A stationary source emits the sound of frequency `f_0 = 492` Hz. The sound is reflected by a large car approaching the source with a speed of `2 ms^(-1)`. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is `330 ms^(-1)` and the car reflects the sound at the frequency it has received).

Explanation

To solve the problem, we will use the Doppler effect to find the frequency of the sound as it is reflected by the moving car and then calculate the beat frequency.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency of the source, \( f_0 = 492 \, \text{Hz} \)
- Speed of the car, \( v_c = 2 \, \text{m/s} \)
- Speed of sound in air, \( v = 330 \, \text{m/s} \)

2. **Calculate the Frequency Received by the Car:**
The frequency \( f' \) received by the car can be calculated using the Doppler effect formula when the source is stationary and the observer (the car) is moving towards the source:
\[ f' = f_0 \cdot \frac{v + v_c}{v} \]
Substituting the values:
\[ f' = 492 \cdot \frac{330 + 2}{330} \]
\[ f' = 492 \cdot \frac{332}{330} \]
\[ f' \approx 492 \cdot 1.00606 \approx 495.0 \, \text{Hz} \]

3. **Calculate the Frequency of the Reflected Sound:**
The car reflects the sound at the frequency it has received. Therefore, the frequency \( f'' \) emitted by the car is the same as \( f' \):
\[ f'' = f' \approx 495.0 \, \text{Hz} \]

4. **Calculate the Beat Frequency:**
The beat frequency \( f_b \) is the absolute difference between the original frequency \( f_0 \) and the frequency of the reflected sound \( f'' \):
\[ f_b = |f'' - f_0| \]
Substituting the values:
\[ f_b = |495.0 - 492| = 3.0 \, \text{Hz} \]

5. **Final Answer:**
The beat frequency of the resulting signal is \( 3 \, \text{Hz} \).