Question

A train is moving toward the station at a speed of 25 m/s. Its horn emits a sound of frequency 600 Hz. (a) Calculate the frequency detected by a person standing still at the station. (Use 345 m/s for the speed of sound.) (b) As the train moves away, still blowing its horn at the same frequency as before, the observer hears a frequency of 567 Hz. Calculate the new speed of the train.

Answer 1

(a) 646.9 Hz

The formula for the Doppler effect is:

$f'=\frac{v+v_o}{v-v_s}f$

where

f = 600 Hz is the real frequency of the sound

f' is the apparent frequency

v = 345 m/s is the speed of sound

$v_o = 0$ is the velocity of the observer (zero since it is stationary at the station)

$v_s = +25 m/s$ is the velocity of the source (the train), moving toward the observer

Substituting into the formula,

$f'=\frac{345 m/s+0}{345 m/s-25 m/s}(600 Hz)=646.9 Hz$

(b) 20.1 m/s

In this case, we have

f = 600 Hz is the real frequency

f' = 567 Hz is the apparent frequency

Assuming the observer is still at rest,

$v_o = 0$

so we can re-arrange the Doppler formula to find $v_s$, the new velocity of the train:

$f'=\frac{v}{v-v_s}f\\\frac{f}{f'}=\frac{v-v_s}{v}\\\frac{f}{f'}v=v-v_s\\v_s = (1-\frac{f}{f'})v=(1-\frac{600 Hz}{567 Hz})(345 m/s)=-20.1 m/s$

and the negative sign means the train is moving away from the observer at the station.