A train, blowing a whistle of frequency f, is standing in the railway yard A person is running towards the engine. The frequency of the sound of the whistle as heard by the person will be

Options

A. greater than f
B. less than f
C. equal to f
D. greater or less than f depending on the speed of the train.

Explanation

To solve the problem of determining the frequency of the sound of the whistle as heard by a person running towards a stationary train, we can use the Doppler effect formula. Here’s a step-by-step solution:

### Step 1: Understand the scenario
We have a train that is stationary and emitting a sound (whistle) of frequency \( f \). A person is running towards the train.

### Step 2: Identify the variables
- \( f_0 \) (the emitted frequency of the source) = \( f \)
- \( v \) (the speed of sound in air) = \( V \) (a constant value, typically around 343 m/s at room temperature)
- \( v_o \) (the speed of the observer, which is the person running towards the train) = \( V_o \) (a positive value)
- \( v_s \) (the speed of the source, which is the train) = \( 0 \) (since the train is stationary)

### Step 3: Write the Doppler effect formula
The formula for the apparent frequency \( f' \) heard by the observer is given by:

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

### Step 4: Substitute the values into the formula
Since the train is stationary, \( v_s = 0 \). Therefore, the formula simplifies to:

\[ f' = f \frac{v + V_o}{v} \]

### Step 5: Simplify the expression
This can be further simplified to:

\[ f' = f \left(1 + \frac{V_o}{v}\right) \]

### Step 6: Analyze the result
Since \( V_o \) (the speed of the observer) is a positive value, the term \( \frac{V_o}{v} \) is also positive. Thus, \( f' \) is greater than \( f \):

\[ f' > f \]

### Conclusion
The frequency of the sound of the whistle as heard by the person running towards the train will be greater than the frequency \( f \) emitted by the train.

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