Question

The sound intensity level at a point $4m$ from the point source is $10dB$ and the sound level at a distance $2m$ from the same source is $\left(x+10\right)dB.$ Find the value of $x.$ $\left(\log 2=0.3\right)$

Answer 1

${\beta }_1-{\beta }_2=10\left(\log \frac{R_2^2}{R_1^2}\right)$

$10-\left(x+10\right)=10\log \left(\frac{4m^2}{16m^2}\right)$

$-x=10\log \left(\frac{1}{4}\right)$

$-x=-6.02$

$x=6.02$