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? - Physics

Sum

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 vs = 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 vm = 36 \text { kmh}^\(-)1 =$36 \times \frac{5}{18} = 10 {\text { ms }}^{- 1}$

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

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

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

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

$= 458 \text{ Hz }$

Concept: Wave Motion
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 80 | Page 357