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 find the original frequency of the first tuning fork (let's denote it as \( \nu_1 \)). We are given the following information:

1. The frequency of the second tuning fork (\( \nu_2 \)) is 256 Hz.
2. The initial beat frequency is 4 beats per second.
3. After loading the first tuning fork with wax, the beat frequency increases to 6 beats per second.

### Step-by-Step Solution:

**Step 1: Understand the concept of beats.**
The beat frequency is the absolute difference between the frequencies of the two tuning forks. This can be expressed mathematically as:
\[ |\nu_1 - \nu_2| = \text{beat frequency} \]

**Step 2: Set up the equations for the initial and final conditions.**
Initially, the beat frequency is 4 beats per second:
\[ |\nu_1 - 256| = 4 \]
This gives us two possible equations:
1. \( \nu_1 - 256 = 4 \) (Equation A)
2. \( 256 - \nu_1 = 4 \) (Equation B)

After loading the first tuning fork with wax, the beat frequency increases to 6 beats per second:
\[ |\nu_1' - 256| = 6 \]
Since loading the fork decreases its frequency, we denote the new frequency as \( \nu_1' = \nu_1 - \Delta \) (where \( \Delta \) is a small decrease in frequency). Therefore, we have:
\[ |\nu_1 - \Delta - 256| = 6 \]

**Step 3: Analyze the equations.**
From Equation A:
\[ \nu_1 = 260 \, \text{Hz} \]
From Equation B:
\[ \nu_1 = 252 \, \text{Hz} \]

Now, we need to check which of these frequencies leads to an increase in beat frequency when the fork is loaded.

**Step 4: Check the conditions for each case.**
1. If \( \nu_1 = 260 \, \text{Hz} \):
- New frequency after loading: \( \nu_1' = 260 - \Delta \)
- Beat frequency after loading: \( |(260 - \Delta) - 256| = |4 - \Delta| \)
- This would not increase the beat frequency to 6 beats per second.

2. If \( \nu_1 = 252 \, \text{Hz} \):
- New frequency after loading: \( \nu_1' = 252 - \Delta \)
- Beat frequency after loading: \( |(252 - \Delta) - 256| = |252 - \Delta - 256| = | -4 - \Delta| \)
- This can increase to 6 beats per second if \( \Delta \) is small enough.

Thus, the original frequency of the tuning fork is:
\[ \nu_1 = 252 \, \text{Hz} \]

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

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