A train moving at a speed of `220ms^-1` towards a stationary object emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is (speed of sound in air is `330ms^(-1)`)

Options

A. 4000 Hz
B. 5000 Hz
C. 3000 Hz
D. 3500 Hz

Explanation

To solve the problem, we need to calculate the frequency of the echo as detected by the driver of the train. We will use the Doppler effect formula for sound waves.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of the train (VS) = 220 m/s (towards the stationary object)
- Frequency of the sound emitted by the train (F) = 1000 Hz
- Speed of sound in air (V) = 330 m/s

2. **Determine the Frequency of Sound Reaching the Object:**
When the train emits sound towards the stationary object, the frequency detected by the object can be calculated using the Doppler effect formula:
\[ f' = F \cdot \frac{V + VO}{V - VS} \]
Here, VO (the speed of the observer) is 0 since the object is stationary. Thus, the formula simplifies to:
\[ f' = F \cdot \frac{V}{V - VS} \]

3. **Substituting the Values:**
\[ f' = 1000 \cdot \frac{330}{330 - 220} \]
\[ f' = 1000 \cdot \frac{330}{110} \]
\[ f' = 1000 \cdot 3 = 3000 \text{ Hz} \]

4. **Determine the Frequency of the Echo Detected by the Train:**
Now, the sound reflected back from the object is perceived by the train. The train is now the observer moving towards the source of the sound (the object). We will use the Doppler effect formula again:
\[ f'' = f' \cdot \frac{V + VT}{V - VS} \]
Here, VT is the speed of the train (220 m/s). So, we have:
\[ f'' = 3000 \cdot \frac{330 + 220}{330 - 220} \]
\[ f'' = 3000 \cdot \frac{550}{110} \]
\[ f'' = 3000 \cdot 5 = 15000 \text{ Hz} \]

5. **Final Result:**
The frequency of the echo as detected by the driver of the train is **15000 Hz**.