## College Physics (4th Edition)

The correct answer is: (c) $f_1 \gt f_2 \gt 620~Hz$
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) $f_1 \gt f_2 \gt 620~Hz$