Answer
(a) $f_2 \lt f_1$
(b) $f_2-f_1 = -10,320~Hz$
Work Step by Step
(a) The source and observer are moving away from each other so the observed frequency is less than the emitted frequency. Therefore, $f_2 \lt f_1$
(b) Let $f_s$ be the frequency observed by the speeder. We can find an expression for $f_s$:
$f_s = f_1~(1-\frac{v}{c})$
We can find the frequency $f_2$:
$f_2 = f_s~(1-\frac{v}{c})$
$f_2 = [f_1~(1-\frac{v}{c})]~(1-\frac{v}{c})$
$f_2 = f_1~(1-\frac{v}{c})^2$
$f_2 = (36.0~GHz)~(1-\frac{43.0~m/s}{3.0\times 10^8~m/s})^2$
$f_2 = 35.99998968~GHz$
We can find $f_2-f_1$:
$f_2-f_1 = 35.99998968~GHz-36.0~GHz$
$f_2-f_1 = -0.00001032~GHz$
$f_2-f_1 = -10,320~Hz$
