Question

A train is moving at 30ms${}^{-1}$ in still air. The frequency of the locomotive whistle is 500 Hz and the speed of sound is 345 ms${}^{-1}$. The apparent wavelengths of sound infront of and behind the locomotive are respectively

• A. 0.63 m, 0.80 m
• B. 0.63 m, 0.75 m
• C. 0.60 m, 0.85 m
• D. 0.60 m, 0.75 m

Answer 1

Answer: (B)

0.63 m, 0.75 m

Frequency in front will be due to sum of both velocities and behind will be due to difference of velocities.
$\frac{f_1}{f_2}=\frac{v+v_s}{v-v_s}=\frac{345+30}{345-30}$

$f_1=\left(\frac{v+v_s}{v}\right)f$

$\Rightarrow {\lambda }_1=\left(\frac{v}{v+v_s}\right)\lambda =\left(\frac{345}{375}\right)\frac{345}{500}$ $=0.63m$

Similarly: ${\lambda }_2=0.75m$