A train blowing a whistle of frequency `320Hz` approaches an observer standing on the platform at a speed of `66m/s`.The frequency observed by the observer will be (given speed of sound `=330ms^(-1)` ) `Hz`.

Answer

To solve the problem of finding the frequency observed by an observer when a train approaches, we can use the Doppler effect formula for sound. The formula for the observed frequency \( f' \) when the source is moving towards a stationary observer is given by:

\[ f' = f_0 \times \frac{v}{v - v_s} \]

Where:
- \( f' \) = observed frequency
- \( f_0 \) = source frequency (320 Hz)
- \( v \) = speed of sound (330 m/s)
- \( v_s \) = speed of the source (66 m/s)

### Step-by-step solution:

1. **Identify the given values:**
- Source frequency \( f_0 = 320 \, \text{Hz} \)
- Speed of sound \( v = 330 \, \text{m/s} \)
- Speed of the source \( v_s = 66 \, \text{m/s} \)

2. **Substitute the values into the Doppler effect formula:**
\[ f' = 320 \times \frac{330}{330 - 66} \]

3. **Calculate the denominator:**
\[ 330 - 66 = 264 \, \text{m/s} \]

4. **Substitute the denominator back into the formula:**
\[ f' = 320 \times \frac{330}{264} \]

5. **Calculate \( \frac{330}{264} \):**
\[ \frac{330}{264} = 1.25 \]

6. **Multiply by the source frequency:**
\[ f' = 320 \times 1.25 = 400 \, \text{Hz} \]

### Final Answer:
The frequency observed by the observer will be \( 400 \, \text{Hz} \).