In a spring- mass system , the length of the spring is L, and it has a mass M attached to it and oscillates with an angular frequency `omega`. The spring is then cut into two parts, one (i) with relaxed length `alphaL` and the other (ii) with relaxed length `(1-alpha)` L. The force constants of the two spring A and B are

Options

A. `(k)/(1-alpha)` and `(k)/(alpha)`
B. `(k)/(alpha)` and `(k)/(1-alpha)`
C. `alphak,(1-alpha)k`
D. k and k

Explanation

To solve the problem, we need to determine the spring constants of the two parts of the spring after it has been cut. Let's denote the original spring constant as \( K \) and the original length of the spring as \( L \). The spring is cut into two parts: one part has a relaxed length of \( \alpha L \) and the other part has a relaxed length of \( (1 - \alpha)L \).

### Step-by-Step Solution:

1. **Understanding the relationship between spring constant and length**:
The spring constant \( K \) of a spring is inversely proportional to its length when the spring is cut. This means that if a spring is cut into two parts, the spring constant of each part can be found using the formula:
\[ K \cdot L = K_1 \cdot \alpha L \]
for the first part and
\[ K \cdot L = K_2 \cdot (1 - \alpha)L \]
for the second part.

2. **Finding the spring constant \( K_1 \)**:
From the first equation, we can isolate \( K_1 \):
\[ K_1 = \frac{K \cdot L}{\alpha L} \]
Here, \( L \) cancels out:
\[ K_1 = \frac{K}{\alpha} \]

3. **Finding the spring constant \( K_2 \)**:
From the second equation, we can isolate \( K_2 \):
\[ K_2 = \frac{K \cdot L}{(1 - \alpha)L} \]
Again, \( L \) cancels out:
\[ K_2 = \frac{K}{1 - \alpha} \]

4. **Final Results**:
The spring constants for the two parts are:
\[ K_1 = \frac{K}{\alpha} \]
\[ K_2 = \frac{K}{1 - \alpha} \]

### Summary:
The force constants of the two springs after cutting are:
- For the first spring (length \( \alpha L \)): \( K_1 = \frac{K}{\alpha} \)
- For the second spring (length \( (1 - \alpha)L \)): \( K_2 = \frac{K}{1 - \alpha} \)