A stretched wire emits a fundamental note of2 56 Hz. Keeping the stretching force constant and reducing the length of wire by 10 cm, the frequency becomes 320 Hz, the original length of the wire is

Options

A. 100 cm
B. 50 cm
C. 400 cm
D. 200 cm

Explanation

To find the original length of the wire, we can use the relationship between frequency, length, tension, and mass per unit length of the wire. The fundamental frequency \( f \) of a stretched wire is given by the formula:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where:
- \( f \) is the frequency,
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the mass per unit length of the wire.

### Step-by-Step Solution:

1. **Identify the given values:**
- Original frequency \( f_1 = 256 \, \text{Hz} \)
- New frequency \( f_2 = 320 \, \text{Hz} \)
- Length reduction \( \Delta L = 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Set up the equations for both frequencies:**
- For the original length \( L \):
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \quad \text{(1)} \]
- For the new length \( L - 0.1 \):
\[ f_2 = \frac{1}{2(L - 0.1)} \sqrt{\frac{T}{\mu}} \quad \text{(2)} \]

3. **Express \( \sqrt{\frac{T}{\mu}} \) from equation (1):**
\[ \sqrt{\frac{T}{\mu}} = 2L f_1 \quad \text{(3)} \]

4. **Substitute equation (3) into equation (2):**
\[ f_2 = \frac{1}{2(L - 0.1)} (2L f_1) \]
Simplifying gives:
\[ f_2 = \frac{L f_1}{L - 0.1} \]

5. **Rearranging the equation:**
\[ f_2 (L - 0.1) = L f_1 \]
Expanding this:
\[ f_2 L - 0.1 f_2 = L f_1 \]
Rearranging gives:
\[ f_2 L - L f_1 = 0.1 f_2 \]
Factor out \( L \):
\[ L (f_2 - f_1) = 0.1 f_2 \]

6. **Solving for \( L \):**
\[ L = \frac{0.1 f_2}{f_2 - f_1} \]

7. **Substituting the values:**
\[ L = \frac{0.1 \times 320}{320 - 256} \]
\[ L = \frac{32}{64} = 0.5 \, \text{m} = 50 \, \text{cm} \]

### Conclusion:
The original length of the wire is \( \boxed{50 \, \text{cm}} \).