A bullet travels past a person at a speed of 220 m/s . Determine...

Let the actual frequency of the whistling sound be $f_0$.

When the bullet approaches the stationary person, the apparent frequency $f_1$ is given by the Doppler effect formula:

$f_1 = f_0 \left( \dfrac{v}{v - v_s} \right)<br><br>Substituting$v = 330 $\text{ m/s} and$v_s = 220 $\text{ m/s}:<br><br>$f_1 = f_0 \left( \dfrac{330}{330 - 220} \right) =$f_0 \left( \dfrac{330}{110} \right) =$3f_0$<br><br>When the bullet recedes from the person, the apparent frequency$f_2$is:<br><br>$f_2 = f_0 \left( \dfrac{v}{v + v_s} \right)$<br><br>$f_2 = f_0 \left( \dfrac{330}{330 + 220} \right) = f_0 \left( \dfrac{330}{550} \right) = 0.6f_0$<br><br>The change in the perceived frequency as the bullet crosses the person is:<br><br>$\Delta f = f_1 - f_2 = 3f_0 - 0.6f_0 = 2.4f_0$<br><br>The fractional change in the frequency is:<br><br>$\dfrac{\Delta f}{f_0} = \dfrac{2.4f_0}{f_0} = 2.4$<br><br>Answer:$2.4