{dB}}}}{{10}}\\{I_1} = \left( {1 \times {{10}^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}} \right)\left( {{{10}^{14}}} \right)\\{I_1} = 100\;{\rm{W/}}{{\rm{m}}^2}\end{aligned}\)

The intensity of sound due to one engine is calculated below:

\({I_2} = \frac{{{I_1}}}{4}\)

Substitute the values in the above equation.

\(\begin{aligned}{c} = \frac{{100\;{\rm{W/}}{{\rm{m}}^2}}}{4}\\ = 25\;{\rm{W/}}{{\rm{m}}^2}\end{aligned}\)

Evaluating the sound level produced by one individual engine

The sound level produced due to one engine is calculated below:

\(\beta ' = 10\log \left( {\frac{{{I_2}}}{{{I_0}}}} \right)\)

Substitute the values in the above equation.

\(\begin{aligned}{c}\beta ' = 10\log \left( {\frac{{25\;{\rm{W/}}{{\rm{m}}^2}}}{{1 \times {{10}^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}}}} \right)\\\beta ' = 133.98\;{\rm{dB}}\end{aligned}\)

Hence, the sound level produced due to one engine is \(133.98\;{\rm{dB}}\).