A train is approaching with velocity $25ms^{- 1}$towards a pedestrian standing on the track, frequency of horn of train is 1 kHz. Frequency heard by the pedestrian is $\left( v = 30ms^{- 1} \right)$

Question

Dr. Abubacker · Accepted Answer

(a) Since, train (source) is moving towards pedestrian (observer), the perceived frequency will be higher than the original.
$v^{'} = v\left( \frac{v + v_{o}}{v - v_{s}} ight)$
Here, $v_{o} = o$(as observer is stationary)
$v_{s} = 25ms^{- 1}$(velocity of source)
$v = 350\ ms^{- 1}$ (velocity of sound)
And v=1kHz (original frequency)
Hence,
$v^{'} = 1000\left( \frac{350 + 0}{350 - 25} ight)$
$= 1000 	imes \frac{350}{325} = 1077\ Hz$

Q. A train is approaching with velocity $25ms^{- 1}$towards a pedestrian standing on the track, frequency of horn of train is 1 kHz. Frequency heard by the pedestrian is $\left( v = 30ms^{- 1} \right)$

Explanation

Reviewed by Dr. Abubacker

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