Two sound producing sources A and B are moving towards and away from a stationary observer with same speed respectively. If frequency of sound produced by both sources are equal as 400 Hz, then speed of sources (approximately) when observer detects 4 beats per second, is [Given, speed of sound =340 m/s]

Options

A. 1.7 m/s
B. 3.4 m/s
C. 2.4 m/s
D. 1 m/s

Explanation

To solve the problem step by step, we will apply the principles of the Doppler effect for sound waves.

### Step 1: Understand the scenario
We have two sound sources:
- Source A is moving towards the observer with speed \( v_s \).
- Source B is moving away from the observer with the same speed \( v_s \).
Both sources produce sound at a frequency \( f = 400 \, \text{Hz} \).

### Step 2: Use the Doppler effect formula
For a stationary observer, the frequency heard from a source moving towards the observer is given by:
\[ f_1 = \frac{v + v_o}{v - v_s} f \]
Where:
- \( f_1 \) is the frequency heard from source A (moving towards the observer).
- \( v \) is the speed of sound in air (given as \( 340 \, \text{m/s} \)).
- \( v_o \) is the speed of the observer (which is \( 0 \) since the observer is stationary).
- \( v_s \) is the speed of the source A.

Since \( v_o = 0 \), the formula simplifies to:
\[ f_1 = \frac{v}{v - v_s} f \]

For source B (moving away from the observer):
\[ f_2 = \frac{v + v_o}{v + v_s} f \]
Again, since \( v_o = 0 \), it simplifies to:
\[ f_2 = \frac{v}{v + v_s} f \]

### Step 3: Set up the beat frequency equation
The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f_1 - f_2| \]
We know from the problem statement that \( f_b = 4 \, \text{Hz} \).

### Step 4: Substitute the frequencies into the beat frequency equation
Substituting \( f_1 \) and \( f_2 \) into the beat frequency equation gives:
\[ 4 = \left| \frac{340}{340 - v_s} \cdot 400 - \frac{340}{340 + v_s} \cdot 400 \right| \]
Factoring out \( 400 \):
\[ 4 = 400 \left| \frac{340}{340 - v_s} - \frac{340}{340 + v_s} \right| \]

### Step 5: Simplify the equation
Dividing both sides by 400:
\[ \frac{4}{400} = \left| \frac{340}{340 - v_s} - \frac{340}{340 + v_s} \right| \]
\[ 0.01 = \left| \frac{340(340 + v_s) - 340(340 - v_s)}{(340 - v_s)(340 + v_s)} \right| \]
This simplifies to:
\[ 0.01 = \left| \frac{340v_s + 340v_s}{(340 - v_s)(340 + v_s)} \right| \]
\[ 0.01 = \left| \frac{680v_s}{340^2 - v_s^2} \right| \]

### Step 6: Solve for \( v_s \)
Cross-multiplying gives:
\[ 0.01(340^2 - v_s^2) = 680v_s \]
This expands to:
\[ 3400 - 0.01v_s^2 = 680v_s \]
Rearranging gives a quadratic equation:
\[ 0.01v_s^2 + 680v_s - 3400 = 0 \]

### Step 7: Use the quadratic formula
Using the quadratic formula \( v_s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 0.01, \, b = 680, \, c = -3400 \]
Calculating the discriminant:
\[ D = b^2 - 4ac = 680^2 - 4 \cdot 0.01 \cdot (-3400) \]
\[ D = 462400 + 136 = 462536 \]
Now, calculate \( v_s \):
\[ v_s = \frac{-680 \pm \sqrt{462536}}{2 \cdot 0.01} \]
Calculating \( \sqrt{462536} \approx 680 \):
\[ v_s = \frac{-680 \pm 680}{0.02} \]
This gives two solutions:
1. \( v_s = 0 \) (not valid, as sources are moving)
2. \( v_s \approx 34 \, \text{m/s} \)

### Final Answer
Thus, the approximate speed of the sources is:
\[ v_s \approx 34 \, \text{m/s} \]