A truck parked outside a petrol pump blows a horn of frequency 200 \mathrm{~Hz} in still air. The Wind then starts blowing towards the petrol pump at 20 \mathrm{~m} / \mathrm{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 \mathrm{~m} / sand the air is still?
A truck parked outside a petrol pump blows a horn of frequency 200 \mathrm{~Hz} in still air. The Wind then starts blowing towards the petrol pump at 20 \mathrm{~m} / \mathrm{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 \mathrm{~m} / sand the air is still?

For the standing observer:

Frequency is given as \mathrm{v}_{\mathrm{H}}=200 \mathrm{~Hz}

Velocity of sound is given as v=340 \mathrm{~m} / \mathrm{s}

Speed of the wind is given as v_{w}=20 \mathrm{~m} / \mathrm{s}

Because there is no relative motion between the observer and the truck, the observer will hear the horn at 200 Hz..

The wind blows in the observer’s direction at 20 \mathrm{~m} / \mathrm{s}.

Effective velocity of the sound will be, v_{E}=340+20=360 \mathrm{~m} / \mathrm{s}

The wavelength (\lambda) of the sound can be calculated as

\lambda=\mathrm{v}_{\mathrm{E}} / \mathrm{V}_{\mathrm{H}}=360 / 200

\lambda=1.8 \mathrm{~m}

For the observer running towards the train :

Speed of the observer is given as v_{0}=20 \mathrm{~m} / \mathrm{s}

We know,

The apparent frequency of the sound as the observers move towards the truck is :

v^{\prime} = v_{H}\left[\left(v+v_{0}\right) / v\right]

=200[(20+340) / 340]=211.764 \mathrm{~Hz}

As the air is still the effective velocity of sound is still 340 \mathrm{~m} / \mathrm{s}.

As the truck is stationary the wavelength remains 1.8 \mathrm{~m}.

Hence, the two cases are not completely identical.