A train is moving on a straight track with speed `20ms^(-1)`. It is blowing its whistle at the frequency of `1000 Hz`. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound `= 320 ms^(-1)`) close to :

Options

A. `12%`
B. `18%`
C. `24%`
D. `6%`

Answer

To solve the problem, we will use the Doppler effect formula for sound, which describes how the frequency of sound changes for an observer moving relative to the source of sound.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of the train (source) \( V_s = 20 \, \text{m/s} \)
- Frequency of the whistle (source frequency) \( f = 1000 \, \text{Hz} \)
- Speed of sound \( V = 320 \, \text{m/s} \)

2. **Calculate the Frequency Heard by the Observer When the Train Approaches:**
- When the train is approaching the observer, the frequency \( f_1 \) can be calculated using the formula:
\[ f_1 = f \left( \frac{V}{V - V_s} \right) \]
- Substituting the values:
\[ f_1 = 1000 \left( \frac{320}{320 - 20} \right) = 1000 \left( \frac{320}{300} \right) = \frac{320000}{300} \approx 1066.67 \, \text{Hz} \]

3. **Calculate the Frequency Heard by the Observer When the Train Passes:**
- When the train is moving away from the observer, the frequency \( f_2 \) can be calculated using the formula:
\[ f_2 = f \left( \frac{V}{V + V_s} \right) \]
- Substituting the values:
\[ f_2 = 1000 \left( \frac{320}{320 + 20} \right) = 1000 \left( \frac{320}{340} \right) = \frac{320000}{340} \approx 941.18 \, \text{Hz} \]

4. **Calculate the Percentage Change in Frequency:**
- The percentage change in frequency can be calculated using the formula:
\[ \text{Percentage Change} = \left( \frac{f_1 - f_2}{f_1} \right) \times 100 \]
- Substituting the values we calculated:
\[ \text{Percentage Change} = \left( \frac{1066.67 - 941.18}{1066.67} \right) \times 100 \]
- Calculating the difference:
\[ \text{Percentage Change} = \left( \frac{125.49}{1066.67} \right) \times 100 \approx 11.77\% \]
- Rounding this value gives approximately \( 12\% \).

5. **Final Answer:**
- The percentage change in frequency heard by the person standing near the track as the train passes him is approximately **12%**.