A train blowing its whistle moves with a constant velocity `v` away from an observer on the ground. The ratio of the natural frequency of the whistle to that measured by the observer is found to be `1.2`. If the train is not rest and the observer moves away from it at the same velocity, this ratio would be given by

Options

A. `0.51`
B. `1.25`
C. `1.52`
D. `2.05`

Explanation

To solve the problem step by step, we will analyze the situation using the Doppler effect for sound waves.

### Step 1: Understand the Doppler Effect
The Doppler effect describes how the frequency of a wave changes for an observer moving relative to the source of the wave. In this case, we have two scenarios:
1. The train (source) is moving away from a stationary observer.
2. The observer is moving away from the train at the same velocity as the train.

### Step 2: First Condition - Train Moving Away from Observer
In the first condition, the train is moving away from the observer. The formula for the apparent frequency \( f_a \) when the source is moving away is given by:
\[ f_a = f_s \left( \frac{v}{v + v_s} \right) \]
Where:
- \( f_a \) = apparent frequency
- \( f_s \) = source frequency (natural frequency of the whistle)
- \( v \) = speed of sound in air
- \( v_s \) = speed of the source (train)

Given that the ratio of the natural frequency to the apparent frequency is \( \frac{f_s}{f_a} = 1.2 \), we can express this as:
\[ \frac{f_s}{f_a} = \frac{v + v_s}{v} \]
Setting this equal to 1.2, we have:
\[ 1.2 = \frac{v + v_s}{v} \]

### Step 3: Solve for \( \frac{v_s}{v} \)
Rearranging the equation:
\[ 1.2v = v + v_s \]
\[ v_s = 1.2v - v = 0.2v \]
Thus, we find:
\[ \frac{v_s}{v} = 0.2 \]

### Step 4: Second Condition - Observer Moving Away from Train
In the second condition, the observer is moving away from the train at the same velocity \( v_s \). The formula for the apparent frequency when the observer is moving away is:
\[ f_a' = f_s \left( \frac{v - v_o}{v} \right) \]
Where \( v_o = v_s \) (the observer's speed).

Substituting \( v_o = v_s \):
\[ f_a' = f_s \left( \frac{v - v_s}{v} \right) \]
Now, substituting \( v_s = 0.2v \):
\[ f_a' = f_s \left( \frac{v - 0.2v}{v} \right) = f_s \left( \frac{0.8v}{v} \right) = 0.8f_s \]

### Step 5: Find the New Ratio
Now, we can find the new ratio of the natural frequency to the apparent frequency:
\[ \frac{f_s}{f_a'} = \frac{f_s}{0.8f_s} = \frac{1}{0.8} = 1.25 \]

### Conclusion
Thus, the ratio of the natural frequency of the whistle to that measured by the observer when both the train and the observer are moving away from each other at the same speed is:
\[ \frac{f_s}{f_a'} = 1.25 \]

### Final Answer
The answer is \( 1.25 \).

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