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 $Q$.