A train blowing a whistle of frequency 320 Hz approaches an observer standing on the platform at a speed of 66 m/s. The frequency observed by the observer will be (given speed of sound = 330 ms) ________ Hz.
Show answer and solution
Answer: 400
Solution
Using the Doppler formula for a source approaching a stationary observer:
f' = f × v/(v - vs) = 320 × 330/(330 - 66) = 320 × 330/264 = 320 × (5/4) = 400 Hz.
The key step is that (330 - 66) = 264 = 330 × (4/5), so the ratio 330/264 = 5/4, giving exactly 400 Hz.
f' = f × v/(v - vs) = 320 × 330/(330 - 66) = 320 × 330/264 = 320 × (5/4) = 400 Hz.
The key step is that (330 - 66) = 264 = 330 × (4/5), so the ratio 330/264 = 5/4, giving exactly 400 Hz.
[JEE Main 2023 · 25 Jan · Shift 2]