One train is approaching an observer at rest and another train is receding from him with the same velocity `4 m//s` . Both trains blow whistles of same frequency of `243 H_(Z)` . The beat frequency in `H_(Z)` as heard by observer is (speed of sound in air = `320 m//s`)

Options

A. `10`
B. `6`
C. `4`
D. `1`

Explanation

To solve the problem of finding the beat frequency heard by an observer when one train is approaching and another is receding, we can follow these steps:

### Step 1: Understand the problem
We have two trains:
- Train 1 is approaching the observer with a velocity of \( v = 4 \, \text{m/s} \).
- Train 2 is receding from the observer with the same velocity of \( v = 4 \, \text{m/s} \).
Both trains emit sound at a frequency \( f = 243 \, \text{Hz} \). The speed of sound in air is given as \( c = 320 \, \text{m/s} \).

### Step 2: Calculate the apparent frequency for Train 1 (approaching)
The formula for the apparent frequency \( f_1 \) when the source is moving towards the observer is given by:

\[ f_1 = f \cdot \frac{c}{c - v} \]

Substituting the values:

\[ f_1 = 243 \cdot \frac{320}{320 - 4} = 243 \cdot \frac{320}{316} \]

### Step 3: Calculate the apparent frequency for Train 2 (receding)
The formula for the apparent frequency \( f_2 \) when the source is moving away from the observer is given by:

\[ f_2 = f \cdot \frac{c}{c + v} \]

Substituting the values:

\[ f_2 = 243 \cdot \frac{320}{320 + 4} = 243 \cdot \frac{320}{324} \]

### Step 4: Calculate the beat frequency
The beat frequency \( f_b \) is the difference between the two apparent frequencies:

\[ f_b = |f_1 - f_2| \]

Substituting the expressions we found for \( f_1 \) and \( f_2 \):

\[ f_b = \left| 243 \cdot \frac{320}{316} - 243 \cdot \frac{320}{324} \right| \]

Factoring out the common terms:

\[ f_b = 243 \cdot 320 \left| \frac{1}{316} - \frac{1}{324} \right| \]

### Step 5: Simplify the expression
To simplify \( \left| \frac{1}{316} - \frac{1}{324} \right| \):

\[ \frac{1}{316} - \frac{1}{324} = \frac{324 - 316}{316 \cdot 324} = \frac{8}{316 \cdot 324} \]

Thus, substituting back into the beat frequency equation:

\[ f_b = 243 \cdot 320 \cdot \frac{8}{316 \cdot 324} \]

### Step 6: Calculate the numerical value
Now we can calculate the beat frequency:

\[ f_b = 243 \cdot 320 \cdot \frac{8}{316 \cdot 324} \]

Calculating the numerical values:

- \( 316 \cdot 324 = 102384 \)
- \( 243 \cdot 320 = 77760 \)
- \( 77760 \cdot 8 = 622080 \)

Thus,

\[ f_b = \frac{622080}{102384} \approx 6 \, \text{Hz} \]

### Final Answer
The beat frequency heard by the observer is approximately **6 Hz**.

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