A person observes two moving trains. First reaching the station and another leaves the station with equal…

The Doppler effect governs the frequency shift for a moving source relative to a stationary observer. For a source approaching the observer, the observed frequency is $f_1 = f_0 \frac{v}{v - v_s}$, while for a receding source, it is $f_2 = f_0 \frac{v}{v + v_s}$, where $f_0 = 300 \, \text{Hz}$, $v = 330 \, \text{m/s}$, and $v_s = 30 \, \text{m/s}$.

For the approaching train: $f_1 = 300 \times \frac{330}{330 - 30} = 300 \times \frac{330}{300} = 330 \, \text{Hz}$.

For the receding train: $f_2 = 300 \times \frac{330}{330 + 30} = 300 \times \frac{330}{360} = 275 \, \text{Hz}$.

The difference in frequencies is $|f_1 - f_2| = |330 - 275| = 55 \, \text{Hz}$.

Final answer: $\boxed{55}$