Answer
$$589 \mathrm{\ Hz}
$$
Work Step by Step
We once more pick a frame of reference where air looks to be still.
Now, velocity of source is $30.5-30.5=0$ ,
and that of the detector is 2$(30.5)$ . As a result,
$$
f^{\prime}=(500 \mathrm{Hz})\left(\frac{343 \mathrm{m} / \mathrm{s}+2(30.5 \mathrm{m} / \mathrm{s})}{343 \mathrm{m} / \mathrm{s}-0}\right)=589 \mathrm{\ Hz}
$$
