A train approaching a hill at a speed of `40 km//hr` sounds a whistle of frequency `580 Hz` when it is at a distance of `1km` from a hill. A wind with a speed of `40km//hr` is blowing in the direction of motion of the train Find
(i) the frequency of the whistle as heard by an observer on the hill,
(ii) the distance from the hill at which the echo from the hill is heard by the driver and its frequency.
(Velocity of sound in air `= 1, 200 km//hr`)

Explanation

To solve the problem step by step, we will break it down into two parts as per the requirements of the question.

### Given Data:
- Speed of train, \( V_t = 40 \, \text{km/hr} \)
- Frequency of whistle, \( f = 580 \, \text{Hz} \)
- Distance from the hill, \( d = 1 \, \text{km} = 1000 \, \text{m} \)
- Speed of wind, \( V_w = 40 \, \text{km/hr} \)
- Velocity of sound in air, \( V = 1200 \, \text{km/hr} \)

### Part (i): Frequency of the whistle as heard by an observer on the hill

1. **Convert speeds to m/s**:
\[ V_t = 40 \, \text{km/hr} = \frac{40 \times 1000}{3600} \approx 11.11 \, \text{m/s} \]
\[ V_w = 40 \, \text{km/hr} = \frac{40 \times 1000}{3600} \approx 11.11 \, \text{m/s} \]
\[ V = 1200 \, \text{km/hr} = \frac{1200 \times 1000}{3600} \approx 333.33 \, \text{m/s} \]

2. **Use the Doppler Effect formula**:
The formula for apparent frequency \( f' \) when the source is moving towards a stationary observer is:
\[ f' = \frac{V + V_w}{V - V_t} \cdot f \]
Here, \( V_w \) is added to the velocity of sound because the wind is blowing in the direction of the train.

3. **Substitute the values**:
\[ f' = \frac{333.33 + 11.11}{333.33 - 11.11} \cdot 580 \]
\[ f' = \frac{344.44}{322.22} \cdot 580 \]

4. **Calculate \( f' \)**:
\[ f' \approx 1.070 \cdot 580 \approx 620.6 \, \text{Hz} \]

### Part (ii): Distance from the hill at which the echo from the hill is heard by the driver and its frequency

1. **Calculate the time taken for the whistle to reach the hill**:
The time taken \( t_1 \) for the sound to travel 1 km to the hill:
\[ t_1 = \frac{1000}{V + V_w} = \frac{1000}{333.33 + 11.11} \approx \frac{1000}{344.44} \approx 2.9 \, \text{s} \]

2. **Calculate the time taken for the echo to return**:
The time taken \( t_2 \) for the echo to return to the train:
The sound travels back at a speed of \( V - V_w \):
\[ t_2 = \frac{1000}{V - V_w} = \frac{1000}{333.33 - 11.11} \approx \frac{1000}{322.22} \approx 3.1 \, \text{s} \]

3. **Total time for the echo to return**:
\[ t_{total} = t_1 + t_2 \approx 2.9 + 3.1 = 6.0 \, \text{s} \]

4. **Calculate the distance traveled by the train in that time**:
The distance \( d_t \) the train travels in \( t_{total} \):
\[ d_t = V_t \cdot t_{total} = 11.11 \cdot 6.0 \approx 66.67 \, \text{m} \]

5. **Calculate the distance from the hill when the echo is heard**:
The distance from the hill when the echo is heard:
\[ d_{echo} = 1000 - d_t \approx 1000 - 66.67 \approx 933.33 \, \text{m} \]

### Summary of Results:
- (i) The frequency of the whistle as heard by an observer on the hill is approximately **620.6 Hz**.
- (ii) The distance from the hill at which the echo from the hill is heard by the driver is approximately **933.33 m**.