Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

Options

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

Answer

To solve the problem, we need to find the fractional increase in the tension of one of the piano wires that will result in the occurrence of 6 beats per second when both wires vibrate simultaneously.

### Step-by-Step Solution:

1. **Understand the given information:**
- Two identical piano wires have a fundamental frequency \( F_1 = F_2 = 600 \) Hz when under the same tension.
- When the tension of one wire is increased, it leads to 6 beats per second.

2. **Define the frequency change:**
- The beat frequency \( \Delta F \) is given as 6 Hz. This means the difference in frequencies of the two wires after changing the tension in one wire is:
\[ \Delta F = |F_1 - F_2| = 6 \text{ Hz} \]

3. **Use the formula for the fundamental frequency of a wire:**
- The fundamental frequency \( F \) of a wire is given by:
\[ F = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
- Here, \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the mass per unit length (which is constant for both wires).

4. **Relate the change in tension to change in frequency:**
- If we denote the tension in the first wire as \( T \) and the tension in the second wire (after increase) as \( T + \Delta T \), the new frequency \( F_2 \) can be expressed as:
\[ F_2 = \frac{1}{2L} \sqrt{\frac{T + \Delta T}{\mu}} \]
- The change in frequency \( \Delta F \) can be expressed as:
\[ \Delta F = F_2 - F_1 \]

5. **Express the change in frequency in terms of tension:**
- Using the approximation for small changes, we can relate the change in frequency to the change in tension:
\[ \frac{\Delta F}{F} = \frac{1}{2} \frac{\Delta T}{T} \]
- Rearranging gives:
\[ \Delta T = 2 \cdot \Delta F \cdot \frac{T}{F} \]

6. **Substituting the known values:**
- We know \( \Delta F = 6 \) Hz and \( F = 600 \) Hz. Therefore:
\[ \Delta T = 2 \cdot 6 \cdot \frac{T}{600} = \frac{12T}{600} = \frac{T}{50} \]

7. **Finding the fractional increase in tension:**
- The fractional increase in tension \( \frac{\Delta T}{T} \) is:
\[ \frac{\Delta T}{T} = \frac{T/50}{T} = \frac{1}{50} \]

8. **Final result:**
- The fractional increase in tension required to produce 6 beats per second is:
\[ \frac{\Delta T}{T} = 0.02 \]

### Conclusion:
The fractional increase in the tension of one wire that will lead to the occurrence of 6 beats per second is \( 0.02 \) or \( \frac{1}{50} \).