Doppler Effect!!

Classical Mechanics Level 2

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 ms − 1 ^{-1} and the man behind walks at a speed 2.0 ms − 1 ^{-1} . A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air is 330 ms − 1 ^{-1} . At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is

Note: Neglect height of the men.


The answer is 5.00.

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1 solution

f 1 = f × ( V V − V s cos ⁡ θ ) f_1 = f\times (\dfrac {V}{V - V_s \cos \theta})

f 2 = f × ( V V + V s ′ cos ⁡ θ ) f_2 = f\times (\dfrac {V}{V + V'_s \cos \theta})

f B = f 1 − f 2 = 1430 ( 330 330 − 2 × 5 13 ) − 1430 ( 330 330 + 1 × 5 13 ) = 1430 × 330 × 13 × 15 4280 × 4295 = 5.00 . f_B = f_1 - f_2 = 1430(\dfrac {330}{330 - 2\times \frac{5}{13}}) - 1430(\dfrac {330}{330 + 1\times \frac{5}{13}}) = \dfrac {1430\times 330\times 13\times 15}{4280\times 4295} = \boxed{5.00}.

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