The identical piano wires kept under the same tension T have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats//s when both the wires oscillate together would be

Options

A. 0.01
B. 0.02
C. 0.03
D. 0.04

Explanation

To solve the problem, we need to determine the fractional increase in the tension of one of the piano wires that will lead to the occurrence of 6 beats per second when both wires oscillate together.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two identical piano wires under the same tension \( T \) with a fundamental frequency \( f_1 = 600 \) Hz.
- We need to find the fractional increase in tension \( \frac{\Delta T}{T} \) for one wire that will cause 6 beats per second when both wires are oscillating.

2. **Beat Frequency**:
- The beat frequency \( f_b \) is given by the difference in frequencies of the two wires:
\[ f_b = |f_1 - f_2| \]
- In this case, \( f_b = 6 \) Hz.

3. **Frequency Relation to Tension**:
- The frequency of a vibrating wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( L \) is the length of the wire and \( \mu \) is the mass per unit length.
- For the second wire with increased tension \( T + \Delta T \), the frequency becomes:
\[ f_2 = \frac{1}{2L} \sqrt{\frac{T + \Delta T}{\mu}} \]

4. **Calculating the Change in Frequency**:
- The change in frequency \( \Delta f \) is:
\[ \Delta f = f_2 - f_1 \]
- Since \( f_1 = 600 \) Hz and \( f_b = 6 \) Hz, we have:
\[ |f_2 - 600| = 6 \]
- This gives us two possible cases:
- Case 1: \( f_2 = 606 \) Hz
- Case 2: \( f_2 = 594 \) Hz

5. **Using the Frequency Relation**:
- For Case 1:
\[ f_2 = 606 \text{ Hz} \]
\[ \Delta f = f_2 - f_1 = 606 - 600 = 6 \text{ Hz} \]
- For Case 2:
\[ f_2 = 594 \text{ Hz} \]
\[ \Delta f = f_1 - f_2 = 600 - 594 = 6 \text{ Hz} \]

6. **Relating Change in Frequency to Change in Tension**:
- From the formula for the change in frequency:
\[ \frac{\Delta f}{f} = \frac{1}{2} \frac{\Delta T}{T} \]
- Rearranging gives:
\[ \frac{\Delta T}{T} = 2 \frac{\Delta f}{f} \]

7. **Calculating the Fractional Increase in Tension**:
- Using \( \Delta f = 6 \) Hz and \( f = 600 \) Hz:
\[ \frac{\Delta T}{T} = 2 \cdot \frac{6}{600} = 2 \cdot 0.01 = 0.02 \]

### Final Answer:
The fractional increase in the tension of one of the wires that will lead to the occurrence of 6 beats per second is \( \frac{\Delta T}{T} = 0.02 \).