The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound intensity level goes up by\(40.0\;{\rm{dB}}\)?

The intensity of the sound is \(d = 40.0\;{\rm{dB}}\) more than the reference level.

The watt per square meter is the SI measure of intensity which depends on the wave's amplitude. The loudness is the effect of sound intensity in the human ear when it exceeds the threshold.

Use the intensity of the sound,

\(\begin{array}{c}{d_1} = 10\log \frac{{{I_1}}}{{{{10}^{ - 12}}}}\\{10^{\frac{{{d_1}}}{{10}}}} \times {10^{ - 12}} = {I_1}\end{array}\)

Now

\(\begin{array}{c}{d_1} + 40.0 = 10\log \frac{{{I_2}}}{{{{10}^{ - 12}}}}\\{I_2} = {10^{\frac{{{d_1} + 40.0}}{{10}}}} \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{array}\)

The ratio of the intensities is,

\(\begin{array}{c}\frac{{{I_2}}}{{{I_1}}} = \frac{{{{10}^{\frac{{{d_1} + 40.0}}{{10}}}} \times {{10}^{ - 12}}}}{{{{10}^{\frac{{{d_1}}}{{10}}}} \times {{10}^{ - 12}}}}\\\frac{{{I_2}}}{{{I_1}}} = {10^4}\end{array}\)

Therefore, the factor is \({10^4}\) when the amplitude of a sound wave increase if the sound intensity level goes up by\(40.0\;{\rm{dB}}\)