Answer
$$1.58 \times 10^{3} \mathrm{Hz}
$$
Work Step by Step
We use Eq. $17-47$ with $f=1200 \mathrm{Hz}$ and $v=329 \mathrm{m} / \mathrm{s}$
In this case, $v_{D}=65.8 \mathrm{m} / \mathrm{s}$ and $v_{S}=29.9 \mathrm{m} / \mathrm{s}$ , and we choose signs so that $f^{\prime}$ is larger than $f :$
$$
f^{\prime}=f\left(\frac{329 \mathrm{m} / \mathrm{s}+65.8 \mathrm{m} / \mathrm{s}}{329 \mathrm{m} / \mathrm{s}-29.9 \mathrm{m} / \mathrm{s}}\right)=1.58 \times 10^{3} \mathrm{Hz}
$$
