Two cars are approaching each other at an equal speed of 7.2 km hr$$^{-1}$$. When they see each other, both blow horns having a frequency of 676 Hz. The beat frequency heard by each driver will be ______ Hz. [Velocity of sound in air is 340 m s$$^{-1}$$.]

The speed of each car is $$v_s = 7.2$$ km hr$$^{-1} = 7.2 \times \frac{1000}{3600} = 2$$ m s$$^{-1}$$. The frequency of each horn is $$f_0 = 676$$ Hz, and the velocity of sound is $$v = 340$$ m s$$^{-1}$$.

Consider the situation from the perspective of driver A. Car B (the source) is approaching driver A (the observer), and both are moving toward each other. Using the Doppler effect formula, the frequency heard by driver A from car B's horn is $$f' = f_0 \times \frac{v + v_{\text{observer}}}{v - v_{\text{source}}} = 676 \times \frac{340 + 2}{340 - 2} = 676 \times \frac{342}{338}$$.

Driver A also hears their own horn at the original frequency $$f_0 = 676$$ Hz. The beat frequency is the difference between the two frequencies: $$f_{\text{beat}} = f' - f_0 = 676 \times \frac{342}{338} - 676 = 676 \times \left(\frac{342 - 338}{338}\right) = 676 \times \frac{4}{338}$$.

Computing this: $$f_{\text{beat}} = \frac{676 \times 4}{338} = \frac{2704}{338} = 8$$ Hz.

Therefore, the beat frequency heard by each driver is $$8$$ Hz.