Question

Two identical sound waves $S_1$ and $S_2$ reach a point $P$ in phase. The resultant loudness at point is $\sqrt{n}\text{dB}$ higher than the loudness of $S_1$. Then, the value of $n$ is
[Take ${\log }_{10}2=0.3$]

• A. $9$
• B. $6$
• C. $36$
• D. $3$

Answer 1

The correct option is C $36$

Let $A$ be the amplitude due to $S_1$ and $S_2$ individually.
Since intensity $I\propto A^2$
So, intensity due to $S_1,I_1\propto A^2$
Since, waves are in phase, they undergo constructive interference and resultant amplitude $=2A$
Thus, Intensity due to $S_1+S_2,I_2\propto (2A{)}^2$

So, $\frac{I_1}{I_2}=\frac{A^2}{(2A{)}^2}=\frac{1}{4}$
$\Rightarrow I_2=4I_1$

Ratio of loudness of resultant wave to loudness of $S_1$
$\beta =10{\log }_{10}\left(\frac{I_2}{I_1}\right)$
$=10{\log }_{10}\left(\frac{4I_1}{I_1}\right)$
$=10{\log }_{10}\left(4\right)=10\times 2{\log }_{10}\left(2\right)=6\text{dB}$
So, $\beta =\sqrt{36}\text{dB}$
i.e $n=36$