The frequency of oscillations of a mass m suspended by spring of `v_(1)`. If the length of the spring is cut to one-half, the same mass oscillates with frequency `v_(2)`. Determine the value of `v_(2)//v_(1)`

Options

A. `2`
B. `sqrt(2)`
C. `4`
D. `sqrt(3)`

Explanation

To solve the problem, we need to determine the ratio of the frequencies of oscillation \( \frac{v_2}{v_1} \) when the length of the spring is cut in half.

### Step-by-Step Solution:

1. **Understanding the Frequency of a Spring-Mass System**:
The frequency of oscillation \( v \) for a mass \( m \) attached to a spring with spring constant \( k \) is given by the formula:
\[ v = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
For the initial spring, we denote the spring constant as \( k_1 \) and the frequency as \( v_1 \):
\[ v_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}} \]

2. **Effect of Cutting the Spring**:
When the spring is cut to half its original length, the spring constant changes. The spring constant \( k \) is inversely proportional to the length of the spring. If the original length is \( L \), then cutting it in half results in a new length of \( \frac{L}{2} \). The new spring constant \( k_2 \) becomes:
\[ k_2 = 2k_1 \]

3. **Frequency of the Shortened Spring**:
The new frequency \( v_2 \) for the shortened spring with the new spring constant \( k_2 \) is:
\[ v_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k_1}{m}} \]

4. **Finding the Ratio of Frequencies**:
To find the ratio \( \frac{v_2}{v_1} \):
\[ \frac{v_2}{v_1} = \frac{\frac{1}{2\pi} \sqrt{\frac{2k_1}{m}}}{\frac{1}{2\pi} \sqrt{\frac{k_1}{m}}} \]
Simplifying this expression:
\[ \frac{v_2}{v_1} = \frac{\sqrt{2k_1/m}}{\sqrt{k_1/m}} = \sqrt{2} \]

5. **Final Result**:
Therefore, the ratio of the frequencies is:
\[ \frac{v_2}{v_1} = \sqrt{2} \]

### Summary:
The value of \( \frac{v_2}{v_1} \) is \( \sqrt{2} \).