Question

Two identical sound waves each of loudness $\beta$interfere constructively at a point to produce a sound level of
A. $6\beta$
B. $3\beta$
C. $\beta + 3$
D. $\beta + 6$

Answer 1

The above problem can be resolved using the formula for the loudness of the sound and the sound level. Then by applying the substitution of the values, the desired result is obtained. Moreover, the sound level has its relation with the intensity of sound at the initial and at final.

Complete step by step answer:
Let I be the intensity of the sound wave from the source.
Then the intensity of the sound wave is formed by constructive interference. And the value for the interference is, $4I$.
We know the expression for the loudness of sound is given as,
$\beta = {\log _{10}}{\left( {\dfrac{I}{{{I_0}}}} \right)^{10}}$
Here, I is the final intensity of sound and ${I_0}$ is the initial intensity of the sound.
Solve by substituting the values in above relation as,

\beta = {\log _{10}}{\left( {\dfrac{I}{{{I_0}}}} \right)^{10}}\\\ \beta = 10{\log _{10}}I - 10\log \left( {{I_0}} \right) \end{array}$$……… (1) Now, the expression for the sound level is given as, $$L = {\log _{10}}{\left( {\dfrac{{4I}}{{{I_0}}}} \right)^{10}}$$……..(2) Solve by substituting the values of equation 1 in 2 as, $$\begin{array}{l} L = {\log _{10}}{\left( {\dfrac{{4I}}{{{I_0}}}} \right)^{10}}\\\ L = 10\left( {\log 4 + \log I} \right) - 10\log {I_0}\\\ L = 10\log 4 + \left( {10\log I - 10\log {I_0}} \right)\\\ L = 6 + \beta \end{array}$$ Therefore, the sound level is of $$\beta + 6$$ and option (D) is correct. **Note:** To resolve the given problem, the concepts and fundamentals of the sound level are required to be taken into consideration. Along with this, the importance of the loudness of sound is also to be remembered. These variables are needed to remember because they are used to design various sound-amplifying devices. Moreover, the concepts of the intensity of sound are also required to remember, along with factors affecting the intensity of sound.