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Q15.21:
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 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 $s^{-1}$

Solution :

Part 1: Analysis of Sound Waves with Wind

Given parameters for the source (whistle) and the medium (air with wind):

  • Frequency of source ($f_0$): $400 \text{ Hz}$
  • Speed of sound in still air ($v$): $340 \text{ m/s}$
  • Speed of wind ($w$): $10 \text{ m/s}$ (in the direction of the observer)
  • Velocity of source ($v_s$): $0 \text{ m/s}$ (train is stationary)
  • Velocity of observer ($v_o$): $0 \text{ m/s}$ (standing on the platform)

Step 1: Calculating the Effective Speed of Sound

Since the wind blows from the source to the observer, the wind velocity adds vectorially to the speed of sound. [Per the Galilean transformation of velocities in a medium]

$v' = v + w = 340 \text{ m/s} + 10 \text{ m/s} = 350 \text{ m/s}$

Step 2: Calculating Wavelength ($\lambda$)

The wavelength is determined by the source and the medium it travels through. [Per the definition of wavelength $\lambda = \frac{v_{medium}}{f}$]

$\lambda = \frac{v'}{f_0} = \frac{350 \text{ m/s}}{400 \text{ Hz}} = 0.875 \text{ m}$

Step 3: Calculating Observed Frequency ($f'$)

As both source and observer are stationary relative to the ground, but the effective wave speed has increased, the frequency perceived by the observer is:

$f' = \frac{v'}{\lambda} = \frac{350 \text{ m/s}}{0.875 \text{ m}} = 400 \text{ Hz}$

Source Wind $w = 10 \text{ m/s}$ Observer

Part 2: Comparison with Observer Moving Towards the Yard

Consider the scenario where air is still ($w = 0$) and the observer moves towards the source with $v_o = 10 \text{ m/s}$.

Step 4: Calculations for Moving Observer

  • Speed of sound: $v = 340 \text{ m/s}$
  • Wavelength in still air: $\lambda = \frac{340}{400} = 0.85 \text{ m}$
  • Observed frequency ($f''$): Using the Doppler Effect formula $f'' = f_0 \left( \frac{v + v_o}{v} \right)$
  • $f'' = 400 \left( \frac{340 + 10}{340} \right) = 400 \left( \frac{350}{340} \right) \approx 411.76 \text{ Hz}$

Step 5: Comparative Analysis

Parameter Wind-blown (Wind = 10 m/s) Observer Moving (10 m/s)
Speed of Sound $350 \text{ m/s}$ $340 \text{ m/s}$
Wavelength $0.875 \text{ m}$ $0.85 \text{ m}$
Observed Frequency $400 \text{ Hz}$ $411.76 \text{ Hz}$

Conclusion: The situations are not identical. In the case of wind, the wavelength changes because the medium itself is in motion relative to the source, keeping the frequency constant. In the case of a moving observer, the speed of sound relative to the observer changes, causing a Doppler shift in frequency. [Per the Doppler Effect Principle for moving observers vs. moving medium]

Final Solution: For wind blowing: Frequency = 400 Hz, Wavelength = 0.875 m, Speed = 350 m/s. For the second case: Frequency = 411.76 Hz, Wavelength = 0.85 m, Speed = 340 m/s. The situations are physically distinct.


More Questions from Class 11 Physics Waves EXERCISES


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