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

Q: Why is the angle $\theta$ the same for both speakers?

Because the car moves along the perpendicular bisector of the line segment joining the two speakers, the geometry is perfectly symmetric. The distance from the car to speaker M is always equal to the distance from the car to speaker N, making the angles identical.

Q: Why does the beat frequency decrease as the car moves from P to R?

As the car approaches Q, the component of its velocity towards the speakers ($v\cos\theta$) decreases because $\theta$ increases from $0^\circ$ to $90^\circ$. After passing Q, the car moves away, so the velocity component becomes negative, further reducing the apparent frequencies and thus the beat frequency.

Q: What is an inflection point and why does it occur at Q?

An inflection point is a point on a curve where the concavity changes. At Q, the rate of change of the beat frequency is at its maximum magnitude (steepest slope). This is where the curve transitions from curving downwards to curving upwards.

Q: Does the speed of sound affect the qualitative shape of the graph?

No, the speed of sound $C$ only acts as a scaling constant in the denominator. The qualitative S-shape of the curve and the location of the maximum rate of change at Q remain entirely unchanged.

Q: Why is the beat frequency at Q exactly equal to the difference in natural frequencies?

At Q, the car's motion is perpendicular to the direction of wave propagation from both speakers. Therefore, there is no Doppler shift for either frequency ($v\cos\theta = 0$), and the passenger hears the exact natural frequencies, resulting in a beat frequency of $121 - 118 = 3\text{ Hz}$.

Question

Two loudspeakers $M$ and $N$ are located $20	ext{ m}$ apart and emit sound at frequencies $118	ext{ Hz}$ and $121	ext{ Hz}$, respectively. A car is initially at a point $P$, $1800	ext{ m}$ away from the mid-point $Q$ of the line $MN$ and moves towards $Q$ constantly at $60	ext{ km/h}$ along the perpendicular bisector of $MN$. It crosses $Q$ and eventually reaches a point $R$, $1800	ext{ m}$ away from $Q$. Let $v(t)$ represent the beat frequency measured by a person sitting in the car at time $t$. Let $v_P$, $v_Q$ and $v_R$ be the beat frequencies measured at locations $P$, $Q$ and $R$ respectively. The speed of sound in air is $330	ext{ ms}^{-1}$. Which of the following statement(s) is (are) true regarding the sound heard by the person?

Options:

(A) The plot below represents schematically the variation of beat frequency with time (Plot A)

(B) The rate of change in beat frequency is maximum when the car passes through $Q$

(C) $v_P + v_R = 2v_Q$

(D) The plot below represents schematically the variations of beat frequency with time (Plot D)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a straight road, listening to two distinct musical notes playing simultaneously.

As you move, the pitch of each note shifts due to the **Doppler Effect**.

In this problem, we have two speakers, $M$ and $N$, separated by a distance of $20	ext{ m}$.

A car travels along the perpendicular bisector of the line segment $MN$, starting from a very distant point $P$, passing through the midpoint $Q$, and heading towards another distant point $R$.

Because the car is moving, the frequencies of the sound waves reaching the passenger are Doppler-shifted.

Since the sources are stationary and the observer is moving, the apparent frequency $f'$ is given by:

$$f' = f \left(1 \pm \frac{v_{	ext{app}}}{C}ight)$$

where $v_{	ext{app}}$ is the component of the car's velocity directed towards the source, and $C = 330	ext{ ms}^{-1}$ is the speed of sound.

Let's define $	heta$ as the angle between the car's velocity vector and the line connecting the car to either speaker.

By symmetry, this angle is identical for both speakers at any given instant.

Therefore, the component of the car's velocity directed towards each speaker is $v \cos	heta$.

### The Master Equation for Beat Frequency

As the car approaches the line $MN$, the apparent frequencies heard by the passenger from speakers $M$ and $N$ are:

$$f'_M = f_M \left(1 + \frac{v \cos	heta}{C}ight)$$

$$f'_N = f_N \left(1 + \frac{v \cos	heta}{C}ight)$$

Beats are produced due to the superposition of these two waves.

