 # You live on a busy street, but as a music lover, you want to reduce the traffic noise. • What would be the fractional impact on lowering the sound intensity (in W/m^2 if the level of sound intensity (in dB) is reduced by 40 dB by the installation of unique windows with sound-reflecting properties?
• What would be the change in the sound intensity level (in dB) if the intensity is reduced by half?

The aim of this question is to find the impact of sound intensity (in $\dfrac{W}{m^2}$) by reducing the sound intensity level (in $dB$). The basic concept behind this article is Sound Intensity and Sound Intensity Level.

Sound Intensity is defined as the energy or power that exists in a sound wave per unit area. It is a vector quantity whose direction is perpendicular to the surface area. As sound intensity is the power of sound waves, hence, it is represented by the SI unit of Watt per square meter $(\dfrac{W}{m^2})$ and expressed as follows:

$Sound\ Intensity\ I=pv$

Where:

$p$ is the sound pressure

$v$ is the particle velocity

Sound intensity level (SIL) is the ratio of the loudness of the given intensity of a sound to the standard intensity. It is represented by the SI unit of Decibels $(dB)$ and expressed as follows:

$Sound\ Intensity\ Level\ SIL\ (dB)=\ 10\log_{10}{\left(\frac{I}{I_0}\right)}$

Where:

$I$ is the sound intensity of a given sound

$I_0$ is the reference sound intensity

$I_0$ Reference sound intensity is generally defined as standard sound level measurement corresponding to hearing by a human ear having a standard threshold at $1000$ $Hz$

$I_0=\ {10}^{-12}\ \frac{W}{m^2}$

Given that:

$Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 40\ dB$

Part-1 Solution

We will substitute the value of given $SIL$ and Reference sound intensity $I_0$ in equation of $SIL$:

$Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{I}{I_0}\right)}$

$40\ dB\ =\ 10\log_{10}{\left(\frac{I}{{10}^{-12}}\right)}$

$\log_{10}{\left(\frac{I}{{10}^{-12}}\right)}\ =\ \frac{40}{10}\ =\ 4$

By applying log formula:

$\log_a{b=x}\ \Rightarrow\ a^x=b$

$\frac{I}{{10}^{-12}}\ =\ {10}^4$

$I\ =\ {10}^4\times{10}^{-12}$

$I\ =\ {10}^{-8}\ \frac{W}{m^2}$

Part-2 Solution

Given that:

Intensity $I$ is reduced by half.

$Intensity\ =\ \frac{1}{2}I$

We know that:

$Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{I}{I_0}\right)}$

Substituting the value of $I$ and $I_0$ in above equation:

$SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{I}{{2\ timesI}_0}\right)}$

$SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{{10}^{-8}}{2\times{10}^{-12}}\right)}$

$SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{{10}^4}{2}\right)}$

$SIL\ (dB)\ =\ 10\log_{10}{\left(5000\right)}$

$SIL\ (dB)\ =\ 36.989\ dB$

If the level of sound intensity (in $dB) is reduced by$40dB$, the sound intensity will be: $I\ =\ {10}^{-8}\ \frac{W}{m^2}$ If the intensity is reduced by half, the sound intensity level (in$dB$) will be: $SIL\ (dB)\ =\ 36.989\ dB$ ## Example What would be the fractional impact on lowering the sound intensity (in$\dfrac{W}{m^2}$) if the level of sound intensity (in$dB$) is reduced by$10dB$? Solution Given that: $Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 10\ dB$ We will substitute the value of given$SIL$value and Reference sound intensity$I_0$in equation of$SIL\$

$Sound\ Intensity\ Level\ SIL\ (dB)\ =\ 10\log_{10}{\left(\frac{I}{I_0}\right)}$

$40\ dB\ =\ 10\log_{10}{\left(\frac{I}{{10}^{-12}}\right)}$

$\log_{10}{\left(\frac{I}{{10}^{-12}}\right)}\ =\ \frac{10}{10}\ =\ 1$

By applying log formula:

$\log_a{b=x}\ \Rightarrow\ a^x=b$

$\frac{I}{{10}^{-12}}\ =\ 10$

$I\ =\ 10\times{10}^{-12}$

$I\ =\ {10}^{-11}\ \frac{W}{m^2}$