The frequency of a streteched uniform wire of certain length is in resonance with the fundamental frequency of closed tube. If length of wire is decreased by `0.5m`, it is in resonance with first overtone of closed pipe. The initial length of wire is

Options

A. `0.5m`
B. `0.75m`
C. `1m`
D. `1.5m`

Answer

To solve the problem, we need to analyze the relationship between the frequency of a stretched wire and the fundamental frequency of a closed tube, as well as the first overtone of the closed tube.

### Step-by-Step Solution:

1. **Understanding the Frequencies**:
- Let the initial length of the wire be \( L \).
- The fundamental frequency of a closed tube (closed at one end) is given by:
\[ f = \frac{v}{4L} \]
where \( v \) is the speed of sound in air.

2. **Resonance Condition**:
- The wire is initially in resonance with the fundamental frequency of the closed tube, so we have:
\[ f = \frac{v}{4L} \]

3. **Length Decrease**:
- When the length of the wire is decreased by \( 0.5 \, m \), the new length of the wire becomes \( L - 0.5 \, m \).

4. **First Overtone of Closed Tube**:
- The first overtone of a closed tube corresponds to the frequency:
\[ f' = \frac{3v}{4L'} \]
where \( L' \) is the length of the closed tube, which is \( L - 0.5 \).

5. **Setting Up the Frequency Relationship**:
- Since the wire is now in resonance with the first overtone, we can equate the frequencies:
\[ f' = \frac{3v}{4(L - 0.5)} \]

6. **Using the Inverse Relationship**:
- We know that frequency is inversely proportional to length:
\[ \frac{f}{f'} = \frac{L - 0.5}{L} \]
- Substituting the expressions for \( f \) and \( f' \):
\[ \frac{\frac{v}{4L}}{\frac{3v}{4(L - 0.5)}} = \frac{L - 0.5}{L} \]
- Simplifying this gives:
\[ \frac{1}{3} = \frac{L - 0.5}{L} \]

7. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ L = 3(L - 0.5) \]

8. **Expanding and Solving for \( L \)**:
- Expanding the equation:
\[ L = 3L - 1.5 \]
- Rearranging gives:
\[ 1.5 = 3L - L \]
\[ 1.5 = 2L \]
\[ L = \frac{1.5}{2} = 0.75 \, m \]

9. **Final Result**:
- The initial length of the wire is:
\[ L = 1.5 \, m \]