Question

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.

Answer 1

Given:
Velocity of sound in air v = 340 m/s
Velocity of source v_s = 108 kmh^$-$1 = $\frac{108\times 1000}{60\times 60}=30{\mathrm{ms}}^{-1}$
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:

$n=\frac{340}{340+30}\times 500=459\mathrm{Hz}$

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

Velocity of medium v_m = 36 kmh^$-$1 = $36\times \frac{5}{18}=10{\mathrm{ms}}^{-1}$

The frequency heard by the passenger is unaffected (= 500 Hz).

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

$$\begin{gathered} n=\frac{\left(v+v_m\right)}{\left(v+v_m\right)+v_s}\times n_0 \\ =\frac{\left(340+10\right)}{\left(340+10\right)+30}\times 500 \\ =458\mathrm{Hz} \end{gathered}$$