A vertical spring has a time period of oscillations of `T_(1)` with a load `m_(1)` at its end. Its time period is `T_(2)` when the load is replaced by another of mass `m_(2)`. The force constant of the spring is

Options

A. `4 pi^(2) ((m_(1)-m_(2))/(T_(1)^(2)-T_(2)^(2)))`
B. `4 pi^(2) ((m_(1)+m_(2))/(T_(1)^(2)+T_(2)^(2)))`
C. `2 pi^(2) ((m_(1)+m_(2))/(T_(1)^(2)+T_(2)^(2)))`
D. `4 pi^(2) ((m_(1)-m_(2))/(T_(1)^(2)+T_(2)^(2)))`

Explanation

To find the force constant \( k \) of the spring, we can use the relationship between the time period of oscillation and the mass attached to the spring. The time period \( T \) for a mass \( m \) attached to a spring with spring constant \( k \) is given by the formula:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

### Step 1: Write the equations for the two time periods
For the first mass \( m_1 \) with time period \( T_1 \):
\[ T_1 = 2\pi \sqrt{\frac{m_1}{k}} \]

For the second mass \( m_2 \) with time period \( T_2 \):
\[ T_2 = 2\pi \sqrt{\frac{m_2}{k}} \]

### Step 2: Square both equations
Squaring both equations gives:
\[ T_1^2 = 4\pi^2 \frac{m_1}{k} \quad \text{(1)} \]
\[ T_2^2 = 4\pi^2 \frac{m_2}{k} \quad \text{(2)} \]

### Step 3: Rearrange both equations to solve for \( k \)
From equation (1):
\[ k = \frac{4\pi^2 m_1}{T_1^2} \]

From equation (2):
\[ k = \frac{4\pi^2 m_2}{T_2^2} \]

### Step 4: Set the two expressions for \( k \) equal to each other
Since both expressions represent the same spring constant \( k \):
\[ \frac{4\pi^2 m_1}{T_1^2} = \frac{4\pi^2 m_2}{T_2^2} \]

### Step 5: Cross-multiply to eliminate \( k \)
Cross-multiplying gives:
\[ m_1 T_2^2 = m_2 T_1^2 \]

### Step 6: Solve for \( k \) in terms of \( m_1, m_2, T_1, \) and \( T_2 \)
Now, we can express \( k \) in terms of \( T_1 \) and \( T_2 \):
\[ k = \frac{4\pi^2 (m_1 + m_2)}{T_1^2 + T_2^2} \]

### Final Result
Thus, the force constant \( k \) of the spring is given by:
\[ k = \frac{4\pi^2 (m_1 + m_2)}{T_1^2 + T_2^2} \]