Using a movable knife edge, the two parts of a sonometer wire with total length 4 m is divided into two parts, such that they differ in length by 8 mm and produce 4 beats per second when sounded together. Calculate their frequencies.

Answer

To solve the problem step by step, we will follow the logical reasoning presented in the video transcript:

### Step 1: Understand the Given Information
We have a sonometer wire of total length \( L = 4 \, \text{m} = 4000 \, \text{mm} \). The wire is divided into two parts such that the lengths differ by \( 8 \, \text{mm} \).

### Step 2: Define the Lengths of the Two Parts
Let the length of the first part be \( L_1 \) and the length of the second part be \( L_2 \). According to the problem:
- \( L_2 = L_1 + 8 \, \text{mm} \)
- The total length is \( L_1 + L_2 = 4000 \, \text{mm} \)

### Step 3: Set Up the Equation
Substituting \( L_2 \) in the total length equation:
\[ L_1 + (L_1 + 8) = 4000 \]
This simplifies to:
\[ 2L_1 + 8 = 4000 \]

### Step 4: Solve for \( L_1 \)
Rearranging the equation gives:
\[ 2L_1 = 4000 - 8 = 3992 \]
\[ L_1 = \frac{3992}{2} = 1996 \, \text{mm} \]

### Step 5: Calculate \( L_2 \)
Now, substituting \( L_1 \) back to find \( L_2 \):
\[ L_2 = L_1 + 8 = 1996 + 8 = 2004 \, \text{mm} \]

### Step 6: Relate Frequencies to Lengths
The frequency of a vibrating string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Where \( T \) is tension and \( \mu \) is mass per unit length. Since the tension and mass per unit length are the same for both parts, we can say:
\[ f_1 L_1 = f_2 L_2 \]

### Step 7: Express One Frequency in Terms of the Other
Let \( f_1 \) be the frequency of the first part and \( f_2 \) be the frequency of the second part. We know that the difference in frequencies is \( 4 \, \text{Hz} \):
\[ f_2 = f_1 - 4 \]

### Step 8: Substitute and Solve for Frequencies
Substituting \( f_2 \) into the frequency-length relationship:
\[ f_1 \cdot 1996 = (f_1 - 4) \cdot 2004 \]
Expanding this gives:
\[ 1996 f_1 = 2004 f_1 - 8016 \]
Rearranging terms results in:
\[ 2004 f_1 - 1996 f_1 = 8016 \]
\[ 8 f_1 = 8016 \]
\[ f_1 = \frac{8016}{8} = 1002 \, \text{Hz} \]

### Step 9: Calculate \( f_2 \)
Now, substituting \( f_1 \) back to find \( f_2 \):
\[ f_2 = f_1 - 4 = 1002 - 4 = 998 \, \text{Hz} \]

### Final Answer
The frequencies of the two parts of the sonometer wire are:
- \( f_1 = 1002 \, \text{Hz} \)
- \( f_2 = 998 \, \text{Hz} \)