When the observer moves towards the stationary source with velocity, `v_1`, the apparent frequency of emitted note is `f_1`. When the observer moves away from the source with velocity `v_1`, the apparent frequency is `f_2`. If v is the velocity of sound in air and `f_1/f_2` = 2,then `v/v_1` = ?

Options

A. (a)2
B. (b)3
C. (c)4
D. (d)5

Answer

To solve the problem, we need to analyze the situation using the Doppler effect for sound. We have two scenarios: one when the observer moves towards the stationary source and another when the observer moves away from the source.

### Step-by-Step Solution:

1. **Understanding the Doppler Effect**:
- When the observer moves towards the source, the apparent frequency \( f_1 \) is given by:
\[ f_1 = \frac{v}{v - v_1} f_0 \]
- When the observer moves away from the source, the apparent frequency \( f_2 \) is given by:
\[ f_2 = \frac{v}{v + v_1} f_0 \]
Where \( f_0 \) is the original frequency of the source, \( v \) is the speed of sound, and \( v_1 \) is the speed of the observer.

2. **Setting Up the Ratio**:
- We know from the problem statement that:
\[ \frac{f_1}{f_2} = 2 \]
- Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{\frac{v}{v - v_1} f_0}{\frac{v}{v + v_1} f_0} = 2 \]
- The \( f_0 \) cancels out:
\[ \frac{v}{v - v_1} \cdot \frac{v + v_1}{v} = 2 \]

3. **Simplifying the Equation**:
- This simplifies to:
\[ \frac{v + v_1}{v - v_1} = 2 \]

4. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ v + v_1 = 2(v - v_1) \]
- Expanding the right side:
\[ v + v_1 = 2v - 2v_1 \]

5. **Rearranging the Equation**:
- Bringing all terms involving \( v \) to one side and \( v_1 \) to the other side:
\[ v + v_1 + 2v_1 = 2v \]
- This simplifies to:
\[ 3v_1 = 2v - v \]
- Thus:
\[ 3v_1 = v \]

6. **Finding the Ratio**:
- Dividing both sides by \( v_1 \):
\[ \frac{v}{v_1} = 3 \]

### Final Answer:
\[ \frac{v}{v_1} = 3 \]