Two train whistles have identical frequencies of 1.85 102 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the station platform hears beats with a frequency of 8.00 beats/s when the whistles operate together. What are the two possible speeds that the moving train can have

frequency of beat heard = 8 beat /s . No of beat produced is equal to difference of frequencies of two sound source . Here difference is created due to Doppler effect . One of the train is moving so it will have apparent frequency which is different one from its original frequency .

When the moving train is approaching the observer , its frequency will be higher . As beat is heard at the rate of 8 beats / s , apparent frequency of approaching train will be 185 + 8 = 193 Hz .

Applying Doppler's formula of apparent frequency ,

193 = 185 x V / ( V - v ) , where V is velocity of sound and v is velocity of train .

193 V - 193 v = 185 V

193 v = 8 V

v = 8 x V / 193

= 8 x 343 / 193

= 14.21 m /s

Second possibility is that apparent velocity is less ie 185 - 8 = 177 Hz

In that case moving train will be moving away from observer . If its velocity be v

177 = 185 x V / ( V + v )

177 V + 177 v = 185 V

v = 8 x 343 / 177

= 15.50 m /s .