A car emitting sound of frequency 500 Hz speeds towards a fixed wall at `4(m)/(s)`. An observer in the car hears both the source frequency as well as the frequency of sound reflected from the wall. If he hears 10 beats per second between the two sounds, the velocity of sound in air will be

Options

A. `330(m)/(s)`
B. `387(m)/(s)`
C. `404(m)/(s)`
D. `340(m)/(s)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a car emitting sound at a frequency of 500 Hz and moving towards a wall at a speed of 4 m/s. The observer inside the car hears both the original sound and the sound reflected from the wall. The difference in frequency (beat frequency) between these two sounds is given as 10 beats per second.

### Step 2: Identify the Frequencies
1. **Frequency of the sound emitted by the car (f1):** This is the original frequency, which is given as \( f_1 = 500 \, \text{Hz} \).
2. **Frequency of the sound reflected from the wall (f2):** We need to calculate this frequency using the Doppler effect.

### Step 3: Calculate the Frequency Reflected from the Wall
To find the frequency \( f_2 \) that the observer hears after the sound reflects off the wall, we first need to find the frequency \( f_3 \) that hits the wall.

Using the Doppler effect formula for a source moving towards a stationary observer (the wall):
\[ f_3 = f_0 \frac{v}{v - v_s} \]
Where:
- \( f_0 = 500 \, \text{Hz} \) (source frequency),
- \( v \) = speed of sound in air (unknown),
- \( v_s = 4 \, \text{m/s} \) (speed of the car).

So,
\[ f_3 = 500 \frac{v}{v - 4} \]

### Step 4: Calculate the Frequency Heard by the Observer
Now, the observer in the car hears the frequency \( f_2 \) after the sound reflects off the wall. The formula for this is:
\[ f_2 = f_3 \frac{v + v_o}{v} \]
Where:
- \( v_o = 4 \, \text{m/s} \) (speed of the observer, which is the same as the speed of the car).

Thus,
\[ f_2 = \left(500 \frac{v}{v - 4}\right) \frac{v + 4}{v} \]
This simplifies to:
\[ f_2 = 500 \frac{(v + 4)}{(v - 4)} \]

### Step 5: Set Up the Beat Frequency Equation
The beat frequency is the absolute difference between the two frequencies:
\[ |f_2 - f_1| = 10 \, \text{Hz} \]
Since \( f_2 > f_1 \):
\[ f_2 - f_1 = 10 \]
Substituting the values we have:
\[ 500 \frac{(v + 4)}{(v - 4)} - 500 = 10 \]

### Step 6: Solve for the Speed of Sound \( v \)
1. Factor out 500:
\[ 500 \left(\frac{(v + 4)}{(v - 4)} - 1\right) = 10 \]
2. Simplify:
\[ \frac{(v + 4) - (v - 4)}{(v - 4)} = \frac{8}{(v - 4)} \]
3. Set the equation:
\[ 500 \cdot \frac{8}{(v - 4)} = 10 \]
4. Cross-multiply:
\[ 500 \cdot 8 = 10(v - 4) \]
5. Simplify:
\[ 4000 = 10v - 40 \]
6. Rearranging gives:
\[ 10v = 4040 \implies v = 404 \, \text{m/s} \]

### Conclusion
The speed of sound in air is \( v = 404 \, \text{m/s} \).

---