A train running at 108 km h^{−1} towards east whistles at a dominant frequency of 500 Hz. Speed of sound in air is 340 m/s. 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 east at a speed of 36 km h^{−1}. Calculate the frequencies heard by the passenger in the train and by the person standing near the track.

#### Solution

Given:

Velocity of sound in air *v* = 340 m/s

Velocity of source *v*_{s} = 108 `\text{ kmh}^\(-)`^{1} =\[\frac{108 \times 1000}{60 \times 60} = 30 {\text { ms }}^{- 1}\]

Frequency of the source \[n_0\]= 500 Hz

(a) Since the velocity of the passenger with respect to the train is zero, he will hear at a frequency of 500 Hz.

(b) Since the observer is moving away from the source while the source is at rest:

Velocity of observer \[v_o\]*=* 0

Frequency of sound heard by person standing near the track is given by:

\[n = \left( \frac{v}{v + v_s} \right) n_0 \]

Substituting the values, we get:

\[n = \frac{340}{340 + 30} \times 500 = 459 \text{ Hz }\]

(c) When medium (wind) starts blowing towards the east:

Velocity of medium *v*_{m} = 36 `\text { kmh}^\(-)`^{1} =\[36 \times \frac{5}{18} = 10 {\text { ms }}^{- 1}\]

However, frequency heard by person standing near the track is given by:

\[n = \frac{\left( v + v_m \right)}{\left( v + v_m \right) + v_s} \times n_0 \]

\[ = \frac{\left( 340 + 10 \right)}{\left( 340 + 10 \right) + 30} \times 500\]

\[ = 458 \text{ Hz }\]