The difference between the apparent frequency of a sound of soun as perceived by an observer during its approach and recession is `2%` of the natural frequency of the source. If the velocity of sound in air is `300m//s`, the velocity of the source is (It is given that velocity of source `ltlt velocity of sound )

Options

A. `6sm//sec`
B. `3 m//sec`
C. `1.5m//sec`
D. `12 m//sec`

Explanation

To solve the problem, we need to find the velocity of the source (Vs) given that the difference between the apparent frequencies during approach and recession is 2% of the natural frequency (n). The velocity of sound in air (V) is given as 300 m/s.

### Step-by-Step Solution:

1. **Understanding the Doppler Effect**:
- When a source of sound approaches an observer, the apparent frequency (n1') increases, and when it recedes, the apparent frequency (n2') decreases.
- The formulas for the apparent frequencies are:
- \( n1' = n \left( \frac{V}{V - V_s} \right) \) (when approaching)
- \( n2' = n \left( \frac{V}{V + V_s} \right) \) (when receding)

2. **Setting Up the Equation**:
- The difference between the apparent frequencies is given as:
\[ n1' - n2' = 0.02n \]
- Substituting the expressions for \( n1' \) and \( n2' \):
\[ n \left( \frac{V}{V - V_s} \right) - n \left( \frac{V}{V + V_s} \right) = 0.02n \]

3. **Simplifying the Equation**:
- Dividing through by \( n \) (assuming \( n \neq 0 \)):
\[ \frac{V}{V - V_s} - \frac{V}{V + V_s} = 0.02 \]
- Finding a common denominator:
\[ \frac{V(V + V_s) - V(V - V_s)}{(V - V_s)(V + V_s)} = 0.02 \]
- This simplifies to:
\[ \frac{V^2 + V V_s - V^2 + V V_s}{(V - V_s)(V + V_s)} = 0.02 \]
- Further simplifying gives:
\[ \frac{2V V_s}{(V - V_s)(V + V_s)} = 0.02 \]

4. **Cross-Multiplying**:
- Cross-multiplying to eliminate the fraction:
\[ 2V V_s = 0.02 (V^2 - V_s^2) \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 2V V_s + 0.02 V_s^2 = 0.02 V^2 \]
- This is a quadratic equation in \( V_s \):
\[ 0.02 V_s^2 + 2V V_s - 0.02 V^2 = 0 \]

6. **Using the Quadratic Formula**:
- The quadratic formula is given by:
\[ V_s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 0.02 \), \( b = 2V \), and \( c = -0.02 V^2 \):
\[ V_s = \frac{-2V \pm \sqrt{(2V)^2 - 4(0.02)(-0.02V^2)}}{2(0.02)} \]
- Simplifying gives:
\[ V_s = \frac{-2V \pm \sqrt{4V^2 + 0.0016V^2}}{0.04} \]
\[ V_s = \frac{-2V \pm \sqrt{4.0016V^2}}{0.04} \]
- Since \( V_s \) must be positive, we take the positive root:
\[ V_s = \frac{-2V + 2V}{0.04} = \frac{2V}{0.04} \]

7. **Substituting the Value of V**:
- Given \( V = 300 \, m/s \):
\[ V_s = \frac{2 \times 300}{0.04} = \frac{600}{0.04} = 15000 \, m/s \]

8. **Final Calculation**:
- However, since we are given that \( V_s << V \), we need to consider the approximation:
\[ V_s \approx \frac{0.02 \times 300}{2} = 3 \, m/s \]

### Conclusion:
The velocity of the source \( V_s \) is approximately **3 m/s**.