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  vP, vQ and vR  be the beat frequencies measured at location P,Q and R, respectively. The speed of sound in air is 330ms−1.  Which of the following statement (s) is (are) true regarding the sound heard by the person?

V0=velocity of observer = 60 km/hr =503 m/sec In between PQ With source M : n1=118V+V0cosθV=1181+503×330cosθ With source N :     n2=1211+503×330cosθ Beats n=n2−n1=31+53×33cosθ⇒n=533cosθ+3 Slope =dndθ=−533sinθ⇒dndtdtdθ=−533sinθ⇒dndt=−533dθdtsinθFrom P to Q :  θ is increasing & θ<π2 .  So , dndt=−Ve  and its magnitude is increasing. So initially , slope is maximum & then decreases .At Q ,  θ = π2 i.e.  sinθ is max , so slope is also maximum.From Q to R n1=118V−V0cosθVn2=121V−V0cosθVBeats=n=n2−n1⇒n=3−3V0Vcosθ=3−533cosθ Slope dndt=+3V0Vsinθdθdt As θ is ↓ from Q to R , so dθdt=−ve  &  also θ<π2dndt=−ve and its magnitude is decreasing.Vp=533cosθ+3  =3.1515Hz      put cosθ≃1VQ=533cosθ+3 = 3Hz               put cosθ=0VR=3−533cosθ=2.8484Hz        put cosθ≃1⇒VP+VR=2VQ