A 55 dB sound wave strikes an eardrum whose area is \(5.0 \times {10^{ - 5}}\;{{\rm{m}}^2}\). (a) How much energy is received by the eardrum per second? (b) At this rate, how long would it take your eardrum to receive a total energy of 1.0 J?

In this problem, the power of the sound wave is calculated by using the product of sound intensity and the area covered by sound.

The level of sound wave is \(\beta = 55\;{\rm{dB}}\).

The surface area of eardrum is \(A = 5.0 \times {10^{ - 5}}\;{{\rm{m}}^2}\).

The total energy received by eardrum is \(E = 1.0\;{\rm{J}}\).

The standard value of threshold intensity is \({I_0} = 1.0 \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}\).

The intensity of sound is calculated below:

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

Substitute the values in the above equation.

\(\begin{aligned}{c}I = {10^{\left( {\frac{{55\;{\rm{dB}}}}{{10}}} \right)}} \times \left( {1.0 \times {{10}^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}} \right)\\I = 3.16 \times {10^{ - 7}}\;{\rm{W/}}{{\rm{m}}^2}\end{aligned}\)

The energy received by the eardrum per second is calculated below:

\(P = IA\)

Substitute the values in the above equation.

\(\begin{aligned}{c}P = \left( {3.16 \times {{10}^{ - 7}}\;{\rm{W/}}{{\rm{m}}^2}} \right)\left( {5.0 \times {{10}^{ - 5}}\;{{\rm{m}}^2}} \right)\\P = 1.58 \times {10^{ - 11}}\;{\rm{J/s}}\end{aligned}\)

Hence, the energy received by the eardrum per second is \(1.58 \times {10^{ - 11}}\;{\rm{J/s}}\).

The time taken by the eardrum to take complete energy is calculated below:

\(t = \frac{E}{P}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}t = \frac{{1.0\;{\rm{J}}}}{{1.58 \times {{10}^{ - 11}}\;{\rm{J/s}}}}\\t = 6.32 \times {10^{10}}\;{\rm{s}} \times \left( {\frac{{1\;{\rm{yr}}}}{{3.154 \times {{10}^7}\;{\rm{s}}}}} \right)\\t = 2005.25\;{\rm{yr}}\end{aligned}\)

Hence, the time taken by the eardrum to take complete energy is \(2005.25\;{\rm{yr}}\).