Question

Two identical sounds \[{S_1}\] and ${S_2}$ reach at a point P in phase. The resultant loudness at point P is $n\,dB$ higher than the loudness of ${S_1}$ . The value of $n$ is
A. 2
B. 4
C. 5
D. 6

Answer 1

Hint: Start with finding the relation between loudness of two sound waves and the resultant loudness of the two sound waves that are reaching at a point P in phase then equate it with the information provided in the question about the resultant loudness. Use the logarithmic formula of resultant loudness and the intensity of the sound waves.

Formula used:
The intensity of sound is,
$I \propto {a^2}$
Where, $I$ is the intensity of the sound wave and $a$ is the amplitude.
Logarithmic formula of resultant loudness is,
$\Delta L = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)$
Where, $I_1$ and $I_2$ is the intensity of sound.

Complete step by step solution:
Let $a$ be the amplitude sound ${S_1}$ and ${S_2}$. Now we know that;
Intensity of the sound is directly proportional to amplitude,
$I \propto {a^2}$
Hence the loudness of the two sound waves will be:
Loudness due to first sound,
${S_1} = {I_1} = K{a^2}$
Loudness due to second sound,
${S_2} = {I_2} = K(2{a^2}) = 4K{a^2} = 4{I_1}$

Resultant loudness will be
$\Delta L = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right) \\
\Rightarrow \Delta L= 10{\log _{10}}\left( {\dfrac{{4{I_1}}}{{{I_1}}}} \right)$........(by putting values from above two equation)
By solving, we get;
$\Delta L = 10{\log _{10}}(4) = 10 \times 0.6 = 6$
Here $\Delta L = n$ that is resultant loudness.

Hence the correct answer is option D.

Note: Here all the values of the intensity of the sound waves are in variables if the numerical values provided in the question put the same in the respective equations. Also note that the sound waves reaching at point P here is in phase if it is not then the answer will be different according to the phase.