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

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

Given data:

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

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

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 \(87.97\;{\rm{dB}}\).