Question

A train moving at a constant speed is passing a stationary observer on a platform. On one of the train cars, a flute player is continually playing the note known as concert A (f = 440 Hz). After the flute has passed, the observer hears the sound with a frequency of 415 Hz. What is the speed of the train? The speed of sound in air is 343 m/s. A) 7.3 m/s B) 12 m/s C) 21 m/s D) 37 m/s E) 42 m/s

Answer 1

Answer:

C) 21 m/s

Explanation:

The general formula of the Doppler effect is:

$f'=(\frac{v+v_r}{v+v_s})f$

where

f' is the apparent frequency

f is the original frequency

v is the speed of the wave in the medium

$v_r$ is the velocity of the receiver, positive if the receiver is moving towards the source

$v_s$ is the velocity of the source, positive if the source is moving away from the receiver

Here we have

f = 440 Hz

f' = 415 Hz

v = 343 m/s

$v_r = 0$ (the observer is stationary)

$v_s$ is positive since we are considering when the train has passed the observer, so it is moving away from him

So we can rewrite the formula as

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

And solving for $v_s$, we find the speed of the train

$v_s = v(\frac{f}{f'}-1)=(343 m/s)(\frac{440 Hz}{415 Hz}-1)=20.7 m/s \sim 21 m/s$