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 frequency, speed of sound, and distance traveled.

### Step-by-Step Solution:

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

2. **Use the Doppler Effect Formula:**
The frequency detected by the observer moving towards the source can be expressed using the Doppler effect formula:
\[ f = \left( \frac{c + v_o}{c} \right) f_0 \]
where \( v_o \) is the velocity of the observer (detector).

3. **Rearranging the Formula:**
Rearranging the formula to find \( v_o \):
\[ f = \left( \frac{c + v_o}{c} \right) f_0 \implies \frac{f}{f_0} = \frac{c + v_o}{c} \]
\[ c + v_o = \frac{f}{f_0} \cdot c \]
\[ v_o = \left( \frac{f}{f_0} - 1 \right) c \]

4. **Substituting Values:**
Substitute the known values into the equation:
\[ v_o = \left( \frac{275}{262} - 1 \right) \cdot 330 \]
Calculate \( \frac{275}{262} \):
\[ \frac{275}{262} \approx 1.0496 \implies v_o \approx (1.0496 - 1) \cdot 330 \approx 0.0496 \cdot 330 \approx 16.368 \, \text{m/s} \]

5. **Calculate the Duration of the Beep:**
The duration \( T \) of the beep can be calculated using the formula:
\[ T = \frac{L}{v_o} \]
Substitute \( L = 78 \, \text{m} \) and \( v_o \approx 16.368 \, \text{m/s} \):
\[ T = \frac{78}{16.368} \approx 4.77 \, \text{s} \]

### Final Answer:
The duration of the beep emitted by the source is approximately \( T \approx 4.77 \, \text{s} \).