A violin player travelling on a slow train sounds a 440 Hz note,...

Let $f_0$ be the actual frequency of the violin, $v$ be the speed of sound, and $v_s$ be the speed of the train.

Given $f_0 = 440 \text{ Hz}$ and $v = 340 \text{ m/s}$.

The apparent frequency $f'$ heard by the stationary person on the ground from the approaching train is given by the Doppler effect formula:

$f' = f_0 \left( \dfrac{v}{v - v_s} \right)<br><br>The person hears a beat frequency of$4.0 $\text{ Hz} between the stationary violin and the approaching violin. Thus, the apparent frequency is:<br><br>$f' = f_0 + f_b = 440 + 4 = 444 $\text{ Hz}<br><br>Substituting the values into the Doppler effect formula:<br><br>$444 = 440 \left( \dfrac{340}{340 - v_s} \right)$<br><br>$444(340 - v_s) = 440 \times 340$<br><br>$150960 - 444 v_s = 149600$<br><br>$444 v_s = 1360$<br><br>$v_s = \dfrac{1360}{444} = \dfrac{340}{111} \text{ m/s}$<br><br>To convert the speed from$\text{m/s}$to$\text{km/h}$, we multiply by$\dfrac{18}{5}$:<br><br>$v_s = \dfrac{340}{111} \times \dfrac{18}{5} \text{ km/h} = \dfrac{1224}{111} \text{ km/h} \approx 11.03 \text{ km/h}$<br><br>Rounding to the nearest integer, the speed of the train is$11 \text{ km/h}$.$