One train is approaching an observer at rest and another train is receding him with same velocity ` 4 ms^(-1)` . Both the trains blow whistles of same frequency of 243 Hz. The beat frequency in Hz as heard by the observer is (speed of sound in air=`320ms^(-1)` )

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Frequency of the whistle, \( F = 243 \, \text{Hz} \)
- Speed of sound in air, \( V = 320 \, \text{m/s} \)
- Speed of the trains (source), \( V_s = 4 \, \text{m/s} \)

### Step 2: Calculate the frequency of the whistle as the train approaches the observer
When the source is approaching the observer, the frequency heard by the observer is given by the formula:
\[ F_1 = \frac{V}{V - V_s} \cdot F \]
Substituting the values:
\[ F_1 = \frac{320}{320 - 4} \cdot 243 \]
\[ F_1 = \frac{320}{316} \cdot 243 \]

### Step 3: Calculate the frequency of the whistle as the train recedes from the observer
When the source is receding from the observer, the frequency heard by the observer is given by the formula:
\[ F_2 = \frac{V}{V + V_s} \cdot F \]
Substituting the values:
\[ F_2 = \frac{320}{320 + 4} \cdot 243 \]
\[ F_2 = \frac{320}{324} \cdot 243 \]

### Step 4: Calculate the beat frequency
The beat frequency is the absolute difference between the two frequencies:
\[ \text{Beat Frequency} = |F_1 - F_2| \]
Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ \text{Beat Frequency} = \left| \frac{320}{316} \cdot 243 - \frac{320}{324} \cdot 243 \right| \]
Factoring out \( 243 \):
\[ \text{Beat Frequency} = 243 \left| \frac{320}{316} - \frac{320}{324} \right| \]

### Step 5: Simplify the expression
To simplify:
\[ \frac{320}{316} - \frac{320}{324} = 320 \left( \frac{1}{316} - \frac{1}{324} \right) \]
Finding a common denominator (which is \( 316 \times 324 \)):
\[ \frac{1}{316} - \frac{1}{324} = \frac{324 - 316}{316 \times 324} = \frac{8}{316 \times 324} \]
Thus,
\[ \text{Beat Frequency} = 243 \cdot 320 \cdot \frac{8}{316 \times 324} \]

### Step 6: Calculate the numerical value
Calculating the above expression:
1. Calculate \( 316 \times 324 \).
2. Substitute back to find the beat frequency.

After performing the calculations, we find:
\[ \text{Beat Frequency} \approx 6 \, \text{Hz} \]

### Final Answer
The beat frequency as heard by the observer is \( 6 \, \text{Hz} \).

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