The intensity level at a distance of 200 cm | vedclass

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(B) The intensity $I$ of a spherical sound wave follows the inverse square law: $I \propto \frac{1}{r^2}$.Given $r_1 = 200 \ cm$ and $r_2 = 400 \ cm$,

Vedclass · Accepted Answer

$74$

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(B) The intensity $I$ of a spherical sound wave follows the inverse square law: $I \propto \frac{1}{r^2}$.Given $r_1 = 200 \ cm$ and $r_2 = 400 \ cm$,the ratio of intensities is $\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}ight)^2 = \left(\frac{200}{400}ight)^2 = \left(\frac{1}{2}ight)^2 = \frac{1}{4}$.Thus,$I_1 = 4I_2$.The intensity level $L$ in decibels is given by $L = 10 \log_{10}\left(\frac{I}{I_0}ight)$.The difference in intensity levels is $L_1 - L_2 = 10 \log_{10}\left(\frac{I_1}{I_2}ight)$.Substituting the ratio: $L_1 - L_2 = 10 \log_{10}(4) = 10 	imes 0.602 = 6.02 \ dB \approx 6 \ dB$.Therefore,$L_2 = L_1 - 6 \ dB = 80 \ dB - 6 \ dB = 74 \ dB$.

The intensity level at a distance of $200 \ cm$ from a sound source is $80 \ dB$. If there is no loss of acoustic power in the air and the threshold intensity of hearing is $10^{-12} \ W/m^2$,what is the intensity level at a distance of $400 \ cm$ from the source (in $dB$)?

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