Q. A train approaches a stationary observer, the velocity of train being $\frac{1}{20}$ of the velocity of sound. A sharp blast is blown with the whistle of the engine at equal intervals of a second. The interval between the successive blasts as heard by the observer is
A) $\frac{1}{20}s$
B) $\frac{1}{20}\min$
C) $\frac{19}{20}s$
D) $\frac{10}{20}\min$
Explanation
(c) The blast is blown at an interval of $1s$, so frequency $= 1Hz$
Frequency heard by the observer
$n^{'} = \frac{v}{v - v_{s}}.n = \frac{v}{v - \frac{v}{20}} \times 1 = \frac{v}{\frac{19v}{20}} = \frac{20}{19}Hz$
Therefore, observed time interval between two successive blasts $= \frac{1}{20/19} = \frac{19}{20}s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.