The two parts of a sonometer wire divided by a movable knife edge , differ in length by `2 mm` and produce `1 beat//s` , when sounded together . Find their frequencies if the whole length of wire is `1.00 m`.

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Understand the Problem
We have a sonometer wire of total length \( L = 1.00 \, m \) divided into two parts by a knife edge. The lengths of these two parts differ by \( 2 \, mm \) (or \( 0.2 \, cm \)). We need to find the frequencies of the two parts of the wire, given that they produce \( 1 \, beat/s \) when sounded together.

### Step 2: Define the Lengths
Let the lengths of the two parts of the wire be \( l_1 \) and \( l_2 \).
Given that:
\[ l_1 + l_2 = 100 \, cm \]
\[ l_1 - l_2 = 0.2 \, cm \]

### Step 3: Solve for Lengths
We can set up the equations based on the above information:
1. \( l_1 + l_2 = 100 \)
2. \( l_1 - l_2 = 0.2 \)

Adding these two equations:
\[ 2l_1 = 100.2 \]
\[ l_1 = \frac{100.2}{2} = 50.1 \, cm \]

Now, substituting \( l_1 \) back into the first equation to find \( l_2 \):
\[ l_2 = 100 - l_1 = 100 - 50.1 = 49.9 \, cm \]

### Step 4: Frequency Relation
The frequency of a vibrating string is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Since tension \( T \) and linear density \( \mu \) are constant for both parts, we can relate their frequencies to their lengths:
\[ f_1 \cdot l_1 = f_2 \cdot l_2 \]

### Step 5: Beat Frequency
The beat frequency is given as \( 1 \, beat/s \), which means:
\[ |f_2 - f_1| = 1 \]
Assuming \( f_2 = f_1 + 1 \):
Substituting \( f_2 \) into the frequency relation:
\[ f_1 \cdot 50.1 = (f_1 + 1) \cdot 49.9 \]

### Step 6: Solve for \( f_1 \)
Expanding the equation:
\[ 50.1 f_1 = 49.9 f_1 + 49.9 \]
Rearranging gives:
\[ 50.1 f_1 - 49.9 f_1 = 49.9 \]
\[ 0.2 f_1 = 49.9 \]
\[ f_1 = \frac{49.9}{0.2} = 249.5 \, Hz \]

### Step 7: Find \( f_2 \)
Now, using \( f_2 = f_1 + 1 \):
\[ f_2 = 249.5 + 1 = 250.5 \, Hz \]

### Final Answer
The frequencies of the two parts of the sonometer wire are:
- \( f_1 = 249.5 \, Hz \)
- \( f_2 = 250.5 \, Hz \)