An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle

Question

2 years ago

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An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle

is 447 Hz when the train is approaching him and 420 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find?

Physics

1 answer:

stira [4] · Answer 1 · 2023-06-28T05:45:43+03:00

Answer:velocity of train (source of sound) =10.68m/s

Explanation: this is a question under doppler effect.

The doppler effect formulae gives the value of sound wave detected by an observer from a source when there is a relative motion between both the observer and the source.

The formulae below

f'= $\frac{v+v'}{v-vs}$ * f

f'=observed frequency

f= actual frequency

v'= velocity of observer

v= velocity of sound in air=343m/s

vs= velocity of source (train)

from the question, the observer is standing thus, v'=0

The sound source (train) is moving towards the observer thus making vs positive relative to the motion of the observer(+vs), at this condition, we have that f'=447hz, v=343m/s v'=0 with the formulae

f' = $\frac{v}{v-vs}$ * f

putting all of these in the doppler effect formulae, we have that

447= $\frac{343}{343-vs}$ * f

thus we have that

447= $\frac{343f}{343-vs}$

by cross mutiplying, we have that

447(343-vs)=343f

we call this equation 1

for the next condition, the source (train) is moving away from the observer thus making vs negative relative to the motion of the observer (-vs), at this condition, we have that

f'=420hz, v'=0, v=343m/s .

since vs is now negative and the formulae it was positive then the resulting sign is positive as shown below

f'= $\frac{v+v'}{v-(-vs)}$