Question

What is the intensity of 60dB sound? Answer in terms of $${10}^{-6}{}^W/{}_{m^2}$$.

Answer 1

Hint : First of all find the ratio of the sound intensity to the threshold intensity. Where the threshold level of sound has the intensity of ${10}^{-12}{}^W/{}_m$. Then take the logarithm of the ratio. After that, multiply it by 10. Then divide the decibel value by 10. And find the exponent of that ratio. Using that power calculate the intensity in Watt per square meter.

Complete step-by-step solution:
The sound in decibel is related to the intensity as,
$\beta \left(dB\right)=10{\log }_{10}\left({\displaystyle \frac{I}{I_0}}\right)$
Where, $I_0$ is the threshold intensity and has the value $I_0={10}^{-12}{}^W/{}_{m^2}$
$\beta \left(dB\right)=10{\log }_{10}\left({\displaystyle \frac{I}{I_0}}\right)$
Where $I_0={10}^{-12}{}^W/{}_{m^2}$
We have the loudness of sound,
$L=\log \left({\displaystyle \frac{I}{I_0}}\right)$
First of all find the ratio of the sound intensity to the threshold intensity.
${\displaystyle \frac{I}{I_0}}={10}^L$
Given that,
L=60dB
$I_0={10}^{-12}{}^W/{}_{m^2}$
For the sound intensity level 60 decibel the corresponding intensity is $1\times {10}^{-6}$. Then take the logarithm of the ratio.
${\displaystyle \frac{I}{I_0}}={10}^6$
Thus the intensity becomes,
$I={10}^{-12}\times {10}^6$
$I={10}^{-6}{}^W/{}_{m^2}$
Thus the intensity of 60dB sound in terms of $${10}^{-6}{}^W/{}_{m^2}$$ is 1.

Note: We can define the intensity of sound as the power per unit area carried by a wave. Since the power is the rate at which the energy is transferred by the wave. Sound intensity levels are usually denoted by decibels than the sound intensities in watt per meter squared. The decibel is a unit less quantity as the ratio intensities. The threshold level of sound has the intensity of ${10}^{-12}{}^W/{}_m$. Although the technical units for sound intensity is watts per meter square the commonly used unit is decibel.