A spring has force constant K and a mass m is suspended from it. The spring is cut in half and the same is suspended from one of the havles. IF the frequency of osicllation in the frst case is `alpha`, then the frequency in the second case will be

Options

A. 2 a
B. a
C. `a//2`
D. `alpha//sqrt(2)`

Explanation

To solve the problem, we need to determine the frequency of oscillation when a spring is cut in half and a mass is suspended from one of the halves.

### Step-by-Step Solution:

1. **Understand the Original Spring System:**
- The original spring has a force constant \( K \).
- The frequency of oscillation for a mass \( m \) suspended from this spring is given by the formula:
\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{K}{m}} \]
- We denote this frequency as \( \alpha \):
\[ \alpha = \frac{1}{2\pi} \sqrt{\frac{K}{m}} \]

2. **Cut the Spring in Half:**
- When the spring is cut in half, the spring constant of each half becomes:
\[ K' = 2K \]
- This is because the spring constant is inversely proportional to the length of the spring. Cutting the spring in half doubles the spring constant.

3. **Determine the New Frequency:**
- The frequency of oscillation for the mass \( m \) suspended from one half of the spring is given by:
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{K'}{m}} = \frac{1}{2\pi} \sqrt{\frac{2K}{m}} \]

4. **Relate the New Frequency to the Original Frequency:**
- We can express \( f_2 \) in terms of \( \alpha \):
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{2K}{m}} = \sqrt{2} \cdot \frac{1}{2\pi} \sqrt{\frac{K}{m}} = \sqrt{2} \cdot \alpha \]

5. **Final Result:**
- Therefore, the frequency of oscillation when the spring is cut in half is:
\[ f_2 = \alpha \sqrt{2} \]

### Conclusion:
The frequency of oscillation in the second case will be \( \alpha \sqrt{2} \).