Question

A train whistle is heard at 350 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 340 Hz.
a. What is the speed of the train before slowing down?
b. What is the speed of the train after slowing down?

Answer 1

Answer:

(a). 19.0 m/s

(b). 9.5m/s

Explanation:

Assuming speed of sound is 343m/s.

(a).

As the train approaches, from the the Doppler equation we have

$f_{obs} = \dfrac{vf_{source}}{v-v_{source}}$

$350Hz = \dfrac{343f_{source}}{343-v_{source}}$

solving for $f_{source}$ we get:

$f_{source} = \dfrac{350(343-v_{source})}{343}$.

And as the train cuts its speed in half, the equation gives

$340Hz = \dfrac{343f_{source}}{343-\dfrac{v_{source}}{2} }$

substituting the value of $f_{source}$ we get:

$340Hz = \dfrac{343\dfrac{350(343-v_{source})}{343}}{343-\dfrac{v_{source}}{2} }$

$340Hz = \dfrac{350(343-v_{source})}{343-\dfrac{v_{source}}{2} }$

$340Hz = \dfrac{700(343-v_{source})}{686-v_{source}}$

$340 (686-v_{source}) = 700(343-v_{source})$

$\boxed{v_{source} = 19.0m/s}$

or 68.6 km per hour, which is the speed of the train before slowing down.

(b).

The speed of the train after slowing down is half the previous speed; therefore,

$v_{after} = \dfrac{v_{source}}{2}$

$\boxed{v_{after} =9.5m/s}$

or 34.3 km per hour.