Two cars are approaching each other at an equal speed of 7.2 km/hr. When they see each other, both blow horns having frequency of 676 Hz. The beat frequency heard by each driver will be ______ Hz. [Velocity of sound in air is 340 m/s.]

CONCEPT:

According to the Doppler's frequency in which we measure the apparent difference between the frequency at which the light waves leave a source and it is written as;

\(f = f_o[\frac{v-v_o}{v+v_o}]\)

Here, f_o be the observer frequency, v is the observer velocity, v is the source velocity and f is the actual frequency.

Given:

Speed of the car, v_o = 7.2 km/hr = 2 m/s

Apparent frequency, f_o = 676 Hz

The velocity of sound in air is = 340 m/s

Here we are given that the two cars approaching each other with a speed of 2 m/s as shown in the figure below,

according to the Doppler's frequency, we have;

\(f = f_o[\frac{v-v_o}{v+v_o}]\)

 \(f = 676\times[\frac{340-2}{340+2}]\\ \Rightarrow f = 676\times \frac{338}{342}\\ \Rightarrow f = 676\times 0.9883\)

⇒ f = 668.0908 Hz

Now the beat frequency is,

f_beat = f_o - f

       = 676-  668.0908 

       =7.9092

Hence, fbeat = 7.9092 Hz