Question

What is the sound level of a sound whose intensity is\(1.5 \times {10^{ - 6}}\;{\rm{W/}}{{\rm{m}}^2}\)?

Short answer

The level of a sound is \(61.76\;{\rm{dB}}\).

Step-by-step solution

Understanding the concept of sound level

The sound level is determined by the relation of intensity and threshold intensity of sound.

Given data

The intensity of the sound is \(I = 1.5 \times {10^{ - 6}}\;{\rm{W/}}{{\rm{m}}^2}\).

Evaluating the sound level by using a relation among intensity and threshold intensity

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

The sound level is calculated below:

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

Substitute the values in the above equation.

\(\begin{aligned}{c}\beta = 10\log \left( {\frac{{1.5 \times {{10}^{ - 6}}\;{\rm{W/}}{{\rm{m}}^2}}}{{1 \times {{10}^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}}}} \right)\\\beta = 61.76\;{\rm{dB}}\end{aligned}\)

Hence, the sound level is \(61.76\;{\rm{dB}}\).