Answer:
-58.84 Hz.
Explanation:
This is a question under Doppler effect.
The mathematical equation of doppler effect is given as
fo = (v + vo)/(v -vs) × fs
Where fo = frequency perceived by observer = ?
v = speed of sound in air = 343m/s
vo = velocity of obseerver relative to the source = 0 (since he is not moving)
vs = velocity of source relative to the observer = 30.8 m/s
fs = frequency of sound produced by source = 325 Hz
For the condition where the train is approaching the observer, velocity of source relative to the observer (vs) is positive, hence vs = 30.8 m/s
fo = (343+0)/(343 -30.8) × 325
fo = 357.06 Hz.
For the condition where the train is moving away from the observer, velocity of source relative to the observer (vs) is negative, hence vs = -30.8 m/s
fo = (343+0)/(343 -(-30.8)) × 325
fo = (343/ 343 +30.8) × 325
fo = (343/373.8) × 325
= 298.22 Hz
Change in frequency as the train moves from approaching to receeding is = 298.22 - 357.06
= -58.84 Hz.
