Afrain approaches a stationary observer, the velocity of train being (1/20)th of the velocity of sound. A sharp blast is blown with the whistle of the engine at equal intervals of a second. Find the interval between the successive blasts as heard by the observer.

Answer

To solve the problem step by step, we will use the concepts of the Doppler effect in sound waves.

### Step 1: Understand the given information
- The train is approaching a stationary observer.
- The speed of sound is denoted as \( V \).
- The speed of the train is \( \frac{V}{20} \).
- The train blows a whistle at intervals of 1 second.

### Step 2: Identify the relevant formula
We will use the formula for the apparent frequency (\( f' \)) as heard by the observer:
\[ f' = f \cdot \frac{V + V_o}{V - V_s} \]
where:
- \( f \) is the actual frequency of the sound (1 Hz, since the whistle is blown every second).
- \( V_o \) is the velocity of the observer (0, since the observer is stationary).
- \( V_s \) is the velocity of the source (the train), which is \( \frac{V}{20} \).

### Step 3: Substitute the values into the formula
Since the observer is stationary (\( V_o = 0 \)), the formula simplifies to:
\[ f' = f \cdot \frac{V}{V - V_s} \]
Substituting \( f = 1 \) Hz and \( V_s = \frac{V}{20} \):
\[ f' = 1 \cdot \frac{V}{V - \frac{V}{20}} = \frac{V}{V \left(1 - \frac{1}{20}\right)} = \frac{V}{V \cdot \frac{19}{20}} = \frac{20}{19} \]

### Step 4: Calculate the time period for the apparent frequency
The time period (\( T' \)) is the reciprocal of the apparent frequency:
\[ T' = \frac{1}{f'} = \frac{1}{\frac{20}{19}} = \frac{19}{20} \text{ seconds} \]

### Final Answer
The interval between the successive blasts as heard by the observer is \( \frac{19}{20} \) seconds.

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