Hence by using \(n' = n\left( {\frac{{v + {v_0}}}{v}} \right)\)
When passenger is sitting in train \(A\), then
\(5.5 = 5\left( {\frac{{v + {v_A}}}{v}} \right)\) …\((i)\)
when passenger is sitting in train \(B\), then
\(6 = 5\left( {\frac{{v + {v_B}}}{v}} \right)\) …\((ii)\)
On solving equation \((i)\) and \((ii)\) we get \(\frac{{{v_B}}}{{{v_A}}} = 2\)
$ {y_1} = {10^{ - 6}}\sin [100\,t + (x/50) + 0.5]\;m $
$ {y_2} = {10^{ - 6}}\cos \,[100\,t + (x/50)]\;m $
જ્યાં $x$ મીટરમાં હોય અને $t$ સેકન્ડમાં છે
$y=\left(10 \cos \pi x \sin \frac{2 \pi t }{ T }\right) cm$
$x=\frac{4}{3} cm$ આગળ રહેલા કણનો કંપવિસ્તાર.....$cm$ હશે