A mass m is suspended from a spring. Its frequency of oscillation is f. The spring is cut into two halves and the same mass is suspended from one of the pieces of the spring. The frequency of oscillation of the mass will be

Options

A. `sqrt(2) f`
B. `(f)/(2)`
C. `f`
D. `2f`

Explanation

To solve the problem step by step, we will analyze the situation of the spring mass system before and after the spring is cut into two halves.

### Step-by-Step Solution:

1. **Understanding the Original System**:
- The frequency of oscillation \( f \) of a mass \( m \) suspended from a spring with spring constant \( k \) is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

2. **Cutting the Spring**:
- When the spring is cut into two equal halves, each half will have a length of \( \frac{l}{2} \).
- The spring constant of a spring is inversely proportional to its length. Therefore, if the original spring constant is \( k \), the spring constant of each half \( k' \) will be:
\[ k' = 2k \]
- This is because the spring constant \( k \) is related to the length \( l \) as follows:
\[ k \cdot l = k' \cdot \frac{l}{2} \implies k' = 2k \]

3. **Calculating the New Frequency**:
- Now, when the mass \( m \) is suspended from one of the halves of the spring, the new frequency \( f' \) can be calculated using the new spring constant \( k' \):
\[ f' = \frac{1}{2\pi} \sqrt{\frac{k'}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}} \]
- We can factor out the square root of 2:
\[ f' = \frac{1}{2\pi} \sqrt{2} \sqrt{\frac{k}{m}} = \sqrt{2} \cdot \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \sqrt{2} \cdot f \]

4. **Final Result**:
- Thus, the frequency of oscillation of the mass when suspended from one half of the spring will be:
\[ f' = \sqrt{2} \cdot f \]

### Summary:
The frequency of oscillation of the mass when suspended from one of the pieces of the spring after cutting it into two halves will be \( \sqrt{2} \) times the original frequency \( f \).