A sound detector is moving with a constant velocity towards a stationary sound source. The source emits a beep of sound of frequency f=262 Hz that propogates with speec c=330m/s. The detector detects the beep of sound over a distance L=78 m and registers a frequency f=275 Hz. Find thhe duration of the beep emitted by the source.

Options

A. 2s
B. 4s
C. 5s
D. 6s

Explanation

To solve the problem, we need to find the duration of the beep emitted by the sound source. We will use the relationship between the frequencies and the speed of sound to derive the duration.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency of the source, \( f_0 = 262 \, \text{Hz} \)
- Speed of sound, \( c = 330 \, \text{m/s} \)
- Frequency detected by the moving detector, \( f = 275 \, \text{Hz} \)
- Distance over which the beep is detected, \( L = 78 \, \text{m} \)

2. **Understand the Relationship Between Frequencies:**
The relationship between the emitted frequency \( f_0 \) and the observed frequency \( f \) when the detector is moving towards the source is given by:
\[ f = \frac{c + v_0}{c} f_0 \]
where \( v_0 \) is the velocity of the detector.

3. **Rearranging the Equation:**
We can rearrange the equation to find \( v_0 \):
\[ v_0 = c \left( \frac{f}{f_0} - 1 \right) \]

4. **Substituting the Values:**
Substitute \( f = 275 \, \text{Hz} \) and \( f_0 = 262 \, \text{Hz} \):
\[ v_0 = 330 \left( \frac{275}{262} - 1 \right) \]

5. **Calculating \( v_0 \):**
\[ v_0 = 330 \left( 1.049 - 1 \right) = 330 \times 0.049 \approx 16.17 \, \text{m/s} \]

6. **Finding the Duration of the Beep:**
The duration of the beep can be calculated using the relationship:
\[ t = \frac{L}{v_0} \]
where \( L = 78 \, \text{m} \).

7. **Substituting the Values:**
\[ t = \frac{78}{16.17} \approx 4.82 \, \text{s} \]

8. **Final Calculation:**
Rounding off, we can say the duration of the beep is approximately \( t \approx 5 \, \text{s} \).

### Conclusion:
The duration of the beep emitted by the source is approximately **5 seconds**.