A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approa

Q: Why does the speed of sound $v$ cancel out in the numerator of the first stage and denominator of the second stage?

In the first stage (car as observer), the frequency is multiplied by $(v + v_{\text{car}})/v$. In the second stage (car as source), the frequency is multiplied by $v/(v - v_{\text{car}})$. When multiplied together, the $v$ in the denominator of the first term and the $v$ in the numerator of the second term cancel out, leaving a simplified ratio.

Q: Can we use binomial approximation directly on the master equation?

Yes! You can write $f_{\text{reflected}} = f_0 (1 + v_{\text{car}}/v)(1 - v_{\text{car}}/v)^{-1} \approx f_0 (1 + v_{\text{car}}/v)(1 + v_{\text{car}}/v) \approx f_0 (1 + 2v_{\text{car}}/v)$. This yields the same result and is mathematically equivalent for small speeds.

Q: Why did we convert the final answer to km/h?

The question explicitly asks for the difference in the speeds of the cars 'in km per hour'. Always read the units requested in the final answer carefully to avoid silly mistakes!

Q: What if the cars were moving away from the source instead of approaching?

If the cars were moving away, the signs of their velocities would flip. The master equation would become $f_{\text{reflected}} = f_0 \left(\frac{v - v_{\text{car}}}{v + v_{\text{car}}}\right)$.

Q: Is the reflection of sound always treated as a two-stage Doppler shift?

Yes, whenever sound reflects off a moving boundary or target, it is physically equivalent to a moving observer receiving the wave and then immediately re-emitting (reflecting) it as a moving source.

Question

A stationary source is emitting sound at a fixed frequency $f_0$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is $1.2\%$ of $f_0$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is $330	ext{ ms}^{-1}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing right next to a stationary sound source $S$ that is continuously emitting a pure tone of frequency $f_0$.

On either side of you, two cars are driving directly towards you.

As they approach, they act as moving observers, intercepting the sound waves at a higher rate than if they were stationary.

This is the **first stage** of the Doppler shift.

But the story doesn't end there.

The sound waves don't just stop at the cars; they bounce off them.

As the cars reflect the sound back towards the source, they act as moving sources themselves, compressing the reflected wave crests in front of them.

This is the **second stage** of the Doppler shift.

Our goal is to find the difference in the speeds of these two cars, given that the difference in their reflected frequencies is exactly $1.2\%$ of the original frequency $f_0$.

---

### The Master Equation for Double Doppler Shift

Let's derive the formula for the frequency reflected by a car moving towards a stationary source with speed $v_{	ext{car}}$.

In the first stage, the car acts as an observer moving towards a stationary source with speed $v_{	ext{car}}$.

The frequency received by the car, $f_{	ext{received}}$, is:

$$f_{	ext{received}} = f_0 \left( \frac{v + v_{	ext{car}}}{v} ight)$$

where $v$ is the speed of sound in air.

In the second stage, the car acts as a moving source emitting $f_{	ext{received}}$ and moving towards the stationary detector at the source.

The reflected frequency $f_{	ext{reflected}}$ detected back at the source is:

$$f_{	ext{reflected}} = f_{	ext{received}} \left( \frac{v}{v - v_{	ext{car}}} ight)$$

Substituting the expression for $f_{	ext{received}}$ into this equation, we get:

$$f_{	ext{reflected}} = f_0 \left( \frac{v + v_{	ext{car}}}{v} ight) \left( \frac{v}{v - v_{	ext{car}}} ight)$$

Notice how 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 our **Master Double-Shift Equation**:

$$f_{	ext{reflected}} = f_0 \left( \frac{v + v_{	ext{car}}}{v - v_{	ext{car}}} ight)$$

---

### Setting up the Difference

Let the speeds of the two cars be $v_1$ and $v_2$.

