Question

The loudness level at a distance R from a long linear source of sound found to be $40{\rm{ dB}}$. At this point, the amplitude of oscillations of air molecules is $0.01{\rm{ cm}}$. Then find the loudness level (in dB) at a point located at a distance $'10R'$ from the source?

Answer 1

We will be using the expression for the loudness of sound, which gives the relation between the intensity of the sound of the given linear source and intensity of reference sound. The intensity of sound provides us with the relation between the intensity of reference sound and distance of measurement point from the source.

Complete step by step answer:
Given:
The initial loudness of sound is ${L_1} = 40{\rm{ dB}}$.
The initial distance between the source of sound and point at which loudness is measured is ${r_1} = R$.
The final distance between the source of sound and point at which loudness is measured is ${r_2} = 10R$.
The amplitude of oscillations of air molecules is $A = 0.01{\rm{ cm}}$.
We have to find the value of loudness level at a point located at a distance ${r_2}$ from the source.
Let us write the expression for the loudness of sound when it is measured at a point whose distance from the source is ${r_1}$.
${L_1} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)$……(1)
Here, I is the intensity of the sound of the linear source and ${I_0}$ is the intensity of reference sound.
We know that the relation between the initial intensity of the linear sound source and that of reference sound can be expressed as:
${I_1} = \dfrac{{{I_0}}}{{r_1^2}}$
On substituting $\dfrac{{{I_0}}}{{r_1^2}}$ for ${I_1}$ in equation (1), we can write:

{L_1} = 10{\log _{10}}\left( {\dfrac{{\dfrac{{{I_0}}}{{r_1^2}}}}{{{I_0}}}} \right)\\\ = 20{\log _{10}}\left( {\dfrac{1}{{{r_1}}}} \right) \end{array}$$……(2) The expression for the loudness of sound when it is measured at a point whose distance from the source is $${r_2}$$. $${L_2} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)$$ We know that the relation between the final intensity of the linear sound source and that of reference sound can be expressed as: $${I_2} = \dfrac{{{I_0}}}{{r_1^2}}$$ On substituting $$\dfrac{{{I_0}}}{{r_2^2}}$$ for $${I_2}$$ in equation (1), we can write: $$\begin{array}{l} {L_2} = 10{\log _{10}}\left( {\dfrac{{\dfrac{{{I_0}}}{{r_2^2}}}}{{{I_0}}}} \right)\\\ = 20{\log _{10}}\left( {\dfrac{1}{{{r_2}}}} \right) \end{array}$$……(3) On subtracting equation (2) and equation (3), we get: $$\begin{array}{l} {L_2} - {L_1} = 20{\log _{10}}\left( {\dfrac{1}{{{r_2}}}} \right) - 20{\log _{10}}\left( {\dfrac{1}{{{r_1}}}} \right)\\\ = 20{\log _{10}}\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right) \end{array}$$ Substitute $$40{\rm{ dB}}$$ for $${L_1}$$, R for $${r_1}$$ and $$10R$$ for $${r_2}$$ in the above expression. $$\begin{array}{l} {L_2} - 40{\rm{ dB}} = 20{\log _{10}}\left( {\dfrac{R}{{10R}}} \right)\\\ {L_2} - 40{\rm{ dB}} = 20\left( { - 1} \right) \end{array}$$ As the expression of loudness gives the value of loudness in dB so we can write: $$\begin{array}{l} {L_2} - 40{\rm{ dB}} = - 20{\rm{ dB}}\\\ {L_2} = 20{\rm{ dB}} \end{array}$$ Therefore, the loudness level at a point located at a distance $$'10R'$$ from the source is $$20{\rm{ dB}}$$. **Note:** While calculating the final value of loudness level at a point located at a distance $${r_2}$$ from the source, do not confuse with the unit of the term on the right-hand side because loudness is measured in dB, hence take it accordingly.