A block of mass `m`, when attached to a uniform ideal apring with force constant `k` and free length `L` executes SHM. The spring is then cut in two pieces, one with free length n `L` and other with free length `(1 - n)L`. The block is also divided in the same fraction. The smaller part of the block attached to longer part of the spring executes SHM with frequency `f_(1)` . The bigger part of the block attached to smaller part of the spring executes SHM with frequency `f_(2)`. The ratio `f_(1)//f_(2)` is

Options

A. `1`
B. `(n)/(1 - n)`
C. `(1 + n)/(n)`
D. `(n)/(1 + n)`

Explanation

To solve the problem, we need to analyze the situation step by step, focusing on the properties of the spring and the masses involved in the Simple Harmonic Motion (SHM).

### Step-by-Step Solution:

1. **Understanding the Spring Constant**:
- The spring constant \( k \) is inversely proportional to the length of the spring. If the original spring has a length \( L \) and spring constant \( k \), when the spring is cut into two parts, the spring constants of the new springs can be calculated.
- For the longer part of the spring with length \( nL \), the spring constant \( k_1 \) is given by:
\[ k_1 = \frac{k}{n} \]
- For the shorter part of the spring with length \( (1-n)L \), the spring constant \( k_2 \) is given by:
\[ k_2 = \frac{k}{1-n} \]

2. **Dividing the Mass**:
- The mass \( m \) is also divided in the same ratio. The smaller part of the mass \( m_1 \) (attached to the longer spring) is:
\[ m_1 = (1-n)m \]
- The bigger part of the mass \( m_2 \) (attached to the shorter spring) is:
\[ m_2 = n m \]

3. **Calculating Frequencies**:
- The frequency of SHM is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
- For the smaller mass \( m_1 \) attached to the longer spring (with spring constant \( k_1 \)):
\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m_1}} = \frac{1}{2\pi} \sqrt{\frac{\frac{k}{n}}{(1-n)m}} = \frac{1}{2\pi} \sqrt{\frac{k}{n(1-n)m}} \]
- For the bigger mass \( m_2 \) attached to the shorter spring (with spring constant \( k_2 \)):
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m_2}} = \frac{1}{2\pi} \sqrt{\frac{\frac{k}{1-n}}{n m}} = \frac{1}{2\pi} \sqrt{\frac{k}{n(1-n)m}} \]

4. **Finding the Ratio of Frequencies**:
- Now, we can find the ratio \( \frac{f_1}{f_2} \):
\[ \frac{f_1}{f_2} = \frac{\sqrt{\frac{k}{n(1-n)m}}}{\sqrt{\frac{k}{n(1-n)m}}} = 1 \]

### Final Answer:
The ratio \( \frac{f_1}{f_2} \) is \( 1 \).