QJEEAdv16Ph9b9d: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph9b9d Solution: Two loudspeakers M and N are located 20m apart and emit sound…

Consider the time $t\left(\leq t_Q\right)$ when car is at $S_1$ between $P$ and $Q$ . The distance travelled by the car in time $t$ is $\mathrm{PS}_1=u t$ . At this instant, the lines $\mathrm{S}_1{N}$ and $\mathrm{S}_1 \mathrm{M}$ both make angle $\theta$ with the velocity vector $\vec{u}$ . The component of observer (person sitting in the car) velocity towards the sources $N$ and $M$ is $u_o=u \cos \theta$ . The sources $N$ and $M$ are at rest i.e., $u_s=0$ . Apply Doppler's effect equation to get frequencies of the sources $N$ and $M$ heard by the observer as

$\begin{aligned}\nu_N^{\prime} & =\frac{v+u_o}{v-u_s} \nu_N=\frac{v+u \cos \theta}{v} \nu_N \\\\& =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right) \nu_N, \\\\\nu_M^{\prime} & =\frac{v+u_o}{v-u_s} \nu_M=\frac{v+u \cos \theta}{v} \nu_M \\\\& =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right) \nu_M .\end{aligned}$

The beat frequency heard by the observer at time $t(\leq$ $\left.t_Q\right)$ is

$\begin{aligned}\nu(t) & =\nu_N^{\prime}-\nu_M^{\prime} \\\\& =\left(1+\frac{u}{v} \frac{D-u t}{\sqrt{d^2+(D-u t)^2}}\right)\left(\nu_N-\nu_M\right) ........(1)\end{aligned}$

Now, consider the time $t\left(\geq t_Q\right)$ when car is at $S_2$ between $Q$ and $R$ . The distance travelled by the car in time $t$ is $\mathrm{PS}_2=u t$ . At this instant, the lines $\mathrm{S}_2 \mathrm{~N}$ and $\mathrm{S}_2 \mathrm{M}$ both make angle $\left(180^{\circ}-\theta\right)$ with the velocity vector $\vec{u}$ . The component of observer velocity towards the sources $N$ and $M$ is $u_o=-u \cos \theta$ . Apply Doppler's effect equation to get

$\begin{aligned}\nu_N^{\prime} & =\frac{v+u_o}{v-u_s} \nu_N=\frac{v-u \cos \theta}{v} \nu_N \\\\& =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right) \nu_N, \\\\\nu_M^{\prime} & =\frac{v+u_o}{v-u_s} \nu_M=\frac{v-u \cos \theta}{v} \nu_M \\\\& =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right) \nu_M .\end{aligned}$

The beat frequency heard by the observer at time $t(\geq$ $\left.t_Q\right)$ is

$\begin{aligned}\nu(t) & =\nu_N^{\prime}-\nu_M^{\prime} \\\\& =\left(1-\frac{u}{v} \frac{u t-D}{\sqrt{d^2+(u t-D)^2}}\right)\left(\nu_N-\nu_M\right) ..........(2)\end{aligned}$

Substitute $t=0$ and $t=t_Q={D \over u}$ in equation (1) to get $\nu_P$ and $\nu_Q$ and substitute $t=t_R=2 {D \over u}$ in equation (2) to get $\nu_R$ i.e.,

$\begin{aligned}& \nu_P=\nu(t=0)=\left(1+\frac{u}{v} \frac{D}{\sqrt{d^2+D^2}}\right)\left(\nu_N-\nu_M\right) \\\\& \nu_Q=\nu(t=D / u)=\left(\nu_N-\nu_M\right) \\\\& \nu_R=\nu(t=2 D / u)=\left(1-\frac{u}{v} \frac{D}{\sqrt{d^2+D^2}}\right)\left(\nu_N-\nu_M\right)\end{aligned}$

$\therefore$ $\nu_P+\nu_R=2 \nu_Q$

Differentiate equations (1) w.r.t. time t to get rate of change of beat frequency

$\frac{\mathrm{d} \nu(t)}{\mathrm{d} t}=-\left(\nu_N-\nu_M\right) \frac{u^2}{v} \frac{d^2}{\left(d^2+(D-u t)^2\right)^{3 / 2}}$ ...........(3)

From equation (3), the slope is negative and its magnitude is maximum when $t={D \over u}=t_Q$ (denominator is minimum). Thus the rate of change of beat frequency is maximum when car passes through $Q$ .