Two tuning forks having frequency 256 Hz (A) and 262 Hz (B) tuning fork. A produces some beats per second with unknown tuning fork, same unknown tuning fork produce double beats per second from B tuning fork then the frequency of unknown tuning fork is :-

Options

A. 262
B. 260
C. 250
D. 300

Explanation

To solve the problem step by step, we will analyze the information given about the tuning forks and apply the concept of beat frequency.

### Step 1: Understand the given frequencies
We have two tuning forks:
- Tuning Fork A has a frequency of \( f_A = 256 \, \text{Hz} \)
- Tuning Fork B has a frequency of \( f_B = 262 \, \text{Hz} \)

We need to find the frequency of an unknown tuning fork, which we will denote as \( f_n \).

### Step 2: Define the beat frequencies
When the unknown tuning fork \( f_n \) is sounded with:
- Tuning Fork A, it produces \( x \) beats per second.
- Tuning Fork B, it produces \( 2x \) beats per second.

### Step 3: Set up equations for beat frequencies
The beat frequency is given by the absolute difference between the frequencies of the two tuning forks. Therefore, we can write:

1. For Tuning Fork A:
\[ |f_A - f_n| = x \]
This can be expressed as:
\[ 256 - f_n = x \quad \text{(1)} \]
or
\[ f_n - 256 = x \quad \text{(2)} \]

2. For Tuning Fork B:
\[ |f_B - f_n| = 2x \]
This can be expressed as:
\[ 262 - f_n = 2x \quad \text{(3)} \]
or
\[ f_n - 262 = 2x \quad \text{(4)} \]

### Step 4: Analyze the relationships
Since we know that \( f_n \) is less than both \( f_A \) and \( f_B \) (as explained in the video transcript), we will use equations (1) and (3):

From equation (1):
\[ x = 256 - f_n \quad \text{(5)} \]

Substituting equation (5) into equation (3):
\[ 262 - f_n = 2(256 - f_n) \]

### Step 5: Solve for \( f_n \)
Now, simplify the equation:
\[ 262 - f_n = 512 - 2f_n \]
Rearranging gives:
\[ 2f_n - f_n = 512 - 262 \]
\[ f_n = 250 \, \text{Hz} \]

### Conclusion
The frequency of the unknown tuning fork is \( f_n = 250 \, \text{Hz} \).

### Verification
To verify, we can check the beat frequencies:
- Beats with A: \( |256 - 250| = 6 \, \text{Hz} \) (which is \( x \))
- Beats with B: \( |262 - 250| = 12 \, \text{Hz} \) (which is \( 2x \))

This confirms that the calculations are consistent.