A man is watching two trains, one leaving and the other coming in with equal speed of 4 m/s. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air is `320(m)/(s)`) will be equal to

Options

A. 6
B. 3
C. 0
D. 12

Explanation

To solve the problem, we need to calculate the apparent frequencies of the sound heard by the observer from both trains and then find the beat frequency. Here’s how to do it step by step:

### Step 1: Identify the known values
- Frequency of the whistle (source frequency, \( f_0 \)) = 240 Hz
- Speed of sound in air (\( v \)) = 320 m/s
- Speed of the trains (\( v_s \)) = 4 m/s (both trains have the same speed)

### Step 2: Calculate the apparent frequency from the train coming towards the observer
For the train approaching the observer, we use the formula for apparent frequency:
\[ f_a = f_0 \times \frac{v + v_o}{v - v_s} \]
Where:
- \( v_o \) = speed of the observer = 0 m/s (the observer is stationary)
- \( v_s \) = speed of the source (train) = 4 m/s

Substituting the values:
\[ f_a = 240 \times \frac{320 + 0}{320 - 4} \]
\[ f_a = 240 \times \frac{320}{316} \]
\[ f_a \approx 240 \times 1.01266 \approx 243.04 \text{ Hz} \]

### Step 3: Calculate the apparent frequency from the train moving away from the observer
For the train moving away from the observer, we use the formula for apparent frequency:
\[ f_b = f_0 \times \frac{v + v_o}{v + v_s} \]
Substituting the values:
\[ f_b = 240 \times \frac{320 + 0}{320 + 4} \]
\[ f_b = 240 \times \frac{320}{324} \]
\[ f_b \approx 240 \times 0.98765 \approx 237.04 \text{ Hz} \]

### Step 4: Calculate the beat frequency
The beat frequency is given by the absolute difference between the two frequencies:
\[ \text{Beat frequency} = |f_a - f_b| \]
Substituting the values:
\[ \text{Beat frequency} = |243.04 - 237.04| = 6 \text{ Hz} \]

### Conclusion
The number of beats heard by the man is **6 Hz**.

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