Question

A train approaching a hill at a speed of 40 km/hr sounds a whistle of frequency 600 Hz when it is at a distance of 1 km from the hill. A wind with a speed of 40 km/hr is blowing in the direction of motion of the train. Find the frequency of the whistle as heard by an observer on the hill.

Answer 1

Hint: Here, the moving train produces a whistle and is heard by an observer on the hill. So the train is the source of the sound and it is moving towards the observer. Since the wind also moves in the same direction, the velocity of sound will be the sum of the speed of sound in air and the speed of the wind.

Formula used:
-The frequency heard by a stationary observer is given by, $f = {f_0}\left( {\dfrac{{v + {v_s}}}{v}} \right)$ where ${f_0}$ is the frequency of the sound produced by the source moving at velocity ${v_s}$ and $v$ is the velocity of the sound in air.

Complete step by step answer.
Step 1: List the key parameters involved in the problem.
In the problem at hand, the source of the whistle is a moving train and its listener is stationary on a hill. The wind also blows in the direction in which the train moves. Thus the velocity of sound will be the sum of the velocities of sound in air and that of the blowing wind.
The speed of the train is given to be ${v_s} = 40{\text{km/hr}}$ and the frequency of the whistle produced is given to be ${f_0} = 600{\text{Hz}}$ .
The speed of sound in the air can be taken as $v = 330{\text{m/s}} = 1188{\text{km/hr}}$ . Also, it is given that the wind blows at a speed of $w = 40{\text{km/hr}}$ in the direction in which the train moves.
Step 2: Express the equation for the frequency of the whistle heard by the listener on the hill.
The frequency of the sound heard by a stationary observer is given by, $f = {f_0}\left( {\dfrac{{v + {v_s}}}{v}} \right)$ where ${f_0}$ is the frequency of the sound produced by the source moving at velocity ${v_s}$ and $v$ is the velocity of the sound in air.
In our scenario, the above equation takes the form $f = {f_0}\left( {\dfrac{{v + w + {v_s}}}{{v + w}}} \right)$ --------- (1) since the wind also moves.
Step 3: Substitute the values in equation (1) to find the frequency heard by the listener.
Substituting values for ${f_0} = 600{\text{Hz}}$ , ${v_s} = - 40{\text{km/hr}}$ , $v = 1188{\text{km/hr}}$ and $w = 40{\text{km/hr}}$ in equation (1) we get, $f = 600 \times \left( {\dfrac{{1188 + 40 - 40}}{{1188 + 40}}} \right) = 580.45{\text{Hz}}$
$\therefore $ the frequency heard by the observer on the hill is $f = 580.45{\text{Hz}}$ .

Note: The train is mentioned to move towards the stationary observer on the hill. So, the velocity of the train is taken to be negative i.e., ${v_s} = - 40{\text{km/hr}}$ while substituting it in equation (1). Also the velocity of speed given as $v = 330{\text{m/s}}$ is converted into $v = 1188{\text{km/hr}}$ for convenience by using the conversion relation $1{\text{m}} = \dfrac{5}{{18}}{\text{km/hr}}$ .