A train moves towards a stationary observer with speed 34ms^ - 1 ....

According to Doppler's effect, the approximate frequency heard by the stationary observer,

$ν = \dfrac{C}{C - {v}_{s}}{ν}_{0}$

Case (I) ${v}_{s} = 34ms^{- 1}$, where& $C =$Speed of sound in air,

${v}_{s} =$ Speed of source, and& ${ν}_{0} =$ frequency of the source.

∴ {ν}_{1} = \left(\dfrac{340}{340 - 34}\right){ν}_{0}</p><p>=\dfrac{340}{306} {ν}_{0} . . . . \left( i\right)$</p><p>Case (ii)${v}_{s} = 17ms^{- 1}$<br />$∴ {ν}_{2} = $\dfrac{340}{340 - 17} {ν}_{0} = \dfrac{340}{323} {ν}_{0} . . . .\left(ii\right)</p><p>From Eq. (i)$ and (ii), we get

∴ $\dfrac{ν1}{ν2} = \dfrac{340 / 306}{340 / 323} = \dfrac{323}{306} = \dfrac{19}{18}$