Question

Two loudspeakers $M$ and $N$ are located $20\, m$ apart and emit sound at frequencies $118 \,Hz$ and $121 \,Hz$, respectively. A car is initially at a point $P, 1800\, m$ away from the midpoint $Q$ of the line $MN$ and moves towards $Q$ constantly at $60\, km / hr$ along the perpendicular bisector of $MN$. It crosses $Q$ and eventually reaches a point $R , 1800\, m$ 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 $330\, ms ^{-1} .$ Which of the following statement(s) is(are) true regarding the sound heard by the person?

• A.

  

• B. The rate of change in beat frequency is maximum when the car passes through Q
• C. $V_P + V_R = 2V_Q$
• D.

  

Answer 1

The Correct Option is D

Beat frequencies in $f _{ B }=\frac{ V + V _{0} \cos \theta}{ V }\left[ f _{2}- f _{1}\right]$ $\because \cos \theta$ decreases with time because $\theta$ increases. So $f _{ B }$ decreases. $\frac{ df _{ B }}{ dt } =\frac{ d }{ dt }\left[1+\frac{ V _{0}}{ V } \cos \theta\right]\left( f _{2}- f _{1}\right) $ $=\left[0+\frac{ V _{0}}{ V }(\sin \theta)\left(\frac{ d \theta}{ dt }\right)\right]\left( f _{2}- f _{1}\right)$ $\because \theta$ increases so $\sin \theta$ also increase from 0 to $90^{\circ}$ and slope increases. So graph is (D). Rate of change in beat frequencies is maximum where $\theta=90^{\circ}$ or at $Q \cdot( B )$ is correct. At $P$ beats frequencies $v_{p}=\frac{V+V_{0}}{V}\left[f_{2}-f_{1}\right]$ At $R$ beats frequencies $v_{ k }=\frac{ V - V _{0}}{ V }\left[ f _{2}- f _{1}\right]$ $v_{ P }+ v _{ R } =\left(\frac{ V + V _{0}}{ V }+\frac{ V - V _{0}}{ V }\right)\left[ f _{2}- f _{1}\right] $ $=2\left( f _{2}- f _{1}\right) $ $v _{ p }+ v _{ R } =2{ }^{v}_{ Q }$ (C) Also correct.