Doppler Effect Physics Example 2

Solution

1. 1
   The observed frequency ($475 \text{ Hz}$) is lower than the emitted frequency ($500 \text{ Hz}$), so the train is moving away.
2. 2
   For a receding source: $f' = f \times \frac{v}{v + v_s}$.
3. 3
   $$475 = 500 \times \frac{340}{340 + v_s} \implies \frac{475}{500} = \frac{340}{340 + v_s}$$
4. 4
   $$0.95(340 + v_s) = 340 \implies 323 + 0.95v_s = 340 \implies v_s = \frac{17}{0.95} \approx 17.9 \text{ m/s}$$