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Academic session 2026-27, Chapter 14 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Physics - Chapter Waves. This page offers a step-by-step solution to the specific question from Exercise 1, Question 21:
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 m s-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station's platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s-1? The speed of sound in still air can be taken as 340 m s-1.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 m s-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station's platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s-1? The speed of sound in still air can be taken as 340 m s-1.
For the stationary observer: 400 Hz; 0.875 m; 350 m/s
For the running observer: Not exactly identical
For the stationary observer:
Frequency of the sound produced by the whistle, V = 400 Hz
Speed of sound = 340 m/s
Velocity of the wind, v = 10 m/s
As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e., 400 Hz.
The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,
Effective speed of the sound, ve = 340 + 10 = 350 m/s
The wavelength (λ) of the sound heard by the observer is given by the relation:
λ = ve / V = 350 / 400 = 0.875 m
For the running observer:
Velocity of the observer, vo = 10 m/s
The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency (v').
This is given by the relation:
v' = (v+vo / v ) V
= (340+10 / 340) x 400 = 411.76 Hz
Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s
The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875 m.
Hence, the given two situations are not exactly identical.
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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 21: A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind star....
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