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 solve the problem, we need to use the relationship between the frequency of a stretched wire and its length. The frequency of the fundamental mode of a stretched wire is given by the formula:

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

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

### Step 1: Write the equations for both frequencies

For the first case, where the frequency \( f_1 = 256 \, \text{Hz} \) and the original length is \( L \):

\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{m}} \implies 256 = \frac{1}{2L} \sqrt{\frac{T}{m}} \tag{1} \]

For the second case, where the frequency \( f_2 = 320 \, \text{Hz} \) and the length is reduced by 10 cm, the new length becomes \( L - 10 \, \text{cm} \):

\[ f_2 = \frac{1}{2(L - 10)} \sqrt{\frac{T}{m}} \implies 320 = \frac{1}{2(L - 10)} \sqrt{\frac{T}{m}} \tag{2} \]

### Step 2: Rearranging the equations

From equation (1):

\[ \sqrt{\frac{T}{m}} = 512L \tag{3} \]

From equation (2):

\[ \sqrt{\frac{T}{m}} = 640(L - 10) \tag{4} \]

### Step 3: Set equations (3) and (4) equal to each other

Since both equations equal \( \sqrt{\frac{T}{m}} \):

\[ 512L = 640(L - 10) \]

### Step 4: Expand and simplify the equation

Expanding the right side:

\[ 512L = 640L - 6400 \]

Rearranging gives:

\[ 640L - 512L = 6400 \]

\[ 128L = 6400 \]

### Step 5: Solve for \( L \)

Dividing both sides by 128:

\[ L = \frac{6400}{128} = 50 \, \text{cm} \]

### Conclusion

The original length of the wire is \( L = 50 \, \text{cm} \).

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