A body is in `SHM` with period `T` when oscillated from a freely suspended spring. If this spring is cut in two parts of length ratio `1 : 3 &` again oscillated from the two the two parts separatedly, then the periods are `T_(1) & T_(2)` then find `T_(1)//T_(2)`.

Answer

To solve the problem, we need to find the ratio of the periods \( T_1 \) and \( T_2 \) of two parts of a spring that has been cut into a ratio of 1:3. We will use the properties of springs and the formula for the period of oscillation in simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Understanding the Spring Constant**:
The spring constant \( k \) is inversely proportional to the length of the spring. If the original length of the spring is \( L \), and it is cut into two parts with lengths \( L_1 \) and \( L_2 \) in the ratio 1:3, we can express:
\[ L_1 = \frac{L}{4} \quad \text{and} \quad L_2 = \frac{3L}{4} \]

2. **Finding the Spring Constants**:
The spring constant \( k \) for a spring is given by the formula:
\[ k = \frac{K}{L} \]
where \( K \) is a constant. Thus, for the two parts:
- For \( L_1 \):
\[ k_1 = \frac{K}{L_1} = \frac{K}{\frac{L}{4}} = \frac{4K}{L} \]
- For \( L_2 \):
\[ k_2 = \frac{K}{L_2} = \frac{K}{\frac{3L}{4}} = \frac{4K}{3L} \]

3. **Using the Period Formula**:
The period \( T \) of a mass \( m \) attached to a spring is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Therefore, we can find \( T_1 \) and \( T_2 \):
- For \( T_1 \):
\[ T_1 = 2\pi \sqrt{\frac{m}{k_1}} = 2\pi \sqrt{\frac{m}{\frac{4K}{L}}} = 2\pi \sqrt{\frac{mL}{4K}} = \frac{T}{2} \]
- For \( T_2 \):
\[ T_2 = 2\pi \sqrt{\frac{m}{k_2}} = 2\pi \sqrt{\frac{m}{\frac{4K}{3L}}} = 2\pi \sqrt{\frac{3mL}{4K}} = \sqrt{3} \cdot \frac{T}{2} \]

4. **Finding the Ratio \( \frac{T_1}{T_2} \)**:
Now we can find the ratio of the two periods:
\[ \frac{T_1}{T_2} = \frac{\frac{T}{2}}{\sqrt{3} \cdot \frac{T}{2}} = \frac{1}{\sqrt{3}} \]

### Final Result:
Thus, the ratio of the periods \( T_1 \) and \( T_2 \) is:
\[ \frac{T_1}{T_2} = \frac{1}{\sqrt{3}} \]