A train running at 108 km/h towards east whistles at a dominant freque

1.	A train running at 108 km/h towards east whistles at a dominant frequency of 500 Hz. The speed of sound in air is 340 m/s. (a) What frequency will a passenger sitting near the open window hear? (b) What frequency will a person standing near the track hear whom the train has just passed? (c) A wind starts blowing towards the east at a speed of 36 km/h. Calculate the frequencies heard by the passenger in the train and by the person standing near the track.
Answer» (a) The frequency of the sound ν = 500 Hz. The speed of sound in air V = 340 m/s. Speed of the source uₛ = 108 km/h =108000/3600 m/s =30 m/s. The speed of the observer uₒ = 108 km/k =30 m/s. In such case apparent frequency ν' = (V+uₒ)ν/(V-uₛ) But here the source is leaving hence ν' = (V+uₒ)ν/(V+uₛ) = ν = 500 Hz {since uₒ = uₛ} It can also be concluded by the fact that the relative motion between the source and the observer is zero. (b) For the person standing near the track uₒ = 0 and the source is leaving. ν' = Vν/(V+uₛ) =340500/(340+30) Hz =340500/370 Hz =459 Hz (c) The speed of the medium uₘ =36 km/h =36000/3600 m/s = 10 m/s towards east. Hence the effective speed of source = uₛ+uₘ and the effective speed of the passenger = uₒ+uₘ Both are in the same direction hence no relative motion. Thus the frequency observed by the passenger = ν = 500 Hz. The effective speed of the sound in the air for the person standing due to the wind towards the east = V-uₘ =340-10 =330 m/s. Now the apparent frequency for the observer can be calculated as usual ν' = Vν/(V+uₛ) =330500/(330+30) Hz =330500/360 Hz =458 Hz

A train running at 108 km/h towards east whistles at a dominant frequency of 500 Hz. The speed of sound in air is 340 m/s. (a) What frequency will a passenger sitting near the open window hear? (b) What frequency will a person standing near the track hear whom the train has just passed? (c) A wind starts blowing towards the east at a speed of 36 km/h. Calculate the frequencies heard by the passenger in the train and by the person standing near the track.

Answer»

(a) The frequency of the sound ν = 500 Hz. The speed of sound in air V = 340 m/s. Speed of the source uₛ = 108 km/h =108000/3600 m/s =30 m/s.

The speed of the observer uₒ = 108 km/k =30 m/s.

In such case apparent frequency ν' = (V+uₒ)ν/(V-uₛ)

But here the source is leaving hence

ν' = (V+uₒ)ν/(V+uₛ) = ν = 500 Hz {since uₒ = uₛ}

It can also be concluded by the fact that the relative motion between the source and the observer is zero.

(b) For the person standing near the track uₒ = 0 and the source is leaving.

ν' = Vν/(V+uₛ) =340*500/(340+30) Hz =340*500/370 Hz

=459 Hz

=36000/3600 m/s = 10 m/s towards east.

Hence the effective speed of source = uₛ+uₘ and the effective speed of the passenger = uₒ+uₘ Both are in the same direction hence no relative motion. Thus the frequency observed by the passenger = ν = 500 Hz.

The effective speed of the sound in the air for the person standing due to the wind towards the east = V-uₘ =340-10 =330 m/s. Now the apparent frequency for the observer can be calculated as usual

ν' = Vν/(V+uₛ)

=330*500/(330+30) Hz

=330*500/360 Hz

=458 Hz

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