Let `T_(1) and T_(2)` be the periods of springs A and B when mass M is suspended from one end of each spring. If both springs are taken in series and the same mass M is, suspended from the séries combination, the time period is T, then

Options

A. ` T = T _ 1 + T _ 2 `
B. ` ( 1 ) / (T ) = ( 1 ) / ( T _ 1 ) + ( 1 ) /( T _ 2 ) `
C. ` T^ 2 = T _ 1^ 2 + T _ 2 ^ 2 `
D. ` (1 ) /( T^ 2 ) = ( 1 ) /( T_ 1 ^ 2 ) + ( 1 ) /( T ^ 2 ) `

Explanation

To solve the problem, we need to find the relationship between the periods of the springs when they are used in series. Let's break down the solution step by step.

### Step 1: Understanding the Period of a Spring
The period \( T \) of a mass \( M \) attached to a spring with spring constant \( k \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{M}{k}} \]

### Step 2: Finding the Spring Constants
For spring A, the period is \( T_1 \):
\[ T_1 = 2\pi \sqrt{\frac{M}{k_1}} \implies k_1 = \frac{4\pi^2 M}{T_1^2} \]

For spring B, the period is \( T_2 \):
\[ T_2 = 2\pi \sqrt{\frac{M}{k_2}} \implies k_2 = \frac{4\pi^2 M}{T_2^2} \]

### Step 3: Combining Springs in Series
When two springs are combined in series, the equivalent spring constant \( k \) is given by:
\[ \frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} \]

Substituting the expressions for \( k_1 \) and \( k_2 \):
\[ \frac{1}{k} = \frac{T_1^2}{4\pi^2 M} + \frac{T_2^2}{4\pi^2 M} \]

### Step 4: Simplifying the Equation
Combining the fractions:
\[ \frac{1}{k} = \frac{T_1^2 + T_2^2}{4\pi^2 M} \]

Taking the reciprocal gives:
\[ k = \frac{4\pi^2 M}{T_1^2 + T_2^2} \]

### Step 5: Finding the Period of the Series Combination
Now, the period \( T \) for the series combination of the springs is:
\[ T = 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{M}{\frac{4\pi^2 M}{T_1^2 + T_2^2}}} \]

This simplifies to:
\[ T = 2\pi \sqrt{\frac{T_1^2 + T_2^2}{4\pi^2}} = \sqrt{T_1^2 + T_2^2} \]

### Conclusion
Thus, we have derived the relationship:
\[ T^2 = T_1^2 + T_2^2 \]

### Final Answer
The correct relationship is:
\[ T^2 = T_1^2 + T_2^2 \]