Step-by-step Derivations

1. Setting up the coordinate system and variables:

Let the midpoint Q of the line segment MN be the origin (0,0). The car moves along the x-axis. The loudspeakers M and N are located on the y-axis at (0, a) and (0, -a) respectively, where $2a = 20$ m, so $a=10$ m.

• Position of source M: $(0, a) = (0, 10)$
• Position of source N: $(0, -a) = (0, -10)$
• Position of the car at any point: $(x, 0)$
• The car moves from P ($x = -1800$ m) to R ($x = 1800$ m).
• Speed of the car: $v = 60 \text{ km/h} = 60 \times \frac{1000}{3600} \text{ m/s} = \frac{50}{3} \text{ m/s}$.
• Frequencies of sources: $f_M = 118$ Hz, $f_N = 121$ Hz.
• Speed of sound in air: $v_s = 330$ m/s.

2. Applying the Doppler Effect:

The apparent frequency $f'$ heard by an observer moving with velocity $\vec{v}_{obs}$ from a stationary source emitting frequency $f_0$ is given by $f' = f_0 \left(1 + \frac{v_{comp}}{v_s}\right)$, where $v_{comp}$ is the component of the observer's velocity towards the source.

• The car's velocity is $\vec{v}_{car} = (v, 0)$.
• The vector from the car at $(x,0)$ to the source M at $(0,a)$ is $\vec{r}_{CM} = (0-x, a-0) = (-x, a)$.
• The unit vector in this direction is $\hat{r}_{CM} = \frac{(-x, a)}{\sqrt{x^2+a^2}}$.
• The component of the car's velocity towards M is $v_{comp, M} = \vec{v}_{car} \cdot \hat{r}_{CM} = (v,0) \cdot \frac{(-x, a)}{\sqrt{x^2+a^2}} = \frac{-vx}{\sqrt{x^2+a^2}}$.
• Similarly, for source N, the vector from the car to N is $\vec{r}_{CN} = (-x, -a)$, and the component of velocity towards N is $v_{comp, N} = \frac{-vx}{\sqrt{x^2+a^2}}$.

The apparent frequencies heard from M and N are: $f'_M = f_M \left(1 + \frac{v_{comp, M}}{v_s}\right) = f_M \left(1 - \frac{vx}{v_s\sqrt{x^2+a^2}}\right)$ $f'_N = f_N \left(1 + \frac{v_{comp, N}}{v_s}\right) = f_N \left(1 - \frac{vx}{v_s\sqrt{x^2+a^2}}\right)$

3. Deriving the Beat Frequency:

The beat frequency $v(x)$ at position $x$ is the magnitude of the difference between the apparent frequencies: $v(x) = |f'_N - f'_M| = |(f_N - f_M) \left(1 - \frac{vx}{v_s\sqrt{x^2+a^2}}\right)|$ Let $\Delta f = f_N - f_M = 121 - 118 = 3$ Hz. The term $\frac{v}{v_s} = \frac{50/3}{330} = \frac{50}{990} \approx 0.05$, and $\frac{|x|}{\sqrt{x^2+a^2}} < 1$. Thus, the term inside the parenthesis is always positive. $v(x) = \Delta f \left(1 - \frac{vx}{v_s\sqrt{x^2+a^2}}\right)$ Since $x$ increases with time $t$, we can analyze the function $v(x)$ to understand $v(t)$.

Evaluating the Options

C: $v_P + v_R = 2v_Q$

• At point P, $x = -1800$ m. $v_P = v(-1800) = \Delta f \left(1 + \frac{1800v}{v_s\sqrt{(-1800)^2+a^2}}\right)$.
• At point Q, $x = 0$ m. $v_Q = v(0) = \Delta f (1 - 0) = \Delta f = 3$ Hz.
• At point R, $x = 1800$ m. $v_R = v(1800) = \Delta f \left(1 - \frac{1800v}{v_s\sqrt{(1800)^2+a^2}}\right)$.

Summing $v_P$ and $v_R$: $v_P + v_R = \Delta f \left(1 + \frac{1800v}{v_s\sqrt{1800^2+a^2}} + 1 - \frac{1800v}{v_s\sqrt{1800^2+a^2}}\right) = 2\Delta f$ Since $v_Q = \Delta f$, we have $v_P + v_R = 2v_Q$. Thus, statement C is true.

B: The rate of change in beat frequency is maximum when the car passes through Q.

The rate of change is $\frac{dv}{dt}$. Using the chain rule, $\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$. First, we find $\frac{dv}{dx}$: $\frac{dv}{dx} = -\frac{\Delta f v}{v_s} \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+a^2}}\right) = -\frac{\Delta f v}{v_s} \left(\frac{a^2}{(x^2+a^2)^{3/2}}\right)$ So, the rate of change with time is: $\frac{dv}{dt} = -\frac{\Delta f v^2 a^2}{v_s (x^2+a^2)^{3/2}}$ The magnitude of this rate is $|\frac{dv}{dt}| = \frac{\Delta f v^2 a^2}{v_s (x^2+a^2)^{3/2}}$. This magnitude is maximum when the denominator $(x^2+a^2)^{3/2}$ is minimum. This occurs at $x=0$, which is point Q. Thus, statement B is true.

A and D: The plot of beat frequency with time.

To determine the shape of the plot of $v(t)$ vs $t$, we analyze its first and second derivatives.

• First derivative (slope): $\frac{dv}{dt}$ is always negative, so $v(t)$ is a monotonically decreasing function.
• Second derivative (concavity): $\frac{d^2v}{dt^2} = \frac{d}{dt}\left(\frac{dv}{dt}\right) = v \frac{d}{dx}\left(\frac{dv}{dt}\right)$. $\frac{d}{dx}\left( -\frac{C}{(x^2+a^2)^{3/2}} \right) = -C \left(-\frac{3}{2}\right)(x^2+a^2)^{-5/2}(2x) = \frac{3Cx}{(x^2+a^2)^{5/2}}$ where $C = \frac{\Delta f v^2 a^2}{v_s}$ is a positive constant. $\frac{d^2v}{dt^2} = v \left( \frac{3Cx}{(x^2+a^2)^{5/2}} \right) = \frac{3vCx}{(x^2+a^2)^{5/2}}$ The sign of $\frac{d^2v}{dt^2}$ is the same as the sign of $x$.
• For the first half of the journey (P to Q), $x < 0$, so $\frac{d^2v}{dt^2} < 0$. The graph is concave down.
• For the second half of the journey (Q to R), $x > 0$, so $\frac{d^2v}{dt^2} > 0$. The graph is concave up.
• At Q ($x=0$), the second derivative is zero, indicating an inflection point.

Now we check the plots:

• Plot D: Shows a decreasing function. The first half is concave down (slope becomes more negative), and the second half is concave up (slope becomes less negative). This matches our analysis. Thus, statement D is true.
• Plot A: Shows a decreasing function. The first half is concave up, and the second half is concave down. This contradicts our analysis. Thus, statement A is false.

Conclusion: Statements B, C, and D are true.