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 find the ratio of the frequencies \( f_1 \) and \( f_2 \) of two blocks attached to different parts of a spring after it has been cut. Here’s a step-by-step breakdown of the solution:

### Step 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 \( k \), then when the spring is cut into two parts:
- The longer part with length \( nL \) will have a spring constant:
\[ k_1 = \frac{k}{n} \]
- The shorter part with length \( (1-n)L \) will have a spring constant:
\[ k_2 = \frac{k}{1-n} \]

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

### Step 3: Finding the Frequencies
The frequency of a mass-spring system executing simple harmonic motion (SHM) is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Using this formula, we can find the frequencies \( f_1 \) and \( f_2 \):

- For the smaller mass \( m_1 \) attached to the longer spring:
\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m_1}} = \frac{1}{2\pi} \sqrt{\frac{\frac{k}{n}}{n \cdot m}} = \frac{1}{2\pi} \sqrt{\frac{k}{n^2 m}} \]

- For the larger mass \( m_2 \) attached to the shorter spring:
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m_2}} = \frac{1}{2\pi} \sqrt{\frac{\frac{k}{1-n}}{(1-n) \cdot m}} = \frac{1}{2\pi} \sqrt{\frac{k}{(1-n)^2 m}} \]

### Step 4: Finding the Ratio of Frequencies
Now we can find the ratio \( \frac{f_1}{f_2} \):
\[ \frac{f_1}{f_2} = \frac{\sqrt{\frac{k}{n^2 m}}}{\sqrt{\frac{k}{(1-n)^2 m}}} = \frac{\sqrt{(1-n)^2}}{\sqrt{n^2}} = \frac{1-n}{n} \]

Thus, the ratio of the frequencies is:
\[ \frac{f_1}{f_2} = \frac{1-n}{n} \]

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

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