A tuning fork A of frequency f gives 5 beats/s with a tuning fork B of frequency 512 Hz. If on filing A, the number of beats remains same, then find the value of f.

Answer

To solve the problem, we need to determine the frequency \( f \) of tuning fork A, given that it produces 5 beats per second with tuning fork B, which has a frequency of 512 Hz.

### Step-by-Step Solution:

1. **Understanding Beats**: The number of beats per second is given by the absolute difference in frequencies of the two tuning forks. Thus, we can write:
\[ |f - 512| = 5 \]

2. **Setting Up the Equations**: This absolute value equation leads to two possible cases:
- Case 1: \( f - 512 = 5 \)
- Case 2: \( f - 512 = -5 \)

3. **Solving Case 1**:
- From Case 1:
\[ f - 512 = 5 \implies f = 512 + 5 = 517 \text{ Hz} \]

4. **Solving Case 2**:
- From Case 2:
\[ f - 512 = -5 \implies f = 512 - 5 = 507 \text{ Hz} \]

5. **Analyzing the Effect of Filing**: The problem states that when tuning fork A is filed, the number of beats remains the same (5 beats/s). Filing increases the frequency of A.
- If \( f = 517 \) Hz, filing it would increase the frequency further, resulting in a greater beat frequency than 5 beats/s.
- If \( f = 507 \) Hz, filing it would also increase its frequency, but it could still maintain a beat frequency of 5 beats/s if it increases to 512 Hz (the frequency of fork B).

6. **Conclusion**: The only value of \( f \) that allows the beat frequency to remain constant at 5 beats/s after filing is:
\[ f = 507 \text{ Hz} \]

### Final Answer:
The value of \( f \) is **507 Hz**.