Answer
$1050Hz,2.10KHz$
Work Step by Step
We know that
$f^{\prime}-f=\frac{fu}{c}$
We plug in the known values to obtain:
$f^{\prime}-f=\frac{(9.00\times 10^9)(35.0)}{3.00\times 10^8}=1050Hz$
Now $f^{\prime \prime}-f=2(f^{\prime}-f)=2(1050)=2.10KHz$
