Question

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

• A. The plot below represents schematically the variation of beat frequency with time
• B. The rate of change in beat frequency is maximum when the car passes through Q
• C. $V_P+V_R=2V_Q$
• D. The plot below represents schematically the variation of beat frequency with time

Answer 1

Answer: (B), (C), (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.
$\frac{d{f}_B}{dt}=\frac{d}{dt}\left[1+\frac{V_0}{V}\cos \theta \right]\left(f_2-f_1\right)$
$=\left[0+\frac{V_0}{V}(\sin \theta )\left(\frac{d\theta }{dt}\right)\right]\left(f_2-f_1\right)$
$\because \theta$ 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 $v_k=\frac{V-V_0}{V}\left[f_2-f_1\right]$
$v_P+v_R=\left(\frac{V+V_0}{V}+\frac{V-V_0}{V}\right)\left[f_2-f_1\right]$
$=2\left(f_2-f_1\right)$
$v_p+v_R=2{}_Q^v$
(C) Also correct.