A body at the end of a spring executes SHM with a period `t_(1),` while the corresponding period for another spring is `t_(2),` If the period of oscillation with the two spring in series is T, then

Options

A. `T=t_(1)+t_(2)`
B. `T^(-2)=T_(1)^(-2)+T_(2)^(-2)`
C. `T^(-1)=T_(1)^(-1)+T_(2)^(-1)`
D. `T^(2)=t_(1)^(2)+t_(2)^(2)`

Explanation

To solve the problem, we need to find the relationship between the time periods \( t_1 \), \( t_2 \), and \( T \) when two springs are connected in series.

### Step-by-Step Solution:

1. **Understanding the Time Period of a Spring:**
The time period \( t \) of a spring executing simple harmonic motion (SHM) is given by the formula:
\[ t = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass attached to the spring and \( k \) is the spring constant.

2. **Expressing Time Periods for Individual Springs:**
For the first spring with time period \( t_1 \):
\[ t_1 = 2\pi \sqrt{\frac{m}{k_1}} \]
For the second spring with time period \( t_2 \):
\[ t_2 = 2\pi \sqrt{\frac{m}{k_2}} \]

3. **Finding the Effective Spring Constant for Springs in Series:**
When two springs are connected in series, the effective spring constant \( k_c \) can be found using:
\[ \frac{1}{k_c} = \frac{1}{k_1} + \frac{1}{k_2} \]
This can be rearranged to:
\[ k_c = \frac{k_1 k_2}{k_1 + k_2} \]

4. **Finding the Time Period for the Springs in Series:**
The time period \( T \) for the combined system of springs in series is:
\[ T = 2\pi \sqrt{\frac{m}{k_c}} \]

5. **Substituting for \( k_c \):**
Substitute \( k_c \) into the equation for \( T \):
\[ T = 2\pi \sqrt{\frac{m}{\frac{k_1 k_2}{k_1 + k_2}}} = 2\pi \sqrt{\frac{m (k_1 + k_2)}{k_1 k_2}} \]

6. **Relating \( T \) to \( t_1 \) and \( t_2 \):**
Now, we can express \( T^2 \) in terms of \( t_1^2 \) and \( t_2^2 \):
\[ T^2 = 4\pi^2 \frac{m (k_1 + k_2)}{k_1 k_2} \]
From the expressions for \( t_1 \) and \( t_2 \):
\[ t_1^2 = 4\pi^2 \frac{m}{k_1}, \quad t_2^2 = 4\pi^2 \frac{m}{k_2} \]
Rearranging gives:
\[ k_1 = \frac{4\pi^2 m}{t_1^2}, \quad k_2 = \frac{4\pi^2 m}{t_2^2} \]
Substitute these into the equation for \( T^2 \):
\[ T^2 = \frac{(t_1^2 + t_2^2)}{t_1^2 t_2^2} \cdot 4\pi^2 m \]

7. **Final Relationship:**
Thus, we can conclude that:
\[ \frac{1}{T^2} = \frac{1}{t_1^2} + \frac{1}{t_2^2} \]

### Conclusion:
The relationship between the time periods \( t_1 \), \( t_2 \), and \( T \) when two springs are in series is given by:
\[ \frac{1}{T^2} = \frac{1}{t_1^2} + \frac{1}{t_2^2} \]