A tuning fork A produces `4` beats `s^(-1)` with another tuning fork B of frequency `320Hz`. `ON` filing one of the prongs of A, `4` beats `s^(-1)` are again heard when sounded with the same fork B. Then the frequency of the fork a before filling is

Options

A. `328Hz`
B. `316Hz`
C. `324Hz`
D. `320Hz`

Explanation

To solve the problem, we need to determine the frequency of tuning fork A before it was filed. Let's break down the steps:

### Step-by-Step Solution:

1. **Understanding Beats**:
When two tuning forks produce beats, the number of beats per second (bps) is equal to the absolute difference between their frequencies. In this case, tuning fork A produces 4 beats per second with tuning fork B, which has a frequency of 320 Hz.

2. **Setting Up the Equation**:
The frequency of tuning fork A (let's denote it as \( f_A \)) can either be:
\[ f_A = f_B + 4 \quad \text{or} \quad f_A = f_B - 4 \]
Given that \( f_B = 320 \, \text{Hz} \), we can calculate:
- \( f_A = 320 + 4 = 324 \, \text{Hz} \)
- \( f_A = 320 - 4 = 316 \, \text{Hz} \)

3. **Considering the Effect of Filing**:
Filing one of the prongs of tuning fork A will decrease its mass, which in turn increases its frequency. This means that after filing, the frequency of A must be higher than its original frequency.

4. **Analyzing the Two Cases**:
- If \( f_A = 324 \, \text{Hz} \) before filing, filing would increase the frequency further, leading to a larger beat frequency with fork B, which contradicts the fact that we still hear 4 beats per second.
- If \( f_A = 316 \, \text{Hz} \) before filing, filing would increase the frequency, and it is possible to still have 4 beats per second with fork B (320 Hz).

5. **Conclusion**:
Since the only viable option that allows for 4 beats per second after filing is when the original frequency of tuning fork A was 316 Hz, we conclude that:
\[ \text{Frequency of tuning fork A before filing} = 316 \, \text{Hz} \]

### Final Answer:
The frequency of tuning fork A before filing is **316 Hz**.

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