A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car appr

Q: Why do we apply the Doppler formula twice? Can't we just do it in one step?

We must apply it twice because there are two distinct physical events. First, the car (observer) moves relative to the stationary source to receive the sound. Second, the car (now acting as a source) moves relative to the stationary observer at the origin to reflect the sound. Each event shifts the frequency.

Q: Does the speed of sound in the air change when it reflects off the car?

No, the speed of sound is a property of the medium (air) and remains constant at $330\text{ ms}^{-1}$ throughout the entire process, regardless of the motion of the source or observer.

Q: Why did we subtract $f_0$ from $f_2$ to find the beats? Why not $f_2 - f_1$?

Beats are produced by the superposition of two waves reaching the same point at the same time. At the source, the two waves present are the original emitted wave ($f_0$) and the returning reflected wave ($f_2$). The frequency $f_1$ is only perceived by the driver inside the car, not by the observer at the source.

Q: What if the car was moving away from the source? How would the formula change?

If the car were moving away, both Doppler shifts would be downward. The reflected frequency would be $f_2 = f_0 \left( \frac{v - v_c}{v + v_c} \right)$, and the beat frequency would be $f_b = f_0 \left( \frac{2v_c}{v + v_c} \right)$.

Q: Is the beat frequency always an integer?

Not necessarily, but JEE questions are designed with clean numbers so that the final answer simplifies to a neat integer, in this case, exactly $6\text{ Hz}$.

Question

A stationary source emits sound of frequency $f_0 = 492	ext{ Hz}$. The sound is reflected by a large car approaching the source with a speed of $2	ext{ ms}^{-1}$. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is $330	ext{ ms}^{-1}$ and the car reflects the sound at the frequency it has received).

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing next to a stationary sound source that is humming at a constant frequency of $f_0 = 492	ext{ Hz}$.

Suddenly, a car approaches this source at a steady speed of $v_c = 2	ext{ ms}^{-1}$.

The sound waves emitted by the source travel through the air at $v = 330	ext{ ms}^{-1}$, strike the moving car, reflect off its surface, and travel all the way back to your ears.

Because the car is moving, the reflected sound waves undergo a double Doppler shift.

When these shifted waves return and superpose with the original $492	ext{ Hz}$ sound waves, they interfere to produce a rhythmic pulsing sound known as **beats**.

Our goal is to calculate the frequency of these beats.

### The First Doppler Shift: Car as a Moving Observer

Let's break down the journey of the sound wave.

First, the sound travels from the stationary source to the moving car.

In this phase, the car acts as a **moving observer** approaching a stationary source.

Because the car is moving towards the incoming wavefronts, it intercepts them more frequently than a stationary observer would.

Thus, the frequency $f_1$ perceived by the car is shifted upwards:

$$f_1 = f_0 \left( \frac{v + v_c}{v} ight)$$

Substituting our known values into this equation:

$$f_1 = 492 \left( \frac{330 + 2}{330} ight)$$

To avoid early rounding errors, we will keep this expression in its fractional form for now.

### The Second Doppler Shift: Car as a Moving Source

Now, the car reflects this sound.

According to the problem, the car reflects the sound at the exact frequency it receives, which is $f_1$.

Since the car is moving towards our stationary position while emitting this reflected wave, it now acts as a **moving source** approaching a stationary observer.

This causes a second upward Doppler shift.

The frequency $f_2$ received back at the source is:

$$f_2 = f_1 \left( \frac{v}{v - v_c} ight)$$

Now, let's substitute our expression for $f_1$ into this equation:

$$f_2 = f_0 \left( \frac{v + v_c}{v} ight) \left( \frac{v}{v - v_c} ight)$$

Notice how beautifully the speed of sound $v$ in the denominator of the first term cancels out with the $v$ in the numerator of the second term!

This leaves us with a highly elegant combined formula for the twice-shifted reflected frequency:

$$f_2 = f_0 \left( \frac{v + v_c}{v - v_c} ight)$$

### Calculating the Beat Frequency

When the reflected wave of frequency $f_2$ arrives back at the source, it superposes with the original wave of frequency $f_0$.

The beat frequency $f_b$ is simply the absolute difference between these two frequencies:

$$f_b = f_2 - f_0$$

Let's substitute our combined formula for $f_2$ into this expression:

$$f_b = f_0 \left( \frac{v + v_c}{v - v_c} ight) - f_0$$

Factoring out $f_0$:

$$f_b = f_0 \left( \frac{v + v_c}{v - v_c} - 1 ight)$$

Finding a common denominator inside the parentheses:

$$f_b = f_0 \left( \frac{v + v_c - (v - v_c)}{v - v_c} ight)$$

$$f_b = f_0 \left( \frac{2v_c}{v - v_c} ight)$$

This is our master formula! It is incredibly clean and prevents any intermediate decimal approximations.

### Final Computation

Now, let's plug in our values: $f_0 = 492	ext{ Hz}$, $v = 330	ext{ ms}^{-1}$, and $v_c = 2	ext{ ms}^{-1}$:

$$f_b = 492 \left( \frac{2 	imes 2}{330 - 2} ight)$$

$$f_b = 492 \left( \frac{4}{328} ight)$$

Notice that $492$ and $328$ share a common factor. In fact, $\frac{492}{328} = 1.5$.

$$f_b = 1.5 	imes 4 = 6	ext{ Hz}$$

Thus, the beat frequency heard back at the source is exactly **$6	ext{ Hz}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Doppler Effect for Moving Observer

Substituting Values for $f_{1}$

Car as a Moving Source

Combining the Two Shifts

Defining Beat Frequency

Algebraic Simplification

Substituting Values into the Simplified Formula

Calculating the Final Value

Final Answer

The Way Forward

The Sigma Insight: Doppler Effect

Analyzing the Setup

The First Doppler Shift

The Second Doppler Shift

Calculating the Beat Frequency

Final Computation

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