Question

Two trains $A$ and $B$ simultaneously start moving along parallel tracks from a station along same direction. A starts with constant acceleration $2{\text{m/s}}^2$ from rest while $B$ with the same acceleration but with initial velocity of $40\text{m/s}$. Twenty seconds after the start, passenger of $A$ hears whistle of $B$. If frequency of whistle is $1194\text{Hz}$ and velocity of sound in air is $322\text{m/s}$, then the frequency observed by the passenger is

• A. $1186\text{Hz}$
• B. $1086\text{Hz}$
• C. $986\text{Hz}$
• D. $886\text{Hz}$

Answer 1

The correct option is B $1086\text{Hz}$


$f_{app}=f_0\left(\frac{v\pm v_0}{v\pm v_s}\right)$

$PR=0+\frac{1}{2}.2{.20}^2=400m$

$PQ=t^2$

$PS=40+\frac{1}{2}2t^2$

$QS=QR+RS$

$⟹PS-PQ=PR-PQ+x_{sound}(20-t)$
So,
$(40t+t^2)-t^2=400-t^2+322(20-t)$
$40t=400-t^2+6440-322t$
$t^2+362t-6840=0t=18$

$f_{app}=f_0\left(\frac{v\pm v_0}{v\pm v_s}\right)v_0\to 20s,v_s\to t=18s.$

$\begin{array}{cc}{v}_0& =0+2\times 20=40m/s\\ v_s& =u+at\\ & =40+2\times 18\\ & =76m/s\end{array}$

$f_{app}=1194\left(\frac{322+40}{322+76}\right)$
$⟹f_{app}=1086Hz$