Answer
(a) $104.4dB$
(b) $99.6dB$.
(c) $2.0\times 10^6m$
Work Step by Step
(a) We know that
$\beta=10log(\frac{P}{4\pi I_{\circ}r^2})$....eq(1)
We plug in the known values to obtain:
$120dB=10log(\frac{P}{4(3.14)(10^{-12}W/m^2)(2m)^2})$
This simplifies to:
$P=50.3W$
Now we plug in the known values in eq(1) to obtain:
$\beta=10 log(\frac{50.3W}{4(3.14)(10^{-12}W/m^2)(12m)^2})$
$\implies \beta=104.4dB$
Thus, the sound intensity heard by an observer at a distance of $12m$ is $104.4dB$.
(b) We know that
$\beta=10 log(\frac{50.3W}{4(3.14)(10^{-12}W/m^2)(21m)^2})$
$\implies \beta=104.4dB$
Thus, the sound intensity heard by an observer at a distance of $21m$ is $99.6dB$.
(c) We know that
$r=\sqrt{\frac{P}{4\pi I_{\circ}}}$
We plug in the known values to obtain:
$r=\sqrt{\frac{50.3W}{4(3.14)(10^{-12}W/m^2)}}$
$\implies r=2.0\times 10^6m$
