Question

A woman is riding a bicycle at 18.0 m/s along a straight road that runs parallel to and right next to some railroad tracks. She hears the whistle of a train that is behind. The frequency emitted by the train is 840 Hz, but the frequency the woman hears is 778 Hz. Take the speed of sound to be 340 m/s. What is the train’s speed?

Answer 1

Answer:

$v_s=7.66066838m/s$

Explanation:

We need to use the doppler effect equation:

$f_o=\frac{(v\pm v_o)}{(v\pm v_s)}*f_s$

Where:

$f_s=Source\hspace{1 mm} frequency=840Hz$

$f_o=Observed\hspace{1 mm} frequency=778Hz$

$v=Speed\hspace{1 mm}of\hspace{1 mm}the\hspace{1 mm}sound=340m/s$

$v_o=Velocity\hspace{1 mm}of\hspace{1 mm}the\hspace{1 mm}observer=18m/s$

$v_s=Velocity\hspace{1 mm}of\hspace{1 mm}the\hspace{1 mm}source$

In this case, we can assume that the observer is moving away from the source and the source is moving towards the observer, then:

$f_o=\frac{(v-v_o)}{(v-v_s)}*f_s$

Isolating $v_s$

$v_s=v-\frac{f_s}{f_o}*(v-v_o)$  (1)

Replacing the data provided in (1)

$v_s=340-\frac{840}{778}*(340-18)=-7.66066838m/s$

The minus in the result is actually telling us that the source is really moving away from the observer.