A submarine A travelling at 18 km/h is being chased along the line of its velocity by another submar

Q: Why do we need to apply the Doppler effect twice?

Because the sound travels from B to A, gets received by A (first Doppler shift), and then A acts as a moving source reflecting that exact received frequency back to B (second Doppler shift).

Q: How do I decide the signs in the Doppler formula?

Always think about the physical effect. If the motion tends to bring the source and observer closer, the frequency must increase (numerator +, denominator -). If the motion separates them, the frequency must decrease (numerator -, denominator +).

Q: Why did we convert km/h to m/s?

The speed of sound is given in m/s (1500 m/s). To add or subtract velocities in the Doppler formula, all velocities must be in the exact same units.

Q: What if submarine A was faster than submarine B?

If A was faster, the distance between them would increase over time. The overall received frequency $\nu$ would be less than the original 500 Hz.

Q: Can we use an approximation formula here?

Since the velocities of the submarines (5 m/s and 7.5 m/s) are much smaller than the speed of sound (1500 m/s), you could use binomial approximation, but exact calculation is safer and straightforward enough here.

Question

A submarine A travelling at $18 	ext{ km/h}$ is being chased along the line of its velocity by another submarine B travelling at $27 	ext{ km/h}$. B sends a sonar signal of $500 	ext{ Hz}$ to detect A and receives a reflected sound of frequency $
u$. The value of $
u$ is close to (Speed of sound in water $= 1500 	ext{ ms}^{-1}$)

Options:

(A) $504 	ext{ Hz}$

(B) $507 	ext{ Hz}$

(C) $499 	ext{ Hz}$

(D) $502 	ext{ Hz}$

Sigma Percentile · Accepted Answer

### The Thrill of the Chase

Imagine a high-stakes underwater chase. Submarine B is relentlessly hunting down Submarine A. To pinpoint A's exact location and speed, B fires a sonar pulse. Our mission is to figure out the exact frequency of the echo that bounces back to B.

This is a classic Doppler effect problem, but with a twist—it's a **double Doppler effect**! The master formula tells us that the observed frequency depends on the relative velocities of the source and the observer with respect to the speed of sound.

$$f_{obs} = f_{source} \left( \frac{v \pm v_{obs}}{v \pm v_{source}} ight)$$

Before we plug anything in, we must ensure our units are consistent. The speed of sound is given in meters per second ($1500 	ext{ m/s}$), but our submarine speeds are in kilometers per hour. Let's quickly multiply by $\frac{5}{18}$.

Submarine A is moving at $v_A = 18 	imes \frac{5}{18} = 5 	ext{ m/s}$, and Submarine B is moving at $v_B = 27 	imes \frac{5}{18} = 7.5 	ext{ m/s}$.

### Phase 1: The Signal Reaches Submarine A

The sound travels from B to A. Let's find the frequency that submarine A actually hears. We will call this $f_1$.

Here, B is the source moving towards A, so the denominator decreases, making the sound pitch higher. But A is running away, so the numerator decreases, trying to lower the pitch. Plugging in our values, we get:

$$f_1 = f_0 \left( \frac{v - v_A}{v - v_B} ight)$$

$$f_1 = 500 \left( \frac{1500 - 5}{1500 - 7.5} ight) = 500 \left( \frac{1495}{1492.5} ight)$$

We won't calculate this just yet; let's keep it as a fraction to avoid rounding errors.

### Phase 2: The Echo Returns to Submarine B

Submarine A reflects this sound back. Now, A becomes the source, moving away from B, so the denominator increases. B is the observer, moving towards A, so the numerator increases.

The new frequency $f_2$ is $f_1$ multiplied by the new Doppler factor. Notice how the roles have reversed!

$$f_2 = f_1 \left( \frac{v + v_B}{v + v_A} ight)$$

$$f_2 = f_1 \left( \frac{1500 + 7.5}{1500 + 5} ight) = f_1 \left( \frac{1507.5}{1505} ight)$$

### The Final Calculation

Time for the final execution. We substitute $f_1$ into our $f_2$ equation. We get a product of two fractions slightly greater than one.

$$f_2 = 500 \left( \frac{1495}{1492.5} ight) \left( \frac{1507.5}{1505} ight)$$

When we multiply $500$ by $\frac{1495}{1492.5}$ (which is $\approx 1.00167$), and then by $\frac{1507.5}{1505}$ (which is $\approx 1.00166$), we get:

$$f_2 \approx 500 	imes 1.00167 	imes 1.00166 \approx 501.67 	ext{ Hz}$$

This makes perfect physical sense. Because the chaser (B) is faster than the target (A), the overall distance between them is decreasing, which means the overall frequency must increase. The closest value in the options is **$502 	ext{ Hz}$**.

Select Answer:

Visualized Solution

$The Chase$

$Doppler Effect Formula$

$Velocity Conversion$

$Frequency Received by A$

$Frequency Reflected to B$

$Final Calculation$

$Conclusion$

The Sigma Insight: Doppler Effect

The Thrill of the Chase

Phase 1

Phase 2

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Chase

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Doppler Effect Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity Conversion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frequency Received by A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frequency Reflected to B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion