A train moving towards a hill at a speed of 7 | vedclass

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(A) Given: Speed of train (source) $v_s = 72\ km/hr = 72 	imes (5/18) = 20\ m/s$. Speed of wind $w = 36\ km/hr = 36 	imes (5/18) = 10\ m/s$. Speed o

Vedclass · Accepted Answer

$532.5$

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(A) Given: Speed of train (source) $v_s = 72\ km/hr = 72 	imes (5/18) = 20\ m/s$. Speed of wind $w = 36\ km/hr = 36 	imes (5/18) = 10\ m/s$. Speed of sound $v = 340\ m/s$. Frequency of source $n = 500\ Hz$. Since the wind is blowing from the hill towards the train,the effective speed of sound relative to the ground is $(v - w)$ for the sound traveling towards the hill. However,for the Doppler effect formula,we consider the velocity of sound relative to the medium. The observer is on the hill (at rest,$v_o = 0$). The source is moving towards the observer. The effective velocity of sound relative to the observer is $v_{eff} = v - w = 340 - 10 = 330\ m/s$. The formula for apparent frequency is $n' = n \left[ \frac{v_{eff}}{v_{eff} - v_s} ight]$. Substituting the values: $n' = 500 \left[ \frac{330}{330 - 20} ight] = 500 \left[ \frac{330}{310} ight] = 500 	imes 1.0645 \approx 532.25\ Hz$. Rounding to the nearest option,the frequency is $532.5\ Hz$.

$A$ train moving towards a hill at a speed of $72\ km/hr$ sounds a whistle of frequency $500\ Hz$. $A$ wind is blowing from the hill at a speed of $36\ km/hr$. If the speed of sound in air is $340\ m/s$,the frequency heard by a man on the hill (nearly) is ... $Hz$

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