Two loudspeakers M and N are located 20m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively…

1. The plot below represents schematically the variation of beat frequency with time
   
2. The plot below represents schematically the variations of beat frequency with time
   
3. The rate of change in beat frequency is maximum when the car passes through Q
4. ${\nu }_P+{\nu }_R=2{\nu }_Q$

Solution

$f_M=\frac{C+V\cos \theta }{C} f_1$
$f_N=\frac{C+V\cos \theta }{C} f_2$
$\mathrm{\Delta }f=f_N-f_M$
$=\frac{C+V\cos \theta }{C} \left(f_2-f_1\right)$
$\frac{d\left(\mathrm{\Delta }f\right)}{dt}= -\frac{V}{C} \left(f_2-f_1\right)\sin \theta \frac{d\theta }{dt}$
$\therefore$ and $\frac{d\left(\mathrm{\Delta }f\right)}{dt}$ is maximum when $\theta ={90}^o$
$\left[\therefore \left(iii\right) is correct\right]$
$v_P=\left(1+\frac{V}{C}\cos \theta \right)\mathrm{\Delta }f$
$v_Q=\mathrm{\Delta }f$
$v_R=\left(1-\frac{V}{C}\cos \theta \right)\mathrm{\Delta }f$
$\therefore$ $v_P+v_r=2v_Q$ `

Asked in: JEE Advanced 2016 (Paper 1)

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