A train sounds its horn before it sets off from the station and an observer waiting on the platform estimates its frequency at 1200 Hz. The trains then moves off and accelerates steadily. Fifty seconds after departure, the driver sounds the horn again and the platform observer estimates the frequency of 1140 Hz. Calculate the train speed 50 s after departure. How far from the station is the train after 50 s? (Speed of sound = 340 m s⁻¹). (Ans: 17.9 m s⁻¹, 448 m)



Data Given:

Frequency of the source (train) = f = 1200 Hz

Frequency observed by (Observer) = f ' = 1140 Hz

Time = t = 50 s

Speed of the sound = v = 340 m s⁻¹

To Find:


Speed of the train (source) (at t = 50 ) = `\v_s` = ?

Distance Traveled (at t = 50 ) = S = ?


Solution:


The formula for apparent frequency f ' when the train moving away from the stationary person (observer)  is


f ' = `\frac {v}{v + v_s}` f -------------(1)

or (by simplifying) 

`\v_s` = `\frac {v f}{f '}` - v

by putting the corresponding values

`\v_s` = `\frac {340 m s⁻¹ ｘ 1200 Hz }{1140 Hz}` - 340 m s⁻¹

`\v_s` = 357.895 - 340 m s⁻¹ 

`\v_s` = 17.895 m s⁻¹ 

or

`\v_s` = 17.9 m s⁻¹  ------------Ans.(1)


Now to find the distance travelled by train we know that 

Distance = Average speed x time

S = `\v_{ave}` t -------(2)

where `\v_{ave}` = `\ frac {0 m s⁻¹ + 17.9 m s⁻¹}{2}` = 8.95 m s⁻¹

by putting values

S = 8.95 m s⁻¹ ï½˜ 50 s

S = 447.5 m 

S = 448 m ----------------Ans.

Hence after 50 s the train has traveled a distance of 448 m

Similar Questions:

A sound source vibrates at 200 Hz and is receding from a stationary observer at 18 ms⁻¹. If the speed of sound is 331 ms⁻¹ then what frequency does the observer hear? (189.68Hz) 




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