Two tuning forks `A and B` give ` 4 beats//s` when sounded together . The frequency of `A is 320 Hz`. When some wax is added to `B` and it is sounded with `A , 4 beats//s per second` are again heard . The frequency of `B` is

Options

A. `312 Hz`
B. `316 Hz`
C. `324 Hz`
D. `328 Hz`

Explanation

To solve the problem, let's break it down step by step:

### Step 1: Understand the concept of beats
When two tuning forks are sounded together, the phenomenon of beats occurs due to the interference of sound waves from the two sources. The beat frequency is equal to the absolute difference between the frequencies of the two tuning forks.

### Step 2: Set up the equation for the initial situation
Given:
- Frequency of tuning fork A, \( f_A = 320 \, \text{Hz} \)
- Beat frequency when A and B are sounded together, \( f_{beat} = 4 \, \text{beats/s} \)

We can express this relationship as:
\[ |f_A - f_B| = 4 \]
This gives us two possible equations:
1. \( f_A - f_B = 4 \)
2. \( f_B - f_A = 4 \)

### Step 3: Solve for \( f_B \)
Using the first equation:
\[ 320 - f_B = 4 \implies f_B = 320 - 4 = 316 \, \text{Hz} \]

Using the second equation:
\[ f_B - 320 = 4 \implies f_B = 320 + 4 = 324 \, \text{Hz} \]

Thus, the possible frequencies for tuning fork B are \( 316 \, \text{Hz} \) or \( 324 \, \text{Hz} \).

### Step 4: Analyze the effect of adding wax to tuning fork B
When wax is added to tuning fork B, its frequency decreases. The problem states that after adding wax, the beat frequency remains the same at \( 4 \, \text{beats/s} \).

### Step 5: Determine the new frequency of B
If we assume:
- Case 1: \( f_B = 316 \, \text{Hz} \) initially
- After waxing, \( f_B \) would decrease, making it less than \( 316 \, \text{Hz} \), which would increase the beat frequency with A, contradicting the information given.

- Case 2: \( f_B = 324 \, \text{Hz} \) initially
- After waxing, \( f_B \) would decrease, making it less than \( 324 \, \text{Hz} \). If it decreases to \( 320 \, \text{Hz} \), the beat frequency would still be \( |320 - 320| = 0 \), which is not the case. However, if it decreases to \( 316 \, \text{Hz} \), the beat frequency remains \( |320 - 316| = 4 \, \text{beats/s} \), which is consistent with the problem statement.

### Conclusion
Thus, the frequency of tuning fork B before adding wax is:
\[ f_B = 324 \, \text{Hz} \]

### Final Answer
The frequency of tuning fork B is **324 Hz**.

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