The reflected frequencies received back at the source are:

$$f_1 = f_0 \left( \frac{v + v_1}{v - v_1} ight)$$

$$f_2 = f_0 \left( \frac{v + v_2}{v - v_2} ight)$$

We are given that the difference between these two reflected frequencies is $1.2\%$ of $f_0$:

$$f_1 - f_2 = 0.012 f_0$$

Substituting our expressions for $f_1$ and $f_2$:

$$f_0 \left[ \frac{v + v_1}{v - v_1} - \frac{v + v_2}{v - v_2} ight] = 0.012 f_0$$

Dividing both sides by $f_0$ simplifies the equation to:

$$\frac{v + v_1}{v - v_1} - \frac{v + v_2}{v - v_2} = 0.012$$

---

### Algebraic Simplification and Approximation

To subtract the fractions, we find a common denominator:

$$\frac{(v + v_1)(v - v_2) - (v + v_2)(v - v_1)}{(v - v_1)(v - v_2)} = 0.012$$

Let's expand the numerator carefully:

$$(v^2 - v v_2 + v_1 v - v_1 v_2) - (v^2 - v v_1 + v_2 v - v_1 v_2)$$

$$= v^2 - v v_2 + v_1 v - v_1 v_2 - v^2 + v v_1 - v_2 v + v_1 v_2$$

$$= 2v(v_1 - v_2)$$

So, our equation becomes:

$$\frac{2v(v_1 - v_2)}{(v - v_1)(v - v_2)} = 0.012$$

Now, we apply a crucial physical approximation.

The problem states that the cars are moving at speeds **much smaller** than the speed of sound ($v_1, v_2 \ll v$).

Therefore, we can approximate:

$$v - v_1 \approx v$$

$$v - v_2 \approx v$$

This simplifies the denominator to $v^2$:

$$\frac{2v(v_1 - v_2)}{v^2} \approx 0.012$$

$$\frac{2(v_1 - v_2)}{v} = 0.012$$

---

### Final Calculation

Now, we solve for the difference in speeds, $\Delta v = v_1 - v_2$:

$$\Delta v = \frac{0.012 	imes v}{2} = 0.006 	imes v$$

Substituting the given speed of sound, $v = 330	ext{ ms}^{-1}$:

$$\Delta v = 0.006 	imes 330 = 1.98	ext{ ms}^{-1}$$

To convert this speed difference into kilometers per hour ($	ext{km/h}$), we multiply by $3.6$ (or $\frac{18}{5}$):

$$\Delta v_{	ext{km/h}} = 1.98 	imes 3.6 = 7.128	ext{ km/h}$$

Rounding to the nearest integer, we get:

$$\Delta v_{	ext{km/h}} \approx \mathbf{7}$$

Thus, the difference in the speeds of the two cars is **$7	ext{ km/h}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Double Doppler Shift

Stage 1: Frequency Received by the Moving Car

Stage 2: Frequency Reflected Back to the Source

The Master Double-Shift Equation

Expressing the Reflected Frequencies for Both Cars

Setting up the Algebraic Difference

Expanding and Canceling Terms

Applying the Low-Speed Approximation

Calculating the Speed Difference in $ms^{- 1}$

Converting Units to Kilometers per Hour

Rounding to the Nearest Integer

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Master Equation for Double Doppler Shift

Setting up the Difference

Algebraic Simplification and Approximation

Final Calculation

Similar Questions

A stationary source emits sound waves of frequency $500 Hz$ . Two observers moving along a line passing through the source detect sound to be of frequencies $480 Hz$ and $530 Hz$ . Their respective speeds are in $ms^{- 1}$ , (Take, speed of sound $= 300 m/s$ )

Two cars are approaching each other at an equal speed of $7.2 km/h$ . When they see each other, both blow horns having frequency of $676 Hz$ . The beat frequency heard by each driver will be ......... $Hz$ . [Velocity of sound in air is $340 m/s$ .]

Two sources of sound $S_{1}$ and $S_{2}$ produce sound waves of same frequency $660 Hz$ . A listener is moving from source $S_{1}$ towards $S_{2}$ with a constant speed $u m/s$ and he hears $10 beats/s$ . The velocity of sound is $330 m/s$ . Then, $u$ equal to

A police car with a siren of frequency $8 kHz$ is moving with uniform velocity $36 km/h$ towards a tall building which reflects the sound waves. The speed of sound in air is $320 m/s$ . The frequency of the siren heard by the car driver is

A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from $440 Hz$ to $480 Hz$ , when it gets reflected from the wall. If the speed of sound in air is $345 m/s$ , then the speed of the car is