A train moves towards a stationary observer with speed $34 \mathrm{m} {\mathrm{s}}^{-1}$. The train blows whistle and its frequency is registered by the observer as ${\nu }_1$. If the train's speed is reduced to $17 \mathrm{m} {\mathrm{s}}^{-1},$ then the frequency registered is ${\nu }_2$. If the speed of sound is $340 \mathrm{m} {\mathrm{s}}^{-1}$, then the ratio $\frac{{\nu }_1}{{\nu }_2}$ is

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

$\nu =\frac{C}{C-v_s}{\nu }_0$

Case (I) $v_s=34 \mathrm{m} {\mathrm{s}}^{-1}$, where $C=$Speed of sound in air,

$v_s=$Speed of source, and ${\nu }_0=$ frequency of the source.

$\therefore {\nu }_1=\left(\frac{340}{340-34}\right){\nu }_0$

$=\frac{340}{306}{\nu }_{0 }....\left(\mathrm{i}\right)$

Case (ii) $v_s=17 \mathrm{m} {\mathrm{s}}^{-1}$
$\therefore {\nu }_2=\frac{340}{340-17}{\nu }_0=\frac{340}{323}{\nu }_0 ....\left(\mathrm{ii}\right)$

From Eq. (i) and (ii), we get

$\therefore \frac{{\nu }_1}{{\nu }_2}=\frac{340/306}{340/323}=\frac{323}{306}=\frac{19}{18}$