Two perfectly identical wires are in unison. When the tension in one wire is increased by 1% then on sounding them together 3 beats are heard in 2 seconds. Calculate the intial frequency of each other?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between tension and frequency
The frequency of a vibrating wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{m}} \]
where:
- \( f \) is the frequency,
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( m \) is the mass per unit length of the wire.

### Step 2: Define the initial conditions
Let the initial tension in the wire be \( T \) and the initial frequency be \( f \). Thus, we can write:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{m}} \quad \text{(Equation 1)} \]

### Step 3: Calculate the new tension
When the tension in one wire is increased by 1%, the new tension \( T' \) becomes:
\[ T' = T + 0.01T = \frac{101}{100}T \]

### Step 4: Calculate the new frequency
The new frequency \( f' \) corresponding to the new tension \( T' \) can be expressed as:
\[ f' = \frac{1}{2L} \sqrt{\frac{T'}{m}} = \frac{1}{2L} \sqrt{\frac{101T}{100m}} \quad \text{(Equation 2)} \]

### Step 5: Relate the frequencies using the beat frequency
The beat frequency is given as 3 beats in 2 seconds, which translates to:
\[ \text{Beat frequency} = \frac{3}{2} = 1.5 \text{ beats per second} \]
This means:
\[ |f' - f| = 1.5 \]
Assuming \( f' > f \), we can write:
\[ f' - f = 1.5 \]

### Step 6: Substitute \( f' \) from Equation 2 into the beat frequency equation
From Equation 2, we can express \( f' \) as:
\[ f' = f + 1.5 \]
Substituting \( f' \) from Equation 2 gives:
\[ \frac{1}{2L} \sqrt{\frac{101T}{100m}} = f + 1.5 \]

### Step 7: Divide Equation 2 by Equation 1
Now, divide Equation 2 by Equation 1:
\[ \frac{f' }{f} = \frac{\frac{1}{2L} \sqrt{\frac{101T}{100m}}}{\frac{1}{2L} \sqrt{\frac{T}{m}}} \]
This simplifies to:
\[ \frac{f' }{f} = \sqrt{\frac{101}{100}} = \frac{1.005}{1} \]
Thus:
\[ f' = f \cdot 1.005 \]

### Step 8: Substitute \( f' \) back into the equation
Now substituting \( f' = f + 1.5 \) into the equation gives:
\[ f \cdot 1.005 = f + 1.5 \]
Rearranging gives:
\[ 1.005f - f = 1.5 \]
\[ 0.005f = 1.5 \]
\[ f = \frac{1.5}{0.005} = 300 \text{ Hz} \]

### Conclusion
The initial frequency of each wire is:
\[ \boxed{300 \text{ Hz}} \]