Two instruments having stretched strings are being played in unison . When the tension in one of the instruments is increases by ` 1% , 3` beats are produced in ` 2 s`. The initial frequency of vibration of each wire is

Options

A. 600Hz
B. 300Hz
C. 200Hz
D. 150Hz

Explanation

To solve the problem step by step, we will follow the concepts of wave motion and frequency related to tension in strings.

### Step 1: Understand the Problem
We have two instruments with stretched strings that are played in unison. When the tension in one instrument is increased by 1%, 3 beats are produced in 2 seconds. We need to find the initial frequency of vibration of each wire.

### Step 2: Define the Frequency Formula
The frequency \( n \) of a vibrating string is given by the formula:
\[ n = \frac{1}{2L} \sqrt{\frac{T}{m}} \]
where:
- \( n \) = frequency
- \( L \) = length of the string
- \( T \) = tension in the string
- \( m \) = mass per unit length of the string

### Step 3: Determine the New Tension
When the tension in one instrument is increased by 1%, the new tension \( T_1 \) can be expressed as:
\[ T_1 = T + 0.01T = 1.01T \]

### Step 4: Calculate the New Frequency
The new frequency \( n_1 \) when the tension is increased is given by:
\[ n_1 = \frac{1}{2L} \sqrt{\frac{T_1}{m}} = \frac{1}{2L} \sqrt{\frac{1.01T}{m}} \]

### Step 5: Relate the Frequencies
The difference in frequencies causes beats. The number of beats produced is given as 3 beats in 2 seconds, which means:
\[ \text{Beat Frequency} = \frac{3 \text{ beats}}{2 \text{ seconds}} = 1.5 \text{ Hz} \]
The beat frequency is equal to the difference in frequencies:
\[ |n_1 - n| = 1.5 \text{ Hz} \]

### Step 6: Express the Frequencies
From the previous steps, we can express the new frequency \( n_1 \) as:
\[ n_1 = n + 1.5 \quad \text{(if } n_1 > n\text{)} \]
Substituting for \( n_1 \):
\[ \frac{1}{2L} \sqrt{\frac{1.01T}{m}} = n + 1.5 \]

### Step 7: Set Up the Equations
Now we have two equations:
1. \( n = \frac{1}{2L} \sqrt{\frac{T}{m}} \)
2. \( n + 1.5 = \frac{1}{2L} \sqrt{\frac{1.01T}{m}} \)

### Step 8: Solve the Equations
Substituting equation 1 into equation 2:
\[ n + 1.5 = \sqrt{1.01} \cdot n \]
This simplifies to:
\[ n + 1.5 = 1.005 \cdot n \]
Rearranging gives:
\[ 1.005n - n = 1.5 \]
\[ 0.005n = 1.5 \]
\[ n = \frac{1.5}{0.005} = 300 \text{ Hz} \]

### Step 9: Conclusion
The initial frequency of vibration of each wire is:
\[ \boxed{300 \text{ Hz}} \]