Two men are walking along a horizontal straight line in the same direction. The man in front walks a

Q: Why does the motion of the men affect the frequency heard by a stationary observer?

When a source of sound moves, it either compresses the wave crests ahead of it (increasing the frequency) or stretches them behind it (decreasing the frequency). This is the classic Doppler effect. Even though the observer is stationary, the relative motion along the line of sight causes a frequency shift.

Q: How do we find the component of velocity along the line of sight?

The sound waves travel along the straight line connecting the source to the observer (the line of sight). Therefore, only the component of the source's velocity along this line of sight contributes to the Doppler shift. This component is given by $v_s \cos\theta$, where $\theta$ is the angle between the velocity vector and the line of sight.

Q: Why is one frequency shifted up and the other shifted down?

The man behind is moving to the right, which is towards the observer's horizontal position, so he is approaching the observer, causing a blue-shift (higher frequency). The man in front is also moving to the right, which is away from the observer's horizontal position, so he is receding from the observer, causing a red-shift (lower frequency).

Q: Is the binomial approximation absolutely necessary here?

While you can calculate the exact frequencies and subtract them, the numbers are extremely close (e.g., $1433.34\text{ Hz}$ and $1428.34\text{ Hz}$). Using the binomial approximation not only makes the calculation much simpler and elegant by hand, but it also reveals the underlying physical dependence of the beat frequency on the sum of the speeds.

Q: What would happen if the two men were walking in opposite directions?

If they were walking in opposite directions, the signs of their velocities along the line of sight would change depending on whether they are approaching or receding. You would need to re-evaluate the components $v_s \cos\theta$ for each man based on their new directions of motion.

Question

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed $1.0	ext{ ms}^{-1}$ and the man behind walks at a speed $2.0	ext{ ms}^{-1}$. A third man is standing at a height $12	ext{ m}$ above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency $1430	ext{ Hz}$. The speed of sound in air $330	ext{ ms}^{-1}$. At the instant, when the moving men are $10	ext{ m}$ apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is ............. .

Sigma Percentile · Accepted Answer

### The Magic of Wave Interference and Doppler Shift

Imagine standing on a quiet street when an ambulance speeds past. You hear the pitch of the siren rise as it approaches and drop as it recedes. This is the **Doppler Effect**, a cornerstone of wave mechanics. But what happens when we have not one, but two moving sound sources, and an observer suspended high above their path?

This problem invites us to explore the beautiful intersection of geometry, wave kinematics, and the phenomenon of **beats**. When two sound waves of slightly different frequencies superimpose, they interfere constructively and destructively, creating a pulsating sound. The frequency of this pulsation is the beat frequency, and our goal is to find it.

### Decoding the Geometry: The Isosceles Triangle of Sound

Before diving into the physics of sound, we must establish the spatial relationships. At the instant of interest, the two men are $10	ext{ m}$ apart. The stationary observer is $12	ext{ m}$ directly above the midpoint of the line segment joining them. This symmetry forms a perfect **isosceles triangle**.

Let's calculate the direct distance from each man to the observer. Using the Pythagorean theorem:

$$d = \sqrt{12^2 + 5^2} = 13	ext{ m}$$

This $13	ext{ m}$ line is the path along which the sound waves travel to reach the observer. The angle $	heta$ between the horizontal direction of motion and this line of sight is crucial because only the component of velocity along this line causes a Doppler shift. From our right-angled triangle, we find:

$$\cos	heta = \frac{5}{13}$$

### The Physics of Doppler Shift: Moving Sources

Now, let's look at the motion. Both men are walking to the right. The man behind ($B$) is moving at $2.0	ext{ m/s}$, while the man in front ($A$) is moving at $1.0	ext{ m/s}$.

Because $B$ is on the left and moving to the right, he is moving **towards** the observer's horizontal position. The component of his velocity along the line of sight is:

$$v_{s,B} = v_B \cos	heta = 2.0 	imes \frac{5}{13} = \frac{10}{13}	ext{ m/s}$$

Since he is approaching, the frequency he emits is blue-shifted (increased):

$$f'_B = f_0 \left( \frac{v}{v - v_{s,B}} ight)$$

Conversely, $A$ is on the right and moving to the right, meaning he is moving **away** from the observer's horizontal position. The component of his velocity along the line of sight is:

$$v_{s,A} = v_A \cos	heta = 1.0 	imes \frac{5}{13} = \frac{5}{13}	ext{ m/s}$$

Since he is receding, the frequency he emits is red-shifted (decreased):

$$f'_A = f_0 \left( \frac{v}{v + v_{s,A}} ight)$$

### The Power of Approximation: Binomial Expansion

We could plug the numbers directly into these equations, but the resulting calculation would be tedious and prone to arithmetic errors. Instead, we can use a beautiful mathematical shortcut: the **binomial approximation**.

Since the speeds of the men ($< 1	ext{ m/s}$ along the line of sight) are tiny compared to the speed of sound ($330	ext{ m/s}$), the ratio $x = v_s/v$ is extremely small. We can write:

$$(1 \mp x)^{-1} \approx 1 \pm x$$

Applying this to our frequency equations:

$$f'_B \approx f_0 \left( 1 + \frac{v_B \cos	heta}{v} ight)$$

$$f'_A \approx f_0 \left( 1 - \frac{v_A \cos	heta}{v} ight)$$

### The Final Synthesis: Calculating the Beats

The beat frequency is simply the difference between these two apparent frequencies:

$$f_{	ext{beat}} = f'_B - f'_A$$

Substituting our approximations:

$$f_{	ext{beat}} \approx f_0 \left[ \left( 1 + \frac{v_B \cos	heta}{v} ight) - \left( 1 - \frac{v_A \cos	heta}{v} ight) ight]$$

Notice how the constant term $1$ cancels out perfectly! We are left with:

$$f_{	ext{beat}} \approx f_0 \left( \frac{(v_B + v_A) \cos	heta}{v} ight)$$

This is incredibly elegant. The beat frequency depends directly on the sum of the speeds of the two sources! Let's substitute our values:

$$f_{	ext{beat}} \approx 1430 	imes \left( \frac{(2.0 + 1.0) 	imes \frac{5}{13}}{330} ight)$$

$$f_{	ext{beat}} \approx 1430 	imes \left( \frac{3 	imes 5}{13 	imes 330} ight) = 1430 	imes \frac{15}{4290}$$

Now, observe the magic of the numbers chosen by the examiners. $4290$ is exactly $3 	imes 1430$! Therefore:

$$f_{	ext{beat}} \approx \frac{15}{3} = 5	ext{ Hz}$$

Thus, the stationary observer hears exactly **$5$ beats per second**.

Enter Numerical Value:

Visualized Solution

Visualize the Physical Setup

Determine the Geometry and Distances

The Doppler Effect for Moving Sources

Analyze the Motion of Man B (Behind)

Analyze the Motion of Man A (In Front)

Apply Binomial Approximation

Calculate the Beat Frequency

Substitute Values and Find the Answer

The Sigma Insight: Doppler Effect

The Magic of Wave Interference and Doppler Shift

Decoding the Geometry

The Physics of Doppler Shift

The Power of Approximation

The Final Synthesis

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