A mass is suspended separately by two springs and the time periods in the two cases are `T_(1)` and `T_(2)`. Now the same mass is connected in parallel `(K = K_(1) + K_(2))` with the springs and the time is suppose `T_(P)`. Similarly time period in series is `T_(S)`, then find the relation between `T_(1),T_(2)` and `T_(P)` in the first case and `T_(1),T_(2)` and `T_(S)` in the second case.

Explanation

To solve the problem, we need to find the relationships between the time periods \( T_1, T_2 \) (for the individual springs) and \( T_P \) (for the parallel connection) as well as \( T_S \) (for the series connection).

### Step-by-Step Solution:

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

2. **Expressing Spring Constants in Terms of Time Periods**:
Rearranging the formula, we can express the spring constant \( k \) in terms of the time period \( T \):
\[ k = \frac{4\pi^2 m}{T^2} \]
For each spring, we can write:
- For spring 1: \( k_1 = \frac{4\pi^2 m}{T_1^2} \)
- For spring 2: \( k_2 = \frac{4\pi^2 m}{T_2^2} \)

3. **Finding the Equivalent Spring Constant for Parallel Connection**:
When the two springs are connected in parallel, the equivalent spring constant \( K_P \) is given by:
\[ K_P = k_1 + k_2 \]
Substituting the expressions for \( k_1 \) and \( k_2 \):
\[ K_P = \frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2} \]
This can be simplified to:
\[ K_P = 4\pi^2 m \left( \frac{1}{T_1^2} + \frac{1}{T_2^2} \right) \]

4. **Relating \( T_P \) to \( T_1 \) and \( T_2 \)**:
The time period \( T_P \) for the parallel connection can be expressed as:
\[ T_P = 2\pi \sqrt{\frac{m}{K_P}} \]
Substituting for \( K_P \):
\[ T_P = 2\pi \sqrt{\frac{m}{4\pi^2 m \left( \frac{1}{T_1^2} + \frac{1}{T_2^2} \right)}} \]
Simplifying this gives:
\[ T_P = 2\pi \sqrt{\frac{1}{4\pi^2 \left( \frac{1}{T_1^2} + \frac{1}{T_2^2} \right)}} \]
\[ T_P = \frac{1}{\sqrt{\frac{1}{T_1^2} + \frac{1}{T_2^2}}} \]
Thus, we have:
\[ T_P^{-2} = T_1^{-2} + T_2^{-2} \]

5. **Finding the Equivalent Spring Constant for Series Connection**:
When the springs are connected in series, the equivalent spring constant \( K_S \) is given by:
\[ \frac{1}{K_S} = \frac{1}{k_1} + \frac{1}{k_2} \]
Substituting:
\[ \frac{1}{K_S} = \frac{T_1^2}{4\pi^2 m} + \frac{T_2^2}{4\pi^2 m} \]
This simplifies to:
\[ \frac{1}{K_S} = \frac{T_1^2 + T_2^2}{4\pi^2 m} \]

6. **Relating \( T_S \) to \( T_1 \) and \( T_2 \)**:
The time period \( T_S \) for the series connection can be expressed as:
\[ T_S = 2\pi \sqrt{\frac{m}{K_S}} \]
Substituting for \( K_S \):
\[ T_S = 2\pi \sqrt{\frac{m \cdot 4\pi^2 m}{T_1^2 + T_2^2}} \]
This gives:
\[ T_S^2 = \frac{4\pi^2 m}{K_S} \]
Thus, we have:
\[ T_S^2 = T_1^2 + T_2^2 \]

### Final Relations:
- For parallel connection:
\[ T_P^{-2} = T_1^{-2} + T_2^{-2} \]
- For series connection:
\[ T_S^2 = T_1^2 + T_2^2 \]