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

Question

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=(n' +n/2) (B) n=√n' n'' (C) n=(2 n' n''/n' +n'') (D) n=(2 n' n''/n'-n'')

Tardigrade Admin · Accepted Answer

As, n'=(v π/v-vs) and n''=(v n/v+ vs) ∴ (n/n')=1-(v5/v), (n/n'')=1+(vs/v) Adding the two, we get (n/n')+(n/n'')=2 ∴ n=(2 n' n''/n' +n'')

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

$n = \frac{n ^{'} + n}{2}$

$n = n^{'} n^{''}$

$n = \frac{2 n ^{'} n ^{''}}{n ^{'} + n ^{''}}$

$n = \frac{2 n ^{'} n ^{''}}{n ^{'} - n ^{''}}$

As, $n^{'} = \frac{v π}{v - v _{s}}$ and $n^{''} = \frac{v n}{v + v _{s}}$
$∴ \frac{n}{n ^{'}} = 1 - \frac{v _{5}}{v},$
$\frac{n}{n ^{''}} = 1 + \frac{v _{s}}{v}$
Adding the two, we get
$\frac{n}{n ^{'}} + \frac{n}{n ^{''}} = 2$
$∴ n = \frac{2 n ^{'} n ^{''}}{n ^{'} + n ^{''}}$

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

n=2n′+n​

n=n′n′′​

n=n′+n′′2n′n′′​

n=n′−n′′2n′n′′​