The string of a sonometer is divided into two parts using wedge. Total length of string is `1m` and two parts differ by `2mm`. When sounded together they produce `2` beats/sec. The frequencies of two parts are

Options

A. `501Hz`, `503Hz`
B. `501Hz`, `499Hz`
C. `499Hz`, `497Hz`
D. `497Hz`, `495Hz`

Explanation

To solve the problem, we need to find the frequencies of two parts of a string that are producing beats when sounded together. Here’s a step-by-step breakdown of the solution:

### Step 1: Define the lengths of the two parts of the string
Let the lengths of the two parts of the string be \( L_1 \) and \( L_2 \). According to the problem:
- The total length of the string is \( L_1 + L_2 = 1 \, \text{m} = 1000 \, \text{mm} \)
- The difference in lengths is \( L_1 - L_2 = 2 \, \text{mm} \)

### Step 2: Set up the equations
From the information given, we can set up the following equations:
1. \( L_1 + L_2 = 1000 \)
2. \( L_1 - L_2 = 2 \)

### Step 3: Solve the equations
We can solve these two equations simultaneously.

Adding the two equations:
\[ (L_1 + L_2) + (L_1 - L_2) = 1000 + 2 \]
\[ 2L_1 = 1002 \implies L_1 = 501 \, \text{mm} \]

Now, substituting \( L_1 \) back into the first equation:
\[ 501 + L_2 = 1000 \implies L_2 = 499 \, \text{mm} \]

### Step 4: Relate lengths to frequencies
The frequency of a vibrating string is inversely proportional to its length. Therefore, we can express the frequencies \( f_1 \) and \( f_2 \) corresponding to lengths \( L_1 \) and \( L_2 \) as:
\[ f_1 \propto \frac{1}{L_1} \quad \text{and} \quad f_2 \propto \frac{1}{L_2} \]

### Step 5: Set up the frequency ratio
Since the velocity of the wave in the string is the same for both parts (same tension and same material), we can write:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} = \frac{499}{501} \]

### Step 6: Use the beat frequency
The problem states that the two parts produce 2 beats per second, which means:
\[ |f_1 - f_2| = 2 \]

Assuming \( f_2 > f_1 \), we can write:
\[ f_2 - f_1 = 2 \]

### Step 7: Express \( f_1 \) in terms of \( f_2 \)
From the frequency ratio, we can express \( f_1 \) in terms of \( f_2 \):
\[ f_1 = \frac{499}{501} f_2 \]

### Step 8: Substitute into the beat frequency equation
Now substitute \( f_1 \) into the beat frequency equation:
\[ f_2 - \frac{499}{501} f_2 = 2 \]
\[ \left(1 - \frac{499}{501}\right) f_2 = 2 \]
\[ \frac{2}{501} f_2 = 2 \implies f_2 = 501 \, \text{Hz} \]

### Step 9: Find \( f_1 \)
Now, substitute \( f_2 \) back to find \( f_1 \):
\[ f_1 = f_2 - 2 = 501 - 2 = 499 \, \text{Hz} \]

### Final Answer
Thus, the frequencies of the two parts of the string are:
- \( f_1 = 499 \, \text{Hz} \)
- \( f_2 = 501 \, \text{Hz} \)