Beat Frequency Variation for Moving Observer between Sound Sources

To determine the correct statements, let us analyze the Doppler effect on the frequencies heard by the observer moving in the car.

1. Geometry and Coordinate Setup

Let the midpoint $Q$ of the line segment $MN$ be at the origin $(0,0)$. The two sound sources $M$ and $N$ are positioned at coordinates: $M = (0, d) \quad \text{and} \quad N = (0, -d)$ where $2d = 20\text{ m} \implies d = 10\text{ m}$.

The car moves along the perpendicular bisector (the $x$-axis) from $P(-1800\text{ m}, 0)$ through $Q(0,0)$ to $R(1800\text{ m}, 0)$ at a constant speed $v_0 = 60\text{ km/h} = \frac{50}{3}\text{ m s}^{-1}$.

At any instant, let the position of the observer be $(x, 0)$. The speed of sound is $v_s = 330\text{ m s}^{-1}$.

2. Doppler Shifted Frequencies

The unit vector directed from $M$ to the observer at $(x,0)$ is: $\hat{r}_M = \frac{x\hat{i} - d\hat{j}}{\sqrt{x^2 + d^2}}$

The velocity vector of the observer is: $\vec{v} = v_0 \hat{i}$

The apparent frequency $f_M'$ received from source $M$ (frequency $f_M = 118\text{ Hz}$) is: $f_M' = f_M \left( 1 - \frac{\vec{v} \cdot \hat{r}_M}{v_s} \right) = f_M \left( 1 - \frac{v_0}{v_s} \frac{x}{\sqrt{x^2 + d^2}} \right)$

Similarly, for source $N$ (frequency $f_N = 121\text{ Hz}$): $f_N' = f_N \left( 1 - \frac{v_0}{v_s} \frac{x}{\sqrt{x^2 + d^2}} \right)$

3. Beat Frequency

The beat frequency detected by the observer at position $x$ is given by: $\nu(x) = |f_N' - f_M'| = (f_N - f_M) \left( 1 - \frac{v_0}{v_s} \frac{x}{\sqrt{x^2 + d^2}} \right)$

Let $\Delta f = f_N - f_M = 121 - 118 = 3\text{ Hz}$. Thus, $\nu(x) = \Delta f \left( 1 - \frac{v_0}{v_s} \frac{x}{\sqrt{x^2 + d^2}} \right)$

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Evaluation of Statements

Statement (A):

At position $P$, $x = -x_0 = -1800\text{ m}$: $\nu_P = \Delta f \left( 1 + \frac{v_0}{v_s} \frac{x_0}{\sqrt{x_0^2 + d^2}} \right)$

At position $Q$, $x = 0$: $\nu_Q = \Delta f (1 - 0) = \Delta f$

At position $R$, $x = x_0 = 1800\text{ m}$: $\nu_R = \Delta f \left( 1 - \frac{v_0}{v_s} \frac{x_0}{\sqrt{x_0^2 + d^2}} \right)$

Adding $\nu_P$ and $\nu_R$: $\nu_P + \nu_R = \Delta f \left( 1 + \frac{v_0}{v_s} \frac{x_0}{\sqrt{x_0^2 + d^2}} \right) + \Delta f \left( 1 - \frac{v_0}{v_s} \frac{x_0}{\sqrt{x_0^2 + d^2}} \right) = 2 \Delta f = 2 \nu_Q$ Thus, Statement (A) is correct.

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Statement (B):

The rate of change of beat frequency with respect to time $t$ is: $\frac{d\nu}{dt} = \frac{d\nu}{dx} \cdot \frac{dx}{dt} = v_0 \frac{d\nu}{dx}$

Differentiating $\nu(x)$ with respect to $x$: $\frac{d\nu}{dx} = -\Delta f \frac{v_0}{v_s} \frac{d}{dx} \left( \frac{x}{\sqrt{x^2 + d^2}} \right) = -\Delta f \frac{v_0}{v_s} \frac{d^2}{(x^2 + d^2)^{3/2}}$

Therefore: $\frac{d\nu}{dt} = -\Delta f \frac{v_0^2 d^2}{v_s (x^2 + d^2)^{3/2}}$

elements in the magnitude of rate of change: $\left| \frac{d\nu}{dt} \right| = \frac{\Delta f v_0^2 d^2}{v_s (x^2 + d^2)^{3/2}}$

This magnitude attains its maximum value when the denominator $(x^2 + d^2)^{3/2}$ is minimized, which occurs at $x = 0$ (point $Q$). Thus, Statement (B) is correct.

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Statements (C) and (D):

From the expression of $\frac{d\nu}{dt}$:

1. Far away from $Q$ (at points $P$ and $R$), $x = \pm 1800\text{ m} \gg d$, making $\left| \frac{d\nu}{dt} \right| \approx 0$. Hence, the slope of $\nu(t)$ vs $t$ curve is nearly horizontal near $P$ and $R$.
2. At $Q$ ($x = 0$), the negative slope $\frac{d\nu}{dt}$ has its maximum magnitude (steepest decay).

This indicates an $S$-shaped (sigmoid) curve with an inflection point at $Q$, which matches the schematic curve shown in option C.

Thus, Statement (C) is correct and Statement (D) is incorrect.