Time period of a block with a spring is `T_(0)`. Now ,the spring is cut in two parts in the ratio `2 : 3`. Now find the time period of same block with the smaller part of the spring.

Explanation

To solve the problem step by step, we will follow the principles of simple harmonic motion and the properties of springs.

### Step-by-Step Solution:

1. **Understanding the Time Period Formula**:
The time period \( T \) of a block attached to a spring is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass of the block and \( k \) is the spring constant.

2. **Relationship Between Spring Constant and Length**:
The spring constant \( k \) is inversely proportional to the length \( L \) of the spring:
\[ k \propto \frac{1}{L} \]
This means that if the length of the spring decreases, the spring constant increases.

3. **Cutting the Spring**:
The spring is cut into two parts in the ratio \( 2:3 \). If we assume the total length of the spring is \( 5L \), then:
- Length of the smaller part \( L_1 = 2L \)
- Length of the larger part \( L_2 = 3L \)

4. **Finding the Spring Constant for the Smaller Part**:
Since the spring constant \( k \) is inversely proportional to the length, we can express the spring constant for the smaller part:
\[ k_1 = \frac{k_0 \cdot 5L}{2L} = \frac{5k_0}{2} \]
where \( k_0 \) is the spring constant of the original spring.

5. **Calculating the Time Period for the Smaller Part**:
Now we can find the time period \( T_1 \) for the smaller part of the spring:
\[ T_1 = 2\pi \sqrt{\frac{m}{k_1}} = 2\pi \sqrt{\frac{m}{\frac{5k_0}{2}}} = 2\pi \sqrt{\frac{2m}{5k_0}} \]

6. **Relating to the Original Time Period**:
The original time period \( T_0 \) is:
\[ T_0 = 2\pi \sqrt{\frac{m}{k_0}} \]
To express \( T_1 \) in terms of \( T_0 \):
\[ T_1 = \sqrt{\frac{2}{5}} T_0 \]

### Final Answer:
The time period of the block with the smaller part of the spring is:
\[ T_1 = \sqrt{\frac{2}{5}} T_0 \]