A 1 cm long string vibrates with fundamental frequency of 256 Hz . If the length is reduced to `1/4` cm keeping the tension unaltered, the new fundamental frequency will be

Answer

To solve the problem, we will use the relationship between the fundamental frequency of a vibrating string and its length. The fundamental frequency (n) of a string is given by the formula:

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

Where:
- \( n \) = fundamental frequency
- \( L \) = length of the string
- \( T \) = tension in the string
- \( m \) = mass per unit length of the string

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Initial length of the string \( L_1 = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Initial fundamental frequency \( n_1 = 256 \, \text{Hz} \)

2. **Determine the new length of the string**:
- New length \( L_2 = \frac{1}{4} \, \text{cm} = \frac{1}{4} \times 0.01 \, \text{m} = 0.0025 \, \text{m} \)

3. **Use the relationship between frequency and length**:
- Since the tension \( T \) and mass per unit length \( m \) remain constant, we can use the relationship:
\[ n_1 L_1 = n_2 L_2 \]
- Rearranging gives:
\[ n_2 = \frac{n_1 L_1}{L_2} \]

4. **Substitute the known values**:
- Substitute \( n_1 = 256 \, \text{Hz} \), \( L_1 = 0.01 \, \text{m} \), and \( L_2 = 0.0025 \, \text{m} \):
\[ n_2 = \frac{256 \times 0.01}{0.0025} \]

5. **Calculate \( n_2 \)**:
- First, calculate \( \frac{0.01}{0.0025} = 4 \)
- Then, \( n_2 = 256 \times 4 = 1024 \, \text{Hz} \)

6. **Final result**:
- The new fundamental frequency \( n_2 \) is \( 1024 \, \text{Hz} \).

### Answer:
The new fundamental frequency will be **1024 Hz**.