The frequency of oscillation of amass m suspended by a spring is `v_(1)`. If length of spring is cut to one third then the same mass oscillations with frequency `v_(2)`. Then

Options

A. `v_(2) = 3v_(1)`
B. `3v_(2)=v_(1)`
C. `v_(2)= sqrt(3)v_(1)`
D. `sqrt(3)v_(2)=v_(1)`

Answer

To solve the problem, we need to find the relationship between the frequencies \( v_1 \) and \( v_2 \) of a mass \( m \) suspended by a spring, where the spring is cut to one-third of its original length.

### Step-by-Step Solution:

1. **Understand the Frequency of the Spring-Mass System**:
The frequency \( v_1 \) of a mass \( m \) attached to a spring with spring constant \( k \) is given by the formula:
\[ v_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

2. **Cut the Spring to One-Third**:
When the spring is cut to one-third of its original length, the spring constant changes. The spring constant \( k \) of a spring is inversely proportional to its length. If the original length is \( L \), the new length is \( \frac{L}{3} \). Therefore, the new spring constant \( k' \) becomes:
\[ k' = 3k \]

3. **Calculate the New Frequency**:
The new frequency \( v_2 \) with the modified spring constant \( k' \) is given by:
\[ v_2 = \frac{1}{2\pi} \sqrt{\frac{k'}{m}} = \frac{1}{2\pi} \sqrt{\frac{3k}{m}} \]

4. **Relate the Two Frequencies**:
Now we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = \frac{1}{2\pi} \sqrt{\frac{3k}{m}} = \sqrt{3} \cdot \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \sqrt{3} \cdot v_1 \]

5. **Final Relation**:
Thus, we find the relationship between the two frequencies:
\[ v_2 = \sqrt{3} \cdot v_1 \]

### Conclusion:
The relationship between the frequencies is:
\[ v_2 = \sqrt{3} \cdot v_1 \]