Question

A small sound source has an intensity of $20{\text{dB}}$ at a distance of ${\text{1m}}$. Upto what distance from the source is the sound audible? Neglect attenuation.

Answer 1

Here we observe that the intensity is given in decibel so, recall the formula to convert intensity into decibel. Use this formula to write an expression to have intensity in decibel in terms of intensity of sound. Recall the dependency of intensity on distance from the source, use this concept to get the required answer.

Complete step by step answer:
Given, intensity of the source in decibel scale is ${L_1} = 20{\text{dB}}$ at a distance ${r_1} = {\text{1m}}$.Intensity of a source in decibel scale is also known as loudness. We have the formula for loudness of a source as,
$L = 10{\log _{10}}\left( {\dfrac{I}{{{I_o}}}} \right)$ (i)
where $I$ is the sound intensity and ${I_o}$ is the reference intensity.
Let ${I_1}$ be the sound intensity at the distance ${r_1} = {\text{1m}}$
Now, for loudness ${L_1} = 20{\text{dB}}$ we use formula from equation (i) and we get
$20 = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_o}}}} \right)$ (ii)
We are asked upto what distance from the source is the sound audible. So let us assume from a distance ${r_2}$from the source the sound is not audible and let ${I_2}$ be the sound intensity at that point. From a distance ${r_2}$ sound is not audible, that is loudness is zero so, using the formula from equation (i) we can form an equation.
$0 = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_o}}}} \right)$ (iii)
We subtract equation (iii) from (ii) and we get
$20 - 0 = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_o}}}} \right) - 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_o}}}} \right)$
$\Rightarrow 20 = 10\left( {{{\log }_{10}}\left( {\dfrac{{{I_1}}}{{{I_o}}}} \right) - {{\log }_{10}}\left( {\dfrac{{{I_2}}}{{{I_o}}}} \right)} \right)$
$\Rightarrow 20 = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_2}}}} \right)$

We know intensity from a source is inversely proportional to the square of the distance from the source. Using this fact here we can write
${I_1} \propto \dfrac{1}{{{r_1}^2}}$ and ${I_2} \propto \dfrac{1}{{{r_2}^2}}$
$\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{r_2}^2}}{{{r_1}^2}}$ (iv)
Using equation (iv) in (iii), we get

\Rightarrow 20 = 10{\log _{10}}{\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right)^2} \\\ \Rightarrow 20 = 10 \times 2 \times {\log _{10}}\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right) $$ $$\Rightarrow 20 = 20{\log _{10}}\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right) \\\ \Rightarrow {\log _{10}}\left( {\dfrac{{{r_2}}}{{{r_1}}}} \right) = 1 $$ $$ \Rightarrow \left( {\dfrac{{{r_2}}}{{{r_1}}}} \right) = 10$$ Now, putting the value of $${r_1}$$, we get $$\dfrac{{{r_2}}}{1} = 10 \\\ \therefore {r_2} = 10{\text{m}} \\$$ **Therefore, the distance upto which the sound from the source is audible is $$10{\text{m}}$$.** **Note:** Loudness of sound tells us about the energy of the pressure waves created. It also relates the intensity of the sound and the intensity with which a person hears. Many times students get confused between pitch and loudness, pitch is related to the frequency of the sound whereas loudness is related to the energy of the sound wave.