A train approaching a hill at a speed of `60 km//hour` sounds a whistle of frequency 600Hz when it is distance of 1km from the hill. Wind is blowing in the direction of the train with a speed of `60 km//h` Find the frequency of the whistle heard by an observer on the hill:speed of sound in air 1200km/h)

Options

A. 610Hz
B. 620Hz
C. 630Hz
D. 650Hz

Answer

To solve the problem of finding the frequency of the whistle heard by an observer on the hill, we can use the Doppler effect formula for sound. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Speed of the train (source) \( v_s = 60 \, \text{km/h} \)
- Frequency of the whistle (source frequency) \( f = 600 \, \text{Hz} \)
- Speed of sound in air \( c = 1200 \, \text{km/h} \)
- Speed of wind \( v_w = 60 \, \text{km/h} \)

### Step 2: Convert all speeds to the same unit
Since all speeds are already in km/h, we can use them directly in our calculations.

### Step 3: Determine the effective speed of sound
The wind is blowing in the same direction as the train, which means it adds to the speed of sound. Therefore, the effective speed of sound \( c' \) is:
\[ c' = c + v_w = 1200 \, \text{km/h} + 60 \, \text{km/h} = 1260 \, \text{km/h} \]

### Step 4: Determine the frequency heard by the observer
The observer is stationary, and the source (train) is moving towards the observer. The formula for the frequency heard by the observer when the source is moving towards the observer is given by:
\[ f' = f \left( \frac{c' + v_o}{c' - v_s} \right) \]
Where:
- \( f' \) is the frequency heard by the observer,
- \( f \) is the source frequency,
- \( c' \) is the effective speed of sound,
- \( v_o \) is the speed of the observer (which is 0 since the observer is stationary),
- \( v_s \) is the speed of the source.

Substituting the values:
\[ f' = 600 \left( \frac{1260 + 0}{1260 - 60} \right) \]

### Step 5: Simplify the equation
Now, simplify the equation:
\[ f' = 600 \left( \frac{1260}{1200} \right) \]
Calculating \( \frac{1260}{1200} \):
\[ \frac{1260}{1200} = 1.05 \]
Thus,
\[ f' = 600 \times 1.05 = 630 \, \text{Hz} \]

### Final Answer
The frequency of the whistle heard by the observer on the hill is **630 Hz**.

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