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 the two tuning forks. When tuning fork A is sounded with tuning fork B (384 Hz), it produces 6 beats per second. This means:
\[ |f_A - f_B| = 6 \]
Where \( f_A \) is the frequency of tuning fork A and \( f_B = 384 \, \text{Hz} \).

### Step 2: Set up the equations
From the information given, we can set up two possible equations based on the beat frequency:
1. \( f_A - 384 = 6 \) (if \( f_A > 384 \))
2. \( 384 - f_A = 6 \) (if \( f_A < 384 \))

From these equations, we can solve for \( f_A \):
1. If \( f_A - 384 = 6 \):
\[ f_A = 390 \, \text{Hz} \]
2. If \( 384 - f_A = 6 \):
\[ f_A = 378 \, \text{Hz} \]

### Step 3: Analyze the effect of wax
After applying wax to tuning fork A, the number of beats produced when sounded with tuning fork B decreases to 4 beats per second. This indicates that the frequency of tuning fork A has decreased.

### Step 4: Determine the new frequency of A
When wax is applied, the frequency of tuning fork A will decrease. We need to check which of the two possible frequencies (390 Hz or 378 Hz) can lead to a decrease in the number of beats to 4 beats per second.

1. If \( f_A = 390 \, \text{Hz} \):
After applying wax, let’s assume the new frequency becomes \( f_A' \). The beat frequency with tuning fork B would be:
\[ |f_A' - 384| = 4 \]
This gives us two cases:
- \( f_A' - 384 = 4 \) → \( f_A' = 388 \, \text{Hz} \)
- \( 384 - f_A' = 4 \) → \( f_A' = 380 \, \text{Hz} \)

Both cases are possible since they indicate a decrease in frequency.

2. If \( f_A = 378 \, \text{Hz} \):
After applying wax, the frequency would decrease further, leading to a frequency lower than 378 Hz, which would not yield a beat frequency of only 4 beats with 384 Hz.

### Step 5: Conclusion
Since only the frequency of 390 Hz allows for a decrease in beats from 6 to 4 after applying wax, we conclude that the frequency of tuning fork A is:
\[ f_A = 390 \, \text{Hz} \]

### Final Answer
The frequency of tuning fork A is **390 Hz**.

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