Two sources of sound moving with same speed v and emitting frequency of 1400 Hz are moving such that one source `s_1` is moving towards the observer and `s_2` is moving away from observer. If observer hears beat frequency of 2 Hz. Then find the speed of source. Given `v_(sound) gt gt v_(source)` and `v_(sound) = 350m/s`

Options

A. 4
B. `1/4`
C. 2
D. `1/2`

Explanation

To solve the problem, we will use the concept of the Doppler effect for sound waves. The beat frequency is the difference between the frequencies heard from two sources moving towards and away from the observer.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two sound sources, \(S_1\) moving towards the observer and \(S_2\) moving away from the observer.
- Both sources emit a frequency \(f_0 = 1400 \, \text{Hz}\).
- The speed of sound in air, \(c = 350 \, \text{m/s}\).
- The beat frequency heard by the observer is \(f_b = 2 \, \text{Hz}\).

2. **Applying the Doppler Effect**:
- For the source \(S_1\) moving towards the observer, the observed frequency \(f_1\) is given by:
\[ f_1 = f_0 \frac{c}{c - v} \]
- For the source \(S_2\) moving away from the observer, the observed frequency \(f_2\) is given by:
\[ f_2 = f_0 \frac{c}{c + v} \]

3. **Finding the Beat Frequency**:
- The beat frequency \(f_b\) is the absolute difference between the two frequencies:
\[ f_b = |f_1 - f_2| \]
- Substituting the expressions for \(f_1\) and \(f_2\):
\[ f_b = \left| f_0 \frac{c}{c - v} - f_0 \frac{c}{c + v} \right| \]

4. **Simplifying the Expression**:
- Factor out \(f_0\):
\[ f_b = f_0 \left| \frac{c}{c - v} - \frac{c}{c + v} \right| \]
- Combine the fractions:
\[ f_b = f_0 \left| \frac{c(c + v) - c(c - v)}{(c - v)(c + v)} \right| \]
- Simplifying the numerator:
\[ f_b = f_0 \left| \frac{cv + cv}{(c - v)(c + v)} \right| = f_0 \left| \frac{2cv}{(c - v)(c + v)} \right| \]

5. **Setting Up the Equation**:
- We know \(f_b = 2 \, \text{Hz}\), so we set up the equation:
\[ 2 = 1400 \left| \frac{2 \cdot 350 \cdot v}{(350 - v)(350 + v)} \right| \]

6. **Solving for \(v\)**:
- Rearranging gives:
\[ 2 = 1400 \cdot \frac{700v}{(350 - v)(350 + v)} \]
- Simplifying further:
\[ 2 = \frac{980000v}{(350 - v)(350 + v)} \]
- Cross-multiplying:
\[ 2(350 - v)(350 + v) = 980000v \]
- Expanding the left side:
\[ 2(122500 - v^2) = 980000v \]
- This simplifies to:
\[ 245000 - 2v^2 = 980000v \]
- Rearranging gives:
\[ 2v^2 + 980000v - 245000 = 0 \]

7. **Using the Quadratic Formula**:
- We can use the quadratic formula \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = 980000\), and \(c = -245000\).
- Calculate the discriminant:
\[ b^2 - 4ac = (980000)^2 - 4 \cdot 2 \cdot (-245000) \]
- Solve for \(v\).

8. **Final Calculation**:
- After calculating the values, we find that:
\[ v \approx 0.25 \, \text{m/s} \]

Thus, the speed of the source is approximately **0.25 m/s**.