Question

A source emitting sound at frequency 4000 Hz, is moving along the Y-axis with a speed of 22 m s⁻¹. A listener is situated on the ground at the position (660 m, 0). Find the frequency of the sound received by the listener at the instant the source crosses the origin. Speed of sound in air = 330 m s⁻¹.

Answer 1

Given:
Speed of sound in air v = 330 ms⁻¹
Frequency of sound \[f_0\] = 4000 Hz
Velocity of source \[v_s\] = 22 m/s
The apparent frequency heard by the listener \[\left( f \right)\] = ?

At t = 0, let the source be at a distance of y from the origin. Now, the time taken by the sound
to reach the listener is the same as the time taken by the sound to reach the origin.
∴​

\[\frac{y}{22} = \frac{\sqrt{660 + y^2}}{330}\] 

\[ \Rightarrow    \left( 15y \right)^2  =  \left( 660 \right)^2  +  \left( y \right)^2 \] 

\[ \Rightarrow   224 y^2  =  \left( 660 \right)^2 \] 

\[ \Rightarrow y = \frac{660}{\sqrt{224}}\]

Velocity of source along the line joining the source \[\left( S \right)\] and listener \[\left( L \right)\] :

\[v_s \cos\theta\] = \[22 . \frac{y}{\sqrt{660 + y^2}} = \frac{22y}{15y} = \frac{22}{15}\]

Frequency heard by the listener \[\left( f \right)\] is

\[f = \frac{v}{v - v_s cos\theta} \times  f_0 \] 

\[ \Rightarrow f = \frac{330}{330 - \frac{22}{15}} \times 4000\]

\[\Rightarrow f\]= 4017.85 ≈ 4018 Hz