On sounding fork `A` with another tuning fork `B` of frequency `384 Hz , 6 beats` are produced per second .After loading the prongs of `A` with wax and then sounding it again with `B , 4 beats` are produced per second. What is the frequency of the tuning fork `A`.

Options

A. `388 Hz`
B. `80 Hz`
C. `378 Hz`
D. `390 Hz`

Explanation

To find the frequency of tuning fork A, we can follow these steps:

### Step 1: Understand the Beat Frequency
The beat frequency is the absolute difference between the frequencies of two tuning forks. If tuning fork A has a frequency \( f_A \) and tuning fork B has a frequency \( f_B = 384 \, \text{Hz} \), then the beat frequency \( f_{beat} \) can be expressed as:
\[ f_{beat} = |f_A - f_B| \]

### Step 2: Set Up the Equations for the First Scenario
Initially, when tuning fork A is sounded with B, 6 beats are produced per second. This gives us two possible equations:
1. \( f_A - 384 = 6 \)
2. \( 384 - f_A = 6 \)

From the first equation:
\[ f_A = 384 + 6 = 390 \, \text{Hz} \]

From the second equation:
\[ f_A = 384 - 6 = 378 \, \text{Hz} \]

### Step 3: Set Up the Equations for the Second Scenario
After loading the prongs of A with wax, the frequency of A decreases, and now 4 beats are produced per second. Let’s denote the new frequency of A as \( f'_A \). We can set up the following equations:
1. \( f'_A - 384 = 4 \)
2. \( 384 - f'_A = 4 \)

From the first equation:
\[ f'_A = 384 + 4 = 388 \, \text{Hz} \]

From the second equation:
\[ f'_A = 384 - 4 = 380 \, \text{Hz} \]

### Step 4: Analyze the Results
We have two possible frequencies for tuning fork A:
- From the first scenario: \( f_A = 390 \, \text{Hz} \) or \( f_A = 378 \, \text{Hz} \)
- From the second scenario: \( f'_A = 388 \, \text{Hz} \) or \( f'_A = 380 \, \text{Hz} \)

Since the problem states that after waxing, the frequency of A decreases, we need to check which of the initial frequencies leads to a decrease:
- If \( f_A = 390 \, \text{Hz} \), then \( f'_A \) could be \( 388 \, \text{Hz} \) (decrease).
- If \( f_A = 378 \, \text{Hz} \), then \( f'_A \) could be \( 380 \, \text{Hz} \) (increase).

### Conclusion
The only consistent solution is:
\[ f_A = 390 \, \text{Hz} \]

Thus, the frequency of tuning fork A is **390 Hz**.

---