15.21 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

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Vishal Baghel | Contributor-Level 10

5 months ago

(a) For the stationary observer:

Frequency of the sound produced by the whistle, = 400 Hz

Speed of sound = 340 m/s

Velocity of 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 towards the observer. Hence, the effective speed of the sound increases by 10 units, i.e.

Effective speed of the sound, $v_{e}$ = 340 + 10 = 350 m/s

The wavelength ( $λ) o f t h e s o u n d$ heard by the observer is given by the relation:

$λ = \frac{v_{e}}{ν}$ = $\frac{350}{400}$ = 0.875 m

(b) For the running Observer:

$V e l o c i t y o f t h e o b s e r v e r$ ,&nb

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(a) For the stationary observer:

Frequency of the sound produced by the whistle, = 400 Hz

Speed of sound = 340 m/s

Velocity of wind, v = 10 m/s

Effective speed of the sound, $v_{e}$ = 340 + 10 = 350 m/s

The wavelength ( $λ) o f t h e s o u n d$ heard by the observer is given by the relation:

$λ = \frac{v_{e}}{ν}$ = $\frac{350}{400}$ = 0.875 m

(b) For the running Observer:

$V e l o c i t y o f t h e o b s e r v e r$ , $v_{o}$ = 10 m/s

$T h e o b s e r v e r$ is moving towards the source. As a result of the relative motions of the source and the observer, there is a change in frequency (f’’’)

$T h i s i s g i v e n b y t h e r e l a t i o n :$ f’’’ = ( $\frac{v + v_{o}}{v}) ν$ = ( $\frac{340 + 10}{340}) \times 400 =$ 411.76 Hz

Since the air is still, the effective speed of sound = 340 + 0 = 340 m/s

The source is at rest. Therefore, 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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(a) For the stationary observer:Frequency of the sound produced by the whistle,  = 400 HzSpeed of sound = 340 m/sVelocity of wind, v = 10 m/sAs 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 towards the observer. Hence, the effective speed of the sound increases by 10 units, i.e.Effective speed of the sound, <math><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></math> = 340 + 10 = 350 m/sThe wavelength ( <math><mi>λ</mi><mo>)</mo><mi></mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">f</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">u</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">d</mi></math> heard by the observer is given by the relation:<math><mi>λ</mi><mo>=</mo><mi></mi><mfrac><mrow><mrow><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></mrow></mrow><mrow><mrow><mi>ν</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>350</mn></mrow></mrow><mrow><mrow><mn>400</mn></mrow></mrow></mfrac></math> = 0.875 m (b) For the running Observer:<math><mi mathvariant="normal">V</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">y</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">f</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">b</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">v</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">r</mi></math> , <math><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>o</mi></mrow></mrow></msub></math> = 10 m/s<math><mi mathvariant="normal">T</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">b</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">v</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">r</mi></math> is moving towards the source. As a result of the relative motions of the source and the observer, there is a change in frequency (f’’’)<math><mi mathvariant="normal">T</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">g</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">v</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">b</mi><mi mathvariant="normal">y</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">n</mi><mo>:</mo></math> f’’’ = ( <math><mfrac><mrow><mrow><mi>v</mi><mo>+</mo><msub><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>o</mi></mrow></mrow></msub></mrow></mrow><mrow><mrow><mi>v</mi></mrow></mrow></mfrac><mo>)</mo><mi></mi><mi>ν</mi></math> = ( <math><mfrac><mrow><mrow><mn>340</mn><mo>+</mo><mn>10</mn></mrow></mrow><mrow><mrow><mn>340</mn></mrow></mrow></mfrac><mo>)</mo><mo>×</mo><mn>400</mn><mo>=</mo></math> 411.76 HzSince the air is still, the effective speed of sound = 340 + 0 = 340 m/sThe source is at rest. Therefore, the wavelength of the sound will not change, i.e. <math><mi></mi><mi>λ</mi></math> remains 0.875 m.Hence, the given two situations are not exactly identical.

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