Answer
The correct answer is:
(c) $f_1 \gt f_2 \gt 620~Hz$
Work Step by Step
Let $v_s$ be the speed of sound. Let $v$ be the speed of the moving source. We can write the equation for the observed frequency $f_1$:
$f_1 = \frac{f_0}{1-\frac{v}{v_s}} = (\frac{v_s}{v_s-v})~f_0$
Let $v_s$ be the speed of sound. Let $v$ be the speed of the moving observer. We can write the equation for the observed frequency $f_2$:
$f_2 = (1+\frac{v}{v_s})~f_0 = (\frac{v_s+v}{v_s})~f_0$
We need to determine which expression is larger. We can multiply both expressions by $v_s(v_s-v)$:
$v_s(v_s-v)\times \frac{v_s}{v_s-v} = v_s^2$
$v_s(v_s-v)\times \frac{v_s+v}{v_s} = v_s^2-v^2$
We can see that $f_1 \gt f_2 \gt f_0$
The correct answer is:
(c) $f_1 \gt f_2 \gt 620~Hz$
