Question

Question: (II) What would be the sound level (in dB) of a sound wave in air that corresponds to a displacement amplitude of vibrating air molecules of 0.13 mm at 440 Hz?

Short answer

The sound level of a sound wave is \(134.6\;{\rm{dB}}\).

Step-by-step solution

Understanding the concept that include various terms that can change the sound intensity

The intensity of the sound wave is a function of the density, velocity of sound, frequency, and amplitude of a sound.

Given data

The displacement amplitude is \(A = 0.13\;{\rm{mm}}\).

The sound frequency is \(f = 440\;{\rm{Hz}}\).

The standard value for the density of air, and speed of sound in air is \(\rho = 1.3\;{\rm{kg/}}{{\rm{m}}^3}\) and \(v = 343\;{\rm{m/s}}\) respectively. The standard value for threshold or relative intensity is \({I_0} = 1.0 \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}\).

Evaluating the intensity of sound wave

The intensity of sound waves is calculated below:

\(I = 2\rho v{\pi ^2}{f^2}{A^2}\)

Substitute the values in the above equation.

\(\begin{array}{c}I = 2\left( {1.3\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {343\;{\rm{m/s}}} \right){\pi ^2}{\left( {440\;{\rm{Hz}}} \right)^2}{\left( {0.13\;{\rm{mm}} \times \frac{{{{10}^{ - 3}}\;{\rm{mm}}}}{{1\;{\rm{m}}}}} \right)^2}\\I = 28.8\;{\rm{W/}}{{\rm{m}}^2}\end{array}\)

Evaluating the sound level of sound wave

The sound level of sound wave is calculated below:

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

Substitute the values in the above equation.

\(\begin{array}{l}\beta = 10\log \left( {\frac{{28.8\;{\rm{W/}}{{\rm{m}}^2}}}{{1.0 \times {{10}^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}}}} \right)\\\beta = 134.6\;{\rm{dB}}\end{array}\)

Hence, the sound level of sound wave is \(134.6\;{\rm{dB}}\).