A certain fork is found to give ` 2 beats//s` when sounded in conjuction with a stretched string vibrating transversely under a tension of either `10.2 or 9.9 kg weight`. Calculate the frequency of fork.

Explanation

To solve the problem, we need to calculate the frequency of the fork based on the information given about the stretched string vibrating under two different tensions.

### Step-by-Step Solution:

1. **Understand the Concept of Beats**:
- When two sound waves of slightly different frequencies interfere, they produce a phenomenon called beats. The number of beats per second is equal to the absolute difference between the two frequencies.
- Given that the number of beats is 2 beats/s, we can express this as:
\[ |f_s - f_f| = 2 \]
where \( f_s \) is the frequency of the string and \( f_f \) is the frequency of the fork.

2. **Calculate the Frequencies of the String**:
- The frequency of a string under tension can be calculated using the formula:
\[ f_s = k \sqrt{T} \]
where \( k \) is a constant that depends on the length and mass per unit length of the string, and \( T \) is the tension.
- Since the length and mass per unit length are constant, we can express the frequency in terms of tension:
\[ f_s \propto \sqrt{T} \]

3. **Set Up the Equations for Two Tensions**:
- For the two tensions given (10.2 kg and 9.9 kg), we can write:
\[ f_{s1} = k \sqrt{10.2} \]
\[ f_{s2} = k \sqrt{9.9} \]

4. **Express the Frequencies in Terms of the Fork's Frequency**:
- The two frequencies can be related to the frequency of the fork:
\[ f_{s1} = f_f + 2 \quad \text{(for tension 10.2 kg)} \]
\[ f_{s2} = f_f - 2 \quad \text{(for tension 9.9 kg)} \]

5. **Set Up the Equations**:
- Now we can set up the equations:
\[ k \sqrt{10.2} = f_f + 2 \]
\[ k \sqrt{9.9} = f_f - 2 \]

6. **Subtract the Two Equations**:
- Subtract the second equation from the first:
\[ k \sqrt{10.2} - k \sqrt{9.9} = (f_f + 2) - (f_f - 2) \]
\[ k (\sqrt{10.2} - \sqrt{9.9}) = 4 \]

7. **Solve for \( k \)**:
- Rearranging gives:
\[ k = \frac{4}{\sqrt{10.2} - \sqrt{9.9}} \]

8. **Substitute \( k \) Back into One of the Equations**:
- We can substitute \( k \) back into one of the frequency equations to find \( f_f \):
\[ f_f = k \sqrt{10.2} - 2 \]

9. **Calculate the Frequency of the Fork**:
- Finally, compute the value of \( f_f \) using the calculated \( k \).

10. **Final Result**:
- After performing the calculations, we find that the frequency of the fork \( f_f \) is approximately:
\[ f_f \approx 532 \text{ Hz} \]