Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of `20 ms^(-1)` with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B? (Speed of sound in air =`340 ms^(-1)`)

Options

A. 2250 Hz
B. 2060 Hz
C. 2300 Hz
D. 2150 Hz

Answer

To solve the problem, we will use the Doppler effect formula for sound. The Doppler effect describes how the frequency of a wave changes for an observer moving relative to the source of the wave.

### Step-by-Step Solution:

1. **Identify the Known Values**:
- Speed of sound in air, \( V = 340 \, \text{m/s} \)
- Speed of car A (observer), \( V_o = 20 \, \text{m/s} \)
- Speed of car B (source), \( V_s = 20 \, \text{m/s} \)
- Observed frequency, \( f = 2000 \, \text{Hz} \)

2. **Doppler Effect Formula**:
The formula for the observed frequency when both the source and observer are moving away from each other is given by:
\[ f = f_0 \left( \frac{V - V_o}{V + V_s} \right) \]
where:
- \( f_0 \) is the natural frequency of the source (what we need to find),
- \( V \) is the speed of sound,
- \( V_o \) is the speed of the observer (moving away),
- \( V_s \) is the speed of the source (also moving away).

3. **Substituting the Known Values**:
Substitute the known values into the formula:
\[ 2000 = f_0 \left( \frac{340 - 20}{340 + 20} \right) \]

4. **Simplifying the Equation**:
Calculate the terms inside the parentheses:
\[ 340 - 20 = 320 \quad \text{and} \quad 340 + 20 = 360 \]
Thus, the equation becomes:
\[ 2000 = f_0 \left( \frac{320}{360} \right) \]

5. **Solving for \( f_0 \)**:
Rearranging the equation to isolate \( f_0 \):
\[ f_0 = 2000 \times \frac{360}{320} \]

6. **Calculating the Value**:
Simplifying the fraction:
\[ \frac{360}{320} = \frac{9}{8} \]
Therefore:
\[ f_0 = 2000 \times \frac{9}{8} = 2000 \times 1.125 = 2250 \, \text{Hz} \]

### Final Answer:
The natural frequency of the sound source in car B is \( f_0 = 2250 \, \text{Hz} \).

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