Q. When a train approaches a stationary observer, the apparent frequency of the whistle is $n^{'}$ and when the same train recedes away from the observer, the apparent frequency is$n"$. Then the apparent frequency n when the observer moves with the train is
A) $n = \frac{n^{'} + n}{2}$
B) $n = \sqrt{n^{'}n^{''}}$
C) $n = \frac{2n^{'}n^{''}}{n^{'} + n^{''}}$
D) $n = \frac{2n^{'}n^{''}}{n^{'} - n^{''}}$
Explanation
(c) $n^{'} = \frac{vn}{v - v_{s}},n^{''} = \frac{vn}{v + v_{s}}$
$\therefore\frac{n}{n^{'}} = 1 - \frac{v_{s}}{v},\frac{n}{n^{''}} = 1 + \frac{v_{s}}{v}$
Adding the two, we get
$\frac{n}{n^{'}} + \frac{n}{n^{''}} = 2\ \ \therefore\ \ n = \frac{2n^{'}n^{''}}{n^{'} + n^{''}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.