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 perpendicular bisector of $MN$. It crosses $Q$ and eventually reaches a point $R$, $1800m$ away from $Q$. Let $\nu \left(t\right)$ represent the beat frequency measured by a person sitting in the car at time $t$. Let ${\nu }_p$, ${\nu }_Q$ and ${\nu }_R$ be the beat frequencies measured at locations$P$, $Q$ 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. A 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. ${\nu }_P+{\nu }_R=2{\nu }_Q$
• D. The plot below represents schematically the variation of beat frequency with time
  

Answer 1

Answer: (B), (C), (D)

The plot below represents schematically the variation of beat frequency with time




Apparent frequency of $M$ and $N$ at $P$,

$f_m^{\prime }=\frac{v+v_0}{v}{f}_m{f}_m^{\prime }=\frac{330+\frac{50}{3}}{330}118=\frac{1040}{990}118f_N^{\prime }=\frac{v+v_0}{v}{f}_N=\frac{1040}{990}121f_P=f_N^{\prime }-{f^{\prime }}_m=\frac{1040}{330}Hz$

Apparent frequency of $M$ and $N$ at $R$,

$f_m^{\prime }=\frac{v-v_0}{v}{f}_m{f}_m^{\prime }=\frac{330-\frac{50}{3}}{330}118=\frac{940}{990}118f_N^{\prime }=\frac{v-v_0}{v}{f}_N=\frac{940}{990}121f_R=f_N^{\prime }-{f^{\prime }}_m=\frac{940}{330}Hz$
Beat frequency at $Q$,

$f_Q=121-118=3Hz$
$f_P+f_R=2f_Q$
hence, option (c) is correct.

velocity of car at point $M$.
$v=3\left[\frac{110+vcos\theta }{110}\right]|\frac{dv}{dt}|=\frac{v}{110}sin\theta \frac{d\theta }{dt}$
$\theta =1$ is maximum at ${90}^o$, so option (b) is correct.

Now, $tan\alpha =\frac{10}{x}\frac{d\alpha }{dt}=\frac{10v}{x^2{sec}^2\alpha }x=10cot\alpha |\frac{dv}{dt}|=\frac{v}{110}sin\alpha \left(\frac{10v}{x^2{sec}^2\alpha }\right)|\frac{dv}{dt}|=k{sin}^3\alpha$
at $|\frac{dv}{dt}|={90}^o$ is maximum at $Q$.
So, option (d) is correct.
Hence, option B, C and D matches correctly.