Two tunning forks A and B produce notes of frequencies 256 Hz & 262 Hz respectively. An unknown note sounded at the same time with A produce beats. When the note is sounded with B, beats frequency is twice as large. The unknown frequency could be

Options

A. 268 Hz
B. 250 Hz
C. 260 Hz
D. 252 Hz

Explanation

To solve the problem step by step, we will denote the unknown frequency as \( F \).

### Step 1: Understand the beat frequency with tuning fork A
The beat frequency when the unknown note \( F \) is sounded with tuning fork A (256 Hz) is given by:
\[ \text{Beat frequency with A} = |256 - F| \]

### Step 2: Understand the beat frequency with tuning fork B
The beat frequency when the unknown note \( F \) is sounded with tuning fork B (262 Hz) is given by:
\[ \text{Beat frequency with B} = |262 - F| \]

### Step 3: Set up the relationship between the beat frequencies
According to the problem, the beat frequency with B is twice that with A:
\[ |262 - F| = 2 |256 - F| \]

### Step 4: Solve the equation for the positive case
First, we will consider the case where both expressions are positive:
\[ 262 - F = 2(256 - F) \]
Expanding this gives:
\[ 262 - F = 512 - 2F \]
Rearranging the equation:
\[ 2F - F = 512 - 262 \]
\[ F = 250 \text{ Hz} \]

### Step 5: Solve the equation for the negative case
Now, we will consider the case where the expression for tuning fork A is negative:
\[ 262 - F = -2(256 - F) \]
Expanding this gives:
\[ 262 - F = -512 + 2F \]
Rearranging the equation:
\[ 262 + 512 = 2F + F \]
\[ 774 = 3F \]
\[ F = \frac{774}{3} = 258 \text{ Hz} \]

### Step 6: Determine the valid frequencies
We have two possible frequencies for \( F \):
1. \( F = 250 \text{ Hz} \)
2. \( F = 258 \text{ Hz} \)

Since the problem states that the unknown frequency must be less than 262 Hz and greater than 256 Hz to satisfy the conditions of the beat frequencies, the valid unknown frequency is:
\[ F = 258 \text{ Hz} \]

### Final Answer
The unknown frequency could be **258 Hz**.

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