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, we need to find the time period of a block attached to the smaller part of a spring after the spring has been cut into two parts in the ratio of 2:3. Let's break down the solution step by step.

### Step 1: Understand the Initial Conditions
The initial time period \( T_0 \) of the block attached to the spring is given by the formula:
\[ T_0 = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass of the block and \( k \) is the spring constant of the original spring.

### Step 2: Determine the Lengths of the Cut Springs
The spring is cut into two parts in the ratio of 2:3. Let the total length of the spring be \( L \). Then, the lengths of the two parts will be:
- Length of the smaller part: \( L_1 = \frac{2}{5}L \)
- Length of the larger part: \( L_2 = \frac{3}{5}L \)

### Step 3: Find the Spring Constants of the New Springs
The spring constant \( k \) is inversely proportional to the length of the spring. Therefore, if we denote the spring constants of the smaller and larger parts as \( k' \) and \( k'' \) respectively, we can use the relationship:
\[ k' \cdot L = k \cdot L \quad \text{and} \quad k'' \cdot L = k \cdot L \]
From this, we can derive:
\[ k' \cdot \frac{2}{5}L = k \cdot L \implies k' = \frac{5k}{2} \]
\[ k'' \cdot \frac{3}{5}L = k \cdot L \implies k'' = \frac{5k}{3} \]

### Step 4: Calculate the Time Period for the Smaller Spring
Now we need to find the time period \( T' \) for the block attached to the smaller part of the spring with spring constant \( k' = \frac{5k}{2} \):
\[ T' = 2\pi \sqrt{\frac{m}{k'}} \]
Substituting \( k' \):
\[ T' = 2\pi \sqrt{\frac{m}{\frac{5k}{2}}} = 2\pi \sqrt{\frac{2m}{5k}} \]

### Step 5: Relate the New Time Period to the Initial Time Period
We can express \( T' \) in terms of the initial time period \( T_0 \):
\[ T' = \sqrt{\frac{2}{5}} \cdot 2\pi \sqrt{\frac{m}{k}} = \sqrt{\frac{2}{5}} \cdot T_0 \]

### Final Result
Thus, the time period of the block with the smaller part of the spring is:
\[ T' = \sqrt{\frac{2}{5}} T_0 \]