The intensity due to a number of independent sound sources is the sum of the individual intensities. (a) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries? (b) To increase the sound intensity level again by the same number of decibels as in part (a), how many more crying babies are required?

The intensity level of the sound is given as $\mathbf{\Delta \beta }\mathbf{=}\mathbf{10}\mathbf{log}\left(\frac{I_2}{I_1}\right)$where,$\mathbf{\Delta \beta }$is the increase in the intensity level of the sound,${\mathbf{I}}_2$is the final intensity of the sound,${\mathbf{I}}_1$is the initial intensity of the sound.

The total intensity because of multiple sound sources is the sum of their individual intensities. If the intensity of the sound of one quadruplet is ${\mathrm{I}}_1$, then the intensity of the sound when the four quadruplets cry is $4I_1$

Substitute$4I_1$for$I_2$ to find$\mathrm{\Delta \beta }$

$\begin{array}{c}\Delta \beta =10\log \left(\frac{I_2}{I_1}\right)\\ =10\log \left(4\right)\\ =6.02dB\end{array}$

Therefore, the increase in decibel of sound intensity level is$6.02\text{\hspace{0.17em}dB}$.

The total intensity because of multiple sound sources is the sum of their individual intensities. The initial intensity level of the sound is $6.02\text{\hspace{0.17em}dB}$. From the calculation of part (a), the increase in the sound intensity level is $6.02\text{\hspace{0.17em}dB}$If the sound intensity level is doubled, it will equal $12.04\text{\hspace{0.17em}dB}$.

Substitute $12.04\text{\hspace{0.17em}dB}$for $\Delta \beta$in the above equation to find $I_3$

$\begin{array}{c}\left(12.04\right)dB=10\log \left(\frac{I_3}{I_1}\right)\\ \frac{I_3}{I_1}={10}^{\frac{12.04dB}{10}}\\ I_3=\left({10}^{1.204}\right)I_1\\ I_3=16I_1\end{array}$

Therefore,the number of required crying babies of quadruplets is 16.