A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork ?

Explanation

To solve the problem, we need to determine the original frequency of the first tuning fork based on the information provided about the beat frequencies.

### Step-by-Step Solution:

1. **Understanding Beat Frequency**:
The beat frequency is the absolute difference between the frequencies of two tuning forks. If two tuning forks have frequencies \( f_1 \) and \( f_2 \), the beat frequency \( f_b \) is given by:
\[ f_b = |f_1 - f_2| \]

2. **Identifying Given Frequencies**:
We know one tuning fork has a frequency \( f_2 = 256 \) Hz. The first tuning fork has an unknown frequency \( f_1 \).

3. **Initial Beat Frequency**:
The initial beat frequency is given as 4 beats per second. Therefore, we can write:
\[ |f_1 - 256| = 4 \]
This gives us two possible equations:
\[ f_1 - 256 = 4 \quad \text{(1)} \]
\[ 256 - f_1 = 4 \quad \text{(2)} \]

4. **Solving the Equations**:
- From equation (1):
\[ f_1 = 256 + 4 = 260 \text{ Hz} \]
- From equation (2):
\[ f_1 = 256 - 4 = 252 \text{ Hz} \]

Thus, the possible frequencies for the first tuning fork are 260 Hz or 252 Hz.

5. **Effect of Loading with Wax**:
When the first tuning fork is loaded with wax, the beat frequency increases to 6 beats per second. This indicates that the frequency of the first tuning fork has decreased (since loading with wax typically lowers the frequency). Therefore, we need to consider the lower frequency option:
\[ f_1 = 252 \text{ Hz} \]

6. **Verifying the Beat Frequency After Loading**:
If \( f_1 = 252 \) Hz, the new beat frequency after loading with wax can be calculated as follows:
\[ |f_1' - 256| = 6 \]
where \( f_1' \) is the new frequency after loading. Since \( f_1 \) decreases:
\[ f_1' = 252 - x \quad \text{(where \( x \) is a small frequency decrease)} \]
Thus:
\[ |(252 - x) - 256| = 6 \]
This leads to:
\[ |252 - 256 - x| = 6 \]
Simplifying gives:
\[ |-4 - x| = 6 \]
This results in two cases:
- Case 1: \( -4 - x = 6 \) → \( x = -10 \) (not possible)
- Case 2: \( -4 - x = -6 \) → \( x = 2 \)

Therefore, the first tuning fork's frequency after loading is \( 250 \) Hz, confirming that the original frequency was indeed \( 252 \) Hz.

### Final Answer:
The original frequency of the tuning fork was **252 Hz**.