Answer
$$598 \mathrm{Hz}
$$
Work Step by Step
We use Eq. $17-47$ with $f=500 \mathrm{Hz}$
and $v=343 \mathrm{m} / \mathrm{s}$ .
We select a signs to make $f^{\prime}>f$
Therefore the frequency heard in still air is will be:
$$
f^{\prime}=(500 \mathrm{Hz})\left(\frac{343 \mathrm{m} / \mathrm{s}+30.5 \mathrm{m} / \mathrm{s}}{343 \mathrm{m} / \mathrm{s}-30.5 \mathrm{m} / \mathrm{s}}\right)=598 \mathrm{Hz}
$$
