Q50E - University Physics with Modern Physics

Question

A railroad train is traveling at 30.0 m/s in still air. The frequency of the note emitted by the train whistle is 352 Hz. What frequency is heard by a passenger on a train moving in the opposite direction to the first at 18.0 m/s and (a) approaching the first and (b) receding from the first?

Step-by-Step Solution

Answer

a) 405.8Hz is the frequency heard by the passengers when approaching the first.

b) 306.9Hz is the frequency heard by the passengers when receding the first.

Step 1:

Given data:

As by the doppler effect:

$f_L=\frac{v+v_L}{v+v_S}{f}_S$

$f_{L=}$Frequency observed by the listener

$v=$speed of sound

$v_L=$speed of listener

$f_s=$frequency of source

$v_s$= speed of the source of sound

And here $v_L$ is positive as the velocity of listener from Listener to Source and $v_s$ is positive as the velocity of source from Listener to Source and velocity is negative.

The frequency of the sound heard by the listener is not the same as the source frequency when the source and listener are moving relative to each other. 

Step 2:

a) Here, $v_L$ is positive as listener moving towards the source and $v_s$ is negative as source is moving towards the listener.

So, putting the formula;

$$\begin{gathered} f_L=\frac{v+v_L}{v+v_S}{f}_S \\ =\frac{344m/s+18m/s}{344m/s-30m/s}\left(352Hz\right) \\ =405.8Hz \end{gathered}$$

Hence, 405.8Hz is the frequency heard by the passengers when approaching the first.

Step 3:

b) Here, $v_L$ is negative as listener moving away from the source and $v_s$ is positive as source is moving away from the listener.

so, putting the value;

$$\begin{gathered} f_L=\frac{v+v_L}{v+v_S}{f}_S \\ =\frac{344m/s-18m/s}{344m/s+30m/s}\left(352Hz\right) \\ =306.9Hz \end{gathered}$$

Hence, 306.9Hz is the frequency heard by the passengers when receding the first.