A truck parked outside a petrol pump blows a horn of frequency 200 Hz in still air.  The Wind then starts blowing towards the petrol pump at 20 m/s. Calculate the wavelength, speed, and frequency of the horn’s sound for a man standing at the petrol pump. Is this situation completely identical to a situation when the observer moves towards the truck at 20 m/s and the air is still?

Asked by Sudhanshu | 1 year ago |  271

1 Answer

Solution :-

For the standing observer:
Frequency, νH = 200 Hz
Velocity of sound, v = 340 m/s
Speed of the wind, vW = 20 m/s

The observer will hear the horn at 200 Hz itself because there is no relative motion between the observer and the truck.

Given that the wind blows in the observer’s direction at 20 m/s.
Effective velocity of the sound, vE = 340 + 20 = 360 m/s

The wavelength ( λ ) of the sound :
λ = \(\dfrac{v_E} {v_H}\)  =\( \dfrac{360} {200}\)
λ = 1.8 m

For the observer running towards the train :
Speed of the observer, vo = 20 m/s

We know,
The apparent  frequency of the sound as the observers move towards the truck is :
v’ =   \( \dfrac{ v_H (v + v_o)}{v}\)
\( \dfrac{200 (20 + 340 )}{340}\) =  211.764 Hz

As the air is still the effective velocity of sound is still 340 m/s.
As the truck is stationary the wavelength remains 1.8 m.

Thus, the two cases are not completely identical.

Answered by Sudhanshu | 1 year ago

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