The beat frequency $v(t)$ is simply the difference between these two apparent frequencies:

$$v(t) = f'_N - f'_M = (f_N - f_M) \left(1 + \frac{v \cos	heta}{C}ight)$$

Substituting the given frequencies $f_N = 121	ext{ Hz}$ and $f_M = 118	ext{ Hz}$, we get:

$$v(t) = 3 \left(1 + \frac{v \cos	heta}{C}ight)$$

This is our master equation! It tells us how the beat frequency evolves as a function of the angle $	heta$.

### Evaluating Key Points

Let's evaluate the beat frequency at the three key locations: $P$, $Q$, and $R$.

At point $P$, which is extremely far away ($1800	ext{ m}$ compared to the $10	ext{ m}$ half-width of the speakers), the angle $	heta$ is virtually $0^\circ$.

Thus, $\cos	heta \approx 1$, giving:

$$v_P = 3 \left(1 + \frac{v}{C}ight)$$

At the midpoint $Q$, the car is directly between the speakers.

Here, the velocity vector is perpendicular to the lines connecting the car to the speakers, meaning $	heta = 90^\circ$ and $\cos	heta = 0$.

This yields:

$$v_Q = 3	ext{ Hz}$$

At point $R$, which is far away on the other side, the car is moving away from the speakers.

The component of velocity is now directed away from the sources, which flips the sign in our Doppler formula:

$$v_R = 3 \left(1 - \frac{v}{C}ight)$$

If we sum the beat frequencies at the symmetric extreme points $P$ and $R$, we find:

$$v_P + v_R = 3 \left(1 + \frac{v}{C}ight) + 3 \left(1 - \frac{v}{C}ight) = 6	ext{ Hz}$$

Since $v_Q = 3	ext{ Hz}$, we have:

$$v_P + v_R = 2v_Q$$

This mathematically proves that **Option (c) is correct**.

### Analyzing the Rate of Change

To find how rapidly the beat frequency changes, we can express $\cos	heta$ in terms of the car's position $y$ along the perpendicular bisector:

$$\cos	heta = \frac{y}{\sqrt{y^2 + d^2}}$$

where $d = 10	ext{ m}$ is the half-distance between the speakers.

Writing $y = -vt$ (where $t = 0$ at $Q$), we get:

$$v(t) = 3 \left(1 - \frac{v^2 t}{C \sqrt{v^2 t^2 + d^2}}ight)$$

Differentiating this with respect to time $t$ gives:

$$\frac{dv}{dt} = -\frac{3 v^2 d^2}{C (v^2 t^2 + d^2)^{3/2}}$$

To find where this rate of change is maximum, we look for where the magnitude of the derivative is maximized.

The denominator $(v^2 t^2 + d^2)^{3/2}$ is minimized when $t = 0$, which corresponds to point $Q$.

Thus, the rate of change of the beat frequency is indeed maximum at the midpoint $Q$.

This confirms that **Option (b) is correct**.

### Sketching the Schematic Curve

Since the derivative $\frac{dv}{dt}$ is negative for all $t$, the beat frequency decreases continuously from $P$ to $R$.

At $t = 0$ (point $Q$), the slope is at its steepest negative value, representing an **inflection point** on the graph.

Before $Q$, the curve is concave down, and after $Q$, it becomes concave up.

This classic S-shaped transition is perfectly represented in **Plot D** (Option d), while Plot A (Option a) is incorrect.

Hence, the correct options are **(b), (c), and (d)**.

Select Answer:

Visualized Solution

Visualizing the Setup

The Doppler Effect for a Moving Observer

Formulating Apparent Frequencies

The Beat Frequency Equation

Evaluating Beat Frequencies at Key Points

Verifying Option C

Finding the Rate of Change of Beat Frequency

Determining the Correct Schematic Plot

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Master Equation for Beat Frequency

Evaluating Key Points

Analyzing the Rate of Change

Sketching the Schematic Curve

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