Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 `ms^(-1)` and the man behind walks at a speed `2.0 ms^(-1)`. A third mad is standing at a height 12m 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 Hz. The speed of sound in air is 330 `ms^(-1)`. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is

Explanation

To solve the problem step by step, we need to calculate the apparent frequencies of the sounds heard by the stationary man from both moving men and then find the beat frequency.

### Step 1: Understand the setup
We have:
- Man A (in front) walking at speed \( v_A = 1.0 \, \text{m/s} \)
- Man B (behind) walking at speed \( v_B = 2.0 \, \text{m/s} \)
- Both are emitting sound at frequency \( f_0 = 1430 \, \text{Hz} \)
- The speed of sound in air \( v = 330 \, \text{m/s} \)
- The height of the stationary man above the horizontal line \( h = 12 \, \text{m} \)
- The distance between the two moving men \( d = 10 \, \text{m} \)

### Step 2: Calculate the distance from the stationary man to each moving man
Since the stationary man is equidistant from both moving men, we can find the horizontal distance from the stationary man to each moving man using the Pythagorean theorem.

Let \( x \) be the horizontal distance from the stationary man to either moving man. The distance from the stationary man to either man can be calculated as:
\[ d_{A} = \sqrt{x^2 + h^2} \]
\[ d_{B} = \sqrt{(x + 10)^2 + h^2} \]

### Step 3: Calculate the angle \( \theta \)
Using the geometry of the situation, we can find the angle \( \theta \) using the right triangle formed by the height and the horizontal distance:
\[ \cos \theta = \frac{x}{\sqrt{x^2 + h^2}} \]

### Step 4: Calculate the apparent frequencies
Using the Doppler effect, the apparent frequency \( f' \) heard by the stationary man from each moving man can be calculated using the formula:
\[ f' = f_0 \left( \frac{v + v_o}{v - v_s} \right) \]
where:
- \( v_o = 0 \) (the observer is stationary)
- \( v_s \) is the speed of the source (the moving man)

For Man A:
\[ f'_A = 1430 \left( \frac{330}{330 - 1} \right) \]

For Man B:
\[ f'_B = 1430 \left( \frac{330}{330 - 2} \right) \]

### Step 5: Calculate the beat frequency
The beat frequency \( f_{beat} \) is given by the absolute difference between the two apparent frequencies:
\[ f_{beat} = |f'_A - f'_B| \]

### Step 6: Substitute values and calculate
1. Calculate \( f'_A \):
\[ f'_A = 1430 \left( \frac{330}{329} \right) \approx 1430 \times 1.003 \approx 1434.29 \, \text{Hz} \]

2. Calculate \( f'_B \):
\[ f'_B = 1430 \left( \frac{330}{328} \right) \approx 1430 \times 1.006 \approx 1438.58 \, \text{Hz} \]

3. Calculate the beat frequency:
\[ f_{beat} = |1434.29 - 1438.58| \approx 4.29 \, \text{Hz} \]

### Final Answer
The frequency of beats heard by the stationary man is approximately **4.29 Hz**.

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