Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118Hz and 121 Hz, respectively. A car is initially at a point P, 1800m away from the midpoint Q of the line MN and moves towards Q and constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800m away from Q. Let v(t) represent the beat frequency measured by a person sitting in the car at time t. Let  $v_P$, $v_Q$ and $v_R$  be the beat frequencies measured at location P,Q and R, respectively. The speed of sound in air is 330$m{s}^{-1}$.  Which of the following statement (s) is (are) true regarding the sound heard by the person?
 

 

 

$V_0=$velocity of observer = 60 km/hr $=\frac{50}{3}m/\sec$

$\begin{array}{l}\begin{array}{l}\text{ In between }PQ\\ \text{ With source M :\hspace{1em}}{n}_1=118\left[\frac{V+V_0\cos \theta }{V}\right]\end{array}\\ \begin{array}{l}=118\left[1+\frac{50}{3\times 330}\cos \theta \right]\\ \text{ With source N} : n_2=121\left[1+\frac{50}{3\times 330}\cos \theta \right]\end{array}\\ \begin{array}{l}\text{ Beats n}=n_2-n_1=3\left[1+\frac{5}{3\times 33}\cos \theta \right]\\ \Rightarrow n=\frac{5}{33}\cos \theta +3\\ \text{ Slope }=\frac{dn}{d\theta }=\frac{-5}{33}\sin \theta \end{array}\\ \begin{array}{l}\Rightarrow \frac{dn}{dt}\frac{dt}{d\theta }=\frac{-5}{33}\sin \theta \\ \Rightarrow \frac{dn}{dt}=\frac{-5}{33}\frac{d\theta }{dt}\sin \theta \end{array}\end{array}$

From P to Q :  $\theta$ is increasing & $\theta <\frac{\pi }{2} . So , \frac{dn}{dt}=-Ve$  and its magnitude is increasing. So initially , slope is maximum & then decreases .

At Q ,  $\theta$ = $\frac{\pi }{2}$ i.e.  $\sin \theta$ is max , so slope is also maximum.
From Q to R 
$\begin{array}{l}\begin{array}{l}{n}_1=118\left[\frac{V-V_0\cos \theta }{V}\right]\\ n_2=121\left[\frac{V-V_0\cos \theta }{V}\right]\\ Beats=n=n_2-n_1\end{array}\\ \begin{array}{l}\Rightarrow n=3-3\left[\frac{V_0}{V}\cos \theta \right]=3-\frac{5}{33}\cos \theta \\ \text{ Slope }\frac{dn}{dt}=+\frac{3V_0}{V}\sin \theta \frac{d\theta }{dt}\end{array}\\ \begin{array}{l}\text{ As }\theta \text{ is }\downarrow \text{ from Q to R} , so \frac{d\theta }{dt}=-ve \text{\&} \text{ also }\theta <\frac{\pi }{2}\\ \frac{dn}{dt}=-ve and its magnitude is decrea\sin g.\end{array}\\ \begin{array}{l}{V}_p=\frac{5}{33}\cos \theta +3 =3.1515Hz \left[put \cos \theta \simeq 1\right]\\ V_Q=\frac{5}{33}\cos \theta +3 = 3Hz \left[put \cos \theta =0\right]\end{array}\\ \begin{array}{l}{V}_R=3-\frac{5}{33}\cos \theta =2.8484Hz \left[put \cos \theta \simeq 1\right]\\ \Rightarrow V_P+V_R=2V_Q\end{array}\end{array}$