Question

A point source emits sound waves of constant power with intensity inversely proportional to the square of the distance from the source. By how many decibels does the sound loudness drop, when you move from point $A$ to $B$ ? Distance of $B$ from the source is two times the distance of $A$ from the source. [Take ${\log }_{10}2=0.3$ ]

• A. $4\text{dB}$
• B. $2\text{dB}$
• C. $6\text{dB}$
• D. $8\text{dB}$

Answer 1

The correct option is C $6\text{dB}$

We know that for a point source,

Intensity $I\propto \frac{1}{r^2}$

From the data given in the question,

$\frac{I_B}{I_A}={\left(\frac{r_A}{r_B}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

$\Rightarrow I_A=4I_B$
From the relation between loudness and intensity of a sound wave,
$\beta =10\log \frac{I}{I_0}$

We can write that,

${\beta }_B-{\beta }_A=10\log \frac{I_B}{I_0}-10\log \frac{I_A}{I_0}$
$\Rightarrow {\beta }_B-{\beta }_A=10\log [\frac{I_B}{I_0}\times \frac{I_0}{I_A}]=10\log \left(\frac{I_B}{I_A}\right)$
$\Rightarrow {\beta }_B-{\beta }_A=10\log \left(\frac{1}{4}\right)$
$=10\times (-2{\log }_{10}2)=10\times -0.6=-6\text{dB}$

i.e loudness drops by $6\text{dB}$

Thus, option (c) is the correct answer.