Q. 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 $345ms^{- 1}.$ The apparent wavelength of sound in front of and behind the locomotive are respectively
A) $0.80m,\ 0.63m$
B) $0.63m,\ 0.80m$
C) $0.50m,\ 0.85m$
D) $0.63m,\ 0.75m$
Explanation
(d) If front of locomotive, $\lambda^{'} = \frac{\lambda(v - v_{s})}{v} = \frac{v - v_{s}}{n}$
$\therefore\lambda^{'} = \frac{345 - 30}{500} = \frac{315}{500} = 0.63m$
Behind locomotive, $\lambda^{''} = \frac{\lambda}{v}\left( v + v_{s} \right) = \frac{v + v_{s}}{n}$
$\therefore\lambda^{''} = \frac{345 + 30}{500} = \frac{375}{500} = 0.75m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.