A tuning fork of unknown frequency produoed 4 beats per second when sounded with a standard fork of frequency 256 Hz. The beat frequency decreased when a little wax was pot on a prong of the first fork. What was the frequency of the fork before loading it with wax ?

Answer

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

### Step-by-Step Solution:

1. **Understanding Beat Frequency**:
The beat frequency is given by the formula:
\[ \text{Beat Frequency} = |f_1 - f_2| \]
where \( f_1 \) is the frequency of the unknown tuning fork and \( f_2 \) is the frequency of the standard tuning fork (256 Hz).

2. **Given Information**:
- Beat frequency = 4 beats per second
- Frequency of standard tuning fork, \( f_2 = 256 \) Hz

3. **Setting Up the Equation**:
Since the beat frequency is 4 Hz, we can write:
\[ |f_1 - 256| = 4 \]
This gives us two possible equations:
- \( f_1 - 256 = 4 \) or
- \( 256 - f_1 = 4 \)

4. **Solving the Equations**:
- From \( f_1 - 256 = 4 \):
\[ f_1 = 256 + 4 = 260 \text{ Hz} \]
- From \( 256 - f_1 = 4 \):
\[ f_1 = 256 - 4 = 252 \text{ Hz} \]

5. **Considering the Effect of Wax**:
When wax is applied to the prong of the first fork, its frequency decreases. This means that the original frequency \( f_1 \) must be greater than 256 Hz, because the beat frequency decreased when the frequency of the first fork was lowered. Thus, we discard the solution \( f_1 = 252 \) Hz.

6. **Final Answer**:
Therefore, the frequency of the tuning fork before loading it with wax is:
\[ f_1 = 260 \text{ Hz} \]

### Summary:
The frequency of the tuning fork before loading it with wax is **260 Hz**.