A train moves towards a stationary observer with speed 20 m/s. The...

Using Doppler's effect, the apparent frequency when the source moves towards a stationary observer is given by:

$f_1 = f_0 \left( \dfrac{v}{v - v_s} \right)<br><br>Substituting$v = 340$m/s and$v_s = 20$m/s:<br><br>$f_1 = f_0 \left( \dfrac{340}{340 - 20} \right) = f_0 \left( \dfrac{340}{320} \right)$<br><br>When the source moves away from the stationary observer, the apparent frequency is given by:<br><br>$f_2 = f_0 \left( \dfrac{v}{v + v_s} \right)

Substituting the values:

$f_2 = f_0 \left( \dfrac{340}{340 + 20} \right) = f_0 \left( \dfrac{340}{360} \right)$

Taking the ratio $\dfrac{f_2}{f_1}$:

$\dfrac{f_2}{f_1} = \dfrac{f_0 \left( \dfrac{340}{360} \right)}{f_0 \left( \dfrac{340}{320} \right)} = \dfrac{320}{360} = \dfrac{8}{9}$

Answer: $\dfrac{8}{9}$