Two similar sonometer wires of the same material under the same tension produces 3 beats per second. The length of one wire is 40 cm and that of the other is 40.1 cm. Calculate the frequency of the two wires ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have two sonometer wires of the same material under the same tension. The lengths of the wires are:
- Length of wire 1 (L1) = 40 cm
- Length of wire 2 (L2) = 40.1 cm

The number of beats produced per second is given as:
- Beats per second = 3

### Step 2: Relate frequency and length
According to the law of vibrating strings, the frequency (f) of a wire is inversely proportional to its length (L). This can be expressed as:
\[ f \propto \frac{1}{L} \]
From this, we can write the relationship between the frequencies and lengths of the two wires:
\[ f_1 \cdot L_1 = f_2 \cdot L_2 \]
Where:
- \(f_1\) is the frequency of the first wire
- \(f_2\) is the frequency of the second wire

### Step 3: Set up the equations
From the beats produced, we know:
\[ |f_1 - f_2| = 3 \]
We can express \(f_1\) in terms of \(f_2\):
\[ f_1 = f_2 + 3 \quad \text{(1)} \]
Now substituting \(f_1\) in terms of \(f_2\) into the length-frequency relationship:
\[ (f_2 + 3) \cdot L_1 = f_2 \cdot L_2 \]
Substituting the lengths:
\[ (f_2 + 3) \cdot 40 = f_2 \cdot 40.1 \]

### Step 4: Solve for \(f_2\)
Expanding the equation:
\[ 40f_2 + 120 = 40.1f_2 \]
Rearranging gives:
\[ 120 = 40.1f_2 - 40f_2 \]
\[ 120 = 0.1f_2 \]
Now, solving for \(f_2\):
\[ f_2 = \frac{120}{0.1} = 1200 \text{ Hz} \]

### Step 5: Find \(f_1\)
Now we can find \(f_1\) using equation (1):
\[ f_1 = f_2 + 3 = 1200 + 3 = 1203 \text{ Hz} \]

### Final Answer
The frequencies of the two wires are:
- Frequency of wire 1 (\(f_1\)) = 1203 Hz
- Frequency of wire 2 (\(f_2\)) = 1200 Hz

---