A tuning fork produces `4` beats/s with a sonometer wire when its lengths are `50cm` , `51cm` . The frequency of that tuning fork is

Options

A. `400Hz`
B. `404Hz`
C. `408Hz`
D. `412Hz`

Explanation

To solve the problem, we need to find the frequency of the tuning fork based on the information provided about the beats produced with the sonometer wire at two different lengths.

### Step-by-Step Solution:

1. **Understanding the Concept of Beats:**
- When two sound waves of slightly different frequencies interfere, they produce a phenomenon known as beats. The number of beats per second is equal to the absolute difference between the two frequencies.

2. **Given Information:**
- The tuning fork produces 4 beats per second with the sonometer wire when its lengths are 50 cm and 51 cm.
- Let the frequency of the tuning fork be \( n \) Hz.
- The frequency of the sonometer wire when its length is 50 cm will be \( n + 4 \) Hz (since it is higher than the tuning fork's frequency).
- The frequency of the sonometer wire when its length is 51 cm will be \( n - 4 \) Hz (since it is lower than the tuning fork's frequency).

3. **Using the Relationship Between Frequency and Length:**
- The frequency of a vibrating string (sonometer wire) is inversely proportional to its length. Thus, we can express the frequencies as:
\[ f_1 = \frac{k}{L_1} \quad \text{and} \quad f_2 = \frac{k}{L_2} \]
- Here, \( k \) is a constant that depends on the tension and mass per unit length of the wire, and \( L_1 \) and \( L_2 \) are the lengths of the wire.

4. **Setting Up the Equations:**
- For \( L_1 = 50 \) cm:
\[ f_1 = \frac{k}{50} = n + 4 \]
- For \( L_2 = 51 \) cm:
\[ f_2 = \frac{k}{51} = n - 4 \]

5. **Equating the Two Frequencies:**
- From the equations above, we can write:
\[ \frac{k}{50} = n + 4 \quad \text{(1)} \]
\[ \frac{k}{51} = n - 4 \quad \text{(2)} \]

6. **Solving for \( k \):**
- Rearranging equation (1):
\[ k = (n + 4) \times 50 \]
- Rearranging equation (2):
\[ k = (n - 4) \times 51 \]

7. **Setting the Two Expressions for \( k \) Equal:**
- Equating the two expressions for \( k \):
\[ (n + 4) \times 50 = (n - 4) \times 51 \]

8. **Expanding and Simplifying:**
- Expanding both sides:
\[ 50n + 200 = 51n - 204 \]
- Rearranging gives:
\[ 200 + 204 = 51n - 50n \]
\[ 404 = n \]

9. **Conclusion:**
- The frequency of the tuning fork is \( n = 404 \) Hz.

### Final Answer:
The frequency of the tuning fork is **404 Hz**.