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The frequency of the whlstle of a train Is 512 Hz . Thetrain crosses a station at a speed of 72 km . h^(-1) Calculate the frequency of the sound heard by a Iistener, standing on the platform , before and after the train crosses the station. Neglect the effect of wind . Velocity of sound is 336 , m s^(-1) .

Answer» <html><body><p></p>Solution :Velocity of the <a href="https://interviewquestions.tuteehub.com/tag/train-1425202" style="font-weight:bold;" target="_blank" title="Click to know more about TRAIN">TRAIN</a> ` = 72 km . `h^(-1)`<br/> ` = (72 xx1000)/(60xx60) m. s^(-1)` <br/> `= 20 m. s^(-1)`<br/> When the train approaches the station , the <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> covered hy <a href="https://interviewquestions.tuteehub.com/tag/512-1895092" style="font-weight:bold;" target="_blank" title="Click to know more about 512">512</a> sound waves = 336 - 20 = <a href="https://interviewquestions.tuteehub.com/tag/316-307105" style="font-weight:bold;" target="_blank" title="Click to know more about 316">316</a>` m<br/> ` therefore ` Apparent wavelength , ` lambda . = (316)/(512) m ` <br/> ` therefore ` Apparent frequencydue to Doppler effect<br/> ` = (" velocity of sound (V))/(lambda .)`<br/> `(336)/((316)/(512)) = (336xx512)/(316) = 544.4 Hz ` <br/> ` Again , when the train recedes from the station , the distance covered y 512 sound waves= 366 + 20 = <a href="https://interviewquestions.tuteehub.com/tag/356-1858306" style="font-weight:bold;" target="_blank" title="Click to know more about 356">356</a> m <br/>` therefore ` Apparent wavelength , ` lambda . = (356)/(512) m`<br/> ` therefore ` Apparent freaquency= `(V) /(lambda .) = (336)/((356)/(512)) = (336xx 512)/(356) = 483.2 Hz .`</body></html>


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