A train moves towards a stationary observer with speed `34 m//s`. The train sounds a whistle and its frequency registered by the observer is `f_(1)`. If the train's speed is reduced to `17 m//s`, the frequency registered is `f_(2)`. If the speed of sound of `340 m//s`, then the ratio `f_(1)//f_(2)`is

Options

A. `18//19`
B. `1//2`
C. `2`
D. `19//18`

Explanation

To solve the problem, we will use the Doppler effect formula for sound waves. The formula for the frequency observed when the source is moving towards a stationary observer is given by:

\[ f' = f \frac{c}{c - v_s} \]

Where:
- \(f'\) is the frequency observed,
- \(f\) is the frequency of the source,
- \(c\) is the speed of sound,
- \(v_s\) is the speed of the source (train).

### Step 1: Calculate \(f_1\) when the train moves at 34 m/s
Given:
- Speed of sound, \(c = 340 \, m/s\)
- Speed of the train, \(v_s = 34 \, m/s\)

Using the formula for \(f_1\):

\[ f_1 = f \frac{c}{c - v_s} = f \frac{340}{340 - 34} = f \frac{340}{306} \]

This simplifies to:

\[ f_1 = f \frac{340}{306} = \frac{170}{153} f \quad \text{(by dividing numerator and denominator by 2)} \]

### Step 2: Calculate \(f_2\) when the train moves at 17 m/s
Now, when the speed of the train is reduced to \(v_s = 17 \, m/s\):

\[ f_2 = f \frac{c}{c - v_s} = f \frac{340}{340 - 17} = f \frac{340}{323} \]

### Step 3: Calculate the ratio \(\frac{f_1}{f_2}\)
Now we can find the ratio of \(f_1\) to \(f_2\):

\[ \frac{f_1}{f_2} = \frac{\frac{170}{153} f}{\frac{340}{323} f} \]

The \(f\) cancels out:

\[ \frac{f_1}{f_2} = \frac{170}{153} \cdot \frac{323}{340} \]

### Step 4: Simplify the ratio
Calculating the ratio:

\[ \frac{f_1}{f_2} = \frac{170 \times 323}{153 \times 340} \]

Calculating the values:

- \(170 \times 323 = 54910\)
- \(153 \times 340 = 52020\)

Thus,

\[ \frac{f_1}{f_2} = \frac{54910}{52020} \]

This simplifies to approximately:

\[ \frac{f_1}{f_2} \approx \frac{19}{18} \]

### Final Answer
The ratio \(\frac{f_1}{f_2}\) is:

\[ \frac{f_1}{f_2} = \frac{19}{18} \]

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