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 natural frequency of 240 Hz, the number of beats beard by the man (velocity of sound in air = 300 m/s ) will be equal to

Options

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

Explanation

To solve the problem, we will use the Doppler effect to find the apparent frequencies of the sound heard by the man from both the approaching and receding trains. Then, we will calculate the number of beats heard by the man.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Natural frequency of the whistles (N) = 240 Hz
- Speed of sound in air (V) = 300 m/s
- Speed of the trains (Vs) = 4 m/s (both trains have the same speed)

2. **Calculate the Frequency from the Approaching Train (Na):**
- The formula for the frequency heard from an approaching source is:
\[ N_a = N \frac{V}{V - V_s} \]
- Substitute the values:
\[ N_a = 240 \frac{300}{300 - 4} \]
- Calculate the denominator:
\[ 300 - 4 = 296 \]
- Now calculate \(N_a\):
\[ N_a = 240 \frac{300}{296} \approx 240 \times 1.0135 \approx 243.24 \text{ Hz} \]

3. **Calculate the Frequency from the Receding Train (Nr):**
- The formula for the frequency heard from a receding source is:
\[ N_r = N \frac{V}{V + V_s} \]
- Substitute the values:
\[ N_r = 240 \frac{300}{300 + 4} \]
- Calculate the denominator:
\[ 300 + 4 = 304 \]
- Now calculate \(N_r\):
\[ N_r = 240 \frac{300}{304} \approx 240 \times 0.9868 \approx 236.80 \text{ Hz} \]

4. **Calculate the Number of Beats:**
- The number of beats heard by the man is the absolute difference between the two frequencies:
\[ \text{Number of Beats} = |N_a - N_r| \]
- Substitute the values:
\[ \text{Number of Beats} = |243.24 - 236.80| \approx 6.44 \text{ Hz} \]

5. **Final Answer:**
- The number of beats heard by the man is approximately 6 Hz.